How to Compute the Gradients of Image Using Python

Can I use numpy gradient function with images

Central differences in the interior and first differences at the boundaries.

15 - 10
13 - 10 / 2
24 - 15 / 2
...
39 - 28

Gradient for Color Image in Python

Find each channel's gradient separately like this

gradR = np.sqrt(np.square(Rx) + np.square(Ry))
gradG = np.sqrt(np.square(Gx) + np.square(Gy))
gradB = np.sqrt(np.square(Bx) + np.square(By))

make a new image

grad = np.dstack((gradR,gradG,gradB))

How to apply gradient/magnitude to an image using OpenCV?

I think there might be an issue with ndimage.filters.convolve. I got similar results as you. But the following seems to work fine using Python/OpenCV

Input:

import cv2
import numpy as np
import skimage.exposure

img = cv2.imread('black_dress.png', cv2.IMREAD_GRAYSCALE)

img = cv2.GaussianBlur(img, (0,0), sigmaX=1.5, sigmaY=1.5)

Kx = np.array([[-1, 0, 1], 
               [-2, 0, 2], 
               [-1, 0, 1]])
Ky = np.array([[1,   2,  1], 
               [0,   0,  0], 
              [-1,  -2, -1]])

Ix = cv2.filter2D(img, -1, Kx)
Iy = cv2.filter2D(img, -1, Ky)

G = np.hypot(Ix, Iy)
G = skimage.exposure.rescale_intensity(G, in_range='image', out_range=(0,255)).astype(np.uint8)

theta = np.arctan2(Iy, Ix)
theta = skimage.exposure.rescale_intensity(theta, in_range='image', out_range=(0,255)).astype(np.uint8)
   
cv2.imwrite('black_dress_gradient_magnitude.png', G)
cv2.imwrite('black_dress_gradient_direction.png', theta)

cv2.imshow("magnitude", G)
cv2.imshow("direction", theta)
cv2.waitKey(0)

Magnitude:

Direction:

Gradient of edges Python

If what you want is the total number of pixels comprising horizontal vs vertical edges, I would suggest defining some threshold for horizontal vs vertical (say 15 degrees). So you can count the number of elements of theta for which
abs(theta) < pi/12 (horizontal)
or abs(theta) > pi-pi/12 (horizontal)
or pi/2 - pi/12 < abs(theta) < pi/2+pi/12 (vertical)

What you're storing in v and h are the vertical and horizontal components of the gradient at each point and what you need is to compare the values of v and h to determine for each point if the gradient vector should count as horizontal or vertical. Comparing theta is probably the most intuitive way to do this.

In order to get the number of elements of theta that satisfy a particular condition, I would suggest using a generator expression:

sum(1 for i in theta if (abs(i)<pi/12) or (abs(i)>pi-pi/12))

would give you the number of horizontal edge pixels for example.

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