Python regex speed - Greedy vs. non-greedy

Question

I am making several regex substitutions in Python along the lines of

  \w\s+\w

over many large documents. Obviously if I make the regex non-greedy (with a ?) it won't change what it matches (as \w != \s) but will it make the code run any faster? In other words, with non-greedy regexes does Python work its way from the first character matched onwards rather than from the end of the document back to that character, or is this a naive view?

Measure it. Most likely not - but measure. – Emil Ivanov Nov 30 '12 at 19:10 — Emil Ivanov, Nov 30 '12 at 19:10

score 2 · Accepted Answer · answered Nov 30 '12 at 19:44

Is this the pattern you implied?

In [15]: s = 'some text   with \tspaces  between'

In [16]: timeit re.sub(r'(\w)(\s+)(\w)', '\\1 \\3', s)
10000 loops, best of 3: 30.5 us per loop

In [17]: timeit re.sub(r'(\w)(\s+?)(\w)', '\\1 \\3', s)
10000 loops, best of 3: 24.9 us per loop

Seems to be a pretty small difference here. Only 5 microseconds with the non-greedy,

Using a 500 word lorem-ipsum, with multiple mixed whitespace between every word, I get an 8 ms difference.

Python regex speed - Greedy vs. non-greedy

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