[Numpy-discussion] How to compare an array of arrays elementwise to None in Numpy 1.13 (was easy before)?

[Sebastian Berg]

On Mon, 2017-07-17 at 09:13 +0000, Martin.Gfeller at swisscom.com wrote:
> Dear all
> 
> I have object array of arrays, which I compare element-wise to None
> in various places:
> 
> > > > a =
> > > > numpy.array([numpy.arange(5),None,numpy.nan,numpy.arange(6),Non
> > > > e],dtype=numpy.object)
> > > > a
> 
> array([array([0, 1, 2, 3, 4]), None, nan, array([0, 1, 2, 3, 4, 5]),
> None], dtype=object)
> > > > numpy.equal(a,None)
> 
> FutureWarning: comparison to `None` will result in an elementwise
> object comparison in the future.
> 
> 
> So far, I always ignored the warning, for lack of an idea how to
> resolve it. 
> 
> Now, with Numpy 1.13, I have to resolve the issue, because it fails
> with: 
> 
> ValueError: The truth value of an array with more than one element is
> ambiguous. Use a.any() or a.all() 
> 
> It seem that the numpy.equal is applied to each inner array,
> returning a Boolean array for each element, which cannot be coerced
> to a single Boolean.
> 
> The expression 
> 
> > > > numpy.vectorize(operator.is_)(a,None)
> 
> gives the desired result, but feels a bit clumsy. 
> 

Yes, I guess ones bug is someone elses feature :(, if it is very bad,
we could delay the deprecation probably. For a solutions, maybe we
could add a ufunc  for elementwise `is` on object arrays (dunno about
the name, maybe `object_identity`.

Just some quick thoughts.

- Sebastian


> Is there a cleaner, efficient way to do an element-wise (but shallow)
> comparison? 
> 
> Thank you and best regards,
> Martin Gfeller, Swisscom
> 
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