parsing nested parentheses in python, grab content by level

Question

Apparently this problem comes up fairly often, after reading

Regular expression to detect semi-colon terminated C++ for & while loops

and thinking about

Answer 1

#!/usr/bin/env python
import re

def ParseNestedParen(string, level):
    """
    Generate strings contained in nested (), indexing i = level
    """
    if len(re.findall("\(", string)) == len(re.findall("\)", string)):
        LeftRightIndex = [x for x in zip(
        [Left.start()+1 for Left in re.finditer('\(', string)], 
        reversed([Right.start() for Right in re.finditer('\)', string)]))]

    elif len(re.findall("\(", string)) > len(re.findall("\)", string)):
        return ParseNestedParen(string + ')', level)

    elif len(re.findall("\(", string)) < len(re.findall("\)", string)):
        return ParseNestedParen('(' + string, level)

    else:
        return 'fail'

    return [string[LeftRightIndex[level][0]:LeftRightIndex[level][1]]]

Tests:

if __name__ == '__main__':

    teststring = "outer(first(second(third)second)first)outer"

    print(ParseNestedParen(teststring, 0))
    print(ParseNestedParen(teststring, 1))
    print(ParseNestedParen(teststring, 2))

    teststring_2 = "outer(first(second(third)second)"

    print(ParseNestedParen(teststring_2, 0))
    print(ParseNestedParen(teststring_2, 1))
    print(ParseNestedParen(teststring_2, 2))

    teststring_3 = "second(third)second)first)outer"

    print(ParseNestedParen(teststring_3, 0))
    print(ParseNestedParen(teststring_3, 1))
    print(ParseNestedParen(teststring_3, 2))

output:

Running tool: python3.1

['first(second(third)second)first']
['second(third)second']
['third']
['first(second(third)second)']
['second(third)second']
['third']
['(second(third)second)first']
['second(third)second']
['third']
>>> 

Answer 2

You don't make it clear exactly what the specification of your function is, but this behaviour seems wrong to me:

>>> ParseNestedParen('(a)(b)(c)', 0)
['a)(b)(c']
>>> nested_paren.ParseNestedParen('(a)(b)(c)', 1)
['b']
>>> nested_paren.ParseNestedParen('(a)(b)(c)', 2)
['']

Other comments on your code:

• Docstring says "generate", but function returns a list, not a generator.
• Since only one string is ever returned, why return it in a list?
• Under what circumstances can the function return the string fail?
• Repeatedly calling re.findall and then throwing away the result is wasteful.
• You attempt to rebalance the parentheses in the string, but you do so only one parenthesis at a time:

>>> ParseNestedParen(')' * 1000, 1)
RuntimeError: maximum recursion depth exceeded while calling a Python object

As Thomi said in the question you linked to, "regular expressions really are the wrong tool for the job!"

---

The usual way to parse nested expressions is to use a stack, along these lines:

def parenthetic_contents(string):
    """Generate parenthesized contents in string as pairs (level, contents)."""
    stack = []
    for i, c in enumerate(string):
        if c == '(':
            stack.append(i)
        elif c == ')' and stack:
            start = stack.pop()
            yield (len(stack), string[start + 1: i])

>>> list(parenthetic_contents('(a(b(c)(d)e)(f)g)'))
[(2, 'c'), (2, 'd'), (1, 'b(c)(d)e'), (1, 'f'), (0, 'a(b(c)(d)e)(f)g')]

Answer 3

Parentheses matching requires a parser with a push-down automaton. Some libraries exist, but the rules are simple enough that we can write it from scratch:

def push(obj, l, depth):
    while depth:
        l = l[-1]
        depth -= 1

    l.append(obj)

def parse_parentheses(s):
    groups = []
    depth = 0

    try:
        for char in s:
            if char == '(':
                push([], groups, depth)
                depth += 1
            elif char == ')':
                depth -= 1
            else:
                push(char, groups, depth)
    except IndexError:
        raise ValueError('Parentheses mismatch')

    if depth > 0:
        raise ValueError('Parentheses mismatch')
    else:
        return groups

print(parse_parentheses('a(b(cd)f)')) # ['a', ['b', ['c', 'd'], 'f']]