我想获取np.array的所有邻居值。
数组如下:
x = np.array([ [1, 2, 3, 4 ],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16] ])
我拥有的是:
i = 2
j = 2
n = x[i,j-1], x[i,j], x[i,j+1], x[i-1,j], x[i+1,j], x[i-1,j-1], x[i+1,j+1], x[i+1,j-1], x[i-1,j+1]
这返回(我想要的)
(10, 11, 12, 7, 15, 6, 16, 14, 8)
但是当我想要
的neightbour值时也出现了错误i = 3
j = 3
给出:
Exception has occurred: IndexError
index 4 is out of bounds for axis 1 with size 4
另一个灵魂是
def find_neighbors(m, i, j, dist=1):
return [row[max(0, j-dist):j+dist+1] for row in m[max(0,-1):i+dist+1]]
和
n = find_neighbors(x, i, j)
哪一个给我一系列的海格特,但我设置时也不能给我所有的海格特
i = 0
j = 0
因为它只会给我:
[array([1, 2]), array([5, 6])]
有人对此有解决方案吗?
谢谢!
答案 0 :(得分:1)
# function to find the start row and column
def find_start(x):
start = x-1 if x-1 >= 0 else 0
return start
# function to find the end row and column
def find_end(x, shape):
end = x+1 if x+1 <= shape else shape
return end
def find_neighbors(a, i, j):
neighbors = []
row_start, row_end = find_start(i), find_end(i, a.shape[0])
col_start, col_end = find_start(j), find_end(j, a.shape[1])
for y in range(a.shape[0]):
for z in range(a.shape[1]):
if y >= row_start and y <= row_end:
if z >= col_start and z <= col_end:
neighbors.append(a[y][z])
return neighbors
i, j = 0, 0
neighbors = find_neighbors(a, i, j)
print(neighbors)
输出:[1, 2, 5, 6]
i, j = 3, 3
neighbors = find_neighbors(a, i, j)
neighbors
输出:[11, 12, 15, 16]
i, j = 2, 2
neighbors = find_neighbors(a, i, j)
neighbors
输出:[6, 7, 8, 10, 11, 12, 14, 15, 16]
这将涵盖所有边缘情况。
答案 1 :(得分:1)
您可以利用python索引环绕来获取负索引。
def wrap_nb(x,i,j):
return x[np.ix_(*((z-1, z, z+1-S) for z,S in zip((i,j), x.shape)))].ravel()
这要求i
和j
为非负且小于x
的形状。
如果不能保证:
def wrap_nb(x,i,j):
return x[np.ix_(*(np.r_[z-1:z+2]%S for z,S in zip((i,j), x.shape)))].ravel()
示例:
>>> wrap_nb(x,1,-2)
array([ 2, 3, 4, 6, 7, 8, 10, 11, 12])
>>> wrap_nb(x,0,-1)
array([15, 16, 13, 3, 4, 1, 7, 8, 5])
>>> wrap_nb(x,0,0)
array([16, 13, 14, 4, 1, 2, 8, 5, 6])
答案 2 :(得分:0)
我从伴侣那里得到了以下解决方案:
新数组:
homes = np.array([ [1, 2, 3, 4 ],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16] ])
用于返回邻居值的代码:
neighbour = []
neighbour += [homes[i][j]] # value itself
neighbour += [homes[i][(j + 1) % n]] # value right
neighbour += [homes[i][(j - 1) % n]] # value left
neighbour += [homes[(i + 1) % n][j]] # value down
neighbour += [homes[(i + 1) % n][(j + 1) % n]] # value right down
neighbour += [homes[(i + 1) % n][(j - 1) % n]] # value left down
neighbour += [homes[(i - 1) % n][j]] # vlaue up
neighbour += [homes[(i - 1) % n][(j + 1) % n]] # vlaue right up
neighbour += [homes[(i - 1) % n][(j - 1) % n]] # value left up
哪个还给我:
i = 0
j = 0
[16, 13, 15, 4, 1, 3, 12, 9, 11]
那是我所需要的,但我仍然对解决方案如阿卜杜勒的解决方案充满兴趣