如何使用 str.replace 一次替换多个对？

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，我的理想情况是这样的：

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

这样我就可以输入/更改我想为特定列替换的内容，而无需调整多行代码。

这可能吗？

Answer 1

您可以创建字典并将其传递给函数，replace()而无需多次链接或命名函数。

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