推迟python中的函数

Question

在JavaScript中,我习惯于能够调用稍后要执行的函数,就像这样

function foo() {
    alert('bar');
}

setTimeout(foo, 1000);

这不会阻止其他代码的执行.

我不知道如何在Python中实现类似的东西.我可以睡觉

import time
def foo():
    print('bar')

time.sleep(1)
foo()

但这会阻止其他代码的执行.(实际上在我的情况下阻塞Python本身并不是问题,但我无法对该方法进行单元测试.)

我知道线程是为不同步执行而设计的,但我想知道是否更容易,类似setTimeout或setInterval存在.

Answer 1

要在延迟后执行函数或使用事件循环(无线程)在给定的秒数内重复函数,您可以:

Tkinter的

#!/usr/bin/env python
from Tkinter import Tk

def foo():
    print("timer went off!")

def countdown(n, bps, root):
    if n == 0:
        root.destroy() # exit mainloop
    else:
        print(n)
        root.after(1000 / bps, countdown, n - 1, bps, root)  # repeat the call

root = Tk()
root.withdraw() # don't show the GUI window
root.after(4000, foo) # call foo() in 4 seconds
root.after(0, countdown, 10, 2, root)  # show that we are alive
root.mainloop()
print("done")

产量

10
9
8
7
6
5
4
3
timer went off!
2
1
done

GTK

#!/usr/bin/env python
from gi.repository import GObject, Gtk

def foo():
    print("timer went off!")

def countdown(n): # note: a closure could have been used here instead
    if n[0] == 0:
        Gtk.main_quit() # exit mainloop
    else:
        print(n[0])
        n[0] -= 1
        return True # repeat the call

GObject.timeout_add(4000, foo) # call foo() in 4 seconds
GObject.timeout_add(500, countdown, [10])
Gtk.main()
print("done")

产量

10
9
8
7
6
5
4
timer went off!
3
2
1
done

扭曲

#!/usr/bin/env python
from twisted.internet import reactor
from twisted.internet.task import LoopingCall

def foo():
    print("timer went off!")

def countdown(n):
    if n[0] == 0:
        reactor.stop() # exit mainloop
    else:
        print(n[0])
        n[0] -= 1

reactor.callLater(4, foo) # call foo() in 4 seconds
LoopingCall(countdown, [10]).start(.5)  # repeat the call in .5 seconds
reactor.run()
print("done")

产量

10
9
8
7
6
5
4
3
timer went off!
2
1
done

ASYNCIO

Python 3.4 为异步IO 引入了新的临时API - asyncio模块:

#!/usr/bin/env python3.4
import asyncio

def foo():
    print("timer went off!")

def countdown(n):
    if n[0] == 0:
        loop.stop() # end loop.run_forever()
    else:
        print(n[0])
        n[0] -= 1

def frange(start=0, stop=None, step=1):
    while stop is None or start < stop:
        yield start
        start += step #NOTE: loss of precision over time

def call_every(loop, seconds, func, *args, now=True):
    def repeat(now=True, times=frange(loop.time() + seconds, None, seconds)):
        if now:
            func(*args)
        loop.call_at(next(times), repeat)
    repeat(now=now)

loop = asyncio.get_event_loop()
loop.call_later(4, foo) # call foo() in 4 seconds
call_every(loop, 0.5, countdown, [10]) # repeat the call every .5 seconds
loop.run_forever()
loop.close()
print("done")

产量

10
9
8
7
6
5
4
3
timer went off!
2
1
done

注意:这些方法之间的界面和行为略有不同.

Answer 2

你想要一个模块中的Timer对象threading.

from threading import Timer
from time import sleep

def foo():
    print "timer went off!"
t = Timer(4, foo)
t.start()
for i in range(11):
    print i
    sleep(.5)

如果你想重复一遍,这里有一个简单的解决方案:不使用Timer,只需使用Thread但传递一个有点像这样的函数:

def call_delay(delay, repetitions, func, *args, **kwargs):             
    for i in range(repetitions):    
        sleep(delay)
        func(*args, *kwargs)

这不会做无限循环,因为这可能导致一个不会死的线程和其他不愉快的行为,如果做得不对.更复杂的方法可能会使用Event基于方法的方法,就像这个方法一样.