How to multiply two matrices

Hello,
For example I would like to multiply the following two matrices:
A = rand([10,10,10]);
B = rand([10,10,10]);
defined by:
for i=1:size(A,3)
C(:,:,i)= A(:,:,i)*B(:,:,i);
end
The question is: is this possible without the use of a for loop? Kind regards, Carlas

2 commentaires

umar siyab
umar siyab le 24 Nov 2011
no its not possible without the use of loop because u have to multiply every element in the matrix with other matrix..
Jan
Jan le 24 Nov 2011
What are the real dimensions in your problem? It matters if size(A, 3) is much larger or much smaller than size(A, 1) and size(A, 2).

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 Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 24 Nov 2011

0 votes

[m n p]= size(A);
o1 = ones(p,1));
out = cell2mat(cellfun(@mtimes,mat2cell(A,m,n,o1),mat2cell(B,m,n,o1),'un',0));
ADD
[m,n,p] = size(A);
out0 = bsxfun(@times,reshape(A,m,[]),reshape(permute(B,[1 3 2]),1,[],n));
out = reshape(sum(reshape(permute(out0,[1 3 2]),m,m,n,[]),3),m,n,[]);
or for out
out = permute(blockproc(out0,[m n],@(block_struct)sum(block_struct.data,2)),[1 3 2]);

Plus de réponses (3)

Titus Edelhofer
Titus Edelhofer le 24 Nov 2011

1 vote

Hi Carlas,
probably yes. But in this case probably the loop is not your worst option, as long as you initialize
C = zeros(size(A,2), size(B,1), size(A,3));
before the loop.
Titus

1 commentaire

Jan
Jan le 24 Nov 2011
I do not completely agree with you, Titus: Instead of "not the worst" I'd claim, it is at least "very good". +1
James Tursa's MTIMESX (http://www.mathworks.com/matlabcentral/fileexchange/25977-mtimesx-fast-matrix-multiply-with-multi-dimensional-support ) can handle such tasks efficiently.

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carlas
carlas le 24 Nov 2011

0 votes

A = rand([10,10,10]);
[m n p]= size(A);
tic
o1 = ones(p,1);
out = cell2mat(cellfun(@mtimes,mat2cell(A,m,n,o1),mat2cell(A,m,n,o1),'un',0));
t1 = toc;
tic
C = zeros([10,10,10]);
for i=1:size(A,3)
C(:,:,i) = A(:,:,i)'*A(:,:,i);
end
t2 = toc;
xFaster = t2/t1
xFaster =
0.5072
The for loop is faster. Based on the above answers I assume that the for loop is the most optimal implementation if the final matrix is initialized. However, it is possible to do without.

1 commentaire

Jan
Jan le 24 Nov 2011
I definitely prefer the loop. Especially for large inputs it has a much smaller memory footprint than the CELL method. For A=rand(100,100,1000) the loop is 3 times faster.

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carlas
carlas le 24 Nov 2011

0 votes

Thanks Titus and Andrei!

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