另一种解决方法：

Question

我有一个索引编号为'l1'的列表，并且我想根据这些索引编号将NaA列中的值替换为

当前数据

Index                A
0        Reviewer: Newbie | 35-44 on Treatment for 1 
1        Reviewer: 45-54 on Treatment for less than 1 
2        Reviewer: Ocetech| 65-74 Male on Treatment 
3        Reviewer: virleo| 55-64 Female on Treatment 
4        Reviewer: Diane perrin| 65-74 on Treatment for

l1 = [1，3，4]

预期产量

Index                A
0        Reviewer: Newbie | 35-44 on Treatment for 1 
1        NaN 
2        Reviewer: Ocetech| 65-74 Male on Treatment 
3        NaN
4        NaN

Answer 1

使用.loc在数据帧中的l1中定位索引，然后像这样分配np.nan：

import numpy as np
df.loc[l1,'A'] = np.nan

Answer 2

在复制您的DataFrame之后，只需了解细节即可。

也导入numpy，将Nan的值设置为所需的索引。

import pandas as pd
import numpy as np

您的数据框：

$ df
                                                A
0     Reviewer: Newbie | 35-44 on Treatment for 1
1    Reviewer: 45-54 on Treatment for less than 1
2      Reviewer: Ocetech| 65-74 Male on Treatment
3     Reviewer: virleo| 55-64 Female on Treatment
4  Reviewer: Diane perrin| 65-74 on Treatment for

您的索引列表：

$ l1
[1, 3, 4]

基于loc对于列A使用Numpy ..将索引设置为Nan。

df.loc[l1,'A'] = np.nan

结果：

print(df)
                                             A
0  Reviewer: Newbie | 35-44 on Treatment for 1
1                                          NaN
2   Reviewer: Ocetech| 65-74 Male on Treatment
3                                          NaN
4                                          NaN

请注意，如果您没有一长串要替换为NaN的索引，则可以直接指定它们，而不用传递列表索引。

$ df.loc[[1,3,4],'A'] = np.nan
$ print(df)
                                             A
0  Reviewer: Newbie | 35-44 on Treatment for 1
1                                          NaN
2   Reviewer: Ocetech| 65-74 Male on Treatment
3                                          NaN
4                                          NaN

另一种解决方法：

$ df.rename(index={1:np.nan, 3:np.nan, 4:np.nan}, inplace=True)

根据索引值在列“ A”中插入Null

2 个答案:

另一种解决方法：