Azzera filtri
Azzera filtri

Vectorize a double loop

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AC
AC il 10 Mag 2021
Commentato: Walter Roberson il 13 Mag 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 Commenti
Walter Roberson
Walter Roberson il 10 Mag 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 il 13 Mag 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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Risposte (1)

Bob Thompson
Bob Thompson il 10 Mag 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 Commento
AC
AC il 10 Mag 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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