如何解析字符串并返回嵌套数组? [英] How to parse a string and return a nested array?

问题描述

我想要一个 Python 函数，它接受一个字符串并返回一个数组，其中数组中的每个项目要么是一个字符，要么是另一个此类数组.嵌套数组在输入字符串中标记为以("开头并以)"结尾.

因此，该函数将如下所示:

1) foo("abc") == ["a", "b", "c"]2) foo("a(b)c") == ["a", ["b"], "c"]3) foo("a(b(c))") == ["a", ["b", ["c"]]]4) foo("a(b(c)") == 错误:缺少右括号5) foo("a(b))c") == 错误:缺少左括号6) foo("a)b(c") == 错误:缺少左括号

注意:我更喜欢纯粹功能性的解决方案.

解决方案

def foo(s):def foo_helper(level=0):尝试:令牌 = 下一个(令牌)除了停止迭代:如果级别 != 0:raise Exception('缺少结束括号')别的:返回 []如果令牌 == ')':如果级别 == 0:raise Exception('缺少开头括号')别的:返回 []elif 令牌 == '(':返回 [foo_helper(level+1)] + foo_helper(level)别的:返回 [令牌] + foo_helper(级别)令牌 = iter(s)返回 foo_helper()

还有，

<预><代码>>>>foo('a((b(c))d)(e)')['a', [['b', ['c']], 'd'], ['e']]

I want a Python function that takes a string, and returns an array, where each item in the array is either a character, or another array of this kind. Nested arrays are marked in the input string by starting with '(' and ending with ')'.

Thus, the function would act like this:

1) foo("abc") == ["a", "b", "c"]
2) foo("a(b)c") == ["a", ["b"], "c"]
3) foo("a(b(c))") == ["a", ["b", ["c"]]]
4) foo("a(b(c)") == error: closing bracket is missing
5) foo("a(b))c") == error: opening bracket is missing
6) foo("a)b(c") == error: opening bracket is missing

Note: I'd prefer a solution that's purely functional.

解决方案

def foo(s):
    def foo_helper(level=0):
        try:
            token = next(tokens)
        except StopIteration:
            if level != 0:
                raise Exception('missing closing paren')
            else:
                return []
        if token == ')':
            if level == 0:
                raise Exception('missing opening paren')
            else:
                return []
        elif token == '(':
            return [foo_helper(level+1)] + foo_helper(level)
        else:
            return [token] + foo_helper(level)
    tokens = iter(s)
    return foo_helper()

And,

>>> foo('a((b(c))d)(e)')
['a', [['b', ['c']], 'd'], ['e']]

这篇关于如何解析字符串并返回嵌套数组?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！