I need to implement the algorithm using asyncio with the following conditions (behaviour):
check the list of arguments is not empty, if empty finish execution
pop next argument from the list of arguments create coroutine with
this argument and schedule it for execution "at same time" can't be
executed no more than 'async_level' coroutines when coroutine finish
execution -> go to the step 1
It is necessary not to plan to complete all tasks at once (as with asyncio.gather), but to do it in parts. When the next task finishes execution, a new one takes its place.
I tried to do it with asyncio.as_completed() but it doesn't actually work as expected:
async_level = 4
params_count = 10
params = [i for i in range(1, params_count + 1)]
tasks = {asyncio.create_task(job(param)) for param in params[0: async_level]}
params = iter(params[async_level:])
while True:
# NOTE: It wont work, because you can't add task in 'tasks' after 'as_completed' is invoked, so execution actually ends when the last coroutine in the 'as_completed' ends
for task in asyncio.as_completed(tasks):
print(f"len(tasks) = {len(tasks)}")
await task
try:
param = next(params)
tasks.add(asyncio.create_task(job(param)))
except StopIteration:
print("StopIteration")
break
Also, I tried to implement it using asyncio.BoundedSemaphore, but first two conditions are not met:
async_level = 4
params_count = 10
params = [i for i in range(1, params_count + 1)]
async def semaphore_job(name, _asyncio_semaphore):
async with _asyncio_semaphore:
await job(name)
asyncio_semaphore = asyncio.BoundedSemaphore(async_level)
jobs = []
# NOTE: This variant schedule all jobs at ones and it's significant drawback because the count of jobs can be overwhelmed
for param in params:
jobs.append(asyncio.ensure_future(semaphore_job(param, asyncio_semaphore)))
await asyncio.gather(*jobs)
I would be grateful for any of your help.
It seems I found the solution myself:
import asyncio
from typing import Callable
from random import randrange
from asyncio import Semaphore, ensure_future, get_event_loop
async def job(name, time_range=10):
timeout = randrange(time_range)
print(f"Task '{name}' started with timeout {timeout}")
await asyncio.sleep(timeout)
print(f"Task '{name}' finished")
return name
async def custom_executor(func: Callable, args: list, async_level: int = 4):
""" Asynchronously executes no more that 'async_level' callables specified by 'func' with corresponding 'args' """
loop = get_event_loop()
sync = Semaphore()
todo = set(args)
doing = set()
def _schedule_task():
if todo:
arg = todo.pop()
fr = func(*arg) if isinstance(arg, (tuple, list, set)) else func(arg)
f = ensure_future(fr, loop=loop)
f.add_done_callback(_on_completion)
doing.add(f)
def _on_completion(f):
doing.remove(f)
sync.release()
_schedule_task()
for _ in range(min(async_level, len(todo))):
_schedule_task()
while True:
if not doing:
break
await sync.acquire()
async def main():
await custom_executor(job, [(1, 3), 7, (8, 2), 12, 5])
if __name__ == '__main__':
asyncio.run(main())
But if you know a better way, please share!
You can create a fixed number of workers and give them tasks using a queue. It's a bit shorter and I find it a bit easier to reason about than your code, which uses callbacks. But YMMV.
async def custom_executor(func, args, async_level=4):
queue = asyncio.Queue(1)
async def worker():
while True:
arg = await queue.get()
fr = func(*arg) if isinstance(arg, (tuple, list, set)) else func(arg)
await fr
queue.task_done()
# create the workers
workers = [asyncio.create_task(worker()) for _ in range(async_level)]
