Is it worth using Python's re.compile?

Question

I've had a lot of experience running a compiled regex 1000s of times versus compiling on-the-fly, and have not noticed any perceivable difference. Obviously, this is anecdotal, and certainly not a great argument against compiling, but I've found the difference to be negligible.

EDIT: After a quick glance at the actual Python 2.5 library code, I see that Python internally compiles AND CACHES regexes whenever you use them anyway (including calls to re.match()), so you're really only changing WHEN the regex gets compiled, and shouldn't be saving much time at all - only the time it takes to check the cache (a key lookup on an internal dict type).

From module re.py (comments are mine):

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

def _compile(*key):

    # Does cache check at top of function
    cachekey = (type(key[0]),) + key
    p = _cache.get(cachekey)
    if p is not None: return p

    # ...
    # Does actual compilation on cache miss
    # ...

    # Caches compiled regex
    if len(_cache) >= _MAXCACHE:
        _cache.clear()
    _cache[cachekey] = p
    return p

I still often pre-compile regular expressions, but only to bind them to a nice, reusable name, not for any expected performance gain.

For me, the biggest benefit to re.compile is being able to separate definition of the regex from its use.

Even a simple expression such as 0|[1-9][0-9]* (integer in base 10 without leading zeros) can be complex enough that you'd rather not have to retype it, check if you made any typos, and later have to recheck if there are typos when you start debugging. Plus, it's nicer to use a variable name such as num or num_b10 than 0|[1-9][0-9]*.

It's certainly possible to store strings and pass them to re.match; however, that's less readable:

num = "..."
# then, much later:
m = re.match(num, input)

Versus compiling:

num = re.compile("...")
# then, much later:
m = num.match(input)

Though it is fairly close, the last line of the second feels more natural and simpler when used repeatedly.

FWIW:

$ python -m timeit -s "import re" "re.match('hello', 'hello world')"
100000 loops, best of 3: 3.82 usec per loop

$ python -m timeit -s "import re; h=re.compile('hello')" "h.match('hello world')"
1000000 loops, best of 3: 1.26 usec per loop

so, if you're going to be using the same regex a lot, it may be worth it to do re.compile (especially for more complex regexes).

The standard arguments against premature optimization apply, but I don't think you really lose much clarity/straightforwardness by using re.compile if you suspect that your regexps may become a performance bottleneck.

Update:

Under Python 3.6 (I suspect the above timings were done using Python 2.x) and 2018 hardware (MacBook Pro), I now get the following timings:

% python -m timeit -s "import re" "re.match('hello', 'hello world')"
1000000 loops, best of 3: 0.661 usec per loop

% python -m timeit -s "import re; h=re.compile('hello')" "h.match('hello world')"
1000000 loops, best of 3: 0.285 usec per loop

% python -m timeit -s "import re" "h=re.compile('hello'); h.match('hello world')"
1000000 loops, best of 3: 0.65 usec per loop

% python --version
Python 3.6.5 :: Anaconda, Inc.

I also added a case (notice the quotation mark differences between the last two runs) that shows that re.match(x, ...) is literally [roughly] equivalent to re.compile(x).match(...), i.e. no behind-the-scenes caching of the compiled representation seems to happen.

Here's a simple test case:

~$ for x in 1 10 100 1000 10000 100000 1000000; do python -m timeit -n $x -s 'import re' 're.match("[0-9]{3}-[0-9]{3}-[0-9]{4}", "123-123-1234")'; done
1 loops, best of 3: 3.1 usec per loop
10 loops, best of 3: 2.41 usec per loop
100 loops, best of 3: 2.24 usec per loop
1000 loops, best of 3: 2.21 usec per loop
10000 loops, best of 3: 2.23 usec per loop
100000 loops, best of 3: 2.24 usec per loop
1000000 loops, best of 3: 2.31 usec per loop

with re.compile:

~$ for x in 1 10 100 1000 10000 100000 1000000; do python -m timeit -n $x -s 'import re' 'r = re.compile("[0-9]{3}-[0-9]{3}-[0-9]{4}")' 'r.match("123-123-1234")'; done
1 loops, best of 3: 1.91 usec per loop
10 loops, best of 3: 0.691 usec per loop
100 loops, best of 3: 0.701 usec per loop
1000 loops, best of 3: 0.684 usec per loop
10000 loops, best of 3: 0.682 usec per loop
100000 loops, best of 3: 0.694 usec per loop
1000000 loops, best of 3: 0.702 usec per loop

So, it would seem to compiling is faster with this simple case, even if you only match once.

I just tried this myself. For the simple case of parsing a number out of a string and summing it, using a compiled regular expression object is about twice as fast as using the re methods.

As others have pointed out, the re methods (including re.compile) look up the regular expression string in a cache of previously compiled expressions. Therefore, in the normal case, the extra cost of using the re methods is simply the cost of the cache lookup.

However, examination of the code, shows the cache is limited to 100 expressions. This begs the question, how painful is it to overflow the cache? The code contains an internal interface to the regular expression compiler, re.sre_compile.compile. If we call it, we bypass the cache. It turns out to be about two orders of magnitude slower for a basic regular expression, such as r'\w+\s+([0-9_]+)\s+\w*'.

Here's my test:

#!/usr/bin/env python
import re
import time

def timed(func):
    def wrapper(*args):
        t = time.time()
        result = func(*args)
        t = time.time() - t
        print '%s took %.3f seconds.' % (func.func_name, t)
        return result
    return wrapper

regularExpression = r'\w+\s+([0-9_]+)\s+\w*'
testString = "average    2 never"

@timed
def noncompiled():
    a = 0
    for x in xrange(1000000):
        m = re.match(regularExpression, testString)
        a += int(m.group(1))
    return a

@timed
def compiled():
    a = 0
    rgx = re.compile(regularExpression)
    for x in xrange(1000000):
        m = rgx.match(testString)
        a += int(m.group(1))
    return a

@timed
def reallyCompiled():
    a = 0
    rgx = re.sre_compile.compile(regularExpression)
    for x in xrange(1000000):
        m = rgx.match(testString)
        a += int(m.group(1))
    return a


@timed
def compiledInLoop():
    a = 0
    for x in xrange(1000000):
        rgx = re.compile(regularExpression)
        m = rgx.match(testString)
        a += int(m.group(1))
    return a

@timed
def reallyCompiledInLoop():
    a = 0
    for x in xrange(10000):
        rgx = re.sre_compile.compile(regularExpression)
        m = rgx.match(testString)
        a += int(m.group(1))
    return a

r1 = noncompiled()
r2 = compiled()
r3 = reallyCompiled()
r4 = compiledInLoop()
r5 = reallyCompiledInLoop()
print "r1 = ", r1
print "r2 = ", r2
print "r3 = ", r3
print "r4 = ", r4
print "r5 = ", r5
</pre>
And here is the output on my machine:
<pre>
$ regexTest.py 
noncompiled took 4.555 seconds.
compiled took 2.323 seconds.
reallyCompiled took 2.325 seconds.
compiledInLoop took 4.620 seconds.
reallyCompiledInLoop took 4.074 seconds.
r1 =  2000000
r2 =  2000000
r3 =  2000000
r4 =  2000000
r5 =  20000

The 'reallyCompiled' methods use the internal interface, which bypasses the cache. Note the one that compiles on each loop iteration is only iterated 10,000 times, not one million.