如何从c3中的3x3 Homography矩阵获取旋转，平移，剪切

Question

我计算了3x3单应矩阵，我需要获得旋转，平移，剪切和缩放以将它们用作windows8媒体元素属性中的参数吗？

Answer 1

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)