在 3D 中拟合一条线 [英] Fitting a line in 3D

问题描述

是否有任何算法可以从一组 3D 数据点返回直线方程?我可以找到很多来源，它们可以从 2D 数据集中给出一条线的方程，但在 3D 中没有.

Are there any algorithms that will return the equation of a straight line from a set of 3D data points? I can find plenty of sources which will give the equation of a line from 2D data sets, but none in 3D.

谢谢.

推荐答案

如果你想从另外两个中预测一个值，那么你应该使用 lstsq 和 a 参数作为您的自变量(加上一列 1 来估计截距)和 b 作为您的因变量.

If you are trying to predict one value from the other two, then you should use lstsq with the a argument as your independent variables (plus a column of 1's to estimate an intercept) and b as your dependent variable.

另一方面，如果您只想获得数据的最佳拟合线，即如果您将数据投影到数据上，将最小化真实点与其投影之间的平方距离的线，则你想要的是第一个主成分.

If, on the other hand, you just want to get the best fitting line to the data, i.e. the line which, if you projected the data onto it, would minimize the squared distance between the real point and its projection, then what you want is the first principal component.

定义它的一种方法是其方向向量是对应于最大特征值的协方差矩阵的特征向量的线，该线通过数据的均值.也就是说，eig(cov(data)) 是一种非常糟糕的计算方法，因为它进行了大量不必要的计算和复制，并且可能不如使用 svd 准确>.见下文:

One way to define it is the line whose direction vector is the eigenvector of the covariance matrix corresponding to the largest eigenvalue, that passes through the mean of your data. That said, eig(cov(data)) is a really bad way to calculate it, since it does a lot of needless computation and copying and is potentially less accurate than using svd. See below:

import numpy as np

# Generate some data that lies along a line

x = np.mgrid[-2:5:120j]
y = np.mgrid[1:9:120j]
z = np.mgrid[-5:3:120j]

data = np.concatenate((x[:, np.newaxis], 
                       y[:, np.newaxis], 
                       z[:, np.newaxis]), 
                      axis=1)

# Perturb with some Gaussian noise
data += np.random.normal(size=data.shape) * 0.4

# Calculate the mean of the points, i.e. the 'center' of the cloud
datamean = data.mean(axis=0)

# Do an SVD on the mean-centered data.
uu, dd, vv = np.linalg.svd(data - datamean)

# Now vv[0] contains the first principal component, i.e. the direction
# vector of the 'best fit' line in the least squares sense.

# Now generate some points along this best fit line, for plotting.

# I use -7, 7 since the spread of the data is roughly 14
# and we want it to have mean 0 (like the points we did
# the svd on). Also, it's a straight line, so we only need 2 points.
linepts = vv[0] * np.mgrid[-7:7:2j][:, np.newaxis]

# shift by the mean to get the line in the right place
linepts += datamean

# Verify that everything looks right.

import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d as m3d

ax = m3d.Axes3D(plt.figure())
ax.scatter3D(*data.T)
ax.plot3D(*linepts.T)
plt.show()

如下所示:

这篇关于在 3D 中拟合一条线的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！