Efficient number occurence count
显示 更早的评论
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?
3 个评论
Jan Siegmund
2020-10-17
编辑：Jan Siegmund
2020-10-17
Adam Danz
2020-10-18
"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
2020-10-18
采纳的回答
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.
更多回答（2 个）
Bruno Luong
2020-10-18
ac = accumarray(num(:),1);
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 );
3 个评论
Jan Siegmund
2020-10-18
编辑：Jan Siegmund
2020-10-18
Jan Siegmund
2020-10-18
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.
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