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Python 高效索引numpy数组邻居_Python_Numpy - Fatal编程技术网

Python 高效索引numpy数组邻居

Python 高效索引numpy数组邻居,python,numpy,Python,Numpy,我正在尝试为特定numpy数组项的邻居建立索引。例如,如果我有如下所示的数组,并且我检查了超过某个值的项,那么我如何在不使用循环等的情况下有效地为真项的上、下、左、右单元格建立索引 In [34]: x Out[34]: array([[ 10., 10., 10., 10., 10.], [ 10., 10., 10., 10., 10.], [ 10., 20., 10., 10., 10.], [ 10., 10., 10., 20.,

我正在尝试为特定numpy数组项的邻居建立索引。例如,如果我有如下所示的数组,并且我检查了超过某个值的项,那么我如何在不使用循环等的情况下有效地为真项的上、下、左、右单元格建立索引

In [34]: x
Out[34]: 
array([[ 10.,  10.,  10.,  10.,  10.],
   [ 10.,  10.,  10.,  10.,  10.],
   [ 10.,  20.,  10.,  10.,  10.],
   [ 10.,  10.,  10.,  20.,  10.],
   [ 10.,  10.,  10.,  10.,  10.]])

In [37]: ans = x > 10

In [38]: ans
Out[38]: 
array([[False, False, False, False, False],
   [False, False, False, False, False],
   [False,  True, False, False, False],
   [False, False, False,  True, False],
   [False, False, False, False, False]], dtype=bool)

这将为您提供邻居的索引:

>>> def neighbors(x, y):
...    return np.array([(x-1, y), (x, y-1), (x+1, y), (x, y+1)])
...
>>> ind = zip(*np.where(x > 10))
>>> neighb = np.concatenate([neighbors(*i) for i in ind])
array([[1, 1],
       [2, 0],
       [3, 1],
       [2, 2],
       [2, 3],
       [3, 2],
       [4, 3],
       [3, 4]])