如何生成通知并从函数返回结果?(Python)
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 我想要这样的东西:如何生成通知并从函数返回结果?(Python),python,generator,yield,server-sent-events,Python,Generator,Yield,Server Sent Events,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 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'
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
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
我找到了一个解决方案,它包含一个字典,可以告诉我们函数是否完成了,但是执行此操作的代码非常脏
还有别的解决办法吗
谢谢大家!
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)
但我收到了这个错误:
TypeError: can't send non-None value to a just-started generator
Python3.5之前:生成器
此解决方案也适用于较新的Python版本,尽管Python3.5中新增的async def
,似乎更适合您的用例。见下一节
生成器生成的值通过迭代或使用next
获得。结束时返回的值存储在表示生成器结束的StopIteration
异常的value
属性中。幸运的是,恢复并不难
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
例子
输出
Python3.5+:async def
如果您正在运行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
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
我的建议是将您的功能嵌入到类中
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您的逻辑。Python 3.3介绍了这一点,这正是您想要的yield from
允许主生成器将功能委托给另一个函数,该函数也包含yield
语句,如下所示:
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
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
请看这个问题:我认为您应该真正了解异步函数,因为这与您正在做的非常接近
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