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RSS Feed for Wolfram Community showing any discussions in tag Wolfram Scienceattachment.php?postid=2617 sorted by activeWhy do grid marks appear when exporting SVG but not EPS?
http://community.wolfram.com/groups/-/m/t/1225125
Observe the following Mathematica Code:
nd[y_, a_, b_, n_] := (Binomial[b - a + 1, y]*Sum[((-1)^i)*Binomial[y, i]*((y - i)/(b - a + 1))^n, {i, 0, y}]);
Unprotect[ColorData];
ColorData["My_Rainbow"] = Function[x, Blend[{Black, Purple, Blue, Cyan, Green, Yellow, Orange, Red}, x]];
Protect[ColorData]
Manipulate[
ArrayPlot[Table[nd[y, 1, K, n], {K, 1, 50}, {n, 1, 50}],
ColorFunction -> Function[{y}, ColorData["My_Rainbow"][y]],
ColorFunctionScaling -> False, DataReversed -> True, Frame -> True,
FrameTicks -> Automatic,
FrameLabel -> {Rotate["K", -90 Degree], "N"}], {y, 1, 50, 1}]
I've noticed some very unusual results attempting to export the graphics as both an SVG and EPS vector image. When I export it as EPS, it behaves exactly as anticipated; it looks just like it does in Mathematica, and nothing gets rasterized, which is fine. But I'd like an SVG image so I can display it on the web. However, when I export as SVG, I get some really strange grid lines that appear! Take a look below:
[![enter image description here][1]][1]
[![enter image description here][2]][2]
As you can see, there are some very noticeable grid lines that appear in the SVG version, which change inconsistently with the zoom level, making them rather annoying. Obviously, I'd prefer to not have the grid lines, so the reasonable solution seems to be to convert the EPS to an SVG, right? Unfortunately, whenever I do that, I get this blurry mess!
[![enter image description here][3]][3]
[1]: https://i.stack.imgur.com/ipFu5.png
[2]: https://i.stack.imgur.com/oeJe4.png
[3]: https://i.stack.imgur.com/fRmwT.png
It's very frustrating, because no matter what I try I can't get what I want. I just don't understand what's going on. If somebody could please shed some light on the situation, I would greatly appreciate it. Thank you very much.John Travolski2017-11-20T05:31:36ZRational functions graphics
http://community.wolfram.com/groups/-/m/t/1225057
Please, could any body help me?
I want to plot graphics of elementary rational functions like f(x)=1/x, or g(x)= (x^2+1)/(x^2-4), what I got is not correct. I have tried to indicate the intervals of the domain of the function with inequalities but it does not work, for instance for g(x): -6<x<-2, -2<x<2, 2<x<6. I understand is a problema with x values that make zero the denominator. I attach a file with the examples. Thank you very much.Salvador Benjamin P. R.2017-11-20T08:20:42ZStrange result in a cubic equation?
http://community.wolfram.com/groups/-/m/t/1222449
Please look at this outcome:
![enter image description here][1]
As you see, all 3 roots are complex numbers. But this equition has a real root at least 1.
This is 0.1736 ... (exactly sin 10 degrees).
![enter image description here][2]
Where is my fault?
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ucuncuDD_forumSoru1.PNG&userId=1222434
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ucuncuDD_forumSoru2.PNG&userId=1222434Metin Guney2017-11-15T09:21:51ZHappy 32nd birthday Windows, and some synchronicity thrown in...
http://community.wolfram.com/groups/-/m/t/1225041
The first version of Microsoft Windows came out 32 years ago today...My life has been full of Synchronicities and I was married exactly to the day at the midpoint between when the first Apple Macintosh came out and when the first version of Windows came out...(Windows came out 666 days after the Macintosh, maybe that brings new meaning to "666"?}...Also, I was born exactly to the day at the midpoint between when my mom was born and when Mathematica first came out...Anyone else have any "Synchronicities" regarding Mathematica or Windows? The introduction dates for important Microprocessors runs in my family also...The first 32 bit CPU for the PC the 80386 came out on my Birthday, and the first 32 bit CPU for the Macintosh the 68020 came out on my sister's Birthday, and the 8086 CPU that is still the basis for the architecture of PCs came out on my Grandfather's Birthday!!!Roy Scott2017-11-20T06:30:02Z[GIF] Microcosm (Stereographic projection of cube grid)
http://community.wolfram.com/groups/-/m/t/1225032
![Stereographic projection of cube grid][1]
**Microcosm**
This is conceptually very simple: take a $7 \times 7$ grid of unit cubes in space, normalize to the unit sphere, stereographically project to the plane, then apply a rotation to the original grid of cubes. Here's the code:
Stereo[p_] := 1/(1 - p[[-1]]) p[[;; 2]];
With[{n = 3, cols = RGBColor /@ {"#4EEAF6", "#291F71"}},
Manipulate[
Graphics[
{Opacity[.5], CapForm[None], cols[[1]], Thickness[.006],
Line[
Flatten[
Transpose[
Table[
Stereo[Normalize[#]] & /@
{{t,
y Cos[θ] - z Sin[θ],
z Cos[θ] + y Sin[θ]},
{y, t Cos[θ] - z Sin[θ],
z Cos[θ] + t Sin[θ]},
{y, z Cos[θ] - t Sin[θ],
t Cos[θ] + z Sin[θ]}},
{z, -n - 1/2, n + 1/2}, {y, -n - 1/2, n + 1/2}, {t, -n - 1/2.,
n + 1/2, 1/20}],
{2, 3, 4, 1, 5}],
2]
]
},
PlotRange -> 1, Axes -> False, ImageSize -> 540, Background -> cols[[-1]]],
{θ, 0, π/2}]
]
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=rotateStereo12c.gif&userId=610054Clayton Shonkwiler2017-11-20T05:51:30ZWhy here is crooked line in $\log(\frac{a+1}{c+1})$ contour plot?
http://community.wolfram.com/groups/-/m/t/1224136
I plotted contour plot of this formula by Wolfram Alpha
$$\log(\frac{a+1}{c+1}), 0 < a < 1000, 0 < c < 1000$$
![enter image description here][1]
In the above figure, there is a crooked line in the bottom part. If I plot this, there is no crooked line.
$$\log(\frac{a}{c}), 0 < a < 1000, 0 < c < 1000$$
Why there is crooked line in the first case?
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=a7xLc.gif&userId=11733Jayong Kim2017-11-18T08:36:17ZSyntax problem
http://community.wolfram.com/groups/-/m/t/1224968
Good evening! I am working on a quantum mechanics problem using Mathematica and when I typed in the functions I need, I keep getting syntax problem saying " Expression imcomplete, more input needed". I have no idea what is going on. Will anyone help me with it? Thank you so much!!!
Input:
\!\(\*SuperscriptBox[\(J\), \('\)]\)[D_] :=
N[E, 10]^2/(
4*\[Pi]*Quantity[1, "ElectricConstant"]*
Quantity["BohrRadius"]) (1/D -
Exp[-2*D] (1/D + 11/8 + (3 D)/4 + D^2/6))
Getting:Syntax::sntxi
Input:
(J^')[D_]:=N[E,10]^2/(4*\[Pi]*Quantity[1, "ElectricConstant"]*Quantity["BohrRadius"]) (1/D-Exp[-2D]*(1/D+11/8+(3D)/4+D^2/6))
\!\(\*SuperscriptBox[\(K\), \('\)]\)[D_] :=
N[E, 10]^2/(
20*\[Pi]*Quantity[1, "ElectricConstant"]*
Quantity[
"BohrRadius"]) (-Exp[-2*D] (-25/8 + 23/4 D + 3 D^2 + 1/3 D^3) +
6/D (\[CapitalDelta]^2 (N[EulerGamma, 10] + Log[D]) +
\!\(\*SuperscriptBox[\(\[CapitalDelta]\), \('\)]\)^2*
ExpIntegralEi[-4 D] - 2*\[CapitalDelta]*
\!\(\*SuperscriptBox[\(\[CapitalDelta]\), \('\)]\)*
ExpIntegralEi[-2 D]))
Getting:Syntax::sntxiMinghui Wang2017-11-20T03:29:49ZTurn this system of differential equations into a single equation?
http://community.wolfram.com/groups/-/m/t/1224952
{
0 == -voltageC0[t] - 2*voltageC0''[t] + 2*voltageC1''[t],
0 == -voltageC1[t] - 2*voltageC1''[t] + 2*voltageC0''[t]
}
I am able to solve this system using DSolve, but now I want to do something different. I want to convert it into a single differential equation, expressed in terms of voltageC0 without reference to voltageC1, in this form:
0 == a0*voltageC0[t] + a1*voltageC0'[t] + a2*voltageC0''[t] + a3*voltageC0'''[t]
I show a 3rd-order solution as an example, but that is just for example of the target form, and the actual order may turn out to be more or less than 3 for all I know.
I want to find what the values of a0, a1, etc. are. Does Mathematica have facilities to help with this? I believe the conversion should be possible, but I don't understand the process for converting.Joe Donaldson2017-11-20T01:58:40ZIs there a way to get Mathematica to evaluate this integral?
http://community.wolfram.com/groups/-/m/t/1224875
Is there a way to help Mathematica evaluate the following integral?
Integrate[((4*Log[((p1 - p3)^2 + \[Omega]^2)/((p1 + p3)^2 + \[Omega]^2)]*
Log[((p1 - p4)^2 + \[Omega]^2)/((p1 + p4)^2 + \[Omega]^2)])/(p3*p4))*(Sin[p1]/p1),
{p1, 0, Infinity}, Assumptions -> Element[p1 | p3 | p4 | \[Omega], Reals],
Assumptions -> p1 > 0 && p3 > 0 && p4 > 0]
or alternatively
Integrate[Limit[Integrate[((4*Log[(p1 - p3 + \[Omega])/(p1 + p3 + \[Omega])]*
Log[(p1 - p4 + \[Omega])/(p1 + p4 + \[Omega])])/(p3*p4))*(Sin[p1]/p1), {p1, 0, Infinity},
Assumptions -> Element[p1 | p3 | p4, Reals]], \[Omega] -> 0], {p1, 0, Infinity},
Assumptions -> Element[p1 | p3 | p4 | \[Omega], Reals]]Arny Toynbee2017-11-19T23:42:57ZColor Function not correctly mapping to range
http://community.wolfram.com/groups/-/m/t/1224856
Observe the following code:
nd[y_, a_, b_, n_] := (Binomial[b - a + 1, y]*Sum[((-1)^i)*Binomial[y, i]*((y - i)/(b - a + 1))^n, {i, 0, y}]);
mycolorfun = Function[Blend[{Black, Purple, Blue, Green, Yellow, Orange, Red, White}, Rescale[#, {0, 1}]]]
Manipulate[
ArrayPlot[Table[nd[y, 1, K, n], {K, 1, 25}, {n, 1, 25}],
ColorFunction -> mycolorfun, DataReversed -> True, Frame -> True,
FrameTicks -> Automatic,
FrameLabel -> {Rotate["K", -90 Degree], "N"}], {y, 1, 15, 1}]
Notice how when the slider is moved up to y=15, there is a white square that appears at N = 25 and K = 21:
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/8Qutc.png
My goal is for the color function to map the individual value of the output of my function to a color as dependent on its closeness to 0 or 1 (since my function is a probability density function, this makes sense).
However, look what happens when you check the actual value of the function for N = 25 and b=K=21:
nd[15, 1, 21, 25.0]
(out) 0.268495
I get 26%! 26% should be mapping to a color somewhere around Purple/Blue, but it's getting mapped all the way up to the color white! White is what's supposed to be mapped to when the output of my function is really close to 1! Clearly something is amiss here, and I cannot figure out what it is. Can somebody please explain what is going on and what I need to do to get the desired color output?John Travolski2017-11-19T17:59:19ZHow do I implement an external function to a descretionary notebook?
http://community.wolfram.com/groups/-/m/t/1224634
Hey guys,
I've got a function which computes the propagation of uncertainty for a given function. It is called "fehler" and an example would be:
fehler[Sin[2*a], a]
which gives me
2 Sqrt[\[CapitalDelta]a^2 Cos[2 a]^2]
So basically the syntax is fehler[function, variable1, variable2,...]. The whole file is attached to the post, the credit goes to some random guy from my university, unluckily I could not figure out who made this function.
Now I'd like to make this function usable inside every notebook, because right now I always use it externally which is not really nice. So for example if i have a random notebook it should work like this:
(*random notebook*)
fehler= (*this is the code i need*)
a = 1; b=2; const=9;
fehler[a*const+b*2, a, b]
(*the output should like this*) Sqrt[const^2 \[CapitalDelta]a^2 + 4 \[CapitalDelta]b^2]
So that in every notebook I use i can simply use the "fehler"-function to calculate to propagation of uncertainty without having to use it globally. I tried to make this possible for a long time, but the main problem is that the existing fehler-function requires to clear all Variables with ClearAll["Global`*"]; before executing fehler. It is important that the output can be further used as a function, and not is simply a (numerical) result.
Do you guys have an idea how I can solve this problem? It would help me (and other students) a lot, if we could the function inside our notebooks and not externally.
Thanks you and best regards,
TobiasTobias Mitterdorfer2017-11-19T00:27:13ZHow to Solve/Reduce systems with an array of variables?
http://community.wolfram.com/groups/-/m/t/1224718
As a simple example I want to reduce the equation d1+2*d2==2, where d1 and d2 are either 0 or 1. This is easy:
In[147]:= Reduce[{d1 + 2 d2 == 2, d1 == 0 || d1 == 1,
d2 == 0 || d2 == 1}, {d1, d2}]
Out[147]= d1 == 0 && d2 == 1
But if I use an array d[i] instead of d1, d2 for the unknown variables, I cannot get it to work. The following seems to be syntactically correct, but it does not work:
In[152]:= Reduce[{d[1] + 2 d[2] == 2,
ForAll[i, d[i] == 0 || d[i] == 1]}, d]
During evaluation of In[152]:= Reduce::nsmet: This system cannot be solved with the methods available to Reduce.
Out[152]= Reduce[{d[1] + 2 d[2] == 2, \!\(
\*SubscriptBox[\(\[ForAll]\), \(i\)]\((d[i] == 0 || d[i] == 1)\)\)},
d]
Of course this example ist extremely simplified. I need some solution for a large d-Array of variables and some equation containing them.Werner Geiger2017-11-19T13:54:34ZHow to replace some terms in an expression by an abbreviation?
http://community.wolfram.com/groups/-/m/t/1224451
I have a symbolic expression (don't ask why, that doesn't matter), say:
exp1 = 2 n (r + h) Sin[\[Pi]/n]
Since I know, that n*Sin[\[Pi]/n] is something special (namely \[Pi] for large n), I want to replace that terms by an abbreviation pn:
pn = n Sin[\[Pi]/n]
and get something like exp2 from exp1 using that abbreviation. I.e.:
exp2 = 2 (r+h) pn
I tried several things like Replace, Reduce, Evaluate and the like, but could get no reasonable result. Does anybody know, how to do that kind of things?Werner Geiger2017-11-18T19:22:14ZWhy does the WolframKernel exit on OS X without a trace during Integrate?
http://community.wolfram.com/groups/-/m/t/1222503
I am trying to calculate a quantity from its density in 3D over a rectangular 2D window frame the following way:
mflxx = Integrate[
mi3ret[[1, 1]], {y, -w/2,
w/2}, {z, -(c/2 + zf + d/2), -(c/2 + zf + d/2 + h)}]
where `mi3ret[[1,1]]` is a fairly long expression containing only elementary functions of the `x,y` and `z` variables, and it is the x-component of a 3D vector function. It also contains some constant parameters, like `a,b,c`. `zf,d,h` are additional constants for the definite integral. After a long calculation the `WolframKernel` is killed by the OS without any message or CrashReport. So, after rebooting the machine and running just Mathematica and the Activity Monitor, I made some screenshuts to see, how the memory is used and put the memory related value into lists. See notebook attached, where the 0 minute is really after about one and half hours into the run. The machine is an early 2011 Early MacPro with 64GB memory and a 500GB Flashdrive partitioned into two 250GB partition. So, the questions are:
1. Why the `WolframKernel` is killed without a trace?
2. Can the OS tuned in such a way that `Integrate` can finish its run?
3 Is there a command or function in Mathematica that would alert to the coming
doom and would allow the examination of the status of the machine, including the status of the `WolframKernel`?
4. Is it possible to configure Mathematica to use, for disk based intermediate storage, the partition that have more free space, /the boot drive itself has 150GB, the other partition has 230GB free space/ ?
Thanks ahead,
JánosJanos Lobb2017-11-15T04:29:13ZConvert DiscretePlot3D to 2D plot with a Color Function
http://community.wolfram.com/groups/-/m/t/1224586
Observe the following wolfram Mathematica code:
nd[y_, a_, b_, n_] := (Binomial[b - a + 1, y]*Sum[((-1)^i)*Binomial[y, i]*((y - i)/(b - a + 1))^n, {i, 0, y}]);
Animate[DiscretePlot3D[nd[i, 1, n, j], {j, 1, 15}, {i, 1, 15}, ExtentSize -> Full, ColorFunction -> "DarkRainbow"], {n, 1, 50, 1}]
nd is just the name of a discrete probability density function of 4 variables that I have defined earlier on. What I would really like to do is make this a 2 dimensional grid of squares where the value of nd is shown only by the color as opposed to both the color and the height; in other words, I don't want it to be 3D. Of course, I can rotate what I have to get a top down view, but this isn't exactly ideal:
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/W83BS.png
I would much prefer to eliminate the third dimension altogether and have a simple grid where the output of nd is shown by a change in color of the squares in the 2D grid. Is this possible in Mathematica? If so, how would I do this? Thank you.John Travolski2017-11-19T04:41:20ZDynamical Systems with Applications using Mathematica, 2nd Edition
http://community.wolfram.com/groups/-/m/t/1224393
The book "Dynamical Systems with Applications using Mathematica, 2nd Edition" has been published:
http://www.springer.com/us/book/9783319614847
The notebook (and other files) to accompany the book can be downloaded here:
http://library.wolfram.com/infocenter/Books/9563/
Best wishes
StephenStephen Lynch2017-11-19T08:52:45ZReplacing an "Except" element of a matrix under a Table
http://community.wolfram.com/groups/-/m/t/1224574
Dear Friends,
In a "For" operation, I have a table defined by indices i & j. Any (i, j)th element of this table consist of the elements of matrix f[m] (6X1 matrix), where m is the element (Integer) on which "For" loop is running.
During the "For" loop I want to replace the terms f[m][[k] by f[m-1][[k]], where k is a number other than the indices i & j for the (i, j)th element.
When I tried to use
`/.f[m][[X-]] -> f[m-1][[X]]`,
I am getting the following error message --
Part::pkspec1: The expression x$_ cannot be used as a part specification.
Could there be an alternative way to achieve this ??
thanks for the help in advance !S G2017-11-19T01:59:09ZIs Mathematica 11.0.1.0 still compatible with OS X 10.13.1 ?
http://community.wolfram.com/groups/-/m/t/1224565
Hi,
Recently I upgraded to OS X 10.13.1 and since I have been having strange freezes of the machine caused by Mathematica. For example, I have a fairly large `List`, about 65000 points, for `ListDensityPlot3D`. The command to show the List is this:
ListDensityPlot3D[Partition[Flatten[tblmi3retx], 4],
PlotLabel -> "delta = " <> ToString[N[Pi/500/7]], PlotRange -> All,
PlotLegends -> Automatic]
, so after the `Partition` I have as elements the `x,y,z` coordinates and the density of the quantity I want to display, one after another. `ListDensityPlot3D` actually does draw the list as it should do. The problems start when I want to enlarge the plot or rotate it into another position. Then Mathematica freeze and gives me a spinning beachball. In earlier OS X versions I could kill Mathematica, restart the machine and start over. Not now. Despite every claim by Apple that no user program can crash the OS, Mathematica is able to do it. The machine is locked up. Interestingly, the cursor is still available, but nothing responds from the OS for any attempt to select a Finder window, or to bring up the Force Quit window. I also cannot ssh into the machine from another computer. If I try to save the desktop with <cmd><shift><3>, I can hear the snap, but it produces only an empty grayish picture
![grayish picture][1]
Ultimately I have to do a hard reset via the Power Button. After restart, the OS is bringing up Mathematica automagically and the strangeness continues, because Mathematica gives a few empty windows on the top of each other, moderately scattered, with the topmost having a spinning cursor in it.
![spinning cursor in empty window][2]
So I have to quit from this automatically booted Mathematica and launch it from the Finder to bring things back to something normal.
This episode shows me that the earlier compatibility with OS X is gone with OS X 10.13.1. Im wondering if Mathematica 10.2 also has this problem, but I do not want to find it out by shelling out my hard earned tax dollars.
Thanks ahead,
János
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-18at19.37.12.png&userId=277053
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-18at19.52.10.png&userId=277053Janos Lobb2017-11-19T01:42:29ZProper use of PredictorMeasurements[]?
http://community.wolfram.com/groups/-/m/t/1208606
I've been playing around with `Predict[]` with multi-dimensional datasets and, for small training sets anyway, things seem to work correctly. For example,
trainingset = {<|"age" -> 47, "sex" -> "M", "height" -> 100,
"weight" -> 60|>, <|"age" -> 22, "sex" -> "M", "height" -> 90,
"weight" -> 55|>, <|"age" -> 43, "sex" -> "M", "height" -> 110,
"weight" -> 61|>, <|"age" -> 23, "sex" -> "F", "height" -> 100,
"weight" -> 41|>, <|"age" -> 33, "sex" -> "F", "height" -> 80,
"weight" -> 50|>, <|"age" -> 43, "sex" -> "F", "height" -> 70,
"weight" -> 51|>};
testset = {<|"age" -> 37, "sex" -> "M", "height" -> 100|>, <|
"age" -> 22, "sex" -> "M", "height" -> 90|>, <|"age" -> 43,
"sex" -> "F", "height" -> 80|>, <|"age" -> 33, "sex" -> "F",
"height" -> 70|>};
p1 = Predict[trainingset -> "weight", PerformanceGoal -> "Quality",
Method -> "RandomForest"];
We can get predictions from the p1 `PredictorFunction` with
Map[Append[#, "prediction" -> p1[#]] &, testset] (* this works *)
I can then compute residuals, etc., myself.
