问题描述
我正在通过Doug Hellman的Python标准库示例，并遇到了这个问题：
I'm working through Doug Hellman's "The Python Standard Library by Example" and came across this:
1.3.2编译表达式
re模块级函数，用于将正则表达式作为文本字符串使用，但是对于程序经常使用的表达式进行编译更为高效。
"1.3.2 Compiling Expressionsre includes module-level functions for working with regular expressions as text strings, but it is more efficient to compile the expressions a program uses frequently."
我不能跟随他的解释为什么是这样的情况。他说模块级函数维护编译表达式的缓存，并且由于缓存大小有限，使用编译表达式直接避免了缓存查找开销。
I couldn't follow his explanation for why this is the case. He says that the "module-level functions maintain a cache of compiled expressions" and that since the "size of the cache" is limited, "using compiled expressions directly avoids the cache lookup overhead."
我非常感谢，如果有人可以请解释或指示我一个解释，我可以更好地理解为什么更有效地编译程序常用的正则表达式，以及这个过程如何实际工作。
I'd greatly appreciate it if someone could please explain or direct me to an explanation that I could better understand for why it is more efficient to compile the regular expressions a program uses frequently, and how this process actually works.
推荐答案
Hm。这很奇怪。我的知识迄今为止（从其他来源获得，） ）建议我的初始答案：
Hm. This is strange. My knowledge so far (gained, among other source, from this question) suggested my initial answer:
Python缓存您使用的最后100个正则表达式，因此即使您不明确编译它们，也不必在每次使用时重新编译。
Python caches the last 100 regexes that you used, so even if you don't compile them explicitly, they don't have to be recompiled at every use.
但是，有两个缺点：当达到100个正则表达式的限制时，整个缓存被拒绝，因此如果在一行中使用101个不同的正则表达式，每次都将重新编译。
However, there are two drawbacks: When the limit of 100 regexes is reached, the entire cache is nuked, so if you use 101 different regexes in a row, each one will be recompiled every time. Well, that's rather unlikely, but still.
其次，为了找出regex是否已经编译，解释器需要在缓存中查找正则表达式每次都需要一点额外的时间（但不是很多，因为字典查找非常快）。
Second, in order to find out if a regex has been compiled already, the interpreter needs to look up the regex in the cache every time which does take a little extra time (but not much since dictionary lookups are very fast).
因此，如果您明确编译正则表达式，就可以避免这个额外的查找步骤。
So, if you explicitly compile your regexes, you avoid this extra lookup step.
我只是做了一些测试（Python 3.3）：
I just did some testing (Python 3.3):
>>> import timeit
>>> timeit.timeit(setup="import re", stmt='''r=re.compile(r"\w+")\nfor i in range(10):\n r.search(" jkdhf ")''')
18.547793477671938
>>> timeit.timeit(setup="import re", stmt='''for i in range(10):\n re.search(r"\w+"," jkdhf ")''')
106.47892003890324
所以看来没有进行缓存。也许这是 timeit.timeit（）运行的特殊条件的一个怪癖？
So it would appear that no caching is being done. Perhaps that's a quirk of the special conditions under which timeit.timeit() runs?
另一方面，在Python 2.7中，区别不是很明显：
On the other hand, in Python 2.7, the difference is not as noticeable:
>>> import timeit
>>> timeit.timeit(setup="import re", stmt='''r=re.compile(r"\w+")\nfor i in range(10):\n r.search(" jkdhf ")''')
7.248294908492429
>>> timeit.timeit(setup="import re", stmt='''for i in range(10):\n re.search(r"\w+"," jkdhf ")''')
18.26713670282241
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