Count occurences of row in matrix faster than by using nnz

Question

RR 2022-7-20

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1764245-count-occurences-of-row-in-matrix-faster-than-by-using-nnz

评论： Bruno Luong 2022-7-20

Hi there,

I have an matrix M of about 500k x 4 and I would like to count, how often each row occurs and the output should look like

[M(1,1), M(1,2), M(1,3), number of occurences;

M(2,1), M(2,2), M(2,3), number of occurences;

.... ]

Currently, I am using

for i=1:1:length(M)

M(i,4)=nnz(all(M(:,1:3)==[M(i,1) M(i,2) M(i,3)],2));

end

which does the job but it's very slow with this matrix size. I read a lot about accumarray for this purpose and it's supposed to be much faster but so far my efforts to get it running weren't successful. Could you help me make it work? Or is there maybe an even more suitable function for this job? Thanks so much in advance! :-)

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Answer 1

DGM 2022-7-20

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https://ww2.mathworks.cn/matlabcentral/answers/1764245-count-occurences-of-row-in-matrix-faster-than-by-using-nnz#answer_1011590

编辑：DGM 2022-7-20

在 MATLAB Online 中打开

If you know there are repeated rows, then you know that you're performing redundant operations. One thing you could do is to use

[C,IA,IC] = unique(A,'rows')

to reduce the size of the set. Then instead of counting instances in A or C, count the instances in IC, since they're all scalars.

Consider:

A = [1 2 3; 4 5 6; 1 2 3; 5 9 3; 1 2 3]
A = 5×3
     1     2     3
     4     5     6
     1     2     3
     5     9     3
     1     2     3
[C,~,IC] = unique(A,'rows');
urows = size(C,1);
instances = zeros(urows,1);
for r = 1:urows
    instances(r) = nnz(IC==r);
end
[C instances]
ans = 3×4
     1     2     3     3
     4     5     6     1
     5     9     3     1