How to repeatedly execute a function every x seconds?


I want to repeatedly execute a function in Python every 60 seconds forever (just like an NSTimer in Objective C or setTimeout in JS). This code will run as a daemon and is effectively like calling the python script every minute using a cron, but without requiring that to be set up by the user.

In this question about a cron implemented in Python, the solution appears to effectively just sleep() for x seconds. I don’t need such advanced functionality so perhaps something like this would work

while True:
    # Code executed here

Are there any foreseeable problems with this code?

Asked By: davidmytton



If your program doesn’t have a event loop already, use the sched module, which implements a general purpose event scheduler.

import sched, time

def do_something(scheduler): 
    # schedule the next call first
    scheduler.enter(60, 1, do_something, (scheduler,))
    print("Doing stuff...")
    # then do your stuff

my_scheduler = sched.scheduler(time.time, time.sleep)
my_scheduler.enter(60, 1, do_something, (my_scheduler,))

If you’re already using an event loop library like asyncio, trio, tkinter, PyQt5, gobject, kivy, and many others – just schedule the task using your existing event loop library’s methods, instead.

Answered By: nosklo

The main difference between that and cron is that an exception will kill the daemon for good. You might want to wrap with an exception catcher and logger.

Answered By: bobince

You might want to consider Twisted which is a Python networking library that implements the Reactor Pattern.

from twisted.internet import task, reactor

timeout = 60.0 # Sixty seconds

def doWork():
    #do work here

l = task.LoopingCall(doWork)
l.start(timeout) # call every sixty seconds

While “while True: sleep(60)” will probably work Twisted probably already implements many of the features that you will eventually need (daemonization, logging or exception handling as pointed out by bobince) and will probably be a more robust solution

Answered By: Aaron Maenpaa

The easier way I believe to be:

import time

def executeSomething():
    #code here

while True:

This way your code is executed, then it waits 60 seconds then it executes again, waits, execute, etc…
No need to complicate things 😀

Answered By: Itxaka

I faced a similar problem some time back. May be might help?

For v0.2, the following snippet works

import cronus.beat as beat

beat.set_rate(2) # run twice per second
while beat.true():
    # do some time consuming work here
    beat.sleep() # total loop duration would be 0.5 sec
Answered By: Anay

Lock your time loop to the system clock like this:

import time
starttime = time.time()
while True:
    time.sleep(60.0 - ((time.time() - starttime) % 60.0))
Answered By: Dave Rove

If you want a non-blocking way to execute your function periodically, instead of a blocking infinite loop I’d use a threaded timer. This way your code can keep running and perform other tasks and still have your function called every n seconds. I use this technique a lot for printing progress info on long, CPU/Disk/Network intensive tasks.

Here’s the code I’ve posted in a similar question, with start() and stop() control:

from threading import Timer

class RepeatedTimer(object):
    def __init__(self, interval, function, *args, **kwargs):
        self._timer     = None
        self.interval   = interval
        self.function   = function
        self.args       = args
        self.kwargs     = kwargs
        self.is_running = False

    def _run(self):
        self.is_running = False
        self.function(*self.args, **self.kwargs)

    def start(self):
        if not self.is_running:
            self._timer = Timer(self.interval, self._run)
            self.is_running = True

    def stop(self):
        self.is_running = False


from time import sleep

def hello(name):
    print "Hello %s!" % name

print "starting..."
rt = RepeatedTimer(1, hello, "World") # it auto-starts, no need of rt.start()
    sleep(5) # your long-running job goes here...
    rt.stop() # better in a try/finally block to make sure the program ends!


  • Standard library only, no external dependencies
  • start() and stop() are safe to call multiple times even if the timer has already started/stopped
  • function to be called can have positional and named arguments
  • You can change interval anytime, it will be effective after next run. Same for args, kwargs and even function!
Answered By: MestreLion

Here’s an update to the code from MestreLion that avoids drifiting over time.

