Select个像素在数组中使用python 3中的结构元素

Question

I'm searching a way to select "pixels" in an array using a structuring element: Imagine we have that array a, and that structuring element s,

a=np.array([[ 0,  1,  2,  3,  4,  5,  6],
         [ 7,  8,  9, 10, 11, 12, 13],
         [14, 15, 16, 17, 18, 19, 20],
         [21, 22, 23, 24, 25, 26, 27],
         [28, 29, 30, 31, 32, 33, 34],
         [35, 36, 37, 38, 39, 40, 41],
         [42, 43, 44, 45, 46, 47, 48]])
s=np.array([[0,1,0],
         [1,1,1],
         [0,1,0]])

I'm then searching for a function that will act like

f(a, position=(3,3), structure=s) = [17,23,24,25,31]

It looks like scipy.ndimage morphology functions can do that internally. A workaround would be to create a np.zeros array with the same shape as a, put a 1 at a position of interest and the dilate it, but that would be very resource consuming - especially since my arrays are not 7 * 7.

Answer 1

Here is an answer using view_as_windows (which uses numpy strides under the hood):

from skimage.util import view_as_windows
def f(a, position, structure):
  return view_as_windows(a,structure.shape)[tuple(np.array(position)-1)][structure.astype(bool)]

output:

f(a, position=(3,3), structure=s)
#[17 23 24 25 31]

if you feed the position as numpy array instead of tuple and structure as boolean array instead of int, the answer will even be shorter since you will not need conversions:

def f(a, position, structure):
      return view_as_windows(a,structure.shape)[tuple(position-1)][structure]