Get the diagonal without calculating the explicit matrix
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Dear all:
I am trying to calculate a diagonal of a matrix (denoted A), which is formed by multiplying two large-dimensional matrices (denoted as B*C).
A naive way to do it is: first, calculating explicitly A = B*C, then get diagonal out from A. However, the first step takes forever to run due to the high-dimension of B and C. But the only thing I need is the diagonal of A.
Another straightforward way in my mind is: I could create a loop by calculating each element of the diagonal of A one by one. It will surely save a lot of time, but I am not sure if this is the most efficient way.
I am wondering if anyone knows a faster/smarter way to calculate it.
Thank you very much in advance!
Best,
Long
1 个评论
The best approach will depend on the dimensions of the matrices, and whether they are of sparse-type or not.
采纳的回答
Assuming B*C results in a square matrix,
diagonal=sum(B.' .* C, 1);
8 个评论
If B*C is not square, the same concept applies, just a bit messier:
N=min(size(B,1), size(C,2));
diagonal=sum(B(1:N,:).' .* C(:,1:N), 1);
Long Hong
2020-3-26
Note that if your matrices are sparse, the looping method will be much slower.
N=7000;
B=sprand(N,10*N,10/N); C=B.';
tic; sum(B.'.*C,1); toc %Elapsed time is 0.021923 seconds.
tic;
for i=1:N
B(i,:)*C(:,i);
end
toc; %Elapsed time is 11.467097 seconds.
the cyclist
2020-3-26
If the transpose in Matt's algorithm is really the time-consuming step, would it be possible to go upstream in your code and do all prior calculations in a way that the B you end up with is the transpose of the current one?
Long Hong
2020-3-26
Long Hong
2020-3-26
更多回答（1 个）
the cyclist
2020-3-26
Here is one way:
% Make up some inputs
N = 4;
B = rand(N);
C = rand(N);
% Calculate the diagonal
A_diag = 0;
for nr = 1:N
A_diag = A_diag + B(:,nr).*C(nr,:)';
end
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