Optimize my for loop with vectors

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olabaz
olabaz 2017-7-28
编辑： Andrei Bobrov 2017-7-29
Hi, I read that vectorizing your matlab program will speed it up so I am trying to vectorize the following for loop:
mysum = 0;
for kx = -1:dk:1
kyMax = sqrt(1^2-kx^2);
for ky = 0:dk:kyMax
myeps = kx^2 + ky^2;
mysum = mysum + myeps;
end
end
I'm not sure how to do it since {kx,ky} is not a rectangular array but rather one who's ky dimension changes depending on the kx.

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2017-7-28
编辑：Andrei Bobrov 2017-7-29
kx = -1:dk:1;
[xx,yy] = ndgrid(0:dk:1,kx);
t = xx <= sqrt(1-kx.^2);
m = [xx(:),yy(:)];
a = m(t(:),:).^2;
out = sum(a(:));
or
kx = -1:dk:1;
B = arrayfun(@(x)0:dk:x,sqrt(1-kx.^2),'un',0);
n = cellfun(@numel,B);
out = sum(repelem(kx,n).^2 + [B{:}].^2);
  2 个评论
olabaz
olabaz 2017-7-28
This is what I was looking for. Thanks.
Chad Greene
Chad Greene 2017-7-28
Can you explain these steps Andrei? When I try it with any value of dk I get a dimension mismatch at the t= line.

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更多回答（2 个）

Chad Greene
Chad Greene 2017-7-28
Hi olabaz,
It looks like eps is getting overwritten each time through the loop, so you'll just end up with eps = 1. As I understand it you want eps to be a 2D matrix for all the values of kx and ky?
A minor note as an aside, eps is actually a different thing in Matlab. The way you're overwriting it works fine here, but it's probably best to avoid using that variable name.
  1 个评论
olabaz
olabaz 2017-7-28
编辑：olabaz 2017-7-28
I did not know about eps being used, I will change the name. Also, I left part of the code out that does a calculation with eps so I will change the code. But ideally, I would like to have two corresponding vectors kx,ky that satisfy that relationship. So I can just do eps(1) = kx(1)^2 + ky(1)^2, eps(2) = kx(2)^2 + ky(2)^2, etc. And then I could easily do eps = sum(eps).

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Chad Greene
Chad Greene 2017-7-28
I really appreciate your interesting in vectorizing. It will take you far in life. First things first, 1^2 will always be 1, so we can remove a calculation by rewriting
kyMax = sqrt(1^2-kx^2);
as
kyMax = sqrt(1-kx^2);
Second, it can be helpful to get in the habit of rewriting this:
for kx = -1:dk:1
...
end
as
kx = -1:dk:1;
for k = 1:length(kx)
...
end
and then inside the loop, when you want to reference the kth value of kx, use kx(k). The difference may seem subtle, abstract, or even like an unnecessary extra step. But it means we can pull
kyMax = sqrt(1-kx^2);
out of the loop. Just calculate it once for all the values of kx. That gives us this:
mysum = 0;
kx = -1:dk:1;
kyMax = sqrt(1-kx.^2);
for k = 1:length(kx)
for ky = 0:dk:kyMax(k)
myeps = kx(k)^2 + ky^2;
mysum = mysum + myeps;
end
end
Note the .^ instead of just ^ which lets us square all the values at once. We can then do the same thing with the next for statement. That becomes this:
kx = -1:dk:1;
kyMax = sqrt(1-kx.^2);
for k = 1:length(kx)
ky = 0:dk:kyMax(k);
myeps = kx(k)^2 + ky.^2;
for kk = 1:length(ky)
mysum = mysum + myeps(kk);
end
end
and now there's really no need for that inner loop because we can just use sum. Taking out the inner loop it looks like this:
mysum = 0;
kx = -1:dk:1;
kyMax = sqrt(1-kx.^2);
for k = 1:length(kx)
ky = 0:dk:kyMax(k);
myeps = kx(k)^2 + ky.^2;
mysum = mysum + sum(myeps);
end
Already we've made some great progress because we no longer have a loop within a loop. Loops within loops have a way of slowing things down tremendously because everything within the inner loop needs to be done not just N times, but N*M times. Now I have to think about how to get rid of that other loop.
  1 个评论
olabaz
olabaz 2017-7-28
Oh, that's interesting. I would've never thought about pulling the kyMax out of the loop.

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