Python中的石头、布、剪刀游戏的问题

我目前正在用Python编写一个小石头、布、剪刀游戏，但我似乎遇到了一个问题。虽然代码有点粗糙，但游戏仍然可以运行，但我尝试让程序通知玩家他们犯了错误，当我测试出来时，它会随机通知玩家他们犯了错误，而他们没有。这是我的问题代码块，它不是整个游戏

def game(self):

    print "This is rock, paper, scissors!"

    rps = ('rock', 'paper', 'scissors')

    comp1 = raw_input("Rock, paper or scissors?\n> ")
    comp2 = random.choice(rps)

    if comp1 == 'rock' and comp2 == 'scissors' or comp1 == 'scissors' and comp2 == 'paper' or comp1 == 'paper' and comp2 == 'rock':
        print "You won! The computer chose %s" % comp2
        return "game"
    elif comp1 == 'rock' and comp2 == 'rock':
        print "In a stalemate, there are no winners, only losers :)\nThe computer also chose %s" % comp2
        return "game"
    elif comp1 == 'scissors' and comp2 == 'scissors':
        print "In a stalemate, there are no winners, only losers :)\nThe computer also chose %s" % comp2
        return "game"
    elif comp1 == 'paper' and comp2 == 'paper':
        print "In a stalemate, there are no winners, only losers :)\nThe computer also chose %s" % comp2
        return "game"
    elif comp1 != 'rock' or 'scissors' or 'paper':
        print "Try choosing between rock, paper and scissors next time. It might help.
        return "game"
    else:
        print "The computer %s. You have failed. Problem?" % comp2
        return "game"

改变

到

您之前所做的工作相当于：

elif (comp1 != 'rock') or 'scissors' or 'paper':

那么，为什么总是满足这些条件呢

仔细观察“剪刀”或“布”部分

1在Python中，非空字符串被视为True，空字符串被视为False。请看一下此交互式会话：

>>> bool('')
False
>>> bool('a')
True

2同样在Python中，没有比较的if语句，例如if var1:正在检查表达式是否为真。所以

if var1:

与

if var1 == True:

如果您将这两个想法结合在一起：

if 'rock':
    # Always executed
if '':
    # Never executed

回到原始的if语句：

elif comp1 != 'rock' or 'scissors' or 'paper':

“剪刀”和“纸”都将始终返回True，因此将始终对包含的语句进行评估

那么in操作符是什么呢

elif comp1 not in rps:中的in运算符将查看comp1的内容是否是元组rps中的一项，该项等于“岩石”、“布”、“剪刀”。前面的not将否定它，检查comp1的内容是否是元组rps中的项。因此，仅当comp1中存储的用户输入无效时，才会执行包含的语句。

应该是这样的

comp1 not in ['rock', 'scissors', 'paper']

“剪刀”和“布”的评估结果总是正确的，或者是正确的，或者是正确的

comp1 != 'rock' or 'scissors' or 'paper'

---

另外，使用comp1==comp2，它要简单得多。

这里的问题在于您的逻辑elif comp1！='“石头”或“剪刀”或“布”：。字符串“剪刀”和“纸”将被计算为布尔值，这是真的，因为它们不为null

---

你想要的是elif comp1！='摇滚乐和comp1！='剪刀和comp1！='纸张“：或者，由于您已经在rps元组中有了它，您可以在rps中执行elif comp1:

我认为这是一个更干净的版本，尽管不是最好的实现。我还添加了一个选项来继续播放，并对其进行了更改，使用户输入不区分大小写

def game():
    import random
    import string
    print "This is rock, paper, scissors!"

    rps = ('rock', 'paper', 'scissors')

    comp1 = raw_input("Rock, paper or scissors? ")
    comp1 = comp1.lower()
    comp2 = random.choice(rps)

    if comp1 == 'rock' and comp2 == 'scissors' or comp1 == 'scissors' and comp2 == 'paper' or comp1 == 'paper' and comp2 == 'rock':
        print "You won! The computer chose %s" % comp2
    elif comp1 == comp2:
        print "In a stalemate, there are no winners, only losers :)\nThe computer also chose %s" % comp2
    elif comp1 not in rps:
        print "Try choosing between rock, paper and scissors next time. It might help."
    else:
        print "The computer chose %s. You have failed. Problem?" % comp2
    new_game = raw_input('Would you like to play again? ')
    new_game = new_game.lower()
    if new_game == 'yes':
        game()

---

在我的手机上，但尝试将comp1转换为小写。你也可以通过comp1==comp2来简化逻辑，因为僵局你没有复制真正的代码，因为你复制的是无效的字符串文字。为什么你总是返回字符串游戏，这看起来很奇怪。。。可能与你的问题无关。使用更简单的世界，它会更简单P

comp1 != 'rock' or 'scissors' or 'paper'

def game():
    import random
    import string
    print "This is rock, paper, scissors!"

    rps = ('rock', 'paper', 'scissors')

    comp1 = raw_input("Rock, paper or scissors? ")
    comp1 = comp1.lower()
    comp2 = random.choice(rps)

    if comp1 == 'rock' and comp2 == 'scissors' or comp1 == 'scissors' and comp2 == 'paper' or comp1 == 'paper' and comp2 == 'rock':
        print "You won! The computer chose %s" % comp2
    elif comp1 == comp2:
        print "In a stalemate, there are no winners, only losers :)\nThe computer also chose %s" % comp2
    elif comp1 not in rps:
        print "Try choosing between rock, paper and scissors next time. It might help."
    else:
        print "The computer chose %s. You have failed. Problem?" % comp2
    new_game = raw_input('Would you like to play again? ')
    new_game = new_game.lower()
    if new_game == 'yes':
        game()