python - 嵌套列表中的嵌套循环

Question

我正在尝试用 python 构建一个简单的 Tic-Tac-Toe 游戏，为了检查是否获胜，我使用嵌套循环在嵌套列表中搜索匹配项。由于某种原因，我的代码只会搜索第一个嵌套列表，而不是像我预期的那样搜索其余列表。

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")

根据董事会预计 O 会获胜，这是我得到的输出:

Checking board for X win ...

[0, 1, 2]

Checking board for O win ...

[0, 1, 2]

有人知道为什么会发生这种情况吗？

Answer 1

无论这三个方 block 是否匹配，您都会从第一个嵌套列表测试中返回。相反，仅当 win 为 true 时才返回:

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 为 false，则上面的内容将继续到下一个嵌套列表以进行下一个测试。

更好的是，当 count 设置为 3 时返回，因为您知道在该阶段已找到匹配项:

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() function来代替计数。 :

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

all() 一旦生成器表达式中的某个测试失败，就会提前返回 False。

最终版本添加了any()一行完成测试:

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