Question

我有以下示例df

import pandas as pd
DateChange = [list(['11/04/1911', '03/06/1919']),
          list(['01/27/1918', '03/27/1984']),
          list(['01/26/1912', '01/15/1918', '02/06/1917'])]
DateGrab = [list(['10.31.11', '03-02-1919']), list(['01-23-18', '03-23-1984']),
       list(['1-22-12', '01.11.18', '2.2.17'])]
df= pd.DataFrame({'Text':['Hey 10.31.11  22|1|13 03-02-1919 d',
                              'things here 01-23-18 or 03-23-1984 then ',
                                  'stuff 1-22-12 01.11.18 or 2.2.17 so so '],
                          'ID': ['E1','E2', 'E3'],
                  'DateChange' : DateChange,
                  'DateGrab' : DateGrab

                          })

看起来像

                DateChange                 DateGrab                 ID   Text
0   [11/04/1911, 03/06/1919]             [10.31.11, 03-02-1919]     E1  Hey 10.31.11 22|1|13 03-02-1919 d
1   [01/27/1918, 03/27/1984]             [01-23-18, 03-23-1984]     E2  things here 01-23-18 or 03-23-1984 then    
2   [01/26/1912, 01/15/1918, 02/06/1917] [1-22-12, 01.11.18, 2.2.17]E3  stuff 1-22-12 01.11.18 or 2.2.17 so so

我的目标如下：

1）如果DateGrab中存在10.31.11值，例如Text，则将Text中的该值替换为DateChange，例如。 11/04/1911

2）创建新列

因此，我想要的输出如下

    DateChange DateGrab ID  Text   New_Text
0                                  Hey 11/04/1911 22|1|13 03/06/1919 d
1                                  things here 01/27/1918 or 03/27/1984 then    
2                                  stuff 01/26/1912 01/15/1918 or 02/06/1917 so so

我已经尝试过以下代码，但这不起作用

df['New_Text']= df['Text'].replace(df.Text, df.DateChange)

如何实现所需的输出？

Answer 1

您可以在熊猫0.25.1中使用explode：

s=df['DateChange'].explode()
s.index=df['DateGrab'].explode()
d=s.to_dict()
for key in d:
    df['New_Text']=df['Text'].str.replace(key,d[key])  
print(df)

                                       Text  ID  \
0        Hey 10.31.11  22|1|13 03-02-1919 d  E1   
1  things here 01-23-18 or 03-23-1984 then   E2   
2   stuff 1-22-12 01.11.18 or 2.2.17 so so   E3   

                             DateChange                     DateGrab  \
0              [11/04/1911, 03/06/1919]       [10.31.11, 03-02-1919]   
1              [01/27/1918, 03/27/1984]       [01-23-18, 03-23-1984]   
2  [01/26/1912, 01/15/1918, 02/06/1917]  [1-22-12, 01.11.18, 2.2.17]   

                                      New_Text  
0           Hey 10.31.11  22|1|13 03-02-1919 d  
1     things here 01-23-18 or 03-23-1984 then   
2  stuff 1-22-12 01.11.18 or 02/06/1917 so so

Answer 2

尝试利用字典中的switch menuType { case .about: let profileVC = AboutViewController() self.present(profileVC, animated: true, completion: nil) case .imageList: let imageVC = ImageViewController() self.present(imageVC, animated: true, completion: nil) case default: break }

df.replace

用一列中的值替换另一列中的值

2 个答案: