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 Comment
Matt Kindig
on 18 Oct 2012
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.
Accepted Answer
Sean de Wolski
on 18 Oct 2012
Edited: Sean de Wolski
on 18 Oct 2012
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.
More Answers (1)
Matt J
on 18 Oct 2012
[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 Comments
Sean de Wolski
on 18 Oct 2012
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
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