Re: sort indexing

[Daniel J Sebald]

On 02/15/2016 02:39 PM, Doug Stewart wrote:
> I have a question about indexing.
>
> a=randi(9,5)
> [b i]=sort(a)
> c=a(i)
>
> I would think that c should be the same as b, but it is not.
>
> The index array i has all the correct information in it as can be seen with
>
> for k=1:columns(a)
> w(:,k)=a(i(:,k),k);
> endfor
> now b-w is equal to 0
>
> Is there  some technical reason that if you try and use the index array
> i on an matrix
> of the same dimensions, that it can't work?
> I would think that it should apply each col of the i to a  in a(i)
> as I did in the loop.
>
> I know it can be vectorized:
>
>   a(sub2ind (size(a), i, repmat(1:4, rows(a), 1)))
>   a(i+(0:columns(a)-1)*rows(a))
>
> but I just think that octave should be smart enough to just do a(i)
>
> Doug

Doug,

I looks like a() is being vectorized when accessed as a(i):

octave:1> a = [0:4] + 5*[0:4]'
a =

    0    1    2    3    4
    5    6    7    8    9
   10   11   12   13   14
   15   16   17   18   19
   20   21   22   23   24

octave:2> i = a + 1
i =

    1    2    3    4    5
    6    7    8    9   10
   11   12   13   14   15
   16   17   18   19   20
   21   22   23   24   25

octave:3> a(i)
ans =

    0    5   10   15   20
    1    6   11   16   21
    2    7   12   17   22
    3    8   13   18   23
    4    9   14   19   24

The Octave documentation states that indexing multidimensional arrays 
when indexed by a scalar are treated in column-major order.  Hence in 
the above example because of the way I defined i, a(i) is transposed.

https://www.gnu.org/software/octave/doc/interpreter/Index-Expressions.html

Dan