按索引级别将值分配给Pandas Multiindex DataFrame [英] Assigning values to Pandas Multiindex DataFrame by index level

问题描述

我有一个Pandas多索引数据框，我需要将值分配给一系列中的一列.该系列与数据框索引的第一级共享其索引.

I have a Pandas multiindex dataframe and I need to assign values to one of the columns from a series. The series shares its index with the first level of the index of the dataframe.

import pandas as pd
import numpy as np
idx0 = np.array(['bar', 'bar', 'bar', 'baz', 'foo', 'foo'])
idx1 = np.array(['one', 'two', 'three', 'one', 'one', 'two'])
df = pd.DataFrame(index = [idx0, idx1], columns = ['A', 'B'])
s = pd.Series([True, False, True],index = np.unique(idx0))
print df
print s

退出:

             A    B
bar one    NaN  NaN
    two    NaN  NaN
    three  NaN  NaN
baz one    NaN  NaN
foo one    NaN  NaN
    two    NaN  NaN

bar     True
baz    False
foo     True
dtype: bool

这些无效:

df.A = s # does not raise an error, but does nothing
df.loc[s.index,'A'] = s # raises an error

预期输出:

             A     B
bar one    True   NaN
    two    True   NaN
    three  True   NaN
baz one    False  NaN
foo one    True   NaN
    two    True   NaN

推荐答案

系列(和字典)可以像带map和apply的函数一样使用(感谢@normanius改进了语法):

Series (and dictionaries) can be used just like functions with map and apply (thanks to @normanius for improving the syntax):

df['A'] = pd.Series(df.index.get_level_values(0)).map(s).values

或类似地:

df['A'] = df.reset_index(level=0)['level_0'].map(s).values

结果:

A    B
bar one     True  NaN
    two     True  NaN
    three   True  NaN
baz one    False  NaN
foo one     True  NaN
    two     True  NaN

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