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Find real numbers closest to a multiple of 500 in a list?

David Kirkby

Posted 9 years ago

I have a data file (see attached), which has two columns. The first is a frequency in MHz, and the second a phase in degrees. There are 1601 data points - the frequency spans 50 MHz to 7000 MHz, in steps of 4.34375 MHz. I'd like to extract some of this data to put in a document, but I don't want to put all 1601 points. What I'd like to do is to find those data points that have a frequency nearest to a non-negative integer multiple of 500 MHz, (i.e. 0, 500, 1000, 1500, 2000, 2500 ...7000 MHz) and output only them. There's no data at 0 MHz, nor 500, nor 1000, but I want to find the closest to those. So I want the points 50.,0.21774563818642534 501.75,0.26184339842516735 1001.28125,0.5089995355033423 1500.8125,0.5958289355437785 2000.34375,0.847843737588093 2499.875,0.9325320400864996 2999.40625,1.4433401865247077 .... 7000.,2.1843935555882155 Any ideas? Attachments:

POSTED BY: David Kirkby

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Sander Huisman

Sander Huisman, University of Twente

Posted 9 years ago

Using a Do loop should be avoided in the answer of Paul, and Nearest is not used in a smart way. The answer of Joel has a similar problem as multiple datapoints could be selected; it depends on the factor 0.1 and 0.9 (ok for this test-data, but not a general solution). A faster/better way is as follows: SetDirectory[NotebookDirectory[]]; data = Import["frequency-and-phase-data.csv"]; d = 500 (the displacement ) find = Range[Floor[Min[data[[All, 1]]], \[Delta]], Ceiling[Max[data[[All, 1]]], \[Delta]], \[Delta]] (the points you look for) nf = Nearest[#[[1]] -> # & /@ data] (* make optimised nearest function) selected = First@nf /@ find (* find for each point ) {{50., 0.217746}, {501.75, 0.261843}, {1001.28, 0.509}, {1500.81, 0.595829}, {2000.34, 0.847844}, {2499.88, 0.932532}, {2999.41, 1.44334}, {3498.94, 1.67507}, {3998.47, 1.95012}, {4498., 2.02858}, {5001.88, 2.22979}, {5501.41, 2.45856}, {6000.94, 2.42086}, {6500.47, 2.26449}, {7000., 2.18439}} You might have to interchange Floor* and Ceiling, depending on how you want to handle the 'ends'.

Using a Do loop should be avoided in the answer of Paul, and Nearest is not used in a smart way. The answer of Joel has a similar problem as multiple datapoints could be selected; it depends on the factor 0.1 and 0.9 (ok for this test-data, but not a general solution). A faster/better way is as follows:

SetDirectory[NotebookDirectory[]];
data = Import["frequency-and-phase-data.csv"];
d = 500   (*the displacement *)
find = Range[Floor[Min[data[[All, 1]]], \[Delta]], Ceiling[Max[data[[All, 1]]], \[Delta]], \[Delta]] (*the points you look for*)
nf = Nearest[#[[1]] -> # & /@ data] (* make optimised nearest function*)
selected = First@*nf /@ find (* find for each point *)

{{50., 0.217746}, {501.75, 0.261843}, {1001.28, 0.509}, {1500.81, 
  0.595829}, {2000.34, 0.847844}, {2499.88, 0.932532}, {2999.41, 
  1.44334}, {3498.94, 1.67507}, {3998.47, 1.95012}, {4498., 
  2.02858}, {5001.88, 2.22979}, {5501.41, 2.45856}, {6000.94, 
  2.42086}, {6500.47, 2.26449}, {7000., 2.18439}}

You might have to interchange Floor and Ceiling, depending on how you want to handle the 'ends'.

POSTED BY: Sander Huisman

Joel Gilly

Joel Gilly, None

Posted 9 years ago

Sir, Thank you for your reply. I now understand the use of Nearest better. In analyzing your solution I noticed two items that I hope you can comment on: The use of the `#[[1]] -> #` in `Nearest[#[[1]] -> # & /@ data]`. I was unaware that you could use a Rule to transform the output of an function as the input to function. The use of `@` in `First@nf /@ find` to function like ReleaseHold. Again, I was unaware of the use of `@*` to act as a ReleaseHold in this context. Why does using First in this manner result in the output having the evaluation being held? eg. First@nf /@ find {Hold[Nearest[{50.->{50.,0.217746},54.3438->{54.3438,0.53045},58.6875->{58.6875,0.561783},63.0313->{63.0313,0.336275},67.375->{67.375,0.368827},71.7188->{71.7188,0.417978},76.0625->{76.0625,0.450648},80.4063->{80.4063,0.442057},84.75->{84.75,0.423945},89.0938->{89.0938,0.514353},93.4375->{93.4375,0.415191},97.7813->{97.7813,0.468367},102.125->{102.125,0.450253},106.469->{106.469,0.333008},110.813->{110.813,0.431593},115.156->{115.156,0.395657},119.5->{119.5,0.37876},123.844->{123.844,0.246633},128.188->{128.188,0.376706},132.531->{132.531,0.401067},136.875->{136.875,0.557509},141.219->{141.219,0.282309},145.563->{145.563,0.25467},149.906->{149.906,0.464329},154.25->{154.25,0.321209},158.594->{158.594,0.401958},162.938->{162.938,0.278126},167.281->{167.281,0.281734},171.625->{171.625,0.182552}}]][50], `Flatten /@ nf /@ find` (although incorrect in this context) does not have this effect. I don't mean to turn this thread into a Mathematica tutorial, but as someone who is still uncovering the subtle nuances of Mathematica, your insight is greatly appreciated.

