Regular expression to match outer brackets

Question

I need a regular expression to select all the text between two outer brackets.

Example: some text(text here(possible text)text(possible text(more text)))end text

Result: (text here(possible text)text(possible text(more text)))

Answer 1

I want to add this answer for quickreference. Feel free to update.

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.NET Regex using balancing groups.

\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\)

Where c is used as the depth counter.

Demo at Regexstorm.com

• Stack Overflow: Using RegEx to balance match parenthesis
• Wes' Puzzling Blog: Matching Balanced Constructs with .NET Regular Expressions
• Greg Reinacker's Weblog: Nested Constructs in Regular Expressions

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PCRE using a recursive pattern.

\((?>[^)(]+|(?R))*+\)

Demo at regex101; Or without alternation:

\((?>[^)(]*(?R)?)*+\)

Demo at regex101; Or unrolled for performance:

\([^)(]*(?:(?R)[^)(]*)*+\)

Demo at regex101; The pattern is pasted at (?R) which represents (?0).

Perl, PHP, Notepad++, R: perl=TRUE, Python: Regex package with (?V1) for Perl behaviour.

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Ruby using subexpression calls.

With Ruby 2.0 \g<0> can be used to call full pattern.

\((?>[^)(]+|\g<0>)*\)

Demo at Rubular; Ruby 1.9 only supports capturing group recursion:

(\((?>[^)(]+|\g<1>)*\))

Demo at Rubular  (atomic grouping since Ruby 1.9.3)

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JavaScript  API :: XRegExp.matchRecursive

XRegExp.matchRecursive(str, '\\(', '\\)', 'g');

JS, Java and other regex flavors without recursion up to 2 levels of nesting:

\((?:[^)(]+|\((?:[^)(]+|\([^)(]*\))*\))*\)

Demo at regex101. Deeper nesting needs to be added to pattern.
To fail faster on unbalanced parenthesis drop the + quantifier.

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Java: An interesting idea using forward references by @jaytea.

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_{Reference - What does this regex mean?}

• rexegg.com - Recursive Regular Expressions
• Regular-Expressions.info - Regular Expression Recursion

Answer 2

"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.

This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns.  This is where the re package greatly
assists in parsing. 
"""

import re


# The pattern below recognises a sequence consisting of:
#    1. Any characters not in the set of open/close strings.
#    2. One of the open/close strings.
#    3. The remainder of the string.
# 
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included.  However quotes are not ignored inside
# quotes.  More logic is needed for that....


pat = re.compile("""
    ( .*? )
    ( \( | \) | \[ | \] | \{ | \} | \< | \> |
                           \' | \" | BEGIN | END | $ )
    ( .* )
    """, re.X)

# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.

matching = { "(" : ")",
             "[" : "]",
             "{" : "}",
             "<" : ">",
             '"' : '"',
             "'" : "'",
             "BEGIN" : "END" }

# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.

def matchnested(s, term=""):
    lst = []
    while True:
        m = pat.match(s)

        if m.group(1) != "":
            lst.append(m.group(1))

        if m.group(2) == term:
            return lst, m.group(3)

        if m.group(2) in matching:
            item, s = matchnested(m.group(3), matching[m.group(2)])
            lst.append(m.group(2))
            lst.append(item)
            lst.append(matching[m.group(2)])
        else:
            raise ValueError("After <<%s %s>> expected %s not %s" %
                             (lst, s, term, m.group(2)))

# Unit test.

if __name__ == "__main__":
    for s in ("simple string",
              """ "double quote" """,
              """ 'single quote' """,
              "one'two'three'four'five'six'seven",
              "one(two(three(four)five)six)seven",
              "one(two(three)four)five(six(seven)eight)nine",
              "one(two)three[four]five{six}seven<eight>nine",
              "one(two[three{four<five>six}seven]eight)nine",
              "oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
              "ERROR testing ((( mismatched ))] parens"):
        print "\ninput", s
        try:
            lst, s = matchnested(s)
            print "output", lst
        except ValueError as e:
            print str(e)
    print "done"

Answer 3

It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.

You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.

Angle brackets <> were used because they do not require escaping.

The regular expression looks like this:

<
[^<>]*
(
    (
        (?<Open><)
        [^<>]*
    )+
    (
        (?<Close-Open>>)
        [^<>]*
    )+
)*
(?(Open)(?!))
>

Answer 4

I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing

balanced.matches({
    source: source,
    open: '(',
    close: ')'
});

You can even do replacements:

balanced.replacements({
    source: source,
    open: '(',
    close: ')',
    replace: function (source, head, tail) {
        return head + source + tail;
    }
});

Here's a more complex and interactive example JSFiddle.

Answer 5

This is the definitive regex:

\(
(?<arguments> 
(  
  ([^\(\)']*) |  
  (\([^\(\)']*\)) |
  '(.*?)'

)*
)
\)

Example:

input: ( arg1, arg2, arg3, (arg4), '(pip' )

output: arg1, arg2, arg3, (arg4), '(pip'

note that the '(pip' is correctly managed as string. (tried in regulator: http://sourceforge.net/projects/regulator/)

Answer 6

This one also worked

re.findall(r'\(.+\)', s)