Performance on local constants?
William McBrine
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])
? Or is Python clever enough to see that the value of s will be the same
on every call, and thus only compile it once?
--
09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0 -- pass it on
Paddy
Python RE's do have a cache but telling it to compile multiple times
is going to take time.
Best to do as the docs say and compile your RE's once before use if
you can.
The timeit module: http://www.diveintopython.org/performance_tuning/timeit.html
will allow you to do your own timings.
- Paddy.
John Machin
> 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
Duncan Booth
> 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])
>
> ? Or is Python clever enough to see that the value of s will be the same
> on every call, and thus only compile it once?
>
The best way to answer these questions is always to try it out for
yourself. Have a look at 'timeit.py' in the library: you can run
it as a script to time simple things or import it from longer scripts.
C:\Python25>python lib/timeit.py -s "import re;s=re.compile('whatnot')" "s.search('some long string containing a whatnot')"
1000000 loops, best of 3: 1.05 usec per loop
C:\Python25>python lib/timeit.py -s "import re" "re.compile('whatnot').search('some long string containing a whatnot')"
100000 loops, best of 3: 3.76 usec per loop
C:\Python25>python lib/timeit.py -s "import re" "re.search('whatnot', 'some long string containing a whatnot')"
100000 loops, best of 3: 3.98 usec per loop
So it looks like it takes a couple of microseconds overhead if you
don't pre-compile the regular expression. That could be significant
if you have simple matches as above, or irrelevant if the match is
complex and slow.
You can also try measuring the compile time separately:
C:\Python25>python lib/timeit.py -s "import re" "re.compile('whatnot')"
100000 loops, best of 3: 2.36 usec per loop
C:\Python25>python lib/timeit.py -s "import re" "re.compile('<(?:p|div)[^>]*>(?P<pat0>(?:(?P<atag0>\\<a[^>]*\\>)\\<img[^>]+class\\s*=[^=>]*captioned[^>]+\\>\\</a\\>)|\\<img[^>]+class\\s*=[^=>]*captioned[^>]+\\>)</(?:p|div)>|(?P<pat1>(?:(?P<atag1>\\<a[^>]*\\>)\\<img[^>]+class\\s*=[^=>]*captioned[^>]+\\>\\</a\\>)|\\<img[^>]+class\\s*=[^=>]*captioned[^>]+\\>)')"
100000 loops, best of 3: 2.34 usec per loop
It makes no difference whether you use a trivial regular expression
or a complex one: Python remembers (if I remember correctly) the last
100 expressions it compiled,so the compilation overhead will be pretty
constant.
Steven D'Aprano
> 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])
>
> ? Or is Python clever enough to see that the value of s will be the same
> on every call, and thus only compile it once?
Let's find out:
>>> import re
>>> import dis
>>>
>>> def spam(x):
... s = re.compile('nobody expects the Spanish Inquisition!')
... return s.search(x)
...
>>> dis.dis(spam)
2 0 LOAD_GLOBAL 0 (re)
3 LOAD_ATTR 1 (compile)
6 LOAD_CONST 1 ('nobody expects the Spanish
Inquisition!')
9 CALL_FUNCTION 1
12 STORE_FAST 1 (s)
3 15 LOAD_FAST 1 (s)
18 LOAD_ATTR 2 (search)
21 LOAD_FAST 0 (x)
24 CALL_FUNCTION 1
27 RETURN_VALUE
No, the Python compiler doesn't know anything about regular expression
objects, so it compiles a call to the RE engine which is executed every
time the function is called.
However, the re module keeps its own cache, so in fact the regular
expression itself may only get compiled once regardless.
Here's another approach that avoids the use of a global variable for the
regular expression:
>>> def spam2(x, s=re.compile('nobody expects the Spanish Inquisition!')):
... return s.search(x)
...
>>> dis.dis(spam2)
2 0 LOAD_FAST 1 (s)
3 LOAD_ATTR 0 (search)
6 LOAD_FAST 0 (x)
9 CALL_FUNCTION 1
12 RETURN_VALUE
What happens now is that the regex is compiled by the RE engine once, at
Python-compile time, then stored as the default value for the argument s.
If you don't supply another value for s when you call the function, the
default regex is used. If you do, the over-ridden value is used instead:
>>> spam2("nothing")
>>> spam2("nothing", re.compile('thing'))
<_sre.SRE_Match object at 0xb7c29c28>
I suspect that this will be not only the fastest solution, but also the
most flexible.
--
Steven
Dustan
> t3 = re.compile('whatever').search
Ack! No! Too Pythonic! GETITOFF! GETITOFF!!
Terry Reedy
"Steven D'Aprano" <st...@REMOVE-THIS-cybersource.com.au> wrote in message
news:13mq041...@corp.supernews.com...
| >>> def spam2(x, s=re.compile('nobody expects the Spanish
Inquisition!')):
| ... return s.search(x)
|
| most flexible.
