How can I vectorize this function with nested FOR loop?
4 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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 commentaires
Bob Thompson
le 2 Juil 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
le 2 Juil 2019
Modifié(e) : Jan
le 2 Juil 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.
Réponse acceptée
Matt J
le 2 Juil 2019
Modifié(e) : Matt J
le 2 Juil 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 commentaires
Plus de réponses (1)
Jan
le 4 Juil 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 commentaires
Jan
le 4 Juil 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.
Voir également
Catégories
En savoir plus sur Parallel Computing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!