How can I vectorize this code ?
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Daniel
el 10 de Mayo de 2017
Comentada: Daniel
el 10 de Mayo de 2017
Input:
- A: (m x n)
- B: (k x l) with k,l > m,n respectively
Output:
- C: (p x q) with p=k-m +1 and q=l-n+1
Each element of C is the sum of the "element by element" product of A and a (m x n) submatrix of B.
- For C(1,1) the (m x n) submatrix is located in the bottom right corner of B.
- For C(u<p,v<q) the (m x n) submatrix is shifted by u upwards and v leftwards.
- For C(p,q) the (m x n) submatrix is located in the top left corner of B.
My code:
C = zeros(p,q);
for u = 1:1:p
for v = 1:1:q
C(u,v) = sum(sum(A .* B( p-u+1:k-u+1 , q-v+1:l-v+1 )));
end
end
It works fine but is way too slow (A and B are very large and contain complex values).
Question:
How can I vectorize this code to increase the speed ?
2 comentarios
Respuesta aceptada
Andrei Bobrov
el 10 de Mayo de 2017
Editada: Andrei Bobrov
el 10 de Mayo de 2017
C = rot90(filter2(A,B,'valid'),2);
or
C = conv2(rot90(B,2),A,'valid');
Más respuestas (1)
Jan
el 10 de Mayo de 2017
Editada: Jan
el 10 de Mayo de 2017
For experiments:
function test
m = 100;
n = 100;
k = 1200;
l = 1200;
p = k - m + 1;
q = l - n + 1;
A = rand(m, n);
B = rand(k, l);
tic;
C = zeros(p, q);
for u = 1:p
for v = 1:q
C(u,v) = sum(sum(A .* B(p-u+1:k-u+1, q-v+1:l-v+1)));
end
end
toc
tic;
C = zeros(p, q);
Av = A(:).';
Bv = zeros(numel(Av), 1);
for v = 1:q
for u = 1:p
Bv(:) = B(p-u+1:k-u+1, q-v+1:l-v+1);
C(u, v) = Av * Bv(:);
end
end
toc
The 2nd version uses the summation performed in the DOT product. For the given values it runs in 19 seconds compared to 38 of the original code. But the dimensions are guessed only.
I assume a C-Mex to be more efficient, because it will avoid the explicite creation of B(p-u+1:k-u+1, q-v+1:l-v+1). Do you have a C-compiler installed?
[EDITED] Andrei's filter2 and conv2 approaches need about 5 seconds. I leave the modified loop method also in the forum, because it demonstrates how to increase the speed by a factor 2 with trivial methods.
[EDITED 2] With complex input the timings look different: 48 sec for the modified loop, 21 seconds for the conv2 approach. Interesting! The loop is 2.5 times slower, conv2 4 times.
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