How to eliminate for-loop?

Hi to everybody,

in the for-loop below, is a multiplication of a scalar with twice a matrix column. If two vectors in a for-loop are multiplied it should be possible to reformulate it into a matrix multiplication. How to do this? The task is to speed up the processing, especially if dimensions of matrix are becoming large.

A = [1 2; 3 4]
B = [5 6 7; 8 9 10; 11 12 13]
C = [1 0 0; 0 2 0; 0 0 0]

for indx = 1:size(A,2)
      D(indx,:,:) = C(indx,indx) .* A(:,indx) * B(:,indx)';
end

Thanks Andy

Answer 1

[Andrei Bobrov]

A = [1 2; 3 4];
B = [5 6 7; 8 9 10; 11 12 13];
C = [1 0 0; 0 2 0; 0 0 0];

n = size(A,2);
Cd = diag(C);
D = bsxfun(@times,A,reshape(Cd(1:n),1,[]));
D = bsxfun(@times,D,permute(B(:,1:n),[3,2,1]));
D = permute(D,[2,1,3]);

[Andreas Schwager]

Dear Andrei,

thanks for this answer!

However, if I do:

clear A B C D E F Cd n

A = [1 2; 3 4];
B = [5 6 7; 8 9 10; 11 12 13];
C = [1 0 0; 0 2 0; 0 0 0];
D = zeros(size(A,2),size(A,2),size(B,2));

for indx = 1:size(A,2)
    D(indx,:,:) = C(indx,indx) .* A(:,indx) * B(:,indx)';
end

n  = size(A,2);
Cd = diag(C);
E  = bsxfun(@times,A,reshape(Cd(1:n),1,[]));
F  = bsxfun(@times,E,permute(B(:,1:n),[3,2,1]));

D - F

... there are a couple of non zero elements. Your code ignores the last column of B.

Thanks! Andy

[Andrei Bobrov]

Thank you Stephen! I am corrected too.

Answer 2

[Stephen23]

This gives the correct output, unlike the accepted answer:

>> N = size(A,2);
>> Cd = diag(C);
>> G = bsxfun(@times,A.',permute(B(:,1:N),[2,3,1]));
>> G = bsxfun(@times,Cd(1:N),G);
>> isequal(D,G)
ans =  1

[Andrei Bobrov]

+1

[Andreas Schwager]

Dear Stephen, Dear Andrei,

Thanks a lot for your support!