字符串regexp中多次出现相同字符 - Python

Given a string made up of 3 capital letters, 1 small caps and another 3 capital ones, e.g. AAAaAAA

给定一个由3个大写字母组成的字符串,1个小写字母和3个大写字母,例如AAAAAAA

I can't seem to find a regexp that would find a string which matches a string that has:

我似乎无法找到一个正则表达式,它会找到一个匹配字符串的字符串:

first 3 capital letters all different

前三个大写字母都不同

any small caps letter

任何小型大写字母

first 2 same capital letters as the very first one

前两个大写字母与第一个相同

last capital letter the same as the last capital letter in the first "trio"

最后一个大写字母与第一个“三重奏”中的最后一个大写字母相同

e.g. A B C a AA C (no spaces)

例如A B C a AA C(无空格)

EDIT:

Turns out I needed something slightly different e.g. ABCaAAC where 'a' is the small caps version of the very fist character, not just any character

结果我需要一些略有不同的东西,例如: ABCaAAC其中'a'是第一个角色的小型大写版本,而不仅仅是任何角色

1 个解决方案

#1

The following should work:

以下应该有效:

^([A-Z])(?!.?\1)([A-Z])(?!\2)([A-Z])[a-z]\1\1\3$

For example:

>>> 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>

Explanation:

^          # 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

If you want to pull these matches out from a larger chunk of text, just get rid of the ^ and $ and use regex.search() or regex.findall().

如果你想从更大的文本块中拉出这些匹配,只需删除^和$并使用regex.search()或regex.findall()。

You may however find the following approach easier to understand, it uses regex for the basic validation but then uses normal string operations to test all of the extra requirements:

但是,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

#1