is there any way that can speedup the process of following code... A and B are 50000x20 matrices..?

indx=1;
for f=1:length(A)
    for b=1:length(B)
    if (A(f,1:6)==B(b,1:6))&(A(f,7)==B(b,7))
    AA(indx,:)=A(f,[1:7 11]);
    BB(indx,:)=B(b,1:11); 
    else
    end
end
indx=indx+1;
end

Answer 1

[Jos (10584)]

Your looking for corresponding rows in A and B. intersect might help you

[~,ia,ib] = intersect(A(:,1:7), B(:,1:7), 'rows') ;
AA = A(ia,:) ;
BB = B(ib,:) ;

btw, you loop using length. But length only returns the longest of the dimensions. If you want to loop over rows, use size with the dimension argument, as in size(..., 1)

[Guillaume]

Yes, never use length on 2D matrices. The loop would fail on 10x20 matrices for example.

In fact never use length at all. For vectors use numel instead.

Note that to replicate the loop the output should be:

AA = A(ia, [1:7 11]);
BB = B(ib, 1:11);

Also note that intersect removes duplicates. so if the keys are present multiple times in AA or BB, only one instance will be present in the result. On the other hand with the loop, the duplicate instances would be replicated many times (times present in A x times present in B).

Finally, with regards to the code in the question, many people do not know how if behaves with vector inputs. I suspect that it is by luck that the loop behaves properly, and even if it isn't luck, it's better to make the implicit all explicit:

if all(A(f, 1:6) == B(b, 1:6) & A(f, 7) == B(b, 7))

[Jos (10584)]

useful additional comments, Guillaume! +1

[Lakshmi Chodavarapu]

Thanks a lot for valuable suggestions and inputs Guillaume. The code is running really well with intercept Jos. Thanks a lot!