how to vectorize nested for loops
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Hello everyone
I want to vectorize the following code to reduce operation time. Can you guys help me out?
thanks
% X is a double matrix
s=size(X);
RepSol=repmat(X,2,2);
% Y is a binary matrix
Y=repmat(Y,2,2);
% Preallocation
p=zeros(s);
for m=1:s(1)
for n=1:s(2)
for i=1:s(1)
for j=1:s(2)
if RepSol(i,j)==RepSol(m+i-1,n+j-1)
p(m,n)=p(m,n)+Y(i,j)*Y(m+i-1,n+j-1);
end
end
end
end
end
回答（2 个）
darova
2019-9-10
IS it correct?
for m=1:s(1)
for n=1:s(2)
cond = RepSol(1:s(1),1:s(2)) == RepSol((1:s(1))+m-1,(1:s(2))+n-1);
% cond = X - RepSol((1:s(1))+m-1,(1:s(2))+n-1);
P = Y(1:s(1),1:s(2)) .* Y((1:s(1))+m-1,(1:s(2))+n-1);
res = cond .* P;
p(m,n) = sum(res(:));
end
end
4 个评论
Bob Thompson
2019-9-10
cond = RepSol(1:s(1),1:s(2)) == RepSol((1:s(1))+m-1,(1:s(2))+n-1);
A couple of things:
RepSol(1:s(1),1:s(2))
This is just the entirety of RepSol, you shouldn't need to index it
RepSol((1:s(1))+m-1,(1:s(2))+n-1)
Your indexing here doesn't make sense. (1:s(1)) is just calling all rows, but then you're trying to adjust that range to +m-1. So, suppose you have ten rows, and you are looking at m = 5, then you are saying you want to look at rows 5:15, which isn't possible because you only have 10 rows.
Unfortunately, I don't have a way to solve your problem off the top of my head, but it might help us come up with a solution if you explain what you're trying to do with your loops. There might be a command which does it already.
darova
2019-9-10
s(1) and s(2) are size of RepSol before repmat (two times less)
s=size(X);
RepSol=repmat(X,2,2);
AliHg
2019-9-11
AliHg
2019-9-17
0 个投票
4 个评论
darova
2019-9-17
Can you tell something more about your problem? What are yuo working on now?
Maybe there is some solution
AliHg
2019-9-17
darova
2019-9-17
Unfortunately is explains nothing
The script does the follwoing if i understood correctly:
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Rik
2019-9-17
I agree with darova that your explanation is minimal. Please edit your question, adding what you're trying to achieve. Maybe your problem can be solved with a well-chosen convolution, but with the current level of detail it is impossible to give you a better solution than you already have been given.
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