Prevent await from blocking async function from going on || Python - python

So I have an async function which calls another function which should never end but I want function one to move on.
async def funtion2():
#do stuff
await function2():
async def function1():
#do stuff
await function2()
#do more stuff
In this sample "do more stuff" will never happen because function2 never ends, how do I still make "do more stuff" happen?
I tried building loops to avoid this but those either failed due to the same problem or the maximum recursion depth.

async def funtion2():
#do stuff
await function2():
async def function1(loop):
#do stuff
loop.create_task(function2())
# do more is now hit
if __name__ == "__main__":
loop = asyncio.get_event_loop()
loop.run(function1(loop))
Is one way you could add it to the currently running event loop

Related

Python doesn't want to run function asynchronous

I just want to run async function like really async, but it blocks main thread for 5 seconds
import asyncio
async def sideTask():
await asyncio.sleep(5) # this will pause main thread for no reason
print("doing side job")
global test
test = 0
# main work
while True:
test += 1
time.sleep(1)
if test > 3:
test = 0
asyncio.run(sideTask())
I am expecting async function being async
What are you trying to achieve by running sideTask() in the loop?
If you're trying to see output of "doing side job" while asyncio.sleep(5) is executing, you should simply put "doing side job" before asyncio.sleep(5)
asyncio.sleep() is not pausing the main thread, you are just doing sleep function which serves no purpose and makes it look like it's not functioning in the main time.

In Python, kick off an async function and return early

I have a function in python that should kick off an asynchronous 'slow' function, and return before that 'slow' function completes.
async def doSomethingSlowButReturnQuickly():
asyncio.create_task(_mySlowFunction())
return "Returning early! Slow function is still running."
async def _mySlowFunction():
// Do some really slow stuff that isn't blocking for our other function.
Running this still seems to cause the 'return' to not happen until AFTER my async task has completed.
How do I correct this?
(And apologies, this is day #2 writing Python)
This isn't exactly what you asked, but you can run code in an async executor and continue while it is still running:
from concurrent.futures import ThreadPoolExecutor
from time import sleep
def blocking_code():
sleep(2)
print('inner')
return 'result'
def main():
with ThreadPoolExecutor(max_workers=1) as executor:
future = executor.submit(blocking_code)
print('after')
print(future.result())
return
if __name__ == '__main__':
main()
Output:
after
inner
result
Maybe it's a dull answer, but if you want to stick with asyncio, I would advise to be sure using await in everyasync def functions.
async def doSomethingSlowButReturnQuickly():
task = asyncio.create_task(_mySlowFunction())
# must await something here!
return "Returning early! Slow function is still running."
async def _mySlowFunction():
# must await something here
If you don't need to get the slow function return value, why don't you directly call the two coroutines from a main function?
import asyncio
async def doSomethingSlowButReturnQuickly():
await asyncio.sleep(0.1) # must await something here!
return "Returning early! Slow function is still running."
async def _mySlowFunction():
await asyncio.sleep(10) # must await something here
async def main():
tasks = [_mySlowFunction(), doSomethingSlowButReturnQuickly()]
asyncio.gather(*tasks)
asyncio.run(main())

Python asyncio wait and notify

I am trying to do something similar like C# ManualResetEvent but in Python.
I have attempted to do it in python but doesn't seem to work.
import asyncio
cond = asyncio.Condition()
async def main():
some_method()
cond.notify()
async def some_method():
print("Starting...")
await cond.acquire()
await cond.wait()
cond.release()
print("Finshed...")
main()
I want the some_method to start then wait until signaled to start again.
This code is not complete, first of all you need to use asyncio.run() to bootstrap the event loop - this is why your code is not running at all.
Secondly, some_method() never actually starts. You need to asynchronously start some_method() using asyncio.create_task(). When you call an "async def function" (the more correct term is coroutinefunction) it returns a coroutine object, this object needs to be driven by the event loop either by you awaiting it or using the before-mentioned function.
Your code should look more like this:
import asyncio
async def main():
cond = asyncio.Condition()
t = asyncio.create_task(some_method(cond))
# The event loop hasn't had any time to start the task
# until you await again. Sleeping for 0 seconds will let
# the event loop start the task before continuing.
await asyncio.sleep(0)
cond.notify()
# You should never really "fire and forget" tasks,
# the same way you never do with threading. Wait for
# it to complete before returning:
await t
async def some_method(cond):
print("Starting...")
await cond.acquire()
await cond.wait()
cond.release()
print("Finshed...")
asyncio.run(main())

