Re: [Numpy-discussion] possible bug with numpy.object_

[Charles R H. <cha...@gm...>]

On 8/31/06, Fernando Perez <fpe...@gm...> wrote:
>
> On 8/31/06, Travis Oliphant <oli...@ie...> wrote:
>
> > What about
> >
> > N.array(3).size
> >
> > N.array([3]).size
> >
> > N.array([3,3]).size
> >
> > Essentially, the [] is being treated as an object when you explicitly
> > ask for an object array in exactly the same way as 3 is being treated as
> > a number in the default case.  It's just that '[' ']' is "also" being
> > used as the dimension delimiter and thus the confusion.
> >
> > It is consistent.   It's a corner case, and I have no problem fixing the
> > special-case code running when dtype=object so that array([],
> > dtype=object) returns an empty array, if that is the consensus.
>
> I wasn't really complaining: these are corner cases I've never seen in
> real use, so I'm not really sure how critical it is to worry about
> them.  Though I could see code which does automatic size/shape checks
> tripping on some of them.  The shape tuples shed a bit of light on
> what's going on for the surprised (like myself):
>
> In [8]: N.array(3).shape
> Out[8]: ()
>
> In [9]: N.array([3]).shape
> Out[9]: (1,)
>
> In [10]: N.array([3,3]).shape
> Out[10]: (2,)
>
> In [11]: N.array([]).shape
> Out[11]: (0,)
>
> In [12]: N.array([[]]).shape
> Out[12]: (1, 0)
>
> In [13]: N.array([[],[]]).shape
> Out[13]: (2, 0)
>
>
> I won't really vote for any changes one way or another, as far as I'm
> concerned it's one of those 'learn the library' things.  I do realize
> that the near-ambiguity between '[]' as an empty object and '[]' as
> the syntactic delimiter for a container makes this case a bit of a
> gotcha.
>
> I guess my only remaining question is: what is the difference between
> outputs #8 and #11 above?  Is an empty shape tuple == array scalar,
> while a (0,) shape indicates a one-dimensional array with no elements?
> If this interpretation is correct, what is the usage of the latter
> kind of object, given how it can't even be indexed?
>
> In [15]: N.array([])[0]
>
> ---------------------------------------------------------------------------
> exceptions.IndexError                                Traceback (most
> recent call last)
>
> /home/fperez/research/code/mjmdim/pycode/<ipython console>
>
> IndexError: index out of bounds
>
>
> And is this really expected?
>
> In [18]: N.array([]).any()
> Out[18]: False


This could be interpreted as : exists x, x element of array, s.t. x is true.

In [19]: N.array([]).all()
> Out[19]: True


Seems right: for all x, x element of array, x is true.

It's a bit funny to have an array for which 'no elements are true'
> (any==false), yet 'all are true' (all==true), isn't it?


Fun with empty sets! The question is, is a zero dimensional array an empty
container or does it contain its value. The numpy choice of treating  zero
dimensional arrays as both empty containers and scalar values makes the
determination a bit ambiguous although it is consistent with the indexing
convention.

Chuck