岩石剪刀 - Python 3 - 初学者

Question

我想要模拟一个石头剪刀游戏,这是我到目前为止.它不是让我在scoregame函数中输入字母.我怎样才能解决这个问题？

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

你需要测试字符串 ; 您现在正在测试变量名称:

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

但是你可以通过测试两个玩家是否相同来简化两个玩家选择相同选项的情况:

if player1 == player2:
    scoregame = "It's a tie, nobody wins."

接下来,我将使用一个映射,一个字典,来编写什么节拍:

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2个测试中进行测试.全部放在一起:

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)