等待特定的延迟后如何在python中启动线程 [英] How to start a thread in python after waiting for a specific delay

问题描述

我有功能detection()，timeralg()

i have functions detection(),timeralg()

在运行timeralg()函数时，我希望detection()在特定延迟后并行开始. 目前，我尝试过这样

while running the timeralg() function, i want detection() to start in parallel after a specific delay. currently i tried like this

def timeralg(c1,c2,c3,c4):
    t=[4,4,4,6,6,20,24,28,32,36,40]#delay determining array
    for y in range(0,3):
        print 'y is ',y
        if((c1>=c2)and(c1>=c3)):
            print 'timer1 on for'
            x=t[c1]        
            print x
            c1=0

        GPIO.output(5,False)#Red1
        GPIO.output(13,True)#red2
        GPIO.output(12,True)#red3
        GPIO.output(7,True)#green1
        if(y==2):
            t = threading.Thread(detection())
            t.start()
            print 'processing strtd in from 1'

        time.sleep(x-3)
        GPIO.output(7,False)
        GPIO.output(3,True)#Yellow1
        time.sleep(3)
        GPIO.output(3,False)#Yellow1
        GPIO.output(5,True)#Red1

与此不同，我不想在我指定的特定延迟后开始.

Unlike this i want 't' to start after a specific delay specified by me.

推荐答案

您可以如下包装detection():

def delayed_detection():
    time.sleep(3)
    detection()

然后使用以下内容启动线程:

Then start your thread with:

t = threading.Thread(delayed_detection)
t.start()

您并没有延迟线程的产生，但是您仍然可以在三秒钟后实现调用detecton()

You're not delaying the spawning of the thread, but you are still achieving calling detecton() after three seconds

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