Hi folks! We've been enjoying your Level 2 certification submissions so far. If you're curious about whether an idea you have would be fit to submit, please feel free to ask in this thread. (If you have technical questions, however, please wait for the next Programming Proficiency Office Hour session, which is currently scheduled for July 25.)

Thanks for pointing this out, Lori! When I updated the question, the number of answer choices changed and messed up the key—but I'd failed to notice it. It should be fixed soon.

Could you please advise if this program would be suitable for me if I only know how to set up and configure development environments and have just started learning programming languages?

Understood, thank you for the advice. Special thanks for "Most importantly, have fun! Write original code about things you know and things you're interested in."

I would do something like this

Cases[listA, x_List /; First[x] == b]

{{b, a, a}, {b, a, b, a}, {b, b, a}, {b, a, b, a, b}}

Cases[listA, x_List /; First[x] == b] /. {a -> b, b -> a}

{{a, b, b}, {a, b, a, b}, {a, a, b}, {a, b, a, b, a}}

blank

Thanks Doug. Works great.

I wonderful (and sometimes confusing) to see in how many different ways something can be coded in Wolfram.

Yes but it is still confusing

listA /. x_List /; First[x] == b :> ReplaceAll[x , {a -> b, b -> a}]

{{a, b, a}, {a, b, b}, {a, b, a, b}, {a, b, b, a, a}, {a, b, a, 
  b}, {a, a, b}, {a, b, a, b}, {a, b, a, b, a}, {a, a, a, a}}

works
but

listA /. x_List /; First[x] == b :> x /. {a -> b, b -> a}

{{b, a, b}, {a, b, b}, {b, a, b, a}, {b, a, a, b, b}, {a, b, a, 
  b}, {a, a, b}, {b, a, b, a}, {a, b, a, b, a}, {b, b, b, b}}

does not .......

Maybe Arben can tell us why ?????

Thanks Arbin

Yes it is the Parenthesis (good to know )

listA /. x_List /; First[x] == b :> (x /. {a -> b, b -> a})

{{a, b, a}, {a, b, b}, {a, b, a, b}, {a, b, b, a, a}, {a, b, a, 
  b}, {a, a, b}, {a, b, a, b}, {a, b, a, b, a}, {a, a, a, a}}

The issue is here:

/; _First

The /; means that you write some test afterward—not only is _First not a component of a test, but it's not the correct pattern syntax. Recall that what follows a Blank (_) should be a head to be matched; i.e. you can't put an arbitrary function there and take it to be acting on whatever the blank is.

If you have

listA= {{a, b, a}, {b, a, a}, {a, b, a, b}, {a, b, b, a, a}, {b, a, b, a}, {b, b, a}, {a, b, a, b}, {b, a, b, a, b},{ a, a, a, a}} (*i removed the errant comma*)

and want to do {a -> b, b -> a} on any sublist that starts with b, you could do something like

listA /. l_List /; First[l] == b :> (l /. {a -> b, b -> a})

if you wanted to use Condition. Better yet, just match the pattern directly:

listA /. {b, rest__} :> ({b, rest} /. {a -> b, b -> a})

Hi Updating Name (Rohit?),

I suspected that I did not use First correctly :-)

Your solution works excellent. Also if I nest the sublists deeper and irregular (different nesting levels). That is great.

Do I understand it correctly, if I say :

{b, rest__} looks at all the innermost sublists and if there is such a sublist that starts with a 'b' and has one or more elements, then the replacement takes place?

Thanks again for your valuable help,

Best, Bert

You reply faster than I can edit. Wow :-)

Hey Mike, we'll get back to you soon about this. In the meantime, please don't worry about any timeline issues—as far as we're concerned, you've submitted your project within the deadline and that's that.

Hey Michael—I'm sorry about that! I finished grading it 2-3 weeks ago; I'm not sure what got blocked up in the process but we're on it. (Spoiler: you passed)

Edit: you should have it later today.

Right, thanks Arben :-)

I just got my first connection through a chat Nb with chatGPT and asked the same question. After two misses, and restating my question more precise (and partly double), I gotv a working answer :

Map[If[First[#] === b, # /. {a -> b, b -> a}, #] &, listA]

It's nice to see it work, although your solution is more refined, I think, and probably faster :-)

But the A.I. solution shows me again the pure function, which I understand in theory, but in practice I often seem to do it wrong. It will come, some day.

