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python中的括号匹配代码

[英]Parenthesis matching code in python

I have written following code and it was working fine.But One of the case it is failing.I tried but not able to fix this issue. 我写了下面的代码,它工作正常。但其中一个案例是失败。我尝试但无法解决这个问题。

#!/usr/bin/env py
import itertools
import sys
import sympy
import re
import pprint
def extract_next_state(s):
    p = re.compile('(\()|(\))')
    depth = 0
    startindex = None
    start_point = False
    for m in p.finditer(s):
        if m.group(1):          # (
            depth += 1
            print "depth (", depth
            if not start_point:
                startindex = m.start()
                start_point = True
        elif m.group(2):          # )
            depth -= 1
            print "depth )", depth
            if depth == 0:
                return s[startindex:m.end()]
    return s[startindex:] + ')' * depth

if __name__ == "__main__":
    #data = ['next_state=(~SE&((~B2&D)|(B2&lq))|(SE&SI))']
    #data = ['next_state=((~SE&((A1&A2)|(B1&B2)))|(SE&SI)))']
    #data = ['next_state=((((~SE&((A1&A2)|(B1&B2)))|(SE&SI)))']
    data = ['next_state=(D1&S&!SE)|(!S&(!SE&D0))|(SE&SI))']
    data_1 = data[0].split(',')
    com = None
    for item in data_1:
        if item.find('next_state=')!= -1:
            item_list = item.split('=')
            item_op = extract_next_state(item_list[1])
            print item_op

output: 输出:

(D1&S&!SE) (D1&S&!SE)

Expected : 预期:

(D1&S&!SE)|(!S&(!SE&D0))|(SE&SI) (D1&S&SE!)|(S&(SE&D0)!)|(SE&SI)

You checking for depth == 0 as condition for returning from extract_next_state() . 检查depth == 0作为从extract_next_state()返回的条件。 That is extract_next_state() returns once the matching closing parenthesis to the first opening parenthesis is found. 也就是说, extract_next_state()找到匹配的右括号到第一个左括号, extract_next_state()返回。 The rest of the string is then not checked for any further parenthesis of course. 然后,当然不检查字符串的其余部分是否有任何进一步的括号。

It's hard to recommend a solution without knowing what rules for "next_state" or the grammar for allowd expressions. 在不知道“next_state”的规则或允许表达式的语法的情况下,很难推荐解决方案。 But from the last line extract_next_state it seems you wish to close any open parenthesis. 但是从最后一行extract_next_state看来你似乎想要关闭任何打开的括号。 So a possible solution may be: 所以可能的解决方案可能是:

def extract_next_state(s):
    p = re.compile('(\()|(\))')
    depth = 0
    startindex = None
    endindex = None
    for m in p.finditer(s):
        if m.group(1):          # (
            depth += 1
            print "depth (", depth
            if not startindex:
                startindex = m.start()
        elif m.group(2):          # )
            depth -= 1
            print "depth )", depth
            if depth == 0:
                endindex = m.end()
    return s[startindex:(endindex if depth == 0 else None)] + ')' * depth

If with the last closing parenthesis all pairs are matched, the rest of the string is discarded, else a matching number of closing parentheses will be added. 如果使用最后一个右括号匹配所有对,则丢弃该字符串的其余部分,否则将添加匹配数量的右括号。

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