Borrar filtros
Borrar filtros

Info

La pregunta está cerrada. Vuélvala a abrir para editarla o responderla.

Optimize the FOR loop

1 visualización (últimos 30 días)
Sam
Sam el 28 de Jun. de 2012
Cerrada: MATLAB Answer Bot el 20 de Ag. de 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 comentarios
Ryan
Ryan el 28 de Jun. de 2012
Editada: Ryan el 28 de Jun. de 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 el 28 de Jun. de 2012
yes

Respuestas (2)

Sean de Wolski
Sean de Wolski el 28 de Jun. de 2012
Editada: Sean de Wolski el 28 de Jun. de 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 comentarios
Tom
Tom el 28 de Jun. de 2012
I just compared arrayfun and bsxfun, the latter is a whole order of magnitude faster.
Ryan
Ryan el 28 de Jun. de 2012
Thank you Sean and Tom!

Tom
Tom el 28 de Jun. de 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 comentario
Tom
Tom el 28 de Jun. de 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.

La pregunta está cerrada.

Community Treasure Hunt

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

Start Hunting!

Translated by