嵌套列表中的嵌套循環

Question

我正在嘗試在python中構建一個簡單的井字游戲，以檢查是否有獲勝。我正在使用嵌套循環在嵌套列表中搜索匹配項。 由於某種原因，我的代碼將只搜索第一個嵌套列表，而不搜索我期望的其余列表。

board = [ 'O', 'X', ' ', 'O', ' ', 'X', 'O', 'X', 'X' ]
wins = [[0,1,2],[3,4,5],[6,7,8],[0,3,6],[1,4,7],[2,5,8],[0,4,8],[2,4,6]]

def checkWin(player):
    win = False
    for test in wins:
        print (test)
        count = 0
        for squares in test:
            if board[squares] == player:
                count = count + 1
            if count == 3:
                win = True
        return win

if __name__ == '__main__':
    print ("\nChecking board for X win ...\n")
    if checkWin("X"): 
        print ("Game over, X wins!")

    print ("\nChecking board for O win ...\n")
    if checkWin("O"):
        print ("Game over, O wins")

基於提供O獲勝的董事會，這是我得到的輸出：

Checking board for X win ...

[0, 1, 2]

Checking board for O win ...

[0, 1, 2]

有誰知道為什么會這樣嗎？

Answer 1

無論這三個正方形是否匹配，您都將從第一個嵌套列表測試中返回。 相反，只有在win為true時才返回：

def checkWin(player):
    win = False
    for test in wins:
        count = 0
        for squares in test:
            if board[squares] == player:
                count = count + 1
            if count == 3:
                win = True
        if win:
            return True
    return False

如果win為false，則上面的內容繼續到下一個嵌套列表，以進行下一個測試。

更好的是，在count設置為3時返回即可，因為您知道在該階段找到了一個匹配項：

def checkWin(player):
    for test in wins:
        count = 0
        for squares in test:
            if board[squares] == player:
                count = count + 1
            if count == 3:
                return True
    return False

您可以使用all()函數來代替計數：

def checkWin(player):
    for test in wins:
        if all(board[square] == player for square in test):
            return True
    return False

生成器表達式中的測試之一失敗后， all()盡早返回False 。

最終版本添加了any()可以在一行中完成測試：

def checkWin(player):
    return any(all(board[square] == player for square in test)
               for test in wins)

演示：

>>> board = [ 'O', 'X', ' ', 'O', ' ', 'X', 'O', 'X', 'X' ]
>>> wins = [[0,1,2],[3,4,5],[6,7,8],[0,3,6],[1,4,7],[2,5,8],[0,4,8],[2,4,6]]
>>> def checkWin(player):
...     return any(all(board[square] == player for square in test)
...                for test in wins)
... 
>>> checkWin('X')
False
>>> checkWin('O')
True