The RepeatedTimer class here calls the given function every “interval” seconds as requested by the OP; the schedule doesn’t depend on how long the function takes to execute. I like this solution since it doesn’t have external library dependencies; this is just pure python.

import threading 
import time

class RepeatedTimer(object):
  def __init__(self, interval, function, *args, **kwargs):
    self._timer = None
    self.interval = interval
    self.function = function
    self.args = args
    self.kwargs = kwargs
    self.is_running = False
    self.next_call = time.time()

  def _run(self):
    self.is_running = False
    self.function(*self.args, **self.kwargs)

  def start(self):
    if not self.is_running:
      self.next_call += self.interval
      self._timer = threading.Timer(self.next_call - time.time(), self._run)
      self.is_running = True

  def stop(self):
    self.is_running = False

Sample usage (copied from MestreLion’s answer):

from time import sleep

def hello(name):
    print "Hello %s!" % name

print "starting..."
rt = RepeatedTimer(1, hello, "World") # it auto-starts, no need of rt.start()
    sleep(5) # your long-running job goes here...
    rt.stop() # better in a try/finally block to make sure the program ends!
Answered By: eraoul

I use this to cause 60 events per hour with most events occurring at the same number of seconds after the whole minute:

import math
import time
import random

TICK = 60 # one minute tick size
TICK_TIMING = 59 # execute on 59th second of the tick
TICK_MINIMUM = 30 # minimum catch up tick size when lagging

def set_timing():

    now = time.time()
    elapsed = now - info['begin']
    minutes = math.floor(elapsed/TICK)
    tick_elapsed = now - info['completion_time']
    if (info['tick']+1) > minutes:
        wait = max(0,(TICK_TIMING-(time.time() % TICK)))
        print ('standard wait: %.2f' % wait)
    elif tick_elapsed < TICK_MINIMUM:
        wait = TICK_MINIMUM-tick_elapsed
        print ('minimum wait: %.2f' % wait)
        print ('skip set_timing(); no wait')
    drift = ((time.time() - info['begin']) - info['tick']*TICK -
        TICK_TIMING + info['begin']%TICK)
    print ('drift: %.6f' % drift)

info['tick'] = 0
info['begin'] = time.time()
info['completion_time'] = info['begin'] - TICK

while 1:


    print('hello world')

    #random real world event

    info['tick'] += 1
    info['completion_time'] = time.time()

Depending upon actual conditions you might get ticks of length:


but at the end of 60 minutes you’ll have 60 ticks; and most of them will occur at the correct offset to the minute you prefer.

On my system I get typical drift of < 1/20th of a second until need for correction arises.

The advantage of this method is resolution of clock drift; which can cause issues if you’re doing things like appending one item per tick and you expect 60 items appended per hour. Failure to account for drift can cause secondary indications like moving averages to consider data too deep into the past resulting in faulty output.

Answered By: litepresence

One possible answer:

import time

while True:
    if time.time()-t>10:
        #run your task here
Answered By: sks

e.g., Display current local time

import datetime
import glib
import logger

def get_local_time():
    current_time ="%H:%M")"get_local_time(): %s",current_time)
    return str(current_time)

def display_local_time():"Current time is: %s", get_local_time())
    return True

# call every minute
glib.timeout_add(60*1000, display_local_time)
Answered By: rise.riyo

I use Tkinter after() method, which doesn’t “steal the game” (like the sched module that was presented earlier), i.e. it allows other things to run in parallel:

import Tkinter

def do_something1():
  global n1
  n1 += 1
  if n1 == 6: # (Optional condition)
    print "* do_something1() is done *"; return
  # Do your stuff here
  # ...
  print "do_something1() "+str(n1)
  tk.after(1000, do_something1)

def do_something2(): 
  global n2
  n2 += 1
  if n2 == 6: # (Optional condition)
    print "* do_something2() is done *"; return
  # Do your stuff here
  # ...
  print "do_something2() "+str(n2)
  tk.after(500, do_something2)

tk = Tkinter.Tk(); 
n1 = 0; n2 = 0

do_something1() and do_something2() can run in parallel and in whatever interval speed. Here, the 2nd one will be executed twice as fast.Note also that I have used a simple counter as a condition to terminate either function. You can use whatever other contition you like or none if you what a function to run until the program terminates (e.g. a clock).

