Parentheses pairing ({}[]()<>) issue

Question

I want to be able to pair up all parentheses in a string, if they aren\'t paired then then they get their index number and False. It seems like it is repeating some values over

Answer 1

Thanks hughdbrown your code was a breeze to get working and it's really short! You've just saved me a headache :D

converted it to pep8 if thats ok :)

Edit

• Added support for comments and strings, it will not match inside them.
• Added support for easy language brace checking, modify the charset dict.
• Correctly paires up, i.e right to left

HTML

charset = dict(opening='{[(<',\
    closing='}])>',\
    string = ('"', "'"),\
    comment=(('<!--', '-->')))

Python

charset = dict(opening='{[(<',\
    closing='}])>',\
    string = ('"', "'"),\
    comment=(("'''", "'''"), ('"""', '"""'), ('#', '\n')))

C++

charset = dict(opening='{[(<',\
    closing='}])>',\
    string = ('"', "'"),\
    comment=(('/*', '*/'), ('//', '\n')))

you get the point? :)

charset = dict(opening='{[(<',\
    closing='}])>',\
    string = ('"', "'"),\
    comment=(('<!--', '-->'), ('"""', '"""'), ('#', '\n')))

allowed = ''.join([x[0][0] + x[1][0] for x in charset['comment']])
allowed += ''.join(charset['string'])
allowed += charset['opening']
allowed += charset['closing']

def brace_check(text):
    o = []
    c = []
    notr = []
    found = []
    busy = False
    last_pos = None
    for i in xrange(len(text)):
        ch = text[i]
        if not busy:
            cont = True
            for comment in charset['comment']:
                if ch == comment[0][0]:
                    como = text[i:len(comment[0])]
                    if como == comment[0]:
                        busy = comment[1]
                        if ch in charset['opening']:
                            last_pos = i
                        cont = False
                        break
            if cont:
                if ch in charset['string']:
                    busy = ch
                elif ch in charset['opening']:
                    o.append((ch, i))
                elif  ch in charset['closing']:
                    c.append((ch, i))
        else:
            if ch == busy[0]:
                if len(busy) == 1:
                    comc = ch
                else:
                    comc = text[i:i + len(busy)]
                if comc == busy:
                    if last_pos is not None:
                        if busy[-1] in charset['closing']:
                            found.append((last_pos, i))
                        last_pos = None
                        text = text[:i] + '\n' * len(comc) +\
                            text[i + len(comc):]
                    busy = not busy
            elif busy in charset['string']:
                if ch == '\n':
                    busy = not busy
    for t, e in reversed(o):
        try:
            n = next((b, v) for b, v in c\
                if b == charset['closing'][\
                    charset['opening'].find(t)] and v > e)
            c.remove(n)
            n = n[1]
            if found != []:
                if e < found[-1][0] and n > found[-1][0] and n < found[-1][1]\
                or e < found[-1][1] and n > found[-1][1] and e > found[-1][0]:
                    found.append((n, False))
                    n = False
        except StopIteration:
            n = False
        found.append((e, n))
    for t, e in c:
        found.append((e, False))
    return found

Answer 2

An understandable solution in Python 3:

def check_balanced_string(str):
  stack = []
  dicc = {'(': ')', '[': ']', '{': '}'}
  for char in str:
    if char in dicc.keys():  # opening char
      stack.append(char)
    elif char in dicc.values():  # closing char
      if dicc[stack[-1]] == char:  # check if closing char corresponds to last opening char
        stack.pop()
      else:
        return False
  return not len(stack)  # returns True when len == 0

eq = '{1+[3*5+(2+1)]}'
print(check_balanced_string(eq))

Answer 3

The below code will display the missing parentheses and the no of times missing in the given string.

from collections import Counter

def find_missing(str):
    stack1 = []
    stack2 = []
    result = []
    res_dict = {}
    open_set = '<[{('
    closed_set = '>]})'
    a = list(str)
    for i in a:
        if i in open_set:
            stack1.append(i)
        elif i in closed_set:
            stack2.append(i)
    dict1 = Counter(stack1)
    dict2 = Counter(stack2)
    print(dict1)
    print(dict2)
    for i in open_set:
        if dict1[i] > dict2[closed_set[open_set.index(i)]]:
            res_dict[closed_set[open_set.index(i)]] = dict1[i] - dict2[closed_set[open_set.index(i)]]
            result.append(closed_set[open_set.index(i)])
    for i in closed_set:
        if dict2[i] > dict1[open_set[closed_set.index(i)]]:
            res_dict[open_set[closed_set.index(i)]] = dict2[i] - dict1[open_set[closed_set.index(i)]]
            result.append(open_set[closed_set.index(i)])
    return res_dict
    # return result

if __name__ == '__main__':
    str1 = '{This ((()bracket {[function]} <<going> crazy}'
    x = find_missing(str1)
    if len(x) > 0:
        print("Imbalanced")
        print(x)
    else:
        print("Balanced")

Answer 4

BRACES = { '(': ')', '[': ']', '{': '}' }

def group_check(s):
    stack = []
    for b in s:
        c = BRACES.get(b)
        if c:
            stack.append(c)
        elif not stack or stack.pop() != b:
            return False
    return not stack

Answer 5

I needed something for a recent project and figured I could build on the OP's solution a bit. It allows for comment patterns, quotes and brackets to be checked, whilst ignoring the surrounding text. I've purposefully made it more generic than it needs to be so that others can take what they want and cut out what they don't.

