# Count common elements of two vectors, including repeats

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Paolo Binetti on 19 Apr 2017
Commented: Paolo Binetti on 20 Apr 2017
I would like to count the common elements of two vectors a and b of integers, including repeats. For example, for:
a = [9 12 8 7 8 3 3]; b = [3 2 9 9 3 5 8]; % result should be 4
a = [9 8 7 8 6]; b = [5 9 9 6 8]; % result should be 3
a = [8 1 2 3]; b = [8 8 8 4]; % result should be 1
a = [9 9 9 8 7 8 6]; b = [5 9 9 6 8 ]; % result should be 4
After several attempts with functions "intersect", "ismember", and "histc", I found a way to do it:
range_ab = [min([a b]):max([a b])];
result = sum(min(histc(a, range_ab), histc(b, range_ab)))
This solution can also be easily vectorised to compare not only a single vector a to vector b, but multiple vectors a1, a2, ... an with vector b at the same time (not shown here), which is what I will use it for.
But isn't there a simpler / more efficient way to perform this apparently simple operation of finding the common elements of two vectors?
Paolo Binetti on 20 Apr 2017
Apologies, my original example was not general enough, I will edit the question in a few minutes, adding these examples:
a= [8 1 2 3]; b = [8 8 8 4]; % result should be 1
a= [9 9 9 8 7 8 6]; b = [5 9 9 6 8 ]; % result should be 4

Stephen on 19 Apr 2017
Edited: Stephen on 19 Apr 2017
Edited using all examples in comments or the original question.
>> a = [9,8,7,8,6]; b = [8,9,9,6,8];
>> idx = bsxfun(@eq,sort(a),sort(b(:)));
>> nnz(cumsum(idx,2) == cumsum(idx,1) & idx)
ans = 4
and
>> a = [9,8,7,8,6]; b = [5,9,9,6,8];
>> idx = bsxfun(@eq,sort(a),sort(b(:)));
>> nnz(cumsum(idx,2) == cumsum(idx,1) & idx)
ans = 3
and
>> a = [8,1,2,3]; b = [8,8,8,4];
>> idx = bsxfun(@eq,sort(a),sort(b(:)))
>> nnz(cumsum(idx,2) == cumsum(idx,1) & idx)
ans = 1
and
>> a = [9 9 9 8 7 8 6]; b = [5 9 9 6 8 ];
>> idx = bsxfun(@eq,sort(a),sort(b(:)));
>> nnz(cumsum(idx,2) == cumsum(idx,1) & idx)
ans = 4
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Paolo Binetti on 20 Apr 2017
Thank you Stephen, your solution works. Since you proposed it however I have found a different way which seems somewhat faster with long vectors. But I suspect even better ways must exist, and this is how I have recasted my post.