我的代码有什么问题？递归查找嵌套列表的总和

Question

我正在尝试创建一个函数，以以下格式返回树列表中所有整数的总和：

element 1 is an integer
element 2 is another treelist or None
element 3 is another treelist or None

ex: [1,[1,None,None],None]

所以基本上我希望我的函数将该列表中的所有整数相加并返回 2。

这是我到目前为止所做的......

def sum_nested(t):
    sum = 0
    for i in t:
        if type(i) == type(None):
           sum += 0
        if type(i) == list:
           return sum_nested(i)
        else:
            sum = sum + i

    return sum  

但是，我得到了错误：

builtins.TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'

当我点击 None 类型时似乎无法弄清楚该怎么做..有什么建议吗？

Answer 1

试试这个：

t = [1, [1, None, None], None]
def sum_nested(t):
    sum = 0
    for i in t:
        if type(i) is int:
            sum += 1
        elif type(i) == list:
           sum += sum_nested(i)

    return sum

print(sum_nested(t))

如果你想测试某个东西是否是None，最短的形式是：

if i:

但在你的情况下，这并不是真正必要的，因为无论如何你都没有改变 sum

Answer 2

怎么样：

def sum_nested(seq):

    # Don't use `sum` as a variable name, as it shadows the builtin `sum()`
    total = 0
    for item in seq:

        # This will allow you to sum *any* iterable (tuples, for example)
        if hasattr(item, '__iter__'): 

            # Use the `+=` syntactic sugar
            total += sum_nested(item) 
        else:

            # `None or 0` evaluates to 0 :)
            total += item or 0 

    return total

Answer 3

def sum_nested(seq):
    total = 0
    for item in seq:
        if hasattr(item, '__iter__'): 
            total += sum_nested(item)
            continue
        try:
            total += item
        except TypeError:
            continue
    return total

print sum_nested([2, ['text', [None, 1], [int, [True, None], 5]]])
# True counts as 1