is it possible to use "find" to process every element of an array without loop

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

[Daniel Shub]

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.

Answer 1

[Andrei Bobrov]

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});

[Jonathan Sullivan]

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});

[Andrei Bobrov]

Thank you, yes!

Answer 2

[Daniel Shub]

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.

Answer 3

[Honglei Chen]

I don't know what your intention is, but for an array, you want to use arrayfun