python队列多线程

2024-04-19 04:26:28 发布

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铁路上只有两列火车(同时)最简单的方法是什么。我的英语不好。这是我唯一能解释的方法。我知道我应该排队吗?我在我的语言里找不到信息

谢谢你!你知道吗

1>;开始,2>;开始。3,4等等。1>;完成,3>;开始(第4个仍在等待)。。你知道吗

from threading import Thread
import time
import random

def trains(city):
    print city, 'start'

    for count in range(1,3):
        delay = random.randrange(5,10)
        print city, 'delay', delay
        time.sleep(delay)

    print city, 'end'


cities = ['prague', 'london', 'berlin', 'moscow']
threadlist = []

for city in cities:                             
    t = Thread(target=trains, args=(city,))
    t.start()
    threadlist.append(t)


for b in threadlist:
    b.join()

Tags: 方法inimportgtcityfortimerandom
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1楼 · 发布于 2024-04-19 04:26:28

我将在这里猜测一些事情:

from threading import Thread, Lock, BoundedSemaphore
import time
import random

def trains(city):
    with railroads:
        with iolock:
            print city, 'start'

        for count in range(1,3):
            delay = random.randrange(5,10)
            with iolock:
                print city, 'delay', delay
            time.sleep(delay)

        with iolock:
            print city, 'end'


cities = ['prague', 'london', 'berlin', 'moscow']
threadlist = []

iolock = Lock()
railroads = BoundedSemaphore(2)

for city in cities:                             
    t = Thread(target=trains, args=(city,))
    t.start()
    threadlist.append(t)


for b in threadlist:
    b.join()

iolock的目的是阻止终端中的混合输出:一次只打印一个线程。railroads的目的是允许最多两个线程同时进入代码体。下面是示例输出。请注意,“布拉格”和“伦敦”刚开始时正好运行,但“柏林”不会在“伦敦”结束前启动。那么“莫斯科”直到“布拉格”结束才开始:

prague start
london start
prague delay 8
london delay 5
london delay 6
prague delay 5
london end
berlin start
berlin delay 8
prague end
moscow start
moscow delay 8
berlin delay 6
moscow delay 7
berlin end
moscow end

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