Trigger a Python function exactly on the minute

Question

I have a function that I want to trigger at every turn of the minute — at 00 seconds. It fires off a packet over the air to a dumb display that will be mounted on the wall.

I know I can brute force it with a while loop but that seems a bit harsh.

I have tried using sched but that ends up adding a second every minute.

What are my options?

Answer 1

You might try APScheduler, a cron-style scheduler module for Python.

From their examples:

from apscheduler.scheduler import Scheduler

# Start the scheduler
sched = Scheduler()
sched.start()

def job_function():
    print "Hello World"

sched.add_cron_job(job_function, second=0)

will run job_function every minute.

Answer 2

What if you measured how long it took your code to execute, and subtracted that from a sleep time of 60?

import time

while True:
    timeBegin = time.time()

    CODE(.....)

    timeEnd = time.time()
    timeElapsed = timeEnd - timeBegin
    time.sleep(60-timeElapsed)

Answer 3

The simplest solution would be to register a timeout with the operating system to expire when you want it to.

Now there are quite a few ways to do so with a blocking instruction and the best option depends on your implementation. Simplest way would be to use time.sleep():

import time

current_time = time.time()
time_to_sleep = 60 - (current_time % 60)
time.sleep(time_to_sleep)

This way you take the current time and calculate the amount of time you need to sleep (in seconds). Not millisecond accurate but close enough.

Answer 4

APScheduler is the correct approach. The syntax has changed since the original answer, however.

As of APScheduler 3.3.1:

def fn():
    print("Hello, world")

from apscheduler.schedulers.background import BackgroundScheduler

scheduler = BackgroundScheduler()
scheduler.start()
scheduler.add_job(fn, trigger='cron', second=0)

Answer 5

The syntax has been changed, so in APScheduler of version 3.6.3 (Released: Nov 5, 2019) use the following snippet:

from apscheduler.schedulers.blocking import BlockingScheduler
from apscheduler.triggers.cron import CronTrigger

def fn():
    print('Hello, world!')


sched = BlockingScheduler()

# Execute fn() at the start of each minute.
sched.add_job(fn, trigger=CronTrigger(second=00))
sched.start()

Answer 6

You can try Threading.Timer

See this Example

from threading import Timer

def job_function():
    Timer(60, job_function).start ()
    print("Running job_funtion")

It will print "Running job_function" every Minute

Edit: If we are critical about the time at which it should run

from threading import Timer
from time import time


def job_function():
    Timer(int(time()/60)*60+60 - time(), job_function).start ()
    print("Running job_funtion")

It will run exactly at 0th second of every minute.

Answer 7

The Python time module is usually what I use for events like this. It has a method called sleep(t), where t equals the time in seconds you want to delay.

Combined with a while loop, you can get what you're looking for:

import time

while condition:
    time.sleep(60)
    f(x)

Answer 8

May use this example:

   def do():
        print("do do bi do")

   while True:
        alert_minutes= [15,30,45,0]
        now=time.localtime(time.time())
        if now.tm_min in alert_minutes:
        do()
        time.sleep(60)

Answer 9

you could use a while loop and sleep to not eat up the processor too much