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histcounts error in place of histc

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Pramit Biswas
Pramit Biswas on 19 Jan 2018
Edited: Jan on 2 Feb 2018
Objective: finding the frequency of unique elements in an array.
A = [0]
F = histc(A,unique(A))
% F = histcounts(A,unique(A))
Expected output is 1 for A=[0] or 2 for A=[0 0], which is perfectly done by histc. As currently histc is not recommended. Use histcounts instead. I tried, but encountered error. Notes for change

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Answers (3)

ANKUR KUMAR
ANKUR KUMAR on 1 Feb 2018
Try this.
A=randi(5,1,15)
a=histcounts(A)
bar(unique(A),a)
If you are facing problem using histcounts, then you can find the frequency of unique elements without using histcounts
clc
clear
A=randi(5,1,15)
B=unique(A)
for ii=1:length(B)
id(ii)=length(find(A==B(ii)))
end
bar(B,id)

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Jan
Jan on 2 Feb 2018
Faster:
id = zeros(1, length(B)); % Pre-allocate
for ii = 1:length(B)
id(ii) = sum(A==B(ii)); % Without FIND
end

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Sean de Wolski
Sean de Wolski on 1 Feb 2018
The edges needs to be at least two elements. I usually do this to make one side (either negative or positive depending on context) open ended:
A = 0
F = histc(A,unique(A))
F2 = histcounts(A,[unique(A), inf])

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Jan
Jan on 2 Feb 2018
+1: This is the best translation of the arguments for the old histc to the new histcounts.
What a pity: For large arrays, creating the temporary vector [unique(A), inf] wastes time compared to the histc method. After all these years of using this function successfully, I'm running my own C-Mex function now instead of histcounts. A bad decision of TMW.

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Jan
Jan on 2 Feb 2018
Edited: Jan on 2 Feb 2018
What a pity that the handy histc is deprecated now.
A = randi(5,1,15);
[uniqA, ~, iA] = unique(A);
N = histcounts(iA, 'BinMethod', 'integers');
Now the element uniqA(k) appears N(k) times.
This is fast, but limited to 65536 elements.

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