Also, in the If statement (which I also used, I first used == instead of what I should have done : ===. I dodged my error by using SameQ[]. So I think this A.I. coding help can be useful in the learning process.

Arben, I tried using Timing[...] and AbsoluteTiming [...] on 3 different instructions to accomplish the same thing: Power, an anonymous function using Keys, and an anonymous function using indexing. I used a large enough list of associations so that Timing was non-zero. All three had Timing between about 2.1 and 2.8 seconds, and AbsoluteTiming between about 3.7 and 3.9 seconds. I ran this in Mathematica 14 on an HP Envy running Windows 11 Home 64-bit OS. My notebook is attached as file TimingExample_1mar24.nb .

I expected Power to be clearly faster but it wasn't. Please explain these timing results.

Using

shortListOfAssoc = {<|"a" -> 1, "b" -> 2|>, <|"a" -> 3, "b" -> 4|>, <|    "a" -> 5, "b" -> 6|>}; 

I added Timing, AbsoluteTiming, and Repeated timing to the file I uploaded before, TimingExample_1mar24,nb , which is attached. (I also added RepeatedTiming for listOfAssoc that I ran previously.) My respective results for shortListOfAssoc using Power, anonymous function with tags, and anonymous function with indices are as follows:

Timing: all three were 0.

AbsoluteTiming: 0.0000124, 0.0000111, 0.0000101

RepeatedTiming: 1.3700510^-6, 2.1609110^-6, 4.3709*10^-6

Thanks Arben. Do you mean by "if the cases are 1:1 . . ." that the rules are of equal length?

So {1->{1,0,1} and 0->{0,0,1}} is 1:1 and {1->{1,0,1} and 0->{0,1}} is Not?

Reverse the nesting rules is not quite possible, I guess, as I do not know which rules might have been used. Also, the nesting might be 2 or 3 deep, so I do not know the length to parse. It feels a bit like backwards running a C.A., which is possible in some cases, but not very often.

Checking for repetitions in a rhythm line is easy, but this Nesting looks like a different animal :-)

As the computation comes into a repeated process of measuring the amount of Nesting in a rhythm ( so thousands of these detection measurements), it might be an idea to create a lookUp table with all the possible nestings for two elements (0 and 1). If a take as a max of number of initial elements, as well max number of elements in a rule to be 3 (which I think servers my purpose) then there would be 14 possible initial conditions and 12 rules for 0 and twelve rules for 1. As these rules are 'mirrored, I need only to take into account half of them. Then I create the lookup table for all Nestings from level 1 to say 3, which should be enough. From there on I could use SequenceCases or SequencePositions.

This will still leave the challenge for detecting Nesting in a complete rhythm, where I use 16 possible values for the possible simultaneous occurances of hits from 4 instruments together. This will explode with lookUp tables, I am afraid.

Best, Bert

What is the deadline for using the Level 2 discount code I received? (I took the first WSG24 this year)

I have a project idea but haven't had much time since the course ended and probably won't for a couple weeks longer.

Never mind, (content removed)

Today's lecture should do you well :).

In[53]:= 
listA = {{{{1, 2}, {3, 4}}, {{5, 8}, {9, 12}, {14, 17}}}, {{{11, 13}, {15, 17}}, {{18, 20}, {22, 24}}, {{26, 31}}}};
listA /. {x_Integer, y_Integer} :> {y - x}

Out[54]= {{{{1}, {1}}, {{3}, {3}, {3}}}, {{{2}, {2}}, {{2}, {2}}, {{5}}}}

More to the style you'd imagined—and faster for a large list—is:

In[57]:= Map[{-Subtract @@ #} &, listA, {-2}]

