Python Regex slower than expected

Question

I read a cool article on how to avoid creating slow regular expressions. Generally speaking it looks like the longer and more explicit and regex is the faster it will complete. A greedy regex can be exponentially slower.

I thought I would test this out by measuring the time it takes to complete a more complex/explicit statement with a less complex/greedy statement. For the most part everything seems to hold true, but I have one greedy statement that clocked in slower. Here are two examples:

import re
from timeit import timeit

# This works as expected, the explicit is faster than the greedy.
# http_x_real_ip explicit 
print(timeit(setup="import re", stmt='''r = re.search(r'(\d{1,3}\.\d{1,3}.\d{1,3}.\d{1,3})', '192.168.1.1 999.999.999.999')''', number=1000000))
1.159849308001867

# http_x_real_ip greedy
print(timeit(setup="import re", stmt='''r = re.search(r'((?:\d{1,3}\.){3}\d{1,3})', '192.168.1.1 999.999.999.999')''', number=1000000))
1.7421739230003368

# This does not work as expected, greedy is faster.
# time_local explicit
print(timeit(setup="import re", stmt='''r = re.search(r'(\d{1,2}/\w{3}/[2][0]\d{2}:\d{2}:\d{2}:\d{2}\s[+][0]{4})', "[23/Jun/2015:11:10:57 +0000]")''', number=1000000))
1.248802040994633

# time_local greedy
print(timeit(setup="import re", stmt='''r = re.search(r'\[(.*)\]', "[23/Jun/2015:11:10:57 +0000]")''', number=1000000))
1.0256699790043058

Is the local_time explict regex just poorly written?

Answer 1

The more a regular expression has to backtrack, the slower it is.

_{This might not hold for very small input data. However, who would care about the performance on small data? :D}

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This topic is well covered in this article:

• http://www.regular-expressions.info/catastrophic.html

Also there are interesting contributions in this question:

• Greedy vs. Reluctant vs. Possessive Quantifiers