脚本专栏 发布日期:2025/10/31 浏览次数:1
Python 的 Queue 模块中提供了同步的、线程安全的队列类,包括FIFO(先入先出)队列Queue,LIFO(后入先出)队列LifoQueue,和优先级队列 PriorityQueue。
这些队列都实现了锁原语,能够在多线程中直接使用,可以使用队列来实现线程间的同步。
模块中的常用方法如下:
#!/usr/bin/python3
import queue
import threading
import time
exitFlag = 0
class myThread (threading.Thread):
  def __init__(self, threadID, name, q):
    threading.Thread.__init__(self)
    self.threadID = threadID
    self.name = name
    self.q = q
  def run(self):
    print ("开启线程:" + self.name)
    process_data(self.name, self.q)
    print ("退出线程:" + self.name)
def process_data(threadName, q):
  while not exitFlag:
    queueLock.acquire()
    if not workQueue.empty():
      data = q.get()
      queueLock.release()
      print ("%s processing %s" % (threadName, data))
    else:
      queueLock.release()
    time.sleep(1)
threadList = ["Thread-1", "Thread-2", "Thread-3"]
nameList = ["One", "Two", "Three", "Four", "Five"]
queueLock = threading.Lock()
workQueue = queue.Queue(10)
threads = []
threadID = 1
# 创建新线程
for tName in threadList:
  thread = myThread(threadID, tName, workQueue)
  thread.start()
  threads.append(thread)
  threadID += 1
# 填充队列
queueLock.acquire()
for word in nameList:
  workQueue.put(word)
queueLock.release()
# 等待队列清空
while not workQueue.empty():
  pass
# 通知线程是时候退出
exitFlag = 1
# 等待所有线程完成
for t in threads:
  t.join()
print ("退出主线程")
知识点扩展:
问题
怎样实现一个按优先级排序的队列? 并且在这个队列上面每次 pop 操作总是返回优先级最高的那个元素
解决方案
下面的类利用 heapq 模块实现了一个简单的优先级队列:
import heapq class PriorityQueue: def __init__(self): self._queue = [] self._index = 0 def push(self, item, priority): heapq.heappush(self._queue, (-priority, self._index, item)) self._index += 1 def pop(self): return heapq.heappop(self._queue)[-1]
下面是它的使用方式:
> class Item:
... def __init__(self, name):
...  self.name = name
... def __repr__(self):
...  return 'Item({!r})'.format(self.name)
...
> q = PriorityQueue()
> q.push(Item('foo'), 1)
> q.push(Item('bar'), 5)
> q.push(Item('spam'), 4)
> q.push(Item('grok'), 1)
> q.pop()
Item('bar')
> q.pop()
Item('spam')
> q.pop()
Item('foo')
> q.pop()
Item('grok')
>