Out[57]= {{{{1}, {1}}, {{3}, {3}, {3}}}, {{{2}, {2}}, {{2}, {2}}, {{5}}}}

And the best version, I think:

``` In[69]:= Apply[{#2 - #1} &, listA, {-2}]

Out[69]= {{{{1}, {1}}, {{3}, {3}, {3}}}, {{{2}, {2}}, {{2}, {2}}, {{5}}}}```

Since you're Apply-ing, you get rid of the List structure. In that context, what does #[[2]] or #[[1]] mean?

Suppose there is an external database or other data source. Unfortunately, in this external data source, the keys are in lower case with underscores ("snake_case"), e.g. company_name. To work properly in Wolfram Language, the keys must be in camel case, e.g. companyName.

Here is an example that works on a List of Associations:

AssociationThread[
   StringReplace[Keys[#], 
     MapApply[Rule, 
      Transpose@{"_" <> # & /@ Alphabet[], 
        ToUpperCase[Alphabet[]]}]] -> Values[#]] & /@ data

To be honest, from my perspective it looks ugly, especially the "Transpose" part. Any ideas on how to make this transformation more concise?

... or is there a built-in function for this that I don't know about? I saw this resource function https://resources.wolframcloud.com/FunctionRepository/resources/ToCamelCase/, but I wasn't able to make it run in this "key of association" problem.

Notebook with sample data attached.

Thank you, Arben, for your super-fast reply.

Interesting - so it works with quotation marks. - Thank you!

Thank you very much, R, impressive! I was looking for something like that.

I think you just do exactly what you said you want to do, i.e. assign the variables before you take the derivative:

D[
Log[x^2 + y^2 + z^2] /. {x -> Sin[t], y -> Cos[t], z -> Tan[t]}
, t]

Or even before you take the Log, but I think that gives you the same answer:

D[Log[(x^2 + y^2 + z^2) /. {x -> Sin[t], y -> Cos[t], z -> Tan[t]}],
  t]

Take Rohit's answer, but also note that if you were to write x'[t] and so on—that is, explicitly note the time dependence—D will take those derivatives symbolically and you can plug in values for the time-dependent functions after the fact, at which point the derivatives will be evaluated.

Will there be another one of these classes?

I'm having problems understanding number three which works with Set Delayed Rule and Blank.

I couldn't catch on C++ a long time ago in college. Maybe that's why I don't understand this concept. Object orientated programming. My problem seems that there's too much overlap in definitions. There's no similarly to compare it to. The concept is Invisible.

There's a few things happening with the blank, and the "f" and the "x" those are all condensed things or overlapping definitions. Left to right replacement. Got that. what is "f" and "x" or x or x you could call them "Head" or a function. of the [ ]. I'm struggling to find an explanation of what x or x is Now I do understand that is is just a blank to be filled in but the x or f is a distraction. why not just _ without f or x that make more sense.

Now I tried to make this work and nope. nothing works. Is it because of the Delayed rule? That's a new idea. I guess I haven't tested that out yet. changing :> to -> instead. I can try that next. I don't understand what to do to get the replacement happening it seems that it should be just happening if there are blanks in the right place but maybe it's because of the Delayed rule. I tried changing the delay to a rule and it's the same. I used Trace. With Trace I just get {} and without Trace I get the same out as I put in f[1,2,3] I don't know if the rule is even working or able to work on f[1,2,3] that's my problem I don't see how the problem is applied to f[1,2,3] it seems it is not.

Thanks for the useful reference. Fills in some of the blanks in Arden's excellent lectures. I always like several points of view.

I worked on a exercise to from EIWL which was just adding up a list and I wanted to do it with Plus, Range. I failed and spent a hour learning. Then came back to it.

Recap. I was working on trying to use a symbol to do the adding or multiplying with @@ I could not get it to work. I could not figure out what was happening. I tried a lot of things. Nothing seemed to work. I looked up Trace, TreeForm, and a whatever I came across. I played around with the documentation changing the code around.

I figured out how to do it, after my nap. Checking with TreeForm told me the "path" of the code that I could not see. Checking the documentation on how to properly write it over and over. I could see that Plus was not getting worked on the List. So I knew I wrote it wrong.

The hour of work was not wasted. From that hour of work. That is how I got the code to be "visible". I worked with TreeForm and the documentation trying out Map, ThreadMap, Funny because I was using Apply in the first place I just wrote it wrong.

I found a you-tube video on Lists that was helpful. I also read some tutorials on how to properly write wolfram code. I read the Filled in Notebooks, and whatever I came across. So it was a bunch of hopping around sources then a long break and back to finally finding the correct way to write it. It helps that I realize that I'm trying to write it the way I think it works and that is wrong. I need to find how it is supposed to be written.