Since version 10, Wolfram Language has included the function `PredictorMeasurements[]`, and the documentation suggests that I should be able to get the predictions above, plus residual reports and other information, with
PredictorMeasurements[p1, testset]
But this does not work. I get the following error: `PredictorMeasurements::bdfmt: Argument {<|age->37,sex->M,height->100,weight->60|>,<|age->22,sex->M,height->90|>,<|age->43,sex->F,height->80|>,<|age->33,sex->F,height->70|>} should be a rule or a list of rules.`
What am I missing?Michael Stern2017-10-25T16:44:44ZSimplify of -Sqrt[r] >= 0 where r is real
http://community.wolfram.com/groups/-/m/t/1224276
Am I missing something? Assuming r is real, the only r that can satisfy -Sqrt[r]>=0 is zero.
$Assumptions = {r \[Element] Reals};
FullSimplify[-Sqrt[r] >= 0]
yields `Sqrt[r] <= 0`
rather than `r==0`Paul Reiser2017-11-18T18:56:53ZPredicting Winning Odds of Italian Serie A Soccer 1934-2017
http://community.wolfram.com/groups/-/m/t/1218215
Dear Community! Although gambling is not my hobby, the validation of predictor engine inside Wolfram Mathematica 11.2 is. Sooner or later someone may get huge benefits in gambling using predictors as they have been developed already today. I got LaLiga "Outcomes" 100 % but my example will be Italian Serie A analysis, yrs 1934-2017. It is downloadable from
https://github.com/jalapic/engsoccerdata
Note that the source was found by google'ing. I am avoiding to be too lengthy and therefore only minimum amount of outputs are given. For necessary prediction only **goals** and **dates** are needed. I have always wondered whether it is useful to put into prediction as much statistical data as possible and it turns out that this must be avoided i.e. the right data for right prediction should be selected for best results. The CSV file should be first downloaded and a little-bit modified in order to use it as Mher Davtyan used it in his original example:
http://community.wolfram.com/groups/-/m/t/908804
then the path may be seen from the import function below:
oddsData =
SemanticImport[
"C:\\Users\\user\\Documents\\engsoccerdata-master\\data-raw\\italy3.\
csv", <|1 -> "Integer", 2 -> "String", 3 -> "Integer", 4 -> "String",
5 -> "String", 7 -> Automatic, 8 -> Automatic|>]
giving an output:
![Input data][1]
Yes, it consists of 25784 games. The collection of additions to Dataset are listed below and explained later:
scoreAdded =
oddsData[All,
Append[#,
"Score" ->
ToString[Slot["hgoal"]] <> ":" <> ToString[Slot["vgoal"]]] &]
homeAndAway =
scoreAdded[
Association["Home" -> GroupBy[#home &],
"Away" -> GroupBy[#visitor &]]];
data01 = (Flatten[(Transpose[{homeAndAway["Home",
scoreAdded[#]["home"], All, "Nr"] // Normal,
Thread[Rule[
homeAndAway["Home", scoreAdded[#]["home"], All,
"home"] // Normal,
homeAndAway["Home", scoreAdded[#]["home"], All,
"hgoal"] // Normal // Accumulate]]}] & /@
Range[1, 25784, 1]) // Union, 1] // Sort)[[All, 2]][[All,
2]];
data02 = (Flatten[(Transpose[{homeAndAway["Away",
scoreAdded[#]["visitor"], All, "Nr"] // Normal,
Thread[Rule[
homeAndAway["Away", scoreAdded[#]["visitor"], All,
"visitor"] // Normal,
homeAndAway["Away", scoreAdded[#]["visitor"], All,
"vgoal"] // Normal // Accumulate]]}] & /@
Range[1, 25784, 1]) // Union, 1] // Sort)[[All, 2]][[All,
2]];
data03 = (Flatten[(Transpose[{homeAndAway["Away",
scoreAdded[#]["visitor"], All, "Nr"] // Normal,
Thread[Rule[
homeAndAway["Away", scoreAdded[#]["visitor"], All,
"visitor"] // Normal,
homeAndAway["Away", scoreAdded[#]["visitor"], All,
"hgoal"] // Normal // Accumulate]]}] & /@
Range[1, 25784, 1]) // Union, 1] // Sort)[[All, 2]][[All,
2]];
data04 = (Flatten[(Transpose[{homeAndAway["Home",
scoreAdded[#]["home"], All, "Nr"] // Normal,
Thread[Rule[
homeAndAway["Home", scoreAdded[#]["home"], All,
"home"] // Normal,
homeAndAway["Home", scoreAdded[#]["home"], All,
"vgoal"] // Normal // Accumulate]]}] & /@
Range[1, 25784, 1]) // Union, 1] // Sort)[[All, 2]][[All,
2]];
addIndividualGoalsCumul[d_] :=
Join[d, Association["homeTeamHomeRating" -> (data01[[d["Nr"]]]),
"VisitorAwayRating" -> (data02[[d["Nr"]]]),
"homeTeamHomeConceded" -> (data04[[d["Nr"]]]),
"VisitorAwayConceded" -> (data03[[d["Nr"]]])]];
ratingsAdded = Map[addIndividualGoalsCumul[#] &, scoreAdded]
extractDate[a_] :=
Append[a, <|
"DateSuccessive" ->
FromDigits[StringSplit[a["Date"], "-"][[1]]] 100 - 190000 +
FromDigits[StringSplit[a["Date"], "-"][[2]]]|>]
dateNew = Map[extractDate[#] &, ratingsAdded]
addOutcome[a_Association] :=
If[a["hgoal"] == a["vgoal"], Append[a, Association["Outcome" -> 0]],
If[a["hgoal"] > a["vgoal"], Append[a, Association["Outcome" -> 1]],
Append[a, Association["Outcome" -> -1]]]]
outcomeAdded = Map[addOutcome[#] &, dateNew]
where data01, data02, data03 and data04 compute the cumulative scores and conceded scores for every team, the "DateSuccessive" modifies the "Date": 1934-06-23 to 03406 and 2017-05-28 to 11705 i.e. year and month, and "Outcome" is similar as in original post. The result table is:
![Final Dataset 1][2]
![Final Dataset 2][3]
The test data starts from index:
In[96]:= Round[0.99218 Length[outcomeAdded]]
Out[96]= 25582
and consists of all 202 games played on yr 2017 for test set for Classifier:
![Testset][4]
Above we see the Testset and needed data using the code:
n = Round[0.99218 Length[outcomeAdded]];
{trainingSet, testSet} =
TakeDrop[outcomeAdded[
All, {"DateSuccessive", "homeTeamHomeRating", "VisitorAwayRating",
"homeTeamHomeConceded", "VisitorAwayConceded", "hgoal", "vgoal",
"Outcome"}], n];
c = Classify[trainingSet -> "Outcome", Method -> "LogisticRegression"] cm = ClassifierMeasurements[c, testSet -> "Outcome"]
with following results:
![enter image description here][5]
Visualisation:
In[138]:= actaulValue = testSet[All, "Outcome"] // Normal
Out[138]= {1, 1, -1, -1, 1, 1, 1, 0, -1, -1, 1, 1, 1, 1, 1, 0, 1, -1, \
0, -1, -1, 1, 1, 1, 0, 1, -1, -1, 1, 1, -1, 0, 1, 0, 0, 1, -1, 0, 1, \
-1, -1, 1, 0, 0, -1, 1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, \
-1, 1, 0, 1, 1, -1, -1, -1, 1, 1, 1, 0, -1, 1, -1, 1, -1, 0, -1, 1, \
0, -1, 0, 1, -1, 1, 0, -1, -1, 0, -1, 1, 0, 1, -1, 1, 1, 1, 1, -1, \
-1, -1, 1, 1, 0, 1, 1, 0, -1, -1, 1, -1, 1, 1, -1, -1, 1, -1, 0, -1, \
0, 0, -1, 0, 0, 1, -1, -1, 1, -1, 1, 0, 1, 1, -1, 0, 0, 1, 0, -1, 0, \
1, 0, 1, 1, -1, 1, 1, -1, -1, 0, 1, -1, 0, 0, 1, 1, 0, -1, -1, -1, 1, \
-1, 0, 1, 0, 1, 1, 1, -1, -1, 0, 0, 0, 1, 1, 1, 1, -1, 1, 1, 0, -1, \
-1, 1, -1, 1, 1, -1, 1, 1, 0, 1, 1, -1, 1, 1, 0, 1, 1, 1, -1, 1}
In[139]:= predictedValue =
c[Normal[testSet[[All, Values]]][[All, ;; 7]]]
Out[139]= {1, 1, -1, -1, 1, 1, 1, 0, -1, -1, 1, 1, 1, 1, 1, 0, 1, -1, \
0, -1, -1, 1, 1, 1, 0, 1, -1, -1, 1, 1, -1, 0, 1, 0, 0, 1, -1, 0, 1, \
-1, -1, 1, 0, 0, -1, 1, -1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, \
-1, 1, 0, 1, 1, -1, -1, -1, 1, 1, 1, 0, -1, 1, -1, 1, -1, 0, -1, 1, \
0, -1, 0, 1, -1, 1, 0, -1, -1, 0, -1, 1, 0, 1, -1, 1, 1, 1, 1, -1, \
-1, -1, 1, 1, 0, 1, 1, 0, -1, -1, 1, -1, 1, 1, -1, -1, 1, -1, 0, -1, \
0, 0, -1, 0, 0, 1, -1, -1, 1, -1, 1, 0, 1, 1, -1, 0, 0, 1, 0, -1, 0, \
1, 0, 1, 1, -1, 1, 1, -1, -1, 0, 1, -1, 0, 0, 1, 1, 0, -1, -1, -1, 1, \
-1, 0, 1, 0, 1, 1, 1, -1, -1, 0, 0, 0, 1, 1, 1, 1, -1, 1, 1, 0, -1, \
-1, 1, -1, 1, 1, -1, 1, 1, 0, 1, 1, -1, 1, 1, 0, 1, 1, 1, -1, 1}
In[140]:= difference = actaulValue - predictedValue
Out[140]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
In[141]:= N@Normalize[Counts[Abs[difference]], Total]
Out[141]= <|0 -> 1.|>
Concluding that all yr 2017 game winners were correctly classified.
Now, prediction with slightly larger set:
In[143]:= n = Round[0.99 Length[outcomeAdded]]
Out[143]= 25526
In[144]:= {trainingSetFull, testSetFull} = TakeDrop[outcomeAdded, n];
{trainingSet1,
testSet1} = {trainingSetFull[
All, {"DateSuccessive", "homeTeamHomeRating", "VisitorAwayRating",
"homeTeamHomeConceded", "VisitorAwayConceded", "hgoal", "vgoal",
"Outcome"}],
testSetFull[
All, {"DateSuccessive", "homeTeamHomeRating", "VisitorAwayRating",
"homeTeamHomeConceded", "VisitorAwayConceded", "hgoal", "vgoal",
"Outcome"}]};
![enter image description here][6]
In[156]:= predictedValue =
Round[p[Normal[testSet1[[All, Values]]][[All, ;; 7]]], 1]
Out[156]= {1, 1, -1, -1, 1, 0, 1, -1, 1, -1, 1, 0, 1, -1, 1, 1, 0, 1, \
1, 1, -1, 1, 0, 1, 1, 0, 1, 1, -1, -1, 0, -1, 1, -1, -1, 1, 1, 1, 1, \
1, 0, 1, 0, 1, 1, -1, -1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, -1, -1, 1, \
1, 1, 0, -1, -1, 1, 1, 1, 1, 1, 0, 1, -1, 0, -1, -1, 1, 1, 1, 0, 1, \
0, -1, 1, 1, -1, 0, 1, 0, 0, 0, -1, 0, 1, -1, -1, 1, 0, 0, -1, 1, -1, \
1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, -1, 1, 0, 1, 1, -1, -1, -1, \
1, 1, 1, 0, -1, 1, -1, 1, -1, 0, -1, 1, 0, -1, 0, 1, -1, 1, 0, -1, \
-1, 0, -1, 1, 0, 1, -1, 1, 1, 1, 1, -1, -1, -1, 1, 1, 0, 1, 1, 0, -1, \
-1, 1, -1, 1, 1, -1, -1, 1, -1, 0, -1, 0, 0, -1, 0, 0, 1, -1, -1, 1, \
-1, 1, 0, 1, 1, -1, 0, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 1, 1, -1, -1, \
0, 1, -1, 0, 0, 1, 1, 0, -1, -1, -1, 1, -1, 0, 1, 0, 1, 1, 1, -1, -1, \
0, 0, 0, 1, 1, 1, 1, -1, 1, 1, 0, -1, -1, 1, -1, 1, 1, -1, 1, 1, 0, \
1, 1, -1, 0, 1, 0, 1, 1, 1, -1, 1}
In[157]:= actaulValue = testSet1[All, "Outcome"] // Normal
Out[157]= {1, 1, -1, -1, 1, 0, 1, -1, 1, -1, 1, 0, 1, -1, 1, 1, 0, 1, \
1, 1, -1, 1, 0, 1, 1, 0, 1, 1, -1, -1, 0, -1, 1, -1, -1, 1, 1, 1, 1, \
1, 0, 1, -1, 1, 1, -1, -1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, -1, -1, \
1, 1, 1, 0, -1, -1, 1, 1, 1, 1, 1, 0, 1, -1, 0, -1, -1, 1, 1, 1, 0, \
1, -1, -1, 1, 1, -1, 0, 1, 0, 0, 1, -1, 0, 1, -1, -1, 1, 0, 0, -1, 1, \
-1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, -1, 1, 0, 1, 1, -1, -1, \
-1, 1, 1, 1, 0, -1, 1, -1, 1, -1, 0, -1, 1, 0, -1, 0, 1, -1, 1, 0, \
-1, -1, 0, -1, 1, 0, 1, -1, 1, 1, 1, 1, -1, -1, -1, 1, 1, 0, 1, 1, 0, \
-1, -1, 1, -1, 1, 1, -1, -1, 1, -1, 0, -1, 0, 0, -1, 0, 0, 1, -1, -1, \
1, -1, 1, 0, 1, 1, -1, 0, 0, 1, 0, -1, 0, 1, 0, 1, 1, -1, 1, 1, -1, \
-1, 0, 1, -1, 0, 0, 1, 1, 0, -1, -1, -1, 1, -1, 0, 1, 0, 1, 1, 1, -1, \
-1, 0, 0, 0, 1, 1, 1, 1, -1, 1, 1, 0, -1, -1, 1, -1, 1, 1, -1, 1, 1, \
0, 1, 1, -1, 1, 1, 0, 1, 1, 1, -1, 1}
In[158]:= difference = Round[actaulValue - predictedValue, 0.01]
Out[158]= {0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., -1., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
-1., 0., 0., 0., 0., 0., 0., 0., 0., 1., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., \
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 0., \
0., 0., 0., 0., 0., 0.}
In[159]:= N@Normalize[Counts[Abs[difference]], Total]
Out[159]= <|0. -> 0.98062, 1. -> 0.0193798|>
showing 98 % accuracy.
## 10 Random Samples ##
randomMatches = RandomSample[testSetFull, 10]
![enter image description here][7]
Goals prediction is not so exact:
![enter image description here][8]
and for visiting team:
![enter image description here][9]
giving 60 % of accuracy.
Having "Outcome" correct, I will just show two tables of Predicted and Actual scores:
![enter image description here][10]
## Conclusions ##
Clearly, game scores are harder to predict, almost 60 % are correct and 38 % differs with 1 goal. The predictor seems to work better when adding time-scale. Over the time the team's performance is changing and this must be reflected.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture1.JPG&userId=94243
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture2.JPG&userId=94243
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture3.JPG&userId=94243
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture4.JPG&userId=94243
[5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture5.JPG&userId=94243
[6]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture6.JPG&userId=94243
[7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture7.JPG&userId=94243
[8]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture8.JPG&userId=94243
[9]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture9.JPG&userId=94243
[10]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Capture11.JPG&userId=94243Tanel Telliskivi2017-11-08T01:18:39ZHow to get Mathematica to simplify powers?
http://community.wolfram.com/groups/-/m/t/1224078
Mathematica fails to simplify the following, even with the Assumptions I give it:
In[59]:= FullSimplify[(xe^n)^(1/n),
Assumptions -> {xe \[Element] Reals, n \[Element] integers, n > 0}]
Out[59]= (xe^n)^(1/n)
Does anyone know how to get Mathematica to simplify (x^n)^(1/n) to x?Kenneth Miller2017-11-18T07:01:16ZRandom variables transformation
http://community.wolfram.com/groups/-/m/t/1224436
Hi,
I am trying to transform a PDF function of x to a PDF function of r using Mathematica, noting that r=x^2 and the PDF of x is as below:
f[x_] := (( (4 Sqrt[π]) (μ^(μ +
0.5)) (h^(μ)) (x^(2 μ)))/((Gamma[μ]) (H^(μ -
0.5)) ((Ω)^(μ +
0.5)))) (E^((-2 (μ) (h) (x^2))/(Ω))) *
BesselI[μ - 0.5, 2 h x^2 μ/Ω]
Anyone can help of how could I do this?
BR,Wisam Abbasi2017-11-18T18:23:34ZRepair a corrupted notebook
http://community.wolfram.com/groups/-/m/t/1224314
Just for everybodies information: I had some problems with corrupted notebooks, which I could not really solve. Now I found a Wolfram tool and procedure to repair such a notebook. It does the job:
See: [How do I repair corrupted notebooks using the AuthorTools package?][1]
Like any files, Wolfram Language notebooks can be corrupted, making them unreadable. In general, we recommend backing up important work on different media. In case notebooks become corrupted, Mathematica provides a tool to help restore them using the AuthorTools package. Restoring a notebook removes any corrupted sections of the notebook, allowing it to be opened and read. Follow these steps to o repair a corrupted notebook:
1) In a new notebook, load the AuthorTools package and open the NotebookRestore dialog by executing:
Quiet[<<AuthorTools`]
NotebookRestore[]
Please note that the tick mark after AuthorTools is a back tick character (`), not an apostrophe (‘). The back tick character is available underneath the tilde key on most keyboards.
2) Select the notebook you want to restore in the NotebookRestore dialog. Mathematica will open a new notebook with what could be restored from the original one. We recommend saving the new notebook with a new name.
[1]: http://support.wolfram.com/kb/12423Werner Geiger2017-11-18T17:13:13ZThe 27_4 Configuration in Barycentrics
http://community.wolfram.com/groups/-/m/t/1224304
A configuration is a set of points and lines such that $n$ points are on each line and $n$ lines go through each point. Here are a few 4-configurations:
![4-configurations][1]
In [barycentric coordinates](http://community.wolfram.com/groups/-/m/t/947771), both lines and points can be represented as triples. If point $P$ is on line $L$, then the dot product $P.L=0$. For awhile now I've wondered if a self-dual barycentric configuration existed where the point set and line set used the same set of triples. The answer is YES, they exist. Here's a $24_3$ configuration. I've found about 120 of these.
set243 = {{-1, 1, 2}, {-1, 2, 1}, {-1, 2, 3}, {-1, 2, 4}, {-1, 3, 2}, {-1, 4, 2}, {0, 1, 2}, {0, 2, 1}, {1, 2, -1}, {2, 1, -1}, {2, 3, -1}, {2, 4, -1}, {3, 2, -1}, {4, 2, -1}, {1, 2, 0}, {2, 1, 0}, {2, -1, 1}, {1, -1, 2}, {3, -1, 2}, {4, -1, 2}, {2, -1, 3}, {2, -1, 4}, {2, 0, 1}, {1, 0, 2}};
tri=Reverse[{{Sqrt[3]/2,-(1/2)},{0,1},{-(Sqrt[3]/2),-(1/2)}}];
FromBarycentrics[{m_,n_,o_},{{x1_,y1_},{x2_,y2_},{x3_,y3_}}]:={m*x1+n*x2+(1-m-n)*x3,m*y1+n*y2+(1-m-n)*y3};With[{full=Union[Flatten[{#, RotateRight[#,1],RotateLeft[#,1]}&/@{{-1,1,2},{-1,2,1},{-1,2,3},{-1,2,4},{-1,3,2},{-1,4,2},{0,1,2},{0,2,1}},1]]},
Graphics[{EdgeForm[Black],Tooltip[Line[#[[2]]],Style[Row[Switch[Sign[#],-1,Style[ToString[Abs[#]],Red],0,Style[ToString[Abs[#]],Darker[Green]],1,Style[ToString[Abs[#]],Blue]]&/@#[[1]]],16,Bold]]&/@Table[{full[[k]],Sort[FromBarycentrics[#/Total[#],tri]&/@Select[full,full[[k]].#==0&]]},{k,1,Length[full]}],White,{Disk[FromBarycentrics[#/Total[#],tri],.075],Black, Style[Text[Row[Switch[Sign[#],-1,Style[ToString[Abs[#]],Red],0,Style[ToString[Abs[#]],Darker[Green]],1,Style[ToString[Abs[#]],Blue]]&/@#],FromBarycentrics[#/Total[#],tri]],16,Bold]}&/@full}]]
![Config 24_3][2]
I also found a single $27_4$ configuration.
config274={
{-2,-1,4},{-2,1,3},{-1,1,1},{-1,2,0},{-1,2,1},{-1,3,2},{-1,4,2},{0,1,2},{1,1,2},
{-1,4,-2},{1,3,-2},{1,1,-1},{2,0,-1},{2,1,-1},{3,2,-1},{4,2,-1},{1,2,0},{1,2,1},
{4,-2,-1},{3,-2,1},{1,-1,1},{0,-1,2},{1,-1,2},{2,-1,3},{2,-1,4},{2,0,1},{2,1,1}}
With[{full=Union[Flatten[{#, RotateRight[#,1],RotateLeft[#,1]}&/@{{-2,-1,4},{-2,1,3},{-1,1,1},{-1,2,0},{-1,2,1},{-1,3,2},{-1,4,2},{0,1,2},{1,1,2}},1]]},
Graphics[{EdgeForm[Black],Tooltip[Line[#[[2]]],Style[Row[Switch[Sign[#],-1,Style[ToString[Abs[#]],Red],0,Style[ToString[Abs[#]],Darker[Green]],1,Style[ToString[Abs[#]],Blue]]&/@#[[1]]],16,Bold]]&/@Table[{full[[k]],Sort[FromBarycentrics[#/Total[#],tri]&/@Select[full,full[[k]].#==0&]]},{k,1,Length[full]}],White,{Disk[FromBarycentrics[#/Total[#],tri],.07],Black, Style[Text[Row[Switch[Sign[#],-1,Style[ToString[Abs[#]],Red],0,Style[ToString[Abs[#]],Darker[Green]],1,Style[ToString[Abs[#]],Blue]]&/@#],FromBarycentrics[#/Total[#],tri]],7]}&/@full}]]
![27_4][3]
It's easy to check. The following code gives a list of 27 4's.
