如何从C#中的3×3同构矩阵获取旋转,平移,剪切

我计算了3×3单应性矩阵,需要旋转,平移,剪切和缩放以将它们用作Windows8媒体元素属性中的参数.

解决方法:

查看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)

标签：homography,windows-8,transformation,c
来源： https://codeday.me/bug/20191031/1973392.html