# Feed the workers tasks. Since the queue is bounded, this will also
# wait for previous tasks to finish, similar to what you wanted to
# achieve with as_completed().
for x in args:
await queue.put(x)
await queue.join() # wait for the remaining tasks to finish
# cancel the now-idle workers
for w in workers:
w.cancel()
Related
I am planning to have an asyncio Queue based producer-consumer implementation for a processing of realtime data where sending out data in correct time order is vital. So here is the code snippet of it :
async def produce(Q, n_jobs):
for i in range(n_jobs):
print(f"Producing :{i}")
await Q.put(i)
async def consume(Q):
while True:
n = await Q.get()
print(f"Consumed :{n}")
x = do_sometask_and_return_the_result(n)
print(f"Finished :{n} and Result: {x}")
async def main(loop):
Q = asyncio.Queue(loop=loop, maxsize=3)
await asyncio.wait([produce(Q, 10), consume(Q), consume(Q), consume(Q)])
print("Done")
Here the producer produces data and puts it into the asyncio Queue. I have multiple consumers to consume and process the data. While seeing the outputs, the order is maintained while printing "Consumed :{n}" (as in 1,2,3,4... and so on) , this is completely fine. but, since the function do_sometask_and_return_the_result(n) takes variable time to return the result, the order is not maintained in the next print of n "Finished :{n}" (as in 2,1,4,3,5,...).
Is there any way to synchronize this data as I need to maintain the order of printing the results? I want to see 1,2,3,4,.. sequential prints for 'n' even after do_sometask_and_return_the_result(n).
You could use a priority queue system (using the python heapq library) to reorder your jobs after they are complete. Something like this.
# add these variables at class/global scope
priority_queue = []
current_job_id = 1
job_id_dict = {}
async def produce(Q, n_jobs):
# same as above
async def consume(Q):
while True:
n = await Q.get()
print(f"Consumed :{n}")
x = do_sometask_and_return_the_result(n)
await process_result(n, x)
async def process_result(n, x):
heappush(priority_queue, n)
job_id_dict[n] = x
while current_job_id == priority_queue[0]:
job_id = heappop(priority_queue)
print(f"Finished :{job_id} and Result: {job_id_dict[job_id]}")
current_job_id += 1
async def main(loop):
Q = asyncio.Queue(loop=loop, maxsize=3)
await asyncio.wait([produce(Q, 10), consume(Q), consume(Q), consume(Q)])
print("Done")
For more information on the heapq module: https://docs.python.org/3/library/heapq.html
import asyncio
import random
async def producer(q: asyncio.Queue):
for _ in range(100):
await q.put(random.randint(1, 10))
async def consumer(q: asyncio.Queue):
while True:
num = await q.get()
print("Got From Queue: ", num)
q.task_done()
async def main():
q = asyncio.Queue()
pr = asyncio.create_task(producer(q))
consumers = [asyncio.create_task(consumer(q)) for _ in range(10)]
await pr
await q.join()
for c in consumers:
c.cancel()
asyncio.run(main())
I created this script to replicate a simplified version of my problem.
so this is a very basic producer-consumer type of script, and this works just fine
however if an exception occurs inside the consumer, the script freezes.
for example, If I modify the consumer script like so:
async def consumer(q: asyncio.Queue):
raise Exception
while True:
num = await q.get()
print("Got From Queue: ", num)
q.task_done()
when I run the script now it freezes and no exceptions get thrown.
However, If I change the main script to generate all the consumer tasks in a for loop instead of generating them through a list comprehension like so:
async def main():
q = asyncio.Queue()
pr = asyncio.create_task(producer(q))
# consumers = [asyncio.create_task(consumer(q)) for _ in range(10)]
for _ in range(10):
asyncio.create_task(consumer(q))
await pr
await q.join()
It throws 10 exceptions of type Exception, as expected behavior.
my 2 questions are:
Why does this strange behaviour happen?
In my actual script, I want to use exception handling, so I'm using now the for-loop approach. However, In this approach, I don't have references to all the consumer tasks, so I can't cancel them when I want. How can I do that? (Simply appending them to a list after creation replicates the original problem)
I am completely clueless, any help appreciated!
I don't think that the use of a list comprehension instead of a for loop has any impact on this.
