vectorize a for loop

Hello Community,

i need ur help, to speed up a routine.

i have n points in space and i need the distances between all points.

Here is the primitve script: ( n is normally some in the range of 1e5...)

n = 8;
xc = rand(n,1);
yc = rand(n,1);
r = zeros(n);
for i = 1:n
    for j=1:n
        if i~=j
        r(i,j) = sqrt((xc(i)-xc(j))^2 + (yc(i)-yc(j))^2);
        end
    end
end

I know that the matrix r is symmetic so i need only to compute half of the elements. (This speed up to 50%)

n = 8;
xc = rand(n,1);
yc = rand(n,1);
r = zeros(n);
for i = 1:n
    for j=1:n
        if and(i~=j,i<j)
        r(i,j) = sqrt((xc(i)-xc(j))^2 + (yc(i)-yc(j))^2);
        end
    end
end
toc
r = (r+r');

But it is possible to vectorize the whole routine?

Maybe with permute and a adjoint matrix which could be vectorised A(:) = ....

Thank you in Advance!

[Matt J]

( n is normally some in the range of 1e5...)

That sounds like a non-starter. The result would consume 37 GB in single floats. Even if you had this much free RAM, I suspect computing the matrix is not the most efficient approach for your application.It just doesn't sound like a reasonable thing to have to do.

[Marko]

Hello Matt,

sorry there is a mistake n should be in the order of 1e3!

I am writing a panel method, and estimatet the max. number of n-panels for an multi element airfoil.

So the Ram should not be the limiting factor.

Answer 1

[Matt J]

This should be pretty well optimized already,

r=pdist([xc,yc]);

[Marko]

Hell Matt,

i testet your suggestion. Because I need the r(i,j) as a matrix:

r=squareform(pdist([xc,yc]));

[Marko]

here the fastest way up to n=10000

n= 10000
xc=rand(n,1);
yc=rand(n,1);
r=(xc.*ones(1,n)-ones(n,1).*xc').^2 + (yc.*ones(1,n)-ones(n,1).*yc').^2;

[Matt J]

Faster:

r=(xc-xc').^2 + (yc-yc').^2;

[Marko]

perfect, thank you Matt!