Python - Skipping websocket callback "queue" - python-3.x

Question

I have the following:
def process(msg):
print(msg)
for i in range(10): # This is in place of the
print(i, end=' ') # actual task, which takes
time.sleep(1) # around 10 seconds to run
ws = websocket()
ws.start_socket_abc(callback=process)
The websocket pushes info every second and makes a call to process each time.
Say the websocket is returning "one" for this second, "two" for the next second, "three" for the second after, and so on. Running the above code, the ouput would be:
one
0 1 2 3 4 5 6 7 8 9 <- this line takes 10 seconds to finish
two
0 1 2 3 4 5 6 7 8 9 <- this line takes 10 seconds to finish
...
It's clear that process("two") gets queued somewhere and is waited to be called after process("one") finishes. By the time the program prints out two, the real-time message should be twelve.
What I'm trying to achieve is to completely skip the "queue". I want it to take the message, do the task, skip all the calls in between (this is important, I actually need to skip them), take the next real-time message and do the task. In short, I want it to print out whatever real-time message is at the moment process("one") finishes. i.e.
one
0 1 2 3 4 5 6 7 8 9 <- this line takes 10 seconds to finish
twelve
...
This is what I have in mind.
message = ''
def process(msg):
message = msg
ws = websocket()
ws.start_socket_abc(callback=process)
def task(msg=message):
print(msg)
for i in range(10): # This is in place of the
print(i, end=' ') # actual task, which takes
time.sleep(1) # around 10 seconds to run
start_thread(target=task)
This performs the task with the message variable, which changes in real-time with the messages received from the websocket so after task("one") finishes, message is at "twelve" and task("twelve") will be called, skipping all the messages in between.
I strongly doubt that this is the proper way of doing it. Someone please enlighten me, thanks in advance!