在迭代是列表或具有len方法的情况下,该解决方案是干净和容易的：

def zip_equal(it1,it2):
    if len(it1) != len(it2):
        raise ValueError("Lengths of iterables are different")
    return zip(it1,it2)

但是,如果it1和it2是生成器,则前一个函数失败,因为长度未定义TypeError：类型为“generator”的对象没有len().

我想象,itertools模块提供了一种简单的方法来实现,但到目前为止我还没有找到它.我想出了这个自制的解决方案：

def zip_equal(it1,it2):
    exhausted = False
    while True:
        try:
            el1 = next(it1)
            if exhausted: # in a prevIoUs iteration it2 was exhausted but it1 still has elements
                raise ValueError("it1 and it2 have different lengths")
        except StopIteration:
            exhausted = True
            # it2 must be exhausted too.
        try:
            el2 = next(it2)
            # here it2 is not exhausted.
            if exhausted:  # it1 was exhausted => raise
                raise ValueError("it1 and it2 have different lengths")
        except StopIteration:
            # here it2 is exhausted
            if not exhausted:
                # but it1 was not exhausted => raise
                raise ValueError("it1 and it2 have different lengths")
            exhausted = True
        if not exhausted:
            yield (el1,el2)
        else:
            return

该解决方案可以用以下代码进行测试：

it1 = (x for x in ['a','b','c'])  # it1 has length 3
it2 = (x for x in [0,1,2,3])     # it2 has length 4
list(zip_equal(it1,it2))           # len(it1) < len(it2) => raise
it1 = (x for x in ['a',3])     # it2 has length 4
list(zip_equal(it2,it1))           # len(it2) > len(it1) => raise
it1 = (x for x in ['a','c','d'])  # it1 has length 4
it2 = (x for x in [0,3])          # it2 has length 4
list(zip_equal(it1,it2))                # like zip (or izip in python2)

我是否俯视任何替代解决方案？我的zip_equal函数有更简单的实现吗？

PS：我在Python 3中写了这个问题,但Python 2解决方案也是受欢迎的.