Limiting queue length with PyZMQ

问题:

I want to limit the amount of memory consumed by my ZeroMQ message queues in a Python application. I know that setting the high-water mark will limit the amount that will be queued on the sender side, but is there a way to control how much will be queued on the receiver side? The Python ZeroMQ binding seems to have it set at unlimited.

My test scenario: I have two python terminals that I am using for testing. One is the receiver:

Python 2.5.1 (r251:54863, Aug 25 2008, 20:50:04) 
[GCC 4.1.2 20071124 (Red Hat 4.1.2-42)] on linux2
Type "help", "copyright", "credits" or "license" for more information.

>>> import zmq
>>> context = zmq.Context()
>>> socket = context.socket(zmq.PULL)
>>> socket.setsockopt(zmq.RCVBUF, 256)
>>> socket.bind("tcp://127.0.0.1:12345")

The other is the sender:

Python 2.5.1 (r251:54863, Aug 25 2008, 20:50:04) 
[GCC 4.1.2 20071124 (Red Hat 4.1.2-42)] on linux2
Type "help", "copyright", "credits" or "license" for more information.

>>> import zmq
>>> context=zmq.Context()
>>> socket = context.socket(zmq.PUSH)
>>> socket.setsockopt(zmq.SNDBUF, 2048)
>>> socket.setsockopt(zmq.HWM, 1)
>>> socket.connect("tcp://127.0.0.1:12345")
>>> num = 0
>>> while True:
...  print num
...  socket.send(str(num))
...  num = num + 1
... 

I run socket.recv() on the receiver side a couple times to make sure that the queue works, but other than that, let the two terminals just sit there. The send loop seems to never block and the receive prompt seems to have a growing memory footprint.

回答1:

In contradiction to the documentation of ZeroMQ, the high water mark needs to be set on both the PUSH side and the PULL side. Once I changed the PULL, it worked better. The new PULL code is:

Python 2.5.1 (r251:54863, Aug 25 2008, 20:50:04) 
[GCC 4.1.2 20071124 (Red Hat 4.1.2-42)] on linux2
Type "help", "copyright", "credits" or "license" for more information.

>>> import zmq
>>> context=zmq.Context()
>>> socket = context.socket(zmq.PULL)
>>> socket.setsockopt(zmq.RCVBUF, 256)
>>> socket.setsockopt(zmq.HWM, 1)
>>> socket.bind("tcp://127.0.0.1:12345")

回答2:

Actually, the documentation says this:

"When a ZMQ_PUSH socket enters an exceptional state due to having reached the high water mark for all downstream nodes, or if there are no downstream nodes at all, then any zmq_send(3) operations on the socket shall block until the exceptional state ends or at least one downstream node becomes available for sending; messages are not discarded."

http://api.zeromq.org/2-1:zmq-socket

Which outright states that you can (and should) set the high water mark for downstream nodes (aka pull), and perhaps implies that setting it on the push side will have no effect (although I suspect that's not true, because there is still the case where downstream nodes are available but messages are coming in faster than they can be sent.)

回答3:

With the zmq.SNDBUF and zmq.RCVBUF options you could set a limit on buffer size.

---

Also, I using zmq.CONFLATE option in reciever side to limiting the ZeroMQ queue size to one:

Here's an example with ZMQ PUSH/PULL:

Sender (zmq.PUSH):

def create_pub_socket(ip, port):
    try:
        context = zmq.Context()
        socket = context.socket(zmq.PUSH)
        socket.setsockopt(zmq.SNDHWM, 1)
        zmq_address = "tcp://{}:{}".format(ip, port)
        socket.connect(zmq_address)
        return socket

    except zmq.ZMQError as exp:
        print(exp)
        return False

sock = create_push_socket('127.0.0.1', 5558)
if sock:
    sock.send_json({'a': 1})

---

Getter (zmq.PULL):

def listen(self):
    sock = None
    try:
        context = zmq.Context()
        sock = context.socket(zmq.PULL)
        sock.setsockopt(zmq.RCVHWM, 1)
        sock.setsockopt(zmq.CONFLATE, 1)  # last msg only.
        sock.bind("tcp://*:5558")

    except zmq.ZMQError:
        logger.captureException()

    configs = None
    while configs is None:
        if sock:
            configs = sock.recv_json()
            time.sleep(1e-1)
        else:
            time.sleep(5)
            listen()  # Recursive.
listen()