Merry coding to everyone.

Have a look at https://www.math.unl.edu/~tshores1/Public/Math208F07/Math208DifferentialsHandout.pdf

https://solitaryroad.com/c353.html

etc. Google is your best friend here.

On the L1 exam, do I take a list shown as {1, 2, 3, a, b, c} verbatim or am I to interpret the list as {1, 2, 3, "a", "b", "c"}? I wasn't sure since the color coding as done in Mathematica didn't carry over to the exam.

(confirming here for, well, official confirmation.)

Use Partition. It can be done in one instruction, but to make it clearer, I break it down:

lst = {1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0};

par = Partition[lst, 3, 1]

Cases[par, {0, 1, 0}]

In attached file Symmetries.nb , I've written and executed a module to find the symmetries you are interested in and a function to count the number of subsequences in an ordered list of length at least 2.

The module creates a random list of zeros and ones of length at least 2, finds all its ordered subsequences of length at least 2, and lists all its symmetric subsequences. I assume that subsequences of length 1 are of no interest because they are trivially symmetric.

The simple function counts the number of subsequences of length 2 or more from a list of length at least 2. The code uses the function to print a table of the counts for lists of length 2 through 15.

Although I compute the number of subsequences as a sum, it's easy to derive the formula by just evaluating the sum. Given a list (i.e., an ordered list) of length n, the number of subsequences of it of length k is just the number of leftmost positions of sequences of length k, namely n-(k-1) = (n+1) - k. Now, sum for k = 2 to n. Since (n+1) is a constant, the sum of the first term is (n-1)(n+1) = n^2 - 1. The sum of the second term is just 1 less than the sum of the first n integers, namely n(n+1)/2 - 1 = (n^2)/2 + n/2 -1. So the number of subsequences of length at least 2 of a list of length n is (n^2 - 1) - ((n^2)/2 + n/2 -1) = ((n^2)/2 - n/2 = n(n-1)/2. So, instead of using the sum to define the function numSequences , one could define it as numSequences[n_Integer /; n >= 2] := n(n-1)/2 . I used the sum just to make the code easy to understand.

I worked through your Nb. Thanks for the work you've done. It works well and I learned a thing or two, which I can apply on other places in my project.

Really appreciate your help, as well as Rohit's.

I am happy to see your explanation for the derivation for the formula for the number of (ordered) subsequences in an n-sequence. Very helpful. I tried Select i.s.o Cases and that works fine as well. I guess that you put the parenthesis around # === Reverse[#] in Select[subSequences, (# === Reverse[#]) &]; for clarity. It also works without them.

Thanks again , Gerald

I thought so, but a useful practice :-)

I have not run any benchmarks but I suspect that using the built-in PalindromeQ may be faster than # === Reverse[#] &.

Use the Sequence* family of functions

list = {1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0}

SequenceCases[list, {0, 1, 0}, Overlaps -> True]

SequencePosition[list, {0, 1, 0}]

Overlaps -> True is the default for SequencePosition which is a little inconsistent

Hi B.

In which sense is Overlaps -> True inconsistent? I saw in the docs, that SequencePosition has the same Overlaps option. Is it also inconsistent there?

What I meant is that the default value of Overlaps is True for SequencePosition and False for SequenceCases. That seems inconsistent to me.

For symmetry something like this?

list = {0, 0, 0, 1, 0, 1, 1, 0, 1}
Subsequences[list, {3, Length@list}] // Select[PalindromeQ]

For permutations generate all of them and test each with SequencePosition?

perms = Permutations[{1, 0, 1}]
list = {1,0,1,x,x,x,x 0,1,1,x,x,x,x,1,1,0}
SequencePosition[list, #] & /@ perms

Hi Rohit,

I wanted to take your idea a little further, more general, but I run into a snag I do not understand. SequenceCases[pattern1, {0, 1, 0}, Overlaps -> False] from your previous post. This works great :-)

But now to find more general patterns : list={0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1} patternToFind = {x, y, x_} SequenceCases[list, patternToFind, Overlaps -> False]

I get {{1, 1, 1}, {1, 0, 1}, {1, 0, 1}} as result.

Why is the [1,1,1} there? I specified the first and third to be the same, but the second element to be different. I expected to get {1,0,1} and {0,1,0}. The result also misses the {0,1,0}.

Any good ideas?

Great Rohit, thanks.

I had understood that {x,y,x} was different from {x,x,x} as a pattern .

I had the 'Overlaps -> True' first, but as I did not have x != y , I got about way too much solutions.

Now it works fine :-)

Best , Bert

Precisely! When I talk about pattern matching, I always talk about x_ as "something—call it x" for exactly this reason.