熊猫MultiIndex，按1和2级选择值

您可以使用df.loc：

import numpy as np
import pandas as pd

columns = pd.MultiIndex.from_product([['A','B','C'],['X','Y','Z']])
df = pd.DataFrame(np.random.randint(10, size=(3,len(columns))), columns=columns)
#    A        B        C      
#    X  Y  Z  X  Y  Z  X  Y  Z
# 0  2  7  5  1  6  0  5  0  0
# 1  8  4  7  2  0  8  7  3  9
# 2  0  6  8  8  1  1  8  0  2

# In some cases `sort_index` may be needed to avoid UnsortedIndexError
df = df.sort_index(axis=1)
print(df.loc[:, (['A','B'],['X','Y'])])

产量（类似）：

   A     B   
   X  Y  X  Y
0  2  7  1  6
1  8  4  2  0
2  0  6  8  1

如果只想选择('A','Y')和('B','X')列，那么请注意，您可以将MultiIndexed列指定为元组：

In [37]: df.loc[:, [('A','Y'),('B','X')]]
Out[37]: 
   A  B
   Y  X
0  7  1
1  4  2
2  6  8

or even just df[[('A','Y'),('B','X')]] (which yields the same result).

And in general it is better to use a single indexer such as df.loc[...] instead of double indexing (e.g. df[...][...]). It can be quicker (because it makes fewer calls to __getitem__, and generates fewer temporary sub-DataFrames) and df.loc[...] = value it is the correct way to make assignments to sub-slices of a DataFrame which modify df itself.

The reason why df[['A','B']][['X','Y']] would not work is because df[['A','B']] returns a DataFrame with a MultiIndex:

In [36]: df[['A','B']]
Out[36]: 
   A        B      
   X  Y  Z  X  Y  Z
0  2  7  5  1  6  0
1  8  4  7  2  0  8
2  0  6  8  8  1  1

So indexing this DataFrame with ['X','Y'] fails because there are no top-level column labels named 'X' or 'Y'.

有时，根据DataFrame的构造方式（或由于对DataFrame进行的操作），在对MultiIndex进行切片之前，需要对其进行按顺序排序。在文档中有一个警告框，其中提到了此问题。对列索引进行词法排序

df = df.sort_index(axis=1)