pandas: slice a MultiIndex by range of secondary index
As of pandas 0.14.0 it is possible to slice multi-indexed objects by providing .loc
a tuple containing slice objects:
In [2]: s.loc[('b', slice(2, 10))]
Out[2]:
b 2 -1.206052
4 -0.735682
dtype: float64
As Robbie-Clarken answers, since 0.14 you can pass a slice in the tuple you pass to loc:
In [11]: s.loc[('b', slice(2, 10))]
Out[11]:
b 2 -0.65394
4 0.08227
dtype: float64
Indeed, you can pass a slice for each level:
In [12]: s.loc[(slice('a', 'b'), slice(2, 10))]
Out[12]:
a 5 0.27919
b 2 -0.65394
4 0.08227
dtype: float64
Note: the slice is inclusive.
Old answer:
You can also do this using:
s.ix[1:10, "b"]
(It's good practice to do in a single ix/loc/iloc since this version allows assignment.)
This answer was written prior to the introduction of iloc in early 2013, i.e. position/integer location - which may be preferred in this case. The reason it was created was to remove the ambiguity from integer-indexed pandas objects, and be more descriptive: "I'm slicing on position".
s["b"].iloc[1:10]
That said, I kinda disagree with the docs that ix is:
most robust and consistent way
it's not, the most consistent way is to describe what you're doing:
- use loc for labels
- use iloc for position
- use ix for both (if you really have to)
Remember the zen of python:
explicit is better than implicit