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?

Wiktor Stribiżew · Accepted Answer · 2017-10-19 21:23:55Z

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, 2017 at 21:23

Wiktor Stribiżew

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answered Feb 22, 2011 at 5:23

Drew

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1 Comment

SingleNegationElimination Over a year ago

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)

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