How does await works in a for-loop? - python

I just watch a async/await tutorial video on youtube.
To my understanding of await, if await is in a task, when execute the task it would turn back to the event-loop while it encounter the await inside of the task.
So if await inside a for loop(that's say 10 loops), the task would be paused for 10 times, and I should use 10 await in the event-loop in order to finished the task, like this:
import asyncio
async def print_numbers():
for i in range(10):
print(i)
await asyncio.sleep(0.25)
async def main():
task2 = asyncio.create_task(print_numbers())
for i in range(10):
await task2
asyncio.run(main())
But, in fact the task can be done by using only 1 await, like this:
async def print_numbers():
for i in range(10):
print(i)
await asyncio.sleep(0.25)
async def main():
task2 = asyncio.create_task(print_numbers())
await task2
asyncio.run(main()
What do I missing in this topic?

it would turn back to the event-loop while it encounter the await inside of the task
It does, but you wait for task[0] to finish before you start task[1], so there is simply no other task in the event loop to do. So your code just ends up sleeping and doing nothing
and I should use 10 await in the event-loop in order to finished the task
Yes you will need to await the 10 tasks you started, so your code will only continue once all 10 tasks are done. But you should use asyncio.wait or asyncio.gather so the individual tasks can be parallelized and don't have to wait for the previous one to finish.
import asyncio
import random
async def print_number(i):
print(i, 'start')
await asyncio.sleep(random.random())
print(i, 'done')
async def main():
await asyncio.wait([
asyncio.create_task(print_number(i))
for i in range(10)
])
print('main done')
asyncio.run(main())

Related

Why is my async code running in the incorrect order?

I'm trying to run some code asynchronously. My expectation is that the test coroutine should not block the print(running first) statement. This is because I've dispatched it to the event loop, and should be seeing the output of this command logged first.
import asyncio
async def test():
await asyncio.sleep(5)
print("I should run second")
asyncio.run(test())
print('running first')
Does anyone have any tips on how to how this code run so that print('running first') is ran before print("I should run second")? I believe this code should be non-blocking, so I'm confused as to why the order of print messages isn't matching my expectation.
I believe this is what you want:
import asyncio
async def test():
await asyncio.sleep(5)
print("I should run second")
async def main():
task1 = asyncio.create_task(test())
print('running first')
await task1
asyncio.run(main())
A more detail explaination:
asyncio.run() will try to wait all of the task inside it to finish before it continues.
In your code, you are running asyncio.run(test()) first and it will continue ONLY IF test() IS ENDED and you awaited the sleep. so test() will end after the sleep and run the print then the main print.
This is why your code delay so long before running. The solution to it is simple. Fire the task without waiting, which is what asyncio.create_task() is doing, I created a task, fire it but wait it at the end.
btw normally when you are using async, you will have a ton of task in a list. If you want to wait it as a list you should use gather():
import asyncio
async def test():
await asyncio.sleep(5)
print("I should run second")
async def main():
task_list = []
for _ in range(100):
task_list.append(asyncio.create_task(test()))
print('running first')
await asyncio.gather(*task_list)
asyncio.run(main())

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())

Async Function Call Inside While Loop

I have a queue which stored on Redis lists. I'm trying to create async consumer for this queue. But couldn't call async function inside loop. Its working like sync function when I call.
import asyncio
async def worker():
print("starting sleep")
await asyncio.sleep(2)
print("slept")
async def main():
while True:
await worker()
asyncio.run(main())
Here is a short and simple example of mine implemantation. I'm expecting to see 'starting sleep' messages until first 'slept' message, it means for 2 seconds.
main is literally awaiting the completion of worker. Until worker is done, main won't progress. async tasks don't run in the background like in multithreading.
What you want is to keep launching new workers without awaiting each one of them. However, if you just keep doing this in a loop like this:
while True:
worker()
then you will never see any output of those workers, since this is an endless loop which never gives anything else the chance to run. You'd need to "break" this loop in some way to allow workers to progress. Here's an example of that:
import asyncio
async def worker():
print("starting sleep")
await asyncio.sleep(2)
print("slept")
async def main():
while True:
asyncio.ensure_future(worker())
await asyncio.sleep(0.5)
asyncio.run(main())
This will produce the expected outcome:
starting sleep
starting sleep
starting sleep
starting sleep
slept
starting sleep
slept
...
The await inside main transfers control back to the event loop, which now has the chance to run the piled up worker tasks, When those worker tasks await, they in turn transfer control back to the event loop, which will transfer it back to either main or a worker as their awaited sleep completes.
Note that this is only for illustration purposes; if and when you interrupt this program, you'll see notices about unawaited tasks which haven't completed. You should keep track of your tasks and await them all to completion at the end somewhere.
Here is an example using asyncio.wait:
import asyncio
async def worker():
print("starting sleep")
await asyncio.sleep(2)
print("slept")
async def main():
tasks = [worker() for each in range(10)]
await asyncio.wait(tasks)
asyncio.run(main())
It spawns all the workers together.

How to do some work before async task finish?

I am new to study about asyncio.I don't know how to
describe my question.But here is a minimal example:
import asyncio
async def work():
await asyncio.sleep(3)
async def check_it():
task = asyncio.create_task(work())
await task
while True:
if task.done():
print("Done")
break
print("Trying...")
asyncio.run(check_it())
My idea is very simple:
create a async task in check_it().And await it.
Use a while loop to check whether the task is finished.
If task.done() return True,break the while loop.Then exit the script.
If my question is duplicate, please flag my question.Thanks!
Try asyncio.wait or use asyncio.sleep. Otherwise, your program will output a lot without some pauses.
import asyncio
async def work():
await asyncio.sleep(3)
async def check_it():
task = asyncio.create_task(work())
# "await" block until the task finish. Do not do here.
timeout = 0 # Probably the first timeout is 0
while True:
done, pending = await asyncio.wait({task}, timeout=timeout)
if task in done:
print('Done')
# Do an await here is favourable in case any exception is raised.
await task
break
print('Trying...')
timeout = 1
asyncio.run(check_it())

asyncio task cancel. Is is synchronous?

I understand that task.cancel() arranges an exception to be thrown inside the task function. Is that happen in a synchronous way? (As I don't await task.cancel()). Can code that follows the line task.cancel() assume that the task will no longer run?
A simple example:
async def task1():
await asyncio.sleep(3)
print("after sleep")
async def task2():
t = loop.create_task(task1())
await asyncio.sleep(1)
t.cancel()
# can the following code lines assume that task1 is no longer running?
loop = asyncio.get_event_loop()
loop.run_forever()
Can code that follows the line task.cancel() assume that the task will
no longer run?
No. task.cancel() only marks task to be cancelled later. You should explicitly await task after it and catch CancelledError to be sure task is cancelled.
See example here.

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