如何将矩阵A的每一列乘以矩阵B的每一行，并在Matlab中求和矩阵？

Question

I have a problem which I hope can be easily solved. A is a N G matrix, B is N G matrix. The goal is to get matrix C

which is equal to multiplying each column of transposed A by each row of B and summing resulting matrices; total number of such matrices before summing is N N, their size is G G This can be easily done in MatLab with two for -loops:

N=5;
G=10;
A=rand(N,G);
B=rand(N,G);
C=zeros(G);
for n=1:1:N
    for m=1:1:N
        C=C+A(m,:)'*B(n,:);
    end
end

However, for large matrices it is quite slow.

So, my question is: is there a more efficient way for calculating C matrix in Matlab?

Thank you

Answer 1

If you write it all out for two 3×3 matrices, you'll find that the operation basically equals this:

C = bsxfun(@times, sum(B), sum(A).');

Running each of the answers here for N=50 , G=100 and repeating each method 100 times:

Elapsed time is 13.839893 seconds. %// OP's original method
Elapsed time is 19.773445 seconds. %// Luis' method
Elapsed time is 0.306447 seconds.  %// Robert's method
Elapsed time is 0.005036 seconds.  %// Rody's method

(a factor of ≈ 4000 between the fastest and slowest method...)

Answer 2

I think this should improve the performance significantly

C = zeros(G);
for n = 1:N
   C = C + sum(A,1)'*B(n,:);
end

You avoid one loop, and should also avoid the problems of running out of memory. According to my benchmarking, it's about 20 times faster than the approach with two loops. (Note, I had to benchmark in Octace since I don't have MATLAB on this PC).

Answer 3

使用bsxfun代替循环，然后sum两次：

C = sum(sum(bsxfun(@times, permute(A, [2 3 1]), permute(B,[3 2 4 1])), 3), 4);