使用字典中的值替换列中的值

import pandas as pd
df= pd.DataFrame({'Data':['Hey this is 123456 Jonny B Good',
                              'This is Jonny B Good at 511-233-1137',
                                  'Wow that is Alice N Wonderland A999b',
                                  'Yes hi: Mick E Mouse 1A25629Q88 or ',
                            'Bye Mick E Mouse A13B ok was seen on '], 
                          'E_ID': ['E11','E11', 'E22', 'E33', 'E33'],
                           'N_ID' : ['111', '112', '211', '311', '312'],
                           'Name' : ['JONNY B GOOD', 'JONNY B GOOD', 
                                      'ALICE N WONDERLAND',
                                      'MICK E MOUSE', 'MICK E MOUSE'],        
                          })

df
                      Data                 E_ID N_ID    Name
0   Hey this is 123456 Jonny B Good         E11 111 JONNY B GOOD
1   This is Jonny B Good at 511-233-1137    E11 112 JONNY B GOOD
2   Wow that is Alice N Wonderland A999b    E22 211 ALICE N WONDERLAND
3   Yes hi: Mick E Mouse 1A25629Q88 or      E33 311 MICK E MOUSE
4   Bye Mick E Mouse A13B ok was seen on    E33 312 MICK E MOUSE

我有一个样品df如上所示。我还有样本词典d如下所示

d = {'E11': ['Jonny',
  'B',
  'Good',
   'Jonny',
   'B',
  'Good',
   '123456',
    '511-233-1137'],

'E22': ['Alice',
  'N',
  'Wonderland',
  'A999b'],

'E33': ['Mick', 
        'E' , 
        'Mouse',
        'Mick', 
        'E' , 
        'Mouse',
        '1A25629Q88',
  'A13B',]} 

我想使用来自的值d e.g. Jonny更改相应的值Data。所以例如Jonny in row 0会变成@@@.

为此，我查看了使用字典重新映射 pandas 列中的值 https://stackoverflow.com/questions/20250771/remap-values-in-pandas-column-with-a-dict and 如何用pandas中的字典键替换列值 https://stackoverflow.com/questions/49008179/how-to-replace-column-values-with-dictionary-keys-in-pandas但他们并没有多大帮助。我想我需要使用这样的东西

 df['New'] = df['Data'].str.replace(d[value], '@@@')

我希望我的输出看起来像这样

     Data   E_ID N_ID Name  New
0                           Hey this is @@@ @@@ @@@ @@@             
1                           This is @@@  @@@  @@@  at @@@   
2                           Wow that is @@@  @@@  @@@  @@@  
3                           Yes hi: @@@  @@@  @@@  @@@  or      
4                           Bye @@@  @@@  @@@  @@@  ok was seen on

我需要做什么才能得到这个输出？

您可以生成并使用正则表达式，如下所示：

df['New']= df['Data']
for key, value in d.items():   
    regex='({alternatives})'.format(alternatives='|'.join(value))
    df.loc[df['E_ID']==key, 'New']= df.loc[df['E_ID']==key, 'New'].str.replace(regex, '@@@')

结果如下：

Out[115]: 
                                    Data E_ID N_ID                Name                                  New
0        Hey this is 123456 Jonny B Good  E11  111        JONNY B GOOD          Hey this is @@@ @@@ @@@ @@@
1   This is Jonny B Good at 511-233-1137  E11  112        JONNY B GOOD           This is @@@ @@@ @@@ at @@@
2   Wow that is Alice N Wonderland A999b  E22  211  ALICE N WONDERLAND          Wow that is @@@ @@@ @@@ @@@
3    Yes hi: Mick E Mouse 1A25629Q88 or   E33  311        MICK E MOUSE          Yes hi: @@@ @@@ @@@ @@@ or 
4  Bye Mick E Mouse A13B ok was seen on   E33  312        MICK E MOUSE  Bye @@@ @@@ @@@ @@@ ok was seen on