Count common elements of two vectors, including repeats

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?

[Jan]

What is the wanted result for [8 1 2 3] and [8 8 8 4]?

[Andrei Bobrov]

if:

a = [9 9 9 8 7 8 6]; b = [5 9 9 6 8 ];

result: ?

[Paolo Binetti]

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

Answer 1

[Stephen23]

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

[Andrei Bobrov]

+1

[Paolo Binetti]

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.