Regular expression to match balanced parentheses

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.

---

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)

---

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.

---

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

---

_{Reference - What does this regex mean?}

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

Answer 2

Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.

But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.

Answer 3

You can use regex recursion:

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

Answer 4

[^\(]*(\(.*\))[^\)]*

[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.

Answer 5

(?<=\().*(?=\))

If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible^(*).

This regex just returns the text between the first opening and the last closing parentheses in your string.

---

^(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.

Answer 6

This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.

---

Regular expressions can not do this.

Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.

In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:

      0     1     1     0
-> S1 -> S2 -> S2 -> S2 ->S1

In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.

In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.

However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.

Answer 7

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 8

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 9

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 10

Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.

Lua

Use %b() (%b{} / %b[] for curly braces / square brackets):

• for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)

Raku (former Perl6):

Non-overlapping multiple balanced parentheses matches:

my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* }
say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/;
# => (｢(a(b)c)｣ ｢((d)f(g))｣)

Overlapping multiple balanced parentheses matches:

say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/;
# => (｢(a(b)c)｣ ｢(b)｣ ｢((d)f(g))｣ ｢(d)｣ ｢(g)｣)

See demo.

Python re non-regex solution

See poke's answer for How to get an expression between balanced parentheses.

Java customizable non-regex solution

Here is a customizable solution allowing single character literal delimiters in Java:

public static List<String> getBalancedSubstrings(String s, Character markStart, 
                                 Character markEnd, Boolean includeMarkers) 

{
        List<String> subTreeList = new ArrayList<String>();
        int level = 0;
        int lastOpenDelimiter = -1;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == markStart) {
                level++;
                if (level == 1) {
                    lastOpenDelimiter = (includeMarkers ? i : i + 1);
                }
            }
            else if (c == markEnd) {
                if (level == 1) {
                    subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i)));
                }
                if (level > 0) level--;
            }
        }
        return subTreeList;
    }
}

Sample usage:

String s = "some text(text here(possible text)text(possible text(more text)))end text";
List<String> balanced = getBalancedSubstrings(s, '(', ')', true);
System.out.println("Balanced substrings:\n" + balanced);
// => [(text here(possible text)text(possible text(more text)))]

Answer 11

I was also stuck in this situation where nested patterns comes.

Regular Expression is right thing to solve the above problem. Use below pattern

'/(\((?>[^()]+|(?1))*\))/'

Answer 12

The regular expression using Ruby (version 1.9.3 or above):

/(?<match>\((?:\g<match>|[^()]++)*\))/

Demo on rubular

Answer 13

You need the first and last parentheses. Use something like this:

str.indexOf('('); - it will give you first occurrence

str.lastIndexOf(')'); - last one

So you need a string between,

String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')');

Answer 14

"""
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 15

The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.

If you need to match matching nested brackets, then you need something more than regular expressions. - see @dehmann

If it's just first open to last close see @Zach

Decide what you want to happen with:

abc ( 123 ( foobar ) def ) xyz ) ghij

You need to decide what your code needs to match in this case.

Answer 16

This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:

Parse parmeters from function string (with nested structures) in javascript

Match structures like:

• matches brackets, square brackets, parentheses, single and double quotes

Here you can see generated regexp in action

/**
 * get param content of function string.
 * only params string should be provided without parentheses
 * WORK even if some/all params are not set
 * @return [param1, param2, param3]
 */
exports.getParamsSAFE = (str, nbParams = 3) => {
    const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/;
    const params = [];
    while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine
        str = str.replace(nextParamReg, (full, p1) => {
            params.push(p1);
            return '';
        });
    }
    return params;
};

Answer 17

because js regex doesn't support recursive match, i can't make balanced parentheses matching work.

so this is a simple javascript for loop version that make "method(arg)" string into array

push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)

const parser = str => {
  let ops = []
  let method, arg
  let isMethod = true
  let open = []

  for (const char of str) {
    // skip whitespace
    if (char === ' ') continue

    // append method or arg string
    if (char !== '(' && char !== ')') {
      if (isMethod) {
        (method ? (method += char) : (method = char))
      } else {
        (arg ? (arg += char) : (arg = char))
      }
    }

    if (char === '(') {
      // nested parenthesis should be a part of arg
      if (!isMethod) arg += char
      isMethod = false
      open.push(char)
    } else if (char === ')') {
      open.pop()
      // check end of arg
      if (open.length < 1) {
        isMethod = true
        ops.push({ method, arg })
        method = arg = undefined
      } else {
        arg += char
      }
    }
  }

  return ops
}

// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)

console.log(test)

the result is like

[ { method: 'push', arg: 'number' },
  { method: 'map', arg: 'test(a(a()))' },
  { method: 'bass', arg: 'wow,abc' } ]

[ { method: '$$', arg: 'groups' },
  { method: 'filter',
    arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
  { method: 'pickBy', arg: '_id,type' },
  { method: 'map', arg: 'test()' },
  { method: 'as', arg: 'groups' } ]

Answer 18

I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:

def extract_code(data):
    """ returns an array of code snippets from a string (data)"""
    start_pos = None
    end_pos = None
    count_open = 0
    count_close = 0
    code_snippets = []
    for i,v in enumerate(data):
        if v =='{':
            count_open+=1
            if not start_pos:
                start_pos= i
        if v=='}':
            count_close +=1
            if count_open == count_close and not end_pos:
                end_pos = i+1
        if start_pos and end_pos:
            code_snippets.append((start_pos,end_pos))
            start_pos = None
            end_pos = None

    return code_snippets

I used this to extract code snippets from a text file.

Answer 19

While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.

Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.

Read more @ here

Answer 20

This might help to match balanced parenthesis.

\s*\w+[(][^+]*[)]\s*

Answer 21

This one also worked

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