Multiplying certain dimensions of two 3D matrices

Hi

I've got two matrices, one 30x30x50 and one 30x30x80

I would like to multiply them in such a way that the result is a 50x80 matrix ie. an implicit sum over all of the 30x30 common elements.

eg. A*B = C

C(k,l) = A(i,j,k)*B(i,j,l) summed over all i, j

Is there a simple way to write this? I would rather avoid complications using "reshape".

Thanks, Mike

Answer 1

[Andrei Bobrov]

Idea of Sean de Wolski

C = squeeze(sum(sum(bsxfun(@times,a,permute(b,[1 2 4 3])))))

or

s = size(a)
C = squeeze(sum(bsxfun(@times,reshape(a,s(1)*s(2),[])',reshape(b,1,s(1)*s(2),[])),2))

[Andrei Bobrov]

Hi Michael! Thank you for "accpted", but Teja's variant is better than my.

Answer 2

[Teja Muppirala]

A = rand(30,30,50);
B = rand(30,30,80);
C = reshape(A,900,[])'*reshape(B,900,[]);

[Andrei Bobrov]

+1

[Jan]

+1

[Michael]

I'll use this one, thanks everyone

[Michael]

I have one more related question,

I'm also trying to compute D = C*A, I think A must be transposed for this to make sense: the (i,j,l) element of D is = (k,l) of C times (i,j,k) of A summed over k. I've experimented a bit and found this,

reshape(reshape(A,900,[])*C,30,30,[]);

Is it the best method? I haven't found a general rule to do these automatically yet and I'm not sure if the 'outer' reshape is preserving the form of my 30x30 matrix and not inadvertently flipping it about a diagonal.

Thanks very much for any help

Answer 3

[Comrun Yousefzadeh]

I have a related questoin. I'm trying to multiply a two dimentioanl array to a specific dimension of a 3-D matrix. For example, A=2000*2000*72, B=72*3 and I'm trying to reach C=2000*2000*3. In other words B should be multiplied to the 3rd dimension of A and repeated (element wise) for 2000*2000 points in the first two dimensions. Currently I'm using a loop like this:

for ii=1:size(A,1)

for jj=1:size(A,2)

A1=A(ii,jj,:);

A1=A1(:);

C1=B'*A1; % C1 has dimension of 3*1

C2=D*C1; %D is a 3-3 matrix therefore, C2 has the same dimension as C1

C(ii,jj,:)=C2(:,1);

end

end

What is the better way to avoid the loop?

Thanks,