Count occurences of row in matrix faster than by using nnz

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! :-)

Answer 1

[DGM]

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

Are there ways to speed up the counting of instances? Probably.

[Bruno Luong]

Yes to count quickly and easily

instances = accumarray(IC,1);

[RR]

Ended up using

[C,~,IC] = unique(M(:,1:3),'rows');
instances = accumarray(IC,1);
M_new[C instances]

and it works like charm (and about 1000x faster), thank you two very much!

[DGM]

Thanks @Bruno Luong

I really wish we could upvote comments ...

[Bruno Luong]

@DGM you just did ;-)