Python sleep without interfering with script?

问题:

Hey I need to know how to sleep in Python without interfering with the current script. I've tried using time.sleep() but it makes the entire script sleep.

Like for example

import time
def func1():
    func2()
    print("Do stuff here")
def func2():
    time.sleep(10)
    print("Do more stuff here")


func1()

I want it to immediately print Do stuff here, then wait 10 seconds and print Do more stuff here.

回答1:

Interpreting your description literally, you need to put the print statement before the call to func2().

However, I'm guessing what you really want is for func2() to a background task that allows func1() to return immediately and not wait for func2() to complete it's execution. In order to do this, you need to create a thread to run func2().

import time
import threading

def func1():
    t = threading.Thread(target=func2)
    t.start()
    print("Do stuff here")
def func2():
    time.sleep(10)
    print("Do more stuff here")

func1()
print("func1 has returned")

回答2:

You could use threading.Timer:

from __future__ import print_function
from threading import Timer

def func1():
    func2()
    print("Do stuff here")
def func2():
    Timer(10, print, ["Do more stuff here"]).start()

func1()

But as @unholysampler already pointed out it might be better to just write:

import time

def func1():
    print("Do stuff here")
    func2()

def func2():
    time.sleep(10)
    print("Do more stuff here")

func1()

回答3:

If you're running your script on command line, try using the -u parameter. It runs the script in unbuffered mode and did the trick for me.

For example:

python -u my_script.py