How to avoid for loops when generating index arrays?
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I often find myself coding nested for loops to generate vectors of integer indices. For example:
n = 4;
i = 1;
for L = 0:n
for M = -L:L
l(i) = L;
m(i) = M;
i = i+1;
end
end
All I need are the vectors "l" and "m". I can preallocate to save some speed, but my real problem is having to use the for loops as sometimes the index vectors I need to create have many more nested for loops whose (note that the inner loop index depends on the outer loop index).
Is there a simple way to avoid using loops to generate index vectors like these?
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Matt J
2013년 4월 30일
Here's another method, less memory consuming than NDGRID
mm=sparse(-n:n);
ll=sparse(0:n);
map=bsxfun(@le,abs(mm.'),ll);
idx=nonzeros(bsxfun(@times,map,1:length(ll) ));
l=full(ll(idx));
idx=nonzeros(bsxfun(@times,map,(1:length(mm)).')) ;
m=full(mm(idx));
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추가 답변 (5개)
Roger Stafford
2013년 4월 30일
편집: Matt J
2013년 4월 30일
For your particular problem you can do this:
M = (0:n*(n+2))';
L = floor(sqrt(M));
M = M-L.*(L+1);
(I've used uppercase letters, 'L' and 'M', in place of your lowercase 'l' and 'm'.)
As with Matt Kindig, I am not sure this will be any faster than your for-loops. Time it with a large value for n and see.
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Matt J
2013년 4월 30일
편집: Matt J
2013년 4월 30일
Here's a way to do it using NDGRID. It's not apriori obvious whether for loops would or would not be faster. It depends what you plan to reuse.
[mg,lg]=ndgrid(-n:n,0:n);
idx=abs(mg)<=lg;
l=lg(idx).',
m=mg(idx).',
댓글 수: 6
Matt J
2013년 5월 1일
편집: Matt J
2013년 5월 2일
I'm starting to think Sean's advice about sticking with for-loops is the best one. There can definitely be ways to cut down on the loop nesting (see my newest Answer based on cell arrays), but the required form would depend on the body of the original set of for-loops.
Sean de Wolski
2013년 4월 30일
편집: Sean de Wolski
2013년 4월 30일
doc meshgrid
doc ndgrid %?
:)
And of course, depending on your application, two nested for-loops or bsxfun() might be better.
댓글 수: 2
Sean de Wolski
2013년 4월 30일
Just use the for-loops, they'll be the fastest by far. If you want to disguise it, write a function that takes L and M and returns l and m.
cellfun and arrayfun are slow and converting between cells and numeric types is slow. The above with preallocation will be pretty quick.
Matt Kindig
2013년 4월 30일
It's kind of hack-y, but it gives the same output as your original posting:
n=4;
l = cell2mat(arrayfun(@(x) x*ones(1,2*x+1), 0:n, 'uni', false));
m= cell2mat( arrayfun(@(x) (-x:1:x), 0:n, 'uni', false));
Keep in mind that this may very well be slower than for-loops--I haven't done any timing comparisons.
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Matt J
2013년 5월 1일
편집: Matt J
2013년 5월 1일
Here's a way to eliminate one nested loop
l=cell(1,n+1);
m=l;
for L=0:n
i=L+1;
m{i}=-L:L;
l{i}=m{i};
l{i}(:)=L;
end
l=[l{:}],
m=[m{:}],
댓글 수: 2
Sean de Wolski
2013년 5월 1일
I'd be surprised if this is faster due to the cell array conversions. I guess one of us will have to run a timing test.
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