Question

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).

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

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).

Was it helpful?

Solution

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

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

OTHER TIPS

Besides using islice, if I read the code correctly then it can be made even simpler:

def client(course, take):
    if take % 2 == 0:
        take += 1
    return list(itertools.islice(course, take))
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