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Optimize the FOR loop

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Sam
Sam 2012 年 6 月 28 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
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 件のコメント
Ryan
Ryan 2012 年 6 月 28 日
編集済み: Ryan 2012 年 6 月 28 日
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 2012 年 6 月 28 日
yes

回答 (2 件)

Sean de Wolski
Sean de Wolski 2012 年 6 月 28 日
編集済み: Sean de Wolski 2012 年 6 月 28 日
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 件のコメント
Tom
Tom 2012 年 6 月 28 日
I just compared arrayfun and bsxfun, the latter is a whole order of magnitude faster.
Ryan
Ryan 2012 年 6 月 28 日
Thank you Sean and Tom!

Tom
Tom 2012 年 6 月 28 日
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 件のコメント
Tom
Tom 2012 年 6 月 28 日
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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