numpy indexing: shouldn't trailing Ellipsis be

Question

While trying to properly understand numpy indexing rules I stumbled across the following. I used to think that a trailing Ellipsis in an index does nothing. Trivial isn't it? Except, it's not actually true:

Python 3.5.2 (default, Nov 11 2016, 04:18:53) 
[GCC 4.8.5] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy as np
>>> 
>>> D2 = np.arange(4).reshape((2, 2))
>>>
>>> D2[[1, 0]].shape; D2[[1, 0], ...].shape
(2, 2)
(2, 2)
>>> D2[:, [1, 0]].shape; D2[:, [1, 0], ...].shape
(2, 2)
(2, 2)
>>> # so far so expected; now
... 
>>> D2[[[1, 0]]].shape; D2[[[1, 0]], ...].shape
(2, 2)
(1, 2, 2)
>>> # ouch!
...
>>> D2[:, [[1, 0]]].shape; D2[:, [[1, 0]], ...].shape
(2, 1, 2)
(2, 1, 2)

Now could someone in the know advise me as to whether this is a bug or a feature? And if the latter, what's the rationale?

Thanks in advance, Paul

Answer 1

Evidently there's some ambiguity in the interpretation of the [[1, 0]] index. Possibly the same thing discussed here:

Advanced slicing when passed list instead of tuple in numpy

I'll try a different array, to see if it makes things any clear

In [312]: D2=np.array([[0,0],[1,1],[2,2]])
In [313]: D2
Out[313]: 
array([[0, 0],
       [1, 1],
       [2, 2]])

In [316]: D2[[[1,0,0]]]
Out[316]: 
array([[1, 1],
       [0, 0],
       [0, 0]])
In [317]: _.shape
Out[317]: (3, 2)

Use of : or ... or making the index list an array, all treat it as a (1,3) index, and expand the dimensions of the result accordingly

In [318]: D2[[[1,0,0]],:]
Out[318]: 
array([[[1, 1],
        [0, 0],
        [0, 0]]])
In [319]: _.shape
Out[319]: (1, 3, 2)
In [320]: D2[np.array([[1,0,0]])]
Out[320]: 
array([[[1, 1],
        [0, 0],
        [0, 0]]])
In [321]: _.shape
Out[321]: (1, 3, 2)

Note that if I apply transpose to the indexing array I get a (3,1,2) result

In [323]: D2[np.array([[1,0,0]]).T,:]
...
In [324]: _.shape
Out[324]: (3, 1, 2)

Without : or ..., it appears to strip off one layer of [] before applying it to the 1st axis:

In [330]: D2[[1,0,0]].shape
Out[330]: (3, 2)
In [331]: D2[[[1,0,0]]].shape
Out[331]: (3, 2)
In [333]: D2[[[[1,0,0]]]].shape
Out[333]: (1, 3, 2)
In [334]: D2[[[[[1,0,0]]]]].shape
Out[334]: (1, 1, 3, 2)
In [335]: D2[np.array([[[[1,0,0]]]])].shape
Out[335]: (1, 1, 1, 3, 2)

I think there's a backward compatibility issue here. We know that the tuple layer is 'redundant': D2[(1,2)] is the same as D2[1,2]. But for compatibility for early versions of numpy (numeric) that first [] layer may be treated in the same way.

In that November question, I noted:

So at a top level a list and tuple are treated the same - if the list can't interpreted as an advanced indexing list.

The addition of a ... is another way of separating the D2[[[0,1]]] from D2[([0,1],)].

From @eric/s pull request seburg explains

 The tuple normalization is a rather small thing (it basically checks for a non-array sequence of length <= np.MAXDIMS, and if it contains another sequence, slice or None consider it a tuple).

[[1,2]] is a 1 element list with a list, so it is considered a tuple, i.e. ([1,2],). [[1,2]],... is a tuple already.