过滤二维数组并从中间返回坐标

Question

I have a 2D array of zeros with some positive integers at (1,6) and (2,7):

[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]

And I want to filter the array by a custom kernel:

[[1 0 1]
 [0 1 0]
 [0 1 0]]

I want to filter the array with this kernel and when 2 or 3 of the ones in this kernel are multiplied by a positive integer, I want it to return the co-ordinates of the ones that were multiplied by 0.

I know from image analysis that it's easy to convolve a 2D array by a kernel but it doesn't yield the intermediate results. On the above 2D array, it would return (1,8) and (3,7).

Is there some package functions that I can use to make this process simple and easy, or will I have to implement it myself? As always, all help is appreciated

Answer 1

This is a numpy implementation of it to start with. You can increase performance probably by modifying it.

Here, num_ones is the lower and upper number of ones in the kernel you would like to filter, referring to when 2 or 3 of the ones in this kernel are multiplied by a positive integer

a = np.array([[0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.],
 [0.,0.,0.,0.,0.,0.,2.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.],
 [0.,0.,0.,0.,0.,0.,0.,2.,0.,0.,0.,0.,0.,0.,0.,0.,0.],
 [0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.,0.]])

kernel = np.array([[1.,0.,1.],\
 [0.,1.,0.],\
 [0.,1.,0.]])

sub_shape = kernel.shape
#throshold of number of kernel ones to have non-zero value
num_ones = [2,3]

#divide the matrix into sub_matrices of kernel size
view_shape = tuple(np.subtract(a.shape, sub_shape) + 1) + sub_shape
strides = a.strides + a.strides
sub_matrices = np.lib.stride_tricks.as_strided(a,view_shape,strides)
#convert non_zero elements to 1 (dummy representation)
sub_matrices[sub_matrices>0.] = 1.

#Do convolution
m = np.einsum('ij,klij->kl',kernel,sub_matrices)

#find sub_matrices that satisfy non-zero elements' condition
filt = np.argwhere(np.logical_and(m>=num_ones[0], m<=num_ones[1]))
#for each sub_matix find the zero elements located in non-zero elements of kernel
output = []
for [i,j] in filt:
  output.append(np.argwhere((sub_matrices[i,j,:,:]==0)*kernel) + [i, j])

output is an array of indices arrays where each array is indices where your condition is met per kernel application in each location [i,j] of your image. If you wish to aggregate them all, you can stack all arrays and take a unique list of it. I am not sure how you would like the output be in case of multiple occurrences.

output:

output =
[[1 8]
 [3 7]] 

UPDATE: regarding einsum:

I would recommend this post about einsum to learn: Understanding NumPy's einsum

sub_matrices is a 4-dimensional array. sub_matrices[k,l,:,:] is sub matrix of a starting at position [k,l] and shape of kernel. (later we changed all non-zero values of it to 1 for our purpose)

m = np.einsum('ij,klij->kl',kernel,sub_matrices) multiplies two dimensions i and j of kernel into last two dimensions i and j of sub_matrices array (in other words, it element-wise multiplies kernel to sub matrices sub_matrices[k,l,:,:] ) and sums all elements into m[k,l] . This is known as 2D convolution of kernel into a .