Efficient Row x Collumn multiplication

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Martijn
Martijn 2012-3-20
Hi all,
Consider the vectors: A of size; s x 4 B of size; 4 x s
Now, I am only interested in the product A(i,:)*B(:,i) of size s x 1
for i=1,..,s. I.e. only the row times the collumn with the same index.
The solution I found myself is: diag(A*B);
But I think there must be a faster solution, since I calculate many useless matrix elements if s>>4.
Do you guys have a suggestion?

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采纳的回答

Honglei Chen
Honglei Chen 2012-3-20
sum(A'.*B)
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更多回答（3 个）

Titus Edelhofer
Titus Edelhofer 2012-3-20
Hi,
don't know if it is really more efficient, give it a try:
x = dot(A',B)'
Titus
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Martijn
Martijn 2012-3-20
t=linspace(0,1,100);
tf=[t'.^3 t'.^2 t' ones(size(t'))];
M=[-1 3 -3 1; 3 -6 3 0 ; -3 0 3 0; 1 4 1 0]; %Standard form for matrix uniform cubic spline evaluation
P=randn(4,length(t));
tic; x = 1/6*diag(tf*M*P); toc
Elapsed time is 0.000065 seconds.
tic; y = 1/6*dot((tf*M)',P)'; toc
Elapsed time is 0.002589 seconds.
Hence, too slow.

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Daniel Shub
Daniel Shub 2012-3-20
Maybe I am missing something here, but it seems like you already have a solution ...
x = zeros(s, 1)
for i = 1:s
x(i) = A(i, :)*B(i, :);
end
Daniel Shub
Daniel Shub 2012-3-20
I am posting this as a separate answer. First when using timing, MATLAB has a hard time timing things that take only 0.000065 seconds, so you should put things in a loop. Second, The size of the matrices need to be established before discounting an answer as too slow.
t=linspace(0,1,1e4);
tf=[t'.^3 t'.^2 t' ones(size(t'))];
M=[-1 3 -3 1; 3 -6 3 0 ; -3 0 3 0; 1 4 1 0]; %Standard form for matrix uniform cubic spline evaluation
P=randn(4,length(t));
tic;
for ii = 1:10
x = diag(tf*M*P);
end;
toc
tic;
for ii = 1:10
x = sum((tf*M)'.*P)';
end;
toc
tic;
for ii = 1:10
x = dot((tf*M)', P)';
end;
toc
tic;
s = length(t);
x = zeros(s, 1);
A = (tf*M);
B = P;
for ii = 1:10
for i = 1:s
x(i) = A(i, :)*B(:, i);
end
end
toc
On my machine I get:
  1. Elapsed time is 7.584815 seconds.
  2. Elapsed time is 0.004872 seconds.
  3. Elapsed time is 0.004445 seconds.
  4. Elapsed time is 0.149892 seconds.
with the "too slow" loop being an order of magnitude faster than diag. With a large s, there is essentially no difference between the dot and sum method. For your example matrix sizes, the error checking in dot takes a significant portion of the time.
To really optimize your code you want to think about the order in which operations are occurring and memory access.
http://undocumentedmatlab.com/blog/matrix-processing-performance/
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Martijn
Martijn 2012-3-20
Thanks for the input! I, however, do not get how you would get a +7 second calculation time for the diag() function.
Anyway, for s>>4 diag() is indeed very slow and sum or dot are much much faster.
Daniel Shub
Daniel Shub 2012-3-20
Because I chose s to be very large (1e4) and I looped 10 times. If s ~4, then the diag method doesn't have that many useless calculations. It is only as s >> 4, that the number of useless calculations dominates.

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