Pandas Feature request: Access multiindex by name

Code Sample, a copy-pastable example if possible

I have not found an easy way to access a column of a multi index by name. Consider the following code:

import pandas as pd
df = pd.DataFrame([[1,1,10,200],
                   [1,2,11,201],
                   [1,3,12,202],
                   [2,1,13,210],
                   [2,2,14,230]],
                  columns=list('ABCD')).set_index(['A','B'])
ii = df.index

To access the values of column B i can e.g. do:

df.reset_index('B').B.tolist()

or

[v[1] for v in ii.tolist()]

Both of there are somewhat cumbersome What I'm proposing is to add similar shorcut access to a multiindex just like a dataframe. E.g.

ii.B

or

ii['B']

that would return the same list as in the two examples above.

output of pd.show_versions()

0.18.1

Comment From: jorisvandenbossche

Related issue / possibly duplicate: #10816

Comment From: jreback

this is the idiom, which is not bad actually.

In [3]: df.index.get_level_values('B')
Out[3]: Int64Index([1, 2, 3, 1, 2], dtype='int64', name=u'B')

Returning a list is non-performant / not-idiomatic.

Comment From: dov

Nice! I missed that idiom.

But, still, is there any reason that:

df.index.B 

is not a shortcut for df.index.get_level_values('B')? What makes the dataframe special?

Comment From: jreback

So doing df.index.B looks simple and generally it is, however there are some edge cases that make this untenable:

df.index['B'] needs to work similarly for a smooth user xp, however this is in direct conflict with positional access on a single level for a regular Index and on a MI:

In [7]: df.index[0]
Out[7]: (1, 1)

In [8]: df.index[[True]*len(df.index)]
Out[8]: 
MultiIndex(levels=[[1, 2], [1, 2, 3]],
           labels=[[0, 0, 0, 1, 1], [0, 1, 2, 0, 1]],
           names=[u'A', u'B'])

You might think that is ok, until you realize that 0 also could mean level=0. Then what do you do?