如何生成通知并从函数返回结果？（Python）

Question

我必须创建一个函数，在内部调用中做一些艰苦的工作。此函数必须是生成器，因为我使用的是服务器发送的事件。因此，我希望这个函数通过使用“yield”来通知计算的进度。之后，此函数必须将结果传递给父函数，以便继续进行其他计算。你知道吗

我想要这样的东西：

def hardWork():
    for i in range(N):
        # hard work
        yield 'Work done: ' + str(i)

    # Here is the problem: I can't return a result if I use a yield
    return result             

def generator():
    # do some calculations
    result = hardWork()
    # do other calculations with this result
    yield finalResult

我发现了一个解决方案，它包含一个字典，告诉我们函数是否已经完成，但是执行这个操作的代码非常脏。你知道吗

还有别的解决办法吗？你知道吗

谢谢你！你知道吗

编辑

我的想法是：

def innerFunction(gen):
    calc = 1

    for iteration in range(10):
        for i in range(50000):
            calc *= random.randint(0, 10)
        gen.send(iteration)

    yield calc


def calcFunction(gen):
    gen2 = innerFunction(gen)
    r = next(gen2)

    gen.send("END: " + str(r + 1))
    gen.send(None)


def notifier():
    while True:
        x = yield
        if x is None:
            return
        yield "Iteration " + x


def generator():
    noti = notifier()
    calcFunction(noti)
    yield from noti


for g in generator():
    print(g)

但我收到一个错误：

TypeError: can't send non-None value to a just-started generator

Answer 1

我的建议是将您的功能嵌入到类中。你知道吗

def Worker:

    def hardWork(self):
        self.Finished = False
        for i in range(10):
            yield 'Work done: ' + str(i)
        self.Finished = True
        self.Result = 'result'

    def generator(self):
        while (not self.Finished):
            print(next(self.hardWork()))
        return self.Result

这将具有您想要的功能，而不必担心围绕异常抛出捕获逻辑编程逻辑。你知道吗

Answer 2

Python3.5之前：生成器

这个解决方案也适用于较新的Python版本，尽管Python3.5中的新版本async def似乎更适合您的用例。请参见下一节。你知道吗

生成器生成的值是通过迭代或使用next获得的。最后返回的值存储在StopIteration异常的value属性中，该异常指示生成器的结束。幸运的是，恢复并不太难。你知道吗

def hardWork():
    output = []

    for i in range(10):
        # hard work
        yield 'Doing ' + str(i)
        output.append(i ** 2)

    return output

def generator():
    # do some calculations
    work = hardWork()

    while True:
        try:
            print(next(work))
        except StopIteration as e:
            result = e.value
            break

    yield result

示例

foo = generator()
print(next(foo))

输出

Doing 0
Doing 1
Doing 2
Doing 3
Doing 4
Doing 5
Doing 6
Doing 7
Doing 8
Doing 9
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

Python3.5+：async def

如果您运行的是Python3.5+，那么您所尝试的似乎非常适合使用等待函数的事件循环。你知道吗

import asyncio

async def hardWork():
    output = []

    for i in range(10):
        # do hard work
        print('Doing ', i)
        output.append(i**2)

        # Break point to allow the event loop to do other stuff on the side
        await asyncio.sleep(0)

    return output

async def main():
    result = await asyncio.wait_for(hardWork(), timeout=None)
    print(result)

loop = asyncio.get_event_loop()

loop.run_until_complete(main())

输出

Doing  0
Doing  1
Doing  2
Doing  3
Doing  4
Doing  5
Doing  6
Doing  7
Doing  8
Doing  9
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

Answer 3

python3.3引入了^{} for generator delegation，这正是您所要求的。yield from允许主生成器将功能委托给另一个也包含yield语句的函数，如下所示：

def hardWork():
    for i in range(N):
        # hard work
        yield 'Work done so far: ' + str(i)

    # With this construction, you can still return a result after the `yield`s
    return result

def generator():
    # here's the new construction that makes it all work:
    result = yield from hardWork()

    # do other calculations with this result
    yield finalResult