On 8/31/06, Fernando Perez
On 8/31/06, Travis Oliphant
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