如何使用 str.replace 一次替换多个对? [英] How to use str.replace to replace multiple pairs at once?

问题描述

目前我正在使用以下代码进行替换，这有点麻烦:

Currently I am using the following code to make replacements which is a little cumbersome:

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:

In order to make changing on the fly easier my ideal scenario would be something like:

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.

This is so I could just input/change what I wanted to replace for a particular column without having to adjust multiple lines of code.

这可能吗?

推荐答案

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.

You can create a dictionary and pass it to the function replace() without needing to chain or name the function so many times.

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

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