def is_grid(self, grid, image):
"""
Checks the "gridness" by analyzing the results of a hough transform.
:param grid: binary image
:return: wheter the object in the image might be a grid or not
"""
# - Distance resolution = 1 pixel
# - Angle resolution = 1° degree for high line density
# - Threshold = 144 hough intersections
# 8px digit + 3*2px white + 2*1px border = 16px per cell
# => 144x144 grid
# 144 - minimum number of points on the same line
# (but due to imperfections in the binarized image it's highly
# improbable to detect a 144x144 grid)
lines = cv2.HoughLines(grid, 1, np.pi / 180, 144)
if lines is not None and np.size(lines) >= 20:
lines = lines.reshape((lines.size/2), 2)
# theta in [0, pi] (theta > pi => rho < 0)
# normalise theta in [-pi, pi] and negatives rho
lines[lines[:, 0] < 0, 1] -= np.pi
lines[lines[:, 0] < 0, 0] *= -1
criteria = (cv2.TERM_CRITERIA_EPS, 0, 0.01)
# split lines into 2 groups to check whether they're perpendicular
if cv2.__version__[0] == '2':
density, clmap, centers = cv2.kmeans(
lines[:, 1], 2, criteria,
5, cv2.KMEANS_RANDOM_CENTERS)
else:
density, clmap, centers = cv2.kmeans(
lines[:, 1], 2, None, criteria,
5, cv2.KMEANS_RANDOM_CENTERS)
# Overall variance from respective centers
var = density / np.size(clmap)
sin = abs(np.sin(centers[0] - centers[1]))
# It is probably a grid only if:
# - centroids difference is almost a 90° angle (+-15° limit)
# - variance is less than 5° (keeping in mind surface distortions)
return sin > 0.99 and var <= (5*np.pi / 180) ** 2
else:
return False
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