Answered By: Apostolos
import time, traceback

def every(delay, task):
  next_time = time.time() + delay
  while True:
    time.sleep(max(0, next_time - time.time()))
    except Exception:
      # in production code you might want to have this instead of course:
      # logger.exception("Problem while executing repetitive task.")
    # skip tasks if we are behind schedule:
    next_time += (time.time() - next_time) // delay * delay + delay

def foo():
  print("foo", time.time())

every(5, foo)

If you want to do this without blocking your remaining code, you can use this to let it run in its own thread:

import threading
threading.Thread(target=lambda: every(5, foo)).start()

This solution combines several features rarely found combined in the other solutions:

  • Exception handling: As far as possible on this level, exceptions are handled properly, i. e. get logged for debugging purposes without aborting our program.
  • No chaining: The common chain-like implementation (for scheduling the next event) you find in many answers is brittle in the aspect that if anything goes wrong within the scheduling mechanism (threading.Timer or whatever), this will terminate the chain. No further executions will happen then, even if the reason of the problem is already fixed. A simple loop and waiting with a simple sleep() is much more robust in comparison.
  • No drift: My solution keeps an exact track of the times it is supposed to run at. There is no drift depending on the execution time (as in many other solutions).
  • Skipping: My solution will skip tasks if one execution took too much time (e. g. do X every five seconds, but X took 6 seconds). This is the standard cron behavior (and for a good reason). Many other solutions then simply execute the task several times in a row without any delay. For most cases (e. g. cleanup tasks) this is not wished. If it is wished, simply use next_time += delay instead.
Answered By: Alfe
    ''' tracking number of times it prints'''
import threading

global timeInterval
def printit():
  threading.Timer(timeInterval, printit).start()
  print( "Hello, World!")
  global count

if __name__ == "__main__":
    timeInterval= int(input('Enter Time in Seconds:'))
Answered By: raviGupta

Here’s an adapted version to the code from MestreLion.
In addition to the original function, this code:

1) add first_interval used to fire the timer at a specific time(caller need to calculate the first_interval and pass in)

2) solve a race-condition in original code. In the original code, if control thread failed to cancel the running timer(“Stop the timer, and cancel the execution of the timer’s action. This will only work if the timer is still in its waiting stage.” quoted from, the timer will run endlessly.

class RepeatedTimer(object):
def __init__(self, first_interval, interval, func, *args, **kwargs):
    self.timer      = None
    self.first_interval = first_interval
    self.interval   = interval
    self.func   = func
    self.args       = args
    self.kwargs     = kwargs
    self.running = False
    self.is_started = False

def first_start(self):
        # no race-condition here because only control thread will call this method
        # if already started will not start again
        if not self.is_started:
            self.is_started = True
            self.timer = Timer(self.first_interval,
            self.running = True
    except Exception as e:
        log_print(syslog.LOG_ERR, "timer first_start failed %s %s"%(e.message, traceback.format_exc()))

def run(self):
    # if not stopped start again
    if self.running:
        self.timer = Timer(self.interval,
    self.func(*self.args, **self.kwargs)

def stop(self):
    # cancel current timer in case failed it's still OK
    # if already stopped doesn't matter to stop again
    if self.timer:
    self.running = False
Answered By: dproc

I ended up using the schedule module. The API is nice.

import schedule
import time

def job():
    print("I'm working...")


while True:
Answered By: Union find

Here is another solution without using any extra libaries.

def delay_until(condition_fn, interval_in_sec, timeout_in_sec):
    """Delay using a boolean callable function.

