如何在生成器表达式 Python 中添加限制

Question

我在 Python 中有一个函数，如果网格矩阵中的所有行或列都相同，则返回 true。 但是，我希望我的函数在其中 4 个匹配时停止迭代并返回 True，而不管网格的大小。 如何修改下面的生成器表达式以实现相同的目的？

def check_won(grids, user, n):
    return any(all(cell == user for cell in grid) for grid in grids)

为了进一步说明，我正在分享我的示例输出：

Input the grid size: 5
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
Input a slot player 1 from 1 to 5: 1
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 0, 0, 0, 0]
Input a slot player 2 from 1 to 5: 2
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 1 from 1 to 5: 1
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 2 from 1 to 5: 2
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 1 from 1 to 5: 1
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 2 from 1 to 5: 2
Current board:
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 1 from 1 to 5: 1
Current board:
[0, 0, 0, 0, 0]
[1, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 2 from 1 to 5: 2
Current board:
[0, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
Input a slot player 1 from 1 to 5: 1
Current board:
[1, 0, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
[1, 2, 0, 0, 0]
Player 1 has won

可以看出，这不会以 4 个匹配（匹配 4）退出，而是需要整列（此处显示 5 个元素）或行进行匹配。

Answer 1

在这种情况下，使用 any 而不是 all 实际上有效。 All 只是为所有行/列打印 1。

def check_won(grids, user, n):
    cnt = 0
    for grid in grids:
        if any(cell == user for cell in grid):
            cnt += 1
            if cnt == 4:
                return True
    return False