使用re.compile()还有一个额外的好处，即使用re.VERBOSE向正则表达式模式添加注释

pattern = '''
hello[ ]world    # Some info on my pattern logic. [ ] to recognize space
'''

re.search(pattern, 'hello world', re.VERBOSE)

虽然这不会影响代码的运行速度，但我喜欢这样做，因为这是我注释习惯的一部分。当我想要修改代码时，我完全不喜欢花时间去记住代码背后的逻辑。

有趣的是，编译对我来说确实更有效(Win XP上的Python 2.5.2):

import re
import time

rgx = re.compile('(\w+)\s+[0-9_]?\s+\w*')
str = "average    2 never"
a = 0

t = time.time()

for i in xrange(1000000):
    if re.match('(\w+)\s+[0-9_]?\s+\w*', str):
    #~ if rgx.match(str):
        a += 1

print time.time() - t

按原样运行上述代码一次，并以相反的方式运行两个if行，编译后的正则表达式的速度将提高一倍

我自己刚试过。对于从字符串中解析数字并对其求和的简单情况，使用编译后的正则表达式对象的速度大约是使用re方法的两倍。

正如其他人指出的那样，re方法(包括re.compile)在以前编译的表达式缓存中查找正则表达式字符串。因此，在正常情况下，使用re方法的额外成本只是缓存查找的成本。

然而，检查代码，缓存被限制为100个表达式。这就引出了一个问题，缓存溢出有多痛苦?该代码包含正则表达式编译器的内部接口re.sre_compile.compile。如果我们调用它，就绕过了缓存。结果表明，对于一个基本的正则表达式，例如r'\w+\s+([0-9_]+)\s+\w*'，它要慢两个数量级。

下面是我的测试:

#!/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

'reallyCompiled'方法使用内部接口，绕过缓存。注意，在每个循环迭代中编译的代码只迭代了10,000次，而不是一百万次。

一般来说，我发现在编译模式时使用标志比内联使用标志更容易(至少更容易记住如何使用)，比如re.I。

>>> foo_pat = re.compile('foo',re.I)
>>> foo_pat.findall('some string FoO bar')
['FoO']

vs

>>> re.findall('(?i)foo','some string FoO bar')
['FoO']

大多数情况下，是否使用re.compile没有什么区别。在内部，所有函数都是按照编译步骤实现的:

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

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

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

def sub(pattern, repl, string, count=0, flags=0):
    return _compile(pattern, flags).sub(repl, string, count)

def subn(pattern, repl, string, count=0, flags=0):
    return _compile(pattern, flags).subn(repl, string, count)

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

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

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

此外，re.compile()绕过了额外的间接和缓存逻辑:

_cache = {}

_pattern_type = type(sre_compile.compile("", 0))

_MAXCACHE = 512
def _compile(pattern, flags):
    # internal: compile pattern
    try:
        p, loc = _cache[type(pattern), pattern, flags]
        if loc is None or loc == _locale.setlocale(_locale.LC_CTYPE):
            return p
    except KeyError:
        pass
    if isinstance(pattern, _pattern_type):
        if flags:
            raise ValueError(
                "cannot process flags argument with a compiled pattern")
        return pattern
    if not sre_compile.isstring(pattern):
        raise TypeError("first argument must be string or compiled pattern")
    p = sre_compile.compile(pattern, flags)
    if not (flags & DEBUG):
        if len(_cache) >= _MAXCACHE:
            _cache.clear()
        if p.flags & LOCALE:
            if not _locale:
                return p
            loc = _locale.setlocale(_locale.LC_CTYPE)
        else:
            loc = None
        _cache[type(pattern), pattern, flags] = p, loc
    return p

除了使用re.compile带来的小速度好处外，人们还喜欢命名潜在复杂的模式规范并将其与应用的业务逻辑分离所带来的可读性:

#### Patterns ############################################################
number_pattern = re.compile(r'\d+(\.\d*)?')    # Integer or decimal number
assign_pattern = re.compile(r':=')             # Assignment operator
identifier_pattern = re.compile(r'[A-Za-z]+')  # Identifiers
whitespace_pattern = re.compile(r'[\t ]+')     # Spaces and tabs

#### Applications ########################################################

if whitespace_pattern.match(s): business_logic_rule_1()
if assign_pattern.match(s): business_logic_rule_2()

注意，另一位受访者错误地认为pyc文件直接存储已编译的模式;然而，在现实中，每次PYC加载时，它们都会被重新构建:

>>> from dis import dis
>>> with open('tmp.pyc', 'rb') as f:
        f.read(8)
        dis(marshal.load(f))

  1           0 LOAD_CONST               0 (-1)
              3 LOAD_CONST               1 (None)
              6 IMPORT_NAME              0 (re)
              9 STORE_NAME               0 (re)

  3          12 LOAD_NAME                0 (re)
             15 LOAD_ATTR                1 (compile)
             18 LOAD_CONST               2 ('[aeiou]{2,5}')
             21 CALL_FUNCTION            1
             24 STORE_NAME               2 (lc_vowels)
             27 LOAD_CONST               1 (None)
             30 RETURN_VALUE

上面的分解来自于一个包含tmp.py的PYC文件:

import re
lc_vowels = re.compile(r'[aeiou]{2,5}')

下面是一个简单的测试用例:

~$ 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

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

因此，这种简单的情况下编译似乎更快，即使只匹配一次。