Python 剪刀布归还问题

所以，我遇到的问题是，当我的计数器试图将数字相加时，它会不断重置 我试着想出一种方法把计数器部分变成函数，但我想不出来

winP1=0
winP2=0
tie=0

ans = input('Would you like to play ROCK, PAPER, SCISSORS?: ')

while ans == 'y':
    p1c = input('Player 1 enter either R,P, or S: ') #p1c stands for "player 1 choice"
    p2c = input('Player 2 enter either R,P, or S: ')
    ans = input('Would you like to play again: ')

def game(p1c,p2c):
    if p1c == p2c:                  #if its a tie we are going to add to the Tie variable
        return 0
    elif p1c == 'R' and p2c == 'P': #We will only set the Player 1 wins because
        return 1                    #player 2 wins can be set as else
    elif p1c == 'P' and p2c == 'S':
        return 1
    elif p1c == 'S' and p2c == 'R':
        return 1
    else:
        return -1

result = game(p1c,p2c)   
if result == -1:
    winP2 += 1
if result == 0:
    tie += 1
else:
    winP1 +=1

print('Player 1 won {} times. \nPlayer 2 won {} times. \nThere were {} ties.'.format(winP1,winP2,tie))

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欢迎使用StackOverflow和Python编程！我希望你在这里过得愉快

您的计数器不断重置，因为对于系统而言，您只玩一个游戏：

while ans == 'y':
    p1c = input('Player 1 enter either R,P, or S: ') #p1c stands for "player 1 choice"
    p2c = input('Player 2 enter either R,P, or S: ')
    ans = input('Would you like to play again: ')

我将在您的代码中移动一些东西，以获得您想要的结果，并进行最小程度的编辑，希望这将显示您需要做什么。你在正确的轨道上

我们将使用一个名为

state

的字典来跟踪游戏状态并传递它

state = {
   'winP1':0,  
   'winP2':0,
   'tie':0
}

def game(p1c,p2c):
    if p1c == p2c:                  #if its a tie we are going to add to the Tie variable
        return 0
    elif p1c == 'R' and p2c == 'P': #We will only set the Player 1 wins because
        return 1                    #player 2 wins can be set as else
    elif p1c == 'P' and p2c == 'S':
        return 1
    elif p1c == 'S' and p2c == 'R':
        return 1
    else:
        return -1

def process_game(p1c, p2c, state): # Move this into a function
    result = game(p1c,p2c)   
    if result == -1:
        state['winP2'] += 1
    if result == 0:
        state['tie'] += 1
    else:
        state['winP1'] +=1

while ans == 'y':
    p1c = input('Player 1 enter either R,P, or S: ') #p1c stands for "player 1 choice"
    p2c = input('Player 2 enter either R,P, or S: ')
    process_game(p1c, p2c, state)
    ans = input('Would you like to play again: ')

print('Player 1 won {} times. \nPlayer 2 won {} times. \nThere were {} ties.'.format(state['winP1'],state['winP2'],state['tie']))

---

似乎在您的程序中，while循环将永远持续。@alKid No，它不会根据用户输入在while循环的最后一行中进行“ans”更改。@ChristianTernus我尝试使用的是，但现在我遇到了分配前引用的局部变量“tie”问题。我尝试使用的是，但现在遇到了分配前引用的局部变量“tie”问题。对！我加入了一个

global

声明，这通常不是一个好主意，但可以完成工作。实际上，你要做的是将游戏状态移动到它自己的对象中，并传递该对象。我不想问，但我还没有讨论全局的想法，我宁愿不使用它。你能给我看一下比赛状态吗？是的，我记得以前见过这样的场面！我现在开始工作了，非常感谢你，没问题。快乐编码！