You should handle the exception in the consumer and call task_done() for each get(), otherwise you'll experience the current behaviour where q.join() will block/wait for unfinished tasks:
task_done()
If a join() is currently blocking, it will resume when all items have
been processed (meaning that a task_done() call was received for every
item that had been put() into the queue).
test.py:
import asyncio
import random
async def producer(q):
for _ in range(10):
await q.put(random.randint(1, 10))
async def consumer(q):
while True:
num = await q.get()
try:
if num in (5, 8):
raise Exception("ERROR")
print(f"Working on: {num}")
except Exception as exc:
print(f"{num}: {exc}")
finally:
q.task_done()
async def main():
q = asyncio.Queue()
pr = asyncio.create_task(producer(q))
tasks = []
for _ in range(10):
tasks.append(asyncio.create_task(consumer(q)))
await pr
await q.join()
for task in tasks:
task.cancel()
await asyncio.gather(*tasks, return_exceptions=True)
if __name__ == "__main__":
asyncio.run(main())
Test:
$ python test.py
Working on: 7
Working on: 4
Working on: 4
Working on: 4
Working on: 1
Working on: 3
8: ERROR
Working on: 6
5: ERROR
Working on: 4
I'm looking at this piece of code from the example from here
And i want to know at what exact moment does the consumers() coroutine get called?
import asyncio
import itertools as it
import os
import random
import time
async def makeitem(size: int = 5) -> str:
return os.urandom(size).hex()
async def randsleep(a: int = 1, b: int = 5, caller=None) -> None:
i = random.randint(0, 10)
if caller:
print(f"{caller} sleeping for {i} seconds.")
await asyncio.sleep(i)
async def produce(name: int, q: asyncio.Queue) -> None:
n = random.randint(0, 10)
for _ in it.repeat(None, n): # Synchronous loop for each single producer
await randsleep(caller=f"Producer {name}")
i = await makeitem()
t = time.perf_counter()
await q.put((i, t))
print(f"Producer {name} added <{i}> to queue.")
async def consume(name: int, q: asyncio.Queue) -> None:
while True:
await randsleep(caller=f"Consumer {name}")
i, t = await q.get()
now = time.perf_counter()
print(f"Consumer {name} got element <{i}>"
f" in {now-t:0.5f} seconds.")
q.task_done()
async def main(nprod: int, ncon: int):
q = asyncio.Queue()
producers = [asyncio.create_task(produce(n, q)) for n in range(nprod)]
consumers = [asyncio.create_task(consume(n, q)) for n in range(ncon)]
await asyncio.gather(*producers)
await q.join() # Implicitly awaits consumers, too
for c in consumers:
c.cancel()
if __name__ == "__main__":
import argparse
random.seed(444)
parser = argparse.ArgumentParser()
parser.add_argument("-p", "--nprod", type=int, default=5)
parser.add_argument("-c", "--ncon", type=int, default=10)
ns = parser.parse_args()
start = time.perf_counter()
asyncio.run(main(**ns.__dict__))
elapsed = time.perf_counter() - start
print(f"Program completed in {elapsed:0.5f} seconds.")
I only see this line triggering the execution for both producer and consumer coroutines.
await asyncio.gather(*producers)
I don't understand how until the await line mentioned above there's no execution in the background when tasks are defined and created in these lines( because none of the print statements inside the producers and consumers are displayed).:
producers = [asyncio.create_task(produce(n, q)) for n in range(nprod)]
consumers = [asyncio.create_task(consume(n, q)) for n in range(ncon)]
While create_task() doesn't start executing the coroutine immediately, it schedules execution in the background at the first possible opportunity, i.e. at the first await that suspends to the event loop.
gather() is just a helper function that waits for the given awaitables to complete. It doesn't prevent previously scheduled coroutines (such as those started with create_task, but also start_server etc.) from executing.
i want to know at what exact moment does the consumers() coroutine get called?
Since consumers is a coroutine, while it's called once, it can suspend and resume many times, each await serving as a point of suspension/resumption. When you call create_task() it is placed in a queue of runnable coroutines. In each iteration of the event loop asyncio goes through runnable coroutines and executes a "step" of each, where the step executes it until the first await that chooses to suspend. In your code the step happens when your main coroutine suspends in order to wait for gather() to complete.
I would like to stop a python asyncio task from another task and start it again when some condition in the second task happen.
