Python 3：为什么我的“and”在“if”条件下充当“or”

Question

我正在编写一个函数来根据所选单元格的坐标获取正交坐标中某个单元格的邻居的坐标。我的代码是：

def get_neighbours_coordinates (x, y):
    neighbours = []
    for temp_x in [x-1, x, x+1]:
        # condition to drop the case, when cell has the same coordinates as treated
        for temp_y in [y-1, y, y+1]:
            if (temp_x != x) and (temp_y != y):
                neighbours.append((temp_x, temp_y))
    print (neighbours)

那么，如果我称它为（为了举例）：

for i in range (10):
    get_neighbours_coordinates(i, i)

返回：

[(-1, -1), (-1, 1), (1, -1), (1, 1)]
[(0, 0), (0, 2), (2, 0), (2, 2)]
[(1, 1), (1, 3), (3, 1), (3, 3)]
[(2, 2), (2, 4), (4, 2), (4, 4)]
[(3, 3), (3, 5), (5, 3), (5, 5)]
[(4, 4), (4, 6), (6, 4), (6, 6)]
[(5, 5), (5, 7), (7, 5), (7, 7)]
[(6, 6), (6, 8), (8, 6), (8, 8)]
[(7, 7), (7, 9), (9, 7), (9, 9)]
[(8, 8), (8, 10), (10, 8), (10, 10)]

虽然它应该返回：

[(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 2), (2, 0), (2, 1), (2, 2)]
[(1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2), (3, 3)]
[(2, 2), (2, 3), (2, 4), (3, 2), (3, 4), (4, 2), (4, 3), (4, 4)]
[(3, 3), (3, 4), (3, 5), (4, 3), (4, 5), (5, 3), (5, 4), (5, 5)]
[(4, 4), (4, 5), (4, 6), (5, 4), (5, 6), (6, 4), (6, 5), (6, 6)]
[(5, 5), (5, 6), (5, 7), (6, 5), (6, 7), (7, 5), (7, 6), (7, 7)]
[(6, 6), (6, 7), (6, 8), (7, 6), (7, 8), (8, 6), (8, 7), (8, 8)]
[(7, 7), (7, 8), (7, 9), (8, 7), (8, 9), (9, 7), (9, 8), (9, 9)]
[(8, 8), (8, 9), (8, 10), (9, 8), (9, 10), (10, 8), (10, 9), (10, 10)]

看起来and 删除了至少一个条件为真的所有情况，而它必须只删除两个条件都为真的情况。

我的代码有什么问题？

附：如果我将and 替换为or，代码将返回所需的输出。
在 Windows 10 上使用 Python 3.9。

Answer 1

您看到的结果与您的布尔逻辑一致。通过说and，您是说要排除相关单元格的整个行和列。您真正要排除的唯一单元格是查询单元格本身。

也就是说，你想要：

not (temp_x == x and temp_y == y)

等同于：

(temp_x != x) or (temp_y != y)

这个逻辑等价是De Morgan's Laws之一。