On 8/31/06, Travis Oliphant <oliphant.travis@ieee.org> 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 In [19]: N.array([]).all() Out[19]: 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? Regards, f