Display a countdown for the python sleep function

Question

I am using time.sleep(10) in my program. Can display the countdown in the shell when I run my program?

>>>run_my_program()
tasks done, now sleeping for 10 seconds

and then I want it to do 10,9,8,7....

is this possible?

Indium

Thank you for all the wonderful answers. – Indium

related: countdown using using Tkinter, or Gtk, or Twisted or Asyncio event loops – Weiland

Answer 1

you could always do

#do some stuff
print 'tasks done, now sleeping for 10 seconds'
for i in xrange(10,0,-1):
    time.sleep(1)
    print i

This snippet has the slightly annoying feature that each number gets printed out on a newline. To avoid this, you can

import sys
import time
for i in xrange(10,0,-1):
    sys.stdout.write(str(i)+' ')
    sys.stdout.flush()
    time.sleep(1)

Backstretch

thanks, this is great. you have just a small mistake which I fixed. time.sleep(i) should be time.sleep(1) – Indium

Yes, indeed, it is fixed for the answer. – Backstretch

If you're getting NameError: name 'xrange' is not defined use range instead. It was renamed in Python 3. – Cognate

Answer 2

This is the best way to display a timer in the console for Python 3.x:

import time
import sys

for remaining in range(10, 0, -1):
    sys.stdout.write("\r")
    sys.stdout.write("{:2d} seconds remaining.".format(remaining))
    sys.stdout.flush()
    time.sleep(1)

sys.stdout.write("\rComplete!            \n")

This writes over the previous line on each cycle.

Generalist

Answer 3

A simple solution that clears the last number from the console:

import time

for i in range(10,0,-1):
    print(f"{i}", end="\r", flush=True)
    time.sleep(1)

By default, the print function sets end="\n" which means subsequent calls to print will be printed on a new line. You can change this to end="\r" to replace the output after each call. (How does carriage return "\r" work in python).

Here f"{i}" is for printing single digit only. You can modify it based on number of digits. e.g. Here it will work for two digits by just adding one space as a postfix- f"{i} "

Also, using flush means you don't have to worry about buffering issues. (What does print()'s flush do?)

This is how it looks:

Rina

This great, however if you input a large number, say 1200, it will end up printing e.g. 8790 instead of 0879 or 879, any way to fix this? – Aubin

Oris, Updated the answer as per your comment. – Trask

Change the "print(f("{i}..." for this line print("{:2d}".format(i), end="\r", flush=True) Or I will add an edit below this answer – Sublunary

An alternative is to print another line prior to clear it, such as print('\r', ' '*100, end='') – Inalienable

Answer 4

You can do a countdown function like:

import sys
import time

def countdown(t, step=1, msg='sleeping'):  # in seconds
    pad_str = ' ' * len('%d' % step)
    for i in range(t, 0, -step):
        print '%s for the next %d seconds %s\r' % (msg, i, pad_str),
        sys.stdout.flush()
        time.sleep(step)
    print 'Done %s for %d seconds!  %s' % (msg, t, pad_str)

The carriage return \r and the comma , will keep the print in the same line (avoiding one line for each countdown value)

As the number of seconds decreases, the pad_str will ensure the last line is overwritten with spaces instead of leaving the last character(s) behind as the output shortens.

The final print overwrites the last status message with a done message and increments the output line, so there is evidence of the delay.

Ancalin

To prevent a new line inserted every second in the countdown, the statement print '%s for the next %d seconds %s\r' % (msg, i, pad_str), should be changed to sys.stdout.write( '%s for the next %d seconds %s\r' % (msg, i, pad_str),). – Orthogenetic

Answer 5

Here's one I did:

import time
a = input("How long is the countdown?")
while a != 0:
    print a
    time.sleep(1)
    a = a-1

At the end if you and an else you can put an alarm or whatever.

Notepaper

Answer 6

time.sleep() may return earlier if the sleep is interrupted by a signal or later (depends on the scheduling of other processes/threads by OS/the interpreter).

To improve accuracy over multiple iterations, to avoid drift for large number of iterations, the countdown may be locked with the clock:

#!/usr/bin/env python
import sys
import time

for i in reversed(range(1, 1001)):
    time.sleep(1 - time.time() % 1) # sleep until a whole second boundary
    sys.stderr.write('\r%4d' % i)

Weiland

Answer 7

This is something that I've learned at one of my first python lessons, we played with ["/","-","|","\","|"] but the principle is the same:

import time

for i in reversed(range(0, 10)):
    time.sleep(1)
    print "%s\r" %i,

Donaugh

Answer 8

Another easy way, without reinventing the wheel, is to use tqdm, which automatically displays a progress bar:

from time import sleep
from tqdm import tqdm

for _ in tqdm(range(10)):
    sleep(1)

Optionally, you can then modify the display of the loading bar as you wish.

Gebhart

Answer 9

Sure, just write a loop that prints 10 minus the iteration counter, then have it sleep 1 second each iteration and run for 10 iterations. Or, to be even more flexible:

def printer(v):
    print v
def countdown_timer(duration, step=1, output_function=printer,
                    prompt='Waiting {duration} seconds.'):
    output_function(prompt.format(duration=duration))
    for i in xrange(duration/step):
        output_function(duration - step * i)

Kwangtung

Answer 10

This one is subsecond precise:

    print()
    _k = 0.5  # ensure k != _k first time round (as k will be integer)
    print('Starting in ')
    while _k > 0:
        k = round(event_timestamp - time())
        if k != _k:
            print(f'\r{k} ', end='', flush=True)
            _k = k
        sleep(0.1)

    print('boom')

Notice the trailing space in f'\r{k} '. So if we go from 100 to 99 or 10 to 9 it clears the second digit.

Also it doesn't require import sys.

sleep(0.0003) if you want millisecond precision.

Plonk

Answer 11

if you don't limit yourself to sleep, then (courtesy automatetheboringstuff), pyautogui has a nifty countdown function:

import pyautogui
print('Starting in ', end=''); pyautogui.countdown(3)

Malediction

Answer 12

may be this will help

    import turtle
    for i in range(10):
        t2 = turtle.Turtle()
        t2.hideturtle()
        t2.penup()
        t2.goto(0,0)
        t2.write(i,font=("Arial", 16, "normal"))
        i-=1
        sleep(1)
        t2.clear()

Donell

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