Vectorize a simple code
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Hello, Is there more vectorized way to write this code ? ( and avoid the loops)
for j=[1:size(C,1)];
for k=1:size(C,2)
C(j,k)=sum(sum(A([1:j],[1:k]).*B([j:-1:1],[k:-1:1])));
end
end
Thank you in advance
EDIT
I modified the code
j=1:size(C,1);
k=1:size(C,2);
B2=rot90(B,2);
C(j,k)=sum(sum(A(1:j,1:k).*B2(end+1-j:end,end+1-k:end)));
It's normal that I have the same value on all the matrix C ?
P.S: I understood that conv2 is the more performant function but I just want to understand where is the problem (why (j, k) doesn't vary) and what's the solution (without loop)
采纳的回答
Note: See https://www.mathworks.com/matlabcentral/answers/35676-why-not-use-square-brackets : Even omitting the unneeded square brackets will accelerate the copde already.
A = rand(56); % If I understand your inputs correctly:
B = rand(56);
tic;
for k = 1:100
C = zeros(size(A));
for j=[1:size(C,1)]
for k=1:size(C,2)
C(j,k)=sum(sum(A([1:j],[1:k]).*B([j:-1:1],[k:-1:1])));
end
end
end
toc
tic;
for k = 1:100
C = zeros(size(A));
for j=1:size(C,1)
for k=1:size(C,2)
C(j,k)=sum(sum(A(1:j,1:k).*B(j:-1:1,k:-1:1)));
end
end
end
tic
tic;
for k = 1:100
C = zeros(size(A));
for k = 1:size(C,2)
BB = B(end:-1:1, k:-1:1);
for j = 1:size(C,1)
C(j, k) = sum(sum(A(1:j, 1:k) .* BB(end-j+1:end, :)));
end
end
end
toc
Elapsed time is 5.530599 seconds.
Elapsed time is 4.280142 seconds.
Elapsed time is 3.919369 seconds.
I assume conv2 is a much better approach, although I cannot include the reverted parts of B yet.
But it is worth to know this detail in general: avoid unneeded square brackets. You see a corresponding MLint warning in the editor also: the small orange mark under the [.
更多回答（1 个）
Honglei Chen
2017-6-30
I don't know what dimensions you have in C, but this looks like
D = conv2(A,B);
Maybe you can use
D = D(1:size(C,1),1:size(C,2));
to extract the portions you want.
HTH
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