Python 将DataFrame列标题设置为多索引
如何将具有单级列的现有数据帧转换为具有层次索引列(多索引) 数据帧示例:Python 将DataFrame列标题设置为多索引,python,pandas,multi-index,Python,Pandas,Multi Index,如何将具有单级列的现有数据帧转换为具有层次索引列(多索引) 数据帧示例: In [1]: import pandas as pd from pandas import Series, DataFrame df = DataFrame(np.arange(6).reshape((2,3)), index=['A','B'], columns=['one','two','three']) df Out [1]: one two
In [1]:
import pandas as pd
from pandas import Series, DataFrame
df = DataFrame(np.arange(6).reshape((2,3)),
index=['A','B'],
columns=['one','two','three'])
df
Out [1]:
one two three
A 0 1 2
B 3 4 5
我本以为reindex()会起作用,但我得到了NaN的:
In [2]:
df.reindex(columns=[['odd','even','odd'],df.columns])
Out [2]:
odd even odd
one two three
A NaN NaN NaN
B NaN NaN NaN
如果使用DataFrame(),则相同:
如果我指定df.values,最后一种方法实际上是有效的:
In [4]:
DataFrame(df.values,index=df.index,columns=[['odd','even','odd'],df.columns])
Out [4]:
odd even odd
one two three
A 0 1 2
B 3 4 5
正确的方法是什么?为什么reindex()会给出NaN的?您已经接近了,只需将列直接设置为一个新的(大小相等的)索引,如(如果是列表,则会转换为多索引)
Reindex将对现有索引进行重新排序/筛选。得到所有NaN的原因是,查找与新索引匹配的现有列;无匹配项,因此如果要使用numpy数组,则首先将其转换为list:
df.columns=list(a)
In [4]:
DataFrame(df.values,index=df.index,columns=[['odd','even','odd'],df.columns])
Out [4]:
odd even odd
one two three
A 0 1 2
B 3 4 5
In [8]: df
Out[8]:
one two three
A 0 1 2
B 3 4 5
In [10]: df.columns = [['odd','even','odd'],df.columns]
In [11]: df
Out[11]:
odd even odd
one two three
A 0 1 2
B 3 4 5