回复:删除括号中的字符串及其空格 [英] re: remove string in brackets and its whitespace

问题描述

删除括号及其内容以及字符串中的尾随空格的最佳re 方法是什么?请注意，并非每个字符串的格式都相同.

what's the best re way to remove brackets and their content, as well as the trailing whitespace within a string? Note that not every string is formatted equally.

脚本:

import pandas as pd
import re

df = pd.DataFrame({'name':
          ['University of Southampton (UK)', 
          'The College of William and Mary', 
          'University of Reading (UK)', 
          'Queensland University (Australia)']})

def cleaning(text):
    cleaned = re.findall(re.compile('^([^,]+).+'), text)
    cleaned = re.findall(re.compile('\(.*\)'), str(cleaned)) # Why do I have to str() here btw?
    return cleaned

df['name'].apply(lambda x: cleaning(x))

退货:

0    []
1    []
2    []
3    []

所需的输出(末尾没有空格):

0    University of Southampton
1    The College of William and Mary
2    University of Reading
3    Queensland University

感谢您的帮助！

推荐答案

仅适用于这种特定情况，但您可以这样做

Only work for this specific case, but you can do

df.name.str.split('\(',expand=True)[0].str.strip()

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