python - 大于二维数组中某个值的最小相邻单元格的索引

Question

我有一个 numpy 数组 a:

a = np.array([[0,4,3,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,4,3,5,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,8,3,9,2,6,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,9,2,6,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,4,2,9,2,6,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,2,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,9,2,6,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,2,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,4,2,2,6,8,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,4,4,3,4,4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10],
          [4,4,4,3,4,4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]])

我正在寻找最小的索引 neighbours value= 2 (numpy.where(a==value)) 但大于 value 的单元格。我还需要我们找到最小邻居的相应单元格的索引 = value。

本例中的结果 (value = 2) 应该是:

• 相邻单元格的索引:[0,2] 和 [4,3]相应的和

• 相应单元格的索引:[1,2] 和 [3,3]

如果问题不是很清楚，请见谅。

这是我目前所拥有的:

import numpy as np
value = 2
neighbors = np.zeros(4, dtype = np.float)
fdx = np.flatnonzero(a== value)
locations = fdx // a.shape[1], fdx % a.shape[1]
maximums = []
for item in zip(*locations):
    i, j = item[0], item[1]
    neighbors[0], neighbors[1], neighbors[2],neighbors[3] = [a[i-1,j], a[i+1,j], a[i,j-1], a[i,j+1]]
    maximums.append(min(neighbors[neighbors> value]))
print np.where(a==min(maximums))

prints: (array([0, 4]), array([2, 3]))

非常慢，而且还不知道如何找到对应单元格的索引。任何与我的解决方案完全不同的解决方案也将被接受。

Answer 1

您可以使用 scipy.ndimage.morphology.binary_dilation 找到邻居

import numpy as np
from scipy.ndimage import morphology
a = np.array([[0,4,3,9,9,9],
              [4,4,2,2,2,9],
              [4,2,2,9,2,6],
              [4,2,2,2,6,8],
              [4,4,4,3,4,4]])
k = 2

# make mask
eq = a==k
# find greater neighbors (as mask)
nbs = morphology.binary_dilation(eq) & (a>k)
# translate to index
minidx = np.argwhere(nbs & (a == np.min(a[nbs])))

# now find neighbors' neighbors
# pad original mask
m,n = a.shape
eqp = np.zeros((m+2, n+2), bool)
eqp[1:-1,1:-1] = eq
# generate offset vectors for the four major directions (up, down, left, right)
# corrected for padding
offsp = np.array([(0,1),(2,1),(1,0),(1,2)])
# without padding correction
offs = offsp - 1#
# for each minimum, find all (1-4) reference neighbors
refidx = [i + offs[eqp[tuple((i+offsp).T)]] for i in minidx]

print(minidx)
print(refidx)

# [[0 2]
#  [4 3]]
# [array([[1, 2]]), array([[3, 3]])]