python正则表达式之re模块方法介绍

Posted 2020-12-26 zaorunzi

python正则表达式之re模块其他方法

1:search(pattern,string,flags=0)

在一个字符串中查找匹配

2:findall(pattern,string,flags=0)

找到匹配，返回所有匹配部分的列表

In [1]: import re

In [2]: str1 = ‘imoooc videonum = 1000‘

In [3]: str1.find(‘1000‘)
Out[3]: 18

In [4]: info = re.search(r‘d+‘,str1)

In [5]: info
Out[5]: <_sre.SRE_Match object; span=(18, 22), match=‘1000‘>

In [6]: info.gr
info.group      info.groupdict  info.groups     

In [6]: info.group()
Out[6]: ‘1000‘

In [7]: str1 = ‘imoooc videonum = 10000‘

In [8]: info = re.search(r‘d+‘,str1)

In [9]: info
Out[9]: <_sre.SRE_Match object; span=(18, 23), match=‘10000‘>

In [10]: info.group()
Out[10]: ‘10000‘

In [11]: str2 = ‘c++=100, java=90, python=80‘

In [12]: info = re.search(r‘d+‘,str2)

In [13]: info
Out[13]: <_sre.SRE_Match object; span=(4, 7), match=‘100‘>

In [14]: info.group()
Out[14]: ‘100‘

In [15]: info = re.find
re.findall   re.finditer  

In [15]: info = re.findall(r‘d+‘,str2)

In [16]: info
Out[16]: [‘100‘, ‘90‘, ‘80‘]

In [17]: sum([int(x) for x in info])
Out[17]: 270

3.sub(pattern,repl,string,count=0,flags=0)

将字符串中匹配正则表达式的部分替换为其他值

4.split(pattern,string,maxsplit=0,flags=0)

根据匹配分割字符串，返回分割字符串组成的列表

In [22]: str3 = ‘imooc videonum = 1000‘

In [24]: info = re.sub(r‘d+‘,‘1001‘,str3)

In [25]: info
Out[25]: ‘imooc videonum = 1001‘

In [26]: def add1(match):
   ....:     val = match.group()
   ....:     num = int(val)+1
   ....:     return str(num)
   ....: 

In [27]: str3
Out[27]: ‘imooc videonum = 1000‘

In [28]: re.sub(r‘d+‘,add1,str3)
Out[28]: ‘imooc videonum = 1001‘

In [36]: str4 = ‘imooc:C C++ Java Python‘

In [37]: re.s
re.search       re.sre_compile  re.sub          re.sys
re.split        re.sre_parse    re.subn         

In [37]: re.split(r‘:| ‘,str4)
Out[37]: [‘imooc‘, ‘C‘, ‘C++‘, ‘Java‘, ‘Python‘]

In [38]: re.split(r‘:| |,‘,str4)
Out[38]: [‘imooc‘, ‘C‘, ‘C++‘, ‘Java‘, ‘Python‘]

In [39]: