Vectorising cell array operations

1 vue (au cours des 30 derniers jours)
Simon
Simon le 25 Juil 2013
I have currently a series of polygons in my code, stored as m*2 arrays, where each 1*2 section is a point. these polygons are stored in a cell array called polys.
I need to flip all of these vertically, so the code i am currently using is
for i = 1:numel(polys)
polys{i}(:, 1) = m-polys{i}(:, 1)
end
Where m is the height of the region (and the y coords are stored in the first position (ie a pioint is (y, x), not (x, y))
I want to execute this whole loop in one line (probably using cellfun?), but im not sure how to do it??
  2 commentaires
Daniel Shub
Daniel Shub le 25 Juil 2013
Why do you want to do that? The for loop seems nice and clean and self documenting to me.
Simon
Simon le 26 Juil 2013
Various reasons that aren't worth going into. One if which being I'm doing everything I possibly can to remove all loops, for speed, but there are others

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Daniel Shub
Daniel Shub le 26 Juil 2013
Modifié(e) : Daniel Shub le 26 Juil 2013
Blindly removing loops for speed is an antiquated view of MATLAB based on the performance of versions over 10 years old prior to the introduction of the JIT accelerator. To fully optimize your code you will want to make use of the tools like the profile (but be aware that it disables the JIT accelerator).
Given all you want to do is remove the loop you could look into either doing the task recursively or using CELLFUN. Neither are likely to give you a substantial speed up (and are likely to actually slow things down) and will be less readable than your simple loop.
For example, with cellfun
polys{1} = randn(5);
polys{2} = randn(6);
polys{3} = randn(7);
m = 10;
polys = cellfun(@(x)[m-x(:, 1), x(:, 2:end)], polys, 'UniformOutput', false);
  2 commentaires
Simon
Simon le 26 Juil 2013
I did mention that I wanted to use cellfun, I can't get it to work... I had an anonymous function declared but it wasn't working at all
Daniel Shub
Daniel Shub le 26 Juil 2013
See my edit, but again I see no advantage to doing this. It is unlikely to be faster.

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