Length /@ Table[Select[config274, config274[[n]].# == 0 &], {n, 1, 27}]
We can make a graph of these triples.
Graph[#[[1]] <-> #[[2]] & /@ Select[Subsets[Sort[config274], {2}], #[[1]].#[[2]] == 0 &], VertexLabels -> "Name"]
![graph 27_4][4]
Whether there is a 5-configuration with this method is something I'm still running.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=config-4.jpg&userId=21530
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=config24_3.jpg&userId=21530
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=27_4.jpg&userId=21530
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=graph274.jpg&userId=21530Ed Pegg2017-11-18T15:03:30ZSimplify a complicated solution by a simpler way?
http://community.wolfram.com/groups/-/m/t/1176517
eq50 has given a complicated solution, and it should be simplified into the sum of several proper-fraction-alike parts.
I can get a simpler one in eq55, but it takes 5 abstruse steps and isn't the simplest yet.
Can you give a easy and straight way?
In[159]:= eq12 = (cinf \[Theta]c \[Lambda])/(s + \[Theta]c) - (
E^((lh - x) /Sqrt[Dc ] Sqrt[s + \[Theta]c]) F1 Sqrt[
Dc ] \[Lambda])/
Sqrt[ (s + \[Theta]c)] /. {Sqrt[Dc] F1 \[Lambda] -> A1,
cinf \[Theta]c \[Lambda] -> A2}
Out[159]= A2/(s + \[Theta]c) - (
A1 E^(((lh - x) Sqrt[s + \[Theta]c])/Sqrt[Dc]))/Sqrt[s + \[Theta]c]
In[160]:= eq13 = s *q[x, s] - Dp*D[q[x, s], {x, 2}] - pinf + eq12 == 0
Out[160]= -pinf + A2/(s + \[Theta]c) - (
A1 E^(((lh - x) Sqrt[s + \[Theta]c])/Sqrt[Dc]))/Sqrt[
s + \[Theta]c] + s q[x, s] - Dp
\!\(\*SuperscriptBox[\(q\), \*
TagBox[
RowBox[{"(",
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[x, s] == 0
In[161]:= eq14 = \[Alpha]1*(q[x, s] /. x -> lh) - (D[q[x, s], x] /.
x -> lh) == 0
Out[161]= \[Alpha]1 q[lh, s] -
\!\(\*SuperscriptBox[\(q\), \*
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[lh, s] == 0
In[162]:= eq15 = (D[q[x, s], x] /. x -> +\[Infinity]) == 0
Out[162]=
\!\(\*SuperscriptBox[\(q\), \*
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[\[Infinity], s] == 0
In[163]:= eq50 =
Assuming[(lh - x) < 0 && s > 0 && (s + \[Theta]c) > 0 && Dp > 0 &&
Dc > 0 && \[Alpha]1 > 0,
DSolve[{eq13, eq14, eq15}, q[x, s], {x, s},
GeneratedParameters -> B4]] // Simplify // Normal
Out[163]= {{q[x,
s] -> (E^(-(Sqrt[s]/Sqrt[Dp] + Sqrt[s/
Dp]) x) (-A2 (-Dc s +
Dp (s + \[Theta]c)) (-4 E^(Sqrt[s/Dp] (lh + x)) Sqrt[Dp/s]
s Sqrt[s/Dp] Sqrt[s + \[Theta]c] Sqrt[
Dp s (s + \[Theta]c)] -
Dp (E^((2 Sqrt[s] x)/Sqrt[Dp])
s (Sqrt[s/Dp] + \[Alpha]1) (s + \[Theta]c) +
E^(2 Sqrt[s/Dp] x)
s (Sqrt[s/Dp] + \[Alpha]1) (s + \[Theta]c) -
2 E^(Sqrt[s/Dp] (lh + x)) Sqrt[s/
Dp] (2 s^2 +
2 s \[Theta]c + \[Alpha]1 Sqrt[s + \[Theta]c] Sqrt[
Dp s (s + \[Theta]c)]))) +
Sqrt[s/
Dp] (1/(Sqrt[s/Dp] Sqrt[s + \[Theta]c])
Dp pinf (-2 E^(Sqrt[s/Dp] (lh + x)) Sqrt[s/
Dp] (Sqrt[Dp/s] s^4 Sqrt[Dp s (s + \[Theta]c)] +
5 Sqrt[Dp/s] s^3 \[Theta]c Sqrt[
Dp s (s + \[Theta]c)] +
6 Sqrt[Dp/s] s^2 \[Theta]c^2 Sqrt[
Dp s (s + \[Theta]c)] +
2 Sqrt[Dp/s] s \[Theta]c^3 Sqrt[
Dp s (s + \[Theta]c)] +
Sqrt[Dp/s] s^(5/2) Sqrt[s + \[Theta]c] Sqrt[
s (s + \[Theta]c)] Sqrt[Dp s (s + \[Theta]c)] -
s \[Alpha]1 Sqrt[
Dp^3 s^5 (s + \[Theta]c)] - \[Alpha]1 \[Theta]c \
Sqrt[Dp^3 s^5 (s + \[Theta]c)]) -
Dp (s + \[Theta]c) (E^((2 Sqrt[s] x)/Sqrt[Dp])
s (Sqrt[s/Dp] + \[Alpha]1) (s + \[Theta]c)^(5/2) +
E^(2 Sqrt[s/Dp] x)
s (Sqrt[s/Dp] + \[Alpha]1) (s + \[Theta]c)^(5/2) -
2 E^(Sqrt[s/Dp] (lh + x)) Sqrt[s/
Dp] (2 s^3 Sqrt[s + \[Theta]c] +
4 s^2 \[Theta]c Sqrt[
s + \[Theta]c] + \[Alpha]1 \[Theta]c^2 Sqrt[
Dp s (s + \[Theta]c)] +
2 s \[Theta]c (\[Theta]c Sqrt[
s + \[Theta]c] + \[Alpha]1 Sqrt[
Dp s (s + \[Theta]c)])))) +
1/(Sqrt[Dp s])
Dc (2 E^(Sqrt[s/Dp] (lh + x)) Sqrt[Dp/s] s Sqrt[Dp s]
Sqrt[Dp s (s + \[Theta]c)] (pinf s (s Sqrt[
s + \[Theta]c] + 2 \[Theta]c Sqrt[s + \[Theta]c] +
Sqrt[s] Sqrt[s (s + \[Theta]c)]) -
A1 (s + \[Theta]c) (-2 s + Sqrt[(
Dp s (s + \[Theta]c))/Dc])) +
Dp^2 s (-2 A1 E^(Sqrt[s/Dp] (lh + x))
s \[Alpha]1 \[Theta]c Sqrt[s + \[Theta]c] +
E^((2 Sqrt[s] x)/Sqrt[Dp]) pinf Sqrt[s/Dp] Sqrt[
Dp s] \[Alpha]1 (s + \[Theta]c)^2 +
E^(2 Sqrt[s/Dp] x) pinf Sqrt[s/Dp] Sqrt[
Dp s] \[Alpha]1 (s + \[Theta]c)^2 -
A1 E^((2 Sqrt[s] x)/Sqrt[
Dp] + (lh - x) Sqrt[(s + \[Theta]c)/
Dc]) (s + \[Theta]c) Sqrt[
Dp s (s + \[Theta]c)] (Sqrt[s/Dp] Sqrt[(
s + \[Theta]c)/
Dc] + \[Alpha]1 (-Sqrt[(s/Dp)] + Sqrt[(
s + \[Theta]c)/Dc])) +
A1 E^(2 Sqrt[s/Dp] x + (lh - x) Sqrt[(s + \[Theta]c)/
Dc]) (s + \[Theta]c) Sqrt[
Dp s (s + \[Theta]c)] (Sqrt[s/Dp] Sqrt[(
s + \[Theta]c)/
Dc] + \[Alpha]1 (Sqrt[s/Dp] + Sqrt[(
s + \[Theta]c)/Dc]))) +
Dp (E^((2 Sqrt[s] x)/Sqrt[Dp]) pinf s^2 Sqrt[
Dp s] (s + \[Theta]c)^2 +
E^(2 Sqrt[s/Dp] x) pinf s^2 Sqrt[
Dp s] (s + \[Theta]c)^2 + (
A1 E^((2 Sqrt[s] x)/Sqrt[
Dp] + (lh - x) Sqrt[(s + \[Theta]c)/Dc])
s (Dp s (s + \[Theta]c))^(3/2))/Dp + (
A1 E^(2 Sqrt[s/Dp] x + (lh - x) Sqrt[(s + \[Theta]c)/
Dc]) s (Dp s (s + \[Theta]c))^(3/2))/Dp -
2 E^(Sqrt[s/
Dp] (lh +
x)) (A1 (Sqrt[Dp s] Sqrt[Dp s^5] \[Alpha]1 Sqrt[
s + \[Theta]c] + (
2 s (Dp s (s + \[Theta]c))^(3/2))/Dp) +
pinf Sqrt[
Dp s] (2 s^4 + 4 s^3 \[Theta]c +
2 s^2 \[Theta]c^2 +
s \[Alpha]1 \[Theta]c Sqrt[s + \[Theta]c] Sqrt[
Dp s (s + \[Theta]c)] + \[Alpha]1 Sqrt[
s + \[Theta]c] Sqrt[
Dp s^5 (s + \[Theta]c)])))))))/(2 Dc Dp^3 (s/Dp)^(
3/2) (Sqrt[s/Dp] + \[Alpha]1) (s + \[Theta]c)^(3/2) Sqrt[
Dp s (s + \[Theta]c)] (Sqrt[s/Dp] - Sqrt[(s + \[Theta]c)/
Dc]) (Sqrt[s/Dp] + Sqrt[(s + \[Theta]c)/Dc]))}}
In[164]:= eq51 =
Assuming[(lh - x) < 0 && s > 0 && (s + \[Theta]c) > 0 && Dp > 0 &&
Dc > 0 && \[Alpha]1 > 0, FullSimplify[q[x, s] /. eq50[[1, 1]]]]
Out[164]= -(E^(-2 Sqrt[s/Dp]
x) (2 A2 s (-Dp E^(Sqrt[s/Dp] (lh + x)) \[Alpha]1 +
E^(2 Sqrt[s/Dp]
x) (Sqrt[Dp s] + Dp \[Alpha]1)) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c)) +
Sqrt[s/Dp] (-1/(Sqrt[(s (s + \[Theta]c))/Dp])2 Dp E^(
Sqrt[s/Dp] x)
pinf (-Dp (E^(lh Sqrt[s/Dp]) - E^(
Sqrt[s/Dp] x)) s \[Alpha]1 (s + \[Theta]c)^(7/2) +
E^(Sqrt[s/Dp]
x) (\[Theta]c^3 Sqrt[Dp s^3 (s + \[Theta]c)] +
3 \[Theta]c^2 Sqrt[Dp s^5 (s + \[Theta]c)] +
3 \[Theta]c Sqrt[Dp s^7 (s + \[Theta]c)] + Sqrt[
Dp s^9 (s + \[Theta]c)])) +
2 Sqrt[Dc]
Dp s (s + \[Theta]c) (pinf (Sqrt[Dc] E^(2 Sqrt[s/Dp] x)
s + (E^(2 Sqrt[s/Dp] x) - E^(
Sqrt[s/Dp] (lh + x))) Sqrt[
Dc Dp s] \[Alpha]1) (s + \[Theta]c) +
A1 (-Sqrt[Dp] E^(Sqrt[s/Dp] (lh + x)) s^(3/2) -
E^(Sqrt[s/
Dp] (lh + x)) (Sqrt[Dp s] \[Theta]c + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]) +
E^(2 Sqrt[s/Dp] x + (lh - x) Sqrt[(s + \[Theta]c)/
Dc]) (s Sqrt[Dc (s + \[Theta]c)] + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))))))/(2 Dp s^2 (Sqrt[s/
Dp] + \[Alpha]1) (s + \[Theta]c)^2 (-Dc s + Dp (s + \[Theta]c)))
In[165]:= eq52 =
ExpandAll[eq51,
x] /. {(Sqrt[s/Dp] + \[Alpha]1) -> ((
Sqrt[s] + Sqrt[ Dp ] \[Alpha]1)/Sqrt[ Dp ]),
Sqrt[s/Dp] -> Sqrt[s]/Sqrt[ Dp ]} // Normal
Out[165]= (
A2 Sqrt[Dp]
E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/Sqrt[Dp]) \[Alpha]1)/(
s (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)) - (
A2 (Sqrt[Dp s] + Dp \[Alpha]1))/(
Sqrt[Dp] s (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)) - (
Dc pinf Sqrt[
s])/((Sqrt[s] + Sqrt[Dp] \[Alpha]1) (-Dc s +
Dp (s + \[Theta]c))) - (Sqrt[Dc] pinf Sqrt[Dc Dp s] \[Alpha]1)/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (-Dc s +
Dp (s + \[Theta]c))) + (
Sqrt[Dc] E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/Sqrt[Dp]) pinf Sqrt[
Dc Dp s] \[Alpha]1)/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (-Dc s +
Dp (s + \[Theta]c))) + (
A1 Sqrt[Dc] Sqrt[Dp]
E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/Sqrt[
Dp]) s)/((Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) + (Dp pinf \[Alpha]1 (s + \[Theta]c)^(3/2))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) Sqrt[(s (s + \[Theta]c))/
Dp] (-Dc s + Dp (s + \[Theta]c))) - (
Dp E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/Sqrt[Dp])
pinf \[Alpha]1 (s + \[Theta]c)^(3/2))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) Sqrt[(s (s + \[Theta]c))/
Dp] (-Dc s + Dp (s + \[Theta]c))) + (
A1 Sqrt[Dc]
E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/Sqrt[
Dp]) (Sqrt[Dp s] \[Theta]c + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) - (
A1 Sqrt[Dc] E^(
lh Sqrt[(s + \[Theta]c)/Dc] -
x Sqrt[(s + \[Theta]c)/
Dc]) (s Sqrt[Dc (s + \[Theta]c)] + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) + (
pinf (\[Theta]c^3 Sqrt[Dp s^3 (s + \[Theta]c)] +
3 \[Theta]c^2 Sqrt[Dp s^5 (s + \[Theta]c)] +
3 \[Theta]c Sqrt[Dp s^7 (s + \[Theta]c)] + Sqrt[
Dp s^9 (s + \[Theta]c)]))/(
s^(3/2) (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)^2 Sqrt[(
s (s + \[Theta]c))/Dp] (-Dc s + Dp (s + \[Theta]c)))
In[166]:= eq53 = Map[FullSimplify, eq52]
Out[166]= (A2 Sqrt[Dp] E^((Sqrt[s] (lh - x))/Sqrt[Dp]) \[Alpha]1)/(
s (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)) - (
A2 (Sqrt[Dp s] + Dp \[Alpha]1))/(
Sqrt[Dp] s (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)) - (
Dc pinf Sqrt[
s])/((Sqrt[s] + Sqrt[Dp] \[Alpha]1) (-Dc s +
Dp (s + \[Theta]c))) - (Sqrt[Dc] pinf Sqrt[Dc Dp s] \[Alpha]1)/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (-Dc s +
Dp (s + \[Theta]c))) + (
Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[Dp]) pinf Sqrt[
Dc Dp s] \[Alpha]1)/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (-Dc s +
Dp (s + \[Theta]c))) + (
A1 Sqrt[Dc] Sqrt[Dp] E^((Sqrt[s] (lh - x))/Sqrt[
Dp]) s)/((Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) + (
pinf Sqrt[s] \[Alpha]1 (s + \[Theta]c)^(
5/2))/((Sqrt[s] + Sqrt[Dp] \[Alpha]1) ((s (s + \[Theta]c))/Dp)^(
3/2) (-Dc s + Dp (s + \[Theta]c))) - (
E^((Sqrt[s] (lh - x))/Sqrt[Dp]) pinf Sqrt[
s] \[Alpha]1 (s + \[Theta]c)^(
5/2))/((Sqrt[s] + Sqrt[Dp] \[Alpha]1) ((s (s + \[Theta]c))/Dp)^(
3/2) (-Dc s + Dp (s + \[Theta]c))) + (
A1 Sqrt[Dc] E^((Sqrt[s] (lh - x))/Sqrt[
Dp]) (Sqrt[Dp s] \[Theta]c + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) - (
A1 Sqrt[Dc]
E^((lh - x) Sqrt[(s + \[Theta]c)/
Dc]) (s Sqrt[Dc (s + \[Theta]c)] + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) + (
pinf (\[Theta]c^3 Sqrt[Dp s^3 (s + \[Theta]c)] +
3 \[Theta]c^2 Sqrt[Dp s^5 (s + \[Theta]c)] +
3 \[Theta]c Sqrt[Dp s^7 (s + \[Theta]c)] + Sqrt[
Dp s^9 (s + \[Theta]c)]))/(
s^(3/2) (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)^2 Sqrt[(
s (s + \[Theta]c))/Dp] (-Dc s + Dp (s + \[Theta]c)))
In[167]:= eq54 =
Collect[eq52, { E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/Sqrt[Dp]), E^(
lh Sqrt[(s + \[Theta]c)/Dc] - x Sqrt[(s + \[Theta]c)/Dc])},
Simplify]
Out[167]= -((
A1 Sqrt[Dc] E^(
lh Sqrt[(s + \[Theta]c)/Dc] -
x Sqrt[(s + \[Theta]c)/
Dc]) (s Sqrt[Dc (s + \[Theta]c)] + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))/(
Sqrt[s] (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c)))) + (E^((lh Sqrt[s])/Sqrt[Dp] - (Sqrt[s] x)/
Sqrt[Dp]) (A1 Sqrt[Dc]
s (Sqrt[Dp] s^(3/2) +
Sqrt[Dp s] \[Theta]c + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]) + \[Alpha]1 (A2 Sqrt[Dp] Sqrt[
s] (-Dc s + Dp (s + \[Theta]c)) +
pinf (s + \[Theta]c) (Sqrt[Dc] s Sqrt[Dc Dp s] -
Dp^2 Sqrt[s + \[Theta]c] Sqrt[(s (s + \[Theta]c))/
Dp]))))/(s^(
3/2) (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) + (-Sqrt[Dc] Sqrt[Dp] pinf s^2 Sqrt[
Dc Dp s] \[Alpha]1 (s + \[Theta]c)^3 +
Dc s^(5/2) (s + \[Theta]c)^2 (A2 Sqrt[Dp s] + A2 Dp \[Alpha]1 -
Sqrt[Dp] pinf Sqrt[s] (s + \[Theta]c)) +
1/(Sqrt[s + \[Theta]c])
Dp (-A2 s^(3/2) (Sqrt[Dp s] + Dp \[Alpha]1) (s + \[Theta]c)^(
7/2) + Sqrt[Dp] pinf Sqrt[(s (s + \[Theta]c))/
Dp] (Dp s \[Alpha]1 (s + \[Theta]c)^4 +
Sqrt[s + \[Theta]c] (\[Theta]c^3 Sqrt[
Dp s^3 (s + \[Theta]c)] +
3 \[Theta]c^2 Sqrt[Dp s^5 (s + \[Theta]c)] +
3 \[Theta]c Sqrt[Dp s^7 (s + \[Theta]c)] + Sqrt[
Dp s^9 (s + \[Theta]c)]))))/(Sqrt[Dp] s^(
5/2) (Sqrt[s] + Sqrt[Dp] \[Alpha]1) (s + \[Theta]c)^3 (-Dc s +
Dp (s + \[Theta]c)))
In[168]:= eq55 =
Assuming[s > 0 && (s + \[Theta]c) > 0 && Dp > 0 && Dc > 0,
FullSimplify[eq54[[1]]] + FullSimplify[eq54[[2]]] +
FullSimplify[eq54[[3]]]] // Normal
Out[168]= (-A2 + pinf (s + \[Theta]c))/(s (s + \[Theta]c)) - (
A1 E^((lh - x) Sqrt[(s + \[Theta]c)/Dc]) Sqrt[Dc/
s] (s Sqrt[Dc (s + \[Theta]c)] + \[Alpha]1 Sqrt[
Dc Dp s (s + \[Theta]c)]))/((Sqrt[s] +
Sqrt[Dp] \[Alpha]1) (s + \[Theta]c) (-Dc s +
Dp (s + \[Theta]c))) + (E^((s (lh - x))/Sqrt[
Dp s]) (-Sqrt[
Dp s] \[Alpha]1 (-Zhonghui Ou2017-09-04T09:25:34ZAvoid SelectionMove inconsistency?
http://community.wolfram.com/groups/-/m/t/1214368
I believe there is a bug in `SelectionMove` when it is used to select a `CellGroup`. The bug is two-fold: not only `SelectionMove` fails to perform the operation, but also it doesn't return `$Failed` is it should according to the Documentation:
> `SelectionMove` returns `$Failed` if it cannot move the selection in the way you request.
