Question

在铁路上只有两列火车（同时）的最简单方法是什么。我的英语不好。这是我解释它的唯一方式。我知道我应该使用Queue？我用我的语言找不到信息

谢谢！

1＆gt; go，2＆gt; go。 3,4wait。 1＆gt;完成，3＆gt; go（第4个仍然等待）..

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()

Answer 1

我会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的目的是在终端中停止混合输出：一次只打印1个线程。 railroads的目的是允许最多两个线程同时进入代码体。这是示例输出。请注意，“prague”和“london”碰巧首先运行，但“berlin”在“london”结束之前不会启动。然后“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

python队列多线程

1 个答案: