python - 不忙于等待 ZeroMQ？

Question

我有一些使用 ZeroMQ 库的简单 PUB/SUB 代码。本质上只是无限等待消息返回。

但是，这似乎是忙等待，因为在等待接收消息时，python 的 CPU 使用率急剧上升。

有更好的方法吗？我想象这就像在调用接收时在套接字上一样，它不会不断地执行指令和浪费 CPU。

import sys
import zmq

port = "5556"
if len(sys.argv) > 1:
    port =  sys.argv[1]
    int(port)

# Socket to talk to server
context = zmq.Context()
socket = context.socket(zmq.SUB)

print "Collecting updates from weather server..."
socket.connect ("tcp://localhost:%s" % port)

# Subscribe to zipcode, default is NYC, 10001
topicfilter = "10001"
socket.setsockopt(zmq.SUBSCRIBE, topicfilter)

# Process 5 updates
total_value = 0
for update_nbr in range (5):
    string = socket.recv()
    topic, messagedata = string.split()
    total_value += int(messagedata)
    print ('{} {}'.format(topic, message)

Answer 1

Q : "Non busy waiting for ZeroMQ?"

当然可以。

...
for update_nbr in range( 5 ):
    print "{0:} A hunt for message started".format( time.ctime() )
    while True: # .............................................. poll() + wait
          if ( 0 == socket.poll( 1000 ) ):
               print "\n.",
          else:
               break # we've got a message .....................
    print "\n{0:} A message came".format( time.ctime() )
    #_________________________________________
    try:
        string = socket.recv( zmq.NOBLOCK )
        topic, messagedata = string.split()
        total_value += int( messagedata )
        print '{} {}'.format( topic, message )

    except:
        print 'EXC:...'

    finally:
        pass
    #_________________________________________