Using hist in Matlab to compute occurrences

Question

I am using hist to compute the number of occurrences of values in a matrix in Matlab.

I think I am using it wrong because it gives me completely weird results. Could you help me to understand what is going on?

When I run this piece of code I get countsB as desired

rng default; 
B=randi([0,3],10,1);
idxB=unique(B);
countsB=(hist(B,idxB))';

i.e.

B=[3;3;0;3;2;0;1;2;3;3];
idxB=[0;1;2;3];
countsB=[2;1;2;5];

When I run this other piece of code I get wrong results for countsA

A=ones(524288,1)*3418;
idxA=unique(A);
countsA=(hist(A,idxA))';

i.e.

idxA=3148;
countsA=[zeros(1709,1); 524288; zeros(1708,1)];

What am I doing wrong?

you want [`histcounts`](https://uk.mathworks.com/help/matlab/ref/histcounts.html)? — Ander Biguri, Mar 23 '17 at 12:30
Not exactly, I want to count the number of occurrences of each element in `A` — TEX, Mar 23 '17 at 12:34

score 3 · Answer 1 · answered Mar 23 '17 at 12:34

3

idxA is a scalar, which means the number of bins in this context. setting idxA as a vector instead e.g. [0,3418] will get you a hist with bins centered at 0 and 3418, similarly to what you got with idxB, which was also a vector

answered Mar 23 '17 at 12:34

Yuval Harpaz

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score 3 · Accepted Answer · answered Mar 23 '17 at 12:36

3

To add to the other answers: you can replace hist by the explicit sum:

idxA = unique(A);
countsA = sum(bsxfun(@eq, A(:), idxA(:).'), 1);

answered Mar 23 '17 at 12:36

Luis Mendo

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Is it as fast as or faster than `hist` for `size(A,1)` very large? – TEX Mar 23 '17 at 12:39
I always [bet on `bsxfun`](https://meta.stackoverflow.com/a/303542/2586922), but better test yourself for your specific case – Luis Mendo Mar 23 '17 at 12:39
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When `idxA` and `A` have huge length (e.g. 8000 and 1316966) it gives me error because `Error using bsxfun Requested 1316966x8000 (9.8GB) array exceeds maximum array size preference. Creation of arrays greater than this limit may take a long time and cause MATLAB to become unresponsive. See array size limit or preference panel for more information` However, when size limits are not exceeded, it is faster than `hist`. – TEX Mar 23 '17 at 13:01

score 2 · Answer 3 · answered Mar 23 '17 at 12:35

2

I think it has to do with:

N = HIST(Y,M), where M is a scalar, uses M bins.

and I think you are assuming it would do:

N = HIST(Y,X), where X is a vector, returns the distribution of Y
among bins with centers specified by X.

In other words, in the first case matlab is assuming that you are asking for 3418 bins

answered Mar 23 '17 at 12:35

HelloWorld

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Do you know any other better way to count occurrences? – TEX Mar 23 '17 at 12:36

Using hist in Matlab to compute occurrences

3 Answers3