Python：如何选择矩阵值的连续性邻居？

Question

我有一个类似的矩阵：

A = array([[12,  6, 14,  8,  4,  1],
       [18, 13,  8, 10,  9, 19],
       [ 8, 15,  6,  5,  6, 18],
       [ 3,  0,  2, 14, 13, 12],
       [ 4,  4,  5, 19,  0, 14],
       [16,  8,  7,  7, 11,  0],
       [ 3, 11,  2, 19, 11,  5],
       [ 4,  2,  1,  9, 12, 12]])

对于每个单元格，我想在radius of k = 2最近的单元格中选择值。

例如，如果我选择a [3,4]我想要一个像以下功能这样的subsatrix，

array([[18, 13,  8, 10,  9],
       [ 8, 15,  6,  5,  6],
       [ 3,  0,  2, 14, 13],
       [ 4,  4,  5, 19,  0],
       [16,  8,  7,  7, 11]])

我定义了以下功能，

def queen_neighbourhood(Adj, in_row, in_col, k):
    j=k
    k+=1
    neighbourhood = Adj[in_row-j:in_row+k, in_col-j:in_col+k]
    return neighbourhood

例如queen_neighbourhood（a，3，2，2）返回，

array([[18, 13,  8, 10,  9],
       [ 8, 15,  6,  5,  6],
       [ 3,  0,  2, 14, 13],
       [ 4,  4,  5, 19,  0],
       [16,  8,  7,  7, 11]])

​​但是它在边界中不起作用。

例如，对于单元格> [0,0] 我想拥有的

array([[12, 6,  14],
       [18, 13,  8],
       [ 8, 15, 16])

，但它返回queen_neighbourhood（a，0，0，2）

array([], shape=(0, 0), dtype=int64)

Answer 1

您可以避免负面指标：

    neighbourhood = Adj[max(in_row-j, 0) : in_row+k,
                        max(in_col-j, 0) : in_col+k]

Answer 2

添加到上一个答案；考虑到极端价值观

def queen_neighbourhood(Adj, in_row, in_col, k):
j=k
k+=1
neighbourhood = Adj[max(in_row-j, 0) : min(in_row+k,Adj.shape[0]),
                    max(in_col-j, 0) : min(in_col+k,Adj.shape[1])]
return(neighbourhood)

Answer 3

您可以使用 numpy roll 确保您始终处理中间值，

import numpy as np

def queen_neighbourhood(Adj, in_row, in_col, k):
    j=k
    k+=1
    midrow = int(Adj.shape[0]/2.)+1
    midcol = int(Adj.shape[1]/2.)+1
    Ashift = np.roll(Adj,(in_row-midrow,in_col-midcol),(0,1))
    neighbourhood = Ashift[1:k+1, 1:k+1]
    return neighbourhood


A = np.array([[18, 13,  8, 10,  9],
       [ 8, 15,  6,  5,  6],
       [ 3,  0,  2, 14, 13],
       [ 4,  4,  5, 19,  0],
       [16,  8,  7,  7, 11]])

print(A)
An = queen_neighbourhood(A, 0, 0, 2)
print(An)

这给出了，

[[11 16  8]
 [ 9 18 13]
 [ 6  8 15]]