我想在Python中每60秒重复执行一个函数(就像Objective C中的NSTimer或JS中的setTimeout)。这段代码将作为守护进程运行,有效地类似于使用cron每分钟调用python脚本,但不需要用户设置。
在这个关于用Python实现的cron的问题中,解决方案似乎只有效地使()休眠x秒。我不需要这么高级的功能,所以也许这样的东西可以工作
while True:
# Code executed here
time.sleep(60)
这段代码是否存在任何可预见的问题?
如果您的程序还没有事件循环,请使用sched模块,它实现了一个通用的事件调度器。
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,))
my_scheduler.run()
如果您已经在使用事件循环库,如asyncio、trio、tkinter、PyQt5、gobject、kivy等,则只需使用现有事件循环库的方法来调度任务。
如果您的程序还没有事件循环,请使用sched模块,它实现了一个通用的事件调度器。
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,))
my_scheduler.run()
如果您已经在使用事件循环库,如asyncio、trio、tkinter、PyQt5、gobject、kivy等,则只需使用现有事件循环库的方法来调度任务。
我使用Tkinter after()方法,它不会“窃取游戏”(就像之前提出的sched模块),即它允许其他东西并行运行:
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()
do_something2()
tk.mainloop()
Do_something1()和do_something2()可以以任意的间隔速度并行运行。在这里,第2个将以两倍的速度执行。还要注意,我使用了一个简单的计数器作为终止任一函数的条件。你可以使用任何你喜欢的条件,或者不使用,如果你想让一个函数运行到程序终止(例如一个时钟)。
import time, traceback
def every(delay, task):
next_time = time.time() + delay
while True:
time.sleep(max(0, next_time - time.time()))
try:
task()
except Exception:
traceback.print_exc()
# 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)
如果你想在不阻塞剩余代码的情况下这样做,你可以使用这个让它在自己的线程中运行:
import threading
threading.Thread(target=lambda: every(5, foo)).start()
该解决方案结合了其他解决方案中很少结合的几个特性:
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.