сопоставление двух или более символов, которые не совпадают

Question

Можно ли написать шаблон регулярного выражения для соответствия abc, где каждая буква не является литеральной, но означает, что текст, подобный xyz (но не xxy), будет сопоставлен? Я могу пройти до (.)(?!\1), чтобы соответствовать a в ab, но затем я озадачен.

Получив ответ ниже, я смог написать процедуру для генерации этого шаблона. Использование шаблонов re гораздо быстрее, чем преобразование шаблона и текста в каноническую форму с последующим их сочетанием.

def pat2re(p, know=None, wild=None):
    """return a compiled re pattern that will find pattern `p`
    in which each different character should find a different
    character in a string. Characters to be taken literally
    or that can represent any character should be given as
    `know` and `wild`, respectively.

    EXAMPLES
    ========

    Characters in the pattern denote different characters to
    be matched; characters that are the same in the pattern
    must be the same in the text:

    >>> pat = pat2re('abba')
    >>> assert pat.search('maccaw')
    >>> assert not pat.search('busses')

    The underlying pattern of the re object can be seen
    with the pattern property:

    >>> pat.pattern
    '(.)(?!\\1)(.)\\2\\1'    

    If some characters are to be taken literally, list them
    as known; do the same if some characters can stand for
    any character (i.e. are wildcards):

    >>> a_ = pat2re('ab', know='a')
    >>> assert a_.search('ad') and not a_.search('bc')

    >>> ab_ = pat2re('ab*', know='ab', wild='*')
    >>> assert ab_.search('abc') and ab_.search('abd')
    >>> assert not ab_.search('bad')

    """
    import re
    # make a canonical "hash" of the pattern
    # with ints representing pattern elements that
    # must be unique and strings for wild or known
    # values
    m = {}
    j = 1
    know = know or ''
    wild = wild or ''
    for c in p:
        if c in know:
            m[c] = '\.' if c == '.' else c
        elif c in wild:
            m[c] = '.'
        elif c not in m:
            m[c] = j
            j += 1
            assert j < 100
    h = tuple(m[i] for i in p)
    # build pattern
    out = []
    last = 0
    for i in h:
        if type(i) is int:
            if i <= last:
                out.append(r'\%s' % i)
            else:
                if last:
                    ors = '|'.join(r'\%s' % i for i in range(1, last + 1))
                    out.append('(?!%s)(.)' % ors)
                else:
                    out.append('(.)')
                last = i
        else:
            out.append(i)
    return re.compile(''.join(out))

Answer 1

Вы можете попробовать:

^(.)(?!\1)(.)(?!\1|\2).$

Демо

Вот объяснение шаблона регулярного выражения:

^          from the start of the string
(.)        match and capture any first character (no restrictions so far)
(?!\1)     then assert that the second character is different from the first
(.)        match and capture any (legitimate) second character
(?!\1|\2)  then assert that the third character does not match first or second
.          match any valid third character
$          end of string