"""
This module is for testing bracket pairings within a given string
Tested with Python 3.5.4
>>> regexp = getRegexFromList(opening + closing)
>>> print(regexp)
(\\<\\-\\-|\\-\\-\\>|\\/\\*|\\/\\/|\\*\\/|\\#|\\"|\\'|\\(|\\[|\\{|\\<|\\\n|\\\n|\\"|\\'|\\)|\\]|\\}|\\>)
>>> test_string = 'l<--([0])-->1/*{<2>}*/3//<--4 &-->\\n5#"6"\\n7"/*(8)*/"9\'"10"\'11({12\ta})13[<14>]'
>>> patterns = re.findall(regexp, test_string)
>>> print(patterns)
['<--', '(', '[', ']', ')', '-->', '/*', '{', '<', '>', '}', '*/', '//', '<--', '-->', '\\n', '#', '"', '"', '\\n', '"', '/*', '(', ')', '*/', '"', '(', '{', '}', ')', '[', '<', '>', ']']
>>> doBracketsMatch(patterns)
True
>>> doBracketsMatch(['"', ')', '"', '[', ']', '\\''])
False
"""


# Dependencies
import re


# Global Variables
# Provide opening and closing patterns, along with their priorities & whether a priority is nestable
opening =  ['<--', '/*', '//',  '#', '"', '\'', '(', '[', '{', '<']
closing =  ['-->', '*/', '\n', '\n', '"', '\'', ')', ']', '}', '>']
priority = [    1,    1,    1,    1,   1,    1,   0,   0,   0,   0]
nestable = {0: True, 1: False}
bracket_pairs = dict(zip(opening + closing, \
                         [[(closing + opening)[i], (priority + priority)[i]] \
                          for i in range(0, opening.__len__() * 2)]))


def getRegexFromList(listOfPatterns):
    """
    Generate the search term for the regular expression
    :param listOfPatterns:
    :return:
    >>> getRegexFromList(['"', '<--', '##', 'test'])
    '(\\\\t\\\\e\\\\s\\\\t|\\\\<\\\\-\\\\-|\\\\#\\\\#|\\\\")'
    """
    # Longer patterns first to prevent false negatives
    search_terms = sorted(listOfPatterns, key=len, reverse=True)
    regex = ""
    for term in search_terms:
        for char in str(term):
            regex = regex + '\\' + char  # Search for all characters literally
        regex = regex + '|'  # Search pattern = (a|b|c)
    return '(' + regex[:-1] + ')'  # Remove excess '|' and add brackets


def doBracketsMatch(list_of_brackets):
    """
    Determine if brackets match up
    :param list_of_brackets:
    :return:
    """
    stack = []
    for bracket in list_of_brackets:
        # Check empty stack conditions
        if stack.__len__() is 0:
            # Check for openings first to catch quotes
            if bracket in opening:
                stack.append(bracket)
            elif bracket in closing:
                return False
            else:
                continue
        # Check for a matching bracket
        elif bracket == bracket_pairs[stack[-1]][0]:
            stack.pop()
        # Ignore cases:
        #  - False positives
        #  - Lower priority brackets
        #  - Equal priority brackets if nesting is not allowed
        elif bracket not in bracket_pairs or \
                bracket_pairs[bracket][1] < bracket_pairs[stack[-1]][1] or \
                (bracket_pairs[bracket][1] == bracket_pairs[stack[-1]][1] and \
                    not nestable[bracket_pairs[bracket][1]]):
            continue
        # New open bracket
        elif bracket in opening:
            stack.append(bracket)
        # Otherwise, unpaired close bracket
        else:
            return False
    # If stack isn't empty, then there is an unpaired open bracket
    return not bool(stack)


if __name__ == '__main__':
    import doctest
    doctest.testmod()

Answer 6

iparens = iter('(){}[]<>')
parens = dict(zip(iparens, iparens))
closing = parens.values()

def balanced(astr):
    stack = []
    for c in astr:
        d = parens.get(c, None)
        if d:
            stack.append(d)
        elif c in closing:
            if not stack or c != stack.pop():
                return False
    return not stack

Example:

>>> balanced('[1<2>(3)]')
True
>>> balanced('[1<2(>3)]')
False