Suppose we have a Notebook containing a `CellGroup`. Then we can paste the following cell after this `CellGroup` and evaluate it in order to select the previous `CellGroup`:
SelectionMove[PreviousCell[], All, CellGroup]
This works in the latest version of *Mathematica* if `PreviousCell[]` is indeed a member of a `CellGroup`. But what if it is not? In this case this piece of code just does nothing AND nothing is returned as output, the latter directly contradicts the cited Documentation statement.
The described behavior is also inconsistent with other cases, for example
SelectionMove[EvaluationCell[], Previous, CellGroup]
selects the previous cell (which isn't a member of a `CellGroup`). Why then the same approach fail to work with `PreviousCell[]`?
The fact that `SelectionMove` doesn't return `$Failed` when it is unable to perform the requested operation is very unfortunate because it makes programming much more difficult: we must check whether selection is moved after EVERY call to `SelectionMove`, but there is no universal way to do this. In the best case it complicates code a lot and is very inefficient. In the worst case it makes it impossible to perform some simple tasks, as a bright example consider this StackExchange question:
- [Stop notebook from auto-scrolling upon printing](https://mathematica.stackexchange.com/q/157183/280)
The problem raised in this question is so simple that it is hard to believe it can't be solved reliably: we need just to move the selection after the input cell OR after the `CellGroup` containing output generated by this cell, and print new cell without scrolling using `NotebookWrite` (`CellPrint` always scrolls, hence it cannot be used in this case). But careful investigation showed that it is currently impossible to do this in an efficient way mainly due to the described bug.
My main question is:
<br>
**Is it possible to check in an efficient and reliable way, whether `SelectionMove` has failed or not?**
<sub>Reported to the support as [CASE:3968507].</sub>Alexey Popkov2017-11-04T17:05:57ZSelect CUDA C compiler in Mathematica 11.2 with CUDA paclet 11.2 ?
http://community.wolfram.com/groups/-/m/t/1193573
Dear,
I am using Mathematica 11.2 with the CUDA paclet 11.2.22 on Windows 10 64-bit.
I have Microsoft Visual Studio 2015 Community.
Although CUDACCompliers[] gives an empty list, (with only VS 2015 installed) CUDAFunctionLoad works fine in Mathematica.
But I have just installed Visual Studio 2017 Community to program in C++ with the just released nVIDIA CUDA Toolkit 9.0.
The CUDA paclet 11.2 is still based on nVIDIA CUDA Toolkit 8.0 which does not support VS 2017.
CUDACCompliers[] is still an empty list, although VS 2015 and VS 2017 are both installed.
But CUDAFunctionLoad selects VS 2017 to compile, which fails for the reason I just mentioned.
How can I overrule Mathematica's choice for VS 2017? It should select VS 2015 for now...
Kind regards,
BertBert Aerts2017-09-28T11:02:40ZCan not make a proper Classic Klein Bottle in version 11.20 of Mathematica.
http://community.wolfram.com/groups/-/m/t/1224031
Everything works fine in version 10, but in Version 11.20 there is a slice taken out of the Klein Bottle image...It does this no matter what version of the code I use in Version 11.2. It also does it while making a stylized Klein Bottle described here... http://members.wolfram.com/jeffb/visualization/klein.shtml Here is an example of code that makes a faulty output...Can someone come up with code that makes a proper Klein Bottle without a slice taken out out it in Version 11.20?
klein[u_, v_] := Module[{
bx = 6 Cos[u] (1 + Sin[u]),
by = 16 Sin[u],
rad = 4 (1 - Cos[u]/2),
X, Y, Z},
X = If[Pi < u <= 2 Pi, bx + rad Cos[v + Pi], bx + rad Cos[u] Cos[v]];
Y = If[Pi < u <= 2 Pi, by, by + rad Sin[u] Cos[v]];
Z = rad Sin[v];
{X, Y, Z}
]
ParametricPlot3D[klein[u, v], {u, 0, 2 Pi}, {v, 0, 2 Pi},
Axes -> False, Boxed -> False, ViewPoint -> {1.4, -2.6, -1.7}]Roy Scott2017-11-18T01:58:23ZGet FinancialData price history?
http://community.wolfram.com/groups/-/m/t/1097940
Since 16 may 2017, why am I not able to get the price history of US stock market tickers? Please see output below! Has anything changed?
In[9]:= FinancialData["GE", {{2017, 1, 3}, {2017, 5, 15}}]
Out[9]= Missing["NotAvailable"]
In[10]:= FinancialData["IBM", {{2017, 1, 3}, {2017, 5, 15}}]
Out[10]= Missing["NotAvailable"]sridev ramaswamy2017-05-18T12:09:48ZMachine Learning With Weighted Data
http://community.wolfram.com/groups/-/m/t/1223968
It does not appear easy within the current Classify, Predict or NetTrain functions to accommodate weighted data. It I am wrong about this, can anyone tell me how to do so. If anyone has a work around, could you please make a suggestion.
Motivation: I have several projects that would benefit from weighted data, the latest being to explain this really interesting paper using Mathematica: [https://arxiv.org/pdf/1602.04938.pdf][1] . The idea would basically be to emulate one machine learning method -- the complex one such as a neural net -- with a less opaque one (such as a decision tree) but with the emulator only being responsible for producing similar answers within a neighborhood of certain points. This way one could use something close to human language to explain what a more complex and more opaque classifier was doing.
I'm attaching a draft notebook that shows some preliminary work in this field and indicates where having a weighted classifier would be an improvement.
[1]: https://arxiv.org/pdf/1602.04938.pdfSeth Chandler2017-11-17T22:10:07ZFinding yoga-poses constellations in the night sky
http://community.wolfram.com/groups/-/m/t/1207400
![found constellations][9]
## Story Time ##
At its first meeting in 1922, the International Astronomical Union (IAU), officially adopted the list of 88 constellations that we use today.
These include 14 men and women, 9 birds, two insects, 19 land animals, 10 water creatures, two centaurs, one head of hair, a serpent, a dragon, a flying horse, a river and 29 inanimate objects. As many of us have (frustratingly) witnessed first hand while star-gazing - most of these bear little resemblance to their supposed figures. Instead, it is more likely that the ancient constellation-makers meant them to be symbolic, a kind of celestial "Hall of Fame" for their favorite animals or heroes.
This begs two questions I sought to answer with this project:
1. Can we 'do better' now with the WL's StarData[] curated data and Machine Learning functionality?
2. What if the ancient constellation-makers were slightly more creative? Say they looked up at the sky, and only saw yoga-poses!
Some examples of the found yoga-pose constellations, projected on images of the night sky are shown here, with a walk-through of the code below:
[![example yoga-pose constellations][1]][2]
## Yoga Poses ##
First things first, finding images for yoga poses.
Turns out, the WL has a built in YogaPose Entity Class with 216(!) available entities and their schematics.
A lot of these are very similar (e.g. palms facing up/down) and we therefore only select a subset of them, which differ substantially from each other:
yogaposes = EntityClass["YogaPose", All] // EntityList;
chosenPoses =
List /@ {4, 6, 7, 8, 11, 14, 23, 25, 28, 35, 38, 43, 51, 54, 56, 59,
63, 65, 71, 72, 76, 77, 78, 84, 88, 90, 98, 100, 102, 103, 110,
111, 112, 113, 116, 118, 119, 125, 126, 133, 139, 142, 149, 152,
153, 154, 155, 156, 160, 164, 165, 167, 172, 177, 178, 180, 182,
183, 184, 190, 193, 194, 195, 197, 202, 204, 206, 207, 209, 210,
212, 215, 216};
Shallow[Extract[yogaposes, chosenPoses]]
We write a wrapper to ensure the schematics are padded to give a square image and visualize our 73 constellations:
makeSquare[gr_, size_, bool_: False] :=
Block[{range, magnitudes, order, padding, newRanges, res},
range = AbsoluteOptions[gr, PlotRange][[1, 2]];
magnitudes = Abs[Subtract @@@ range];
order = Ordering[magnitudes, 2, Greater];
padding = Subtract @@ magnitudes[[order]]/2;
newRanges = {{-1, 1} padding + Last[range[[order]]],
First[range[[order]]]};
res = Show[gr, PlotRange -> Reverse@newRanges[[order]],
ImageSize -> size];
If[bool, Rasterize[res], res]]
Multicolumn[
makeSquare[#["SimplifiedSchematic"], 64] & /@
Most[Extract[yogaposes, chosenPoses]], 9, Frame -> All]
![constellations graphic][3]
## Neural Network ##
Our problem now can be worded as a classification one: *given 5-10 stars, classify them as one of the 73 yoga-poses constellations*
The problem with that formulation of-course is that 5-10 randomly selected points inside the constellation region isn't specific enough to differentiate between constellations. The image below shows that although 1000 sets of 10 points will define the shape, 10 points alone fall short:
![pointsInMesh][4]
Instead, we compute the Voronoi diagram of the points, using DistanceTransform to leverage already-optimized convolutional neural nets used for classifying images. Note that, even if (naturally) the results are less clear with fewer points (left to right) - the result with only 10 points at the far right is still quite recognizable to a human eye:
![distanceTransforms][5]
With this in-mind, we create a neural network similar to the [VGG16][6] neural network, with successive 5x5 convolutions, ReLu activation functions and Max pooling layers. Finally note that we use an average pooling instead of fully connected layers to reduce the number of parameters and the use of dropout layers.
conv[output_] := ConvolutionLayer[output, {5, 5}, "PaddingSize" -> 2]
pool[size_] := PoolingLayer[{size, size}, "Stride" -> size]
lenet = NetChain[
{conv[32], Ramp, conv[32], Ramp, pool[2], conv[64], Ramp, conv[64],
Ramp, pool[2], conv[128], Ramp, conv[128], Ramp, conv[128], Ramp,
PoolingLayer[{9, 9}, "Stride" -> 9, "Function" -> Mean],
FlattenLayer[], DropoutLayer[], 73, SoftmaxLayer[]},
"Output" -> NetDecoder[{"Class", Extract[yogaposes, chosenPoses]}],
"Input" -> NetEncoder[{"Image", {64, 64}, ColorSpace -> "Grayscale"}]
]
The training set then consists of generating 10-15 points inside the constellation region (resized and padded to allow for rotations later), and taking their DistanceTransform:
meshes[n_] :=
meshes[n] =
ImagePad[ColorNegate@
makeSquare[
yogaposes[[chosenPoses[[n, 1]]]]["SimplifiedSchematic"], 44], 10]
trainingSet[n_, m_, iter_] :=
With[{reg =
ImageMesh[meshes[n], CornerNeighbors -> False,
Method -> "MarchingSquares"]},
Thread[ColorConvert[
ImageAdjust[
DistanceTransform[Image[SparseArray[# -> 0, {64, 64}, 1]]]],
"Grayscale"] & /@
Clip[Round@RandomPoint[reg, {iter, m}], {1, 64}] ->
yogaposes[[chosenPoses[[n, 1]]]]]]
exampleTrainingSet =
Flatten[trainingSet[#, 10, 1] & /@ RandomInteger[{1, 73}, 10], 1]
The net took a couple of hours to train on ~100,000 examples on my CPU.
Here we import the trained net:
lenetTrained = Import["constellationsNet.wlnet"]
## Night Sky ##
We're almost ready to classify the night sky. We first need the location of the 10,000 brightest stars along with their Right Ascension and Declination:
brightest =
StarData[EntityClass[
"Star", {EntityProperty["Star", "ApparentMagnitude"] ->
TakeSmallest[10000]}], {"RightAscension", "Declination",
"ApparentMagnitude"}]
We can plot these on the night sky using no projection (i.e. RA Vs Dec), using the sinusoidal projection (taken by Kuba's excellent [answer][7] in SE) or on the celestial sphere given by the following transformations respectively:
$$sinusoidal:\{(\alpha -\pi ) \cos (\delta
),\delta \}$$
$$map to3D:\{\cos (\alpha ) \cos (\delta ),\sin
(\alpha ) \cos (\delta ),\sin (\delta
)\}$$
![brightestStarsProjections][8]
## Classification ##
Finally, we use these 10,000 brightest stars to compute a multivariate smooth kernel distribution out of which to sample from and a Nearest function to compute neighboring stars. We need to of-course use our own distance function on the celestial sphere:
wrap[list_] := Block[{xs, ys},
{xs, ys} = Transpose[list];
Thread[{Mod[xs, 2 \[Pi]], Mod[ys, \[Pi], -\[Pi]/2]}]]
mapTo3D[\[Alpha]_, \[Delta]_] = {Cos[\[Alpha]] Cos[\[Delta]],
Cos[\[Delta]] Sin[\[Alpha]], Sin[\[Delta]]};
dist[{u_, v_}, {x_,
y_}] := (#1[[1]] - #2[[1]])^2 + (#1[[2]] - #2[[2]])^2 + (#1[[
3]] - #2[[3]])^2 & @@ mapTo3D @@@ {{u, v}, {x, y}}
pts = wrap@QuantityMagnitude[UnitConvert[brightest[[6 ;;, ;; 2]]]];
nf = Nearest[pts, DistanceFunction -> dist]
sm = SmoothKernelDistribution[pts]
Our search algorithm is therefore defined as follows:
1. Pick a random position from the night sky distribution
2. Compute its 5-10 nearest neighbors
3. Classify those stars and their rotations by $\frac{2 \pi }{15}$
4. Select the rotation which gives the highest accuracy
5. Associate constellation to running association and repeat
rescale[list_] := Block[{xs, ys},
{xs, ys} = Thread[list];
Thread[Rescale[#, MinMax[#], {11, 54}] & /@ {xs, ys}]]
rotate[\[Alpha]_, pts_] :=
ImageAdjust@
DistanceTransform[
Image@SparseArray[
Round@RotationTransform[\[Alpha] , {65/2, 65/2}][rescale[pts]] ->
0, {64, 64}, 1]]
sky = <||>;
accumulate[] :=
Block[{pts =
nf[Mod[RandomVariate[sm] + {0, \[Pi]}, 2 \[Pi]] - {0, \[Pi]},
RandomInteger[{5, 10}]], \[Alpha], pred},
{\[Alpha], pred} =
Last[SortBy[
First /@
Table[Thread[{\[Alpha],
lenetTrained[rotate[\[Alpha], pts],
"TopProbabilities"]}], {\[Alpha], 0, 2 \[Pi], (2 \[Pi])/
15}], Last]];
If[Not[KeyExistsQ[sky, pred[[1]]]] ||
TrueQ[sky[pred[[1]], "Accuracy"] < pred[[2]]],
AssociateTo[sky,
pred[[1]] -> <|"Accuracy" -> pred[[2]],
"Image" ->
HighlightImage[
makeSquare[pred[[1]]["SimplifiedSchematic"],
64], {PointSize[Medium], Red,
Round[RotationTransform[\[Alpha], 65/2 {1, 1}][rescale@pts],
0.5]}], "Points" -> pts, "Angle" -> \[Alpha]|>], Nothing];]
We can import a precomputed association with 20 such constellations:
selectedAsc = Import["skyAsc.m.gz"]
## Results ##
Finally, we orient the schematic based on the optimally found angle and (manually) connect the dots.
orient[pts_, \[Alpha]_] :=
Round[RotationTransform[\[Alpha], 65/2 {1, 1}][rescale@pts], 0.5]
lines = {{{10, 7, 3, 1, 4, 9, 2, 5, 8}, {2, 6}}, {{7, 3, 1, 9,
6}, {10, 2, 1, 8}, {4, 3, 5}}, {{7, 8, 9, 4, 3, 2, 10}, {1, 2, 6,
5}}, {{10, 2, 1, 3, 9, 5}, {7, 1, 4, 6, 8}}, {{5, 4, 1, 2,
3}}, {{10, 4, 5, 7, 9, 3, 8}, {6, 1, 2, 3}}, {{9, 5, 6, 3, 1, 2,
7}, {7, 4, 8}}, {{7, 6, 2, 3, 5, 8}, {9, 5, 4}, {3, 1}}, {{6, 2,
1, 3, 4}, {2, 5}}, {{5, 6, 4, 1, 2, 8, 3, 7}}, {{10, 6, 3, 2,
7}, {5, 2, 1, 4, 9, 8}}, {{9, 5, 1, 2, 4, 8, 7}, {6, 3, 8}}, {{6,
10, 4, 2, 1, 3, 9, 7, 5, 8}}, {{6, 7, 3, 2, 1, 4, 5}}, {{10, 8,
3, 6, 7, 2, 5, 9}, {5, 1, 4, 3}}, {{6, 5, 2, 1, 3}, {1, 4}}, {{5,
1, 2, 3}, {1, 4}}, {{6, 3, 2, 1, 4}, {3, 5}, {2, 7}}, {{3, 6, 8,
2, 4, 5, 1, 7}}, {{3, 1, 2, 6, 5}}};
makeSquareAndScale[gr_, \[Alpha]_, size_, opacity_] :=
Block[{range, magnitudes, order, padding, newRanges, s},
range = AbsoluteOptions[gr, PlotRange][[1, 2]];
magnitudes = Abs[Subtract @@@ range];
order = Ordering[magnitudes, 2, Greater];
padding = Subtract @@ magnitudes[[order]]/2;
s = size/First[magnitudes[[order]]];
newRanges = {{-1, 1} padding + Last[range[[order]]],
First[range[[order]]]};
Graphics[{Opacity[opacity],
GeometricTransformation[
GeometricTransformation[
gr[[1]], {ScalingMatrix[{s,
s}], -1 First /@ (s Reverse@newRanges[[order]])}],
RotationTransform[\[Alpha], (size + 1)/2 {1, 1}]]},
ImageSize -> size]]
With[{l = Map[Line, lines, {2}]},
Multicolumn[
Table[Show[
makeSquareAndScale[
Keys[selectedAsc][[i]][
"SimplifiedSchematic"], -selectedAsc[[i, -1]], 64, 0.25],
Graphics[
GraphicsComplex[
rescale[selectedAsc[[i, -2]]], {l[[i]], Red, PointSize[Large],
Point /@ Sort /@ Flatten /@ lines[[i]]}]],
ImageSize -> 150], {i, 20}], 5, Frame -> All,
Appearance -> "Horizontal"]]
![found constellations][9]
These can also be superimposed on the full night-sky:
Graphics[{With[{l = Map[Line, lines, {2}]},
Table[GraphicsComplex[
selectedAsc[[i, -2]], {Orange, Thickness[Large], l[[i]], Red,
PointSize[Medium], Point /@ Sort /@ Flatten /@ lines[[i]]}], {i,
20}]], Opacity[0.25], PointSize[Small],
Point[Complement[pts, Flatten[Values@selectedAsc[[All, -2]], 1]]]},
ImageSize -> 750]
![found overlay][10]
sinusoidal[\[Alpha]_, \[Delta]_] = {(\[Alpha] - \[Pi]) Cos[\[Delta]], \
\[Delta]}
Graphics[{With[{l = Map[Line, lines, {2}]},
Table[GraphicsComplex[
sinusoidal @@@ selectedAsc[[i, -2]], {Orange, Thickness[Large],
l[[i]], Red, PointSize[Medium],
Point /@ Sort /@ Flatten /@ lines[[i]]}], {i, 20}]],
Opacity[0.25], PointSize[Small],
Point[sinusoidal @@@
Complement[pts, Flatten[Values@selectedAsc[[All, -2]], 1]]]},
ImageSize -> 750]
![found sinusoidal][11]
It is then a matter of Overlaying these found constellations to existing images of night skies to produce the images at the beginning of the post:
overlay[{img_, constellation_}, {size_, loc_, opac_}] :=
ImageCompose[imgs[[img]],
ImageResize[
Show[makeSquareAndScale[
Keys[selectedAsc][[constellation]][
"SimplifiedSchematic"], -selectedAsc[[constellation, -1]], 64,
opac], Graphics[
GraphicsComplex[
rescale[selectedAsc[[constellation, -2]]], {White,
Thickness[0.0075], Line /@ lines[[constellation]], White,
PointSize[.025],
Point /@ Sort /@ Flatten /@ lines[[constellation]]}]],
ImageSize -> 1000], size], Scaled[loc]]
## Conclusions / Lessons Learned ##
- It IS possible to find collections of stars in the night sky matching all sorts of shapes. Other built-in entities to try are Pokemons, Dinosaurs etc
- Not all the found constellation work 'perfectly'
- Machine Learning is a powerful tool, and the WL implementation makes it easy to get started
- Reformulating the problem to an easier, already solved problem (e.g. points -> 2D image input) can help Classification Accuracy
- Perhaps the most interesting aspect of neural networks now, is diversifying its applications - so be creative
This work was presented at the 2017 WTC (as part of a larger talk entitled "(De)Generative Art"). <br>
I look forward to any comments/suggestions.
