确定单元二维列表的邻居 [英] Determining neighbours of cell two dimensional list

问题描述

我有一个列表列表，类似

I have a list of lists, something like

[[1, 2, 3,],[4, 5, 6,],[7, 8, 9]].

以图形方式表示为:

1 2 3
4 5 6
7 8 9

我正在寻找一种优雅的方法来水平，垂直和对角地检查单元格的邻居的值.例如，[0] [2]的邻居是[0] [1]，[1] [1]和[1] [2]或数字2、5、6.

I'm looking for an elegant approach to check the value of neighbours of a cell, horizontally, vertically and diagonally. For instance, the neighbours of [0][2] are [0][1], [1][1] and [1][2] or the numbers 2, 5, 6.

现在我意识到，我可以对每个值进行一次暴力攻击:

Now I realise, I could just do a bruteforce attack checking every value a la:

[i-1][j]
[i][j-1]
[i-1][j-1]
[i+1][j]
[i][j+1]
[i+1][j+1]
[i+1][j-1]
[i-1][j+1]

但是那很容易，我想我可以通过看一些更优雅的方法来学到更多东西.

But thats easy, and I figured I can learn more by seeing some more elegant approaches.

推荐答案

# Size of "board"
X = 10
Y = 10

neighbors = lambda x, y : [(x2, y2) for x2 in range(x-1, x+2)
                               for y2 in range(y-1, y+2)
                               if (-1 < x <= X and
                                   -1 < y <= Y and
                                   (x != x2 or y != y2) and
                                   (0 <= x2 <= X) and
                                   (0 <= y2 <= Y))]

>>> print(neighbors(5, 5))
[(4, 4), (4, 5), (4, 6), (5, 4), (5, 6), (6, 4), (6, 5), (6, 6)]

我不知道这是否干净，但是这种单行代码可以遍历所有邻居并丢弃任何边缘情况，从而为您提供所有邻居.

I don't know if this is considered clean, but this one-liner gives you all neighbors by iterating over them and discarding any edge cases.

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