给出一个由3个大写字母组成的字符串,1个小型大写字母和3个大写字母, 例如AAAAAAA
我似乎找不到一个能找到匹配字符串的字符串的正则表达式:
e.g。 A B C a AA C (无空格)
编辑:
原来我需要一些略有不同的东西,例如: ABCaAAC其中'a'是 小拳头版本的非常拳头角色,而不仅仅是任何角色
答案 0 :(得分:11)
以下内容应该有效:
^([A-Z])(?!.?\1)([A-Z])(?!\2)([A-Z])[a-z]\1\1\3$
例如:
>>> regex = re.compile(r'^([A-Z])(?!.?\1)([A-Z])(?!\2)([A-Z])[a-z]\1\1\3$')
>>> regex.match('ABAaAAA') # fails: first three are not different
>>> regex.match('ABCaABC') # fails: first two of second three are not first char
>>> regex.match('ABCaAAB') # fails: last char is not last of first three
>>> regex.match('ABCaAAC') # matches!
<_sre.SRE_Match object at 0x7fe09a44a880>
说明:
^ # start of string
([A-Z]) # match any uppercase character, place in \1
(?!.?\1) # fail if either of the next two characters are the previous character
([A-Z]) # match any uppercase character, place in \2
(?!\2) # fail if next character is same as the previous character
([A-Z]) # match any uppercase character, place in \3
[a-z] # match any lowercase character
\1 # match capture group 1
\1 # match capture group 1
\3 # match capture group 3
$ # end of string
如果你想从更大的文本块中提取这些匹配项,只需删除^
和$
并使用regex.search()
或regex.findall()
。
然而,您可能会发现以下方法更容易理解,它使用正则表达式进行基本验证,然后使用常规字符串操作来测试所有额外要求:
def validate(s):
return (re.match(r'^[A-Z]{3}[a-z][A-Z]{3}$', s) and s[4] == s[0] and
s[5] == s[0] and s[-1] == s[2] and len(set(s[:3])) == 3)
>>> validate('ABAaAAA')
False
>>> validate('ABCaABC')
False
>>> validate('ABCaAAB')
False
>>> validate('ABCaAAC')
True