Sir,

Thank you for your reply.

I now understand the use of Nearest better.

In analyzing your solution I noticed two items that I hope you can comment on:

The use of the #[[1]] -> # in Nearest[#[[1]] -> # & /@ data].
I was unaware that you could use a Rule to transform the output of an function as the input to function.

The use of @* in First@*nf /@ find to function like ReleaseHold. Again, I was unaware of the use of @* to act as a ReleaseHold in this context. Why does using First in this manner result in the output having the evaluation being held? eg.

    First@nf /@ find

    {Hold[Nearest[{50.->{50.,0.217746},54.3438->{54.3438,0.53045},58.6875->{58.6875,0.561783},63.0313->{63.0313,0.336275},67.375->{67.375,0.368827},71.7188->{71.7188,0.417978},76.0625->{76.0625,0.450648},80.4063->{80.4063,0.442057},84.75->{84.75,0.423945},89.0938->{89.0938,0.514353},93.4375->{93.4375,0.415191},97.7813->{97.7813,0.468367},102.125->{102.125,0.450253},106.469->{106.469,0.333008},110.813->{110.813,0.431593},115.156->{115.156,0.395657},119.5->{119.5,0.37876},123.844->{123.844,0.246633},128.188->{128.188,0.376706},132.531->{132.531,0.401067},136.875->{136.875,0.557509},141.219->{141.219,0.282309},145.563->{145.563,0.25467},149.906->{149.906,0.464329},154.25->{154.25,0.321209},158.594->{158.594,0.401958},162.938->{162.938,0.278126},167.281->{167.281,0.281734},171.625->{171.625,0.182552}}]][50],

`Flatten /@ nf /@ find` (although incorrect in this context) does
        not have this effect.

I don't mean to turn this thread into a Mathematica tutorial, but as someone who is still uncovering the subtle nuances of Mathematica, your insight is greatly appreciated.

POSTED BY: Joel Gilly

Sander Huisman

Sander Huisman, University of Twente

Posted 9 years ago

POSTED BY: Sander Huisman

Joel Gilly

Joel Gilly, None

Posted 9 years ago

The "Select" function with a criteria is your friend in this case. The creation of a suitable criteria function would be the complicated part.. Using data supplied from your other post, this is something congruent to your query... Select[thedata, (FractionalPart[IntegerPart[#[[1]]]/50] < 0.1) \|\| (FractionalPart[IntegerPart[#[[1]]]/50] > 0.9) &] As mentioned above, you will need to tweak the criteria function to meet your specific application. The "Nearest" function (see the online documentation) might also be applicable, but I am unfamiliar with it's use. I do not know how you would construct a criteria to narrow your results from "Nearest"...

POSTED BY: Joel Gilly

Paul Cleary

Posted 9 years ago

This would work, but first I opened the csv file with excel and saved it as an xlsx file. and the file is where MMa first looks for files. data = Flatten[Import["frequency-and-phase-data.xlsx", "Data"], 1]; list = {}; Do[x = FromDigits[Nearest[data[[All, 1]], 500 n]]; AppendTo[list, Cases[data, {a_, b_} /; a == x]], {n, 1, 14}]; Column[list] giving {{501.75,0.261843}} {{1001.28,0.509}} {{1500.81,0.595829}} {{2000.34,0.847844}} {{2499.88,0.932532}} {{2999.41,1.44334}} {{3498.94,1.67507}} {{3998.47,1.95012}} {{4498.,2.02858}} {{5001.88,2.22979}} {{5501.41,2.45856}} {{6000.94,2.42086}} {{6500.47,2.26449}} {{7000.,2.18439}} Change the 500 to 50 and the range in {n, 1, 14} to {n, 1, 140} for more points Paul.

This would work, but first I opened the csv file with excel and saved it as an xlsx file. and the file is where MMa first looks for files.

data = Flatten[Import["frequency-and-phase-data.xlsx", "Data"], 1];
list = {}; Do[x = FromDigits[Nearest[data[[All, 1]], 500 n]]; 
 AppendTo[list, Cases[data, {a_, b_} /; a == x]], {n, 1, 
  14}]; Column[list]

giving

{{501.75,0.261843}}
{{1001.28,0.509}}
{{1500.81,0.595829}}
{{2000.34,0.847844}}
{{2499.88,0.932532}}
{{2999.41,1.44334}}
{{3498.94,1.67507}}
{{3998.47,1.95012}}
{{4498.,2.02858}}
{{5001.88,2.22979}}
{{5501.41,2.45856}}
{{6000.94,2.42086}}
{{6500.47,2.26449}}
{{7000.,2.18439}}

Change the 500 to 50 and the range in {n, 1, 14} to {n, 1, 140} for more points

Paul.

POSTED BY: Paul Cleary

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