如何解析一个字符串并返回一个嵌套数组？

Question

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都是一个字符,或者是另一个这样的数组.嵌套数组在输入字符串中以'('和以')'开头标记.

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)") == 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

注意:我更喜欢纯粹功能性的解决方案.

Answer 1

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()

和,

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

Answer 2

迭代.

def foo(xs):
    stack = [[]]
    for x in xs:
        if x == '(':
            stack[-1].append([])
            stack.append(stack[-1][-1])
        elif x == ')':
            stack.pop()
            if not stack:
                return 'error: opening bracket is missing'
                #raise ValueError('error: opening bracket is missing')
        else:
            stack[-1].append(x)
    if len(stack) > 1:
        return 'error: closing bracket is missing'
        #raise ValueError('error: closing bracket is missing')
    return stack.pop()

assert foo("abc") == ["a", "b", "c"]
assert foo("a(b)c") == ["a", ["b"], "c"]
assert foo("a(b(c))") == ["a", ["b", ["c"]]]
assert foo("a((b(c))d)(e)") == ['a', [['b', ['c']], 'd'], ['e']]
assert foo("a(b(c)") == "error: closing bracket is missing"
assert foo("a(b))c") == "error: opening bracket is missing"
assert foo("a)b(c") == 'error: opening bracket is missing'