你重写run()方法了吗?如果你重写了__init__，你是否确保调用了基本线程threading.Thread.__init__()?

在启动两个线程之后，主线程是否继续在子线程上无限地/阻塞/连接工作，以便在子线程完成任务之前主线程的执行不会结束?

最后，是否有任何未处理的异常?

你可以在Thread构造函数中使用target参数直接传入一个被调用的函数，而不是run。

你不需要使用Thread的子类来实现这个工作——看看我下面发布的简单示例，看看如何做到这一点:

from threading import Thread
from time import sleep

def threaded_function(arg):
    for i in range(arg):
        print("running")
        sleep(1)


if __name__ == "__main__":
    thread = Thread(target = threaded_function, args = (10, ))
    thread.start()
    thread.join()
    print("thread finished...exiting")

在这里，我将展示如何使用threading模块创建一个线程，该线程将调用一个普通函数作为其目标。你可以看到我如何在线程构造函数中传递我需要的任何参数。

你的代码有几个问题:

def MyThread ( threading.thread ):

你不能子类化一个函数;只有在课堂上 如果你要使用子类，你需要线程化。Thread，不是threading.thread

如果你真的想只用函数来做这件事，你有两个选择:

线程:

import threading
def MyThread1():
    pass
def MyThread2():
    pass

t1 = threading.Thread(target=MyThread1, args=[])
t2 = threading.Thread(target=MyThread2, args=[])
t1.start()
t2.start()

线程:

import thread
def MyThread1():
    pass
def MyThread2():
    pass

thread.start_new_thread(MyThread1, ())
thread.start_new_thread(MyThread2, ())

Doc for thread.start_new_thread

我尝试添加另一个join()，它似乎有效。这是代码

from threading import Thread
from time import sleep

def function01(arg,name):
    for i in range(arg):
        print(name,'i---->',i,'\n')
        print (name,"arg---->",arg,'\n')
        sleep(1)

def test01():
    thread1 = Thread(target = function01, args = (10,'thread1', ))
    thread1.start()
    thread2 = Thread(target = function01, args = (10,'thread2', ))
    thread2.start()
    thread1.join()
    thread2.join()
    print ("thread finished...exiting")

test01()

Python 3有启动并行任务的功能。这使我们的工作更容易。

它有线程池和进程池。

下面让我们来了解一下:

ThreadPoolExecutor例子

import concurrent.futures
import urllib.request

URLS = ['http://www.foxnews.com/',
        'http://www.cnn.com/',
        'http://europe.wsj.com/',
        'http://www.bbc.co.uk/',
        'http://some-made-up-domain.com/']

# Retrieve a single page and report the URL and contents
def load_url(url, timeout):
    with urllib.request.urlopen(url, timeout=timeout) as conn:
        return conn.read()

# We can use a with statement to ensure threads are cleaned up promptly
with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
    # Start the load operations and mark each future with its URL
    future_to_url = {executor.submit(load_url, url, 60): url for url in URLS}
    for future in concurrent.futures.as_completed(future_to_url):
        url = future_to_url[future]
        try:
            data = future.result()
        except Exception as exc:
            print('%r generated an exception: %s' % (url, exc))
        else:
            print('%r page is %d bytes' % (url, len(data)))

另一个例子

import concurrent.futures
import math

PRIMES = [
    112272535095293,
    112582705942171,
    112272535095293,
    115280095190773,
    115797848077099,
    1099726899285419]

def is_prime(n):
    if n % 2 == 0:
        return False

    sqrt_n = int(math.floor(math.sqrt(n)))
    for i in range(3, sqrt_n + 1, 2):
        if n % i == 0:
            return False
    return True

def main():
    with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
        for number, prime in zip(PRIMES, executor.map(is_prime, PRIMES)):
            print('%d is prime: %s' % (number, prime))

if __name__ == '__main__':
    main()

使用线程实现多线程进程的简单方法

相同的代码片段

import threading

#function which takes some time to process 
def say(i):
    time.sleep(1)
    print(i)

threads = []
for i in range(10):
    
    thread = threading.Thread(target=say, args=(i,))

    thread.start()
    threads.append(thread)

#wait for all threads to complete before main program exits 
for thread in threads:
    thread.join()