python regex replace

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-dd
s = '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'

import re

RE_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']