在3D中拟合线 [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?我可以找到很多资料来源，这些资料可以根据2D数据集给出一条直线方程，而对于3D数据集则没有.

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

如果尝试从其他两个值中预测一个值，则应将lstsq与a参数一起用作自变量(加上1的列以估计截距)，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，如果您仅想获得与数据最佳的拟合线，即，如果将数据投影到该线上，该线将使实点与其投影之间的平方距离最小，则您想要的是第一个主要组成部分.

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.

一种定义它的方法是直线，其方向向量是与最大特征值相对应的协方差矩阵的特征向量，它穿过数据的均值.就是说，eig(cov(data))是一种非常不好的计算方式，因为它执行了许多不必要的计算和复制，并且可能不如使用svd那样准确.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()

外观如下:

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