BUG: DataFrame.loc with a MultiIndex does not always collapse scalar levels

[shoyer]

All of the following should be equivalent:

In [21]: df = pd.DataFrame({'a': [1], 'b': [2], 'c': [3]}).set_index(['a', 'b'])

In [22]: df.loc[1]
Out[22]:
   c
b
2  3

In [23]: df.loc[(1,), slice(None)]
Out[23]:
   c
b
2  3

In [24]: df.loc[(1, slice(None)), slice(None)]
Out[24]:
     c
a b
1 2  3

In the last output, there should no longer be a level a.

[jreback]

The prior impl made this hard to do, but was just refactored so I think it could be fixed up to give the proper semantics (e.g. scalar vs. list-like).

[jorisvandenbossche]

Just to be clear, the general rule we want is: "if the level indexer is a scalar, collapse the level, otherwise keep the level". Correct?

This should then also hold for levels that are not up front?

So this

In [42]: df.loc[(slice(None), 2), :]
Out[42]:
     c
a b
1 2  3

would give

So

In [42]: df.loc[(slice(None), 2), :]
Out[42]:
   c
a
1  3

[jreback]

currently this is ONLY implemented for .xs and .loc with the axes (not multi-levels), though may work for a single multi-level as per the impl. I think @jorisvandenbossche is prob right, should squeeze scalars; this can be done as a post-processing step in .get_loc, so straightforward I think.