Python在第一个StopIteration上退出使用者 [英] Python exit consumer on first StopIteration

问题描述

这是我的 1个生成器-多个消费者的后续操作问题。由于 StopIteration 是生成器用尽信号的方式，因此，不幸的是，我现在在 client各处散布着许多异常处理代码代码（在下面的示例中，每个 next（）语句）。

It's a follow-up to my 1 generator -- multiple consumers question. As StopIteration is the way the generator signals its exhaustion, unfortunately, I now have many exception-handling code littered all over the place in the client code (for every next() statement in the example below).

是否有一个遇到第一个 StopIteration 异常时，使用 meal 内置的任何值退出的更好方法是什么？

Is there a better way to exit with whatever value is built in meal upon hitting the first StopIteration exception?

def client(course, take):
    meal = []
    for _ in range(take):
        try:
            some_meal = next(course)
            meal.append(some_meal)
        except StopIteration:
            pass
    if take % 2 == 0:
        try:
            some_meal = next(course)
            meal.append(some_meal)
        except StopIteration:
            pass
    return meal

更新最终，我最终使用了'itertools.islice'（请参见accepte下面的d解决方案），因为此函数负责处理 StopIteration 本身（请参见 for 循环等效的实现， itertools 文档。与使用 next 默认第二个参数相比，我更喜欢这种解决方案，因为它暗示着检查每个 meal （不过，我还是最好

UPDATE Eventually, I ended up using 'itertools.islice' (see accepted solution below) as this function takes care of the StopIteration itself (see the for-loop equivalent implementation shown in the itertools doc. I prefer this solution over using next default second argument as it would imply checking each meal (still, I'd better use the latter than all the exception handling above).

推荐答案

只需直接在第一个异常处理程序中返回：

Just return directly in the first exception handler:

def client(course, take):
    meal = []
    for _ in range(take):
        try:
            some_meal = next(course)
            meal.append(some_meal)
        except StopIteration:
            return meal
    if take % 2 == 0:
        try:
            some_meal = next(course)
            meal.append(some_meal)
        except StopIteration:
            pass
    return meal

尽管我在这里仍会使用标准库，而不必捕获那些 StopIteration 异常几乎一样多：

although I'd still use the standard library more here and not have to catch those StopIteration exceptions nearly as much:

from itertools import islice


def client(course, take):
    meal = list(islice(course, take))

    if take % 2 == 0:
        some_meal = next(course, None)
        if some_meal is not None:
            meal.append(some_meal)

    return meal

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