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How to eliminate for-loop?

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Andreas Schwager
Andreas Schwager 2016-7-27
评论： Andreas Schwager 2016-8-1
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

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2016-7-27
编辑：Andrei Bobrov 2016-7-27
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]);
  2 个评论
Andreas Schwager
Andreas Schwager 2016-7-27
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
Andrei Bobrov 2016-7-27
Thank you Stephen! I am corrected too.

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更多回答（1 个）

Stephen23
Stephen23 2016-7-27
编辑：Stephen23 2016-7-27
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

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