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?

Accepted Answer

Matt J
Matt J on 30 Apr 2013
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));

More Answers (5)

Roger Stafford
Roger Stafford on 30 Apr 2013
Edited: Matt J on 30 Apr 2013
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.

Matt J
Matt J on 30 Apr 2013
Edited: Matt J on 30 Apr 2013
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).',
  6 Comments
Matt J
Matt J on 1 May 2013
Edited: Matt J on 2 May 2013
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.

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Sean de Wolski
Sean de Wolski on 30 Apr 2013
Edited: Sean de Wolski on 30 Apr 2013
doc meshgrid
doc ndgrid %?
:)
And of course, depending on your application, two nested for-loops or bsxfun() might be better.
  2 Comments
Sean de Wolski
Sean de Wolski on 30 Apr 2013
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.

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Matt Kindig
Matt Kindig on 30 Apr 2013
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.

Matt J
Matt J on 1 May 2013
Edited: Matt J on 1 May 2013
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{:}],
  2 Comments
Matt J
Matt J on 1 May 2013
Edited: Matt J on 2 May 2013
For n=1000 I get this,
Original Approach:
Elapsed time is 0.093130 seconds.
Cell-Based Approach
Elapsed time is 0.027393 seconds.
I think the vectorization inherent in
m{i}=-L:L;
l{i}=m{i};
l{i}(:)=L;
trumps the overhead from the cell conversion.

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