    `condition_fn` is invoked every `interval_in_sec` until `timeout_in_sec`.
    It can break early if condition is met.

        condition_fn     - a callable boolean function
        interval_in_sec  - wait time between calling `condition_fn`
        timeout_in_sec   - maximum time to run

    Returns: None
    start = last_call = time.time()
    while time.time() - start < timeout_in_sec:
        if (time.time() - last_call) > interval_in_sec:
            if condition_fn() is True:
            last_call = time.time()
Answered By: Dat

If drift is not a concern

import threading, time

def print_every_n_seconds(n=2):
    while True:
thread = threading.Thread(target=print_every_n_seconds, daemon=True)

Which asynchronously outputs.

#Tue Oct 16 17:29:40 2018
#Tue Oct 16 17:29:42 2018
#Tue Oct 16 17:29:44 2018

If the task being run takes appreciable amount of time, then the interval becomes 2 seconds + task time, so if you need precise scheduling then this is not for you.

Note the daemon=True flag means this thread won’t block the app from shutting down. For example, had issue where pytest would hang indefinitely after running tests waiting for this thead to cease.

Answered By: Adam Hughes

Alternative flexibility solution is Apscheduler.

pip install apscheduler
from apscheduler.schedulers.background import BlockingScheduler
def print_t():

sched = BlockingScheduler()
sched.add_job(print_t, 'interval', seconds =60) #will do the print_t work for every 60 seconds


Also, apscheduler provides so many schedulers as follow.

  • BlockingScheduler: use when the scheduler is the only thing running in your process

  • BackgroundScheduler: use when you’re not using any of the frameworks below, and want the scheduler to run in the background inside your application

  • AsyncIOScheduler: use if your application uses the asyncio module

  • GeventScheduler: use if your application uses gevent

  • TornadoScheduler: use if you’re building a Tornado application

  • TwistedScheduler: use if you’re building a Twisted application

  • QtScheduler: use if you’re building a Qt application

Answered By: Sivaram Rasathurai

Simply use

import time

while True:
    print("this will run after every 30 sec")
    #Your code here
Answered By: Swadeshi

I think it depends what you want to do and your question didn’t specify lots of details.

For me I want to do an expensive operation in one of my already multithreaded processes. So I have that leader process check the time and only her do the expensive op (checkpointing a deep learning model). To do this I increase the counter to make sure 5 then 10 then 15 seconds have passed to save every 5 seconds (or use modular arithmetic with math.floor):

def print_every_5_seconds_have_passed_exit_eventually():
    opts = argparse.Namespace(start=time.time())
    next_time_to_print = 0
    while True:
        current_time_passed = time.time() - opts.start
        if current_time_passed >= next_time_to_print:
            next_time_to_print += 5
            print(f'worked and {current_time_passed=}')
            print(f'{current_time_passed % 5=}')
            print(f'{math.floor(current_time_passed % 5) == 0}')
starting __main__ at __init__
worked and current_time_passed=0.0001709461212158203
current_time_passed % 5=0.0001709461212158203
worked and current_time_passed=5.0
current_time_passed % 5=0.0
worked and current_time_passed=10.0
current_time_passed % 5=0.0
worked and current_time_passed=15.0
current_time_passed % 5=0.0

To me the check of the if statement is what I need. Having threads, schedulers in my already complicated multiprocessing multi-gpu code is not a complexity I want to add if I can avoid it and it seems I can. Checking the worker id is easy to make sure only 1 process is doing this.

Note I used the True print statements to really make sure the modular arithemtic trick worked since checking for exact time is obviously not going to work! But to my pleasant surprised the floor did the trick.

Answered By: Charlie Parker

timed-count can do that to high precision (i.e. < 1 ms) as it’s synchronized to the system clock. It won’t drift over time and isn’t affected by the length of the code execution time (provided that’s less than the interval period of course).

A simple, blocking example:

from timed_count import timed_count

for count in timed_count(60):
    # Execute code here exactly every 60 seconds

You could easily make it non-blocking by running it in a thread:

from threading import Thread
from timed_count import timed_count

def periodic():
    for count in timed_count(60):
        # Execute code here exactly every 60 seconds

thread = Thread(target=periodic)
Answered By: 101
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