How can I vectorize this function with nested FOR loop?

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abtin irani
abtin irani am 2 Jul. 2019
Bearbeitet: Matt J am 5 Jul. 2019
I have two for loops.L is one matrix of random numbers between 0,1 with dimension 24*100000.I want to vectorize it but i can't. because current code is very slow and take a long time.please help me.
K=zeros(100000,1);
T=zeros(100000,1);
for i=1:100000
for j=1:100000
K(j,1)=exp(-4*norm(L(:,i)-L(:,j))^2/norm(L(:,i))^2);
end
T(i,:)=sum(K)-1;
end
  2 Kommentare
Bob Thompson
Bob Thompson am 2 Jul. 2019
(L(:,i)-L(:,j))
This is what is going to make it difficult to vectorize. I believe there is a command to calculate this for you, but I'm not sure what exactly it is. It might be easier to do some research for this specifically, rather than the vectorization.
Jan
Jan am 2 Jul. 2019
Bearbeitet: Jan am 2 Jul. 2019
Why do you want to vectorize the code? There is no general benefit in doing this. Does the current code run too slow? Then an acceleration is the way to go. Vectorizing can improve the speed, but this is not in general.
To optimize the code, we need the chance to run it. Without meaningful input arguments, this is hard. So please provide L.
norm(x)^2 calculates an expensive square root only to square the result afterwards.
Do oyu have the parallel processing toolbox? A parfor might be very useful.

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Matt J
Matt J am 2 Jul. 2019
Bearbeitet: Matt J am 2 Jul. 2019
Using mat2tiles
chunksize=10000;
Lc=mat2tiles(L.',[chunksize,24]);
normL=mat2tiles(vecnorm(L,2,1), [1, chunksize]);
N=numel(Lc);
Tc=cell(N);
for i=1:N
for j=1:N
E=pdist2(Lc{i},Lc{j})./normL{j};
Tc{i,j}=sum( exp(-4*E.^2) ,1);
end
end
T=sum( cell2mat(Tc) ,1).'-1;
  18 Kommentare
abtin irani
abtin irani am 4 Jul. 2019
so,what can i do if i don't use pdist2.is there any other solution instead of pdist2 that doesn't give this error? i want one solution for chunksize=1000000. thank you for your answers.
Matt J
Matt J am 4 Jul. 2019
Bearbeitet: Matt J am 5 Jul. 2019
Well, maybe the question to ask is, why are you making the chunksize larger, when you toldu us that the code already worked when the chunksize was just 10,000? Why change/fix something that is not broken?

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Jan
Jan am 4 Jul. 2019
n = 1000;
L = rand(24, n);
T = zeros(n, 1);
for i=1:n
K = exp(-4*sum((L(:,i) - L) .^ 2, 1) ./ sum(L(:,i).^2, 1));
T2(i) = sum(K, 2) - 1;
end
This is 100 times faster than the original version for n=1000.
With parfor instead of for a further acceleration is possible.
To my surprise the above code is faster than this, which omits the repeated squaring:
T = zeros(n,1);
L2 = L .^ 2;
for i=1:n
K = exp(-4*sum((L2(:,i) - 2*L(:,i).*L + L2), 1) ./ sum(L2(:,i), 1));
T(i) = sum(K, 2)-1;
end
  2 Kommentare
abtin irani
abtin irani am 4 Jul. 2019
excuse me i got this error. Error using .*
Matrix dimensions must agree.
Error in Untitled8 (line 33)
K=exp(-4*sum((L2(:,i) - 2*L(:,i).*L + L2), 1) ./ sum(L2(:,i), 1));
Jan
Jan am 4 Jul. 2019
You use Matlab < R2016b. Then:
K = exp(-4*sum((L2(:,i) - bsxfun(@times, 2*L(:,i), L) + L2), 1) ...
./ sum(L2(:,i), 1));
But the first version seems to be faster.

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