asyncio: wait for async callback with timeout

I am not much used to asyncio, so perhaps this question is trivial.
I have a code running asynchronously, which will run a callback when done (the callback can be callable or awaitable). I would like to wait for the callback to be called, with timeout. I sense that it is conceptually a task, but I am not sure how to create the task but wait for it somewhere else.
import asyncio, inspect
async def expensivefunction(callback):
# this is something which takes a lot of time
await asyncio.sleep(10)
# but eventually computes the result
result=10
# and calls the callback
callback(result)
if inspect.isawaitable(callback): await callback
# just print the result, for example
async def callback(result): print(result)
# main code async
async def myfunc():
await expensivefunction(callback=callback)
# this will wait for callback to be called within 5 seconds
# if not, exception is thrown
await asyncio.wait_for(...??,timeout=5)
asyncio.run(myfunc())
What would be the right approach to this?
Please find working example:
import asyncio
AWAIT_TIME = 5.0
async def expensive_function():
"""this is something which takes a lot of time"""
await asyncio.sleep(10)
result = 10
return result
def callback(fut: asyncio.Future):
"""just prints result. Callback should be sync function"""
if not fut.cancelled() and fut.done():
print(fut.result())
else:
print("No results")
async def amain():
"""Main async func in the app"""
# create task
task = asyncio.create_task(expensive_function())
task.add_done_callback(callback)
# try to await the task
try:
r = await asyncio.wait_for(task, timeout=AWAIT_TIME)
except asyncio.TimeoutError as ex:
print(ex)
else:
print(f"All work done fine: {r}")
finally:
print("App finished!")
if __name__ == '__main__':
asyncio.run(amain())
If any questions, please let me know.

python Make an async timer without waiting to finish

I want to make a timer which is started in a normal function, but in the timer function, it should be able to call an async function
I want to do something like this:
startTimer()
while True:
print("e")
def startTimer(waitForSeconds: int):
# Wait for `waitForSeconds`
await myAsyncFunc()
async def myAsyncFunc():
print("in my async func")
Where the while True loop should do its stuff and after waitForSeconds the timer the async function should execute an other async function, but waiting shouldn't block any other actions and doesn't need to be awaited
If something isn't understandable, I'm sorry, I'll try to explain it then
Thanks
If you want to run your synchronous and asynchronous code in parallel, you will need to run one of them in a separate thread. For example:
def sync_code():
while True:
print("e")
async def start_timer(secs):
await asyncio.sleep(secs)
await async_func()
async def main():
asyncio.create_task(start_timer(1))
loop = asyncio.get_event_loop()
# use run_in_executor to run sync code in a separate thread
# while this thread runs the event loop
await loop.run_in_executor(None, sync_code)
asyncio.run(main())
If the above is not acceptable for you (e.g. because it turns the whole program into an asyncio program), you can also run the event loop in a background thread, and submit tasks to it using asyncio.run_coroutine_threadsafe. That approach would allow startTimer to have the signature (and interface) like you wanted it:
def startTimer(waitForSeconds):
loop = asyncio.new_event_loop()
threading.Thread(daemon=True, target=loop.run_forever).start()
async def sleep_and_run():
await asyncio.sleep(waitForSeconds)
await myAsyncFunc()
asyncio.run_coroutine_threadsafe(sleep_and_run(), loop)
async def myAsyncFunc():
print("in my async func")
startTimer(1)
while True:
print("e")
I'm pretty sure that you are familiar with concurent processing, but you didn't show exactly what you want. So if I understand you correctly you want to have 2 processes. First is doing only while True, and the second process is the timer(waits e.g. 5s) and it will call async task. I assume that you are using asyncio according to tags:
import asyncio
async def myAsyncFunc():
print("in my async func")
async def call_after(delay):
await asyncio.sleep(delay)
await myAsyncFunc()
async def while_true():
while True:
await asyncio.sleep(1) # sleep here to avoid to large output
print("e")
async def main():
task1 = asyncio.create_task(
while_true())
task2 = asyncio.create_task(
call_after(5))
# Wait until both tasks are completed (should take
# around 2 seconds.)
await task1
await task2
asyncio.run(main())

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