替换前几位生日熊猫

Question

背景

我有以下示例 df

import pandas as pd
df = pd.DataFrame({'Birthdate':['This person was born Date of Birth: 5/6/1950 and other',
                          'no Date of Birth: nothing here',
                          'One Date of Birth: 01/01/2001 last here'], 
                  'P_ID': [1,2,3],
                  'N_ID' : ['A1', 'A2', 'A3']} 

                 )

 df
                                 Birthdate                 N_ID P_ID
    0   This person was born Date of Birth: 5/6/1950 a...   A1  1
    1   no Date of Birth: nothing here                      A2  2
    2   One Date of Birth: 01/01/2001 last here             A3  3

目标

用*BDAY* 替换前几位生日，例如5/6/1950 变为 *BDAY*1950

期望的输出

                                 Birthdate                 N_ID P_ID
    0   This person was born Date of Birth: *BDAY*1950 a... A1  1
    1   no Date of Birth: nothing here                      A2  2
    2   One last Date of Birth: *BDAY*2001 last here        A3  3

试过

来自python - Replace first five characters in a column with asterisks 我尝试了以下代码： df.replace(r'Date of Birth: ^\d{3}-\d{2}', "*BDAY*", regex=True) 但它并没有给我想要的输出

问题

如何实现我想要的输出？

Answer 1

这个表达式也可能有效：

import pandas as pd
df = pd.DataFrame({'Birthdate':['This person was born Date of Birth: 5/6/1950 and other',
                          'no Date of Birth: nothing here',
                          'One Date of Birth: 01/01/2001 last here'], 
                  'P_ID': [1,2,3],
                  'N_ID' : ['A1', 'A2', 'A3']} 

                 )
df= df.replace(r'(?i)date\s+of\s+birth:\s+\d{1,2}/\d{1,2}/', "Date of Birth: *BDAY*", regex=True)

print(df)

表达式在regex101.com 的右上方面板中进行了解释，如果您想探索/简化/修改它，在this link 中，您可以查看它如何与一些示例输入进行匹配，如果您愿意的话。

Answer 2

试试这个：

df['Birthdate'] = df.Birthdate.str.replace(r'[0-9]?[0-9]/[0-9]?[0-9]/', '*BDAY*')


Out[273]:
                                           Birthdate  P_ID N_ID
0  This person was born Date of Birth: *BDAY*1950...     1   A1
1                     no Date of Birth: nothing here     2   A2
2            One Date of Birth: *BDAY*2001 last here     3   A3

Answer 3

错误的正则表达式

您的正则表达式查找 3 位数字，然后是“-”，然后是 2 位数字。您的示例数据有 2 位数字，一个“/”，然后是 2 位数字。

试试：

df.replace(
    r'(Date of Birth:\s+)\d{2}/\d{2}/',
    r"\1*BDAY*",
    regex=True)