Python“石头，纸，剪刀”验证

Question

我一直在做一个石头剪刀游戏，它的工作就像一个梦。然而，当我试图在#中添加一些验证时，我的游戏就不起作用了，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)

Answer 1

再次使用or操作符。

player != r or p or s

应该是

player not in (r, p, s)

或者类似的。

解释：

A or B计算为A，如果A被认为是真的(真)。如果A被认为是虚假的(如False、0、0.0、[]、'')，则A or B评估为B。

player != r or p or s和(player != r) or p or s是一样的。现在，(player != r) or p or s的计算结果为True，如果是player != r，则为p。由于True和p都是“真”，所以这两行是等价的：

if player != r or p or s:
if True:

Answer 2

下面是使用一些更高级python成语的更短版本的代码：

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]))

Answer 3

改变这个

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

到这个

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