Vectorized for loop

2 views (last 30 days)
Barend
Barend on 17 Nov 2011
Good day
Okay, so my problem is the following. I have a massive number of x,y coordinates and need to do some calculations on them. I used for loops, but it takes forever. Below is a simple piece of code that explains what I am trying to do:
x = [ 0.2 0.3 0.2 0.1];
y = [ 0.8 0.4 0.5 0.2];
for i = 1:numel(x)
ind = 0;
for j = i+1:numel(x)-1
ind = ind +1;
d = abs(y(i)-y(j);
c(ind) = d + x(i)+x(j);
end
e = [e c];
end
The idea is to end up with a vector e (that is a function of x and y) that I can pass to another function.
So, my idea is to avoid a double for loop by having long (very long) vectors (sacrifice memory for runtime): thus to calculated d
vectori = [0.8 0.8 0.8 0.4 0.4 0.5];
vectorj = [0.4 0.5 0.2 0.5 0.2 0.2];
d = abs(vectori - vectorj);
The fact that j runs from i+1 to numel(x)-1 makes this also more challenging for me. How can I build vectori and vectorj for this application?
Regards, Barend.

Answers (3)

Robert Cumming
Robert Cumming on 17 Nov 2011
before you spend a lot of time recoding - check where the code is slow. to do this use the profiler
profile on
% run you code
profile viewer
Check that you are pre-allocating your arrays (c and e for example are both growing in your loop above - this is very slow)
  2 Comments
Barend
Barend on 17 Nov 2011
Hi Robert
This is a very simple example - the calculations in the second for loop is actually much more complex (i am calling another function every time in the second loop) So the majority of my time is spent on the iteration of the code in the for loop. If I have the vectors ready in the desired format, I can just call the function once.
Robert Cumming
Robert Cumming on 17 Nov 2011
preallocation is still valid
check where the time is with the profiler.

Sign in to comment.


Jan
Jan on 17 Nov 2011
As Robert has written already: pre-allocation is crucial. In addition all repeated calculatíon should be moved out of the inner loop.
x = [ 0.2 0.3 0.2 0.1];
y = [ 0.8 0.4 0.5 0.2];
nx = numel(x); % Once only!
c = zeros(1, nx); % Once only!
e = NaN(nx, nx);
for i = 1:nx
ind = 0;
xi = x(i); % Once only!
yi = y(i);
for j = i+1:nx-1
ind = ind + 1;
d = abs(yi - y(j));
c(ind) = d + xi + x(j);
end
e(1:ind, i) = c(1:ind); % Do not let e grow
end
e = e(isfinite(e)); % Remove the NaNs finally
Unfortunately you've posted a simplified computation only. In this example the inner loop can be vectorized easily:
c = abs(yi - y(i+1:nx-1)) + xi + x(i+1:nx-1);
I cannot guess if this will work in the original code also.

Andrei Bobrov
Andrei Bobrov on 17 Nov 2011
e = nonzeros(tril(bsxfun(@plus,x1,x1')+abs(bsxfun(@minus,y1,y1')),-1));
variant with loop
nx = numel(x);
ny = nx - 1;
e = nan(nnz(tril(ones(nx-1),-1)),1);
id = ny;
k = 0;
for i1 = 1:ny
j1 = i1+1:ny;
id = id - 1;
k = k(end) + (1:id);
e(k) = abs(y(i1)-y(j1)) + x(i1)+x(j1);
end
  2 Comments
Jan
Jan on 17 Nov 2011
Creating the explicite index vector is slower, because Matlab checks the boundary for each element:
k = k(end) + (1:id); e(k) = ...
Faster:
e(k+1:k+id) = ...; k = k+id;
Andrei Bobrov
Andrei Bobrov on 17 Nov 2011
Thank you, Jan, for helpful advice!

Sign in to comment.

Categories

Find more on Programming in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!