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()

计时计数可以做到高精度(即< 1毫秒)，因为它与系统时钟同步。它不会随着时间的推移而漂移，也不受代码执行时间长度的影响(当然，前提是它小于间隔时间)。

一个简单的阻塞的例子:

from timed_count import timed_count

for count in timed_count(60):
    # Execute code here exactly every 60 seconds
    ...

你可以通过在线程中运行它来让它变得不阻塞:

from threading import Thread
from timed_count import timed_count

def periodic():
    for count in timed_count(60):
        # Execute code here exactly every 60 seconds
        ...

thread = Thread(target=periodic)
thread.start()

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.

我认为更简单的方法是:

import time

def executeSomething():
    #code here
    time.sleep(60)

while True:
    executeSomething()

这样，你的代码被执行，然后等待60秒，然后再次执行，等待，执行，等等…… 没有必要把事情复杂化:D

你可能会考虑Twisted，它是一个实现了Reactor Pattern的Python网络库。

from twisted.internet import task, reactor

timeout = 60.0 # Sixty seconds

def doWork():
    #do work here
    pass

l = task.LoopingCall(doWork)
l.start(timeout) # call every sixty seconds

reactor.run()

虽然“While True: sleep(60)”可能会工作，Twisted可能已经实现了许多你最终需要的功能(如bobince指出的守护进程化、日志记录或异常处理)，并且可能是一个更健壮的解决方案

简单地使用

import time

while True:
    print("this will run after every 30 sec")
    #Your code here
    time.sleep(30)