Question

我是python的新手（并为此编码）并正在进行摇滚，剪纸和剪刀练习。在第二场比赛之后我的代码不会问我是否想要打另一场比赛，而是自动调用游戏功能。有谁知道我哪里出错了？

import sys

print("Welcome to rock, paper, scissors!")

player_a = input("Player A input one of the following; rock, paper or scissors: ")
player_b = input("Player B input one of the following; rock, paper or scissors: ")

def game(p1, p2):
    if p1 == p2:
        print("It is a draw!")
    elif p1 == "rock" and p2 == "scissors":
        print("Player A wins!")
    elif p1 == "rock" and p2 == "paper":
        print("Player B wins!")
    elif p1 == "paper" and p2 == "rock":
        print("Player A wins!")
    elif p1 == "paper" and p2 == "scissors":
        print("Player B wins!")
    elif p1 == "scissors" and p2 == "rock":
        print("Player A wins!")
    elif p1 == "scissors" and p2 == "paper":
        print("Player B wins!")

game(player_a, player_b)

playAgain = input("Would you like to play again? (Insert Y or N): ")

while playAgain == "Y":
    player_a = input("Player A input one of the following; rock, paper or scissors: ")
    player_b = input("Player B input one of the following; rock, paper or scissors: ")
    game(player_a, player_b)
else:
    print("Thanks for playing!")
    sys.exit()

Answer 1

你只需要再次播放＆＃34;再玩一次？＆＃34;你的循环中的问题。

编码的一个重要原则是DRY：Don't Repeat Yourself。通过一点点重组，我们可以摆脱代码中的一些重复。

在下面的代码中，我们假装在程序开始时用户已经回答了＆＃34; Y＆＃34;再次播放？＆＃34;问题

print("Welcome to rock, paper, scissors!")

def game(p1, p2):
    if p1 == p2:
        print("It is a draw!")
    elif p1 == "rock" and p2 == "scissors":
        print("Player A wins!")
    elif p1 == "rock" and p2 == "paper":
        print("Player B wins!")
    elif p1 == "paper" and p2 == "rock":
        print("Player A wins!")
    elif p1 == "paper" and p2 == "scissors":
        print("Player B wins!")
    elif p1 == "scissors" and p2 == "rock":
        print("Player A wins!")
    elif p1 == "scissors" and p2 == "paper":
        print("Player B wins!")

playAgain = "Y"
while playAgain == "Y":
    player_a = input("Player A input one of the following; rock, paper or scissors: ")
    player_b = input("Player B input one of the following; rock, paper or scissors: ")
    game(player_a, player_b)
    playAgain = input("Would you like to play again? (Insert Y or N): ")
else:
    print("Thanks for playing!")

请注意，无需拨打sys.exit()，我们可以让程序自然终止。

我们还可以大大压缩game函数。如果它不是平局，并且A没有赢，那么B 必须获胜。所以我们不需要做所有那些B测试。我们可以使用or运算符将A测试合并到一个测试中。

def game(p1, p2):
    if p1 == p2:
        print("It is a draw!")
    elif ((p1 == "rock" and p2 == "scissors") 
        or (p1 == "paper" and p2 == "rock")
        or (p1 == "scissors" and p2 == "paper")):
        print("Player A wins!")
    else:
        print("Player B wins!")

进行测试的更高级（和更多Pythonic）方法是使用元组：

def game(p1, p2):
    if p1 == p2:
        print("It is a draw!")
    elif (p1, p2) in (("rock", "scissors"), ("paper", "rock"), ("scissors", "paper")):
        print("Player A wins!")
    else:
        print("Player B wins!")

Answer 2

您的代码有几点：

1）我一个人几乎不会拼写＆＃34;剪刀＆＃34;正确。如果主要功能只是期待＆＃39;，＆＃39; p＆＃39;或者＆＃39;，那么为什么不通过第一个字母，这可能会使游戏更具可玩性game()的输入？类似的东西：

player_a = input("Player A input one of the following; rock, paper or scissors: ")
payer_a = player_a.strip().lower()[0]

，同样适用于player_b。

2）我总是将石头剪刀视为一个循环：

    R
  /  \
 S -- P

每个可能性都被顺时针方向的项目所击败，player_2是否在player_1的顺时针方向可以通过模运算来确定：

def game(p1,p2):
    if p1 == p2:
        print("It is a draw!")
    else:
        i = "rps".index(p1)
        if "rps"[(i+1)%3] == p2:
            print("Player B wins!")
        else:
            print("Player A wins!")

这样可以避免案件泛滥。此外，如果你想将游戏扩展到摇滚纸剪刀Spock-lizard＆＃34;这种东西更容易扩展。大爆炸（表演，而不是宇宙的起源）成名。关于指数的模块化算法是要走的路（虽然＆＃34;摇滚剪刀-Spock-lizard＆＃34;需要的不仅仅是简单的方向）。

Answer 3

或

while True:
    if input("Would you like to play again? (Insert Y or N): ") == "Y":
        player_a()
        player_b()
    else:
        print('thanks for playing')
        break

摇滚，纸张，剪刀游戏不按预期循环

3 个答案: