another method to for loop

how i can change for loop to code that is faster because i have a large dataset

for i=1:n,
    for j=1:n,
        if ind(i) == ind(j)
            s(i,j)=abs(z(i,ind(i))-z(j,ind(j))) / (z(i,ind(i))+z(j,ind(j)))
        else
            s(i,j)=abs(z(i,ind(j))-z(j,ind(i))) / (z(i,ind(j))+z(j,ind(i)))
        end
    end
end

where n is the no of rows ind is the cluster number z is the distance from point to cluster no s is the similarity between any two point

[per isakson]

What's ind and z?

Answer 1

[Roger Stafford]

There is one obvious way to speed things up. Your test for ind(i)==ind(j) is totally unnecessary and can be eliminated, because the expressions

   abs(z(i,ind(i))-z(j,ind(j)))/(z(i,ind(i))+z(j,ind(j)))

and

   abs(z(i,ind(j))-z(j,ind(i)))/(z(i,ind(j))+z(j,ind(i)))

are identically equal when ind(i) equals ind(j). Just write

   for i=1:n
    for j=1:n
     s(i,j) = abs(z(i,ind(j))-z(j,ind(i)))/(z(i,ind(j))+z(j,ind(i)));
    end
   end

Even more important is that you should make sure the 's' matrix has been preallocated the proper amount of memory space before entering these for-loops. That make an enormous difference in speed. Doing an initial 'zeros' call with the appropriate dimensions would do the job.

I see a third way to possibly get more speed. You have symmetry in the placement of values in the 's' matrix, so you can cut the number of computations with 'z' by almost half.

   for i = 1:n
    for j = 1:i  % <-- Note the 1:i instead of 1:n
     s(i,j) = abs(z(i,ind(j))-z(j,ind(i)))/(z(i,ind(j))+z(j,ind(i)));
     s(j,i) = s(i,j);
    end
   end