如何找到平面上点投影的坐标

问题描述

希望一切顺利.我有两个 numpy 数组，都是空间中的一些点.使用python，我想首先找到通过第一个数据集(surface_maker)的表面，然后找到第二个数组的投影相邻点的x，y和z(contact_maker)在创建的表面上.surface_maker 总是创建平面表面.对于投影，我只想要从相邻点到表面的垂直方向.实际上，我在这两组中都有很多要点，但我在这里复制一个简单的案例:

surface_maker=np.array([[50., 15., 46.04750574],[50., 5., 45.56400925],[44.83018398, 5., 25.],[44.76296902, 15., 25.],[50., 25., 45.56400925],[44.83018398, 25., 25.],[59.8336792, 5., 75.],[59.71483707, 15., 75.],[59.8336792, 25., 75.]])contact_maker=np.array([[10., 5., 70.00014782],[10., 15., 70.00018358],[10., 25., 70.0001955],[30., 5., 69.99981105],[30., 15., 69.99982297],[30., 25., 69.99985874],[70., 5., 50.00000298],[70., 15., 50.00002682],[70., 25., 50.00005066],[90., 5., 49.99996871],[90., 15., 49.99999255],[90., 25., 50.00001788]])

我尝试了几种解决方案，例如

在此先感谢您的帮助.

我知道你需要解决两个问题:

• 找到适合点集合的平面
• 将第二组点沿特定方向投影到该平面上

第二个问题已在另一个答案中得到充分解决，因此我正在为第一个问题提供更通用的方法.

确实，当您确定所有点都位于一个平面上时，您可以只选择三个未对齐的点并计算平面.但是您的分数可能来自带有一些噪音的真实测量结果，您可能希望找到最适合您的分数的平面.

以下函数解决了寻找最适合点集合的平面的一般问题.请参阅评论中的解释:

将 numpy 导入为 npPRECISION = 1e-8 # 用于现实世界的任意零def plane_from_points(点):# 调整后的平面穿过点集合的质心质心= np.mean(点，轴= 0)# 使用SVD计算点集合的主轴#(特征向量)及其相对大小(特征值)_，值，向量 = np.linalg.svd(点 - 质心)# 每个奇异值都与它的向量配对，并从# 从最大值到最小值.# 调整后的平面必须包含对应的特征向量# 两个最大的特征值.如果只有一个特征向量不同# 从零开始，然后点对齐并且它们不定义平面.如果值[1] <精确:raise ValueError("点对齐，不能定义平面")# 所以平面法线是特征值最小的特征向量正常 = 向量[2]# 计算平面方程ax+by+cz+d=0的系数(a,b,c,d).# 前三个系数由法线给出，第四个# 一 (d) 是平面到坐标原点的有符号距离d = -np.dot(质心，正常)平面 = np.append(正常，d)# 如果最小特征向量接近于零，则集合# 点完全平坦.特征向量越大，越不平坦.# 你可能想知道这一点.厚度 = 值[2]返回平面，厚度

你可以检查一下:

>>>surface_maker=np.array([[50., 15., 46.04750574], [50., 5., 45.56400925], [44.83018398, 5., 25.], [44.76296902, 15., 25.], [50., 25., 45.56400925], [44.83018398, 25., 25.], [59.8336792, 5., 75.], [59.71483707, 15., 75.], [59.8336792, 25., 75.]])>>>平面，厚度 = plane_from_points(surface_maker)>>>打印(平面)[-0.95725318 0. 0.28925136 35.2806339]>>>打印(厚度)1.3825669490602308

所以，事实上，你的点分布并不平坦(厚度明显不同于零)，你不能随便选择三个点来解决你的问题.

Hope doing well. I have two numpy array, both are some points in the space. Using python, I want to firstly find the surface passing the first data set (surface_maker) and then find the x,y and z of the projection adjacent opoints of the second array (contact_maker) on the created surface. surface_maker always created planar surfaces. For projection, I only want a vertical going from adjacent point toward the surface. In reality I have lots of points in both sets but I copies a simple case here:

surface_maker=np.array([[50., 15., 46.04750574],
                        [50., 5., 45.56400925],
                        [44.83018398, 5., 25.],
                        [44.76296902, 15., 25.],
                        [50., 25., 45.56400925],
                        [44.83018398, 25., 25.],
                        [59.8336792, 5., 75.],
                        [59.71483707, 15., 75.],
                        [59.8336792, 25., 75.]])
contact_maker=np.array([[10.,  5., 70.00014782],
                        [10., 15., 70.00018358],
                        [10., 25., 70.0001955 ],
                        [30.,  5., 69.99981105],
                        [30., 15., 69.99982297],
                        [30., 25., 69.99985874],
                        [70., 5., 50.00000298],
                        [70., 15., 50.00002682],
                        [70., 25., 50.00005066],
                        [90., 5., 49.99996871],
                        [90., 15., 49.99999255],
                        [90., 25., 50.00001788]])

I have tried several solutions like 1, 2 and so on. But I was successful to solve my issue. For me it is important to have the location of projection as x, y and z. The figure also shows what I want (as it shows, I need only location six adjacent point of the contact_maker projected on the surface created by surface_maker):

import numpy as np
PRECISION = 1e-8    # Arbitrary zero for real-world purposes

def plane_from_points(points):
    # The adjusted plane crosses the centroid of the point collection
    centroid = np.mean(points, axis=0)

    # Use SVD to calculate the principal axes of the point collection
    # (eigenvectors) and their relative size (eigenvalues)
    _, values, vectors = np.linalg.svd(points - centroid)

    # Each singular value is paired with its vector and they are sorted from
    # largest to smallest value.
    # The adjusted plane plane must contain the eigenvectors corresponding to
    # the two largest eigenvalues. If only one eigenvector is different
    # from zero, then points are aligned and they don't define a plane.
    if values[1] < PRECISION:
        raise ValueError("Points are aligned, can't define a plane")

    # So the plane normal is the eigenvector with the smallest eigenvalue
    normal = vectors[2]

    # Calculate the coefficients (a,b,c,d) of the plane's equation ax+by+cz+d=0.
    # The first three coefficients are given by the normal, and the fourth
    # one (d) is the plane's signed distance to the origin of coordinates
    d = -np.dot(centroid, normal)
    plane = np.append(normal, d)

    # If the smallest eigenvector is close to zero, the collection of
    # points is perfectly flat. The larger the eigenvector, the less flat.
    # You may wish to know this.
    thickness = values[2]

    return plane, thickness

>>> surface_maker=np.array([[50., 15., 46.04750574], [50., 5., 45.56400925], [44.83018398, 5., 25.], [44.76296902, 15., 25.], [50., 25., 45.56400925], [44.83018398, 25., 25.], [59.8336792, 5., 75.], [59.71483707, 15., 75.], [59.8336792, 25., 75.]])
>>> plane, thickness = plane_from_points(surface_maker)
>>> print(plane)
[-0.95725318  0.          0.28925136 35.2806339 ]
>>> print(thickness)
1.3825669490602308