'Most flexible' in a different way is
def searcher(rex):
crex = re.compile(rex)
def _(txt):
return crex.search(txt)
return _
One can then create and keep around multiple searchers based on different
patterns, to be used as needed.
tjr
John Machin
> "Steven D'Aprano" <st...@REMOVE-THIS-cybersource.com.au> wrote in message
>
> news:13mq041...@corp.supernews.com...
> | >>> def spam2(x, s=re.compile('nobody expects the Spanish
> Inquisition!')):
> | ... return s.search(x)
> |
> | I suspect that this will be not only the fastest solution, but also the
> | most flexible.
>
> 'Most flexible' in a different way is
>
> def searcher(rex):
> crex = re.compile(rex)
> def _(txt):
> return crex.search(txt)
> return _
>
I see your obfuscatory ante and raise you several dots and
underscores:
class Searcher(object):
def __init__(self, rex):
self.crex = re.compile(rex)
def __call__(self, txt):
return self.crex.search(txt)
Cheers,
John
Terry Reedy
"John Machin" <sjma...@lexicon.net> wrote in message
news:ab88db50-ce4e-4298...@e25g2000prg.googlegroups.com...
| On Dec 23, 5:38 am, "Terry Reedy" <tjre...@udel.edu> wrote:
| > 'Most flexible' in a different way is
| >
| > def searcher(rex):
| > crex = re.compile(rex)
| > def _(txt):
| > return crex.search(txt)
| > return _
| >
|
| I see your obfuscatory ante and raise you several dots and
| underscores:
I will presume you are merely joking, but for the benefit of any beginning
programmers reading this, the closure above is a standard functional idiom
for partial evaluation of a function (in this this, re.search(crex,txt))
| class Searcher(object):
| def __init__(self, rex):
| self.crex = re.compile(rex)
| def __call__(self, txt):
| return self.crex.search(txt)
while this is, the equivalent OO version. Intermdiate Python programmers
should know both.
tjr
John Machin
> "John Machin" <sjmac...@lexicon.net> wrote in message
Semi-joking; I thought that your offering of this:
def searcher(rex):
crex = re.compile(rex)
def _(txt):
return crex.search(txt)
return _
foo_searcher = searcher('foo')
was somewhat over-complicated, and possibly slower than already-
mentioned alternatives. The standard idiom etc etc it may be, but the
OP was interested in getting overhead out of his re searching loop.
Let's trim it a bit.
step 1:
def searcher(rex):
crexs = re.compile(rex).search
def _(txt):
return crexs(txt)
return _
foo_searcher = searcher('foo')
step 2:
def searcher(rex):
return re.compile(rex).search
foo_searcher = searcher('foo')
step 3:
foo_searcher = re.compile('foo').search
HTH,
John
William McBrine
couldn't think of a better one.)
Gabriel Genellina
escribió:
> On Dec 23, 2:39 pm, "Terry Reedy" <tjre...@udel.edu> wrote:
>> I will presume you are merely joking, but for the benefit of any
>> beginning
>> programmers reading this, the closure above is a standard functional
>> idiom
>> for partial evaluation of a function (in this this, re.search(crex,txt))
>
> Semi-joking; I thought that your offering of this:
>
> def searcher(rex):
> crex = re.compile(rex)
> def _(txt):
> return crex.search(txt)
> return _
> foo_searcher = searcher('foo')
>
> was somewhat over-complicated, and possibly slower than already-
> mentioned alternatives. The standard idiom etc etc it may be, but the
> OP was interested in getting overhead out of his re searching loop.
> Let's trim it a bit.
>
> step 1:
> def searcher(rex):
> crexs = re.compile(rex).search
> def _(txt):
> return crexs(txt)
> return _
> foo_searcher = searcher('foo')
>
> step 2:
> def searcher(rex):
> return re.compile(rex).search
> foo_searcher = searcher('foo')
>
> step 3:
> foo_searcher = re.compile('foo').search
Nice derivation! Like the word-stairs game: love -> rove -> rave -> have
-> hate
--
Gabriel Genellina
Matthew Franz
(except to be intentionally obtuse) one would use the def
searcher(rex) pattern which I assure you would call with
searcher(r)(t) right?
- mdf
John Machin
> I get class Searcher(object) but can't for the life of me see why
> (except to be intentionally obtuse) one would use the def
> searcher(rex) pattern which I assure you would call with
> searcher(r)(t) right?
>
The whole point of the thread was performance across multiple searches
for the one pattern. Thus one would NOT do
searcher(r)(t)
each time a search was required; one would do
s = searcher(r)
ONCE, and then do
s(t)
each time ...
Matthew Franz
returning a function thing and (for a while) thought _ was some sort
of spooky special variable.
- mdf
> --
> http://mail.python.org/mailman/listinfo/python-list
>
--
Matthew Franz
http://www.threatmind.net/