When I have a pandas.DataFrame df with columns ["A", "B", "C", "D"], I can filter it using constructions like df[df["B"] == 2].
How do I do the equivalent of df[df["B"] == 2], if B is the name of a level in a MultiIndex instead? (For example, obtained by df.groupby(["A", "B"]).mean() or df.setindex(["A", "B"]))
I would suggest either:
df.xs(2, level='B')
or
df[df.index.get_level_values('B') == val]
I'd like to make the syntax for the latter operation a little nicer.
I see two ways of getting this, both of which look like a detour – which makes me think there must be a better way which I'm overlooking.
MultiIndex into columns: df[df.reset_index()["B"] == 2]MultiIndex and then use lookup by index: df.swaplevel(0, "B").ix[2]I think you are looking to grouping by index levels (see GroupBy with MultiIndex).
Here's a short, and not very exciting, example:
In [126]: df = DataFrame([[1,2,3,4],[2,2,np.nan,6]],columns=["A", "B", "C", "D"])
In [127]: df1 = df.set_index(['A','B'])
In [128]: df1
Out[128]:
C D
A B
1 2 3 4
2 2 NaN 6
In [129]: df1.groupby(level='B', axis=0).mean()
Out[129]:
C D
B
2 3 5
