问题:“失火” Python异步/等待
有时需要执行一些非关键性的异步操作,但我不想等待它完成。在Tornado的协程实现中,您可以通过简单地省略yield
关键字来“触发并忘记”一个异步函数。
我一直在尝试找出如何使用Python 3.5中发布的新语法async
/ 来“激发并忘记” await
。例如,一个简化的代码片段:
async def async_foo():
print("Do some stuff asynchronously here...")
def bar():
async_foo() # fire and forget "async_foo()"
bar()
但是,发生的事情是bar()
永远不会执行,而是收到运行时警告:
RuntimeWarning: coroutine 'async_foo' was never awaited
async_foo() # fire and forget "async_foo()"
回答 0
更新:
如果您使用的是Python> = 3.7,请在任何地方替换asyncio.ensure_future
为asyncio.create_task
最新的,更好的派生task的方法。
asyncio。“解雇”的任务
根据python docs的asyncio.ensure_future
函数创建的任务不会阻止执行(因此函数将立即返回!)。这似乎是您要求的一种“解雇”的方法。
import asyncio
async def async_foo():
print("async_foo started")
await asyncio.sleep(1)
print("async_foo done")
async def main():
asyncio.ensure_future(async_foo()) # fire and forget async_foo()
# btw, you can also create tasks inside non-async funcs
print('Do some actions 1')
await asyncio.sleep(1)
print('Do some actions 2')
await asyncio.sleep(1)
print('Do some actions 3')
if __name__ == '__main__':
loop = asyncio.get_event_loop()
loop.run_until_complete(main())
输出:
Do some actions 1
async_foo started
Do some actions 2
async_foo done
Do some actions 3
如果事件循环完成后正在执行任务怎么办?
请注意,asyncio期望任务将在事件循环完成时完成。因此,如果您更改main()
为:
async def main():
asyncio.ensure_future(async_foo()) # fire and forget
print('Do some actions 1')
await asyncio.sleep(0.1)
print('Do some actions 2')
程序完成后,您会收到以下警告:
Task was destroyed but it is pending!
task: <Task pending coro=<async_foo() running at [...]
为防止这种情况,您可以在事件循环完成后等待所有待处理的任务:
async def main():
asyncio.ensure_future(async_foo()) # fire and forget
print('Do some actions 1')
await asyncio.sleep(0.1)
print('Do some actions 2')
if __name__ == '__main__':
loop = asyncio.get_event_loop()
loop.run_until_complete(main())
# Let's also finish all running tasks:
pending = asyncio.Task.all_tasks()
loop.run_until_complete(asyncio.gather(*pending))
杀死任务而不是等待任务
有时,您不想等待任务完成(例如,某些任务可能创建为永久运行)。在这种情况下,您可以只取消()而不是等待它们:
import asyncio
from contextlib import suppress
async def echo_forever():
while True:
print("echo")
await asyncio.sleep(1)
async def main():
asyncio.ensure_future(echo_forever()) # fire and forget
print('Do some actions 1')
await asyncio.sleep(1)
print('Do some actions 2')
await asyncio.sleep(1)
print('Do some actions 3')
if __name__ == '__main__':
loop = asyncio.get_event_loop()
loop.run_until_complete(main())
# Let's also cancel all running tasks:
pending = asyncio.Task.all_tasks()
for task in pending:
task.cancel()
# Now we should await task to execute it's cancellation.
# Cancelled task raises asyncio.CancelledError that we can suppress:
with suppress(asyncio.CancelledError):
loop.run_until_complete(task)
输出:
Do some actions 1
echo
Do some actions 2
echo
Do some actions 3
echo
回答 1
谢谢谢尔盖的简洁回答。这是相同的装饰版。
import asyncio
import time
def fire_and_forget(f):
def wrapped(*args, **kwargs):
return asyncio.get_event_loop().run_in_executor(None, f, *args, *kwargs)
return wrapped
@fire_and_forget
def foo():
time.sleep(1)
print("foo() completed")
print("Hello")
foo()
print("I didn't wait for foo()")
产生
>>> Hello
>>> foo() started
>>> I didn't wait for foo()
>>> foo() completed
注意:检查我的其他答案,使用普通线程也可以做到这一点。
回答 2
这不是完全异步执行,但也许run_in_executor()适合您。
def fire_and_forget(task, *args, **kwargs):
loop = asyncio.get_event_loop()
if callable(task):
return loop.run_in_executor(None, task, *args, **kwargs)
else:
raise TypeError('Task must be a callable')
def foo():
#asynchronous stuff here
fire_and_forget(foo)
回答 3
由于某种原因,如果您无法使用,asyncio
那么这里是使用普通线程的实现。检查我的其他答案和谢尔盖的答案。
import threading
def fire_and_forget(f):
def wrapped():
threading.Thread(target=f).start()
return wrapped
@fire_and_forget
def foo():
time.sleep(1)
print("foo() completed")
print("Hello")
foo()
print("I didn't wait for foo()")
声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。