平衡括号挑战：检查不包含括号的字符串

Question

I did an implementation on checking if whether a string has balanced parentheses or not using a STACK only. But I checked for balanced parentheses when the string has words. For example:

>>> is_paren_balanced("[{whoa (this is rough [how do I do this!])}]")
True

I was successful in doing that. However, I am not trying to check for a case if the string has NO parenthesis. And then i got this result:

>>> is_paren_balanced("Hi i love food.")
no parenthesis
True
>>> is_paren_balanced("[{whoa (this is rough [how do I do this!])}]")
no parenthesis
False

For the first result, I don't want the boolean result as thats kind of vague. I'm trying to show the user that the string has no parenthesis so it is neither True or False. I just want the print statement shown saying "no parentheses here" or whatever. For the second result, obviously just the boolean result and it shouldve returned True.

Any suggestions? I'm stumped.

Here is the code I'm practicing with:

from stack import Stack

def is_match(p1, p2):
    if p1 == "(" and p2 == ")":
        return True

    elif p1 == "{" and p2 == "}":
        return True

    elif p1 == "[" and p2 == "]":
        return True

    else:
        return False


def is_paren_balanced(paren_str):
    s = Stack() #initialize a stack object
    is_balanced = True #boolean flag: whether the str has balanced parenthesis or not
    # then set to false if otherwise
    index = 0 # keep track of where we are in the string
    paren_str = paren_str.replace(" ", "") #accounts for possible spaces in the input


    while index < len(paren_str) and is_balanced:
        paren = paren_str[index]       
        if paren in "({[":
            s.push(paren)

        elif paren in ")}]":
            if s.is_empty():
                is_balanced = False

            else:
                top = s.pop()
                if not is_match(top, paren):
                    is_balanced = False

        else:
            if paren not in "({[]})":
                print("no parenthesis")
                break



        index += 1

    if s.is_empty() and is_balanced:
        return True

    else:
        return False

Sorry if it doesn't seem pythonic. I was just mostly exploring with this problem and will edit and improve later on.

Answer 1

You can use either regex or list comprehension to get the total amount of parenthesis first:

import re

def all_parenthesis(msg):
    left = re.findall(r"[\[{(]",msg)
    right = re.findall(r"[\]|})]",msg)
    if not left and not right:
        return "No parenthesis"
#or

def check_balance(msg):
    d = {"[":"]",
         "{":"}",
         "(":")"}
    left = [char for char in msg if char in d]
    right = [char for char in reversed(msg) if char in d.values()]  
    #if not left not right...  

But if you already constructed a list of parenthesis on both sides, why don't go ahead and compare them instead? Could be something like this:

def check_balance(msg):
    d = {"[":"]",
         "{":"}",
         "(":")"}
    left = [char for char in msg if char in d]
    right = [char for char in reversed(msg) if char in d.values()]
    if not left and not right:
        return "No parenthesis"
    elif len(left) != len(right):
        return False
    for a, b in zip(left,right):
        if d.get(a) != b:
            return False
    return True