Row by column multiplication

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John
John 2012-10-15
评论： Matt J 2015-10-22
Consider two matrices A and B defined by
A=rand(10,3);
B=rand(3,10);
I'm interested in multiplying the vectors defined the first, second, and third rows in B by the vectors defined the first, second, and third columns in A, respectively. My intent is to generate ten, 3X3 matrices defined by
Matrix1= B(:,1)*A(1,:)
Matrix2= B(:,2)*A(2,:)
...
Matrix10= B(:,10)*A(10,:)
My initial thought was something like
B(:,1:10)*A(1:10,:)
but this approach populates the matrices B and A prior to multiplication, yielding a single 3X3 matrix. How to I change the "order-of-operations" if you will---call each vector on a term-by-term basis, multiplying in between each call to generate the desired matrix? Of course, I really want to avoid having to use for loops.

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采纳的回答

Matt J
Matt J 2012-10-15
Here's a completely vectorized method, which requires James Tursa's MTIMESX function, available at the link below
A=reshape(A',1,3,10);
B=reshape(B,3,1,10);
C=mtimesx(B,A)
http://www.mathworks.com/matlabcentral/fileexchange/25977-mtimesx-fast-matrix-multiply-with-multi-dimensional-support
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Amandeep Gautam
Amandeep Gautam 2015-10-22
I wonder if there is anything wrong with A' * B? Seems like it does the same thing. Can you comment?
Matt J
Matt J 2015-10-22
That will generate an error
>> A=rand(10,3); B=rand(3,10); A'*B
Error using *
Inner matrix dimensions must agree.

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更多回答（4 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-10-15
编辑：Azzi Abdelmalek 2012-10-15
for k=1:10
matrix{k}=B(:,k)*A(k,:)
end
%or
for k=1:10
matrix(:,:,k)=B(:,k)*A(k,:)
end
Matt Fig
Matt Fig 2012-10-15
编辑：Matt Fig 2012-10-15
I think you can avoid FOR loops, but I see no reason to do so here:
A=randi(10,10,3);
B=randi(10,3,10);
for ii = 10:-1:1,C{ii} = B(:,ii)*A(ii,:);end
Here is a vectorization, but I can almost guarantee you it will be slower than the above FOR loop:
cellfun(@mtimes,mat2cell(B,3,ones(1,10)).',mat2cell(A,ones(1,10),3),'Un',0)
Matt J
Matt J 2012-10-15
Here's a 1-liner. It's syntactically brief, but not superior to using loops directly,
C=cellfun(@(a,b) a(:)*b(:).', num2cell(B,1), num2cell(A,2).', 'uni',0)
Richard Brown
Richard Brown 2012-10-15
编辑：Richard Brown 2012-10-15
There is a built in way to do it quickly if you use sparse multiplication. Essentially, you construct the block diagonal matrix blkdiag(B(:, 1), ... , B(:, N)) directly, and multiply it by A to give Y = [B(:, 1)*A(1, :) ; ... ; B(:, N) * A(N, :)]
N = 10000;
A = rand(N, 3);
B = rand(3, N);
I = 1:3*N;
J = repmat(1:N, 3, 1);
Y = mat2cell(sparse(I, J, B) * A, repmat(3, N, 1), 3);
edit: brief explanation added
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Richard Brown
Richard Brown 2012-10-16
Well, it depends what precisely speaking we mean by "built in" doesn't it?! My definition is that if it comes with a basic MATLAB install it's builtin.
Also, the mat2cell call isn't necessary -- it's just to convert it to the output format in some of the other answers. And the code doesn't touch the branch of repmat that uses for loops. Poor old for loops, they get such a hard time here :)
Matt J
Matt J 2012-10-16
I suppose that's true...

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