[Numpy-discussion] how to use argsort result?
Pau Gargallo
pau.gargallo at gmail.com
Thu Jul 13 07:09:07 CDT 2006
On 7/13/06, Travis Oliphant <oliphant.travis at ieee.org> wrote:
> Pau Gargallo wrote:
> > On 7/12/06, Victoria G. Laidler <laidler at stsci.edu> wrote:
> >
> >> Hi,
> >>
> >> Pardon me if I'm reprising an earlier discussion, as I'm new to the list.
> >>
> >> But is there a reason that this obscure syntax
> >>
> >> A[arange(2)[:,newaxis],indexes]
> >>
> >> A[arange(A.shape[0])[:,newaxis],indexes]
> >>
> >> is preferable to the intuitively reasonable thing that the Original
> >> Poster did?
> >>
> >> A[indexes]
> >>
> >>
> >
> > i don't think so.
> > The obscure syntax is just a way you can solve the problem with the
> > current state of NumPy. Of course, a more clearer syntax would be
> > better, but for this, something in NumPy should be changed.
> >
> > This other syntax is longer but clearer:
> > ind = indices(A.shape)
> > ind[ax] = A.argsort(axis=ax)
> > A[ind]
> >
> > Which brings me to the question:
> >
> > Would it be reasonable if argsort returned the complete tuple of
> > indices, so that
> > A[A.argsort(ax)] would work ?
> >
> >
> I think this is reasonable. We would need a way for the argsort()
> function to work as it does now. I'm not sure if anybody actually uses
> the multidimensional behavior of argsort now, but it's been in Numeric
> for a long time.
>
actually I use the multidimensional behavior of argmin and argmax in
its current state, and found it useful as it is, even if
A[A.argmax(ax)] doesn't work.
May be a keyword could be added, so that A.argxxx( axis=ax,
indices=True ) returns the tuple of indices. (The keyword name and
default should be discussed)
I don't know if that's *the* way, but ...
pau
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