石头，纸，剪刀-蟒蛇-密码。帮助简化

Question

我是Python新手，我只编写了几个程序。下面是我最近为一个rock-paper-scissors游戏编写的代码。我已经测试过了，效果很好。有什么办法可以简化吗？谢谢!

import random

wins=0   
losses=0    
ties=0    
rounds=0

r=1 #rock    
p=2 #paper    
s=3 #scissors

y = "The computer has made its choice, how about you?"

while rounds <= 10:    
 print y    
 x = input('(1)rock, (2)paper, or (3)scissors? :')
 choice = x    
 cpu_choice= random.randint(1, 3)

if (choice, cpu_choice) == (1, 2):    
  rounds += 1    
  losses += 1    
  print 'computer chose paper, you lose'
elif (choice, cpu_choice) == (3, 2):    
  print 'you win'    
  rounds += 1    
  wins += 1    
elif (choice, cpu_choice) == (2, 2):    
  print 'TIE!'    
  rounds += 1    
  ties += 1    
elif (choice, cpu_choice) == (1, 3):    
  print 'you win'    
  rounds += 1    
  wins += 1    
elif (choice, cpu_choice) == (3, 3):   
  print 'TIE!'    
  rounds += 1    
  ties += 1    
elif (choice, cpu_choice) == (2, 3):    
  print 'computer chose scissors, you lose'    
  rounds += 1    
  losses += 1    
elif (choice, cpu_choice) == (1, 1):    
  print 'TIE'    
  rounds += 1    
  ties += 1    
elif (choice, cpu_choice) == (3, 1):    
  print 'computer chose rock, you lose'    
  rounds += 1    
  losses += 1    
elif (choice, cpu_choice) == (2, 1):    
  print 'you win'    
  rounds += 1    
  wins += 1    
else:    
  print 'Please choose 1, 2, or 3'

print 'Game Over'

if wins>losses:
  print 'YOU WON'    
  print 'wins:' , wins   
  print 'losses' , losses    
  print 'ties' , ties    
else:    
 print 'you lose'    
 print 'wins:' , wins    
 print 'losses:' , losses    
 print 'ties:' , ties

Answer 1

虽然堆栈溢出并不是真正意义上的学习平台，但以下是一些建议：

• 阅读ZEN (在您的python控制台中键入import this )。
• 在你的特殊情况下，很多条件通常都是个坏主意。

至少，所有的绑定条件都可以组合在一起：

 if choice == cpu_choice:
    # TIE

加上一些语法：

names = ['rock', 'paper', 'scissors']
print("Computer chooses {}, you loose".format(names[cpu_choice]))

基本上，只有三个条件：

wins, losses = 0, 0

for round in range(10):

    # Your choice and CPU choice

    cpu_wins = (cpu_choice > choice or (choice == 3 and cpu_choice == 1))
    tie = (cpu_choice == choice)

    if cpu_wins:
        # You loose
        print("Computer chooses {}, you loose".format(names[cpu_choice]))
        losses += 1
    if not cpu_wins and tie:
        # tie
    else:
        # you win

而且，您甚至不使用上面定义的变量p、r和s .

Answer 2

一些建议：

1. 除了出现错误的数据输入时，所有条件情况下都包含圆型变量增加，因此可以将循环的+= 1线取上，而在其他情况下则只减少一次循环变量。
2. 如果案件做同样的工作，例如“打成平手！”已经发生了，最好把这类案件分组。“‘TIE！”案例可以按一个条件分组，选择== cpu_choice，从而包含3个elif子句。在其他游戏案例中考虑同样的问题。
3. 使用更好的代码格式，例如佩普-8标准所建议的。

Answer 3

您可以使用模块化算法来确定玩家是否获胜：

player_result = ["tie", "win", "lose"]
player_choice = input('(1)rock, (2)paper, or (3)scissors? :')
cpu_choice= random.randint(1, 3)
result = player_result[(player_choice - cpu_choice) % 3]

print "You " + result
if result == "win":
    wins += 1
elif result == "lose":
    loses += 1