Python 创建一个线程并在按下按键时启动它

Question

我制作了一个 python 脚本，可以将数字分解为其质因数。然而，当处理大数字时，我可能想了解计算的进度。 (我简化了脚本)

import time, sys, threading

num = int(input("Input the number to factor: "))
factors = []

def check_progress():
    but = input("Press p: ")
    if but == "p":
        tot = int(num**(1/2))
        print("Step ", k, " of ", tot, " -- ", round(k*100/tot,5), "%", end="\r", sep="")


t = threading.Thread(target=check_progress) ?
t.daemon = True ?
t.start() ?

k = 1
while(k != int(num**(1/2))):
    k = (k+1)
    if num%k == 0:
        factors.append(int(k))
        num = num//k
        k = 1
print(factors)

我想知道是否有一种方法可以按需显示进度，例如，在循环期间，我按下一个键，它会打印进度？

我如何在我的脚本中实现类似的线程？

感谢并抱歉我的英语

编辑:

def check_progress():
    while True:
        but = input("## Press return to show progress ##")
        tot = int(num**(1/2))
        print("Step ", k, " of ", tot, " -- ", round(k*100/tot,5), "%", sep="")

Answer 1

这是一种可能的设计:

主线程:

• 创建queue和线程
• 启动进度线程
• 等待用户输入
  • 输入时:
  • 从队列中弹出结果(可能是None)
  • 显示它
  • 循环

进度线程:

• 完成工作并将状态放入队列

我可以提供例子，但我觉得你愿意学习。请随时发表评论寻求帮助。

编辑:带有队列的完整示例。

from time import sleep
from Queue import Queue
from threading import Thread


# Main thread:
def main():
    # create queue and thread
    queue = Queue()
    thread = Thread(target=worker, args=(queue,))

    # start the progress thread
    thread.start()

    # wait user input
    while thread.isAlive():
        raw_input('--- Press any key to show status ---')

        # pop result from queue (may be None)
        status = queue.get_nowait()
        queue.task_done()

        # display it
        if status:
            print 'Progress: %s%%' % status
        else:
            print 'No status available'

# Progress thread:
def worker(queue):
    # do the work an put status in queue
    # Simulate long work ...
    for x in xrange(100):
        # put status in queue
        queue.put_nowait(x)
        sleep(.5)

if __name__ == '__main__':
    main()