Please note, than I don't want to cancel the coroutine of the first task (the state of that coroutine when it stopped should be available). Also, I don't care about the exact state the first task is in, I just want the event loop stop running the first task until told otherwise from the second.
I hope this example code helps understanding the problem:
import asyncio
async def coroutine1():
i = 0
while(True):
i += 1
print("coroutine1: " + str(i) )
await asyncio.sleep(1)
async def coroutine2(task1):
i = 0
while(True):
i += 1
if (i > 3) and (i<10):
pass #TODO: stop task1 here
else:
pass #TODO: Maybe check if task1 is running
#and start task1 again if it's not?
print("coroutine2: " + str(i) )
await asyncio.sleep(1)
async def main_coroutine():
loop = asyncio.get_event_loop()
task1 = loop.create_task(coroutine1())
task2 = loop.create_task(coroutine2(task1))
done, pending = await asyncio.wait(
[task1, task2]
, return_when=asyncio.FIRST_COMPLETED,)
loop = asyncio.get_event_loop()
loop.run_until_complete(main_coroutine())
loop.close()
I would like to stop a python asyncio task from another task and start it again when some condition in the second task happen.
I assume you control the task creation, but don't want to touch the implementation of the coroutine. In your case, you control coroutine2 and main_coroutine, but not the insides of coroutine1.
In that case you can wrap the coroutine in a an __await__ that, instead of the normal yield from loop, checkes your stopped flag and waits for a future that tells it when to resume.
class Stoppable:
def __init__(self, coro):
self._coro_iter = coro.__await__()
self._stopped = None
def __await__(self):
while True:
while self._stopped:
print('awaiting stopped')
yield from self._stopped.__await__()
try:
v = next(self._coro_iter)
except StopIteration as e:
return v
yield v
def stop(self):
loop = asyncio.get_event_loop()
self._stopped = loop.create_future()
def start(self):
if self._stopped is not None:
self._stopped.set_result(None)
self._stopped = None
You can use the wrapper to modify coroutine2 to stop and resume the execution of coroutine1 at will:
async def coroutine2(s):
i = 0
while True:
i += 1
if i == 3:
print('stopping coroutine1')
s.stop()
elif i == 10:
print('restarting coroutine1')
s.start()
print("coroutine2: " + str(i) )
await asyncio.sleep(1)
async def main_coroutine():
loop = asyncio.get_event_loop()
s = Stoppable(coroutine1())
fut1 = asyncio.ensure_future(s)
task2 = loop.create_task(coroutine2(s))
done, pending = await asyncio.wait(
[fut1, task2], return_when=asyncio.FIRST_COMPLETED)
The way wrapper works is by unrolling the loop inherent in yield from. For example, to just delegate __await__ to another coroutine, one would write:
def __await__(self):
yield from self._coro_iter
Written like this, you can't implement stopping because the yield from contains an implicit loop that yields all the values produced by the underlying iterator - something like:
def __await__(self):
while True:
try:
v = next(self._coro_iter)
except StopIteration as e:
return e.value
yield v
Taken like this, it is easy enough to add an if that checks for _stopped at each iteration pass, meaning each time we're resumed by the event loop. The remaining hurdle is that one cannot just busy-loop until _stopped is rescinded - we must yield something else to allow the event loop to resume running other coroutines. Fortunately that is easily achieved by making _stopped a future, and yielding from the future. When the future's result is set, we will be automatically resumed and continue executing the wrapped coroutine.
It seems it can't be done.
It's possible to cancel an ongoing task with task1.cancel() and it's possible to create a new task with asyncio.get_event_loop().create_task(newTask).
