我最终使用了时间表模块。API很好。

import schedule
import time

def job():
    print("I'm working...")

schedule.every(10).minutes.do(job)
schedule.every().hour.do(job)
schedule.every().day.at("10:30").do(job)
schedule.every(5).to(10).minutes.do(job)
schedule.every().monday.do(job)
schedule.every().wednesday.at("13:15").do(job)
schedule.every().minute.at(":17").do(job)

while True:
    schedule.run_pending()
    time.sleep(1)

像这样将你的时间循环锁定到系统时钟上:

import time
starttime = time.time()
while True:
    print("tick")
    time.sleep(60.0 - ((time.time() - starttime) % 60.0))

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.

下面是MestreLion对代码的更新，它可以避免随着时间的推移而漂移。

这里的RepeatedTimer类按照OP的请求每隔“间隔”秒调用给定函数;调度并不取决于函数执行的时间。我喜欢这个解决方案，因为它没有外部库依赖关系;这是纯python。

import threading 
import time

class RepeatedTimer(object):
  def __init__(self, interval, function, *args, **kwargs):
    self._timer = None
    self.interval = interval
    self.function = function
    self.args = args
    self.kwargs = kwargs
    self.is_running = False
    self.next_call = time.time()
    self.start()

  def _run(self):
    self.is_running = False
    self.start()
    self.function(*self.args, **self.kwargs)

  def start(self):
    if not self.is_running:
      self.next_call += self.interval
      self._timer = threading.Timer(self.next_call - time.time(), self._run)
      self._timer.start()
      self.is_running = True

  def stop(self):
    self._timer.cancel()
    self.is_running = False

示例用法(摘自MestreLion的回答):

from time import sleep

def hello(name):
    print "Hello %s!" % name

print "starting..."
rt = RepeatedTimer(1, hello, "World") # it auto-starts, no need of rt.start()
try:
    sleep(5) # your long-running job goes here...
finally:
    rt.stop() # better in a try/finally block to make sure the program ends!

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)。

以下是MestreLion代码的改编版本。 除了原来的函数，这段代码:

1)添加用于在特定时间触发计时器的first_interval(调用者需要计算first_interval并传递进来)

2)在原代码中解决一个竞态条件。在原始代码中，如果控制线程未能取消正在运行的计时器(“停止计时器，并取消计时器动作的执行。这只会在计时器仍处于等待阶段时起作用。”引用自https://docs.python.org/2/library/threading.html)，计时器将无休止地运行。

class RepeatedTimer(object):
def __init__(self, first_interval, interval, func, *args, **kwargs):
    self.timer      = None
    self.first_interval = first_interval
    self.interval   = interval
    self.func   = func
    self.args       = args
    self.kwargs     = kwargs
    self.running = False
    self.is_started = False

def first_start(self):
    try:
        # no race-condition here because only control thread will call this method
        # if already started will not start again
        if not self.is_started:
            self.is_started = True
            self.timer = Timer(self.first_interval, self.run)
            self.running = True
            self.timer.start()
    except Exception as e:
        log_print(syslog.LOG_ERR, "timer first_start failed %s %s"%(e.message, traceback.format_exc()))
        raise

def run(self):
    # if not stopped start again
    if self.running:
        self.timer = Timer(self.interval, self.run)
        self.timer.start()
    self.func(*self.args, **self.kwargs)

def stop(self):
    # cancel current timer in case failed it's still OK
    # if already stopped doesn't matter to stop again
    if self.timer:
        self.timer.cancel()
    self.running = False