uncertain number of loop structures
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Here is my problem:
the function has uncertain number of inputs, while the number of loop structures depends on the number of inputs. For example, if there is only one input, then only one loop structure is needed, two inputs, two loop structures, ...
I know how to deal with uncertain inputs, but need help to code uncertain number of loop structures.
Leon
2 个评论
Sean de Wolski
2011-5-5
What are the loops doing? It's quite possible to avoid the loops altogether which makes the whole problem irrelevant. Paste a snippet of code and the problem for us to see!
Walter Roberson
2012-8-24
(Re-opened as the question is of sufficient general interest and the answers give enough information to be useful.)
回答(4 个)
the cyclist
2011-5-5
I agree with Sean de, that you may be able to avoid the loops with some more clever programming, so you might want to post some code.
However, once upon a time, I had to do exactly what you are asking. It can be done with the eval() function. Here is a trivial example that will give you the idea:
NDIM = 4;
forString = [];
doString = [];
endString = [];
for nd = 1:NDIM
forString = ['for i',num2str(nd),'=1:10,',forString];
doString = ['disp(i',num2str(nd),');',doString];
endString = [endString,'end;']; %#ok<AGROW>
end
loopString = [forString,doString,endString];
eval(loopString);
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Sean de Wolski
2011-5-5
Of course to avoid using evil eval, if you know what the maximum number of loops possibly necessary is, you could just have that many written out and skip all of the interior ones you don't need.
Sean de Wolski
2011-5-5
Or one could use fprintf along the same lines as The Cyclist's suggestion to generate a custom m-file for each number of dimensions applicable. Then when the user enters the number of dimensions that m-file is called.
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Liang
2011-5-5
2 个评论
Daniel Shub
2011-5-5
Are you just trying to get to the final "i"? It doesn't even look like you need while loops. You should be able to loop over (or probably even nest) a bunch of "find" commands.
Daniel Shub
2011-5-5
function looper(FcnHandle, NDIM, LoopIndices, Array)
LoopNo = length(Array)+1;
if LoopNo == NDIM
for ii = LoopIndices{LoopNo}
Array{LoopNo} = ii;
FcnHandle(Array)
end
else
for ii = LoopIndices{LoopNo}
Array{LoopNo} = ii;
looper(FcnHandle, NDIM, LoopIndices, Array)
end
end
end
looper(@(x)(disp(num2str([x{:}]))), 3, {1:3, 1:3, 1:3}, {})
1 个评论
Daniel Shub
2011-5-5
Opps, this doesn't answer the revised question. I think it does what Cyclist did without eval or knowing the maximum number of loops.
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