减少设定时间的变量

Question

我试图用5减少一个变量，打印新变量，然后在X时间后再次增加变量并打印变量。它适用于基于文本的小型游戏，您可以将一些人送走，而且他们不会回来一段时间。

import time 

TimeReturned = time.time() + 5 
Survivors = 10 

Test = raw_input("Run Script?") 
if Test == "Yes": 
    print Survivors 
    if TimeReturned > time.time(): 
     Survivors -= 5 
     print Survivors 
    Survivors += 5 
    print Survivors 

所有我得到的输出是10,10,10直接，它不会延迟时间。如果含糊不清，这是我的第一个问题。

1. 我做了什么错误导致此代码立即返回变量？

2. 如果我想在脚本中做其他事情，比如发送剩余的幸存者，那么在脚本运行时还是需要等待5秒钟？

感谢您的耐心:)

Answer 1

为了使你的脚本等待，使用time.sleep(num_seconds)。 time.time()只是返回当前时间;它根本没有做任何等待。

可能有一些代码在睡觉，但有其他代码在同一时间运行，做其他事情。要做到这一点，你必须使用threads，但需要一些时间才能习惯。也许this tutorial会有用。

编辑：哦也可以做到这一点没有线程，但你必须仔细跟踪你的变量。你搞砸了你的数学。说time.time（）是1000时，你的代码就运行了。然后TimeReturned为1005.假设用户输入Yes需要1秒。然后if TimeReturned > time.time()检查if 1005 > 1001，这是真的。你真正想要检查的是if time.time() > TimeReturned - 如果当前时间晚于TimeReturned。

此外，您的脚本不具有互动性，所以很难看到任何进展。尝试运行此脚本：

import time 

survivors = 15 
survivor_return_seconds = 10.0 
time_survivors_left = None 

while True: 
    action = raw_input("Type 'x' to make survivors leave, ENTER to see how many are left: ") 

    #check if survivors returned 
    if time_survivors_left is not None: 
     if time.time() >= time_survivors_left + survivor_return_seconds: 
      survivors += 5 
      time_survivors_left = None 
      print "Survivors came back!" 

    if action == 'x': 
     if time_survivors_left is not None: 
      print "Survivors already left! Wait a bit!" 
     else: 
      survivors -= 5 
      time_survivors_left = time.time() 

    print "There are %s survivors left." % (survivors,) 
    if time_survivors_left is not None: 
     print "5 survivors will return in %.2fs" % (
      time_survivors_left + survivor_return_seconds - time.time()) 

输出示例：

Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 15 survivors left. 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 15 survivors left. 
Type 'x' to make survivors leave, ENTER to see how many are left: x 
There are 10 survivors left. 
5 survivors will return in 9.99s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 9.05s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 7.66s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 6.45s 
Type 'x' to make survivors leave, ENTER to see how many are left: x 
Survivors already left! Wait a bit! 
There are 10 survivors left. 
5 survivors will return in 5.73s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 4.15s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 2.90s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 1.72s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 10 survivors left. 
5 survivors will return in 0.48s 
Type 'x' to make survivors leave, ENTER to see how many are left: 
Survivors came back! 
There are 15 survivors left. 
Type 'x' to make survivors leave, ENTER to see how many are left: 
There are 15 survivors left. 
Type 'x' to make survivors leave, ENTER to see how many are left: 

Answer 2

好...你没做什么，以使脚本停顿和等待。

if TimeReturned > time.time(): 

这只是检查条件是否为真，并据此进行，它并没有指示任何人等待。

您应该使用time.sleep(secs)函数来停止执行所需时间的代码。

尽管如此，在睡眠期间，没有任何执行。所以，如果你想要一些后台进程继续运行，并且只有你提到的那些人等待，你应该在自己的线程中运行脚本的那部分。

Answer 3

你可以用Timer做你想做的。它运行指定的秒数，然后执行您指定的功能。 A Timer只是使用第二个线程的简单方法，这就是为什么它在threading模块中。

from threading import Timer 
import time 

Survivors = 10 

def increase_survivors(): 
    global Survivors 
    Survivors += 5 

# start execution 
print Survivors 
Survivors -= 5 

t = Timer(5, increase_survivors) 
t.start() 
for i in range(10): 
    print Survivors 
    time.sleep(1) 

最后一点只是说明它是如何工作的。尽管可能有更好的方法来管理变量，但我相信我会得到一些建议。