Vectorize a double loop

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AC
AC el 10 de Mayo de 2021
Comentada: Walter Roberson el 13 de Mayo de 2021
Hi everyone
I'm tryng to vectorize the folowing piece of code:
n=30;
d=3;
a=1;
b=2;
cell=3;
for ki = 2:n-1
for kj = 2:n-1
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj-1) = 1/d^2;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj) = -4/d^2-a;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj+1) = 1/d^2;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj-n) = 1/d^2;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj+n) = 1/d^2;
C((cell-1)*n^2+(ki-1)*n+kj,1) = b;
end
end
How can I do it? I will appreciate any help! Thanks!
  3 comentarios
Walter Roberson
Walter Roberson el 10 de Mayo de 2021
(cell-1)*n^2+(ki-1)*n+kj
You are faking 4 dimensional indexing. You should switch to actual 4D indexing. reshape() before and after if you need to.
Walter Roberson
Walter Roberson el 13 de Mayo de 2021
If you feel that your post is unclear, then since you are the one who wrote it, you should clarify it.

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Respuestas (1)

Bob Thompson
Bob Thompson el 10 de Mayo de 2021
Does this work? I haven't been able to test it.
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)-1) = 1/d^2;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)) = -4/d^2-a;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)+1) = 1/d^2;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)-n) = 1/d^2;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1)j,(cell-1)*n^2+(1:n-2)*n+(2:n-1)+n) = 1/d^2;
C((cell-1)*n^2+(1:n-2)*n+(2:n-1),1) = b;
  1 comentario
AC
AC el 10 de Mayo de 2021
Thank you for your answer! I tried this, but didn't work. When using the double for loop, I get (n-2)*(n-2) combinations of ki's and kj's. But (cell-1)*n^2+(1:n-2)*n+(2:n-1) is a vector of length (n-2). I need a vector of length (n-2)*(n-2)

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