将带有嵌套括号的字符串转换为嵌套列表，pyth

Turn a string with nested parenthesis into a nested list, python(将带有嵌套括号的字符串转换为嵌套列表，python)

问题描述

在 Stack Overflow 上还有其他相关问题，例如 如何解析字符串并返回嵌套数组?

There are other questions referring to this on Stack Overflow such as how-to-parse-a-string-and-return-a-nested-array?

但它们都是以((abc)de(fg)))格式引用列表.转到表格:[['a','b','c']'d','e'['f','g',]]] 我有一个列表形式:

But they all refer to lists in the format of ((abc)de(fg))). going to the form:[['a','b','c']'d','e'['f','g',]]] I have a list of the form:

((wordOneWord2)OtherWord(FinalWord)))

通过使用我从嵌套列表的其他问题中学到的方法，形式为:

By using the methods I learnt from the other questions by nested list was of the form:

[['w','o','r','d','O','n','e','W','o','r','d','2']'O','t','h','e','r','W','o','r','d',['F','i','n','a','l','W','o','r','d']]]

而不是想要的

[['wordOneWord2'], 'OtherWord', ['FinalWord']]

我可以通过逐个字母解析列表然后将每个列表中的项目重新连接在一起来获得所需的结果，但它需要的代码比我认为需要的要多，有没有更快的方法?

I can achieve the desired result by parsing the list letter by letter and then concatenating the items within each list back together but it takes more code than I think necessary, is there a faster way of doing this?

推荐答案

基于此falsetru的解决方案:>

import re

def parse_nested(text, left=r'[(]', right=r'[)]', sep=r','):
    """ Based on https://stackoverflow.com/a/17141899/190597 (falsetru) """
    pat = r'({}|{}|{})'.format(left, right, sep)
    tokens = re.split(pat, text)    
    stack = [[]]
    for x in tokens:
        if not x or re.match(sep, x): continue
        if re.match(left, x):
            stack[-1].append([])
            stack.append(stack[-1][-1])
        elif re.match(right, x):
            stack.pop()
            if not stack:
                raise ValueError('error: opening bracket is missing')
        else:
            stack[-1].append(x)
    if len(stack) > 1:
        print(stack)
        raise ValueError('error: closing bracket is missing')
    return stack.pop()

text = '((wordOneWord2)OtherWord(FinalWord))'
print(parse_nested(text))
# [[['wordOneWord2'], 'OtherWord', ['FinalWord']]]

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