如何使用str.replace更简单地一次替换多对？

Question

目前，我正在使用以下代码进行替换，这有点麻烦：

df1['CompanyA'] = df1['CompanyA'].str.replace('.','')
df1['CompanyA'] = df1['CompanyA'].str.replace('-','')
df1['CompanyA'] = df1['CompanyA'].str.replace(',','')
df1['CompanyA'] = df1['CompanyA'].str.replace('ltd','limited')
df1['CompanyA'] = df1['CompanyA'].str.replace('&','and')
df1['Address1A'] = df1['Address1A'].str.replace('.','')
df1['Address1A'] = df1['Address1A'].str.replace('-','')
df1['Address1A'] = df1['Address1A'].str.replace('&','and')
df1['Address1A'].str.replace(r'\brd\b', 'road')
df1['Address2A'] = df1['Address2A'].str.replace('.','')
df1['Address2A'] = df1['Address2A'].str.replace('-','')
df1['Address2A'] = df1['Address2A'].str.replace('&','and')
df1['Address2A'].str.replace(r'\brd\b', 'road')

为了使即时更改更容易，我的理想情况是：

df1['CompanyA'] = df1['CompanyA'].str.replace(('&','and'), ('.', ''), ('-','')....)
df1['Address1A'] = df1['Address1A'].str.replace(('&','and'), ('.', ''), ('-','')....)
df1['Address2A'] = df1['Address2A'].str.replace(('&','and'), ('.', ''), ('-','')....)

因此，我可以输入/更改要替换的特定列的内容，而无需调整多行代码。

这有可能吗？

谢谢！

Answer 1

您可以创建字典并与replace()一起使用，而无需多次链接或命名该函数。

replacers = {',':'','.':'','-':'','ltd':'limited'} #etc....
df1['CompanyA'] = df1['CompanyA'].replace(replacers)

Answer 2

您可以链接替换项：

df1['CompanyA'] = df1['CompanyA'].str.replace('.','').replace('-','').replace(',','').replace('ltd','limited').replace('&','and')
...

Answer 3

替换功能也接受值作为字典。您可以执行以下操作：

df1.replace({'CompanyA' : { '&' : 'and', '.': '' , '-': ''}},regex=True)