3d空间中线曲线的样条插值系数 [英] spline interpolation coefficients of a line curve in 3d space

问题描述

我是python的新手. 我在3D空间中有一条由一组给定点定义的线曲线. 谁能建议我如何使用scipy包的样条函数进行插值来获取曲线的样条系数，就像MATLAB中的spline.coeff函数一样? 谢谢！

I am new to python. I have a line curve in the 3D space defined by a set of given points. Can anyone suggest how I can use the interpolate with spline functions of the scipy package to get the spline coefficients of the curve just like the spline.coeff function in MATLAB? Thank you!

我用过

tck = interpolate.SmoothBivariateSpline(pts2[:,0], pts2[:,1], pts2[:,2])
test_pts = pts2[:,2]-tck.ev(pts2[:,0], pts2[:,1])
print test_pts

但这显然是用于表面的，而不是用于直线曲线的.pts2是包含点坐标的Nx3 numpy array

but this is for surfaces apparently and not for line curves pts2 is a Nx3 numpy array containing the coordinates of the points

好的，我弄清楚我做错了什么.我的输入点太少了.现在我有另一个问题.函数get_coeffs应该返回所有样条系数.这些系数按什么顺序返回?我有一个79 tx和79 ty的数组代表结，当我调用函数来调用结时，得到的数组是1x5625

ok I figured out what I was doing wrong. my input points where too few. now I have another question. The function get_coeffs is supposed to return the spline coefficients at every not. In which order those coefficients are returned? I have an array of 79 tx and 79 ty which represent the knots and I get an array of 1x5625 when I call the function to call the knots

推荐答案

我也是python的新手，但是最近的搜索使我转向

I too am new to python, but my recent searching led me to a very helpful scipy interpolation tutorial. From my reading of this I concur that the BivariateSpline family of classes/functions are intended for interpolating 3D surfaces rather than 3D curves.

对于我的3D曲线拟合问题(我认为这与您的问题非常相似，但还希望消除噪声)，最终我使用了

For my 3D curve fitting problem (which I believe is very similar to yours, but with the addition of wanting to smooth out noise) I ended up using scipy.interpolate.splprep (not to be confused with scipy.interpolate.splrep). From the tutorial linked above, the spline coefficients your are looking for are returned by splprep.

正常输出是一个三元组(t，c，k)，其中包含 结点t，样条系数c和阶数k.

The normal output is a 3-tuple, (t,c,k) , containing the knot-points, t , the coefficients c and the order k of the spline.

与较新的，面向对象的" UnivariateSpline和BivariateSpline类相比，文档始终将这些过程函数称为"FITPACK的较旧的，非面向对象的包装".我本来希望使用更新的，面向对象的"，但据我所知UnivariateSpline仅处理一维情况，而splprep直接处理N-D数据.

The docs keep referring to these procedural functions as an "older, non object-oriented wrapping of FITPACK" in contrast to the "newer, object-oriented" UnivariateSpline and BivariateSpline classes. I would have preferred "newer, object-oriented" myself, but as far as I can tell UnivariateSpline only handles the 1-D case whereas splprep handles N-D data directly.

下面是我用来弄清楚这些功能的简单测试用例:

Below is a simple test-case that I used to figure out these functions:

import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate
from mpl_toolkits.mplot3d import Axes3D


# 3D example
total_rad = 10
z_factor = 3
noise = 0.1

num_true_pts = 200
s_true = np.linspace(0, total_rad, num_true_pts)
x_true = np.cos(s_true)
y_true = np.sin(s_true)
z_true = s_true/z_factor

num_sample_pts = 80
s_sample = np.linspace(0, total_rad, num_sample_pts)
x_sample = np.cos(s_sample) + noise * np.random.randn(num_sample_pts)
y_sample = np.sin(s_sample) + noise * np.random.randn(num_sample_pts)
z_sample = s_sample/z_factor + noise * np.random.randn(num_sample_pts)

tck, u = interpolate.splprep([x_sample,y_sample,z_sample], s=2)
x_knots, y_knots, z_knots = interpolate.splev(tck[0], tck)
u_fine = np.linspace(0,1,num_true_pts)
x_fine, y_fine, z_fine = interpolate.splev(u_fine, tck)

fig2 = plt.figure(2)
ax3d = fig2.add_subplot(111, projection='3d')
ax3d.plot(x_true, y_true, z_true, 'b')
ax3d.plot(x_sample, y_sample, z_sample, 'r*')
ax3d.plot(x_knots, y_knots, z_knots, 'go')
ax3d.plot(x_fine, y_fine, z_fine, 'g')
fig2.show()
plt.show()

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