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Vectorized for loop

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Barend
Barend 2011-11-17
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.

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回答（3 个）

Robert Cumming
Robert Cumming 2011-11-17
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 个评论
Barend
Barend 2011-11-17
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 2011-11-17
preallocation is still valid
check where the time is with the profiler.

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Jan
Jan 2011-11-17
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 2011-11-17
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 个评论
Jan
Jan 2011-11-17
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 2011-11-17
Thank you, Jan, for helpful advice!

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