[Numpy-discussion] Indexing question

[CJ Carey]

Or as a one-liner:

out[np.arange(len(x)), x] = 1

If NEP 21 is accepted (
https://numpy.org/neps/nep-0021-advanced-indexing.html) this would be even
simpler:

out.vindex[:, x] = 1

Was there ever a decision about that NEP? I didn't follow the discussion
too closely at the time.

On Thu, May 20, 2021 at 10:06 AM Neal Becker <ndbecker2 at gmail.com> wrote:

> Thanks!
>
> On Thu, May 20, 2021 at 9:53 AM Robert Kern <robert.kern at gmail.com> wrote:
> >
> > On Thu, May 20, 2021 at 9:47 AM Neal Becker <ndbecker2 at gmail.com> wrote:
> >>
> >> This seems like something that can be done with indexing, but I
> >> haven't found the solution.
> >>
> >> out is a 2D array is initialized to zeros.  x is a 1D array whose
> >> values correspond to the columns of out.  For each row in out, set
> >> out[row,x[row]] = 1.  Here is working code:
> >> def orthogonal_mod (x, nbits):
> >>     out = np.zeros ((len(x), 1<<nbits), dtype=complex)
> >>     for e in range (len (x)):
> >>         out[e,x[e]] = 1
> >>     return out
> >>
> >> Any idea to do this without an explicit python loop?
> >
> >
> >
> > i = np.arange(len(x))
> > j = x[i]
> > out[i, j] = 1
> >
> > --
> > Robert Kern
> > _______________________________________________
> > NumPy-Discussion mailing list
> > NumPy-Discussion at python.org
> > https://mail.python.org/mailman/listinfo/numpy-discussion
>
>
>
> --
> Those who don't understand recursion are doomed to repeat it
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion at python.org
> https://mail.python.org/mailman/listinfo/numpy-discussion
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <https://mail.python.org/pipermail/numpy-discussion/attachments/20210520/4ebabae6/attachment.html>