Efficiency help 2

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Andy
Andy 2011-12-7
for q = 1:length(maxtab2)
for r = 1: length(maxtab2{q})
for s = 1: length(replacewith)
if abs (maxtab2{q}(r) - replacewith(s)) ==1
maxtab2{q}(r) = replacewith(s);
end
end
end
end
maxtab2 is a list of cells, each containing 2 colums of numbers(up to 20 numbers each column).
length(maxtab2)=190 length(maxtab2{q}) would be up to 20
can anyone help me make the 3 loops more effecient please? It is taking way too long right now, Thanks!
here is the data for maxtab2, its just the content of 1 cell
{[705;4000]}
replacewith is a list of over 5000 variables. the code here chooses one of these 5000 variable and replace the corresponding maxtab2 (determined by the for loops) with the chosen variable from replacewith
  2 个评论
Sven
Sven 2011-12-7
Andy, a couple of suggestions:
You've asked a few questions all about one program you have. The questions themselves all involve nested loops with various cells of arrays of numbers. The answers are generally ways to re-write your loops. However, you're only really providing enough information to look at your loop and try to re-write it. We never really see the full picture. It might very well be the case that the underlying data structure you have chosen means that these loops are required. If that's the case, maybe asking a bigger-picture question will help get a better answer. Try the following:
1. Upload a sample data set somewhere
2. Explain in words and code what you are trying to do *on the whole*, maybe even put in why.
I have a feeling from previous questions that some of your cells could be merged into matrices or something else in a way that would make your whole code faster AND your underlying data easier to manage.
Sean de Wolski
Sean de Wolski 2011-12-7
I agree with Sven. Why?
"Because the mathematician knew it was a good idea."

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Sean de Wolski
Sean de Wolski 2011-12-7
for s
if abs (maxtab2{q}(r) - replacewith(s)) ==1
maxtab2{q}(r) = replacewith(s);
end
end
I'm pretty sure this can be replaced with
maxtab{q}(r) = find(abs(maxtab{q}(r)-replacewith)==1,1,'last')
but this operation seems wierd to me. You're going to replace it with the value one less than it or one more than it, you're aware of this right?
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Andy
Andy 2011-12-7
which data do you need?
Andy
Andy 2011-12-7
i ended up using this: works well
for q=1:length(maxtab2)
for r = 1:length(maxtab2{q})
finding = abs(maxtab2{q}(r) - replacewith(1:length(replacewith)));
if ismember (1,finding)==1
maxtab2{q}(r)=replacewith(find(finding==1,1,'last'));
end
end
end

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更多回答（1 个）

Jerry
Jerry 2011-12-7
If this maxtab2 can be stored in a matrix. This whole thing can be done without iteration. But if it's a cell, I think it's pretty hard to speed it up since a lot of operations cannot be applied to cell, like "-"(minus), and "find".
  1 个评论
Andy
Andy 2011-12-7
maxtab2 is a list of cells, and each cell contains two colums, one columns is x-axis coordinate, the other is y-axis coordinate

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