Python 在集合上迭代的类

函数获取一个集合并返回一个列表

布景总是很好

s={1,2,4,8,16}

例如：

class Ordered:
    def __init__(self,aset):
        self.aset = aset
    def __iter__(self):
        for v in sorted(self.aset): # iterate over list of values returned by sorted
            yield v

但当我的功能发挥作用时

s = {1, 2, 4, 8, 16}
i = iter(Ordered(s))
print(next(i))
print(next(i))
s.remove(8)
print(next(i))
s.add(32)
print(next(i))
print(next(i))

it should prints 1 2 4 16 32

应该打印出来

[next(i), next(i), s.remove(8), next(i), next(i), s.add(32), next(i)]

但相反，它打印：

[1, 2, None, 4, 16, None, 32]

有人能告诉我怎么修吗？谢谢

我在下面发布了我得到的错误，以帮助理解：

[1, 2, None, 4, 8, None, 16]

函数对参数进行排序并返回一个列表。修改输入集不会影响排序列表

---

如果您希望迭代器反映对原始集合的更改，迭代器需要在每次迭代时检查集合的状态并做出相应的响应。

使用排序的

时，您需要从集合中创建一个排序列表。该列表与创建它的原始集没有连接，并且不会反映对该集的任何更改。您必须自己跟踪更改，同时按排序顺序迭代元素

跟踪删除的元素很简单：只需检查当前元素是否仍在原始集合中。跟踪新元素有点复杂。您可以使用该模块并从集合中创建一个新的堆，而不是一个排序列表，然后您可以在迭代堆的同时向该堆添加新元素。要查找新元素，请创建原始集合的副本，并在每次迭代中比较两者。另外，根据您的测试用例，您必须检查当前元素是否比前一个元素小，在本例中跳过它

39 *Error: Failed [next(i), next(i), s.remove(8), next(i), next(i), s.add(32), next(i)] == [1, 2, None, 4, 16, None, 32]
      evaluated: [1, 2, None, 4, 8, None, 16] == [1, 2, None, 4, 16, None, 32]
42 *Error: [next(i), next(i), next(i), s.add(3), next(i), s.add(10), s.add(32), next(i), next(i), next(i)] raised exception; unevaluated: [1, 2, 4, None, 8, None, None, 10, 16, 32]
46 *Error: Failed [next(i), s.remove(2), s.remove(4), s.remove(8), next(i)] == [1, None, None, None, 16]
      evaluated: [1, None, None, None, 2] == [1, None, None, None, 16]
49 *Error: Failed [next(i), s.remove(2), next(i), s.remove(4), s.remove(8), next(i)] == [1, None, 4, None, None, 16]
      evaluated: [1, None, 2, None, None, 4] == [1, None, 4, None, None, 16]

删除某些内容后，迭代器不再有效。但是我如何修复它？有人能帮我修复它吗？
import heapq

class Ordered:

    def __init__(self,aset):
        self.aset = aset

    def __iter__(self):
        # memorize original elements
        known = set(self.aset)
        last = None

        # create heap
        heap = list(self.aset)
        heapq.heapify(heap)

        # yield from heap
        while heap:
            v = heapq.heappop(heap)
            if (v in self.aset and  # not removed
                    (last is None or last < v)): # not smaller than last
                yield v
            last = v

            # in case of new elements, update the heap
            diff = self.aset - known
            if diff:
                for x in self.aset - known:
                    heapq.heappush(heap, x)
                known = set(self.aset)

>>> s = {1, 2, 4, 8, 16}
>>> i = iter(Ordered(s))
>>> [next(i), next(i), s.remove(8), next(i), next(i), s.add(32), next(i)]
[1, 2, None, 4, 16, None, 32]
>>> [next(i), next(i), next(i), s.add(3), next(i), s.add(10), s.add(32), next(i), next(i), next(i)]
[1, 2, 4, None, 8, None, None, 10, 16, 32])
>>> [next(i), s.remove(2), s.remove(4), s.remove(8), next(i)]
[1, None, None, None, 16]
>>> [next(i), s.remove(2), next(i), s.remove(4), s.remove(8), next(i)]
[1, None, 4, None, None, 16]