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@gmail.com> wrote:
Thanks!

On Thu, May 20, 2021 at 9:53 AM Robert Kern <robert.kern@gmail.com> wrote:
>
> On Thu, May 20, 2021 at 9:47 AM Neal Becker <ndbecker2@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@python.org
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