Using find with a vector without having to use a for loop

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Matlab2010
Matlab2010 2013-3-14
I have code which does
for t = 1: length(vector2)
idx(t) = find(vector <=vector2(t), 1, 'last');
end
Can I call find.m without using a for loop (or bsxfun.m) and if so how?
Or should I just use histc?
[~,idx]=histc(vector2, vector);
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Jan
Jan 2013-3-14
BSXFUN is not a slower version of a loop. I think ARRAYFUN could earn this description, especially when used with anonymous functions.
Matlab2010
Matlab2010 2013-3-28
Having read more about this, I accept my previous statement is wrong and I was getting bsxfun confused with arrayfun.
it seems like bsxfun is what I need. thank you!

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Friedrich
Friedrich 2013-3-14
编辑：Friedrich 2013-3-14
Hi,
use bsxfun and make sure one vector is a row vector and the other a column vector (a would be vector and b would be vector2)
a = [1 2 3 4 5 6]
b = [ 1 5 8 3 -10 2]
idx = bsxfun(@le,a',b)
Now you need to get a bit tricky I guess:
tmp = idx*(10.^[1:numel(b)]')
floor(log10(tmp))
Not sure if maybe log2 would be faster, you would need to try it:
tmp = idx*(2.^[1:numel(b)]')
floor(log2(tmp))
  2 个评论
Jan
Jan 2013-3-14
POWER and LOG10 are expensive functions. There must be a solution based on the integer indices also.
Jan
Jan 2013-3-14
Yes! When the input vector2 (as in the original question) is sorted (and find(., 1, 'last') implies that it is), the logical matrix created by bsxfun contains zeros on the left and ones on the right (or top and bottom?). Then this matrix can be reshaped to a row vector and strfind searchs for [false true]. Finally the resulting indices must be cleaned up using mod() with the number of columns.

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