Ezio Melotti <ezio.melo...@gmail.com> added the comment:
This is a proof that you can have an equivalent regex without including all the
'letter chars' (tested on both narrow and wide builds):
>>> s = u''.join(unichr(c) for c in range(sys.maxunicode))
>>> diff = set(re.findall(u'[^\W\d]', s, re.U)) ^ set(re.findall(u'[%s_-]' %
>>> makew(), s, re.U))
>>> diff.remove('-')
>>> re.findall(u'(?:[^\W\d%s]|-)' % ''.join(diff), s, re.U) ==
>>> re.findall(u'[%s_-]' % makew(), s, re.U)
True
(I don't like the way I included the '-' but I couldn't find anything better.)
It looks however that most of the time is spent during the findall and from a
quick benchmark it seems that my regex is slower (even if it's shorter and it
compiles faster).
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue8064>
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