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('')
Out[3]: 18
In [4]: info = re.search(r'\d+',str1)
In [5]: info
Out[5]: <_sre.SRE_Match object; span=(18, 22), match=''>
In [6]: info.gr
info.group info.groupdict info.groups
In [6]: info.group()
Out[6]: ''
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=''>
In [10]: info.group()
Out[10]: ''
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=''>
In [14]: info.group()
Out[14]: ''
In [15]: info = re.find
re.findall re.finditer
In [15]: info = re.findall(r'\d+',str2)
In [16]: info
Out[16]: ['', '', '']
In [17]: sum([int(x) for x in info])
Out[17]: 2703.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+','',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]: