Python“；石头、布、剪刀”；验证_Python

Python“；石头、布、剪刀”；验证

python

Python“；石头、布、剪刀”；验证,python,Python,我一直在做一个石头剪刀游戏，它就像一个梦。然而，当我尝试在中添加一些验证时（用#显示），我的游戏无法运行。我不知道为什么会这样我的代码如下： from random import randint from sys import exit computer = randint(1,3) r = "r" p = "p" s = "s" print ("The computer has chosen. Your turn") play

我一直在做一个石头剪刀游戏，它就像一个梦。然而，当我尝试在中添加一些验证时（用

显示），我的游戏无法运行。我不知道为什么会这样

我的代码如下：

    from random import randint
    from sys import exit
    computer = randint(1,3)

    r = "r"
    p = "p"
    s = "s"
    print ("The computer has chosen. Your turn")
    player = input ("r is Rock, p is Paper, and s is Scissors. Put your letter in HERE-----> ")

    #from here

    if (player != r or p or s): 
            player = input ("That wasn't r, p, or s. Please try again. r is Rock, p is Paper, and s is Scissors. Put your letter in HERE-----> ")
            if (player != r or p or s) :
                    print ("Can you srsly not understand that " + player + " is not r, p, or s? I give up")
                    exit()


    #to here



    if (computer == 1):
            AI = ("rock")

    if (computer == 2):
            AI = ("paper")

    if (computer == 3):
            AI = ("scissors")

    if (player == r and computer == 1):
            print ("lol draw")
            exit()

    if (player == p and computer == 2):
            print ("lol draw")
            exit()

    if (player == s and computer == 3):
            print ("lol draw")
            exit()

    if (player == r and computer == 3):
            print ("You WIN!!!!!! AI chose " + AI)

    if (player == p and computer == 1):
            print ("You WIN!!!!!! AI chose " + AI)

    if (player == s and computer == 2):
            print ("You WIN!!!!!! AI chose " + AI)

    if (player == s and computer == 1):
            print ("You LOSE!!!!!! AI chose " + AI)

    if (player == r and computer == 2):
            print ("You LOSE!!!!!! AI chose " + AI)

    if (player == p and computer == 3):
            print ("You LOSE!!!!!! AI chose " + AI)

改变这个

if (player != r or p or s):

对此

if player != r and player != p and player != s:

改变这个

if (player != r or p or s):

对此

if player != r and player != p and player != s:

请重新启动

或

运算符

player != r or p or s

应该是

player not in (r, p, s)

或类似的

说明：

如果认为

为真（真），则

A或B

计算为

。如果

被认为是错误的（例如

False

，

0.0

，

[]

，

），则

A或B

的计算结果为

player！=r或p或s

与

（玩家！=r）或p或s

相同。现在

（player！=r）或p或s的计算结果为True
如果player！=r
和top
否则。由于True
和p
都是“True”，因此这两行是等效的：
if player != r or p or s:
if True:

请重新启动或
运算符
player != r or p or s

应该是
player not in (r, p, s)

或类似的
说明：
如果认为A
为真（真），则A或B
计算为A
。如果A
被认为是错误的（例如False
，0
，0.0
，[]
，'
），则A或B
的计算结果为B

player！=r或p或s
与（玩家！=r）或p或s
相同。现在（player！=r）或p或s的计算结果为True
如果player！=r
和top
否则。由于True
和p
都是“True”，因此这两行是等效的：
if player != r or p or s:
if True:

您拥有的等价于（不是真正的Python代码）：
if player！=r或if p或if s
这不是您想要的：因为p
和s
已经定义为true，所以每次都会触发if语句
你想要如果玩家！=r和玩家！=p和玩家！=s
（即，这三件事都不是真的）
如果播放器不在[r，p，s]
中，如果您愿意，它也可以同等工作——我认为它更像Python。您拥有的是等效的（不是真正的Python代码）：
if player！=r或if p或if s
这不是您想要的：因为p
和s
已经定义为true，所以每次都会触发if语句
你想要如果玩家！=r和玩家！=p和玩家！=s
（即，这三件事都不是真的）
如果播放器不在[r，p，s]
中，如果您愿意，它也可以同等工作——我认为这有点像python。
这不是语法问题的答案，但是我觉得按照前面介绍的类似结构，“石头，布，剪刀”验证可以大大简化
# read player input and return "r", "p" or "s"
def getPlayerChoice():
    # return value read from input, as per your code, with the fixes.

# read computer choice and return "r", "p" or "s"
def getComputerChoice():
    # return "r", "p", or "s" based on the random number

# get the "English" name of a value
def getName(rps):
    # given "r", "p", or "s", return the name like "rock", etc.

# now these two variables will have the value "r", "p" or "s"
# no numbers, no "rock", just some consistent values
player = getPlayerChoice()
computer = getComputerChoice()

# By using consistent values we are able to make the tie check very easy
# and make the win checks easy to read.

# chose same value - "r", "p" or "s"
if player == computer:
    print("Tie")

# all ways player could have won
elif (   (player == "r" and computer == "s")
      or (player == "p" and computer == "r")
      or (player == "s" and computer == "p")):
    print("You win! " + getName(player) + " beats " + getName(computer))

# otherwise the computer won
else:
    print("You lose! " + getName(computer) + " beats " + getName(player))

YMMV，以及上述结构中的任何缺陷都是“免费的”。
这不是语法问题的答案，但是我觉得，遵循类似的结构，可以大大清理“石头、布、剪刀”验证
# read player input and return "r", "p" or "s"
def getPlayerChoice():
    # return value read from input, as per your code, with the fixes.

