Question

我正试图弄清楚Python中的这个同步问题。我有一个生产者线程和（可选）多个消费者线程（取决于命令，即./script sums.txt -c 10）。现在有1个生产者和1个消费者没有问题，因为同步是用Queue处理的。

现在问题是，有超过1个消费者线程，线程1可能从队列中获取一个项目并对其进行处理。虽然线程2执行相同但比线程1更快，并在线程1之前打印。我试图用随机定时器模拟这个问题。

我的输出现在使用随机定时器：“。/ script sommen.txt -c 2” 正如您所注意到的那样，队列中的第二项是在第一项之前处理的，如果没有随机定时器，则不会发生很多操作，因为操作非常简单，因此线程足够快。有没有办法解决这个问题？我想过锁，但这会使程序效率低下吗？

另一件事，清理线程的最佳方法是什么。我知道我的队列何时完成（哨兵值），但是有什么方法可以清理线程？

非常感谢！

Consumers is set to: 2
I'm thread number: 4316991488 Read  (P): 12 + 90
I'm thread number: 4316991488 Read  (P): 420 / 20
I'm thread number: 4316991488 Read  (P): 12 + 90
I'm thread number: 4316991488 Read  (P): 420 / 20
Monitor is done
I'm thread number: 4329586688 Write (C): 420 / 20 = 21.0
I'm thread number: 4324331520 Write (C): 12 + 90 = 102

-

#!/usr/bin/env python

import threading
import operator
import sys
import queue
import optparse
from time import sleep
import random

def optionsparser():
    parser = optparse.OptionParser(
        usage="usage: %prog file [Options]")
    parser.add_option("-c", "--consumer", dest="consumer", type="int",
                      help="consumer <ident> [default: %default]")

    parser.set_defaults(consumer=1)
    opts, files = parser.parse_args()

    filename = files[0]

    try:
        _f = open(filename)
        return(filename, opts.consumer)
    except IOError:
        print ('Oh dear I/O Error')

def readitems(filename):

    print("Read from file: ", filename)
    with open(filename, 'r') as f:
        mylist = [line.rstrip('\n') for line in f]
    f.close()

    try: 
        for _line in mylist:
            data = _line.split(' ')

            qprint.put(data) #write to monitor queue
            qsum.put(data) #write to consumer queue

    except ValueError as e:
        print(e)
    except RuntimeError as err:
        print(err)
    finally:
        qsum.put("Done Flag")
        qprint.put("Done Flag")
def consumer(qsum):

    while qsum:
        sleeptime = random.randint(1,10)
        sleep(sleeptime)
        try:
            if qsum.get() == "Done Flag":
                print("Monitor queue empty", threading.get_ident())
                ## Clean up
                # Put bakc for other consumers
                qsum.put("Done Flag")
                #cleanup here

            else:
                data = qsum.get()
                operator = calc(data)

        except EnvironmentError as Err:
            print(Err)

def calc(data):

    try:
        sleeptime = random.randint(1,10)
        sleep(sleeptime)
        getal1, diff, getal2 = data
        getal1 = int(getal1)
        getal2 = int(getal2)

        if diff == '+':
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=",  operator.add(getal1, getal2))
        elif diff == '-':
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=",  operator.sub(getal1, getal2))
        elif diff == '*':
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=",  operator.mul(getal1, getal2))
        elif diff == '/':
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=",  operator.truediv(getal1, getal2))
        elif diff == '%':
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=",  operator.mod(getal1, getal2))
        elif diff == '**':
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=",  operator.pow(getal1, getal2))
        else:
            print("I'm thread number:", threading.get_ident(), "Write (C):", str(getal1), diff, str(getal2), "=", "Unknown operator!")

    except ZeroDivisionError as Err:
        print(Err)
    except ValueError:
        print("Wrong input")

def producer(reqs):  
    try:
        readitems(reqs)
    except IndexError as e:
        print(e)


def monitor(qprint):

    while qprint:
        try:
            if qprint.get() == "Done Flag":

                print("Monitor is done")
            else:
                data = (qprint.get())
                getal1, diff, getal2 = data
                print("I'm thread number:", threading.get_ident(), "Read  (P):", str(getal1), diff, str(getal2))
        except RuntimeError as e:
            print(e)

if __name__ == '__main__':

    try:
        reqs = optionsparser() 
        #create queu's
        qprint = queue.Queue()
        qsum = queue.Queue()
        #monitor threads
        t2 = threading.Thread(target=monitor, args=(qprint,))
        t2.start()
        #create consumers threads 
        thread_count = reqs[1]
        print("Consumers is set to:", thread_count)
        for i in range(thread_count):
            t = threading.Thread(target=consumer, args=(qsum,))
            t.start()

        #start producer 
        producer(reqs[0])

    except RuntimeError as Err:
        print(Err)
    except AssertionError as e:
        print(e)

Answer 1

当任务可以被拆分并独立威胁时，使用线程是有效的。如果您想使用thead，请记住，当代码中没有锁定点或锁定点很少时，并行化代码会更有效。锁定点可以是共享资源。

在您的情况下，您只需生成/使用数据，并希望它同步。如果以顺序方式运行此代码会更有效，否则您必须更准确地定义哪些任务可以从并行化中受益。

Answer 2

首先：不要使用Python线程来加速CPU绑定任务，比如计算。除了减速之外，你永远不会看到任何东西。 Because GIL。将Python线程用于I / O绑定任务，例如URL提取。

如果您希望结果按发布顺序到达，请为每个队列元素指定一个序列号。这样每个任务都会知道其结果所在的位置。

使用有序集合（例如列表）将序列号作为索引放置工作线程生成的结果。由于您可能会以相反的顺序接收结果，因此您需要将它们全部存储起来（不能流式传输）。

我不明白为什么在这里使用锁定。首先，锁通过阻止其他独立工作者来破坏并行处理的目的。其次，锁很难，容易出现细微的错误。队列更友好。

多个消费者单一生产者队列

2 个答案: