计算python中n维字段的邻居

Question

When given a matrix of n-dimension and a grid size of 1 I'd like to calculate the nearest neighbours of a field. Given below is an example for a 2-dimensional field

P = (1,1)

p_neighbours = [(0,0),(2,2),(0,1),(0,2),(1,0),(2,0),(2,1),(1,2)]

Mathematically this can be as easily described as P +/- 1 in a vectorian system (as far as I understand). The size of the n-dimensional neighbour array is described as (n^3)-1 I've already found a pretty good old topic , anyhow I couldn't understand how any of the presented solutions could be extended to a n-dimensional function..

Answer 1

from itertools import product

def stencil(dim):
    stencils = list(product([-1,0,1], repeat=dim))
    zero = ((0,) * dim)
    stencils.remove(zero)
    return stencils

def neighbours(P):
    stencils = stencil(len(P))
    return [tuple([sum(x) for x in zip(P,s)]) for s in stencils]

P = (4, 4, 4)

print(neighbours(P))

Answer 2

正确地说，n维邻居数组的大小不是（n ^ 3）-1，而是（3 ^ n）-1（如果n> 3，则数组大小要大得多！）生成邻居在任何维度上，您都必须实现一个递归算法，该算法在维度上进行迭代并调用自身（我不知道确切地如何实现此方法，但是我正在研究它）

Answer 3

I guess should be something like this

p_neighbours = []
for x in [-1,0,1]:
  for y in [-1,0,1]:
    p_neighbours.append((P(0)+x,P(1)+y))

Answer 4

I think the clearest way to go here is to use a simple list comprehension:

p = (1,1)
px, py = p
p_neighbours = [(px+x,py+y) for x in range(-1,2) for y in range(-1,2) if (x,y) != (0,0)]

Here we check that (x,y) is not (0,0) to avoid adding p itself to its neighbors.