限制Python线程的并发性和速率

给定一个数字线程,我想将工作函数的调用速率限制为每秒一个速率.

我的想法是跟踪上次在所有线程上进行调用的时间,并将其与每个线程中的当前时间进行比较.然后,如果current_time – last_time<率.我让线程睡了一会儿.我的实现出了点问题 - 我认为我可能对锁的工作原理有了错误的认识. 我的代码：

from Queue import Queue
from threading import Thread, Lock, RLock
import time

num_worker_threads = 2
rate = 1
q = Queue()
lock = Lock()
last_time = [time.time()]

def do_work(i, idx):
    # Do work here, print is just a dummy.
    print('Thread: {0}, Item: {1}, Time: {2}'.format(i, idx, time.time()))

def worker(i):
    while True:
        lock.acquire()
        current_time = time.time()
        interval = current_time - last_time[0]
        last_time[0] = current_time
        if interval < rate:
            time.sleep(rate - interval)
        lock.release()
        item = q.get()
        do_work(i, item)
        q.task_done()

for i in range(num_worker_threads):
     t = Thread(target=worker, args=[i])
     t.daemon = True
     t.start()

for item in xrange(10):
    q.put(item)

q.join()

我期待每秒看到一次do_work呼叫,但是,我同时得到大约2个呼叫(每个线程1个),然后暂停一秒钟.怎么了？

好吧,有些编辑.简单地限制项目放入队列的速率的建议很好,但我记得我必须处理工作人员将项目重新添加到队列的情况.典型示例：网络任务中的分页或后退重试.我想出了以下内容.我想对于实际的网络任务来说,eventlet / gevent库在资源上可能更容易,但这只是一个例子.它基本上使用优先级队列来堆积请求,并使用额外的线程以均匀的速率将项目从堆中铲到实际的任务队列.我模拟了工人重新插入桩中,然后再重新插入物品.

import sys
import os
import time
import random

from Queue import Queue, PriorityQueue
from threading import Thread

rate = 0.1

def worker(q, q_pile, idx):
    while True:
        item = q.get()
        print("Thread: {0} processed: {1}".format(item[1], idx))
        if random.random() > 0.3:
            print("Thread: {1} reinserting item: {0}".format(item[1], idx))
            q_pile.put((-1 * time.time(), item[1]))
        q.task_done()

def schedule(q_pile, q):
    while True:
        if not q_pile.empty():
            print("Items on pile: {0}".format(q_pile.qsize()))
            q.put(q_pile.get())
            q_pile.task_done()
        time.sleep(rate)

def main():

    q_pile = PriorityQueue()
    q = Queue()

    for i in range(5):
        t = Thread(target=worker, args=[q, q_pile, i])
        t.daemon = True
        t.start()

    t_schedule = Thread(target=schedule, args=[q_pile, q])
    t_schedule.daemon = True
    t_schedule.start()

    [q_pile.put((-1 * time.time(), i)) for i in range(10)]
    q_pile.join()
    q.join()

if __name__ == '__main__':
    main()

解决方法:

I get mostly 2 calls at the same time (1 for each thread), followed by
a one second pause. What is wrong?

这正是您对实施的期望.让我们说时间t从0开始,速率为1：

Thread1执行此操作：

    lock.acquire() # both threads wait here, one gets the lock
    current_time = time.time() # we start at t=0
    interval = current_time - last_time[0] # so interval = 0
    last_time[0] = current_time # last_time = t = 0
    if interval < rate: # rate = 1 so we sleep
        time.sleep(rate - interval) # to t=1
    lock.release() # now the other thread wakes up
    # it's t=1 and we do the job

Thread2这样做：

    lock.acquire() # we get the lock at t=1 
    current_time = time.time() # still t=1
    interval = current_time - last_time[0] # interval = 1
    last_time[0] = current_time
    if interval < rate: # interval = rate = 1 so we don't sleep
        time.sleep(rate - interval)
    lock.release() 
    # both threads start the work around t=1

我的建议是限制项目进入队列的速度.

标签：python,multithreading,concurrency,rate-limiting
来源： https://codeday.me/bug/20190704/1379250.html