Python regex replace

Regular match-direct content substitution

s =  ‘dsoheoifsdfscoopaldshfowefcoopasdfjkl;‘ss = s.replace(‘coop‘,‘###‘)print(s,‘\n‘,ss)

dsoheoifsdfscoopaldshfowefcoopasdfjkl;  dsoheoifsdfs###aldshfowef###asdfjkl;

import re regex = re.compile(r‘coop‘)  # 正则匹配替换

regex.sub(‘$$$$$‘,‘sdlafhksdalkfcoopasdhflcoopa;sdhf‘)

‘sdlafhksdalkf$$$$$asdhfl$$$$$a;sdhf‘

# 通过分组替换字符串格式 ，mm/dd/yy -> yy-mm-dds = ‘替换日期格式：10/01/2008，12/25/2018‘re_date = re.compile(r‘(\d+)/(\d+)/(\d+)‘)re_date.sub(r‘\3-\1-\2‘,s)  # 分组 1 2 3 分别对应上一行分组每个（）的位置

‘替换日期格式：2008-10-01，2018-12-25‘

#########

# 替换字符串中多余的空格

s = ‘  coop regex  python  easy to learn,come on  ‘s.strip()re_blank = re.compile(r‘\s+‘)  # 匹配任意空吧字符，相当于[\t\n\r\f\v]re_blank.sub(‘‘,s)  

‘coopregexpythoneasytolearn,comeon‘

Case: Mobile phone number, phone number:, mobile phone number: ^1\d{10}\$ phone number required area code version: \d{3,4}-\d{8} \d{3}-\d{8}|\d{4}-\d{8} mailbox: E-mail authentication:/(\[email protected] (\w+.) +\w{2,3})?/Verify Email Address: ^ (\w+[-+.) \w+) @\w+ ([-.] \w+). \w+ ([-.] \w+) *$ Verify email address:/[email protected]+. [a-z]+/xxx:xxx Number: ^ (\d{15}|\d{17} (\d|x)) \$

import reRE_PHONE = re.compile(r‘\d{3,4}-\d{8} \d{3}-\d{8}|\d{4}-\d{8}‘)def find_phone(text:str)  -> list:    return RE_PHONE.findall(text)def main():    text = ‘what ever number:110-11111111,0372-32122222‘    find_phone(text)    print(find_phone(text))if  __name__ == ‘__main__‘:    main()

[‘0372-32122222‘]

Python regex replace