Executing periodic actions in Python [duplicate]

Question

This question already has an answer here:

• What is the best way to repeatedly execute a function every x seconds in Python? 15 answers

I am working on Windows. I want to execute a function foo() every 10 seconds.

How do I do this?

Answer 1

At the end of foo(), create a Timer which calls foo() itself after 10 seconds.
Because, Timer create a new thread to call foo().
You can do other stuff without being blocked.

import time, threading
def foo():
    print(time.ctime())
    threading.Timer(10, foo).start()

foo()

#output:
#Thu Dec 22 14:46:08 2011
#Thu Dec 22 14:46:18 2011
#Thu Dec 22 14:46:28 2011
#Thu Dec 22 14:46:38 2011

Answer 2

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()
        count = 0
        while True:
            count += 1
            yield max(t + count*period - 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 3

Here's a simple single threaded sleep based version that drifts, but tries to auto-correct when it detects drift.

NOTE: This will only work if the following 3 reasonable assumptions are met:

1. The time period is much larger than the execution time of the function being executed
2. The function being executed takes approximately the same amount of time on each call
3. The amount of drift between calls is less than a second

-

from datetime import timedelta
from datetime import datetime

def exec_every_n_seconds(n,f):
    first_called=datetime.now()
    f()
    num_calls=1
    drift=timedelta()
    time_period=timedelta(seconds=n)
    while 1:
        time.sleep(n-drift.microseconds/1000000.0)
        current_time = datetime.now()
        f()
        num_calls += 1
        difference = current_time - first_called
        drift = difference - time_period* num_calls
        print "drift=",drift

Answer 4

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

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

---

edit: 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()
raw_input('do other things...')

Answer 5

Simply sleeping for 10 seconds or using threading.Timer(10,foo) will result in start time drift. (You may not care about this, or it may be a significant source of problems depending on your exact situation.) There can be two causes for this - inaccuracies in the wake up time of your thread or execution time for your function.

You can see some results at the end of this post, but first an example of how to fix it. You need to track when your function should next be called as opposed to when it actually got called and account for the difference.

Here's a version that drifts slightly:

import datetime, threading

def foo():
    print datetime.datetime.now()
    threading.Timer(1, foo).start()

foo()

Its output looks like this:

2013-08-12 13:05:36.483580
2013-08-12 13:05:37.484931
2013-08-12 13:05:38.485505
2013-08-12 13:05:39.486945
2013-08-12 13:05:40.488386
2013-08-12 13:05:41.489819
2013-08-12 13:05:42.491202
2013-08-12 13:05:43.492486
2013-08-12 13:05:44.493865
2013-08-12 13:05:45.494987
2013-08-12 13:05:46.496479
2013-08-12 13:05:47.497824
2013-08-12 13:05:48.499286
2013-08-12 13:05:49.500232

You can see that the sub-second count is constantly increasing and thus, the start time is "drifting".

This is code that correctly accounts for drift:

import datetime, threading, time

next_call = time.time()

def foo():
  global next_call
  print datetime.datetime.now()
  next_call = next_call+1
  threading.Timer( next_call - time.time(), foo ).start()

foo()

Its output looks like this:

2013-08-12 13:21:45.292565
2013-08-12 13:21:47.293000
2013-08-12 13:21:48.293939
2013-08-12 13:21:49.293327
2013-08-12 13:21:50.293883
2013-08-12 13:21:51.293070
2013-08-12 13:21:52.293393

Here you can see that there is no longer any increase in the sub-second times.

If your events are occurring really frequently you may want to run the timer in a single thread, rather than starting a new thread for each event. While accounting for drift this would look like:

import datetime, threading, time

def foo():
    next_call = time.time()
    while True:
        print datetime.datetime.now()
        next_call = next_call+1;
        time.sleep(next_call - time.time())

timerThread = threading.Thread(target=foo)
timerThread.start()

However your application will not exit normally, you'll need to kill the timer thread. If you want to exit normally when your application is done, without manually killing the thread, you should use

timerThread = threading.Thread(target=foo)
timerThread.daemon = True
timerThread.start()

Answer 6

If you meant to run foo() inside a python script every 10 seconds, you can do something on these lines.

import time

def foo():
    print "Howdy"

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