# 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
Edited: 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

Stephen23 on 19 Apr 2017
Edited: Stephen23 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.