将python`cycle`转换为`list`

cycle 在这里仍然可以很好地工作 set

from itertools import cycle, filterfalse

allobjects = [...]
numuniqueobjects = len(set(allobjects))
inactiveobjects = set()

# Each time we request an item, filterfalse pulls items from the cycle
# until we find one that isn't in our inactive set
for object in filterfalse(inactiveobjects.__contains__, cycle(allobjects)):

    # ... do actual stuff with object ...

    # Any objects that should go active again get removed from the set and will be
    # seen again the next time their turn comes up in the original order
    inactiveobjects -= objects_that_should_become_active()

    # Won't see this object again until it's removed from inactiveobjects
    if object.should_go_inactive():
        inactiveobjects.add(object)
        if len(inactiveobjects) == numuniqueobjects:
            # Nothing is active, continuing loop would cause infinite loop
            break

这种设计的优点是:

主要的缺点是它会给“什么都没有改变”的情况增加一点点开销,特别是如果 设置 inactiveobjects 增长到相当于对象总数的一小部分;您仍然需要 周期 对象即使过滤掉 许多的 他们中的一个。

如果这不适合您的用例,那么 deque 根据建议 wim 可能是最好的通用解决方案:

from collections import deque
from collections.abc import Iterator

class mutablecycle(Iterator):
    def __init__(self, it):
        self.objects = deque(it)
        self.objects.reverse() # rotate defaults to equivalent of appendleft(pop())
                               # reverse so next item always at index -1

    def __next__(self):
        self.objects.rotate() # Moves rightmost element to index 0 efficiently
        try:
            return self.objects[0]
        except IndexError:
            raise StopIteration

    def removecurrent(self):
        # Remove last yielded element
        del self.objects[0]

    def remove(self, obj):
         self.objects.remove(obj)

    def add(self, obj, *, tofront=True):
        if tofront:
            # Putting it on right makes it be yielded on next request
            self.objects.append(obj)
        else:
            # Putting it on left makes it appear after all other elements
            self.objects.appendleft(obj)

mycycle = mutablecycle(allobjects):
for object in mycycle:
    # ... do stuff with object ...

    if object.should_go_inactive():
        mycycle.removecurrent()  # Implicitly removes object currently being iterated

不,你不能,因为 cycle 是一个无限序列。您的计算机“冻结”是因为Python试图迭代一个永无止境的项集合(如果您将其保留足够长的时间,该进程将耗尽内存并崩溃)。

您可以做的是,将预先确定的有限数量的项目收集到一个列表中:

n = 10  # some fixed size 
results = []
for i in range(n):
    results.append(next(my_cycle))

没有通用的方法可以知道一次循环要消耗多少项,因为cycle对象不公开任何关于底层迭代周期的状态,即在重复之前迭代了多少项。

没有任何公共方法可以修改从循环返回的项,只要第一个循环返回一次 StopIteration

>>> L = [0,1,2]
>>> g = itertools.cycle(L)
>>> next(g)
0
>>> L.remove(1)
>>> next(g)
2
>>> next(g)
0
>>> L.remove(2)
>>> next(g)
2

collections.deque 实例作为数据结构 rotate 方法是有效的)。