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

on 14 Mar 2013

Friedrich (view profile)

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);
```

Friedrich

Friedrich (view profile)

on 14 Mar 2013

That is not true! BSXFUN is highly parallized under the hood. Fastest function you can get.

Jan Simon

Jan Simon (view profile)

on 14 Mar 2013

BSXFUN is not a slower version of a loop. I think ARRAYFUN could earn this description, especially when used with anonymous functions.

Matlab2010

Matlab2010 (view profile)

on 28 Mar 2013

it seems like bsxfun is what I need. thank you!

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Friedrich (view profile)

on 14 Mar 2013
Edited by Friedrich

Friedrich (view profile)

on 14 Mar 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))```

Jan Simon

Jan Simon (view profile)

on 14 Mar 2013

POWER and LOG10 are expensive functions. There must be a solution based on the integer indices also.

Jan Simon

Jan Simon (view profile)

on 14 Mar 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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