[Numpy-discussion] Consider improving numpy.outer's behavior with zero-dimensional vectors

[Neil Girdhar]

That sounds good to me.

I can always put np.outer = np.multiply.outer at the start of my code to
get what I want.  Or could that break things?

On Thu, Apr 16, 2015 at 6:28 PM, Matthew Brett <matthew.brett at gmail.com>
wrote:

> Hi,
>
> On Thu, Apr 16, 2015 at 3:19 PM, Neil Girdhar <mistersheik at gmail.com>
> wrote:
> > Actually, looking at the docs, numpy.outer is *only* defined for 1-d
> > vectors.  Should anyone who used it with multi-dimensional arrays have an
> > expectation that it will keep working in the same way?
> >
> > On Thu, Apr 16, 2015 at 10:53 AM, Neil Girdhar <mistersheik at gmail.com>
> > wrote:
> >>
> >> Would it be possible to deprecate np.outer's usage on non
> one-dimensional
> >> vectors for a few versions, and then reintroduce it with definition
> np.outer
> >> == np.multiply.outer?
>
> I think the general idea is that
>
> a) people often miss deprecation warnings
> b) there is lots of legacy code out there, and
> c) it's very bad if legacy code silently gives different answers in
> newer numpy versions
> d) it's not so bad if newer numpy gives an intelligible error for code
> that used to work.
>
> So, how about a slight modification of your proposal?
>
> 1) Raise deprecation warning for np.outer for non 1D arrays for a few
> versions, with depraction in favor of np.multiply.outer, then
> 2) Raise error for np.outer on non 1D arrays
>
> Best,
>
> Matthew
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