Find percentile corresponding to an input value

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I'm trying to find a MATLAB function that is similar to the PERCENTRANK formula in Excel. With this formula you enter an array and a scalar and it tells you what percentile that scalar corresponds to.

Accepted Answer

Walter Roberson
Walter Roberson on 8 Jul 2017
PERCENTRANK = @(YourArray, TheProbes) reshape( mean( bsxfun(@le, YourArray(:), TheProbes(:).') ) * 100, size(TheProbes) )
This accepts a vector or array of data, and an array of probe positions (could be a scalar), and returns an array of percentiles the same size as the probe positions, considered over the entire content of the data array (not per row or per column)
  3 Comments
Wenersamy de Alcantara
Wenersamy de Alcantara on 11 May 2020
Edited: Wenersamy de Alcantara on 11 May 2020
Hi, Alex, the algorithm of the suggested PERCENTRANK function basically averages the result of a logical comparison.
For example, in the vector (6 numbers drawn from a discrete uniform distribution):
[23 45 98 2 81 17]
If you compare it with, say, 30: [23 45 98 2 81 17] <= 30, you'll have:
[1 0 0 1 0 1]
If you avarage the results of the comparison, you'll have:
3/6 = 0.5 or 50% percentile.
Of course, the approximation gets better with a bigger sample.
In other words, you can implement excel function PERCENTRANK.INC(matrix,k) by just using: mean(matrix<=k).
Note that they are not exactly the same algorithm, but they get closer as the size of "matrix" increases.
Miguel Lovino
Miguel Lovino on 22 Jun 2020
Hi Walter,
this code works perfectly for a vector (lat - lon -time varying). I want to adapt it to an array array of 121 * 121 * 2995 (it's lat-lon -time). That is, for a given lat-lon, it works. I try to rewrite as:
A = ncread('ERA5_pentads_sm_l123_1979-2019.nc','sm');
B = ncread('ERA5_pentads_sm_l123_1979-2019.nc','sm');
[n,m,t]=size(A);
As = zeros(n,m,t);
for ii=1:n
for jj=1:m
PER = @(A,B) reshape( mean( bsxfun(@le, A(ii,jj,:), ipermute(A(ii,jj,:),[3 2 1])))...
* 100, t, []);
As (ii,jj,:)= PER (A);
end
end
It works, but it seems that lat-lon points are not correct.
I really appreciate some help,
Best
Miguel

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More Answers (1)

Stefano Crema
Stefano Crema on 15 Oct 2019
Great, compact and handy solution, thanks Walter!

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