use of Queue

#1 08-27-2008, 05:45 AM

how is Queue intended to be used? I found the following code in python
manual, but I don't understand how to stop consumers after all items
have been produced. I tried different approaches but all of them
seemed incorrect (race, deadlock or duplicating queue functionality)

def worker():
while True:
item = q.get()
do_work(item)
q.task_done()

q = Queue()
for i in range(num_worker_threads):
t = Thread(target=worker)
t.setDaemon(True)
t.start()

for item in source():
q.put(item)

q.join() # block until all tasks are done

#2 08-27-2008, 05:56 AM

Alexandru Mosoi wrote:

> how is Queue intended to be used? I found the following code in python
> manual, but I don't understand how to stop consumers after all items
> have been produced. I tried different approaches but all of them
> seemed incorrect (race, deadlock or duplicating queue functionality)
>
>
> def worker():
> while True:
> item = q.get()
> do_work(item)
> q.task_done()
>
> q = Queue()
> for i in range(num_worker_threads):
> t = Thread(target=worker)
> t.setDaemon(True)
> t.start()
>
> for item in source():
> q.put(item)
>
> q.join() # block until all tasks are done

Put a sentinel into the queue that gets interpreted as "terminate" for the
workers. You need of course to put it in there once for each worker.

Diez

#3 08-27-2008, 06:06 AM

Alexandru Mosoi wrote:
> how is Queue intended to be used? I found the following code in python
> manual, but I don't understand how to stop consumers after all items
> have been produced. I tried different approaches but all of them
> seemed incorrect (race, deadlock or duplicating queue functionality)
>
>
> def worker():
> while True:
> item = q.get()

if item is None:
break

> do_work(item)
> q.task_done()
>
> q = Queue()
> for i in range(num_worker_threads):
> t = Thread(target=worker)
> t.setDaemon(True)
> t.start()
>
> for item in source():
> q.put(item)

# stop all consumers
for i in range(num_worker_threads):
q.put(None)

>
> q.join() # block until all tasks are done

This is how I do it.

-- Gerhard

#4 08-27-2008, 06:11 AM

Diez> Put a sentinel into the queue that gets interpreted as "terminate"
Diez> for the workers. You need of course to put it in there once for
Diez> each worker.

Or make the consumers daemon threads so that when the producers are finished
an all non-daemon threads exit, the consumers do as well.

Skip

#5 08-27-2008, 06:41 AM

skip> Or make the consumers daemon threads so that when the producers
skip> are finished an all non-daemon threads exit, the consumers do as
skip> well.

Forget that I wrote this. If they happen to be working on the token they've
consumed at the time the other threads exit, they will as well. Use the
sentinel token idea instead.

Skip

#6 08-27-2008, 07:49 AM

On Aug 27, 1:06*pm, Gerhard Häring <g...@ghaering.de> wrote:
> Alexandru Mosoi wrote:
> > how is Queue intended to be used? I found the following code in python
> > manual, but I don't understand how to stop consumers after all items
> > have been produced. I tried different approaches but all of them
> > seemed incorrect (race, deadlock or duplicating queue functionality)
>
> > * * def worker():
> > * * * * while True:
> > * * * * * * item = q.get()
>
> * * * * * * * *if item is None:
> * * * * * * * * * *break
>
> > * * * * * * do_work(item)
> > * * * * * * q.task_done()
>
> > * * q = Queue()
> > * * for i in range(num_worker_threads):
> > * * * * *t = Thread(target=worker)
> > * * * * *t.setDaemon(True)
> > * * * * *t.start()
>
> > * * for item in source():
> > * * * * q.put(item)
>
> # stop all consumers
> for i in range(num_worker_threads):
> * * *q.put(None)
>
>
>
> > * * q.join() * * * # block until all tasks are done
>
> This is how I do it.
>
> -- Gerhard

Your solution works assuming that you know how many consumer threads
you have

. I don't

. More than that, it's not correct if you have
more than one producer

. Having a sentinel was my very first idea,
but as you see... it's a race condition (there are cases in which not
all items are processed).

#7 08-27-2008, 07:54 AM

> Your solution works assuming that you know how many consumer threads
> you have

. I don't

. More than that, it's not correct if you have
> more than one producer

. Having a sentinel was my very first idea,
> but as you see... it's a race condition (there are cases in which not
> all items are processed).

Queue raises an Empty exception when there are no items left in the
queue. Put the q.get() call in a try block and exit in the except
block.

You can also use a condition variable to signal threads to terminate.

#8 08-27-2008, 07:58 AM

On Aug 27, 2:54*pm, Jeff <jeffo...@gmail.com> wrote:

> Queue raises an Empty exception when there are no items left in the
> queue. *Put the q.get() call in a try block and exit in the except
> block.
Wrong. What if producer takes a long time to produce an item?
Consumers
will find the queue empty and exit instead of waiting.

> You can also use a condition variable to signal threads to terminate.
This is the solution I want to avoid because it duplicates Queue's
functionality.
I prefer having a clean solution with nice design to hacking Queue
class.

#9 08-27-2008, 08:17 AM

On Aug 27, 12:45*pm, Alexandru Mosoi <brtz...@gmail.com> wrote:
> how is Queue intended to be used? I found the following code in python
> manual, but I don't understand how to stop consumers after all items
> have been produced. I tried different approaches but all of them
> seemed incorrect (race, deadlock or duplicating queue functionality)
>
> * * def worker():
> * * * * while True:
> * * * * * * item = q.get()
> * * * * * * do_work(item)
> * * * * * * q.task_done()
>
> * * q = Queue()
> * * for i in range(num_worker_threads):
> * * * * *t = Thread(target=worker)
> * * * * *t.setDaemon(True)
> * * * * *t.start()
>
> * * for item in source():
> * * * * q.put(item)
>
> * * q.join() * * * # block until all tasks are done

ok. I think I figured it out

. let me know what you think

global num_tasks, num_done, queue
num_tasks = 0
num_done = 0
queue = Queue()

# producer
num_tasks += 1
for i in items:
num_tasks += 1
queue.put(i)

num_tasks -= 1
if num_tasks == num_done:
queue.put(None)

# consumer
while True:
i = queue.get()
if i is None:
queue.put(None)
break

# do stuff

num_done += 1
if num_done == num_tasks:
queue.put(None)
break

#10 08-27-2008, 08:40 AM

>
> Your solution works assuming that you know how many consumer threads
> you have

. I don't

. More than that, it's not correct if you have
> more than one producer

. Having a sentinel was my very first idea,
> but as you see... it's a race condition (there are cases in which not
> all items are processed).

If you have several producers, how do you coordinate when to shut down?

Apart from that, you can easily solve the problem of not knowing how many
consumers you have by making a consumer stuff back the sentinel into the
queue. Then it will ripple down until no consumer is left.

Diez

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