使用Python从3D中的六个点确定均匀仿射变换矩阵

我得到三点的位置：

p1 = [1.0, 1.0, 1.0]
p2 = [1.0, 2.0, 1.0]
p3 = [1.0, 1.0, 2.0]

和他们改造的同行：

p1_prime = [2.414213562373094,  5.732050807568877, 0.7320508075688767]
p2_prime = [2.7677669529663684, 6.665063509461097, 0.6650635094610956]
p3_prime = [2.7677669529663675, 5.665063509461096, 1.6650635094610962]

仿射变换矩阵的形式

trans_mat = np.array([[…, …, …, …],
                      […, …, …, …],
                      […, …, …, …],
                      […, …, …, …]])

这样的

import numpy as np

def transform_pt(point, trans_mat):
    a  = np.array([point[0], point[1], point[2], 1])
    ap = np.dot(a, trans_mat)[:3]
    return [ap[0], ap[1], ap[2]]

你会得到：

transform_pt(p1, trans_mat) == p1_prime
transform_pt(p2, trans_mat) == p2_prime
transform_pt(p3, trans_mat) == p3_prime

假设转换是同质的(仅包含旋转和平移),我该如何确定转换矩阵？

从CAD程序中,我知道矩阵是：

trans_mat = np.array([[0.866025403784, -0.353553390593, -0.353553390593, 0],
                      [0.353553390593,  0.933012701892, -0.066987298108, 0],
                      [0.353553390593, -0.066987298108,  0.933012701892, 0],
                      [0.841081377402,  5.219578794378,  0.219578794378, 1]])

我想知道如何找到它.

解决方法:

单独的六点不足以唯一地确定仿射变换.但是,根据您之前在问题中提出的问题(在删除之前不久)以及your comment,您似乎不仅仅在寻找仿射变换,而是在寻找同质仿射变换.

This answer by robjohn提供了解决问题的方法.虽然它解决了许多问题的更普遍的问题,但6点的解决方案可以在答案的最底部找到.我将以更加程序员友好的格式将其转录：

import numpy as np

def recover_homogenous_affine_transformation(p, p_prime):
    '''
    Find the unique homogeneous affine transformation that
    maps a set of 3 points to another set of 3 points in 3D
    space:

        p_prime == np.dot(p, R) + t

    where `R` is an unknown rotation matrix, `t` is an unknown
    translation vector, and `p` and `p_prime` are the original
    and transformed set of points stored as row vectors:

        p       = np.array((p1,       p2,       p3))
        p_prime = np.array((p1_prime, p2_prime, p3_prime))

    The result of this function is an augmented 4-by-4
    matrix `A` that represents this affine transformation:

        np.column_stack((p_prime, (1, 1, 1))) == \
            np.dot(np.column_stack((p, (1, 1, 1))), A)

    Source: https://math.stackexchange.com/a/222170 (robjohn)
    '''

    # construct intermediate matrix
    Q       = p[1:]       - p[0]
    Q_prime = p_prime[1:] - p_prime[0]

    # calculate rotation matrix
    R = np.dot(np.linalg.inv(np.row_stack((Q, np.cross(*Q)))),
               np.row_stack((Q_prime, np.cross(*Q_prime))))

    # calculate translation vector
    t = p_prime[0] - np.dot(p[0], R)

    # calculate affine transformation matrix
    return np.column_stack((np.row_stack((R, t)),
                            (0, 0, 0, 1)))

对于您的样本输入,这将恢复与您从CAD程序获得的完全相同的矩阵：

>>> recover_homogenous_affine_transformation(
        np.array(((1.0,1.0,1.0),
                  (1.0,2.0,1.0),
                  (1.0,1.0,2.0))),
        np.array(((2.4142135623730940, 5.732050807568877, 0.7320508075688767),
                  (2.7677669529663684, 6.665063509461097, 0.6650635094610956),
                  (2.7677669529663675, 5.665063509461096, 1.6650635094610962))))
array([[ 0.8660254 , -0.35355339, -0.35355339,  0.        ],
       [ 0.35355339,  0.9330127 , -0.0669873 ,  0.        ],
       [ 0.35355339, -0.0669873 ,  0.9330127 ,  0.        ],
       [ 0.84108138,  5.21957879,  0.21957879,  1.        ]])

标签：python,linear-algebra,coordinate-transformation,coordinate-systems,homogenous-tr
来源： https://codeday.me/bug/20190612/1223823.html