用下面的例子:

h = re.compile('hello')
h.match('hello world')

上面例子中的匹配方法和下面的不一样:

re.match('hello', 'hello world')

Re.compile()返回一个正则表达式对象，这意味着h是一个正则表达式对象。

regex对象有自己的匹配方法，带有可选的pos和endpos参数:

的。匹配(字符串[线程][线程]])

pos

可选的第二个参数pos给出了字符串中的一个索引 搜寻就要开始了;缺省值为0。这并不完全是 相当于对字符串进行切片;'^'模式字符匹配于 字符串的真正开始和在a之后的位置 换行符，但不一定在搜索到的索引处 开始。

尾部

可选参数endpos限制了字符串的长度 搜索;这就好像字符串有endpos个字符那么长 只搜索从pos到endpos - 1的字符 匹配。如果endpos小于pos，则找不到匹配;否则, 如果rx是编译后的正则表达式对象，则rx。搜索(字符串,0, 50)等于rx。搜索(字符串(:50),0)。

regex对象的search、findall和finditer方法也支持这些参数。

Re.match (pattern, string, flags=0)不支持，如你所见， 它的search、findall和finditer也没有。

match对象具有补充这些参数的属性:

match.pos

的search()或match()方法传递的pos的值 一个正则表达式对象。这是正则表达式所在字符串的索引 引擎开始寻找匹配。

match.endpos

传递给search()或match()方法的endpos值 正则表达式对象的。对象超出的字符串的索引 RE引擎不会去。

---

一个regex对象有两个唯一的，可能有用的属性:

regex.groups

模式中捕获组的数量。

regex.groupindex

将(?P)定义的任何符号组名映射到的字典 组数字。如果没有使用符号组，则字典为空 在模式中。

---

最后，match对象有这个属性:

match.re

其match()或search()方法的正则表达式对象 生成此匹配实例。

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。

下面是一个使用re.compile的示例，在请求时速度超过50倍。

这一点与我在上面的评论中所说的是一样的，即当您的使用从编译缓存中获益不多时，使用re.compile可能是一个显著的优势。这种情况至少发生在一个特定的情况下(我在实践中遇到过)，即当以下所有情况都成立时:

您有很多regex模式(不仅仅是re._MAXCACHE，它目前的默认值是512)，以及 你经常使用这些正则表达式，而且 相同模式的连续使用之间被多个re._MAXCACHE其他正则表达式分隔，因此每个正则表达式在连续使用之间从缓存中刷新。

import re
import time

def setup(N=1000):
    # Patterns 'a.*a', 'a.*b', ..., 'z.*z'
    patterns = [chr(i) + '.*' + chr(j)
                    for i in range(ord('a'), ord('z') + 1)
                    for j in range(ord('a'), ord('z') + 1)]
    # If this assertion below fails, just add more (distinct) patterns.
    # assert(re._MAXCACHE < len(patterns))
    # N strings. Increase N for larger effect.
    strings = ['abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz'] * N
    return (patterns, strings)

def without_compile():
    print('Without re.compile:')
    patterns, strings = setup()
    print('searching')
    count = 0
    for s in strings:
        for pat in patterns:
            count += bool(re.search(pat, s))
    return count

def without_compile_cache_friendly():
    print('Without re.compile, cache-friendly order:')
    patterns, strings = setup()
    print('searching')
    count = 0
    for pat in patterns:
        for s in strings:
            count += bool(re.search(pat, s))
    return count

def with_compile():
    print('With re.compile:')
    patterns, strings = setup()
    print('compiling')
    compiled = [re.compile(pattern) for pattern in patterns]
    print('searching')
    count = 0
    for s in strings:
        for regex in compiled:
            count += bool(regex.search(s))
    return count

start = time.time()
print(with_compile())
d1 = time.time() - start
print(f'-- That took {d1:.2f} seconds.\n')

start = time.time()
print(without_compile_cache_friendly())
d2 = time.time() - start
print(f'-- That took {d2:.2f} seconds.\n')

start = time.time()
print(without_compile())
d3 = time.time() - start
print(f'-- That took {d3:.2f} seconds.\n')

print(f'Ratio: {d3/d1:.2f}')

我在笔记本电脑上获得的示例输出(Python 3.7.7):

With re.compile:
compiling
searching
676000
-- That took 0.33 seconds.

Without re.compile, cache-friendly order:
searching
676000
-- That took 0.67 seconds.

Without re.compile:
searching
676000
-- That took 23.54 seconds.

Ratio: 70.89

I didn't bother with timeit as the difference is so stark, but I get qualitatively similar numbers each time. Note that even without re.compile, using the same regex multiple times and moving on to the next one wasn't so bad (only about 2 times as slow as with re.compile), but in the other order (looping through many regexes), it is significantly worse, as expected. Also, increasing the cache size works too: simply setting re._MAXCACHE = len(patterns) in setup() above (of course I don't recommend doing such things in production as names with underscores are conventionally “private”) drops the ~23 seconds back down to ~0.7 seconds, which also matches our understanding.

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

Ubuntu 22.04:

$ python --version
Python 3.10.6

$ 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 loop, best of 5: 972 nsec per loop
:0: UserWarning: The test results are likely unreliable. The worst time (186 usec) was more than four times slower than the best time (972 nsec).
10 loops, best of 5: 819 nsec per loop
:0: UserWarning: The test results are likely unreliable. The worst time (13.9 usec) was more than four times slower than the best time (819 nsec).
100 loops, best of 5: 763 nsec per loop
1000 loops, best of 5: 699 nsec per loop
10000 loops, best of 5: 653 nsec per loop
100000 loops, best of 5: 655 nsec per loop
1000000 loops, best of 5: 656 nsec per loop

$ 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 loop, best of 5: 985 nsec per loop
:0: UserWarning: The test results are likely unreliable. The worst time (134 usec) was more than four times slower than the best time (985 nsec).
10 loops, best of 5: 775 nsec per loop
:0: UserWarning: The test results are likely unreliable. The worst time (13.9 usec) was more than four times slower than the best time (775 nsec).
100 loops, best of 5: 756 nsec per loop
1000 loops, best of 5: 701 nsec per loop
10000 loops, best of 5: 704 nsec per loop
100000 loops, best of 5: 654 nsec per loop
1000000 loops, best of 5: 651 nsec per loop

尽管这两种方法在速度方面是可以比较的，但是您应该知道，如果您正在处理数百万次迭代，那么仍然存在一些可以忽略不计的时间差。

以下速度测试:

import re
import time

SIZE = 100_000_000

start = time.time()
foo = re.compile('foo')
[foo.search('bar') for _ in range(SIZE)]
print('compiled:  ', time.time() - start)

start = time.time()
[re.search('foo', 'bar') for _ in range(SIZE)]
print('uncompiled:', time.time() - start)

给出了以下结果:

compiled:   14.647532224655151
uncompiled: 61.483458042144775

编译后的方法在我的PC上(使用Python 3.7.0)始终快大约4倍。

如文档中所述:

如果在循环中访问正则表达式，预编译它将节省一些函数调用。在循环之外，由于内部缓存，没有太大区别。