Question

显然，在阅读

之后，这个问题经常出现

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

并且想了一会儿的问题，我写了一个函数来返回包含在任意数量的嵌套（）内的内容

该功能可以轻松扩展到任何正则表达式对象，在此处发布您的想法和注意事项。

任何重构建议都将受到赞赏

（注意，我仍然是python的新手，并且不想弄清楚如何引发异常或其他什么，所以我只是让函数返回'失败'如果它无法弄清楚发生了什么）

编辑功能以考虑评论：

def ParseNestedParen(string, level):
    """
    Return string contained in nested (), indexing i = level
    """
    CountLeft = len(re.findall("\(", string))
    CountRight = len(re.findall("\)", string))
    if CountLeft == CountRight:
        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 CountLeft > CountRight:
        return ParseNestedParen(string + ')', level)

    elif CountLeft < CountRight:
        return ParseNestedParen('(' + string, level)

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

Answer 1

您没有明确说明您的功能规范是什么，但这种行为对我来说似乎不对：

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

对您的代码的其他评论：

Docstring说“生成”，但函数返回一个列表，而不是生成器。
由于只返回一个字符串，为什么要在列表中返回？
在什么情况下函数可以返回字符串fail？
反复拨打re.findall然后丢弃结果是浪费。
您尝试重新平衡字符串中的括号，但一次只能使用一个括号：

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

正如托米在question you linked to中所说，“正则表达式确实是错误的工具！”

解析嵌套表达式的常用方法是使用堆栈，沿着这些行：

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 2

括号匹配需要具有下推自动机的解析器。有些库存在，但规则很简单，我们可以从头开始编写：

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']]

Answer 3

#!/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]]]

<强>试验：

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))

<强>输出：

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 4

下面是我的 Python 解决方案，时间复杂度为 O(N)

str1 = "(a(b(c)d)(e(f)g)hi)"

def content_by_level(str1, l):
    level_dict = {}
    level = 0
    level_char = ''
    for s in str1:
        if s == '(':
            if level not in level_dict:
                level_dict[level] = [level_char]
            elif level_char != '':
                level_dict[level].append(level_char)
            level_char = ''
            level += 1
        elif s == ')':
            if level not in level_dict:
                level_dict[level] = [level_char]
            elif level_char != '':
                level_dict[level].append(level_char)
            level_char = ''
            level -= 1
        else:
            level_char += s
    
    print(level_dict) # {0: [''], 1: ['a', 'hi'], 2: ['b', 'd', 'e', 'g'], 3: ['c', 'f']}
    return level_dict[l]

print(content_by_level(str1,0)) # ['']
print(content_by_level(str1,1)) # ['a', 'hi']
print(content_by_level(str1,2)) # ['b', 'd', 'e', 'g']
print(content_by_level(str1,3)) # ['c', 'f']

在python中解析嵌套括号，按级别获取内容

4 个答案: