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How can I make a time delay in Python?


Question

I would like to know how to put a time delay in a Python script.

2018/06/11
1
2790
6/11/2018 12:05:47 AM


You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds
2019/05/21

How can I make a time delay in Python?

In a single thread I suggest the sleep function:

>>> from time import sleep

>>> sleep(4)

This function actually suspends the processing of the thread in which it is called by the operating system, allowing other threads and processes to execute while it sleeps.

Use it for that purpose, or simply to delay a function from executing. For example:

>>> def party_time():
...     print('hooray!')
...
>>> sleep(3); party_time()
hooray!

"hooray!" is printed 3 seconds after I hit Enter.

Example using sleep with multiple threads and processes

Again, sleep suspends your thread - it uses next to zero processing power.

To demonstrate, create a script like this (I first attempted this in an interactive Python 3.5 shell, but sub-processes can't find the party_later function for some reason):

from concurrent.futures import ThreadPoolExecutor, ProcessPoolExecutor, as_completed
from time import sleep, time

def party_later(kind='', n=''):
    sleep(3)
    return kind + n + ' party time!: ' + __name__

def main():
    with ProcessPoolExecutor() as proc_executor:
        with ThreadPoolExecutor() as thread_executor:
            start_time = time()
            proc_future1 = proc_executor.submit(party_later, kind='proc', n='1')
            proc_future2 = proc_executor.submit(party_later, kind='proc', n='2')
            thread_future1 = thread_executor.submit(party_later, kind='thread', n='1')
            thread_future2 = thread_executor.submit(party_later, kind='thread', n='2')
            for f in as_completed([
              proc_future1, proc_future2, thread_future1, thread_future2,]):
                print(f.result())
            end_time = time()
    print('total time to execute four 3-sec functions:', end_time - start_time)

if __name__ == '__main__':
    main()

Example output from this script:

thread1 party time!: __main__
thread2 party time!: __main__
proc1 party time!: __mp_main__
proc2 party time!: __mp_main__
total time to execute four 3-sec functions: 3.4519670009613037

Multithreading

You can trigger a function to be called at a later time in a separate thread with the Timer threading object:

>>> from threading import Timer
>>> t = Timer(3, party_time, args=None, kwargs=None)
>>> t.start()
>>>
>>> hooray!

>>>

The blank line illustrates that the function printed to my standard output, and I had to hit Enter to ensure I was on a prompt.

The upside of this method is that while the Timer thread was waiting, I was able to do other things, in this case, hitting Enter one time - before the function executed (see the first empty prompt).

There isn't a respective object in the multiprocessing library. You can create one, but it probably doesn't exist for a reason. A sub-thread makes a lot more sense for a simple timer than a whole new subprocess.

2019/11/27

Delays can be also implemented by using the following methods.

The first method:

import time
time.sleep(5) # Delay for 5 seconds.

The second method to delay would be using the implicit wait method:

 driver.implicitly_wait(5)

The third method is more useful when you have to wait until a particular action is completed or until an element is found:

self.wait.until(EC.presence_of_element_located((By.ID, 'UserName'))
2019/12/10

A bit of fun with a sleepy generator.

The question is about time delay. It can be fixed time, but in some cases we might need a delay measured since last time. Here is one possible solution:

Delay measured since last time (waking up regularly)

The situation can be, we want to do something as regularly as possible and we do not want to bother with all the last_time, next_time stuff all around our code.

Buzzer generator

The following code (sleepy.py) defines a buzzergen generator:

import time
from itertools import count

def buzzergen(period):
    nexttime = time.time() + period
    for i in count():
        now = time.time()
        tosleep = nexttime - now
        if tosleep > 0:
            time.sleep(tosleep)
            nexttime += period
        else:
            nexttime = now + period
        yield i, nexttime

Invoking regular buzzergen

from sleepy import buzzergen
import time
buzzer = buzzergen(3) # Planning to wake up each 3 seconds
print time.time()
buzzer.next()
print time.time()
time.sleep(2)
buzzer.next()
print time.time()
time.sleep(5) # Sleeping a bit longer than usually
buzzer.next()
print time.time()
buzzer.next()
print time.time()

And running it we see:

1400102636.46
1400102639.46
1400102642.46
1400102647.47
1400102650.47

We can also use it directly in a loop:

import random
for ring in buzzergen(3):
    print "now", time.time()
    print "ring", ring
    time.sleep(random.choice([0, 2, 4, 6]))

And running it we might see:

now 1400102751.46
ring (0, 1400102754.461676)
now 1400102754.46
ring (1, 1400102757.461676)
now 1400102757.46
ring (2, 1400102760.461676)
now 1400102760.46
ring (3, 1400102763.461676)
now 1400102766.47
ring (4, 1400102769.47115)
now 1400102769.47
ring (5, 1400102772.47115)
now 1400102772.47
ring (6, 1400102775.47115)
now 1400102775.47
ring (7, 1400102778.47115)

As we see, this buzzer is not too rigid and allow us to catch up with regular sleepy intervals even if we oversleep and get out of regular schedule.

2018/06/15

The Tkinter library in the Python standard library is an interactive tool which you can import. Basically, you can create buttons and boxes and popups and stuff that appear as windows which you manipulate with code.

If you use Tkinter, do not use time.sleep(), because it will muck up your program. This happened to me. Instead, use root.after() and replace the values for however many seconds, with a milliseconds. For example, time.sleep(1) is equivalent to root.after(1000) in Tkinter.

Otherwise, time.sleep(), which many answers have pointed out, which is the way to go.

2019/11/27

Source: https://stackoverflow.com/questions/510348
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