Cube Edge detection

Question

I am trying to get the edges of a cube in the image below using the Hough transform in the Scikit-Image library .

Update Here is the code I am working with:

smoothed_image = filters.frangi(gray_image)

# Perform edge detection using the Canny edge detector
low_threshold = 0.1  
high_threshold = 3  
low_threshold = low * smoothed_image.max()
high_threshold = high * low_threshold
edges = canny(smoothed_image, sigma, low_threshold=low_threshold,
             high_threshold=high_threshold)


# Hough transform to detect the filament profile
hspace, theta, dist = hough_line(edges)

# Find the vertical lines
vertical_peaks = hough_line_peaks(hspace, theta, dist, num_peaks=2)

vertical_lines = []

for _, angle, dist in zip(*vertical_peaks):
    x = dist * np.cos(angle)
    y = dist * np.sin(angle)
    vertical_lines.append((x, y))


# Visualize the results
fig, axes = plt.subplots(1, 3, figsize=(15, 5))
axes[0].imshow(smoothed_image)
axes[0].set_title('Smoothed Image')
#axes[0].axis('off')
axes[1].imshow(edges, cmap='gray')
axes[1].set_title('Canny Edge Detection')
axes[1].axis('off')
axes[2].imshow(edges, cmap='gray')
for x, y in vertical_lines:
    axes[2].axvline(x=x, color='red')
axes[2].set_xlim((0, image.shape[1]))
axes[2].set_ylim((image.shape[0], 0))
axes[2].set_title('Detected Lines')
axes[2].axis('off')
plt.tight_layout()
plt.show()

And this is the resulting output

Any ideas on how to make the Hough transform detect the other edge?

Mask out the tube and the cone below the cube if they are going to be in the same place all the time. Threshold and find contours? — Pete, Jul 17 '23 at 16:38
what is that tube doing there anyway? it's occluding the cube. — Christoph Rackwitz, Jul 17 '23 at 23:07
Hough detects lines, you want to detect edges. Apply a filter that transforms the edges into lines, for example the gradient magnitude. — Cris Luengo, Jul 18 '23 at 01:34
@Pete, masking out the tube and burner does not work. I even try to crop out the tube and the burner but still didn;t work — David, Jul 18 '23 at 06:26
@ChristophRackwitz, the cube is used as some form of calibration and to get the pixel-distance correlation of the tube. — David, Jul 18 '23 at 06:28

score 0 · Accepted Answer · answered Jul 19 '23 at 08:15

As suggested by @Pete "Mask out the tube" as it is occluding the cube, and adjust the parameters:

Canny parameters
Gaussian Filter parameters
Hough Line detector parameters

so you can detect the other edge.

P.S: I used OpenCV, but you might repeat the results using Scikit Image

#!/usr/bin/env python3

import cv2 
import numpy as np 
import matplotlib.pyplot as plt 

im_path = "cube.png"

img = cv2.imread(im_path)
  
# Convert the img to grayscale
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

# Gaussian Filter
k = 5
sigma = 0

gauss = cv2.GaussianBlur(gray,(k,k),sigma)

# Apply edge detection method on the image
edges = cv2.Canny(gauss, 50, 150, apertureSize=3)

# Remove unwanted tube
# https://stackoverflow.com/a/64175996/8618242
sum_ = np.sum(edges, axis=1)
mean_ = np.mean(sum_)
edges[sum_> mean_, :] = 0


# This returns an array of r and theta values
lines = cv2.HoughLines(edges, 1, np.pi/180, 150)
  
# The below for loop runs till r and theta values
# are in the range of the 2d array
for r_theta in lines:
    arr = np.array(r_theta[0], dtype=np.float64)
    r, theta = arr
    # Stores the value of cos(theta) in a
    a = np.cos(theta)
  
    # Stores the value of sin(theta) in b
    b = np.sin(theta)
  
    # x0 stores the value rcos(theta)
    x0 = a*r
  
    # y0 stores the value rsin(theta)
    y0 = b*r
  
    # x1 stores the rounded off value of (rcos(theta)-1000sin(theta))
    x1 = int(x0 + 1000*(-b))
  
    # y1 stores the rounded off value of (rsin(theta)+1000cos(theta))
    y1 = int(y0 + 1000*(a))
  
    # x2 stores the rounded off value of (rcos(theta)+1000sin(theta))
    x2 = int(x0 - 1000*(-b))
  
    # y2 stores the rounded off value of (rsin(theta)-1000cos(theta))
    y2 = int(y0 - 1000*(a))
  
    # cv2.line draws a line in img from the point(x1,y1) to (x2,y2).
    # (0,0,255) denotes the colour of the line to be
    # drawn. In this case, it is red.
    cv2.line(img, (x1, y1), (x2, y2), (0, 0, 255), 2)


cv2.namedWindow("output", cv2.WINDOW_NORMAL)
cv2.imshow("output", img)
cv2.waitKey(0)

Cube Edge detection

1 Answers1