Using find with a vector without having to use a for loop
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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);
5 Kommentare
Jan
am 14 Mär. 2013
BSXFUN is not a slower version of a loop. I think ARRAYFUN could earn this description, especially when used with anonymous functions.
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Friedrich
am 14 Mär. 2013
Bearbeitet: Friedrich
am 14 Mär. 2013
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 Kommentare
Jan
am 14 Mär. 2013
POWER and LOG10 are expensive functions. There must be a solution based on the integer indices also.
Jan
am 14 Mär. 2013
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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