Efficient number occurence count

I want to efficiently count the number of occurences of numbers between 1-numel(num) in a Matrix. I came up with two options for that:

sz = [3000 2000];
mx = prod(sz);
num = randi([1 mx],sz);
tic;
% First option
counts = zeros(numel(num),1);
for i = 1:numel(num)
    counts(num(i)) = counts(num(i)) + 1;
end
toc
tic;
% Second option
uni = unique(num);
uni = reshape(uni,[],1);
hc  = histcounts(num,[uni;uni(end)]);
toc

Execution times are:

Elapsed time is 0.098540 seconds.
Elapsed time is 0.342214 seconds.

So option 1 is clearly faster. However the for loop bugs me. Is there any possibility to vectorize this?

[Jan Siegmund]

Note: The reshape is because unique always returns a column vector, but if it receives a row vector as input, is produces a row vector and the concat fails... gotta love that inconsistency

[Adam Danz]

"I want to efficiently count the number of occurences of numbers between 1-numel(num) in a Matrix"

I'm a bit lost. Your matrix contains integers between 1 and 1000 and has 6000000 values (3000x2000). So, why are you looking for 6000000 different values when you only have a max of 1000 values?

[Jan Siegmund]

Sorry, num was a stupid example. A more suitable would be

sz = [3000 2000];
mx = prod(sz);
num = randi([1 mx],sz);
%...

Answer 1

[Matt J]

In this situation, accumarray will be faster than histcounts, but still not as fast as the for-loop,

tic;
hc=accumarray(num(:),1,[mx,1]).';
toc

Elapsed time is 0.172890 seconds. %for-loop
Elapsed time is 0.236013 seconds. %accumarray

unless the values are pre-sorted,

num=sort(num(:));
tic;
hc=accumarray(num(:),1,[mx,1]).';
toc

Elapsed time is 0.168976 seconds. %for-loop
Elapsed time is 0.075965 seconds. %accumarray

I think this is simply one of those situations where Matlab's for-loop optimization has caught up to vectorized code.

[Jan Siegmund]

Alright, accumarray is also a great choice. Thank you for your time and effort. This is the answer I am going to accept.

Answer 2

[Bruno Luong]

ac = accumarray(num(:),1);

Answer 3

[Matt J]

I want to efficiently count the number of occurences of numbers between 1-numel(num) in a Matrix

If that's really what you want, then

hc  = histcounts(num(:), 1:numel(num)+1 );

but as Adam points out, it would make more sense to have

hc  = histcounts(num(:), 1:1001 );

[Jan Siegmund]

Yea, I wanted 1-numel(num). Sorry for the bad example.

Your proposal is great!

So I updated my example.

sz = [3000 2000];
mx = prod(sz);
num = randi([1 mx],sz);
tic;
counts = zeros(numel(num),1);
for i = 1:numel(num)
    counts(num(i)) = counts(num(i)) + 1;
end
toc
tic;
hc  = histcounts(num(:),1:mx+1);
toc

But histcounts is still slower than the subscripted increment:

Elapsed time is 0.296456 seconds.
Elapsed time is 0.984777 seconds.

I think this is due to the tons of branching done in histcounts.m . Is there any more efficient way?

[Jan Siegmund]

Ok no, the branching is not the problem. Calling

 matlab.internal.math.histcounts

directly only results in a minor improvement.

[Matt J]

I think the for-loop is the fastest for integer data ranges that large. Unfortunately (and strangely), histcounts cannot innately recognize that the data consists only of integers and use a simpler binning method for that case. There is an input option 'BinMethod'='integers' that is offered, however, it will not permit more than 65536 bins.