python多線程限制並發數示例
阿新 • • 發佈:2018-02-06
star 使用 str rtp 隊列 -- col bin append
#coding: utf-8 #!/usr/bin/env python import Queue import threading import time prolock = threading.Lock() # 定義同時隊列數 queue = Queue.Queue(maxsize=10) # 定義任務初值值及最大值 taskidx = 0 maxidx = 100 # 生成任務列表 def taskList(): task = [] for i in range(100): task.append("task" + str(i))return task # 把任務放入隊列中 class Producer(threading.Thread): def __init__(self, name, queue): self.__name = name self.__queue = queue super(Producer, self).__init__() def run(self): while True: global taskidx, prolock, maxidx time.sleep(4) prolock.acquire() print ‘Producer name: %s‘ % (self.__name) if maxidx == taskidx: prolock.release() break ips = taskList() ip = ips[taskidx] self.__queue.put(ip) taskidx = taskidx + 1 prolock.release()# 線程處理任務 class Consumer(threading.Thread): def __init__(self, name, queue): self.__name = name self.__queue = queue super(Consumer, self).__init__() def run(self): while True: ip = self.__queue.get() print ‘Consumer name: %s‘ % (self.__name) consumer_process(ip) self.__queue.task_done() def consumer_process(ip): time.sleep(1) print ip def startProducer(thread_num): t_produce = [] for i in range(thread_num): p = Producer("producer"+str(i), queue) p.setDaemon(True) p.start() t_produce.append(p) return t_produce def startConsumer(thread_num): t_consumer = [] for i in range(thread_num): c = Consumer("Consumer"+str(i), queue) c.setDaemon(True) c.start() t_consumer.append(c) return t_consumer def main(): t_produce = startProducer(3) t_consumer = startConsumer(5) # 確保所有的任務都生成 for p in t_produce: p.join() # 等待處理完所有任務 queue.join() if __name__ == ‘__main__‘: main() print ‘------end-------‘
一般生成任務都會比較快,可以使用單線程來生成任務,示例如下:
#coding: utf-8 #!/usr/bin/env python import Queue import threading import time # 定義同時處理任務數 queue = Queue.Queue(maxsize=3) # 生成任務列表 def taskList(): task = [] for i in range(100): task.append("task" + str(i)) return task # 把任務放入隊列中 class Producer(threading.Thread): def __init__(self, name, queue): self.__name = name self.__queue = queue super(Producer, self).__init__() def run(self): for ip in taskList(): self.__queue.put(ip) # 線程處理任務 class Consumer(threading.Thread): def __init__(self, name, queue): self.__name = name self.__queue = queue super(Consumer, self).__init__() def run(self): while True: ip = self.__queue.get() print ‘Consumer name: %s‘ % (self.__name) consumer_process(ip) self.__queue.task_done() def consumer_process(ip): time.sleep(1) print ip def startConsumer(thread_num): t_consumer = [] for i in range(thread_num): c = Consumer(i, queue) c.setDaemon(True) c.start() t_consumer.append(c) return t_consumer def main(): p = Producer("Producer task0", queue) p.setDaemon(True) p.start() startConsumer(9) # 確保所有的任務都生成 p.join() # 等待處理完所有任務 queue.join() if __name__ == ‘__main__‘: main() print ‘------end-------‘
python多線程限制並發數示例