<br>
<br>George
<br>
<br> PS The editor doesn't seem to let me attach the .wlnet and .m.gz files.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=exampleImages.png&userId=11733
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=exampleImages.png&userId=616023
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=4214constellationsGraphic.png&userId=616023
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=randomPointsInmesh.png&userId=616023
[5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=distanceTransforms.png&userId=616023
[6]: https://arxiv.org/pdf/1409.1556v6.pdf
[7]: https://mathematica.stackexchange.com/questions/89668/geoprojection-for-astronomical-data-wrong-ticks/89792#89792
[8]: http://community.wolfram.com//c/portal/getImageAttachment?filename=maps.png&userId=616023
[9]: http://community.wolfram.com//c/portal/getImageAttachment?filename=foundConstellations.png&userId=616023
[10]: http://community.wolfram.com//c/portal/getImageAttachment?filename=overlaidConstellations.png&userId=616023
[11]: http://community.wolfram.com//c/portal/getImageAttachment?filename=overlaidConstellationsSinusoidal.png&userId=616023George Varnavides2017-10-24T04:17:44ZCould not read result file, missing data matrix, the file might be courupt
http://community.wolfram.com/groups/-/m/t/1223831
I'm trying to run a Modelica code in SystemModeler:
model code
//constants
constant Real pi = 2 * Modelica.Math.asin(1.0);
parameter Real mukpc = 0.2;
parameter Real muspc = 0.0;
parameter Real mukps = 0.0;
parameter Real musps = 0.0;
parameter Real mukcs = 0.2;
parameter Real muscs = 0.0;
parameter Real mp = 1.0;
parameter Real js = 2.0;
parameter Real b1 = 0.01;
parameter Real b2 = 0.01;
parameter Real b3 = 0.015;
parameter Real r1 = 0.01;
parameter Real r2 = 0.003;
parameter Real r3 = 0.003;
parameter Real Ts = 1;
parameter Real Tw = 0.9;
parameter Real absTol = 0.01;
//variables
Real X1, X2, X3, Y1, Y2, Y3, tets, V1, V2, V3, omgs, A1, A2, A3, alps, Fp1, Fp2, Fp3, Fncp1, Fncp2, Fncp3, Ffcp1, Ffcp2, Ffcp3, Tcp1, Tcp2, Tcp3, Ffsp1, Ffsp2, Ffsp3, Fnsp1, Fnsp2, Fnsp3, Tfsp1, Tfsp2, Tfsp3, Text, Tfsc, Fscx, Fscy, T1, T2, T3;
initial equation
tets = pi / 3;
omgs = 0.1;
equation
//inpute forces
Fp1 = if mod(time, 3 * Ts) < Tw then 1000 else 0;
Fp2 = if mod(time, 3 * Ts) < Tw + Ts and mod(time, 3 * Ts) > Ts then 1000 else 0;
Fp3 = if mod(time, 3 * Ts) < Tw + 2 * Ts and mod(time, 3 * Ts) > 2 * Ts then 1000 else 0;
Text = 0;
//derivatives
V1 = der(X1);
V2 = der(X2);
V3 = der(X3);
A1 = der(V1);
A2 = der(V2);
A3 = der(V3);
omgs = der(tets);
alps = der(omgs);
//kinematics
X1 = r1 * cos(tets) + r2;
Y1 = r1 * sin(tets);
X2 = r1 * cos(tets + 2 * pi / 3) + r2;
Y2 = r1 * sin(tets + 2 * pi / 3);
X3 = r1 * cos(tets + 4 * pi / 3) + r2;
Y3 = r1 * sin(tets + 4 * pi / 3);
//dynamic equations
//shaft
js * alps = T1 + T2 + T3 - Text - Tfsc;
Fscx = Fnsp1 + Fnsp2 + Fnsp3;
Fscy = Ffsp1 + Ffsp2 + Ffsp3;
//Fsc =sqrt(Fscx^2+Fscx^2);
T1 = Fnsp1 * Y1 - Ffsp1 * X1 - Tfsp1;
T2 = Fnsp1 * Y2 - Ffsp2 * X2 - Tfsp2;
T3 = Fnsp3 * Y3 - Ffsp3 * X3 - Tfsp3;
//p1
mp * A1 = Fp1 - Ffcp1 - Fnsp1;
Ffsp1 = Fncp1;
Tfsp1 + Fnsp1 * (b2 / 2 - Y1) = Ffsp1 * b1 - Fp1 * b2 / 2 + Tcp1;
//p2
mp * A2 = Fp2 - Ffcp2 - Fnsp2;
Ffsp2 = Fncp2;
Tfsp2 + Fnsp2 * (b2 / 2 - Y2) = Ffsp2 * b1 - Fp2 * b2 / 2 + Tcp2;
//p3
mp * A3 = Fp3 - Ffcp3 - Fnsp3;
Ffsp3 = Fncp3;
Tfsp3 + Fnsp3 * (b2 / 2 - Y3) = Ffsp3 * b1 - Fp3 * b2 / 2 + Tcp3;
//friction
Ffcp1 = -mukpc * (abs(Fncp1) + abs(Tcp1 / b3)) * sign(V1);
Ffcp2 = -mukpc * (abs(Fncp2) + abs(Tcp2 / b3)) * sign(V2);
Ffcp3 = -mukpc * (abs(Fncp3) + abs(Tcp3 / b3)) * sign(V3);
Tfsp1 = 0;
Tfsp2 = 0;
Tfsp3 = 0;
Ffsp1 = 0;
Ffsp2 = 0;
Ffsp3 = 0;
Tfsc = mukcs * sqrt(Fscx ^ 2 + Fscy ^ 2) * sign(omgs);
end code;
But I get the error:
> Error
> Simulation of foo exited with an error(-1073741819), see log window for details
![enter image description here][2]
and then:
> Error: Could not read result file "C:/Users/foo/AppData/Local/Temp/WolframSystemModeler-5.0.0/sme.5.0.0_1510933977_23281.mat". Missing data_2 matrix, the file might be corrupt.
![enter image description here][3]
**I would appreciate if you could help me understand what is the problem and how can I solve it?**
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=5892Untitled.png&userId=1014408
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=4053Untitled.png&userId=1014408
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=10469Untitled.png2.png&userId=1014408Foad Sojoodi Farimani2017-11-17T16:05:50ZCan this expression be written in closed form?
http://community.wolfram.com/groups/-/m/t/1222861
Observe the following Wolfram Mathematica expression which results in a table of integer sequences:
b[n_, k_] := b[n, k] = If[n == 0, 1, If[n == 1, If[k > 0, 1, 0], Sum[Binomial[n-1, r-1]*b[r-1, k-1]*b[n-r, k-1], {r, 1, n}] ] ]; t [n_, k_] := b[n, k] - If[k > 0, b[n, k-1], 0]; Table[Table[t[n, k], {k, 0, n}], {n, 0, 10}]
I'm struggling to interpret what this is actually doing. Clearly it's quite a complicated mess of if statements, but I'm not entirely sure if they are necessary or not. I obtained the expression from here:
https://oeis.org/A195581
Because I wanted to determine an expression (dependent on the two variables n and k) which would result in table of integers seen there. Would somebody please write the Mathematica code above in closed form so that it's easier to understand? If this is not possible, can you at least try to explain what's going on, and perhaps give an expression which models the above in *most* situations?John Travolski2017-11-16T03:22:25ZCorrection to Documentation for Cepstrogram & Cepstrogram
http://community.wolfram.com/groups/-/m/t/1218929
Just as an FYI, the default partition size for these two routines is
2^(Round[Log[2,Length[list]]]+1)
rather than
2^Round[Log[2,Length[list]]]+1
as stated in the docs. The default step size is any number between `Round[partition/3]` and the length of the list.
Cheers, ScottScott Guthery2017-11-09T01:11:25ZMy Mathematica keeps asking for Activation Key on startup?
http://community.wolfram.com/groups/-/m/t/1222950
Hi all,
Upon startup, every now and then, my Mathematica program will ask for an activation key. **It is not expired.**I will go on the Wolfram User Portal to find my activation key, and I will enter it. It will not work. I will have to uninstall and redownload Mathematica.
Does anybody have any suggestions or know what may be causing this issue?
Best,
NNish Paul2017-11-16T01:31:52Z[✓] Pasting numerical expressions into Input Cells?
http://community.wolfram.com/groups/-/m/t/1220439
I would like to create a PasteButton that will paste the following expression into an existing Input Cell and display in the Input Cell with the 6 place precision while keeping the full underlying machine precision. From my efforts I suspect that this is not possible because the only way to obtain the shortened display is to paste an entire Output Cell..
expr = N[x/3 + 2 y/3]
giving
0.333333 x + 0.666667 y
The following is just the simplest approach, which does not work. I haven't found any other approach that works either.
PasteButton["1/3", expr]
The resulting button pastes the following where the shorter form would be preferable.
0.3333333333333333` x+0.6666666666666666` y
Does anyone know a way to specify a PasteButton that will paste the 6 place Output display into an Input Cell?David J M Park Jr2017-11-12T06:12:37ZSort a table with respect to other?
http://community.wolfram.com/groups/-/m/t/1221539
Hi,
I have two 36 X 3 tables. First columns in both of them are independent terms/numbers, while second and third columns have same indices (numbers) from 1 - 6, but in different orders. I want to arrange the rows of table B in the same order as in A, with respect to second and third elements of them.
For example if
A = {{X, 2, 1}, {Y, 2, 3}, {Z, 2, 4}...}
and
B = {{x, 2, 3}, {y, 2, 4}, {z, 2, 1}....}
I want B to be rearranged to
B_new = {{z, 2, 1}, {x, 2, 3}, {y, 2, 4}....}
Will be thankful for any suggestion. Attached is the *.nb file having these tables.S G2017-11-14T10:28:55ZTrace path of a point from a body using MultiBody library in SM
http://community.wolfram.com/groups/-/m/t/1222597
In many CAE software (SolidWorks, RecurDyn...) it is possible to trace the curve that a certain point follows during a simulation. This is useful to show the motion of the point and compare the results with the expected ones or even with the output of another software.
Is there a way to do this in SystemModeler? I am working with mutibody mechanical systems using the MultiBody library.
For example, in the following simulation:
![enter image description here][1]
The goal is to show the arc followed by the sphere during the simulation. The curve is to be drawn as the motion occurs and export the data to a plain text file (if possible).
Best regards,
William
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=SMMotion.gif&userId=1190887JOSE OLIVER2017-11-15T11:03:39ZDatabin not updated?
http://community.wolfram.com/groups/-/m/t/1222161
Hi, I recently started to use Data Drop to store sensor data for analysis and visualization,
Creating the bin was easy and my Python program on my Raspberry Pi writes data to it via the Web API.
When I refer to it in Mathematica on my Mac (as Databin["psyLf0X2"] ) it tells me it has 6 entries and every time I evaluate it again that does not change.
However when I look in My Databins it actually has 14 entries... Why is it not updated?
What am I missing?Ted vanderTogt2017-11-15T15:53:18ZThe voters have spoken - but what did they say? (Part 3)
http://community.wolfram.com/groups/-/m/t/1222959
Introduction
----------
Elections are fundamental to our democracies and way of life. The electorate bestows power on the lawmakers and public officials whose legitimisation is derived solely from the fact that they are elected. In recent times, many elections, such as the presidential elections the United States and the Brexit and general elections in the United Kingdom have seen a very pronounced polarisation the electorate. Can mathematics and the Wolfram Language help to better understand this crucial process of the most fundamental decision making process in societies? The main question I want to discuss in a sequence of posts is what can be learnt from elections. So this post is about computational politics.
It is important to notice that democracies rely on a general acceptance of the result of elections. Without that consensus a peaceful co-existence is at risk. Societies agree on certain electoral procedures and the results have to be understood to be final. There are, however, more fundamental questions such as how do we derive the "voters will" from ballots. That is a highly non-trivial problem. Also, elected politicians often claim that they represent the "will of the people". But is that true and what is the "will of the people"? Hence, my question: the voters have spoken - but what did they say?
This is the third of a series of posts:
1. [Electoral systems - interpreting the will of the voters][1]
2. Gerrymandering - shaping the voters' will to your liking
3. **Analysis of the Brexit vote - how do we know what did the voters really want?**
This part of the sequence of posts will analyse the results of the Brexit vote in the United Kingdom. The question asked was basically a yes/no question and with small majority the social choice was to conduct Brexit. The rules of this process have been followed and the decision is binding. There is, however, a related question: Did the outcome in this particular case actually give sufficient evidence that this is the "will of the people" as many politicians state. If we want to evaluate the power of evidence there are mathematically established procedures. Similar procedures are for example applied in medical trials to evaluate the evidence provided by data to support a particular hypothesis. We will apply this type of procedure to the Brexit elections.
All posts were developed with [Bjoern Schelter][2], who is also a member of this community. He is indeed the expert on these methods.
**Disclaimer:** this post might be modified slightly over the next couple of days.
Data download and preparation
----------
We will start by obtaining some of the data required for this post. The [central dataset is provided by the electoral commission][3]. It is attached to this post. We download and them import the data like so:
SetDirectory[NotebookDirectory[]]
votingData = Import["EU-referendum-result-data.csv"];
Here is a short extract of the data:
votingData[[1 ;; 5]] // Transpose // TableForm
![enter image description here][4]
Next, we will need some data on the local authorities/electoral districts from [here][5]. We then import:
localauthorities = Import["Local_Authority_Districts_December_2016_Generalised_Clipped_Boundaries_in_Great_Britain.shp"];
explanationShape = Import["Local_Authority_Districts_December_2016_Full_Clipped_Boundaries_in_Great_Britain.csv"];
These data sets contain information about the shape and location of the local authority boundaries and data on how to match that with the voting outcome in the first file. To get a first glance at the data we re-arrange and organise the data like so:
Monitor[areas =
Table[QuantityMagnitude[UnitConvert[If[Length[#] > 1, Total[GeoArea /@ #], GeoArea[#]]], "SI"] & @(localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]), {k, 1, Length[localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]]]}];, k]
and
translateshpvote = SortBy[(Flatten /@ Transpose[{SortBy[Transpose[{Range[380], areas}], Last], SortBy[explanationShape[[2 ;;]], #[[9]] &]}])[[All, {1, 4}]], First];
The first representation is not very telling. It shows the outcome of the election.
Graphics[Flatten[
Table[{ColorDataFunction["TemperatureMap", "Gradients", {0, 1}, Blend["TemperatureMap", #]& ][(Select[
votingData[[2 ;;]], #[[4]] == translateshpvote[[k, 2]] &][[1]])[[-3]]/100.], localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1, 380}],1], AspectRatio -> GoldenRatio]
![enter image description here][6]
The "blue-r" the colour the higher the percentage of leave votes. Yellow indicates a preference for remain. We will need to clean the data further and will then plot it with a threshold to see which regions decided for Leave and which ones for Remain:
substitute = ({#[[1]][[{1, 4}]] -> #[[2]][[{1, 4}]], #[[2]][[{1,
4}]] -> #[[1]][[{1, 4}]]} &@((Flatten /@
Transpose[{SortBy[Transpose[{Range[380], areas}], Last],
SortBy[explanationShape[[2 ;;]], #[[9]] &]}])[[205 ;; 206]]))
translateshpvoteclean = translateshpvote /. substitute;
Graphics[Flatten[
Table[{ColorDataFunction["TemperatureMap", "Gradients", {0, 1}, Blend["TemperatureMap", #]& ][
If[# > 0.5, 1, 0] & @((Select[Select[votingData[[2 ;;]], #[[1]] < 381 &], #[[4]] == translateshpvoteclean[[k, 2]] &][[1]])[[-3]]/100.)],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1, 380}],1], AspectRatio -> GoldenRatio]
![enter image description here][7]
The strong divide between England and Scotland becomes apparent from that figure. We will use the following style of the image in much of the remainder of the post:
Graphics[Flatten[
Table[{ColorData[
"RedGreenSplit"][((Select[
Select[votingData[[2 ;;]], #[[1]] < 381 &], #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-3]]/100.)],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1, 380}],
1], Background -> Black, AspectRatio -> GoldenRatio]
![enter image description here][8]
What does the Brexit election outcome tell us? What did voters want?
----------
UK wide the percentage of leave votes (with respect to all valid votes) is:
N[Total[votingData[[2 ;;, 13]]]/Total[votingData[[2 ;;, 11]]]]
51.89%. This is in percentage of all valid votes. The percentage of leave with respect to the entire electorate is:
N[Total[votingData[[2 ;;, 13]]]/Total[votingData[[2 ;;, 6]]]]
37.4%. This is the percentage of the entire electorate who voted for remain.
N[Total[votingData[[2 ;;, 12]]]/Total[votingData[[2 ;;, 6]]]]
34.7%. The percentage of non-voters is:
N[(Total[votingData[[2 ;;, 6]]] - Total[Flatten@votingData[[2 ;;, 12 ;; 13]]])/Total[votingData[[2 ;;, 6]]]]
27.8%. This means that we do not know the will of about 28% of all voters. Following the predefined social choice procedure, the vote was narrowly won by leave. That does not necessarily imply that this is "the will of the majority of people".
What would the result be with indirect representation?
----------
If we had a representational vote we would have to count the districts for remain and leave
{Count[#, "Remain"], Count[#, "Leave"]} &@((#1 > #2) & @@@ votingData[[2 ;;, {12, 13}]] //. {False -> "Leave", True -> "Remain"})
{119,263}. This result is even more extreme for leave. This is because the population in the local authorities is very different.
Histogram[votingData[[2 ;;, 6]], PlotRange -> All, PlotTheme -> "Marketing", ImageSize -> Large,
FrameLabel -> {"population", "number of authorities"}, LabelStyle -> Directive[Bold, 15]]
![enter image description here][9]
This shows that there are large fluctuations in the population of the areas. Let's represent that slightly differently. We first collect data about Leave votes, Electorate, Turnout and Electorate.
electionEUresolved =
Transpose@({#[[1]], #[[2]], #[[3]], #[[4]]} &@Transpose@((({#[[13]], #[[11]], #[[9]], #[[6]]} & /@ (Reverse@SortBy[votingData[[2 ;;]], #[[9]] &])))));
BubbleChart[{#[[3]], N@#[[1]]/#[[2]], #[[4]]} & /@ electionEUresolved,
BubbleSizes -> Small,
ChartLabels ->
Placed[(Reverse@SortBy[votingData[[2 ;;]], #[[9]] &])[[
All, {5, 3}]], Tooltip],
Epilog -> {Red, Line[{{50, 0.5}, {100, 0.5}}]},
PlotTheme -> "Marketing", BubbleScale -> "Area", ImageSize -> Large,
FrameLabel -> {"turnout", "percentage leave votes"},
LabelStyle -> Directive[Bold, 15]]
![enter image description here][10]
The bubble sizes (areas) correspond to the size of the electorate. Let's see whether there is a correlation with turnout:
lm = LinearModelFit[{#[[3]], N@#[[1]]/#[[2]]} & /@ electionEUresolved,x, x, Weights -> electionEUresolved[[All, 4]]]
![enter image description here][11]
The following table shows that the positive slope is highly significant.
lm["ParameterTable"]
![enter image description here][12]
We can add that to the figure:
Show[BubbleChart[{#[[3]], N@#[[1]]/#[[2]], #[[4]]} & /@
electionEUresolved, BubbleSizes -> Small,
ChartLabels ->
Placed[(Reverse@SortBy[votingData[[2 ;;]], #[[9]] &])[[
All, {5, 3}]], Tooltip],
Epilog -> {Red, Line[{{50, 0.5}, {100, 0.5}}]},
PlotTheme -> "Marketing", BubbleScale -> "Area", ImageSize -> Large,
FrameLabel -> {"turnout", "percentage leave votes"},
LabelStyle -> Directive[Bold, 15]],
Plot[lm[x], {x, 50, 85}, PlotStyle -> White]]
![enter image description here][13]
Seeing such a linear dependence, raises the natural question to which extent the result is biased by such correlations in the data. Linear Models are a powerful tool to account for various potential influencing factors on the final outcome of the election.
