Avoid for following loops
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Hello everyone,
Can anyone show me how I can avoid following for loops. Thanks!
N1=10;
N2=10;
N3=10;
l1 = 100;
l2 = 100;
t=1;
n1 = [1:10];
n2 = [1:10];
x3 = rand(N3,1);
for ik1=1:N1
for ik2=1:N2
for ix3=1:N3
k1= 2*pi*n1(ik1)/l1;
k2= 2*pi*n2(ik2)/l2;
s = sqrt(k1^2+k2^2);
if s~=0
R11(ik1,ik2,ix3)=1/(exp(s*t)-exp(-s*t))*k1*exp(s*x3(ix3));
end
end
end
end
1 个评论
Matt Kindig
2012-10-18
Perhaps, but it could get a little complicated-- you will need to use meshgrid() to define k1 and k2 over the full n1xn2 (10x10) grid, and defining the third dimension in R11 could get a bit complicated using purely vector operations.
Why are you doing this? Are you doing this for speed reasons? Oftentimes, for loops are actually faster than vectorized loops. Your best bet to improve speed is to preallocate R11 prior to entering the loop. That is, add this line to right before your for loop:
R11 = NaN(N1, N2, N3); %this preallocates memory for R11
Now each iteration of the loop just fills in data in the already existing R11 variable.
采纳的回答
Sean de Wolski
2012-10-18
编辑：Sean de Wolski
2012-10-18
How about this beauty?
k1 = (2*pi.*n1(1:N1).')./l1;
S=bsxfun(@hypot,k1,2*pi*n2(1:N2)./l2);
P=bsxfun(@(s,ix3)(1./(exp(s.*t)-exp(-s.*t))).*exp(s.*ix3),S,x3(reshape(1:N3,1,1,N3)));
R22 = bsxfun(@times,k1,P);
Check to make sure it's close:
norm(R22(:)-R11(:))
The difference arises because I used hypot in places of sqrt(x^2+y^2), it's a more numerically stable implementation of this.
Also note, that just redoing your for loops with some simple preallocation and calculation rearranging would help a lot.
preallocate R11 before the loop:
R11 = zeros(N1,N2,N3);
And move things that don't change into their respective loops
for ix1 = etc
k1 = etc.
for ix2 = etc.
k2 = etc.
Since k1 and k2 are independent of the inner loops.
更多回答（1 个）
Matt J
2012-10-18
[ik1,ik2,ix3]=ndgrid(1:N1,1:N2,1:N3);
k1= 2*pi*n1(ik1)/l1;
k2= 2*pi*n2(ik2)/l2;
s = sqrt(k1.^2+k2.^2);
R11=1./(exp(s.*t)-exp(-s.*t)).*k1.*exp(s.*x3(ix3));
R11(isnan(R11))=0;
6 个评论
hung lequang
2012-10-18
Sean de Wolski
2012-10-18
Matt's solution is certainly easier to understand. However, it will be slower, especially when N1,N2,N3 get large. In fact I would guess that being smart about the for-loops will beat an ndgrid solution in timing no matter what the size of N1,N2,N3.
Matt J
2012-10-18
You might want to tic...toc both approaches. Often, bsxfun leads to faster code, so if what you're doing is speed critical, Sean's method might be better for you.
I do, however, think it is worthwhile understanding ndgrid and what it let's you do. BSXFUN is not wieldy for all situations, whereas ndgrid is a rather flexible/general way of un-looping operations on a grid.
Matt's solution is certainly easier to understand. However, it will be slower, especially when N1,N2,N3 get large
As I said, that's probably true unless possibly, the ngrid output will be recycled later for further operations. Once you've paid the overhead of ndgrid execution time, further operations tend to be fast element-wise operations on the grid variables.
Also, you have not one but several calls to BSXFUN. Not sure how that adds up...
Sean de Wolski
2012-10-18
Well, I stand corrected, the ndgrid solution is still faster up to 150. At this point a smart for-loop wins:
tic
tau = 2*pi;
R11 = zeros(N1,N2,N3);
for ik1=1:N1
k1= tau*n1(ik1)/l1;
for ik2=1:N2
k2 = tau*n2(ik2)/l2;
s = sqrt(k1^2+k2^2);
st = s*t;
R11(ik1,ik2,:)=1/(exp(st)-exp(-st))*k1*exp(s*x3(1:N3));
end
end
hung lequang
2012-10-19
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2012-10-18
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