Does Python support conditional structure in regex?

Question

Does Python support conditional structure in regex?

1. If yes, why I can't have the following (using lookahead in the if part) right? Any way to make Python support it?

   >>> p = re.compile(r'(?(?=regex)then|else)')
   Traceback (most recent call last):
     File "<stdin>", line 1, in <module>
     File "/usr/lib/python2.7/re.py", line 190, in compile
       return _compile(pattern, flags)
     File "/usr/lib/python2.7/re.py", line 242, in _compile
       raise error, v # invalid expression
   sre_constants.error: bad character in group name
   

2. Using backreference as the if part works, however:

   >>> p = re.compile(r'(expr)?(?(1)then|else)')
   

http://www.regular-expressions.info/conditional.html says

Conditionals are supported by the JGsoft engine, Perl, PCRE, Python, and the .NET framework.

What is the closest solution to use conditionals in regex?

My Python is 2.7.3. I don't know how to check the version of re module (how can I?). Thanks.

Answer 1

According to the documentation you referenced:

Python supports conditionals using a numbered or named capturing group. Python does not support conditionals using lookaround, even though Python does support lookaround outside conditionals. Instead of a conditional like (?(?=regex)then|else), you can alternate two opposite lookarounds: (?=regex)then|(?!regex)else).