Gruber’s URL Regular Expression in Python

问题:

How do I rewrite this new way to recognise addresses to work in Python?

\b(([\w-]+://?|www[.])[^\s()<>]+(?:\([\w\d]+\)|([^[:punct:]\s]|/)))

回答1:

The original source for that states "This pattern should work in most modern regex implementations" and specifically Perl. Python's regex implementation is modern and similar to Perl's but is missing the [:punct:] character class. You can easily build that using this:

>>> import string, re
>>> pat = r'\b(([\w-]+://?|www[.])[^\s()<>]+(?:\([\w\d]+\)|([^%s\s]|/)))'
>>> pat = pat % re.sub(r'([-\\\]])', r'\\\1', string.punctuation)

The re.sub() call escapes certain characters inside the character set as required.

Edit: Using re.escape() works just as well, since it just sticks a backslash in front of everything. That felt crude to me at first, but certainly works fine for this case.

>>> pat = pat % re.escape(string.punctuation)

回答2:

I don't think python have this expression

[:punct:]

Wikipedia says [:punct:] is same to

[-!\"#$%&\'()*+,./:;<=>?@\\[\\\\]^_`{|}~]

回答3:

Python doesn't have the POSIX bracket expressions.

The [:punct:] bracket expression is equivalent in ASCII to

[!"#$%&'()*+,\-./:;<=>?@[\\\]^_`{|}~]