Re: sort indexing

[Daniel J Sebald]

On 02/15/2016 06:43 PM, Nicholas Jankowski wrote:
> On Feb 15, 2016 4:10 PM, "Daniel J Sebald" <address@hidden
> <mailto:address@hidden>> wrote:
> >  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
>
> Anyone checked Matlab compatibility yet?

If there is no such thing, we could add a mode "scalar" to both sort() 
and sortrows(), e.g., rather than

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] = sort(a)
i =

   1   1   1   1   1
   2   2   2   2   2
   3   3   3   3   3
   4   4   4   4   4
   5   5   5   5   5

the user could use

octave:2> [~,i] = sort(a, "scalar")
i =

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

in which case a(i) would then equal b.

Dan