pandas 将 DataFrame 列标题设置为 MultiIndex
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Setting DataFrame column headers to a MultiIndex
提问by mqk
How do I convert an existing dataframe with single-level columns to have hierarchical indexcolumns (MultiIndex)?
如何将具有单级列的现有数据框转换为具有分层索引列 (MultiIndex)？
Example dataframe:
示例数据框：
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
I'd have thought that reindex() would work, but I get NaN's:
我原以为 reindex() 会起作用，但我得到了 NaN：
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
Same if I use DataFrame():
如果我使用 DataFrame() 也是一样：
In [3]:
DataFrame(df,columns=[['odd','even','odd'],df.columns])
Out [3]:
odd even odd
one two three
A NaN NaN NaN
B NaN NaN NaN
This last approach actually does work if I specify df.values:
如果我指定 df.values，最后一种方法实际上确实有效：
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
What's the proper way to do this? Why does reindex() give NaN's?
这样做的正确方法是什么？为什么 reindex() 给出 NaN？
回答by Jeff
You were close, just set the columns directly to a new (equal sized) index-like (which if its a list-of-list will convert to a multi-index)
你很接近，只需将列直接设置为一个新的（相同大小的）索引（如果它的列表将转换为多索引）
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
Reindex will reorder / filter the existing index. The reason you get all nans is you are saying, hey find the existing columns that match this new index; none match, so that's what you get
Reindex 将重新排序/过滤现有索引。你得到所有 nans 的原因是你说，嘿，找到与这个新索引匹配的现有列；没有匹配，所以这就是你得到的
