有没有更好的方法在python中执行“解开”功能？

因此,例如：我在8个线程上执行2,000个url请求但只想要前100个结果,但不是所有100个来自同一个潜在线程.

因此,解开：

import itertools

def unravel(*iterables,with_limit = None):
    make_iter = {a:iter(i) for a,i in enumerate(iterables)}

    if not isinstance(with_limit,int):
        with_limit = -1

    resize = False

    while True:
        for iid,take_from in make_iter.items():
            if with_limit == 0:
                raise StopIteration

            try:
                yield next(take_from)
            except StopIteration:
                resize = iid
            else: 
                with_limit -= 1

        if resize:
            resize = False

            if len(make_iter.keys()) > 1:
                make_iter.pop(resize)

            else: raise StopIteration

用法：

>>> a = [1,2,3,4,5]
>>> b = [6,7,8,9,10]
>>> c = [1,5,7]
>>> d = [2,6,8]
>>> 
>>> print([e for e in unravel(c,d)])
[1,8]
>>> print([e for e in unravel(c,d,with_limit = 3)])
[1,3]
>>> print([e for e in unravel(a,b,with_limit = 6)])
[1,8]
>>> print([e for e in unravel(a,with_limit = 100)])
[1,10]

这样的事情是否已经存在,或者这是一个不错的实现？

谢谢

编辑,工作修复

受到@abernert建议的启发,这就是我的选择.谢谢大家！

def unravel(*iterables,limit = None):
    yield from itertools.islice(
            filter(None,itertools.chain.from_iterable(
                    itertools.zip_longest(
                        *iterables
                    )
                ) 
            ),limit)



>>> a = [x for x in range(10)]
>>> b = [x for x in range(5)]
>>> c = [x for x in range(0,20,2)]
>>> d = [x for x in range(1,30,2)]
>>> 
>>> print(list(unravel(a,b)))
[1,1,9]
>>> print(list(unravel(a,limit = 3)))
[1,2]
>>> print(list(unravel(a,c,limit = 20)))
[1,10,11]