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