Does Python optimize regexes?

Peter Otten __peter__ at
Tue Jun 29 14:26:28 CEST 2004

Jason Smith wrote:

> Hi.  I just have a question about optimizations Python does when
> converting to bytecode.
> import re
> for someString in someListOfStrings:
> if re.match('foo', someString):
> print someString, "matched!"
> Does Python notice that re.match is called with the same expression, and
> thus lift it out of the loop?  Or do I need to always optimize by hand
> using re.compile?  I suspect so because the Python bytecode generator
> would hardly know about a library function like re.compile, unlike e.g.
> Perl, with builtin REs.
> Thanks much for any clarification or advice.

Python puts the compiled regular expressions into a cache. The relevant code
is in

def match(pattern, string, flags=0):
    return _compile(pattern, flags).match(string)


def _compile(*key):
    p = _cache.get(key)
    if p is not None:
        return p

So not explicitly calling compile() in advance only costs you two function
calls and a dictionary lookup - and maybe some clarity in your code.


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