在Python中实现preg_match_all

Question

I basically want the same functionality of preg_match_all() from PHP in a Python way.

If I have a regex pattern and a string, is there a way to search the string and get back a dictionary of each occurrence of a vowel, along with its position in the string?

Example:

s = "supercalifragilisticexpialidocious"

Would return:

{
  'u' : 1,
  'e' : 3,
  'a' : 6,
  'i' : 8,
  'a' : 11,
  'i' : 13,
  'i' : 15
}

Answer 1

You can do this faster without regexp

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

Here are some timings:
For 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

For 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

Answer 2

What you ask for can't be a dictionary, since it has multiple identical keys. However, you can put it in a list of tuples like this:

>>> [(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)]

Answer 3

Like this, for example:

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")

Note that re.finditer only finds non-overlapping matches. And the dict keys will be overwritten, so if you want the first match, you'll have to replace the innermost loop by:

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