ThreadPool.spawn is too slow under greenlet.

[Fei Li]

here is my code:

import time

import contextlib

import gevent

from gevent.threadpool import ThreadPool

pool = ThreadPool(1)

@contextlib.contextmanager

def timer(label):

    bt = time.time()

    yield

    print '%s %.2f ms' % (label, 1000 * (time.time() - bt))

def _eat_cpu():

    time.sleep(1)

def handle(i):

    with timer('pool.spawn: %s' % i):

        r = pool.spawn(_eat_cpu)

    return r.get()

def test1():

    for i in xrange(10000):

        gevent.spawn(handle, i)

def test2():

    for i in xrange(10000):

        handle(i)

test1()

# test2()

gevent.wait()

result of test1:

pool.spawn: 0 0.20 ms

pool.spawn: 1 0.10 ms

pool.spawn: 2 1000.98 ms

pool.spawn: 3 2003.38 ms

pool.spawn: 4 3007.69 ms

pool.spawn: 5 4011.77 ms

pool.spawn: 6 5012.74 ms

pool.spawn: 7 6015.41 ms

pool.spawn: 8 7016.91 ms

pool.spawn: 9 8018.87 ms

pool.spawn: 10 9023.03 ms

result of test2:

pool.spawn: 0 0.74 ms

pool.spawn: 1 0.08 ms

pool.spawn: 2 0.08 ms

pool.spawn: 3 0.08 ms

pool.spawn: 4 0.12 ms

pool.spawn: 5 0.07 ms

pool.spawn: 6 0.11 ms

pool.spawn: 7 0.07 ms

pool.spawn: 8 0.14 ms

pool.spawn: 9 0.10 ms

pool.spawn: 10 0.07 ms

I find some lines in gevent/threadpool.py below which describe the reason. i try to remove line 161 and the test works well.

160             # rawlink() must be the last call

161             result.rawlink(lambda *args: self._semaphore.release())

162             # XXX this _semaphore.release() is competing for order with get()

163             # XXX this is not good, just make ThreadResult release the semaphore before doing anything else

I want to know why  _semaphore.release() is competing for order with get() and whether removing the code is OK.

Also how to make a pool's spawn works like queue's put_nowait.

Wait for help and reply, thank you.

[Denis Bilenko]

> pool = ThreadPool(1)

Your pool has size = 1, so it can only run one request at a time. Try
changing 1 into 10. or 1000.

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[Fei Li]

I set size=1 intentional to make the result more obvious.

My point is there is a semaphore in ThreadPool which make the pool process task one by one when the pool is full.

[Fei Li]

From the result I paste in my first post, we can find the time which task cost become larger and larger. This is very bad for a online system.

For example:

Consider the pool can process 10 requests per second, and there are 15 requests per second need to be processed. We should drop 5 requests per second to make 66% requests be well processed because the requests need to be processed in time.

The original threadpool process every submitted task which makes the new tasks wait longer and longer.

[Catstyle Lee]

your `timer` is misused, it should be

with timer('pool.spawn: %s'  % i):

    r = pool.spawn(_eat_cpu)

    return r.get()

or you can try add some log as below

def timer(label):

    bt = time.time()

    print gevent.getcurrent()

    yield

    print '%s %.2f ms' % (label, 1000 * (time.time() - bt))

then you can see something useful

--

BR,

/Catstyle_Lee