減少設定時間的變量

Question

我試圖用5減少一個變量，打印新變量，然後在X時間後再次增加變量並打印變量。它適用於基於文本的小型遊戲，您可以將一些人送走，而且他們不會回來一段時間。

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 

所有我得到的輸出是10,10,10直接，它不會延遲時間。如果含糊不清，這是我的第一個問題。

1. 我做了什麼錯誤導致此代碼立即返回變量？

2. 如果我想在腳本中做其他事情，比如發送剩餘的倖存者，那麼在腳本運行時還是需要等待5秒鐘？

感謝您的耐心:)

Answer 1

爲了使你的腳本等待，使用time.sleep(num_seconds)。 time.time()只是返回當前時間;它根本沒有做任何等待。

可能有一些代碼在睡覺，但有其他代碼在同一時間運行，做其他事情。要做到這一點，你必須使用threads，但需要一些時間才能習慣。也許this tutorial會有用。

編輯：哦也可以做到這一點沒有線程，但你必須仔細跟蹤你的變量。你搞砸了你的數學。說time.time（）是1000時，你的代碼就運行了。然後TimeReturned爲1005.假設用戶輸入Yes需要1秒。然後if TimeReturned > time.time()檢查if 1005 > 1001，這是真的。你真正想要檢查的是if time.time() > TimeReturned - 如果當前時間晚於TimeReturned。

此外，您的腳本不具有互動性，所以很難看到任何進展。嘗試運行此腳本：

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()) 

輸出示例：

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...你沒做什麼，以使腳本停頓和等待。

if TimeReturned > time.time(): 

這只是檢查條件是否爲真，並據此進行，它並沒有指示任何人等待。

您應該使用time.sleep(secs)函數來停止執行所需時間的代碼。

儘管如此，在睡眠期間，沒有任何執行。所以，如果你想要一些後臺進程繼續運行，並且只有你提到的那些人等待，你應該在自己的線程中運行腳本的那部分。

Answer 3

你可以用Timer做你想做的。它運行指定的秒數，然後執行您指定的功能。 A Timer只是使用第二個線程的簡單方法，這就是爲什麼它在threading模塊中。

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) 

最後一點只是說明它是如何工作的。儘管可能有更好的方法來管理變量，但我相信我會得到一些建議。