histcounts error in place of histc

Links: histc, histcounts

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

Answer 1

[ANKUR KUMAR]

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)

[Jan]

Faster:

id = zeros(1, length(B));  % Pre-allocate
for ii = 1:length(B)
  id(ii) = sum(A==B(ii));  % Without FIND
end

Answer 2

[Sean de Wolski]

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])

[Jan]

+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.

Answer 3

[Jan]

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.