python - __next__ 和 next 带参数

Question

我正在编写双向链表的实现。为了遍历列表，我使用了类似的东西:

class Node:
    """ A node in our linked list """
    def __init__(self, value: Any, next: Union['Node', None] =None,
            previous: Union['Node', None] =None) -> None:
        self.value = value
        self.next = next
        self.previous = previous

    ...

    def __next__(self, direction: int =1) -> Union['Node', None]:
        if direction == 1:
            return self.get_next()
        else:
            return self.get_previous()

    ...

get_next 和 get_previous 只是 self.next 和 self.previous 的 getter。

但是，PyCharm 对我大喊大叫，因为我试图将 next 调用为下一个(some_node，direction=-1)。执行此操作的正确方法是什么？

Answer 1

除了__iter__还有__reversed__。两者都需要返回 iterators。 __next__ 方法应该在 iterators 上实现(而不是在节点类上)。请注意，所有魔术方法(当由 next 之类的函数调用而不是直接调用时)都需要实现预期参数，不多也不少。

例如，双向链表可以只实现 __iter__ 和 __reversed__ 并依赖于 next 和 previous 属性节点的:

class Node(object):
    def __init__(self, val, nxt, prv):
        self.val = val
        self.nxt = nxt
        self.prv = prv


class DoublyLinkedList(object):
    def __init__(self, base=None, last=None):
        self.base = base
        self.last = last

    def prepend(self, val):
        new = Node(val, self.base, None)
        if self.base is None:
            self.base = new
            self.last = new
        else:
            self.base.prv = new
            self.base = new

    def append(self, val):
        new = Node(val, None, self.last)
        if self.last is None:
            self.base = new
            self.last = new
        else:
            self.last.nxt = new
            self.last = new

    def __iter__(self):
        current = self.base
        while current is not None:
            yield current
            current = current.nxt

    def __reversed__(self):
        current = self.last
        while current is not None:
            yield current
            current = current.prv

例如:

dl = DoublyLinkedList()
dl.prepend(10)
dl.prepend(20)
dl.prepend(30)

for i in dl:
    print(i.val)

给出:

30
20
10

类似于reversed:

for i in reversed(dl):
    print(i.val)

# prints:
10
20
30