import re
line = "..12345678910111213141516171820212223"
regex = re.compile(r'((?:[a-zA-Z0-9])\1+)')
print ("not coming here")
matches = re.findall(regex,line)
print (matches)
在上面的代码中,我试图捕获重复字符的组。你知道吗
例如,我需要这样的答案: 111 222 等等
但是当我运行上面的代码时,我得到一个错误:
Traceback (most recent call last):
File "First.py", line 3, in <module>
regex = re.compile(r'((?:[a-zA-Z0-9])\1+)')
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\re.py", lin
e 224, in compile
return _compile(pattern, flags)
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\re.py", lin
e 293, in _compile
p = sre_compile.compile(pattern, flags)
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_compile
.py", line 536, in compile
p = sre_parse.parse(p, flags)
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_parse.p
y", line 829, in parse
p = _parse_sub(source, pattern, 0)
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_parse.p
y", line 437, in _parse_sub
itemsappend(_parse(source, state))
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_parse.p
y", line 778, in _parse
p = _parse_sub(source, state)
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_parse.p
y", line 437, in _parse_sub
itemsappend(_parse(source, state))
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_parse.p
y", line 524, in _parse
code = _escape(source, this, state)
File "C:\Users\bhatsubh\AppData\Local\Programs\Python\Python35\lib\sre_parse.p
y", line 415, in _escape
len(escape))
sre_constants.error: cannot refer to an open group at position 16
请有人指导我哪里出了问题。你知道吗
用
.findall
做这个是可能的,但是用.finditer
做这个更简单,如Jan的回答所示。你知道吗输出
我们使用
\2
,因为\1
表示外圆括号中的模式,\2
表示内圆括号中的模式。你知道吗在另一个组中找不到组引用。如果您只想打印出那些重复的字符,那么有一个小技巧可以使用
re.sub
:你(可能)想要
见a demo on regex101.com。
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