pandas: slice a MultiIndex DataFrame by range of secondary index

Question

It has been posted that slicing on the second index can be done on a multi-indexed pandas Series:

import numpy as np
import pandas as pd

buckets = np.repeat(range(3), [3,5,7])
sequence = np.hstack(map(range,[3,5,7]))

s = pd.Series(np.random.randn(len(sequence)), 
              index=pd.MultiIndex.from_tuples(zip(buckets, sequence)))

print s

0  0    0.021362
   1    0.917947
   2   -0.956313
1  0   -0.242659
   1    0.398657
   2    0.455909
   3    0.200061
   4   -1.273537
2  0    0.747849
   1   -0.012899
   2    1.026659
   3   -0.256648
   4    0.799381
   5    0.064147
   6    0.491336

Then to get the first three rows for the first index=1, you simply say:

s[1].ix[range(3)]

0   -0.242659
1    0.398657
2    0.455909

This works fine for 1-dimensional Series, but not for DataFrames:

buckets = np.repeat(range(3), [3,5,7])
sequence = np.hstack(map(range,[3,5,7]))

d = pd.DataFrame(np.random.randn(len(sequence),2), 
                 index=pd.MultiIndex.from_tuples(zip(buckets, sequence)))

print d

            0         1
0 0  1.217659  0.312286
  1  0.559782  0.686448
  2 -0.143116  1.146196
1 0 -0.195582  0.298426
  1  1.504944 -0.205834
  2  0.018644 -0.979848
  3 -0.387756  0.739513
  4  0.719952 -0.996502
2 0  0.065863  0.481190
  1 -1.309163  0.881319
  2  0.545382  2.048734
  3  0.506498  0.451335
  4  0.872743 -0.070985
  5 -1.160473  1.082550
  6  0.331796 -0.366597

d[1].ix[range(3)]

0  0    0.312286
   1    0.686448
   2    1.146196
Name: 1

It gives you the "1th" column of data, and the first three rows, irrespective of the first index level. How can you get the first three rows for the first index=1 for a multi-indexed DataFrame?

Answer 1

d.xs(1)[0:3]


          0         1
0 -0.716206  0.119265
1 -0.782315  0.097844
2  2.042751 -1.116453

Answer 2

.loc is more efficient and is evaluated simultaneously

s.loc[pd.IndexSlice[1],:3] will return 0th level = 1 and [0:3] entry.