How can I vectorize this code ?
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Daniel
am 10 Mai 2017
Kommentiert: Daniel
am 10 Mai 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 Kommentare
Akzeptierte Antwort
Andrei Bobrov
am 10 Mai 2017
Bearbeitet: Andrei Bobrov
am 10 Mai 2017
C = rot90(filter2(A,B,'valid'),2);
or
C = conv2(rot90(B,2),A,'valid');
Weitere Antworten (1)
Jan
am 10 Mai 2017
Bearbeitet: Jan
am 10 Mai 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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