如何在dataframe中用字符串替换endswith

Question

我有一个数据帧df，看起来像这样：

      Company Name     ID
0        Finl Corp    111
1          Fund Tr    222
2          Invt Fd    333           
3          Govt Fd    444
4      Trinity Inc    555

我正在替换通常缩写的字符串：

df['Company Name'] = df['Company Name'].str.replace('Finl', 'Financial')
df['Company Name'] = df['Company Name'].str.replace('Tr', 'Trust')
df['Company Name'] = df['Company Name'].str.replace('Invt', 'Investment')
df['Company Name'] = df['Company Name'].str.replace('Fd', 'Fund')
df['Company Name'] = df['Company Name'].str.replace('Govt', 'Government')

但是，当Tr是字符串的最后两个字符时，如行1（而不是行4），如何仅用Trust替换Tr

我可以做类似的事情吗

df['Company Name'] = df['Company Name'].str.endswith(' Tr').replace(' Tr', ' Trust')以便输出：

      Company Name    ID  
0   Financial Corp   111
1       Fund Trust   222
2  Investment Fund   333
3  Government Fund   444
4      Trinity Inc   555

Answer 1

使用replace替换regex=True一次替换多个内容：

repl_dict = {'Govt': 'Government', 'Fd$': 'Fund',}                                                                                                  
df['Company Name'].replace({k : v for k, v in repl_dict.items()}, 
                           regex=True)                                                            

0          Finl Corp
1            Fund Tr
2          Invt Fund
3    Government Fund
4        Trinity Inc
Name: Company Name, dtype: object