Linear Modelling
----------
Linear Models are a generalisation of linear regression; linear models allow us to model a Gaussian Distributed outcome by various inputs that may or may not be continuous. To demonstrate this we use the simple voting data; the dependent, outcome, variable is the percentage of people who voted to leave the EU.
percentleave = N[votingData[[2 ;;, 13]]/votingData[[2 ;;, 11]]];
collatedData = (Flatten /@
Transpose[{Table[
Select[votingData, explanationShape[[k, 2]] === #[[4]] &], {k,
2, Length[explanationShape[[1 ;;]]]}],
explanationShape[[2 ;;]]}]);
We can then prepare the data for the linear model:
inputLinearModel =
Flatten /@ Transpose[{# - Mean[#] &@ collatedData[[All, 9]], collatedData[[All, 3]], # - Mean[#] &@ (collatedData[[All, 6]]/collatedData[[All, 30]]),
N[collatedData[[All, 13]]/collatedData[[All, 11]]]}];
The steps above generate the table that defines the input into the linear model. All continuous input variables have been centralised such that their mean is zero. This is not needed but simplifies the interpretability of the results. This input variables to the linear model are turnout, i.e., the fraction of people who voted in the given local administrative area, region, i.e., the rough geographical region (non continuous, nominal variable), and a proxy for the population density.
lmodel = LinearModelFit[inputLinearModel, {turnout, region, popdensity}, {turnout, region, popdensity}, NominalVariables -> region];
lmodel["ParameterTable"]
![enter image description here][14]
The model output shows the estimates with respect to the intercept (1) in the table. The missing region is Yorkshire and The Humber such that Mathematica relates all results to this region. Note, that the estimate of 0.548879 means that Yorkshire and The Humber would have voted to leave the EU with 54.89%; this is the model output and does not necessarily match the actual result of the election. Since all continuous variables are mean zero, these 54.89% would be the result for Yorkshire and The Humber under the assumption that the have a typical UK average population density and turnout. The model can obviously be used to approximate the result for Yorkshire and The Humber better by evaluating the function with the correct values for the area. All other influencing factors are estimated with respect to this result. In other words to obtain the result for Scotland, the Scotland result would need to be added to the Yorkshire and The Humber result; so the best model guess for Scotland if otherwise indistinguishable from the while UK would have been -23.93%+54.89%=30.96%. This obviously deviates from the actual result for Scotland, which is likely caused by the other covariates, the turnout and population density in Scotland are different from the average UK value.
Next we want to analyse the same model but with data for Gibraltar and Ni added. We first need the respective areas of the regions:
fixNIGI =
Join[#[[1 ;; -2]],
ConstantArray[Missing[],
8], {#[[-1]]}, {Missing[]}] & /@ (Flatten /@
Transpose[{Select[
votingData, #[[1]] > 380 &], {QuantityMagnitude[
UnitConvert[
Entity["AdministrativeDivision", {"NorthernIreland",
"UnitedKingdom"}]["Area"], Quantity[1, (("Meters")^2)]]],
QuantityMagnitude[
UnitConvert[Entity["Country", "Gibraltar"]["Area"],
Quantity[1, (("Meters")^2)]]]}}]);
Then we can add the data to the table, create the input the linear model and estimate it:
collatedDatacomplete = Join[collatedData, fixNIGI];
inputLinearModelcomplete =
Flatten /@ Transpose[{# - Mean[#] &@ collatedDatacomplete[[All, 9]],
collatedDatacomplete[[All, 3]], # - Mean[#] &@ (collatedDatacomplete[[All, 6]]/collatedDatacomplete[[All, 30]]),
N[collatedDatacomplete[[All, 13]]/collatedDatacomplete[[All, 11]]]}];
lmodelcomplete = LinearModelFit[inputLinearModelcomplete, {turnout, region, popdensity}, {turnout, region, popdensity}, NominalVariables -> region]
lmodelcomplete["ParameterTable"]
![enter image description here][15]
The values match the result above quite well, obviously there is now a result for NI as well. The key thing about linear models is that one can combine the various areas to get a area weighted estimate for the election result. To achieve this we need to obtain the parameters from the table above.
parametermatrix = Delete[lmodelcomplete["ParameterTableEntries"], {{2}, {-1}}]
To achieve a area weighted result, we need to obtain the weights for each region which is defined as the number of entries per area divided by the total number of entries:
weights = N[#/Total[#]] &@RotateRight[SortBy[Tally[inputLinearModelcomplete[[All, 2]]], First]][[All, 2]]
As all parameters are given with respect to one area and present differences to this area, we need to add all estimates to the intercept, see example for Scotland above. This is achieved through linear algebra, by defining a so called contrast matrix.
contrast = IdentityMatrix[Length[weights]];
contrast[[1]] = ConstantArray[1, Length[weights]];
contrast = Transpose[contrast]
Multiplying the matrix to the parameter estimates gives the response functions, the model results for each area.
responsefunctions = contrast.parametermatrix[[All, 1]]
Multiplying this with the weights for each area results in a weighted mean for the UK. It is worth noting that this result is no longer influenced by the turnout and the average population density as these are 0 for the UK, as we subtracted their overall mean in the beginning.
weights.responsefunctions
So the model's referendum result would have been 52.99% in favour of leave, so higher than the actual result. Importantly, Linear Models provide errors for each parameter estimate, such that we can estimate the overall uncertainty of this result. As standard errors are not additive but variances are, we obtain the estimated standard error for this result as follows, assuming independence of the estimates
seresponsefunctions = contrast.(parametermatrix[[All, 2]]^2)
The final standard error is
Sqrt[weights.seresponsefunctions]
0.0255668. This means that best model guess is that 52.99% p/m 2.56% are in favour of leave, using a Gaussian approximation, this implies that there is a
Probability[x < -(weights.responsefunctions - 0.5)/Sqrt[weights.seresponsefunctions], x \[Distributed] NormalDistribution[0, 1]]
12.11% chance that this result was obtained although the majority of the UK electorate does actually not want to leave the EU. We emphasise that by no means are we challenging the result of the referendum, we are just arguing that given model assumptions the results obtained are actually compatible with a population split in which the majority does not want to leave the EU. In many fields such as Science and Medicine, a 12.11% 'risk' that the result is actually a false positive, i.e., the conclusions drawn and actions taken are wrong, is considered too high, 5% or 1% are typical values, in particle Physics the threshold is much lower. So it remains to be investigated if statements that the election results reflect the true opinion of the population. Again, we do not challenge the outcome of the election as there are clear rules to the election that are known in advance such as that it is everybody's free choice to vote or not and how the final result is obtained. There is just a potential non-causal link to people's real opinions, which is often assumed but not necessarily correctly so.
Getting more data...
----------
An interesting thing to check is whether we can learn something from adding additional data on the various voting districts. The data is from [here][16]. We import the file with the data
additionaldata = Import["EU-referendum-results-and-characteristics-data.xlsx"];
We also use the electoral region id data from above:
ids = collatedDatacomplete[[All, 4]];
Then we need to deal with incomplete data/clean the data up:
fillers = ConstantArray[Missing[], #] & /@ Length /@ additionaldata[[2 ;; 11, 4]];
We now can generate a fixed data table:
fulldatatable =
Table[Join[collatedDatacomplete[[k]], Flatten@Table[If[Length[#] > 0, #, fillers[[i - 1]]] &@
Select[additionaldata[[i]], MemberQ[#, ids[[k]]] &], {i, 2, 11}]] /. "" -> Missing[], {k, 1, Length[ids]}];
To better work with the data we then generate a look-up table for the various columns:
datadescriptors =
Rule @@@ (Transpose[{Range[Length[#]], #}] & @
Join[votingData[[1]], explanationShape[[1]],
additionaldata[[2, 9]], additionaldata[[3, 1]],
additionaldata[[4, 8]], additionaldata[[5, 9]],
additionaldata[[6, 8]], additionaldata[[7, 4]],
additionaldata[[8, 8]], additionaldata[[9, 4]],
additionaldata[[10, 3]], additionaldata[[11, 3]]])
![enter image description here][17]
The following shows again a part of the table of all the data we can use now:
![enter image description here][18]
We can then generate some first part of the input for the linear model.
inputLinearModel =
Flatten /@ Transpose[{# - Mean[#] &@ collatedData[[All, 9]], collatedData[[All, 3]], # - Mean[#] &@ (collatedData[[All, 6]]/collatedData[[All, 30]]),
N[collatedData[[All, 13]]/collatedData[[All, 11]]]}];
Again we need to clean the data
fulldatatable[[All, 198]] /. "#" -> Missing[]
subMissing[x_] := If[StringContainsQ[ToString[FullForm[x]], "Missing"], Missing[], x ]
SetAttributes[subMissing, Listable]
This generates the full input for the linear model:
inputLinearModelcomplete =
Flatten /@
Transpose[{fulldatatable[[All, 3]], (# - Mean[#]) &@
fulldatatable[[All,
9]], (# - Mean[#]) &@ (fulldatatable[[All, 6]]/
fulldatatable[[All, 30]]), (# -
Mean[DeleteMissing[#]]) &@ (subMissing /@ (fulldatatable[[
All, 198]]/fulldatatable[[All, 6]])), (# -
Mean[DeleteMissing[#]]) &@ (fulldatatable[[All, 243]] /.
"#" -> Missing[]), (# -
Mean[DeleteMissing[#]]) &@ (subMissing /@ (fulldatatable[[
All, 227]]/fulldatatable[[All, 6]])),
N[fulldatatable[[All, 13]]/fulldatatable[[All, 11]]]}];
Here are a couple of lines of the input data we choose for our linear model:
inputLinearModelcomplete[[1 ;; 20]] // TableForm
![enter image description here][19]
Some more cleaning:
inputLinearModelcomplete = Select[subMissing /@ inputLinearModelcomplete, ! MemberQ[#, Missing[]] &];
Next we fit the linear model:
lmodelfulldata =
LinearModelFit[inputLinearModelcomplete, {region, turnout, popdensity, highqual, earning, nchildren}, {region, turnout, popdensity, highqual,
earning, nchildren}, NominalVariables -> {region}];
lmodelfulldata["ParameterTable"]
![enter image description here][20]
It is easy to see that the income and the number of children in the household significantly influence the preference for/against Brexit. We now use the parameter matrix to compute the weights and contrasts as before:
parametermatrix = lmodelfulldata["ParameterTableEntries"][[1 ;; 12]];
weights = N[#/Total[#]] &@RotateRight[SortBy[Tally[inputLinearModelcomplete[[All, 2]]], First]][[All, 2]];
contrast = IdentityMatrix[Length[weights]];
contrast[[1]] = ConstantArray[1, Length[weights]]; contrast = Transpose[contrast];
With the contrasts we can calculate the response functions:
responsefunctions = contrast[[1 ;; 12, 1 ;; 12]].parametermatrix[[1 ;; 12, 1]]
What is the best estimate for the "will of the voters"?
----------
We an now also use the linear model to estimate what the most likely scenario (based on the model) for the "underlying will of the voter" is given the data from above and the result of the election. We first prepare the data we need, which this time includes the outcome of the election.
inputLinearModelcompletewRcode =
Flatten /@
Transpose[{fulldatatable[[All, 4]],
fulldatatable[[All, 3]], (# - Mean[#]) &@
fulldatatable[[All,
9]], (# - Mean[#]) &@ (fulldatatable[[All, 6]]/
fulldatatable[[All, 30]]), (# -
Mean[DeleteMissing[#]]) &@ (subMissing /@ (fulldatatable[[
All, 198]]/fulldatatable[[All, 6]])), (# -
Mean[DeleteMissing[#]]) &@ (fulldatatable[[All, 243]] /.
"#" -> Missing[]), (# -
Mean[DeleteMissing[#]]) &@ (subMissing /@ (fulldatatable[[
All, 227]]/fulldatatable[[All, 6]])),
N[fulldatatable[[All, 13]]/fulldatatable[[All, 11]]]}];
This is how the data looks:
inputLinearModelcompletewRcode[[1 ;; 20]] // TableForm
![enter image description here][21]
Again we clean the data a bit:
inputLinearModelcompletewRcode = Select[subMissing /@ inputLinearModelcompletewRcode, ! MemberQ[#, Missing[]] &];
Now we can use the linear model to estimate the actual percentage of yes/no votes in the different voting areas:
codepercent = Transpose[{inputLinearModelcompletewRcode[[All, 1]], MapThread[lmodelfulldata, Transpose[inputLinearModelcompletewRcode[[All, 2 ;; -2]]]]}]
First we represent the actual election outcome. Green corresponds to "Remain" and red to "Leave" votes.
Graphics[Flatten[
Table[{ColorData[
"RedGreenSplit"][((Select[
Select[votingData[[2 ;;]], #[[1]] < 381 &], #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-3]]/100.)],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1, 380}],
1], Background -> Black, AspectRatio -> GoldenRatio]
![enter image description here][22]
We can now calculate the best estimate of the Leave/Remain split based on our model:
polygonscoloured =
Table[{ColorData["RedGreenSplit"][1. - ((Select[codepercent, #[[1]] == translateshpvoteclean[[k, 2]] &][[1]])[[-1]]) /. Missing[] -> 0.],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1, Length[translateshpvote]}];
Graphics[polygonscoloured, AspectRatio -> GoldenRatio, Background -> Black]
![enter image description here][23]
The split between England and Scotland has become even more pronounced. Note that we haven't dealt very cleanly with the few missing data cases. Next, we can compare the outcome of the elections with the best estimate of the actual Remain/Leave distribution.
Next, we compare the election outcome to our best model estimate:
GraphicsRow[{
Graphics[
Flatten[Table[{ColorData[
"RedGreenSplit"][((Select[
Select[votingData[[2 ;;]], #[[1]] < 381 &], #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-3]]/100.)],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1,
380}], 1], Background -> Black, AspectRatio -> GoldenRatio],
Graphics[Table[{ColorData["RedGreenSplit"][
1 - ((Select[
codepercent, #[[1]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-1]]) /.
Missing[] -> 0.],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1,
Length[translateshpvote]}], Background -> Black,
AspectRatio -> GoldenRatio],
Graphics[Flatten[
Table[{ColorData[
"RedGreenSplit"][(0.5 +
3.*((1 - ((Select[
codepercent, #[[1]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-1]] /.
Missing[] ->
0.)) - ((Select[
Select[votingData[[2 ;;]], #[[1]] <
381 &], #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-3]]/
100.)))],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1,
380}], 1], Background -> Black, AspectRatio -> GoldenRatio]},
ImageSize -> Full]
![enter image description here][24]
On the left we have the outcome of the elections. In the middle we represent the estimate of the remain/leave ratio based on our linear model. The right figure shows the size of the "correction". The model shows a strong separation between Scotland and most regions in England. The exception in England is the region around London.
Area weighted model
----------
Next we will pool the different regions to better understand regional difference.
codepercent = Transpose[{inputLinearModelcompletewRcode[[All, 1]], MapThread[lmodelfulldata, Transpose[inputLinearModelcompletewRcode[[All, 2 ;; -2]]]]}]
We basically set all the inputs to zero apart from the region data
codepercentregion =
Transpose[{inputLinearModelcompletewRcode[[All, 1]],
MapThread[lmodelfulldata,
Transpose[(Flatten /@
Transpose[{inputLinearModelcompletewRcode[[All, 1 ;; 2]],
ConstantArray[0, {382, 6}]}])[[All, 2 ;; -2]]]]}];
GraphicsRow[{Graphics[
Flatten[Table[{ColorData[
"RedGreenSplit"][((Select[
Select[votingData[[2 ;;]], #[[1]] < 381 &], #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1]])[[-3]]/100.)],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1,
380}], 1], Background -> Black, AspectRatio -> GoldenRatio],
Graphics[Table[{ColorData["RedGreenSplit"][
1 - (((If[# == {}, {Select[
votingData, #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1, {4, -2}]]/{1,
100.}}, #] &@
Select[codepercentregion, #[[1]] ==
translateshpvoteclean[[k, 2]] &])[[1]])[[-1]])],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1,
Length[translateshpvote]}], Background -> Black,
AspectRatio -> GoldenRatio],
Graphics[Flatten[
Table[{ColorData[
"RedGreenSplit"][(0.5 +
3.*((1 - (((If[# == {}, {Select[
votingData, #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1, {4, -2}]]/{1,
100.}}, #] &@
Select[
codepercentregion, #[[1]] ==
translateshpvoteclean[[k, 2]] &])[[
1]])[[-1]])) - (((If[# == {}, {Select[
votingData, #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1, {4, -2}]]/{1,
100.}}, #] &@
Select[
Select[
votingData[[2 ;;]], #[[1]] < 381 &], #[[4]] ==
translateshpvoteclean[[k, 2]] &])[[1]])[[-3]]/
100.)))],
localauthorities[[1, 2, 1, 2, 1, 2, 1, 1, 1]][[k]]}, {k, 1,
380}], 1], Background -> Black, AspectRatio -> GoldenRatio]},
ImageSize -> Full]
![enter image description here][25]
This provides the smoothed picture as regions are pooled and weighted according to their representation in the UK. As before the leftmost image represents the outcome of the election, the middle one is the best region weighted estimate and the right figure shows the differences. The strong divide between the different regions is enhanced in this representation.
Population weighted representation
----------
The representations above have the critical drawback that equal area sizes do not necessarily correspond to equally sized electorates. We can represent everything such that area corresponds to population size. This will decrease the size of the regions with low population densities such as Scotland. The following file (attached) contains a hexagonal lattice that describes the areas in the UK.
hexagonshape = Import["GB_Hex_Cartogram_LAs.shp"]
![enter image description here][26]
The following file contains a lookup table of regions vs area codes.
hexagonlegend = Import["GB_Hex_Cartogram_LAs.dbf"];
hexagonlegend[[1 ;; 10, {3, 2}]] // TableForm
![enter image description here][27]
We now need to translate between the hexagonal structure and the voting areas from our model.
tranlatehexagonvoteclean =
Table[Select[hexagonlegend, #[[2]] == translateshpvoteclean[[i, 2]] &], {i, 1,Length[translateshpvoteclean]}][[All, 1, 1]];
We can then represent the estimate from our model in an population/area adjusted representation.
Graphics[Table[{ColorData["RedGreenSplit"][1 - (( If[# == {}, {Select[votingData, #[[4]] ==
translateshpvoteclean[[k, 2]] &][[1, {4, -2}]]/{1, 100.}}, #] &@
Select[codepercent, #[[1]] == translateshpvoteclean[[k, 2]] &])[[1]][[-1]]) /. Missing[] -> 0.],
hexagonshape[[1, 2, 1, 2]][[tranlatehexagonvoteclean[[k]]]]}, {k, 1, Length[tranlatehexagonvoteclean]}], Background -> Black]
![enter image description here][28]
Preliminary summary
----------
These calculations show that we might have to perform some sort of estimation procedure to better understand the "will of the people". The method we have used is quite standard in for example medical studies. This is, however, not a suitable way of decision making in societies. As said many times before, the decision is based on an agreed upon procedure. It is not based on a statistical analysis of the data, which as in our case is model dependent.
Bootstrapping
----------
Note, that there are different approaches to try to mathematically guesstimate what the people who did not vote think about Brexit. Alternatively to the method described above we could pool all data from all local authorities and scale up to the entire electorate (unnecessary) and bootstrap on that.
The argument of many politicians would be that the non-voters essentially behave like the voters. In that case we basically scale up the results from the voters to the entire electorate. The bootstrapping in this case would look like this:
Monitor[fullpoolresults =
Table[Count[RandomChoice[Total /@ Transpose[{N@#[[13]]/#[[11]]*#[[6]], #[[6]] - N@#[[13]]/#[[11]]*#[[6]]} & /@ (Reverse@
SortBy[votingData[[2 ;;]], #[[9]] &])] -> {"Leave", "Remain"}, 46500001], "Remain"]/46500001., {i, 1, 100}];, i]
We can fit a smooth kernel distribution to the outcomes of the bootstrapping procedure:
distfullpool = SmoothKernelDistribution[fullpoolresults]
As expected the probability that more than 50% of the electorate would have wanted to Remain is (numerically) zero.
Probability[x > 0.50, x \[Distributed] distfullpool]
(*0*)
So, under this model assumption it is certain that people really wanted Brexit.
The assumption that the electorate is that homogeneous is quite questionable. We might therefore choose to bootstrap with respect to the local authorities. In other words, we assume that it is on the authority level that voters are similar.
Monitor[localauthresults =
Table[N@(Total[#[[All, 13]]/#[[All, 11]]*#[[All, 6]]]/
Total[#[[All, 6]]]) &@
RandomChoice[(Reverse@SortBy[votingData[[2 ;;]], #[[9]] &])[[All,
6]] -> (Reverse@SortBy[votingData[[2 ;;]], #[[9]] &]),
Length[votingData[[2 ;;]]]], {i, 1, 1000}], I];
This bootstraps for the percentage of Remain votes. We can then again fit a distribution:
distlocalauthorities = SmoothKernelDistribution[localauthresults]
and plot the result:
Plot[PDF[distlocalauthorities, x], {x, 0.45, 0.55}, PlotRange -> All,
FrameLabel -> {"percent leave", "probablity density"},
LabelStyle -> Directive[Bold, 18], ImageSize -> Large,
PlotTheme -> "Marketing"]
![enter image description here][29]
This means that the average of the bootstrap results for Remain votes is below 50%.
Mean[distlocalauthorities]
0.494498. This is, however, quite close to 50%. There is a substantial area above the 50% level:
Probability[x > 0.50, x \[Distributed] distlocalauthorities]
0.17122. This suggests that there is a 17% probability that the voter actually wanted to tell us that they want to remain. This is a substantial fraction/probability. Under these assumptions it is by no means clear that the electorate wanted to leave. Once again, this is not how we take democratic decisions. The decision making process is determined by the outcome of the election and not by an estimate of the voters' will. The point is that it is questionable whether politicans can claim that the voters' will is clear.
The last approach is to bootstrap over the local authorities according to the number of ballots cast:
localvotersresults =
Table[N@(Total[#[[All, 13]]]/Total[#[[All, 11]]]) &@RandomChoice[(Reverse@SortBy[votingData[[2 ;;]], #[[9]] &])[[All, 11]] -> (Reverse@SortBy[votingData[[2 ;;]], #[[9]] &]), Length[votingData[[2 ;;]]]], {10000}];
Once more we estimate the distribution:
localvotersdist = SmoothKernelDistribution[localvotersresults]
and plot the resulting probability density function:
Plot[PDF[localvotersdist, x], {x, 0.45, 0.55}, PlotRange -> All,
FrameLabel -> {"percent leave", "probablity density"},
LabelStyle -> Directive[Bold, 18], ImageSize -> Large,
PlotTheme -> "Marketing"]
![enter image description here][30]
Probability[x < 0.50, x \[Distributed] localvotersdist]
0.487381. So under this assumption we would interpret the voters' will as nearly completely undecided. There is a 50%/50% split, i.e. it is practically equally likely that the voters wanted to remain as that voters wanted to leave.
Conclusion
----------
"The will of the voters" is difficult to understand in close elections. It is by no means as clear that the winning alternative of an election represents the "will of the people" when elections are very close. Politicians often like to stress that they represent the will of the people after winning an election. Mathematically speaking, this is not necessarily true. Note, that this is irrelevant for the decision making process. As a society we have decided to abide by fixed electoral rules. Only people who actually vote will make their voices heard. We can mathematically try to figure out what the entire electorate wanted, or to be precise make probability statements about what they wanted. These statements are clearly model dependent.
[1]: http://community.wolfram.com/groups/-/m/t/1221950
[2]: http://community.wolfram.com/web/bschelter
[3]: https://www.electoralcommission.org.uk/find-information-by-subject/elections-and-referendums/past-elections-and-referendums/eu-referendum/electorate-and-count-information
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1600.47.30.png&userId=48754
[5]: http://esriuktechnicalsupportopendata-techsupportuk.opendata.arcgis.com/datasets/686603e943f948acaa13fb5d2b0f1275_0
[6]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1600.55.48.png&userId=48754
[7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1600.59.21.png&userId=48754
[8]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.01.56.png&userId=48754
[9]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.09.35.png&userId=48754
[10]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.09.59.png&userId=48754
[11]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.12.00.png&userId=48754
[12]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.12.45.png&userId=48754
[13]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.13.40.png&userId=48754
[14]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.17.37.png&userId=48754
[15]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.20.53.png&userId=48754
[16]: https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwihifGz0cHXAhXHKcAKHdtTDU0QFggmMAA&url=http://researchbriefings.files.parliament.uk/documents/CBP-7639/EU-referendum-results-and-characteristics-data.xlsx&usg=AOvVaw3RFJU1dCOJM-tukE0VZGpx
[17]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.28.13.png&userId=48754
[18]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.29.24.png&userId=48754
[19]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.32.00.png&userId=48754
[20]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.33.34.png&userId=48754
[21]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.37.36.png&userId=48754
[22]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.39.19.png&userId=48754
[23]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.40.41.png&userId=48754
[24]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.42.25.png&userId=48754
[25]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.44.22.png&userId=48754
[26]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.47.25.png&userId=48754
[27]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.49.14.png&userId=48754
[28]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.51.29.png&userId=48754
[29]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.55.45.png&userId=48754
[30]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1601.58.42.png&userId=48754Marco Thiel2017-11-16T02:01:31ZClassifying Japanese characters from the Edo period
http://community.wolfram.com/groups/-/m/t/1221098
Introduction
----------
I recently came across a post for a computer program that in some fields intends to compete with the Wolfram Language and which has a toolbox for Machine Learning. I wanted to compare the work described with the workflow in Mathematica. The challenge is to classify old Japanese Characters from texts from the so-called Edo period:
Style[StringTake[WikipediaData["Edo period"], 601], 16]
![enter image description here][1]
The characters are written in running style and apparently are difficult to read for Japanese speakers today. We will use two approaches of machine learning in the Wolfram Language to tackle this problem.
Historical Background
-----------------------------
Before we proceed, let's have a look at the Tokugawa shogunate. I will use code borrow from [@Vitaliy Kaurov][at0] from his great article on ["Computational history: countries that are gone"][2]. The following lines compute the historic boundaries of the shogunate:
Monitor[tokugawaPOLY =
Table[EntityValue[Entity["HistoricalCountry", "TokugawaShogunate"],
EntityProperty["HistoricalCountry",
"Polygon", {"Date" -> DateObject[{t}]}]], {t, 1600, 1870}];, t]
We then only filter for "changes" in the boundaries:
tokugawaPOLY // Length
tokugawaPOLYcomp =
DeleteMissing[
DeleteDuplicates[Transpose[{Range[1600, 1870], tokugawaPOLY}],
Last[#1] == Last[#2] &], 1, 2];
tokugawaPOLYcomp // Length
In the 271 I consider in the first place there are only 4 changes to the boarders, which we can plot like so:
GeoGraphics[{EdgeForm[Red], GeoStyling[Opacity[.1]], #} & /@
tokugawaPOLYcomp[[All, 2]], GeoProjection -> "Mercator",
ImageSize -> 800, GeoBackground -> GeoStyling["StreetMap"],
GeoRange -> Entity["Country", "Japan"],
GeoRangePadding -> Quantity[800, "Kilometers"], GeoZoomLevel -> 6]
![enter image description here][3]
So the boarders were indeed very stable for a long period reflecting the "no more wars" philosophy of the day.
Arts and culture
----------
The stability led to a very rich art and literature scene. Here are some images that reflect the style of the time:
ImageCollage["Thumbnail" /. Normal[Normal[WebImageSearch["Edo Period"]]], Method -> "Rows"]
![enter image description here][4]
Our dataset
----------
Also literature flourished during this period. Our dataset will be taken from the [Center for Open Data in the Humanities][5]. The keen reader might have noticed that the page is in Japanese. Mathematica can translate the page, and [so can Google][6]. The dataset is called the "Japanese classic registered glyph data set"; 日本古典籍字形データセット.
TextTranslation["日本古典籍字形データセット"]
"Japan classics-type datasets". Following the first link (grey box) on the original Japanese page one gets to t[his page][7], from the bottom of which (look for a download symbol and the text "(TAR+GZ 24.57 MB)" the dataset can be downloaded. This is the google translation of the reference:
> Japanese Classic Statement Character Data Set" (Kokugaku Kenkyu Other
> Collection / CODH Processing)
which in the original is "日本古典籍字形データセット』（国文研ほか所蔵／CODH加工".
The glyphs come from "bamboo letters" from 15 classic texts; the dataset contains more than 20000 glyphs of the 10 most frequent symbols. These letters were annotated by the "National Institute of Letters of Japan" (google). The dataset was also used in a [data/machine learning challenge][8].
After extracting the file to a folder on the desktop I can import the images with:
characterImages = Import["/Users/thiel/Desktop/pmjt_sample_20161116/train_test_file_list.h5", "Data"];
Here is a sample of the glyphs:
Grid[Partition[ImageAdjust[Image[#/256.]] & /@ characterImages[[3, 1 ;; 24]], 6], Frame -> All]
![enter image description here][9]
Similar to the MNIST dataset, they consist of 28 times 28 pixels. They give the grey level as integers, which is why I divide by 256. This shows the pixelation of the glyphs:
Image[characterImages[[3, 12]]/256.] // ImageAdjust
![enter image description here][10]
The dataset conveniently contains a training and a test set. There are
characterImages[[3]] // Length
19909 glyphs in the training set. All of these are annotated:
Flatten[characterImages[[4]]]
![enter image description here][11]
We can convert the training data to images like so:
Monitor[trainingset = Table[ImageAdjust[Image[characterImages[[3, k]]/256.]] -> Flatten[characterImages[[4]]][[k]], {k, 1, 19909}];, k]
The test set contains
characterImages[[1]] // Length
3514 glyphs. The annotation of which are found in
characterImages[[2]] // Flatten // Short
![enter image description here][12]
This is how we prepare the test set data for our machine learning:
Monitor[testset = Table[ImageAdjust[Image[characterImages[[1, k]]/256.]] -> Flatten[characterImages[[2]]][[k]], {k, 1, 3514}];, k]
What are the 10 most frequent symbols?
----------
In one of the annotation files we find the character codes of the symbols so that we can map the classification to actual symbols. These are the glyphs we will consider:
Rasterize /@ (Style[
FromCharacterCode[
ToExpression["16^^" <> StringJoin[Characters[#]]]],
100] & /@ {"3057", "306B", "306E", "3066", "308A", "3092",
"304B", "304F", "304D", "3082"})
![enter image description here][13]
We can also make an image collage of them:
ImageCollage[
Image /@ (Rasterize /@ (Style[
FromCharacterCode[
ToExpression["16^^" <> StringJoin[Characters[#]]]],
100] & /@ {"3057", "306B", "306E", "3066", "308A", "3092",
"304B", "304F", "304D", "3082"})), Method -> "Rows"]
![enter image description here][14]
I though it would be nice if they were in a more calligraphic form. So I tried to used ImageRestyle to try and make them more calligraphic. I think that the attempt was not very successful, but I want to show it anyway.
First I download a couple of calligraphic symbols:
caligraphy = "Thumbnail" /. Normal[Normal[WebImageSearch["japanese calligraphy"]]]
I liked the 5th one, which I will use for the style transfer:
calisymbols =
ImageRestyle[#, caligraphy[[5]],
PerformanceGoal ->
"Quality"] & /@ (Rasterize /@ (Style[
FromCharacterCode[
ToExpression["16^^" <> StringJoin[Characters[#]]]],
100] & /@ {"3057", "306B", "306E", "3066", "308A", "3092",
"304B", "304F", "304D", "3082"}))
![enter image description here][15]
We can also remove the background to get an alternative representation:
![enter image description here][16]
Machine Learning (Classify)
----------
Our first approach to classify the glyphs will be via Classify. The classification is very fast (only a couple of seconds on my machine) and quite accurate:
standardcl = Classify[trainingset, ValidationSet -> testset, PerformanceGoal -> "Quality"]
We can first calculate the ClassifierMeasurements
cmstandard = ClassifierMeasurements[standardcl, testset]
and the accuracy:
cmstandard["Accuracy"]
With 92.5% this is quite accurate and also much faster than the competitor product. The confusion plot shows that it is a very reasonable accuracy - in particular when considering that native speakers have problems identifying the glyphs.
cmstandard["ConfusionMatrixPlot"]
![enter image description here][17]
Here are some glyphs that the Classifier identified as class "0":
cmstandard["Examples" -> {0, 0}][[1 ;; 20]]
![enter image description here][18]
They look quite diverse, but so does the original set:
RandomSample[Select[trainingset, #[[2]] == 0 &], 20]
![enter image description here][19]
It is quite impressive that the Classify function manages to achieve this level of accuracy. We can, however, do better!
Machine Learning (NetTrain)
----------
We can improve the accuracy by training our own custom made network, which in this case will be the standard net that is often used for the MNIST dataset in Mathematica. It turns out that we do not even need a GPU for the training.
lenet = NetChain[{ConvolutionLayer[20, 5], Ramp, PoolingLayer[2, 2],
ConvolutionLayer[50, 5], Ramp, PoolingLayer[2, 2], FlattenLayer[],
500, Ramp, 10, SoftmaxLayer[]},
"Output" -> NetDecoder[{"Class", Range[0, 9]}],
"Input" -> NetEncoder[{"Image", {28, 28}, "Grayscale"}]]
![enter image description here][20]
This we can then train:
lenet = NetTrain[lenet, trainingset, ValidationSet -> testset, MaxTrainingRounds -> 20];
![enter image description here][21]
The training is done in about 4:30 minutes, but it appears that 20 training rounds are not necessary. Here are some results when applying the classifier:
imgs = Keys@RandomSample[testset, 60]; Thread[imgs -> lenet[imgs]]
![enter image description here][22]
Let's see how well it does:
cm = ClassifierMeasurements[lenet, testset]
cm["ConfusionMatrixPlot"]
![enter image description here][23]
This corresponds to an accuracy of
cm["Accuracy"]
95.6%, which is really impressive given the dataset. The following table illustrates the difficulty:
Grid[Table[
Prepend[Select[trainingset, #[[2]] == k &][[1 ;; 10]][[All, 1]],
ImageResize[calisymbols[[k + 1]], 28]], {k, 0, 9}], Frame -> All,
Background -> {{Red}, None}]
![enter image description here][24]
The first (red) column is derived from the glyph based on the character code - using our "calligraphy-style". The remaining columns show items from the training set, i.e. glyphs that were manually classified to belong to the given group.
Conclusion
----------
I do not speak/read Japanese and have no more than Wikipedia knowledge about the Edo period. I still find it quite amazing that using the Wolfram Language's high level Classify function and one of the pre-implemented networks, we can achieve quite a remarkable classification. I am sure that the network can be improved. I would also like to see a more complete dataset with more than 10 different symbols.
[at0]: http://community.wolfram.com/web/vitaliyk
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-13at23.24.03.png&userId=48754
[2]: http://community.wolfram.com/groups/-/m/t/852277
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-13at23.30.18.png&userId=48754
[4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-13at23.33.24.png&userId=48754
[5]: http://codh.rois.ac.jp/char-shape/
[6]: https://translate.google.com/translate?hl=en&sl=auto&tl=en&u=http://codh.rois.ac.jp/char-shape/
[7]: http://codh.rois.ac.jp/char-shape/book/
[8]: https://translate.google.com/translate?hl=en&sl=ja&tl=en&u=http://codh.rois.ac.jp/seminar/japanese-character-challenge-20170210/&sandbox=1
[9]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-13at23.56.19.png&userId=48754
[10]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-13at23.57.57.png&userId=48754
[11]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-13at23.59.36.png&userId=48754
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[17]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.15.09.png&userId=48754
[18]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.17.07.png&userId=48754
[19]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.18.25.png&userId=48754
[20]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.21.43.png&userId=48754
[21]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.22.50.png&userId=48754
[22]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.27.23.png&userId=48754
[23]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.28.27.png&userId=48754
[24]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2017-11-14at00.40.41.png&userId=48754Marco Thiel2017-11-14T00:49:30ZAridity, an arcGRID import problem
http://community.wolfram.com/groups/-/m/t/1222924
There's publicly available data database of aridity and PET, that is precipitation/evaporation/transpiration
http://www.cgiar-csi.org/data/global-aridity-and-pet-database
It's a raster on a 10 arc-second grid. It should be importable using a command like this
Import[ "http://exampledata.wolfram.com/ArcGRID.zip"], which works with the example data set.
But I get this error message with the (several hundred megabyte) aridity dataset downloaded to my machine.
LinkObject::linkd: Unable to communicate with closed link LinkObject['/Applications/Mathematica.app/Contents/SystemFiles/Converters/Binaries/MacOSX-x86-64/GDAL.exe',588130,13].
Import::fmterr: Cannot import data as ArcGRID format.
Any wisdom from this crowd?Daniel Seligson2017-11-15T23:23:16ZHTML Import is not working
http://community.wolfram.com/groups/-/m/t/1222542
Hi. I am trying to use Import to read a webpage, but it doesn't seem to work. Please help.
website = "https://www.cse.lk/home/company-info/JKH.N0000/announcements"
websiteData = Import[website]
websiteData just ends up as a blank string.
Any help would be much appreciated. Thank you.
Priyan.Priyan Fernando2017-11-15T09:39:27ZSpeedup DiscretizePDE when InitializePDECoefficients contains Interpolation
http://community.wolfram.com/groups/-/m/t/1222183
I work on FEM in Mathematica and want to speed up the computation when dealing with very fined mesh (i.e., a huge number of DOF).
I found that a very much longer time is consumed when the Coefficients of PDE contains interpolation functions, compared to the case that the Coefficients are constants. To be more specific, let consider the following exemplary code.
ClearAll["Global`*"]
SetDirectory[NotebookDirectory[]];
Needs["NDSolve`FEM`"];
width = 1.0; heigth = 2.0; ydatum = 0.0;
\[ScriptCapitalR] =
ImplicitRegion[0 <= x <= width && ydatum <= y <= heigth + ydatum , {x, y}];
RegionPlot[\[ScriptCapitalR], AspectRatio -> Automatic, ImageSize -> Small]
pde = {Div[{{\[Lambda] + 2 \[Mu], 0}, {0, \[Mu]}}.Grad[
u[x, y], {x, y}] + {{0, \[Lambda]}, {\[Mu], 0}}.Grad[
v[x, y], {x, y}], {x, y}],
Div[{{0, \[Mu]}, {\[Lambda], 0}}.Grad[
u[x, y], {x, y}] + {{\[Mu], 0}, {0, \[Lambda] + 2 \[Mu]}}.Grad[
v[x, y], {x, y}], {x, y}]};
\[CapitalGamma]g = {DirichletCondition[u[x, y] == 0, x == 0],
DirichletCondition[v[x, y] == 0, y == ydatum],
DirichletCondition[v[x, y] == 0.2, y == heigth + ydatum]};
\[Lambda] = 1.0; \[Mu] = 1.0;
{dPDE, dBC, vd, sd, md} =
ProcessPDEEquations[{pde == {0, 0}, \[CapitalGamma]g}, {u,
v}, {x, y} \[Element] \[ScriptCapitalR],
Method -> {"FiniteElement",
"MeshOptions" -> {"ImproveBoundaryPosition" -> False,
"MaxCellMeasure" -> 0.0001, "MeshOrder" -> 2}}];
mesh2 = md["ElementMesh"];
linearLoad = dPDE["LoadVector"];
linearStiffness = dPDE["StiffnessMatrix"];
diriPos = dBC["DirichletRows"];
offsets = md["IncidentOffsets"];
md["DegreesOfFreedom"]
Show[mesh2["Wireframe"]]
ClearAll[rhs]
tol = 10^-8;
err = 1.0;
iter = 0;
\[Alpha] = 1.0;
rhs[uIn_] := Module[{uOld}, uOld = uIn;
While[err >= tol && iter <= 20,
u0 = ElementMeshInterpolation[{mesh2},
uOld[[offsets[[1]] + 1 ;; offsets[[2]]]]];
v0 = ElementMeshInterpolation[{mesh2},
uOld[[offsets[[2]] + 1 ;; offsets[[3]]]]];
nlPdeCoeff = InitializePDECoefficients[vd, sd,
"LoadCoefficients" -> -{{0}, {-1}},
"LoadDerivativeCoefficients" -> {{{(\[Lambda] + 2 \[Mu]) D[
u0[x, y], x] + \[Lambda] D[v0[x, y],
y], \[Mu] (D[u0[x, y], y] +
D[v0[x, y], x])}}, {{\[Mu] (D[u0[x, y], y] +
D[v0[x, y], x]), \[Lambda] D[u0[x, y],
x] + (\[Lambda] + 2 \[Mu]) D[v0[x, y], y]}}},
"ReactionCoefficients" -> {{0, 0}, {0,
0}}];
t1 = AbsoluteTime[];
nlsys = DiscretizePDE[nlPdeCoeff, md, sd];(*time ~s*)
time1 = AbsoluteTime[] - t1;
nlLoad = nlsys["LoadVector"];
nlStiffness = nlsys["StiffnessMatrix"];
ns = nlStiffness + linearStiffness;
nl = nlLoad + linearLoad;
nl[[diriPos]] = {0.};
ns[[diriPos, All]] = 0.;
ns[[All, diriPos]] = 0.;
(ns[[#, #]] = 1.) & /@ diriPos;
dU = LinearSolve[ns, nl];
Print[iter, " Residual: ", Norm[nl, Infinity], " Correction: ",
Norm[dU, Infinity], " Time: ", time1];
err = Norm[nl, Infinity]; iter++;
uOld = uOld + \[Alpha]*dU;];
uOld]
uOld = ConstantArray[{0.}, md["DegreesOfFreedom"]];
uOld[[diriPos]] = dBC["DirichletValues"];
uNew = rhs[uOld];
The "LoadDerivativeCoefficients" already contain the interpolated function u0 and v0, the other Coefficients are constants. After timing each command during running, I found the DiscretizePDE is most time consuming. For example, in this case, the DiscretizePDE command will cost ~6 seconds during each iteration.
On the other hand, if we set InitializePDECoefficients as:
nlPdeCoeff = InitializePDECoefficients[vd, sd,
"LoadCoefficients" -> -{{0}, {-v0[x, y]^2}},
"LoadDerivativeCoefficients" -> {{{(\[Lambda] + 2 \[Mu]) D[
u0[x, y], x] + \[Lambda] D[v0[x, y],
y], \[Mu] (D[u0[x, y], y] +
D[v0[x, y], x])}}, {{\[Mu] (D[u0[x, y], y] +
D[v0[x, y], x]), \[Lambda] D[u0[x, y],
x] + (\[Lambda] + 2 \[Mu]) D[v0[x, y], y]}}},
"ReactionCoefficients" -> {{0,
0}, {0, -2 v0[x,
y]}}];
Now the time of running DiscretizePDE is ~10 seconds. If we refine the mesh, the consumed time will increase. (The above code is running with 16 cores.)
Since I need to run a lot of such code. Is there a way to cut of the time that the DiscretizedPDE needs when InitializePDECoefficients contains Interpolation functions?Wilhelm Deng2017-11-15T18:19:37ZThe voters have spoken - but what did they say?
http://community.wolfram.com/groups/-/m/t/1221950
Introduction
----------
Elections are fundamental to our democracies and way of life. The electorate bestows power on the lawmakers and public officials whose legitimisation is derived solely from the fact that they are elected. In recent times, many elections, such as the presidential elections the United States and the Brexit and general elections in the United Kingdom have seen a very pronounced polarisation the electorate. Can mathematics and the Wolfram Language help to better understand this crucial process of the most fundamental decision making process in societies? The main question I want to discuss in a sequence of posts is what can be learnt from elections. So this post is about **computational politics**.
