CODEWITHSUNDEEP
matlabmatrixmatrix-multiplication
Math
Math

Reputation: 99

Multiplying two matrices by a vector

I am multiplying two matrices by a vector using loop. Is it possible to do that without using loop?

Something like D1=C.*(A.*B) is not working.

Below sample of code

clear;
clc;
A=rand(5,5);
B=rand(5,5);
C=[0.1 0.3];
for ii=1:2
    D(:,:,ii)=A.*B.*C(ii);
end

Upvotes: 1

Views: 94

Answers (2)

Leander Moesinger
Leander Moesinger

Reputation: 2462

You can do that with mostly matrix indexing:

clear;
clc;
A=rand(5,5);
B=rand(5,5);
C=[0.1 0.3];

% Get matrices to final size
A = A(:,:,ones(length(C),1)); % repeat into third dimension as many times as length(C)
B = B(:,:,ones(length(C),1)); % repeat into third dimension as many times as length(C)
C = C(ones(1,size(A,2)),:,ones(1,size(A,1))); % make it size size(A,2)xlength(C)xsize(A,1) 
C = permute(C,[3 1 2]); % change to correct order

D = A.*B.*C;

Or as one liner (faster,requires less memory and doesn't change input variables):

D = A(:,:,ones(length(C),1)).*B(:,:,ones(length(C),1)).*permute(C(ones(1,size(A,2)),:,ones(1,size(A,1))),[3 1 2]);

Still, i think for most matrices sizes bsxfun is faster (and better readable). But solving stuff with indexing is a lot more fun :P

Upvotes: 1

bla
bla

Reputation: 26069

this how to do it:

 D=bsxfun(@times,A.*B,permute(C,[3 1 2]))

Explanation: the trick is to change C from a row vector (say x-direction) to the 3rd dimension (or z-direction) using permute, which is as if you would have defined C differently:

C(:,:,1)=0.1;
C(:,:,2)=0.3;

Now, bsxfun is a compact way to do the for loop you wrote. that's it.

Upvotes: 3

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