is it possible to use "find" to process every element of an array without loop
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Hi guys,
There is an array, saying a1, a2, ..., an.
For every element in the array, I want to do something like find(X<ai), where "X" is another array.
Is it possible to do it without loop and "cellfun"?
It seems cellfun is very slow.
Or maybe another way even not using find?
Thanks,
Zhong
댓글 수: 1
Daniel Shub
2012년 5월 11일
So for every element of array a you want a list of the elements of array X that are less than the current element of a and you want to do this without a loop? How big are your a and X? I would bet a loop working on a sorted X and a would be super fast.
답변 (3개)
Andrei Bobrov
2012년 5월 11일
eg:
a = randi(9,7,1);
X = randi(9,12,1);
solution:
[I,J] = find(bsxfun(@lt,X,a.'));
out = accumarray(J,I,[],@(x){x});
댓글 수: 2
Jonathan Sullivan
2012년 5월 11일
I must say, that is a very slick solution!
I think you might have missed a close parenthesis at the end of your last line. It should read:
out = accumarray(J,I,[],@(x){x});
Daniel Shub
2012년 5월 11일
I know it has a loop, but what about something like this:
a = randperm(1e5);
X = 2e3*randn(1, 1e5);
[b, ai] = sort(a);
[Y, Xi] = sort(X);
z = cell(size(a));
for iz = 1:length(z)
ii = find(Y<b(iz));
z{ai(iz)} = Xi(ii);
jj = 0;
if ~isempty(ii)
jj = max(ii);
end
Y = Y((jj+1):end);
Xi = Xi+jj;
end
Even with two "big" vectors, the code completes pretty fast. Given I take the time to sort the data, I bet there is a way to optimize the find and get even faster performance.
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Honglei Chen
2012년 5월 11일
I don't know what your intention is, but for an array, you want to use arrayfun
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