It's also possible to get the coroutine of a running task with task1._coro but if we try to create a task again with a previously scheduled coroutine we will get a RuntimeError exception. This the discussion where they decided it: https://bugs.python.org/issue25887
Finally, a possible way of accomplishing the desire effect is using a asyncio.Queue object:
import asyncio
async def coroutine1(stop_queue):
i = 0
while(True):
if stop_queue.empty(): #if the queue is empty keep working.
i += 1
print("coroutine1: " + str(i) )
await asyncio.sleep(1)
async def coroutine2(stop_queue):
i = 0
while(True):
i += 1
if i == 3:
await stop_queue.put("whatever..") #put something in the queue
if i == 11:
await stop_queue.get() #take something from the queue
print("coroutine2: " + str(i) )
await asyncio.sleep(1)
async def main_coroutine():
stop_queue = asyncio.Queue()
done, pending = await asyncio.wait(
[coroutine1(stop_queue), coroutine2(stop_queue)]
, return_when=asyncio.ALL_COMPLETED,)
loop = asyncio.get_event_loop()
loop.run_until_complete(main_coroutine())
loop.close()
I have a program with one producer and two slow consumers and I'd like to rewrite it with coroutines in such way that each consumer will handle only last value (i.e. skip new values generated during processing the old ones) produced for it (I used threads and threading.Queue() but with it blocks on put(), cause the queue will be full most of the time).
After reading answer to this question I decided to use asyncio.Event and asyncio.Queue. I wrote this prototype program:
import asyncio
async def l(event, q):
h = 1
while True:
# ready
event.set()
# get value to process
a = await q.get()
# process it
print(a * h)
h *= 2
async def m(event, q):
i = 1
while True:
# pass element to consumer, when it's ready
if event.is_set():
await q.put(i)
event.clear()
# produce value
i += 1
el = asyncio.get_event_loop()
ev = asyncio.Event()
qu = asyncio.Queue(2)
tasks = [
asyncio.ensure_future(l(ev, qu)),
asyncio.ensure_future(m(ev, qu))
]
el.run_until_complete(asyncio.gather(*tasks))
el.close()
and I have noticed that l coroutine blocks on q.get() line and doesn't print anything.
It works as I expect after adding asyncio.sleep() in both (I get 1,11,21,...):
import asyncio
import time
async def l(event, q):
h = 1
a = 1
event.set()
while True:
# await asyncio.sleep(1)
a = await q.get()
# process it
await asyncio.sleep(1)
print(a * h)
event.set()
async def m(event, q):
i = 1
while True:
# pass element to consumer, when it's ready
if event.is_set():
await q.put(i)
event.clear()
await asyncio.sleep(0.1)
# produce value
i += 1
el = asyncio.get_event_loop()
ev = asyncio.Event()
qu = asyncio.Queue(2)
tasks = [
asyncio.ensure_future(l(ev, qu)),
asyncio.ensure_future(m(ev, qu))
]
el.run_until_complete(asyncio.gather(*tasks))
el.close()
...but I'm looking for solution without it.
Why is it so? How can I fix it? I think I cannot call await l() from m as both of them have states (in original program the first draws solution with PyGame and the second plots results).
The code is not working as expected as the task running the m function is never stopped. The task will continue increment i in the case that event.is_set() == False. Because this task is never suspended, the task running function l will never be called. Therefore, you need a way to suspend the task running function m. One way of suspending is awaiting another coroutine, that is the reason why a asyncio.sleep works as expected.
I think the following code will work as you expect. The LeakyQueue will ensure that only the last value from the producer will be processed by the consumer. As the complexity is very symmetric, the consumer will consume all values produced by the producer. If you increase the delay argument, you can simulate that the consumer only processes the last value created by the producer.
import asyncio
class LeakyQueue(asyncio.Queue):
async def put(self, item):
if self.full():
await self.get()
await super().put(item)
async def consumer(queue, delay=0):
h = 1
while True:
a = await queue.get()
if delay:
await asyncio.sleep(delay)
print ('consumer', a)
h += 2
async def producer(queue):
i = 1
while True:
await asyncio.ensure_future(queue.put(i))
print ('producer', i)
i += 1
loop = asyncio.get_event_loop()
queue = LeakyQueue(maxsize=1)
tasks = [
asyncio.ensure_future(consumer(queue, 0)),
asyncio.ensure_future(producer(queue))
]
loop.run_until_complete(asyncio.gather(*tasks))