比较python中的括号

Question

from collections import Counter
x = '((()))'

print Counter(x)

Counter({')': 3, '(': 3})

I need to compare whether the open brackets and close brackets of my string are equal or not. I used counter to do this. But how should I compare the two values of the counter variable?

Answer 1

You can use a small function which uses .count() :

>>> def checkparen(inp):
...     return inp.count('(') == inp.count(')') and inp.count('[') == inp.count(']') and inp.count('{') == inp.count('}')
... 

As such:

>>> checkparen('((()))')
True
>>> checkparen('((())')
False
>>> checkparen('[ [ { { ( ( ) ) } } ] ]')
True
>>> 

Or, using collections.Counter :

>>> def checkparen(inp):
...     counter = collections.Counter(inp)
...     symbols = {'{':'}','[':']','(':')'}
...     for symbol in symbols:
...             if counter[symbol] != counter[symbols[symbol]]:
...                     return False
...     return True
... 
>>> checkparen('{')
False
>>> checkparen('[ [ { { ( ( ) ) } } ] ]')
True
>>> 

Answer 2

In general, when you try and extend to the larger problem described in the article you linked, the class of problems you are looking for is context-free-grammar parsers . This is a very classic problem

http://en.wikipedia.org/wiki/Context-free_grammar#Well-formed_nested_parentheses_and_square_brackets

even considered canonical as an introduction to compiler theory and related topics.

A simple top-down parser is very easy to implement http://effbot.org/zone/simple-top-down-parsing.htm

A chart parser like Early is tougher, but there are 150 line implementations avaliable https://github.com/tomerfiliba/tau/blob/master/earley3.py

You can also use a parser generator.

Any other approach might seem nice at the beginning, but this will be the only strategy that will scale

Answer 3

I think that you can use a list as a stack to see if all the brackets match, like this:

def checkio(expr):
    a_stack = []
    for i in expr:
        if i in '({[':
            a_stack.append(i)
        elif i in '}])':
            if a_stack == []:
                return False
            else:
                poped = a_stack.pop()
                if poped == '(' and i != ')':
                    return False
                elif poped == '[' and i != ']':
                    return False
                elif poped == '{' and i != '}':
                    return False
    return len(a_stack) == 0

print checkio("((5+3)*2+1)") == True
print checkio("{[(3+1)+2]+}") == True
print checkio("(3+{1-1)}") == False
print checkio("[1+1]+(2*2)-{3/3}") == True
print checkio("(({[(((1)-2)+3)-3]/3}-3)") == False
print checkio("2+3") == True

First push a char into a stack, then pop one and see if they are matched, at last too see if the stack is empty.