一个可能的答案是:

import time
t=time.time()

while True:
    if time.time()-t>10:
        #run your task here
        t=time.time()

一个可能的答案是:

import time
t=time.time()

while True:
    if time.time()-t>10:
        #run your task here
        t=time.time()

我以前也遇到过类似的问题。也许http://cronus.readthedocs.org会有帮助?

对于v0.2，下面的代码片段可以工作

import cronus.beat as beat

beat.set_rate(2) # run twice per second
while beat.true():
    # do some time consuming work here
    beat.sleep() # total loop duration would be 0.5 sec

这里是另一个不使用任何额外库的解决方案。

def delay_until(condition_fn, interval_in_sec, timeout_in_sec):
    """Delay using a boolean callable function.

    `condition_fn` is invoked every `interval_in_sec` until `timeout_in_sec`.
    It can break early if condition is met.

    Args:
        condition_fn     - a callable boolean function
        interval_in_sec  - wait time between calling `condition_fn`
        timeout_in_sec   - maximum time to run

    Returns: None
    """
    start = last_call = time.time()
    while time.time() - start < timeout_in_sec:
        if (time.time() - last_call) > interval_in_sec:
            if condition_fn() is True:
                break
            last_call = time.time()

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.

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)。