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。

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']

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。

作为一个替代答案，正如我看到之前没有提到的，我将继续引用Python 3文档:

您是应该使用这些模块级函数，还是应该获取模式并自己调用它的方法?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。

对我来说，re.compile的最大好处是能够将正则表达式的定义与其使用分开。

即使是一个简单的表达式，如0|[1-9][0-9]*(以10为基数，不带前导零的整数)，也可能非常复杂，以至于您宁愿不重新输入它，检查是否有任何拼写错误，然后在开始调试时重新检查是否有拼写错误。另外，使用像num或num_b10这样的变量名比0|[1-9][0-9]*更好。

当然可以存储字符串并将它们传递给re.match;然而，这就不那么容易读了:

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

与编译:

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

虽然它很接近，但当重复使用时，第二句的最后一行感觉更自然、更简单。

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