Equivalent of setInterval in python

问题

I have recently posted a question about how to postpone execution of a function in Python (kind of equivalent to Javascript setTimeout) and it turns out to be a simple task using threading.Timer (well, simple as long as the function does not share state with other code, but that would create problems in any event-driven environment).

Now I am trying to do better and emulate setInterval. For those who are not familiar with Javascript, setInterval allows to repeat a call to a function every x seconds, without blocking the execution of other code. I have created this example decorator:

import time, threading

def setInterval(interval, times = -1):
    # This will be the actual decorator,
    # with fixed interval and times parameter
    def outer_wrap(function):
        # This will be the function to be
        # called
        def wrap(*args, **kwargs):
            # This is another function to be executed
            # in a different thread to simulate setInterval
            def inner_wrap():
                i = 0
                while i != times:
                    time.sleep(interval)
                    function(*args, **kwargs)
                    i += 1
            threading.Timer(0, inner_wrap).start()
        return wrap
    return outer_wrap

to be used as follows

@setInterval(1, 3)
def foo(a):
    print(a)

foo('bar')
# Will print 'bar' 3 times with 1 second delays

and it seems to me it is working fine. My problem is that

it seems overly complicated, and I fear I may have missed a simpler/better mechanism
the decorator can be called without the second parameter, in which case it will go on forever. When I say foreover, I mean forever - even calling sys.exit() from the main thread will not stop it, nor will hitting Ctrl+c. The only way to stop it is to kill python process from the outside. I would like to be able to send a signal from the main thread that would stop the callback. But I am a beginner with threads - how can I communicate between them?

EDIT In case anyone wonders, this is the final version of the decorator, thanks to the help of jd

import threading

def setInterval(interval, times = -1):
    # This will be the actual decorator,
    # with fixed interval and times parameter
    def outer_wrap(function):
        # This will be the function to be
        # called
        def wrap(*args, **kwargs):
            stop = threading.Event()

            # This is another function to be executed
            # in a different thread to simulate setInterval
            def inner_wrap():
                i = 0
                while i != times and not stop.isSet():
                    stop.wait(interval)
                    function(*args, **kwargs)
                    i += 1

            t = threading.Timer(0, inner_wrap)
            t.daemon = True
            t.start()
            return stop
        return wrap
    return outer_wrap

It can be used with a fixed amount of repetitions as above

@setInterval(1, 3)
def foo(a):
    print(a)

foo('bar')
# Will print 'bar' 3 times with 1 second delays

or can be left to run until it receives a stop signal

import time

@setInterval(1)
def foo(a):
    print(a)

stopper = foo('bar')

time.sleep(5)
stopper.set()
# It will stop here, after printing 'bar' 5 times.

回答1:

Your solution looks fine to me.

There are several ways to communicate with threads. To order a thread to stop, you can use threading.Event(), which has a wait() method that you can use instead of time.sleep().

stop_event = threading.Event()
...
stop_event.wait(1.)
if stop_event.isSet():
    return
...

For your thread to exit when the program is terminated, set its daemon attribute to True before calling start(). This applies to Timer() objects as well because they subclass threading.Thread. See http://docs.python.org/library/threading.html#threading.Thread.daemon

回答2:

Maybe a bit simpler is to use recursive calls to Timer:

from threading import Timer
import atexit

class Repeat(object):

    count = 0
    @staticmethod
    def repeat(rep, delay, func):
        "repeat func rep times with a delay given in seconds"

        if Repeat.count < rep:
            # call func, you might want to add args here
            func()
            Repeat.count += 1
            # setup a timer which calls repeat recursively
            # again, if you need args for func, you have to add them here
            timer = Timer(delay, Repeat.repeat, (rep, delay, func))
            # register timer.cancel to stop the timer when you exit the interpreter
            atexit.register(timer.cancel)
            timer.start()

def foo():
    print "bar"

Repeat.repeat(3,2,foo)

atexit allows to signal stopping with CTRL-C.

回答3:

Maybe these are the easiest setInterval equivalent in python:

 import threading

 def set_interval(func, sec):
    def func_wrapper():
        set_interval(func, sec) 
        func()  
    t = threading.Timer(sec, func_wrapper)
    t.start()
    return t

回答4:

this class Interval

class ali:
    def __init__(self):
        self.sure = True;

    def aliv(self,func,san):
        print "ali naber";
        self.setInterVal(func, san);

    def setInterVal(self,func, san):

        # istenilen saniye veya dakika aralığında program calışır. 
        def func_Calistir():

            func(func,san);  #calışıcak fonksiyon.
        self.t = threading.Timer(san, func_Calistir)
        self.t.start()
        return self.t


a = ali();
a.setInterVal(a.aliv,5);

来源：https://stackoverflow.com/questions/5179467/equivalent-of-setinterval-in-python

标签

python

multithreading

setinterval