3D numpy 数组中的两个非零元素是否“连接”？

Question

Viewing a 3D numpy array as representing a 3D space, two array indices A and B are "connected" if there exists a Path from A to B, where a Path is a list of array indices where every adjacent element pair Path[i] and Path[i+1] are (1) indices to non-zero elements in the input array and (2) "next" to each other, np.max(np.abs(Path[i] - Path[i+1])) <= 1.

Given an initial index A, I'd like to generate the list A_Connected which is the list of all array indices which are "connected" to A.

My slow method would be:

def find_connected_list(array, index, connected_list):
   connected_list.append(index)

   for x in range(index[0]-1, index[0]+2):
      for y in range(index[1]-1, index[1]+2):
         for z in range(index[2]-1, index[2]+2):
            if x>= 0 and x < array.shape[0] and y >= 0 and y < array.shape[1] and z >= 0 and z < array.shape[2] and array[x,y,z] > 0 and not [x,y,z] in connected_list:
               connected_list = find_connected_list(array, [x,y,z], connected_list)

   return connected_list

Answer 1

You need to look at the problem under a completely different angle. It is a graph problem for which you need a transition matrix.

• for each element of your array, you should identify its immediate connected neighbours

• the result of this should be stored in a 2d NxN array where N is the total number of elements in you 3d array, this is the transition matrix T

• your departure point is then a vector V of length N with N-1 zeros and one 1

• compute T^k V and increase T to find all connected points that are k steps away from V

This is only a guideline, you will find details here .

Python coding is not very difficult. You can use the fact that T is symmetric with 1's in the diagonal to fasten the process.

Hope that it helps.