在 Python 中与 finditer() 重叠匹配

Overlapping matches with finditer() in Python(在 Python 中与 finditer() 重叠匹配)

问题描述

I'm using a regex to match Bible verse references in a text. The current regex is

REF_REGEX = re.compile('''
  (?<!w)                        # Not preceded by any words
  (?P<quote>q(?:uote)?s+)?      # Match optional 'q' or 'quote' followed by many spaces
  (?P<book>                           
    (?:(?:[1-3]|I{1,3})s*)?     # Match an optional arabic or roman number between 1 and 3.
    [A-Za-z]+                    # Match any alphabetics
  ).?                           # Followed by an optional dot
  (?:                         
    s*(?P<chapter>d+)          # Match the chapter number
    (?:
      [:.](?P<startverse>d+)   # Match the starting verse number, preceded by ':' or '.'
        (?:-(?P<endverse>d+))?  # Match the optional ending verse number, preceded by '-'
    )?                           # Verse numbers are optional
  )
  (?:
    s+(?:                       # Here be spaces
      (?:froms+)|(?:ins+)|(?P<lbrace>())   # Match 'from[:space:]', 'in[:space:]' or '('
      s*(?P<version>w+)        # Match a word preceded by optional spaces
      (?(lbrace)))              # Close the '(' if found earlier
  )?                             # The whole 'in|from|()' is optional
  ''', re.IGNORECASE | re.VERBOSE | re.UNICODE)

This matches the following expressions fine:

"jn 3:16":                           (None, 'jn', '3', '16', None, None, None),
"matt. 18:21-22":                    (None, 'matt', '18', '21', '22', None, None),
"q matt. 18:21-22":                  ('q ', 'matt', '18', '21', '22', None, None),
"QuOTe jn 3:16":                     ('QuOTe ', 'jn', '3', '16', None, None, None),
"q 1co13:1":                         ('q ', '1co', '13', '1', None, None, None), 
"q 1 co 13:1":                       ('q ', '1 co', '13', '1', None, None, None),
"quote 1 co 13:1":                   ('quote ', '1 co', '13', '1', None, None, None),
"quote 1co13:1":                     ('quote ', '1co', '13', '1', None, None, None),
"jean 3:18 (PDV)":                   (None, 'jean', '3', '18', None, '(', 'PDV'),
"quote malachie 1.1-2 fRom Colombe": ('quote ', 'malachie', '1', '1', '2', None, 'Colombe'),
"quote malachie 1.1-2 In Colombe":   ('quote ', 'malachie', '1', '1', '2', None, 'Colombe'),
"cinq jn 3:16 (test)":               (None, 'jn', '3', '16', None, '(', 'test'),
"Q   IIKings5.13-58   from   wolof": ('Q     ', 'IIKings', '5', '13', '58', None, 'wolof'),
"This text is about lv5.4-6 in KJV only": (None, 'lv', '5', '4', '6', None, 'KJV'),

but it fails to parse:

"Found in 2 Cor. 5:18-21 ( Ministers":                    (None, '2 Cor', '5', '18', '21', None, None),

because it returns (None, 'in', '2', None, None, None, None) instead.

Is there a way to get finditer() to return all matches, even if they overlap, or is there a way to improve my regex so it matches this last bit properly?

Thanks.

A character consumed is consumed, you should not ask the regex engine to go back.

From your examples the verse part (e.g. :1) seems not optional. Removing that will match the last bit.

ref_regex = re.compile('''
(?<!w)                      # Not preceeded by any words
((?i)q(?:uote)?s+)?            # Match 'q' or 'quote' followed by many spaces
(
    (?:(?:[1-3]|I{1,3})s*)?    # Match an arabic or roman number between 1 and 3.
    [A-Za-z]+                   # Match many alphabetics
).?                            # Followed by an optional dot
(?:
    s*(d+)                    # Match the chapter number
    (?:
        [:.](d+)               # Match the verse number
        (?:-(d+))?             # Match the ending verse number
    )                    # <-- no '?' here
)
(?:
    s+
    (?:
        (?i)(?:froms+)|        # Match the keyword 'from' or 'in'
        (?:ins+)|
        (?P<lbrace>()      # or stuff between (...)
    )s*(w+)
    (?(lbrace)))
)?
''', re.X | re.U)

(If you're going to write a gigantic RegEx like this, please use the /x flag.)

If you really need overlapping matches, you could use a lookahead. A simple example is

>>> rx = re.compile('(.)(?=(.))')
>>> x = rx.finditer("abcdefgh")
>>> [y.groups() for y in x]
[('a', 'b'), ('b', 'c'), ('c', 'd'), ('d', 'e'), ('e', 'f'), ('f', 'g'), ('g', 'h')]

You may extend this idea to your RegEx.

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