Do Python regular expressions have an equivalent to Ruby's atomic grouping?

Question

Ruby's regular expressions have a feature called atomic grouping (?>regexp), described here, is there any equivalent in Python's re module?

Answer 1

Python does not directly support this feature, but you can emulate it by using a zero-width lookahead assert ((?=RE)), which matches from the current point with the same semantics you want, putting a named group ((?P<name>RE)) inside the lookahead, and then using a named backreference ((?P=name)) to match exactly whatever the zero-width assertion matched. Combined together, this gives you the same semantics, at the cost of creating an additional matching group, and a lot of syntax.

For example, the link you provided gives the Ruby example of

/"(?>.*)"/.match('"Quote"') #=> nil

We can emulate that in Python as such:

re.search(r'"(?=(?P<tmp>.*))(?P=tmp)"', '"Quote"') # => None

We can show that I'm doing something useful and not just spewing line noise, because if we change it so that the inner group doesn't eat the final ", it still matches:

re.search(r'"(?=(?P<tmp>[A-Za-z]*))(?P=tmp)"', '"Quote"').groupdict()
# => {'tmp': 'Quote'}

You can also use anonymous groups and numeric backreferences, but this gets awfully full of line-noise:

re.search(r'"(?=(.*))1"', '"Quote"') # => None

(Full disclosure: I learned this trick from perl's perlre documentation, which mentions it under the documentation for (?>...).)

In addition to having the right semantics, this also has the appropriate performance properties. If we port an example out of perlre:

[nelhage@anarchique:~/tmp]$ cat re.py
import re
import timeit


re_1 = re.compile(r'''(
                           (
                             [^()]+           # x+
                           |
                             ( [^()]* )
                           )+
                       )
                   ''', re.X)
re_2 = re.compile(r'''(
                           (
                             (?=(?P<tmp>[^()]+ ))(?P=tmp) # Emulate (?> x+)
                           |
                             ( [^()]* )
                           )+
                       )''', re.X)

print timeit.timeit("re_1.search('((()' + 'a' * 25)",
                    setup  = "from __main__ import re_1",
                    number = 10)

print timeit.timeit("re_2.search('((()' + 'a' * 25)",
                    setup  = "from __main__ import re_2",
                    number = 10)

We see a dramatic improvement:

[nelhage@anarchique:~/tmp]$ python re.py
96.0800571442
7.41481781006e-05

Which only gets more dramatic as we extend the length of the search string.