如何将数据发送到正在运行的python线程?

时间:2014-09-18 05:08:36

标签: python multithreading

我有一个在我的应用程序中的不同线程中运行的类。我可以一次运行多个线程,线程是守护进程。经过一段时间后,这些线程的某些需要接收和处理消息。我该怎么做?

我的代码示例如下:

import threading
import time 

class MyThread(threading.Thread):
    def __init__(self, args=(), kwargs=None):
        threading.Thread.__init__(self, args=(), kwargs=None)
        self.daemon = True
        self.receive_messages = args[0]

    def run(self):
        print threading.currentThread().getName(), self.receive_messages

    def do_thing_with_message(self, message):
        if self.receive_messages:
            print threading.currentThread().getName(), "Received %s".format(message)

if __name__ == '__main__':
    threads = []
    for t in range(10):
        threads.append( MyThread(args=(t % 2 == 0,)))
        threads[t].start()
        time.sleep(0.1)

    for t in threads:
        t.do_thing_with_message("Print this!")

输出:

Thread-1 True
Thread-2 False
Thread-3 True
Thread-4 False
Thread-5 True
Thread-6 False
Thread-7 True
Thread-8 False
Thread-9 True
Thread-10 False
MainThread Received %s
MainThread Received %s
MainThread Received %s
MainThread Received %s
MainThread Received %s

但是,我希望最后五行与MainThread无关,而不是%s,我希望它Print this!,就像这样:

Thread-1 True
Thread-2 False
Thread-3 True
Thread-4 False
Thread-5 True
Thread-6 False
Thread-7 True
Thread-8 False
Thread-9 True
Thread-10 False
Thread-1 Received Print this!
Thread-3 Received Print this!
Thread-5 Received Print this!
Thread-7 Received Print this!
Thread-9 Received Print this!

如何正确地将这样的消息发送到正在运行的线程?

附录:

如果我在Print this!阻止之后有这个阻止,并使用@ dano的代码来解决上述问题,它似乎不会响应这些新消息。

for t in threads:
    t.queue.put("Print this again!")
    time.sleep(0.1)

在这种情况下,我希望输出的结尾看起来像这样

Thread-1 Received Print this!
Thread-3 Received Print this!
Thread-5 Received Print this!
Thread-7 Received Print this!
Thread-9 Received Print this!
Thread-1 Received Print this again!
Thread-3 Received Print this again!
Thread-5 Received Print this again!
Thread-7 Received Print this again!
Thread-9 Received Print this again!

1 个答案:

答案 0 :(得分:13)

您可以使用Queue.Queue(或Python 3中的queue.Queue):

import threading
import time 
from Queue import Queue

print_lock = threading.Lock()

class MyThread(threading.Thread):
    def __init__(self, queue, args=(), kwargs=None):
        threading.Thread.__init__(self, args=(), kwargs=None)
        self.queue = queue
        self.daemon = True
        self.receive_messages = args[0]

    def run(self):
        print threading.currentThread().getName(), self.receive_messages
        val = self.queue.get()
        self.do_thing_with_message(val)

    def do_thing_with_message(self, message):
        if self.receive_messages:
            with print_lock:
                print threading.currentThread().getName(), "Received {}".format(message)

if __name__ == '__main__':
    threads = []
    for t in range(10):
        q = Queue()
        threads.append(MyThread(q, args=(t % 2 == 0,)))
        threads[t].start()
        time.sleep(0.1)

    for t in threads:
        t.queue.put("Print this!")

    for t in threads:
        t.join()

我们将Queue个实例传递给每个帖子,并使用Thread将邮件发送到queue.put。我们等待消息到达run方法,该方法是Thread对象的一部分,它实际上是在一个单独的执行线程中运行的。收到消息后,我们调用do_thing_with_message,它将在同一个后台线程中运行。

我还在代码中添加了threading.Lock,因此stdout的打印件不会混淆。

修改

如果您希望能够向线程传递多条消息,只需使用循环:

def run(self):
    print threading.currentThread().getName(), self.receive_messages
    while True:
        val = self.queue.get()
        if val is None:   # If you send `None`, the thread will exit.
            return
        self.do_thing_with_message(val)