我正在使用cv2.fitEllipse()
使椭圆适合轮廓。该函数返回中心坐标,长轴和短轴以及旋转 Angular 。我想知道旋转角是否与此处给出的长轴相对于正水平轴的 Angular 相同(src:Wikipedia):
如果不是,那么有什么方法可以在以下等式中获取椭圆的系数:
然后直接计算 Angular
最佳答案
这将向您展示Python / OpenCV中的fitEllipse Angular 。
输入:
import cv2
import numpy as np
import math
# read input
img = cv2.imread('labrador.jpg')
# convert to gray
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# threshold
thresh = cv2.threshold(gray, 100 , 255, cv2.THRESH_BINARY)[1]
# find largest contour
contours = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
contours = contours[0] if len(contours) == 2 else contours[1]
big_contour = max(contours, key=cv2.contourArea)
# fit contour to ellipse and get ellipse center, minor and major diameters and angle in degree
ellipse = cv2.fitEllipse(big_contour)
(xc,yc),(d1,d2),angle = ellipse
print(xc,yc,d1,d1,angle)
# draw ellipse
result = img.copy()
cv2.ellipse(result, ellipse, (0, 255, 0), 3)
# draw circle at center
xc, yc = ellipse[0]
cv2.circle(result, (int(xc),int(yc)), 10, (255, 255, 255), -1)
# draw vertical line
# compute major radius
rmajor = max(d1,d2)/2
if angle > 90:
angle = angle - 90
else:
angle = angle + 90
print(angle)
xtop = xc + math.cos(math.radians(angle))*rmajor
ytop = yc + math.sin(math.radians(angle))*rmajor
xbot = xc + math.cos(math.radians(angle+180))*rmajor
ybot = yc + math.sin(math.radians(angle+180))*rmajor
cv2.line(result, (int(xtop),int(ytop)), (int(xbot),int(ybot)), (0, 0, 255), 3)
cv2.imwrite("labrador_ellipse.jpg", result)
cv2.imshow("labrador_thresh", thresh)
cv2.imshow("labrador_ellipse", result)
cv2.waitKey(0)
cv2.destroyAllWindows()
结果:
关于python - OpenCV:计算椭圆的长轴和短轴的定向角,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/62698756/