如何从具有周期性边界条件的numpy数组中选择一个窗口? [英] How do I select a window from a numpy array with periodic boundary conditions?

问题描述

假设我像这样制作一个二维数组:

Suppose I make a 2d array like this:

>>> A=np.arange(16).reshape((4,4))
>>> A
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15]])

，并且我希望能够围绕任何给定的元素选择一个3x3的窗口，以便该窗口围绕边界包裹，我该怎么做呢?我知道如果窗口的边界不与原始数组的边界重叠，我可以做到这一点:

and I want to be able to select a 3x3 window around any given element so that the window wraps around the boundaries how would I do that? I know I can do this if the boundaries of the window don't overlap the boundaries of the original array:

>>> A[1:4,0:3]
array([[ 4,  5,  6],
       [ 8,  9, 10],
       [12, 13, 14]])

但是如果我使用像A[i-1:i+2,j-1:j+2]这样的表达式，它只会为i = 0(例如j = 0)返回一个空数组.

but if I use an expression like A[i-1:i+2,j-1:j+2] it only returns an empty array for i=0, j=0 for example.

推荐答案

import numpy as np

A=np.arange(16).reshape((4,4))

def neighbors(arr,x,y,n=3):
    ''' Given a 2D-array, returns an nxn array whose "center" element is arr[x,y]'''
    arr=np.roll(np.roll(arr,shift=-x+1,axis=0),shift=-y+1,axis=1)
    return arr[:n,:n]

print(A)
# [[ 0  1  2  3]
#  [ 4  5  6  7]
#  [ 8  9 10 11]
#  [12 13 14 15]]

print(neighbors(A,0,0))
# [[15 12 13]
#  [ 3  0  1]
#  [ 7  4  5]]

print(neighbors(A,1,0))
# [[ 3  0  1]
#  [ 7  4  5]
#  [11  8  9]]

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