# read computer choice and return "r", "p" or "s"
def getComputerChoice():
    # return "r", "p", or "s" based on the random number

# get the "English" name of a value
def getName(rps):
    # given "r", "p", or "s", return the name like "rock", etc.

# now these two variables will have the value "r", "p" or "s"
# no numbers, no "rock", just some consistent values
player = getPlayerChoice()
computer = getComputerChoice()

# By using consistent values we are able to make the tie check very easy
# and make the win checks easy to read.

# chose same value - "r", "p" or "s"
if player == computer:
    print("Tie")

# all ways player could have won
elif (   (player == "r" and computer == "s")
      or (player == "p" and computer == "r")
      or (player == "s" and computer == "p")):
    print("You win! " + getName(player) + " beats " + getName(computer))

# otherwise the computer won
else:
    print("You lose! " + getName(computer) + " beats " + getName(player))

以上结构中的任何错误都是“免费的”。
以下是使用一些更高级python习惯用法的代码的简短版本：
from random import randint
from sys import exit
computer = randint(0,2)
choices = 'rps'

print ("The computer has chosen. Your turn")
player = raw_input("r is Rock, p is Paper, and s is Scissors. Put your letter in HERE-----> ")

if (player not in choices): 
    player = raw_input("That wasn't r, p, or s. Please try again. r is Rock, p is Paper, and s is Scissors. Put your letter in HERE-----> ")
    if (player not in choices):
        print ("Can you srsly not understand that '%s' is not r, p, or s? I give up" % player)
        exit()

if (player == choices[computer]):
        print ("lol draw, AI also chose %s" % choices[computer])
        exit()

flip = choices.index(player) > computer
result = ("WIN", "LOSE")[(flip + choices.index(player) - computer) % 2] 
print ("You %s!!!!!! AI chose %s" % (result, choices[computer]))

下面是一个简短的代码版本，使用了一些更高级的python习惯用法：
from random import randint
from sys import exit
computer = randint(0,2)
choices = 'rps'

print ("The computer has chosen. Your turn")
player = raw_input("r is Rock, p is Paper, and s is Scissors. Put your letter in HERE-----> ")

if (player not in choices): 
    player = raw_input("That wasn't r, p, or s. Please try again. r is Rock, p is Paper, and s is Scissors. Put your letter in HERE-----> ")
    if (player not in choices):
        print ("Can you srsly not understand that '%s' is not r, p, or s? I give up" % player)
        exit()

if (player == choices[computer]):
        print ("lol draw, AI also chose %s" % choices[computer])
        exit()

flip = choices.index(player) > computer
result = ("WIN", "LOSE")[(flip + choices.index(player) - computer) % 2] 
print ("You %s!!!!!! AI chose %s" % (result, choices[computer]))

你说“不行”是什么意思？现在把“石头、布、剪刀、蜥蜴、斯波克”编码为。。。看来我的问题已经解决了。TYRemember记得在正确的问题旁边打勾。我想你的意思是回答“它不起作用”是什么意思？现在代码是“石头、布、剪刀、蜥蜴、斯波克”。。。看来我的问题已经解决了。TYRemember请记住在正确的问题旁边打勾。我想你的意思是回答解释为什么，因为我猜你是新手：你所拥有的等同于（不是真正的Python代码）：这个答案不正确x！=真或x！=False
将始终产生True
…应该是和而不是或来解释原因，因为我猜你是新手：你所拥有的等同于（不是真正的Python代码）：这个答案不正确x！=真或x！=False
将始终产生True
…应该是和而不是或乐于帮助，请阅读我的解释并尝试消化它。并阅读python逻辑运算符的文档。很高兴能提供帮助，请阅读我的解释并尝试理解它。并阅读python逻辑运算符的文档。不是根据我的repl<代码>玩家='r'p='p's='s'如果（玩家！=r或p或s）：如果（玩家！=r）打印'yes'：打印'no'
给出yes.OK，player！=r或p或s
为真
iffplayer！=r
，否则为p
。所以基本上它总是真的
。不用担心——实际上我的代码中有一个完全不同的错误，我已经修复了。介意取消你的否决票吗？他们让我很伤心。我一离开CD就会。不是根据我的回复<代码>玩家='r'p='p's='s'如果（玩家！=r或p或s）：如果（玩家！=r）打印'yes'：打印'no'
给出yes.OK，player！=r或p或s
为真
iffplayer！=r
，否则为p
。所以基本上它总是真的
。不用担心——实际上我的代码中有一个完全不同的错误，我已经修复了。介意取消你的否决票吗？他们让我很难过。我一离开CD就去。对不起，我必须投反对票，因为这是无效的python。@SethMMorton，没关系，绝对不是。我确实纠正了&&/| |/|的严重误用；和条件表达式。愚蠢的催促了我。对不起，我必须投反对票，因为这是无效的。@SethMMorton好的，绝对不是。我确实纠正了&&/| |/|的严重误用；和c