'Python: how to select contiguity neighbors of matrix values?
I have a matrix like the following:
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]])
For each cell I want to select the values in a radius of k=2 closest cells.
For instance if I select the A[3,4] I would like a submatrix like the following
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]])
I defined the following function
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
such as queen_neighbourhood(A, 3, 2, 2) returns
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]])
However it does not work in borders.
For instance, for the cell [0,0] I would like to have
array([[12, 6, 14],
[18, 13, 8],
[ 8, 15, 16])
but it returns queen_neighbourhood(A, 0, 0, 2)
array([], shape=(0, 0), dtype=int64)
Solution 1:[1]
You could avoid negative indices:
neighbourhood = Adj[max(in_row-j, 0) : in_row+k,
max(in_col-j, 0) : in_col+k]
Solution 2:[2]
Adding to the previous answer; taking into consideration the extreme values
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)
Solution 3:[3]
You can use numpy roll to ensure you are always dealing with the middle value,
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)
which gives,
[[11 16 8]
[ 9 18 13]
[ 6 8 15]]
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Pychopath |
| Solution 2 | |
| Solution 3 | Ed Smith |
