一篇文章彻底搞懂python正则表达式

import re

# 用find和startswith找字符串
str1 = 'imooc python'
print(str1.find('11'))
-1
print(str1.find('imooc'))
0
print(str1.startswith('imooc'))
True

import re
pa = re.compile(r'imooc')   #  compile生成一个pattern对象，r'imooc'读原字符串，否则需要转义
ma = pa.match(str1)  # 匹配不到返回为None，返回一个对象
print(ma)

re.Match object; span=(0, 5), match='imooc'>

print(ma.group())  # 返回一个字符串或字符串组成的元组ma.groups()
imooc

print(ma.span())  # 返回所在字符串的位置
print(ma.string)  # 返回原字符串
print(ma.re)     # 返回实例

(0, 5)
imooc python
re.compile('imooc')

# 匹配大小写，后面加上大写
pa = re.compile(r'imooc', re.I)
print(pa)

re.compile('imooc', re.IGNORECASE)

ma = pa.match('imooc python')
print(ma.group())

imooc
ma = pa.match('Imooc python')
print(ma.group())

Imooc

# 如果只有一个，可以直接生成一个match对象，也可以达到同样的效果
ma = re.match(r'imooc', str1)
print(ma)
print(ma.group())

re.Match object; span=(0, 5), match='imooc'>
imooc

ma = re.match(r'a', 'a')
print(ma.group())

ma = re.match(r'a', 'b')
print(type(ma))  # 再调ma.group()则报错

a
class 'NoneType'>

# 用.匹配字符
ma = re.match(r'.', 'b')
print(ma.group())

ma = re.match(r'.', '0')
print(ma.group())

b
0

# 匹配{}中的字符
ma = re.match(r'{.}', '{0}')
print(ma.group())
ma = re.match(r'{..}', '{01}')
print(ma.group())

{0}
{01}

# []匹配字符集
ma = re.match(r'{[abc]}', '{a}')
print(ma.group())
ma = re.match(r'{[abc]]}', '{d}')  # 匹配不到，则报错
print(ma.group())
ma = re.match(r'{[a-z]]}', '{d}')
print(ma.group()) # 匹配a-z中任意一个字符

{a}
{d}

ma = re.match(r'{[a-zA-Z]}', '{A}')
print(ma.group())

{A}

ma = re.match(r'{[a-zA-Z0-9]}', '{0}')
print(ma.group())

{0}

ma = re.match(r'{[\w]}', '{A}')
print(ma.group())

ma = re.match(r'{[\w]}', '{ }')
print(ma.group())  # 匹配不到

ma = re.match(r'{[\W]}', '{ }')
print(ma.group())

{A}
{ }

ma = re.match(r'[[\w]]', '[a]')
print(ma.group())  # 匹配不到

# 匹配[]需要加转义\

ma = re.match(r'\[[\w]\]', '[a]')
print(ma.group())
[a]

ma = re.match(r'[A-Z][a-z]', 'Aa')
print(ma.group())

ma = re.match(r'[A-Z][a-z]', 'A')
print(ma.group()) # 匹配不到

ma = re.match(r'[A-Z][a-z]*', 'Aa')
print(ma.group())

ma = re.match(r'[A-Z][a-z]*', 'Aagfagsagaha')
print(ma.group())

ma = re.match(r'[A-Z][a-z]*', 'Aa123')
print(ma.group())

Aa
A
Aagfagsagaha
Aa

# 匹配下划线或字符开头的无限次
ma = re.match(r'[_a-zA-Z]+[_\w]*', '10')
print(ma.group())   # 匹配不到
ma = re.match(r'[_a-zA-Z]+[_\w]*', '_hte10')
print(ma.group())

_hte10

# 匹配0-99，01则不对
 ma = re.match(r'[1-9]?[0-9]', '90')
print(ma.group())

ma = re.match(r'[1-9]?[0-9]', '9')
print(ma.group())

ma = re.match(r'[1-9]?[0-9]', '09')
print(ma.group())  # 只匹配到0

90
9
0

# 匹配邮箱，匹配6次
ma = re.match(r'[a-zA-Z0-9]{6}', 'abc123')
print(ma.group())

abc123
ma = re.match(r'[a-zA-Z0-9]{6}', 'abc12')
print(ma.group()) # 少一个则匹配不到

ma = re.match(r'[a-zA-Z0-9]{6}', 'abc1234')
print(ma.group()) # 多则匹配前6个

abc123

ma = re.match(r'[a-zA-Z0-9]{6}@163.com', 'abc123@163.com')
print(ma.group())

abc123@163.com

# 匹配6-10位的邮箱
ma = re.match(r'[a-zA-Z0-9]{6, 10}@163.com', 'abc123@163.com')
print(ma.group())

ma = re.match(r'[0-9][a-z]*', '1abc')
print(ma.group())

ma = re.match(r'[0-9][a-z]*?', '1abc')
print(ma.group())  # 只匹配1

1abc
1

ma = re.match(r'[0-9][a-z]+?', '1abc')
print(ma.group()) # 只匹配一次

1a

ma = re.match(r'[\w]{4,10}@163.com', 'imooc@163.com')
print(ma.group())

imooc@163.com

ma = re.match(r'[\w]{4,10}@163.com', 'imooc@163.comabc')  # 后面加上abc,match从头开始匹配，则可以找到
print(ma.group())

imooc@163.com

ma = re.match(r'[\w]{4,10}@163.com', 'imooc@163.comabc')
print(ma.group())  # 匹配不到
ma = re.match(r'^[\w]{4,10}@163.com', 'imooc@163.comabc')
print(ma.group())  # 匹配不到
ma = re.match(r'^[\w]{4,10}@163.com', 'imooc@163.com')
print(ma.group())  # 加上^和$,限制开头结尾

imooc@163.com

# 指定开头结尾
ma = re.match(r'\Aimooc[\w]*', 'imoocpython')
print(ma.group())
ma = re.match(r'\Aimooc[\w]*', 'iimoocpython')
print(ma.group())  # 匹配不到

imoocpython

# 在一个字符串中查找匹配
1:search(pattern, string, flags=0)

# 找到匹配，返回所有匹配部分的列表
2:findall(pattern, string, flags=0)

# 将字符串中匹配正则表达式得部分替换为其它,repl可以替换函数
3:sub(pattern, repl, string, count=0, flags=0)

# 根据匹配分割字符串，返回分割字符串组成的列表
4:split(pattern, string, maxsplit=0, flags=0)

# search
import re
str1 = 'imooc videonum = 1000'
print(str1.find('1000'))  # 数字改变则失效

17

info = re.search(r'\d+', str1)
print(info.group())

1000

str1 = 'imooc videonum = 10000'
info = re.search(r'\d+', str1)
print(info.group())

10000

# findall
# 当有多个数字时

str2 = 'a=100, b=200, c=300'
info = re.search(r'\d+', str2)
print(info.group())

100

info = re.findall(r'\d+', str2)
print(info.group())

[100, 200, 300]

# sub 替换
str3 = 'imooc videonum = 1000'
info = re.sub(r'\d+', '1001', str3)
print(info)

imooc videonum = 1001

# split
# 分割
str3 = 'imooc：C C++ Java'
print(re.split(r':| ', str4))

[imooc, C, C, Java]