Question

为什么此代码不是并行运行，所以需要20秒才能运行，这意味着它是按顺序运行的。谢谢您的帮助。

import time
from queue import Queue
from threading import Thread
start = time.time()
def f():
    time.sleep(0.5)
    print("yes")
    return 'yes'


def do_stuff(q):
    while True:
        output = q.get()
        q.task_done()


q = Queue(maxsize=100)
for message_nbr in range(40):
    q.put(f())

num_threads = 10

for i in range(num_threads):
    worker = Thread(target=do_stuff, args=(q, ))
    worker.setDaemon(True)
    worker.start()

q.join()
print("time: ", time.time() - start)  # ~20 seconds

Answer 1

这是可行的！

request_threaded_irq()

Answer 2

答案就在这里：

for message_nbr in range(40):
    q.put(f())

您正在将40个None实例放入队列，因为您正在调用f()，该实例返回None而不是传入f（函数对象）。此块占用20运行几秒钟！

更改此代码

def do_stuff(q):
    while True:
        output = q.get()
        q.task_done()

对此

def do_stuff(q):
    while True:
        output = q.get()
        output()
        q.task_done()

也是必需的（您需要调用该函数！）

最终：

import time
from queue import Queue
from threading import Thread
start = time.time()
def f():
    time.sleep(0.5)
    print("yes")
    return 'yes'


def do_stuff(q):
    while True:
        output = q.get()
        output()
        q.task_done()


q = Queue(maxsize=100)
for message_nbr in range(40):
    q.put(f)

num_threads = 10

for i in range(num_threads):
    worker = Thread(target=do_stuff, args=(q, ))
    worker.setDaemon(True)
    worker.start()

q.join()
print("time: ", time.time() - start)  # time:  2.183439254760742

python多线程，队列按顺序运行，而不是并行运行

2 个答案: