python – pandas concat中的levels选项

df1 = DataFrame(np.arange(6).reshape(3, 2), index=['a', 'b', 'c'],
        columns=['one', 'two'])
df2 = DataFrame(5 + np.arange(4).reshape(2, 2), index=['a', 'c'],
        columns=['three', 'four'])

>>> df1
   one  two
a    0    1
b    2    3
c    4    5

>>> df2
   three  four
a      5     6
c      7     8


res = pd.concat([df1, df2], axis=1, levels=['level1', 'level2'],
        names=['upper', 'lower'])
>>> res
   one  two  three  four
a    0    1      5     6
b    2    3    NaN   NaN
c    4    5      7     8

我的问题是为什么级别和名称没有显示在上面的res输出中？如何使用level选项的任何真实示例？

谢谢你的时间和帮助

解决方法:

非常有趣的问题.

我在SO研究但从未使用过:(

但在docs是一个通知的样本：

Yes, this is fairly esoteric, but is actually necessary for implementing things like GroupBy where the order of a categorical variable is meaningful.

docs也说：

levels : list of sequences, default None. Specific levels (unique values) to use for constructing a MultiIndex. Otherwise they will be inferred from the keys.

因此它为MultiIndex添加了新的级别：

res = pd.concat([df1, df2], axis=1,
                keys=['level1','level2'], 
                levels=[['level1', 'level2','level3']], 
                names=['upper', 'lower'])

print (res)
upper level1     level2     
lower    one two  three four
a          0   1    5.0  6.0
b          2   3    NaN  NaN
c          4   5    7.0  8.0

print (res.columns)
MultiIndex(levels=[['level1', 'level2', 'level3'], ['four', 'one', 'three', 'two']],
           labels=[[0, 0, 1, 1], [1, 3, 2, 0]],
           names=['upper', 'lower'])

同样没有参数级别：

res = pd.concat([df1, df2], axis=1,
                keys=['level1','level2'], 
                names=['upper', 'lower'])

print (res)
upper level1     level2     
lower    one two  three four
a          0   1    5.0  6.0
b          2   3    NaN  NaN
c          4   5    7.0  8.0

print (res.columns)
MultiIndex(levels=[['level1', 'level2'], ['four', 'one', 'three', 'two']],
           labels=[[0, 0, 1, 1], [1, 3, 2, 0]],
           names=['upper', 'lower'])