在Python中对正则表达式使用compile有什么好处吗?
h = re.compile('hello')
h.match('hello world')
vs
re.match('hello', 'hello world')
在Python中对正则表达式使用compile有什么好处吗?
h = re.compile('hello')
h.match('hello world')
vs
re.match('hello', 'hello world')
当前回答
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。
其他回答
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.
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。
我有很多运行编译过的regex 1000的经验 与实时编译相比,并没有注意到 任何可感知的差异
对已接受答案的投票导致假设@Triptych所说的对所有情况都是正确的。这并不一定是真的。一个很大的区别是当你必须决定是接受一个正则表达式字符串还是一个编译过的正则表达式对象作为函数的参数时:
>>> timeit.timeit(setup="""
... import re
... f=lambda x, y: x.match(y) # accepts compiled regex as parameter
... h=re.compile('hello')
... """, stmt="f(h, 'hello world')")
0.32881879806518555
>>> timeit.timeit(setup="""
... import re
... f=lambda x, y: re.compile(x).match(y) # compiles when called
... """, stmt="f('hello', 'hello world')")
0.809190034866333
编译正则表达式总是更好的,以防需要重用它们。
请注意,上面timeit中的示例模拟在导入时一次创建已编译的regex对象,而不是在需要匹配时“动态”创建。
这是个好问题。你经常看到人们毫无理由地使用re.compile。它降低了可读性。但是可以肯定的是,很多时候需要预编译表达式。就像你在循环中重复使用它一样。
这就像编程的一切(实际上是生活中的一切)。运用常识。
在无意中看到这里的讨论之前,我运行了这个测试。然而,在运行它之后,我想我至少会发布我的结果。
我剽窃了Jeff Friedl的“精通正则表达式”中的例子。这是在一台运行OSX 10.6 (2Ghz英特尔酷睿2双核,4GB内存)的macbook上。Python版本为2.6.1。
运行1 -使用re.compile
import re
import time
import fpformat
Regex1 = re.compile('^(a|b|c|d|e|f|g)+$')
Regex2 = re.compile('^[a-g]+$')
TimesToDo = 1000
TestString = ""
for i in range(1000):
TestString += "abababdedfg"
StartTime = time.time()
for i in range(TimesToDo):
Regex1.search(TestString)
Seconds = time.time() - StartTime
print "Alternation takes " + fpformat.fix(Seconds,3) + " seconds"
StartTime = time.time()
for i in range(TimesToDo):
Regex2.search(TestString)
Seconds = time.time() - StartTime
print "Character Class takes " + fpformat.fix(Seconds,3) + " seconds"
Alternation takes 2.299 seconds
Character Class takes 0.107 seconds
运行2 -不使用re.compile
import re
import time
import fpformat
TimesToDo = 1000
TestString = ""
for i in range(1000):
TestString += "abababdedfg"
StartTime = time.time()
for i in range(TimesToDo):
re.search('^(a|b|c|d|e|f|g)+$',TestString)
Seconds = time.time() - StartTime
print "Alternation takes " + fpformat.fix(Seconds,3) + " seconds"
StartTime = time.time()
for i in range(TimesToDo):
re.search('^[a-g]+$',TestString)
Seconds = time.time() - StartTime
print "Character Class takes " + fpformat.fix(Seconds,3) + " seconds"
Alternation takes 2.508 seconds
Character Class takes 0.109 seconds