如何使用此命令删除括号？

Question

我想用这种代码或递归 function 从这个列表中删除括号，首先请让我知道这些代码有什么问题，然后请给我正确的代码谢谢

def fun(a) :
    x = []
    while a[0,len(a)+1] is not int :
        for i in range(len(a)):
            b = a[i]
            b = b[0] if b is not int else b
            if b is int :
                x.append(b)
                break
    return x
l = [[1],[2],[[3]],[[[4]]]]
print(fun(l))

或者

L = [[[[1]]], [[2, 3, 4, 5]], [6, 7, 8], [9, 10]]
x = []
b = []
for i in range(len(L)):
    while L[i] is not int:
        if b is int:
            L[i].append(x)
            break
        else:
            b = L[i]
            b = b[0]
            L[i] = b
print(L)

Answer 1

希望这可以帮助：

L = [[[[1]]], [[2, 3, 4, 5]], [6, 7, 8], [9, 10]]

def remove_brackets(L):
    L_cleaned = []
    for i in L:
        j = i
        while type(j[0]) is not int:
            j = j[0]
            
        for k in j:
            L_cleaned.append(k)
    return L_cleaned
        
remove_brackets(L)

Answer 2

要处理由多个嵌套列表组成的列表（例如[[[1, 2], [3, 4]]] ），您需要这样的东西：

def check(l):
    # check if all extra brackets removed
    for x in l:
        if not type(x) is int:
            return False
    return True

def fun(l):
    new_l = []
    for x in l:
        if type(x) is int:
            new_l.append(x)
        else:
            # x is list
            new_l.extend(fun(x))
    return new_l

或者：

def bar(l):
    return str(l).replace("[", "").replace("]", "").split(",")

Answer 3

而不是使用type(x) is...你通常应该使用isinstance 。 这是一个简单的递归解包 function。

def unpack(item):
    if isinstance(item, list):
        for subitem in item:
            yield from unpack(subitem)
    else:
        yield item

L = [[[[1]]], [[2, 3, 4, 5]], [6, 7, 8], [9, 10]]
L2 = list(unpack(L)) # L2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]