DataFrame.loc(axis=1)[..] not working for single-level indexer #13655
Description
When doing .loc(axis=1)[...], this only works when indexing into multiple levels. Here an example with a MultiIndex column:
In [23]: df = pd.DataFrame(np.arange(27).reshape(3, 9), columns=pd.MultiIndex.from_product([['a1', 'a2', 'a3'], ['b1', 'b2', 'b3']])) In [24]: df Out[24]: a1 a2 a3 b1 b2 b3 b1 b2 b3 b1 b2 b3 0 0 1 2 3 4 5 6 7 8 1 9 10 11 12 13 14 15 16 17 2 18 19 20 21 22 23 24 25 26 In [25]: df.loc(axis=1)['a1':'a2', ['b1', 'b2']] Out[25]: a1 a2 b1 b2 b1 b2 0 0 1 3 4 1 9 10 12 13 2 18 19 21 22 Given the above, I would expect leaving out the ['b1', 'b2'] to work as well, but in this case it looks like the axis=1 is ignored, which is a bit surprising IMO:
In [26]: df.loc(axis=1)['a1':'a2'] Out[26]: Empty DataFrame Columns: [(a1, b1), (a1, b2), (a1, b3), (a2, b1), (a2, b2), (a2, b3), (a3, b1), (a3, b2), (a3, b3)] Index: [] In [27]: df.loc(axis=1)['a1'] ... KeyError: 'the label [a1] is not in the [index]' Also when having a simple frame, the same observation applies:
In [28]: df2 = pd.DataFrame(np.arange(9).reshape(3,3), columns=['a', 'b', 'c']) In [29]: df2 Out[29]: a b c 0 0 1 2 1 3 4 5 2 6 7 8 In [30]: df2.loc(axis=1)['a'] ... KeyError: 'the label [a] is not in the [index]'