pandas: slice a MultiIndex by range of secondary index

Question

I have a series with a MultiIndex like this:

import numpy as np import pandas as pd  buckets = np.repeat(['a','b','c'], [3,5,1]) sequence = [0,1,5,0,1,2,4,50,0]  s = pd.Series(     np.random.randn(len(sequence)),      index=pd.MultiIndex.from_tuples(zip(buckets, sequence)) )  # In [6]: s # Out[6]:  # a  0    -1.106047 #    1     1.665214 #    5     0.279190 # b  0     0.326364 #    1     0.900439 #    2    -0.653940 #    4     0.082270 #    50   -0.255482 # c  0    -0.091730 

I'd like to get the s['b'] values where the second index ('sequence') is between 2 and 10.

Slicing on the first index works fine:

s['a':'b'] # Out[109]:  # bucket  value # a       0        1.828176 #         1        0.160496 #         5        0.401985 # b       0       -1.514268 #         1       -0.973915 #         2        1.285553 #         4       -0.194625 #         5       -0.144112 

But not on the second, at least by what seems to be the two most obvious ways:

1) This returns elements 1 through 4, with nothing to do with the index values

s['b'][1:10]  # In [61]: s['b'][1:10] # Out[61]:  # 1     0.900439 # 2    -0.653940 # 4     0.082270 # 50   -0.255482 

However, if I reverse the index and the first index is integer and the second index is a string, it works:

In [26]: s Out[26]:  0   a   -0.126299 1   a    1.810928 5   a    0.571873 0   b   -0.116108 1   b   -0.712184 2   b   -1.771264 4   b    0.148961 50  b    0.089683 0   c   -0.582578  In [25]: s[0]['a':'b'] Out[25]:  a   -0.126299 b   -0.116108 

Answer 1

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

Answer 2

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