Python-re模块深入

import re


'''
字符串切割
'''

str1 = "sunck     is a good man"
print(str1.split())
print(re.split(r" +",str1))




'''
re.finditer函数
原型:finditer(pattern,string,flags=0)
参数:
patter:匹配的正则表达式
string:要匹配的字符串
flags:标志位,用于控制正则表达式的匹配方式
功能:与findall类似,扫描整个字符串,返回的是一个迭代器

'''

str3 = "sunck is a good man !sunck is a nice man! sunck is a handsome man"
d = re.finditer(r"(sunck)",str3)
while True:
    try:
        l = next(d)
        print(d)
    except StopIteration as e:
        break


'''
sunck is a good man
字符串的替换和修改
sub(pattern, repl, string, count=0,)
subn(pattrn, repl, string, count=0,)
pattrn:正则表达式(规则)
repl:  指定的用来替换的字符串
string:目标字符串
count: 最多替换次数
功能:在目标字符串中以正则表达式的规则匹配字符串,再把它们替换成指定的字符串。
可以指定替换的次数,如果不指定,替换所有的匹配字符串

区别:前者返回一个被替换的字符串,后者返回一个元组,第一个元素是被替换的
     字符串,第二元素表示被替换的次数

'''
str5 = "sunck is a good good good man"
print(re.sub(r"(good)","nice",str5))
print(type(re.sub(r"(good)","nice",str5)))
print(re.subn(r"(good)","nice",str5))
print(type(re.subn(r"(good)","nice",str5)))



'''
分组:
概念:除了简单的判断是否匹配之外,正则表达式还有提取字串的功能。用()表示的就是提取分组

'''
str6 = "010-54327654"
m = re.match(r"(?P<first>(\d{3})-(?P<last>\d{8}))",str6)
# 使用序号获取对应组的信息,group(0)一直代表的就是原始字符串
print(m.group(0))
print(m.group(1))
print(m.group("first"))
print(m.group(2))
# 查看匹配的各组的情况



'''
编译:当我们使用正则表达式时,re模块会干两件事
1、编译正则表达式,如果正则表达式本身不合法,会报错
2、用编译后的正则表达式其匹配对象
compile(pattern,flags=0)
pattern:要编译的正则表达式
'''

pat = r"^1(([3578]\d)|(47))d{8}$"
print(re.match(pat,"13600000000"))
# 编译成正则对象
re_telephon = re.compile(pat)
print(re_telephon.match("13600000000"))

# re.match()
# re.match(pattern,string,flags=0)
# re_telephon.match(string)

# re.search()
# re.search(pattern,string,flags=0)
# re_telephon.search(string)

# re.findall(pattern,string,flags=0)
# re_telephon.findall(string)
# re.findall()

# re.finditer(pattern,string,flags=0)
# re_telephon.finditer(string)
# re.finditer()
# re.split(pattern,string,maxsplit=0,flags=0)
# re_telephon.split(string,maxsplit=0)
# re.split()

# re.sub(pattern,repl,string,count=0,flags=0)
# re_telephon.sub(repl,string,count=0)
# re.sub()
# re.subn(pattern,repl,string,count=0,flags=0)
# re_telephon.subn(repl,string,count=0)
# re.subn()

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转载自blog.csdn.net/huaxiawudi/article/details/81489416