How to wait until only the first thread is finished in Python

multithreading, python

Solution

Use a queue: each thread when completed puts the result on the queue and then you just need to read the appropriate number of results and ignore the remainder:

#!python3.3
import queue    # For Python 2.x use 'import Queue as queue'
import threading, time, random

def func(id, result_queue):
    print("Thread", id)
    time.sleep(random.random() * 5)
    result_queue.put((id, 'done'))

def main():
    q = queue.Queue()
    threads = [ threading.Thread(target=func, args=(i, q)) for i in range(5) ]
    for th in threads:
        th.daemon = True
        th.start()

    result1 = q.get()
    result2 = q.get()

    print("Second result: {}".format(result2))

if __name__=='__main__':
    main()

Documentation for `Queue.get()` (with no arguments it is equivalent to `Queue.get(True, None)`:

Queue.get([block[, timeout]])

Remove and return an item from the queue. If optional args block is true and timeout is None (the default), block if necessary until an item is available. If timeout is a positive number, it blocks at most timeout seconds and raises the Empty exception if no item was available within that time. Otherwise (block is false), return an item if one is immediately available, else raise the Empty exception (timeout is ignored in that case).

Problem

The requirement is to start five threads, and wait only in the fastest thread. All five threads went to look for the same data 5 directions, and one is enough to continue the control flow. Actually, I need to wait for the first two threads to return, to verify against each other. But I guess if I know how to wait for the fastest. I can figure out how to wait for the second-fastest. A lot talk about `join(timeout)`, but you don't know in advance which one to wait (which one to apply `join` in advance).

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