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VistasHow to find the number of neighbours pixels in binary array
I am looking for an easy way to count the number of the green pixels in the image below, where the original image is the same but the green pixels are black.
I tried it with numpy.diff(), but then I am counting some pixels twice. I thought about numpy.gradient() – but here I am not sure if it is the right tool.
I know there have to be many solutions to this problem, but I don't know how to google for it. I am looking for a solution in python.
To make it clearer, I have only one image (only black and white pixels). The image with the green pixel is just for illustration.
1 Respuestas
Responde la pregunta0
You can use the edge detection kernel for this problem.
import numpy as np
from scipy.ndimage import convolve
a = np.array([[0, 0, 0, 0],
[0, 1, 1, 1],
[0, 1, 1, 1]])
kernel = np.array([[-1, -1, -1],
[-1, 8, -1],
[-1, -1, -1]])
Then, we will convolve the original array with the kernel. Notice that the edges are all negatives.
>>> convolve(a, kernel)
[[-1 -2 -3 -3]
[-2 5 3 3]
[-3 3 0 0]]
We will count the number of negative values and get the result.
>>> np.where(convolve(a, kernel) < 0, 1, 0)
[[1 1 1 1]
[1 0 0 0]
[1 0 0 0]]
>>> np.sum(np.where(convolve(a, kernel) < 0, 1, 0))
6
Edges-only kernel
There are a lot of things you can do with the kernel. For example, you can modify the kernel if you don't want to include diagonal neighbors.
kernel = np.array([[ 0, -1, 0],
[-1, 4, -1],
[ 0, -1, 0]])
This gives the following output.
>>> np.where(convolve(a, kernel) < 0, 1, 0)
[[0 1 1 1]
[1 0 0 0]
[1 0 0 0]]
>>> np.sum(np.where(convolve(a, kernel) < 0, 1, 0))
5