Vectorized iterative summation of matrices

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Pascal Kiefer
Pascal Kiefer on 3 Dec 2021
Commented: Matt J on 6 Dec 2021
Dear MATLAB community,
I want to iteratively sum over matrices and store my result. I defined a mask where every 1 value defines a position where I want to insert a matrix around it. If two inserted matrices overlap, I want to summarize the value. Doing this with a for loop is very easy but also very time consuming if I am dealing with large matrices. Therefore, I wanted to do this with a vectorized expression. However, it does not work as intended yet. Let me show you what I got:
% This defines the mask I am using for inserting the matrices. At each 1 I
% want to insert another matrix b
a=eye(10);
a(1,1)=0;
a(end,end)=0;
% define matrix b to be inserted
b = [1 2 3; 4 5 6; 7 8 9];
% define result matrix
c = zeros(10,10);
% now iteratively put the b matrix at each defined position by a and
% summarize overlapping parts
[ix,iy] = find(a==1);
c(ix-1:ix+1,iy-1:iy+1) = c(ix-1:ix+1,iy-1:iy+1) + b;
My problem is that MATLAB only inserts the matrix b at the very first pair of indices. For all other positions it does not work. As I said, for computational time reasons I wanted to find a vectorized expression instead of using loops. Does anyone have an idea how to do that?
Thank you in advance!

Accepted Answer

Matt J
Matt J on 3 Dec 2021
Edited: Matt J on 3 Dec 2021
c=conv2(a,rot90(b,2),'same');
  8 Comments
Matt J
Matt J on 6 Dec 2021
Maybe using single floats would help it fit.

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More Answers (2)

David Hill
David Hill on 3 Dec 2021
Edited: David Hill on 3 Dec 2021
What is wrong with using a loop? Why do you think it is going to be slower?
a=eye(10);
a(1,1)=0;
a(end,end)=0;
b = [1 2 3; 4 5 6; 7 8 9];
d = zeros(10,10);
[ix,iy] = find(a==1);
for k=1:length(ix)
c=zeros(10,10);
c(ix(k)-1:ix(k)+1,iy(k)-1:iy(k)+1)=b;
d=d+c;
end
  1 Comment
Pascal Kiefer
Pascal Kiefer on 6 Dec 2021
As far as I have understood it (correct me if I am wrong), vectorized expressions in MATLAB are computed in C++ (or C?) and therefore inherently much faster than "normal" for loops. I have noticed a tremendous difference for other computations where vectorization was easier than in this example.

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VBBV
VBBV on 3 Dec 2021
c = subs(a(ix,iy),1,b)
c = subs(a(ix-1:ix+1,iy-1:iy+1),1,b)
Try these two combinations.
  1 Comment
Pascal Kiefer
Pascal Kiefer on 6 Dec 2021
Thank you! I have tried out your combinations but they do not yield the correct result. I didn't know about subs, though, so thank you for that! Should I maybe modify some indices slightly to get the correct result?

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