Question

我可以使用此代码检查矩阵中的行是否为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!"

如何检查我的矩阵是否具有= x的对角线或列，而不使用Numpy（是否有数学方法如上所示？）

示例：（X =无关紧要的事情）

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{其他人}）

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）

How the matrix should have its "winning conditions"

Answer 1

那么你可以将你的获胜条件的枚举翻译成一个元组的元组，而不是在3x3板块世界中工作。

下面的内容（拿你的示例板）并产生一个平局应该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()

Answer 2

这个怎么样？适用于任意形状的q

def check_col_diag (q, x):
    """
    Returns:
    0 if there was a column,
    1 if there was a row,
    2 if there was a diagonal on 1st direction
    3 if there was a diagonal on 2nd direction
    """

    # Get a mask to store the positions
    # on each row of q where q == x
    mask = q
    for row_ix in range(len(q[0])):
        for elem_ix in range(len(q[1])):
            if q[row_ix][elem_ix] == x:
                mask[row_ix][elem_ix] = 1
            else:
                mask[row_ix][elem_ix] = 0

    # Check rows
    c = [1]*len(q[0])
    for row in mask:
        # element-wise list multiplication
        c = [a*b for a,b in zip(c,row)]
    # Return 0 if there was a column
    if any(c):
        return 0

    # Check columns 
    c = [1]*len(q[1])
    # Iterate through rows of transposed list
    _q = list(map(list, zip(*mask)))
    for row in _q:
        c = [a*b for a,b in zip(c,row)]
    # Return 1 if there was a row
    if any(c):
        return 1

    # Check diagonal 1
    c = 1
    for row_ix in range(len(q[0])):
        c *= mask[row_ix][row_ix]
    # Return 2 if there was a 1st diagonal
    if c == 1:
        return 2

    # Check diagonal 2
    c = 1
    for row_ix in range(len(_q[0])):
        c *= mask[row_ix][row_ix]
    # Return 3 if there was a 2nd diagonal
    if c == 1:
        return 3


q = [[1,2,1],[1,2,1],[2,1,2]]
v = check_col_diag (q, 1)

检查矩阵中的列或对角线= x（没有Numpy）

2 个答案: