8000 numpy.index_exp isn't quite consistent with direct indexing in backward compatibility case · Issue #9797 · numpy/numpy · GitHub
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numpy.index_exp isn't quite consistent with direct indexing in backward compatibility case #9797

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FrozenBob opened this issue Sep 30, 2017 · 1 comment
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numpy.index_exp isn't quite consistent with direct indexing in backward compatibility case #9797

FrozenBob opened this issue Sep 30, 2017 · 1 comment
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@FrozenBob
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The numpy.s_/numpy.index_exp docs claim the following:

For any index combination, including slicing and axis insertion, a[indices] is the same as a[np.index_exp[indices]] for any array a.

However, this appears to only be completely accurate for numpy.s_, due to the weird backward compatibility logic where certain non-tuple sequences are converted to tuples. For example,

In [7]: a = np.zeros([4, 4])

In [8]: indices = [[0, 1], [2, 3]]

In [9]: a[indices].shape
Out[9]: (2,)

In [10]: a[np.s_[indices]].shape
Out[10]: (2,)

In [11]: a[np.index_exp[indices]].shape
Out[11]: (2, 2, 4)

numpy.index_exp's tuple creation doesn't reflect the backward compatibility handling. Either the docs or the behavior should be adjusted to match the other.
@eric-wieser
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eric-wieser commented Sep 30, 2017 •
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There's a PR to deprecate the weird backwards compatibility at #9686.

What's actually true is that:

  • a[np.index_exp[x, y, z]] is the same as a[x, y, z]
  • a[np.index_exp[x]] is the same as a[(x,)]
  • a[(x,)] is usually the same as a[x], unless the latter invokes the compatibility behaviour

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