The Hough transform in its simplest form is a method to detect straight lines but it can also be used to detect circles or ellipses. The algorithm assumes that the edge is detected and it is robust against noise or missing points.
In the following example, the Hough transform is used to detect coin positions and match their edges. We provide a range of plausible radii. For each radius, two circles are extracted and we finally keep the five most prominent candidates. The result shows that coin positions are well-detected.
Given a black circle on a white background, we first guess its radius (or a range of radii) to construct a new circle. This circle is applied on each black pixel of the original picture and the coordinates of this circle are voting in an accumulator. From this geometrical construction, the original circle center position receives the highest score.
Note that the accumulator size is built to be larger than the original picture in order to detect centers outside the frame. Its size is extended by two times the larger radius.
import numpy as np import matplotlib.pyplot as plt from skimage import data, color from skimage.transform import hough_circle, hough_circle_peaks from skimage.feature import canny from skimage.draw import circle_perimeter from skimage.util import img_as_ubyte # Load picture and detect edges image = img_as_ubyte(data.coins()[160:230, 70:270]) edges = canny(image, sigma=3, low_threshold=10, high_threshold=50) # Detect two radii hough_radii = np.arange(20, 35, 2) hough_res = hough_circle(edges, hough_radii) # Select the most prominent 5 circles accums, cx, cy, radii = hough_circle_peaks(hough_res, hough_radii, total_num_peaks=3) # Draw them fig, ax = plt.subplots(ncols=1, nrows=1, figsize=(10, 4)) image = color.gray2rgb(image) for center_y, center_x, radius in zip(cy, cx, radii): circy, circx = circle_perimeter(center_y, center_x, radius) image[circy, circx] = (220, 20, 20) ax.imshow(image, cmap=plt.cm.gray) plt.show()
In this second example, the aim is to detect the edge of a coffee cup. Basically, this is a projection of a circle, i.e. an ellipse. The problem to solve is much more difficult because five parameters have to be determined, instead of three for circles.
The algorithm takes two different points belonging to the ellipse. It assumes that it is the main axis. A loop on all the other points determines how much an ellipse passes to them. A good match corresponds to high accumulator values.
A full description of the algorithm can be found in reference .
|||Xie, Yonghong, and Qiang Ji. “A new efficient ellipse detection method.” Pattern Recognition, 2002. Proceedings. 16th International Conference on. Vol. 2. IEEE, 2002|
import matplotlib.pyplot as plt from skimage import data, color, img_as_ubyte from skimage.feature import canny from skimage.transform import hough_ellipse from skimage.draw import ellipse_perimeter # Load picture, convert to grayscale and detect edges image_rgb = data.coffee()[0:220, 160:420] image_gray = color.rgb2gray(image_rgb) edges = canny(image_gray, sigma=2.0, low_threshold=0.55, high_threshold=0.8) # Perform a Hough Transform # The accuracy corresponds to the bin size of a major axis. # The value is chosen in order to get a single high accumulator. # The threshold eliminates low accumulators result = hough_ellipse(edges, accuracy=20, threshold=250, min_size=100, max_size=120) result.sort(order='accumulator') # Estimated parameters for the ellipse best = list(result[-1]) yc, xc, a, b = [int(round(x)) for x in best[1:5]] orientation = best # Draw the ellipse on the original image cy, cx = ellipse_perimeter(yc, xc, a, b, orientation) image_rgb[cy, cx] = (0, 0, 255) # Draw the edge (white) and the resulting ellipse (red) edges = color.gray2rgb(img_as_ubyte(edges)) edges[cy, cx] = (250, 0, 0) fig2, (ax1, ax2) = plt.subplots(ncols=2, nrows=1, figsize=(8, 4), sharex=True, sharey=True) ax1.set_title('Original picture') ax1.imshow(image_rgb) ax2.set_title('Edge (white) and result (red)') ax2.imshow(edges) plt.show()
Total running time of the script: ( 0 minutes 18.432 seconds)