It is important to notice that democracies rely on a general acceptance of the result of elections. Without that consensus a peaceful co-existence is at risk. Societies agree on certain electoral procedures and the results have to be understood to be final. There are, however, more fundamental questions such as how do we derive the "voters will" from ballots. That is a highly non-trivial problem. Also, elected politicians often claim that they represent the "will of the people". But is that true and what is the "will of the people"? Hence, my question: the voters have spoken - but what did they say?
This is the first of a series of posts (hopefully):
1. **Electoral systems - interpreting the will of the voters**
2. Gerrymandering - shaping the voters' will to your liking
3. [Analysis of the Brexit vote - how do we know what did the voters really want?][1]
I apologise because there won't be any nice graphics in this post, but I find the results relevant anyway. In the following posts we will apply some more sophisticated algorithms to try to figure out what the voters meant to tell us. We will then see that we can use mathematics to estimate what the voters might have wanted to tell us; and then we can correct for that.
![enter image description here][2]
Note, that this cannot be the procedure by which we determine the decision that the electorate have taken. This is all a bit confusing, but I hope that it will become a bit clearer in these posts.
All posts were developed with [Bjoern Schelter][3], who is also a member of this community. The definitions and main ideas for this post were taken from the excellent book ["Mathematics And Politics: Strategy, Voting, Power, and Proof"][4] by Alan Taylor, a text which I highly recommend. It is a treasure trove of ideas and concepts for anyone interested in this ares. It goes far beyond what is described in this post.
Social choice procedure
----------
It turns out that reading the voters' will from a ballot is much more complicated than one might think - in many situations it is impossible (mathematically provable!) to do so in a "good" way. Let's define some words first. Let's assume that each voter ranks some alternatives in their preferred order.
**Definition:** A social choice procedure is a function for which a typical input is a sequence of lists (without ties) of some set A (the set of alternatives) and the corresponding output is either an element of A, a subset of A, of "No Winner".
There is a beautiful theorem by May (1952):
**Theorem:**
If the number of people is odd and each election produces a unique winner, then the majority rule is the only social choice procedure for two alternatives that satisfies the following three conditions:
1. It treats all the voters the same: If any two voters exchange their ballots, the outcome of the election is un-effected.
2. It threats both alternatives the same: If every voter reverses his or her vote (changing a vote for a to a vote for b and vice versa), the the election outcome is reversed as well.
3. It is monotone: If some voter were to change his or her ballot from a vote for the loser to a vote for the winner, then the election outcome would be unchanged.
Six examples of social choice procedures
----------
Let us assume that we have several alternatives to be voted for:
alternatives = {a, b, c}
Then there are the voting lists of the voters. They form the ballots:
ballots = Table[RandomSample[alternatives, Length[alternatives]], {k, 4}]
(*{{a, b, c}, {c, b, a}, {a, b, c}, {a, b, c}}*)
> Condorcet's method:
>
> An alternative x is among the winners if for any other alternative y,
> at least half of the voters rank x over y on their ballots.
This is an implementation of this social choice function:
condorcetindiv[ballots_List, a_, b_] :=
Module[{},
If[a === b, "NA",
If[#[[1, 2]] >= Length[ballots]/2, {a, b}, #[[-1, 1, 1]]] &@
SortBy[Tally[Select[#, # == a || # == b &] & /@ ballots], Last]]]
condorcet[ballots_List, alternatives_] :=
Module[{},
If[Max[#] >= Length[ballots],
alternatives[[
Flatten[Position[#, Max[#]]]]], {}] &@(Count[
Flatten[Outer[condorcetindiv[ballots, #1, #2] &, alternatives,
alternatives], Infinity], #] & /@ alternatives)]
Given the ballots and the alternatives we can now compute the social choice by applying the function condorcet:
condorcet[ballots, alternatives]
{a}
Note, that the social choice is given as a list, because it is not necessarily unique. This will become important later on.
> Plurality:
>
> Declare the social choice(s) the alternative(s) with the largest
> number of first-place rankings in the ballot list.
This is the respective implementation:
plurality[ballots_] :=
Module[{}, GatherBy[SortBy[Tally[ballots[[All, 1]]], Last], Last][[-1, All, 1]]]
The social choice is computed like this:
plurality[ballots]
(*{a}*)
> Borda Count
>
> We award points to the n alternatives in a ballot as follows: the
> alternative at the bottom receives zero points; the one next to it
> receives one point and so on up to the alternative on the top of the
> list which receives (n-1) points.
This is the implementation of the social choice function:
bordacount[ballots_, alternatives_] :=
Module[{}, GatherBy[SortBy[Total[Flatten[{Reverse[Range[0, Length[alternatives] - 1]]*Transpose[ballots]}]] /. {Plus -> List, Times -> List},
First], First][[-1, All, 2]]]
which can be applied like so:
bordacount[ballots, alternatives]
(*{a}*)
> Hare System
>
> The procedure is as follows: we delete the alternative(s) that is on
> top of the fewest of the individual preference lists. The resulting
> individual preference lists are at least one shorter than before. We
> then repeat this process; the last alternative(s) is/are the
> winner(s).
Our Wolfram Language is
haresystem[ballots_] :=
Module[{},
Sort[DeleteDuplicates[Flatten[If[Length[#[[-1, 1]]] > 0, #[[-1, 1]], #[[-2, 1]]] &@
NestWhileList[((# //. (# -> Nothing & /@ Complement[alternatives, #[[All, 1]]])) //. (# -> Nothing & /@
GatherBy[SortBy[Tally[#[[All, 1]]], Last], Last][[1, All, 1]])) &, ballots, Length[DeleteDuplicates[Flatten[#, Infinity]]] > 1 &]]]]]
which is evaluated like so:
haresystem[ballots]
(*{a}*)
> Sequential pairwise voting
>
> We start with a fixed order of alternatives {a,b,c,d,..}. The first
> alternative is evaluated against the second in a one-to-one contest.
> The winning alternative (or both if there is a tie) then is compared
> to the third alternative. An alternative is deleted at the end of any
> round if it loses a one-to-one contest.
We will use a new ballot for the evaluation for clarity:
alternatives = {a, b, c, d}
ballots = Table[RandomSample[alternatives, Length[alternatives]], {k, 4}]
(*{{d, b, a, c}, {d, a, c, b}, {d, c, b, a}, {b, d, c, a}}*)
We also need a fixed agenda:
fixedagenda = alternatives[[RandomSample[RandomSample[Range[Length[alternatives]]]]]]
(*{a, b, c, d}*)
The social choice function is then implemented like so:
concorcetindivloser[ballots_List, bin_, acomp_] :=
Module[{},
If[acomp === bin,
"NA", (Which[#1[[1, 2]] === Length[ballots]/2, {"W",
"W"}, #1[[-1, 1, 1]] === bin, "W", True, "L"] &)[
SortBy[Tally[(Select[#1, #1 == acomp || #1 == bin &] &) /@
ballots], Last]]]]
seqpair[ballots_, fixedagenda_] :=
Module[{winners = {fixedagenda[[1]]}},
Do[AppendTo[winners, fixedagenda[[k]]];
winners =
Pick[winners, ! MemberQ[Flatten[#], "L"] & /@
Outer[concorcetindivloser[ballots, #1, #2] &, #, #] &@
winners], {k, 2, Length[fixedagenda]}]; winners]
We can evaluate this like so:
seqpair[ballots, fixedagenda]
(*{b}*)
> Dictatorship
>
> One of the voters is chosen and called "dictator". The alternative on
> top if their individual preference list is the social choice.
We can implement this function like so:
dictatorhip[ballots_, k_] := {ballots[[k, 1]]}
where $k$ is the index of the person who is the "dictator".
dictatorhip[ballots, 3]
(*{d}*)
A closer look at the social choice procedures
----------
All of these social choice functions are implemented in different parts of the world. It turns out that they have the curious habit of not agreeing on the social choice even if the ballots are the same!
Let's suppose we have the following ballots:
ballots = {{a, b, c, d, e}, {a, d, b, e, c}, {a, d, b, e, c}, {c, b, d, e, a}, {c, d, b, a, e}, {b, c, d, a, e}, {e, c, d, b, a}};
So there are 7 voters and the lists are the preference lists of them. Here is another representation:
Transpose@ballots // MatrixForm
![enter image description here][5]
Let's apply Condorcet's method first:
condorcet[ballots, alternatives]
(*{}*)
That means that Condorcet's method does not produce a social choice. This is obviously a critical problem of a social choice function, because it would not be suitable to lead the decision making process to a conclusion.
Let's try the plurality rule:
plurality[ballots]
(*{a}*)
It determines that the social choice is a. So the electorate has decided for alternative a.
Next up: the Borda Count:
bordacount[ballots, alternatives]
(*{b}*)
So according to the Borda Count b is the winner - in spite of the ballots not having changed!
What about the Hare system?
haresystem[ballots]
(*{c}*)
Next up, the sequential voting system. Let's choose the fixed agenda
fixedagenda = {a, b, c, d, e};
We then obtain:
seqpair[ballots, fixedagenda]
(*{d}*)
So under this social choice function d is the winner.
Last the dictatorship. Let's make person 7 the dictator:
dictatorhip[ballots, 7]
(*{e}*)
Now, e is the winner. Curious...
So in this simple case existing social choice procedures would have led to all different alternatives in spite of the ballots being exactly the same. This is a bit frightening because all these systems are actually being used today.
Optimal social choice procedure
----------
The question is whether there is an "optimal" social choice procedure. The answer to that question turns out to be no. First we define the following five desirable properties:
1. Always a winner. The procedure should always produce a winner. We have already seen that for example Condorcet's method does not give a winner for the last example.
2. Condorcet winner criterion. A social choice procedure is said to satisfy the Condorcet winner criterion (CWC) provided that\[LongDash]if there is a Condorcet winner\[LongDash]then it alone is the social choice.
3. Pareto condition. A social choice procedure is pareto if for every pair x and y of alternatives we have: If everyone prefers x to y, then y is not a social choice.
4. Monotonicity. A social choice procedure is monotone if for any x the following holds: If x is the social choice (or tied for such) and someone changes his or her preference list by moving x up one spot (that is, exchanging x's position with that of the alternative immediately above x on his or her list), then x should still be the social choice (or tied for such).
5. Independence of Irrelevant Alternatives. For every pair of alternative x and y we have: If the social choice set includes x but not y, and one or more voters change their preferences, but no one changes his or her mind about whether x is preferred to y or y to x, then the social choice set should not change so as to include y.
These are very (!) desirable properties and reasonable requirements. Unfortunately, the following theorem holds:
**THEOREM.**
There is no social choice procedure for three or more alternatives that satisfies the always-a-winner criterion, independence of irrelevant alternatives, and the Condorcet winner criterion.
There is also a more general theorem by Arrow, that -roughly speaking- that there cannot be a social choice function fulfilling some basic properties.
What happens for larger constituencies?
----------
Let's create a larger set of voters and hence individual preference lists.
alternatives = {a, b, c, d, e};
and
constituency = Table[RandomSample[alternatives], 200];
Condorcet:
condorcet[constituency, alternatives]
(*{}*)
Plurality:
plurality[constituency]
(*{b}*)
Borda count:
bordacount[constituency, alternatives]
(*{e}*)
Hare system:
haresystem[constituency]
(*{d}*)
Sequential pairwise:
fixedagenda = alternatives[[RandomSample[RandomSample[Range[Length[alternatives]]]]]];
(*{e}*)
That is not very encouraging. But perhaps the outcomes depend on the population size?
Monitor[results =
Table[{pop,
Table[constituency = Table[RandomSample[alternatives], pop];
fixedagenda =
alternatives[[
RandomSample[
RandomSample[Range[Length[alternatives]]]]]]; {Length[
DeleteCases[#, {}]], Length /@ #} &@
Union[Sort /@ {condorcet[constituency, alternatives],
plurality[constituency],
bordacount[constituency, alternatives],
haresystem[constituency],
seqpair[constituency, fixedagenda]}], {50}]}, {pop, 10, 500,
10}];, pop]
I have varied the population size of the electorate from 10 to 500 in steps of 10. The variable results looks like this (here for 500 voters):
results[[-1]]
(*{500, {{2, {0, 1, 1}}, {2, {0, 1, 2}}, {3, {0, 1, 1, 2}}, {1, {0,
1}}, {3, {0, 1, 1, 1}}, {2, {0, 1, 1}}, {2, {0, 1, 1}}, {2, {0, 1,
1}}, {3, {0, 1, 2, 2}}, {1, {0, 1}}, {2, {0, 1, 1}}, {2, {0, 1,
1}}, {1, {0, 1}}, {1, {0, 1}}, {1, {0, 1}}, {3, {0, 1, 1,
2}}, {4, {0, 1, 1, 2, 3}}, {1, {0, 1}}, {1, {0, 1}}, {1, {0,
1}}, {1, {0, 1}}, {1, {0, 1}}, {3, {0, 1, 1, 2}}, {1, {0,
1}}, {2, {0, 1, 1}}, {2, {0, 1, 2}}, {2, {0, 1, 1}}, {2, {0, 1,
1}}, {2, {0, 1, 1}}, {1, {0, 1}}, {2, {0, 1, 3}}, {1, {0,
1}}, {3, {0, 1, 1, 1}}, {1, {0, 1}}, {2, {0, 1, 1}}, {2, {0, 1,
1}}, {2, {0, 1, 2}}, {1, {0, 1}}, {3, {0, 1, 1, 1}}, {2, {0, 1,
1}}, {2, {0, 1, 1}}, {3, {0, 1, 1, 2}}, {3, {0, 1, 1, 1}}, {2, {0,
1, 1}}, {4, {0, 1, 1, 1, 2}}, {1, {0, 1}}, {2, {0, 1,
1}}, {2, {0, 1, 1}}, {1, {0, 1}}, {2, {0, 1, 1}}}}*)
The first 500 is the number of voters. Then there are 50 "realisations", i.e. 50 independent elections where voters make random choices. Each of the 50 sublists has a first entry which gives the number of different results using different social choice functions and ignoring empty sets, i.e. "no-winner" scenarios. Let's plot the histogram of the number of different outcomes using the various social choice functions:
Histogram[results[[-1, 2, All, 1]],
AxesLabel -> {"number of different outcomes", "number of elections"},
ImageSize -> Large, LabelStyle -> Directive[Bold, Medium]]
So in most cases the social choice functions give different results, even if there are 500 voters. Then we can count how many winners there are in cases of different results:
results[[-1, 2, All, 2]]
(*{{0, 1, 1}, {0, 1, 2}, {0, 1, 1, 2}, {0, 1}, {0, 1, 1, 1}, {0, 1,
1}, {0, 1, 1}, {0, 1, 1}, {0, 1, 2, 2}, {0, 1}, {0, 1, 1}, {0, 1,
1}, {0, 1}, {0, 1}, {0, 1}, {0, 1, 1, 2}, {0, 1, 1, 2, 3}, {0,
1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1, 1, 2}, {0, 1}, {0, 1,
1}, {0, 1, 2}, {0, 1, 1}, {0, 1, 1}, {0, 1, 1}, {0, 1}, {0, 1,
3}, {0, 1}, {0, 1, 1, 1}, {0, 1}, {0, 1, 1}, {0, 1, 1}, {0, 1,
2}, {0, 1}, {0, 1, 1, 1}, {0, 1, 1}, {0, 1, 1}, {0, 1, 1, 2}, {0, 1,
1, 1}, {0, 1, 1}, {0, 1, 1, 1, 2}, {0, 1}, {0, 1, 1}, {0, 1,
1}, {0, 1}, {0, 1, 1}}*)
So all cases contain cases of "no winner", when there is a zero in the sublist. This comes mostly from Condorcet's method, which often does not give a winner. The ones indicate that in the respective outcomes there was only one winner. A two or three means that the winner was not unique, i.e. a tie. So basically of all the outcomes only the ones lead to a decision. So only
N[Count[#, 1]/Length[#]] &@Flatten[results[[-1, 2, All, 2]]]
(*0.568*)
57% of the cases lead to a unique winner. If we ignore the no-winner scenarios we see that only in
N[Count[#, 1]/Length[#]] &@(Length[#] - 1 & /@ results[[-1, 2, All, 2]])
(*0.34*)
about 34% of the cases the methods agree on the winner.
Let's see how these numbers change for different sizes of electorates. First, we compute the percentage of "unique winner scenarios" for different sizes of the electorate (ignoring no winner scenarios):
ListLinePlot[Table[{results[[k, 1]], N[Count[#, 1]/Length[#]] &@Flatten[results[[k, 2, All, 2]]]}, {k, 1, 50}],
FrameLabel -> {"size of electorate", "percentage unique winner"}, LabelStyle -> Directive[Bold, Medium], ImageSize -> Large,
PlotTheme -> "Marketing"]
![enter image description here][6]
With a bit more CPU time, we can study whether there is a saturation effect for larger population sizes...
Let's look at the agreement of the different social choice procedures:
ListLinePlot[Table[{results[[k, 1]], N[Count[#, 1]/Length[#]] &@(Length[#] - 1 & /@ results[[k, 2, All, 2]])}, {k, 1, 50}],
FrameLabel -> {"size of electorate", "percentage agreement"},
LabelStyle -> Directive[Bold, Medium], ImageSize -> Large, PlotTheme -> "Marketing"]
![enter image description here][7]
So, more often than not, the social choice procedures to not agree on the winner. This might suggest a type of independence of the population size.
Conclusion
----------
Even if the ballots are fixed deciding what the voters meant to say is by no means trivial. "Reading" the voters' minds even if they have spoken is mathematically speaking very problematic. For the case of only two alternatives (Brexit/No-Brexit) and an odd number of voters the thing appears to be easier, but as we will see later, there are many more issues.
**This is not meant to say that we should give up or ignore the outcome of elections.** To maintain order and piece as a society we need to agree on a particular social choice procedure and accept the outcome. We need to take decisions and these decisions have to be based on established rules. The only issue is that sometimes the social choice ("will of the people") depends on the social choice function and not solely on the ballots!
We have a type of "societal contract" to accept the outcome of elections under the rules (social choice function) that is agreed upon before the elections. This does not necessarily mean, however, that politicians can claim that this is the "will of the people"; it is the decision made based on a ballot and an agreed upon social choice procedure.
[1]: http://community.wolfram.com/groups/-/m/t/1222959
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1423.26.23.png&userId=48754
[3]: http://community.wolfram.com/web/bschelter
[4]: https://www.amazon.co.uk/Mathematics-Politics-Strategy-Voting-Power/dp/1441926615
[5]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1422.42.52.png&userId=48754
[6]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1423.46.02.png&userId=48754
[7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2017-11-1423.47.36.png&userId=48754Marco Thiel2017-11-14T23:53:48Z[✓] Get list of industry segments using FinancialData?
http://community.wolfram.com/groups/-/m/t/1222525
In the helpfile I found two examples to get a list of industry-segments:
Take[FinancialData["DrugManufacturers-Major", "Members"], 20]
FinancialData["Publishing-Books", "Members"]
I would like to get a list of companies from different industrial area's. I tried several names like for example `FinancialData["banks", "Members"]` with the same result: Missing["NotAvailable"].
Is there a list of industrie-segments availableMichiel van Mens2017-11-15T08:17:12ZAvoid warning without disabling it for this plot?
http://community.wolfram.com/groups/-/m/t/1222092
Hi,
I am using this line to plot a magnetic field :
ReplaceAll[
crossSectionFieldPlot["\[ScriptCapitalH]",
nqvrbmmf[mm0, a, b, c, x, 0, z][[{1, 3}]] ], {mm0 -> 10^6,
a -> 0.01, b -> 0.01, c -> 0.01}]
`crossSectionFieldPlot` is from the blog of Michael Trott with a small modification. The plot is created at the end, but I am getting a bunch of warnings before, because Mathematica tries to use the `a, b, c` parameters of `nqvrbmmf` in `crossSectionFieldPlot` without doing the numerical replacements first. I tried also this form:
crossSectionFieldPlot["\[ScriptCapitalH]",
nqvrbmmf[mm0, a, b, c, x, 0, z][[{1, 3}]] ] //
ReplaceAll[#, {mm0 -> 10^6, a -> 0.01, b -> 0.01, c -> 0.01}] &
but the warnings are still there. So the real question is, how to make the substitutions with `ReplaceAll` for the parameters of `nqvrbmmf` BEFORE the evaluation of `crossSectionFieldPlot` ? I also tried a few combinations of `Hold` and `ReleaseHold` here and there with no cigar.
Thanks ahead,
JánosJanos Lobb2017-11-15T01:25:47Z[✓] Area under the curve of a Parametric Plot function?
http://community.wolfram.com/groups/-/m/t/1220905
I am looking to find the area under the curve of the a parametric plot function I tried to use a simple example but it's giving me two answers for each function rather than one when I use NIntegrate. Is there another way to get the area under the curve of the parametric plot ?
Thank you
ParametricPlot[{Sin[t], Sin[2 t]}, {t, 0, 1.57 }]
NIntegrate[{Sin[t], Sin[2 t]}, {t, 0, 1.57 }]Joeseph A2017-11-12T20:41:27Z[✓] Find a simple x^n power kind of block in Modelica library?
http://community.wolfram.com/groups/-/m/t/1222299
I purchased Mathematica and system modeler a while ago to learn design with computers...simulation and the like..and when trying to derive the energy of a system I was attempting to simulate, I required from the math block ( I would think at least) the power of an input...which for the life of me I can't find in the program, the documentation or anywhere on the internet....and I ended up making my own component...basically a modified gain, where k=n n being whatever power I desired...but system modeler complains everything time I try to load a simulation with my custom components...but that is actually another problem for another day...
So my question is. Where do I find a simple x^n power kind of block? Surely something like this isn't missing.Mor Bo2017-11-14T23:01:29Z