python - 如何在二维 numpy 数组中查找行？

Question

我有一个2D numpy 数组，我想找到水平线和垂直线的边界点。

gray_img = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0],
                     [0, 255, 255, 255, 255, 255, 255, 255, 0],
                     [0, 255, 255, 255, 255, 255, 255, 255, 0],
                     [0, 255, 255, 0, 0, 0, 0, 0, 0],
                     [0, 255, 255, 0, 0, 0, 0, 0, 0],
                     [0, 255, 255, 0, 0, 0, 0, 0, 0],
                     [0, 255, 255, 0, 0, 255, 255, 255, 255],
                     [0, 255, 255, 0, 0, 255, 255, 255, 255],
                     [0, 255, 255, 0, 0, 0, 0, 0, 0],
                     [0, 255, 255, 0, 0, 0, 0, 0, 0],
                     [0, 255, 255, 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, 0, 0, 0, 0, 0, 0, 0],
                     [0, 0, 0, 0, 0, 0, 0, 0, 0]])

desired_outcome = [ [[1,1],[10,1]],
                    [[1,2],[10,2]],
                    [[1,3],[2,3]], ...]

以下是我要查找的行:

稍后，我想删除较小的线以仅保留距离超过 2 点的线。

Answer 1

垂直线:

m = np.diff(gray_img, 1, 0) # get discrete difference along the 0-axis
m = np.argwhere(m != 0)     # get indices where value is not zero 
m = m[np.lexsort(m.T)]      # sort indices first 1-column then 0-column
m[::2,0] += 1               #

输出:

[[ 1  1]
 [10  1]
 [ 1  2]
 [10  2]
 [ 1  3]
 [ 2  3]
 [ 1  4]
 [ 2  4]
 [ 1  5]
 [ 2  5]
 [ 6  5]
 [ 7  5]
 [ 1  6]
 [ 2  6]
 [ 6  6]
 [ 7  6]
 [ 1  7]
 [ 2  7]
 [ 6  7]
 [ 7  7]
 [ 6  8]
 [ 7  8]]

水平线:

m = np.diff(gray_img, 1, 1, append=np.zeros((gray_img.shape[0], 1)))
m = np.argwhere(m != 0)
m[::2,1] += 1

输出:

[[ 1  1]
 [ 1  7]
 [ 2  1]
 [ 2  7]
 [ 3  1]
 [ 3  2]
 [ 4  1]
 [ 4  2]
 [ 5  1]
 [ 5  2]
 [ 6  1]
 [ 6  2]
 [ 6  5]
 [ 6  8]
 [ 7  1]
 [ 7  2]
 [ 7  5]
 [ 7  8]
 [ 8  1]
 [ 8  2]
 [ 9  1]
 [ 9  2]
 [10  1]
 [10  2]]