如何从c3中的3x3 Homography矩阵获取旋转,平移,剪切
答
看到https://math.stackexchange.com/questions/78137/decomposition-of-a-nonsquare-affine-matrix
def getComponents(normalised_homography):
'''((translationx, translationy), rotation, (scalex, scaley), shear)'''
a = normalised_homography[0,0]
b = normalised_homography[0,1]
c = normalised_homography[0,2]
d = normalised_homography[1,0]
e = normalised_homography[1,1]
f = normalised_homography[1,2]
p = math.sqrt(a*a + b*b)
r = (a*e - b*d)/(p)
q = (a*d+b*e)/(a*e - b*d)
translation = (c,f)
scale = (p,r)
shear = q
theta = math.atan2(b,a)
return (translation, theta, scale, shear)
您可以通过以下链接开始:找到最旋转和斜对的一矩阵变换] [1] [1] :http://stackoverflow.com/questions/5107134/find-the-rotation-and-skew-of-a-matrix-transformation – 2013-03-14 21:40:45