检查矩阵中的列或对角线是否 = x（不使用 Numpy）

我可以使用此代码来检查矩阵中的一行是否 = x：

q = [[1,2,1],[1,2,1],[2,1,2]]
answer = [sum(row) for row in q]
for i in range(0, len(q)):
    if answer[i] == 6:
        print "Player 2 won!"
    if answer[i] == 3:
        print "Player 1 won!"
if answer[i] != 6 and 3:
    print "It's a tie!"

如何在不使用 Numpy 的情况下检查我的矩阵是否具有 = x 的对角线或列（有如上所示的数学方法吗？）

示例：（X = 无关紧要的东西）

q = [[1,X,X],[1,X,X],[1,X,X]]应打印True

q = [[1,X,X],[X,1,X],[X,X,1]]应打印True（对角线）

q = [[X,X,1],[X,1,X],[1,X,X]]应打印True(对角线{其他})

q = [[1,X,X],[X,1,X],[X,1,X]]应打印False

q = [[X,1,X],[X,1,X],[X,1,X]]应打印True（水平的）

矩阵应该如何具备它的“制胜条件”

好吧，您可以将获胜条件的枚举转换为一个元组，一个对的元组……在 3x3 棋盘世界中不需要太多工作。

像下面这样的事情（拿你的样本板）并导致平局应该可以让你开始进一步学习Python：

#! /usr/bin/env python
"""Check in snaive 3x3 game board world for diagonal,
column, or row all ocupied by one player."""
from __future__ import print_function

players = (1, 2)  # Code for the players
board = [[1, 2, 1],  # Board interpreted as 3 lists rows
         [1, 2, 1],
         [2, 1, 2]]
winning_configs = (  # outer-inner-index pairs that win:
    ((0, 0), (1, 1), (2, 2)),  # TL to BR diagonal
    ((0, 2), (1, 1), (2, 0)),  # TR to BL diagonal
    ((0, 0), (1, 0), (2, 0)),  # L column
    ((0, 1), (1, 1), (2, 1)),  # M column
    ((0, 2), (1, 2), (2, 2)),  # R column
    ((0, 0), (0, 1), (0, 2)),  # L row
    ((1, 0), (1, 1), (1, 2)),  # M row
    ((2, 0), (2, 1), (2, 2)),  # R row
)


def and_the_winner_is(players, board, winning_configs):
    """First one matching rules is returned as winner,
    otherwise None to indicate a tie."""
    for player in players:
        for cfg in winning_configs:
            if all([board[i][j] == player for i, j in cfg]):
                return player
    else:
        return None


def main():
    """Determine the result from board."""
    winner = and_the_winner_is(players, board, winning_configs)

    if winner in players:
        print('Winner is Player({})'.format(winner))
    else:
        print('A tie')


if __name__ == '__main__':
    main()