How do I write function a better function to act as a 3 second countdown?-CodePudding

I need a function to start a countdown from 3, and then run a program once the countdown is complete, I can't use the time.sleep() as that particular function won't start the program.

Currently, my function counts up from zero, and also prints multiple iterations of the same number.

I also feel like the function is overcomplicated, are there any simpler methods?

Code:

def countdown():

     countdown_time = 3
     start_time = time.time()

     while (time.time() - start_time) < countdown_time:
        print(round(time.time() - start_time))
        if (time.time() - start_time) >= countdown_time:
            break

Ideally it should just print:

3
2
1

But it prints something more like this:

CodePudding user response：

I think this what you are looking for:

from time import sleep

def countDown():
    for i in range(3, 0, -1):
        print(i)
        sleep(1)
        
countDown()

You are saying that you can't use the time.sleep() as that particular function won't start the program which is wrong. It works here just fine.

CodePudding user response：

This works:

import time

def countdown():

     countdown_time = 3
     start_time = time.time()
     while countdown_time > 0:
        if countdown_time - round(time.time() - start_time) <= countdown_time and countdown_time != 0:
          print(countdown_time)
          countdown_time -= 1

countdown()

Output:

3
2
1