Match first instance of Python regex search

Question

I'm looking to the first instance of a match two square brackets using regular expressions. Currently, I am doing

regex = re.compile("(?<=(\[\[)).*(?=\]\])")
r = regex.search(line)

which works for lines like

[[string]]

returns string

but when I try it on a separate line:

[[string]] ([[string2]], [[string3]])

The result is

string]] ([[string2]], [[string3

What am I missing?

score 51 · Accepted Answer · edited Oct 19 '17 at 21:23

51

Python *, +, ? and {n,m} quantifiers are greedy by default

Patterns quantified with the above quantifiers will match as much as they can by default. In your case, this means the first set of brackets and the last. In Python, you can make any quantifier non-greedy (or "lazy") by adding a ? after it. In your case, this would translate to .*? in the middle part of your expression.

edited Oct 19 '17 at 21:23

Wiktor Stribiżew

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answered Feb 22 '11 at 05:23

Drew

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Minor nitpick. *regular expressions* are not fundamentally greedy or non-greedy. The python `re` package provides a greedy, recursive backtracking implementation of regular expressions. (+1 for right answer, though) – SingleNegationElimination Feb 22 '11 at 05:31

Match first instance of Python regex search

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