如何基于另一列的NaN值在pandas数据框中设置值? [英] How set values in pandas dataframe based on NaN values of another column?

问题描述

我有一个名为df的数据框，其原始形状为(4361, 15). agefm列的某些值是NaN.看看:

I have dataframe named df with original shape (4361, 15). Some of agefm column`s values are NaN. Just look:

> df[df.agefm.isnull() == True].agefm.shape
(2282,)

然后我创建新列并将其所有值设置为0:

Then I create new column and set all its values to 0:

df['nevermarr'] = 0

所以我想将nevermarr值设置为1，然后在该行中agefm是Nan:

So I would like to set nevermarr value to 1, then in that row agefm is Nan:

df[df.agefm.isnull() == True].nevermarr = 1

什么都没改变:

> df['nevermarr'].sum()
0

我在做什么错了?

推荐答案

最好是使用

The best is use numpy.where:

df['nevermarr'] = np.where(df.agefm.isnull(), 1, 0)
print (df)
   agefm  nevermarr
0    NaN          1
1    5.0          0
2    6.0          0

或使用 loc ， ==True可以省略:

df.loc[df.agefm.isnull(), 'nevermarr'] = 1

或 mask :

df['nevermarr'] = df.nevermarr.mask(df.agefm.isnull(), 1)
print (df)
   agefm  nevermarr
0    NaN          1
1    5.0          2
2    6.0          3

示例:

import pandas as pd
import numpy as np

df = pd.DataFrame({'nevermarr':[7,2,3],
                   'agefm':[np.nan,5,6]})

print (df)
   agefm  nevermarr
0    NaN          7
1    5.0          2
2    6.0          3

df.loc[df.agefm.isnull(), 'nevermarr'] = 1
print (df)
   agefm  nevermarr
0    NaN          1
1    5.0          2
2    6.0          3

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