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Performance on local constants?

John Machin sjmachin at lexicon.net
Sat Dec 22 07:04:17 EST 2007 
On Dec 22, 9:53 pm, William McBrine <wmcbr... at users.sf.net> wrote:
> Hi all,
>
> I'm pretty new to Python (a little over a month). I was wondering -- is
> something like this:
>
> s = re.compile('whatever')
>
> def t(whatnot):
>     return s.search(whatnot)
>
> for i in xrange(1000):
>     print t(something[i])
>
> significantly faster than something like this:
>
> def t(whatnot):
>     s = re.compile('whatever')
>     return s.search(whatnot)
>
> for i in xrange(1000):
>     result = t(something[i])
>
> ?

No.

Or is Python clever enough to see that the value of s will be the same
> on every call,

No. It doesn't have a crystal ball.

> and thus only compile it once?

But it is smart enough to maintain a cache, which achieves the desired
result.

Why don't you do some timings?

While you're at it, try this:

def t2(whatnot):
    return re.search('whatever', whatnot)

and this:

t3 = re.compile('whatever').search

HTH,
John
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