石头布剪刀Python3贝金

Question

我想模拟一个石头，布，剪刀的游戏，这就是我目前所拥有的。它不允许我在scoregame函数中输入字母。我该怎么解决这个问题？在

def scoregame(player1, player2):
    if player1 == R and player2 == R:
        scoregame = "It's a tie, nobody wins."
    if player1 == S and player2 == S:
        scoregame == "It's a tie, nobody wins."
    if player1 == P and player2 == P:
        scoregame = "It's a tie, nobody wins."
    if player1 == R and player2 == S:
        scoregame = "Player 1 wins."
    if player1 == S and player2 == P:
        scoregame = "Player 1 wins."
    if player1 == P and player2 == R:
        scoregame = "Player 1 wins."
    if player1 == R and player2 == P:
        scoregame == "Player 2 wins."
    if player1 == S and player2 == R:
        scoregame == "Player 2 wins."
    if player1 == P and player2 == S:
        scoregame = "Player 2 wins."

print(scoregame)

Answer 1

您使用的是没有引号的字母，因此它需要一个名为p的变量，但您需要的是一个字符串“p”，因此请将字母用引号括起来：

if player1 == "P" and player2 == "S":

Answer 2

您需要针对字符串进行测试；现在针对变量名进行测试：

if player1 == 'R' and player2 == 'R':

但是你可以通过测试是否相等来简化两个玩家选择相同选项的情况：

^{pr2}$

接下来，我会用一个地图，一本字典来编纂什么胜过什么：

beats = {'R': 'S', 'S': 'P', 'P': 'R'}

if beats[player1] == player2:
    scoregame = "Player 1 wins."
else:
    scoregame = "Player 2 wins."

现在你的游戏可以在两个测试中测试。总而言之：

def scoregame(player1, player2):
    beats = {'R': 'S', 'S': 'P', 'P': 'R'}
    if player1 == player2:
        scoregame = "It's a tie, nobody wins."
    elif beats[player1] == player2:
        scoregame = "Player 1 wins."
    else:
        scoregame = "Player 2 wins."
    print(scoregame)