快速修复

Question

我想传递一个值列表，并将每个值作为独立的线程传递：

例如：

import time
import threading

list_args = [arg1, arg2, arg3, ...., argn]

# Code to execute in an independent thread
import time
def countdown(n):
    while n > 0:
        print('T-minus', n)
        n -= 1
        time.sleep(0.5)


# Create and launch a thread
t1 = threading.Thread(target=countdown, args=(arg1,))
t2 = threading.Thread(target=countdown, args=(arg2,))
.
.
.
tn = threading.Thread(target=countdown, args=(argn,))
t1.start(); t2.start(); tn.start()

Answer 1

快速修复

在最后的每个.join()，t1等主题上致电t2。

详细

我怀疑你真正的问题是'为什么不countdown被召唤？ Thread.join()方法导致主线程在继续之前等待其他线程完成执行。没有它，一旦主线程完成，它就会终止整个过程。

在主程序完成执行的程序中，该进程与其所有线程一起终止，然后后者可以调用它们的countdown函数。

其他问题：

最好包含minimum working example。您的代码无法按照书面形式执行。
通常，某种数据结构用于管理线程。这很好，因为它使代码更紧凑，更通用，更可重用。
您无需两次导入time。

这可能接近你想要的：

import time
import threading

list_args = [1,2,3]

def countdown(n):
    while n > 0:
        print('T-minus', n)
        n -= 1
        time.sleep(0.5)


# Create and launch a thread
threads = []
for arg in list_args:
    t = threading.Thread(target=countdown,args=(arg,))
    t.start()
    threads.append(t)

for thread in threads:
    thread.join()

Answer 2

import time
import threading

list_args = [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]

# Code to execute in an independent thread
def countdown(n):
    while n > 0:
        print('T-minus', n)
        n -= 1
        time.sleep(0.5)


# Create and launch a thread
for item in list_args:
    thread = threading.Thread(target=countdown, args=(item,))
    thread.start()

Answer 3

我认为这会做你所描述的：

for a in list_args:
    threading.Thread(target=countdown, args=(a,)).start()

Answer 4

您可以将所有线程添加到列表中，使其与参数数量无关。在主程序结束时，您必须加入所有线程。否则主线程退出并终止所有其他线程。

工作代码：

import time
import threading

# create list [0, 1, ..., 9] as argument list
list_args = range(10)

# Code to execute in an independent thread
def countdown(n):
    while n > 0:
        print('T-minus', n)
        n -= 1
        time.sleep(0.5)

if __name__ == '__main__':
    threads = []

    # create threads
    for arg in list_args:
        threads.append(threading.Thread(target=countdown, args=(arg,)))

    # start threads
    for thread in threads:
        print("start")
        thread.start()

    # join threads (let main thread wait until all other threads ended)
    for thread in threads:
        thread.join()

    print("finished!")

线程参数列表

4 个答案:

快速修复

详细