Executing periodic actions in Python [duplicate]

Answer 1

Surprised to not find a solution using a generator for timing. I just designed this one for my own purposes.

This solution: single threaded, no object instantiation each period, uses generator for times, rock solid on timing down to precision of the time module (unlike several of the solutions I've tried from stack exchange).

Note: for Python 2.x, replace next(g) below with g.next().

import time

def do_every(period,f,*args):
    def g_tick():
        t = time.time()
        while True:
            t += period
            yield max(t - time.time(),0)
    g = g_tick()
    while True:
        time.sleep(next(g))
        f(*args)

def hello(s):
    print('hello {} ({:.4f})'.format(s,time.time()))
    time.sleep(.3)

do_every(1,hello,'foo')

Results in, for example:

hello foo (1421705487.5811)
hello foo (1421705488.5811)
hello foo (1421705489.5809)
hello foo (1421705490.5830)
hello foo (1421705491.5803)
hello foo (1421705492.5808)
hello foo (1421705493.5811)
hello foo (1421705494.5811)
hello foo (1421705495.5810)
hello foo (1421705496.5811)
hello foo (1421705497.5810)
hello foo (1421705498.5810)
hello foo (1421705499.5809)
hello foo (1421705500.5811)
hello foo (1421705501.5811)
hello foo (1421705502.5811)
hello foo (1421705503.5810)

Note that this example includes a simulation of the cpu doing something else for .3 seconds each period. If you changed it to be random each time it wouldn't matter. The max in the yield line serves to protect sleep from negative numbers in case the function being called takes longer than the period specified. In that case it would execute immediately and make up the lost time in the timing of the next execution.

Answer 2

You can execute your task in a different thread. threading.Timer will let you execute a given callback once after some time has elapsed, if you want to execute your task, for example, as long as the callback returns True (this is actually what glib.timeout_add provides, but you might not have it installed in windows) or until you cancel it, you can use this code:

import logging, threading, functools
import time

logging.basicConfig(level=logging.NOTSET,
                    format='%(threadName)s %(message)s')

class PeriodicTimer(object):
    def __init__(self, interval, callback):
        self.interval = interval

        @functools.wraps(callback)
        def wrapper(*args, **kwargs):
            result = callback(*args, **kwargs)
            if result:
                self.thread = threading.Timer(self.interval,
                                              self.callback)
                self.thread.start()

        self.callback = wrapper

    def start(self):
        self.thread = threading.Timer(self.interval, self.callback)
        self.thread.start()

    def cancel(self):
        self.thread.cancel()


def foo():
    logging.info('Doing some work...')
    return True

timer = PeriodicTimer(1, foo)
timer.start()

for i in range(2):
    time.sleep(2)
    logging.info('Doing some other work...')

timer.cancel()

Example output:

Thread-1 Doing some work...
Thread-2 Doing some work...
MainThread Doing some other work...
Thread-3 Doing some work...
Thread-4 Doing some work...
MainThread Doing some other work...

Note: The callback isn't executed every interval execution. Interval is the time the thread waits between the callback finished the last time and the next time is called.

Answer 3

This will insert a 10 second sleep in between every call to foo(), which is approximately what you asked for should the call complete quickly.

import time

while True:
    foo()
    time.sleep(10)

To do other things while your foo() is being called in a background thread

import time
import sys
import threading

def foo():
    sys.stdout.write('({}) foo\n'.format(time.ctime()))

def foo_target():
    while True:
        foo()
        time.sleep(10)

t = threading.Thread(target=foo_target)
t.daemon = True
t.start()
print('doing other things...')