Modify Python code as newbie

[ron.eggler at ecoation.com]

Hi,

I'm semi new to Python but need to modify a program that calls the mqtt_client.publish()  function from aws iot.
Now, if the publish function fails, it raises an exception.  I need to change the code so that when an exception is raised, instead of giving up, it should retry.
Here's some semi pseudo code of what I came up with and what I'm particularly interested in is, if the exception in pub_msg() is raised, will my thread t keep running? 
What else could/should be improved about the below snippet? I'm looking forward to get some inputs.

Thanks!


import retrying
import Queue as queue
import threading as th
NUM_THREADS=1
numth=[]
def worker():
    while not_terminated():
        item = q.get()
        if item is None:
            continue
        do_stuff(item)
        
def enQ(self, msg, topic):
    if len(numth<int(NUM_THREADS):
        t = th.Thread(target=worker)
        t.start()
        numth.append(t)
    q.put([self,msg,topic])
    
def do_stuff(dat):
    self    = dat[0]
    msg     = dat[1]
    topic   = dat[2]
    pub-msg(slef, msg, topic)
    
def send_msg(self, msg,topic):
    enQ(self,msg,topic)
    
def pub_msg(self,msg,topic):
    try:
        if topic == "test" and \
                something[self.xyz]:
            return
        except KeyError as err:
            foo("Exception {}".format(err))

    aws_publish(self,msg,topic)
    
@retry (wait_random_min=1000, wait_random_max=2000)
def aws_publish(self.msg,topic):
    self.__mqtt_client.publish(
        "/{}/{}".format(topic, self._uuid), msg_json, 1)