Effacer les filtres
Effacer les filtres

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Optimize the FOR loop

1 vue (au cours des 30 derniers jours)
Sam
Sam le 28 Juin 2012
Clôturé : MATLAB Answer Bot le 20 Août 2021
I am learning to optimize the following nested FOR loop, any comments are appreciated,
x=[537 558 583 606 631 655 666 700 722 799 823 847];
y=[48 216 384 552 720 888 1056];
z = zeros(1,numel(x));
for j = 1:numel(x)
for i = 1:numel(y)
if(x(j) <= y(i) )
z(j) = i;
break;
end
end
end
%ans
%z = [4 5 5 5 5 5 5 5 6 6 6 6]
  2 commentaires
Ryan
Ryan le 28 Juin 2012
Modifié(e) : Ryan le 28 Juin 2012
To clarify: For each value in 'x', you're trying to locate the position of the first value in 'y' that is greater than the value in 'x' and then return that position in a vector 'z'?
Sam
Sam le 28 Juin 2012
yes

Réponses (2)

Sean de Wolski
Sean de Wolski le 28 Juin 2012
Modifié(e) : Sean de Wolski le 28 Juin 2012
If you can guarantee that the is at least one occurence of x(:) <y (:) then this will work:
x=[537 558 583 606 631 655 666 700 722 799 823 847];
y=[48 216 384 552 720 888 1056];
[~,z] = max(bsxfun(@le,x,y'),[],1)
  4 commentaires
Tom
Tom le 28 Juin 2012
I just compared arrayfun and bsxfun, the latter is a whole order of magnitude faster.
Ryan
Ryan le 28 Juin 2012
Thank you Sean and Tom!

Tom
Tom le 28 Juin 2012
It seems what you're trying to is find the first instance of each value of x being less than each value of y. You can do this in using arrayfun:
arrayfun(@(n) find(n<y,1),x)
the first argument is an anonymous function. For each value in x, the find function is used to find the first instance of that x value being less than the y vector.
  1 commentaire
Tom
Tom le 28 Juin 2012
Seeing what Sean said, this way also only works if there is an occurrence of for all of them- if there isn't then 'UniformOutput' has to be set to false, which means the output will be a cell array.

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