python队列多线程

铁路上只有两列火车（同时）的最简单方式是什么。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？我找不到我语言中的信息

谢谢大家!

1> 开始，2>开始。3,4等等。1> 完成，3>开始（第4个仍在等待）

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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。

RailRails

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”在“伦敦”结束前不会启动。然后“莫斯科”直到“布拉格”结束才开始：

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听起来你想要一个

线程.BoundedSemaphore（2）

，但我对你的问题理解不够透彻，无法建议实际的代码。

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()

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