在python中实现preg_match_all

你要的不能是字典,因为它有多个相同的键。但是,您可以将它放入这样的元组列表中:

>>> [(m.group(0), m.start()) for m in re.finditer('[aeiou]',s)]
[('u', 1), ('e', 3), ('a', 6), ('i', 8), ('a', 11), ('i', 13), ('i', 15), ('i', 18), ('e', 20), ('i', 23), ('a', 24), ('i', 26), ('o', 28), ('i', 30), ('o', 31), ('u', 32)]

你可以做到这一点 更快 没有正则表达式

[(x,i) for i,x in enumerate(s) if x in "aeiou"]

以下是一些时间安排:
为了 s = "supercalifragilisticexpialidocious"

timeit [(m.group(0), m.start()) for m in re.finditer('[aeiou]',s)]
10000 loops, best of 3: 27.5 µs per loop

timeit [(x,i) for i,x in enumerate(s) if x in "aeiou"]
100000 loops, best of 3: 14.4 µs per loop

为了 s = "supercalifragilisticexpialidocious"*100

timeit [(m.group(0), m.start()) for m in re.finditer('[aeiou]',s)]
100 loops, best of 3: 2.01 ms per loop

timeit [(x,i) for i,x in enumerate(s) if x in "aeiou"]
1000 loops, best of 3: 1.24 ms per loop

例如:

import re

def findall(pattern, string):
    res = {}
    for match in re.finditer(pattern, string):
        res[match.group(0)] = match.start()
    return res

print findall("[aeiou]", "Test this thang")

注意 re.finditer 只找到不重叠的匹配项。并且dict键将被覆盖,因此如果您想要第一个匹配项,则必须将最里面的循环替换为:

    for match in re.finditer(pattern, string):
        if match.group(0) not in res: # <-- don't overwrite key
            res[match.group(0)] = match.start()