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()

如果漂移不是一个问题

import threading, time

def print_every_n_seconds(n=2):
    while True:
        print(time.ctime())
        time.sleep(n)
    
thread = threading.Thread(target=print_every_n_seconds, daemon=True)
thread.start()

异步输出。

#Tue Oct 16 17:29:40 2018
#Tue Oct 16 17:29:42 2018
#Tue Oct 16 17:29:44 2018

如果正在运行的任务需要相当多的时间，那么间隔就变成2秒+任务时间，所以如果您需要精确的调度，那么这并不适合您。

注意daemon=True标志意味着这个线程不会阻止应用程序关闭。例如，在运行测试等待此头停止后，pytest将无限期挂起的问题。

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

我认为这取决于你想做什么，你的问题没有详细说明。

对我来说，我想在一个已经多线程的进程中做一个昂贵的操作。所以我让leader流程检查时间，只有她做昂贵的操作(检查点深度学习模型)。为了做到这一点，我增加了计数器，以确保5秒、10秒、15秒过去，每5秒保存一次(或使用math.floor的模块化算术):

def print_every_5_seconds_have_passed_exit_eventually():
    """
    https://stackoverflow.com/questions/3393612/run-certain-code-every-n-seconds
    https://stackoverflow.com/questions/474528/what-is-the-best-way-to-repeatedly-execute-a-function-every-x-seconds
    :return:
    """
    opts = argparse.Namespace(start=time.time())
    next_time_to_print = 0
    while True:
        current_time_passed = time.time() - opts.start
        if current_time_passed >= next_time_to_print:
            next_time_to_print += 5
            print(f'worked and {current_time_passed=}')
            print(f'{current_time_passed % 5=}')
            print(f'{math.floor(current_time_passed % 5) == 0}')

starting __main__ at __init__
worked and current_time_passed=0.0001709461212158203
current_time_passed % 5=0.0001709461212158203
True
worked and current_time_passed=5.0
current_time_passed % 5=0.0
True
worked and current_time_passed=10.0
current_time_passed % 5=0.0
True
worked and current_time_passed=15.0
current_time_passed % 5=0.0
True

对我来说，检查if语句是我所需要的。在我已经复杂的多处理器多gpu代码中拥有线程，调度器并不是我想要添加的复杂性，如果我可以避免它，似乎我可以。检查worker id很容易确保只有一个进程在做这件事。

注意，我使用True print语句来确保模块化的算术技巧有效，因为检查确切的时间显然是行不通的!但令我惊喜的是，地板竟然起了作用。

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)