[英]find tangent vector at a point for discrete data points
I have a vector with a min of two points in space, eg: 我有一个向量,在空间中至少有两个点,例如:
A = np.array([-1452.18133319 3285.44737438 -7075.49516676])
B = np.array([-1452.20175668 3285.29632734 -7075.49110863])
I want to find the tangent of the vector at a discrete points along the curve, gg the beginning and end of the curve. 我想在曲线的离散点处找到向量的切线,gg曲线的起点和终点。 I know how to do it in Matlab but I want to do it in Python. 我知道如何在Matlab中做到这一点,但我想在Python中做到这一点。 This is the code in Matlab: 这是Matlab中的代码:
A = [-1452.18133319 3285.44737438 -7075.49516676];
B = [-1452.20175668 3285.29632734 -7075.49110863];
points = [A; B];
distance = [0.; 0.1667];
pp = interp1(distance, points,'pchip','pp');
[breaks,coefs,l,k,d] = unmkpp(pp);
dpp = mkpp(breaks,repmat(k-1:-1:1,d*l,1).*coefs(:,1:k-1),d);
ntangent=zeros(length(distance),3);
for j=1:length(distance)
ntangent(j,:) = ppval(dpp, distance(j));
end
%The solution would be at beginning and end:
%ntangent =
% -0.1225 -0.9061 0.0243
% -0.1225 -0.9061 0.0243
Any ideas? 有任何想法吗? I tried to find the solution using numpy and scipy using multiple methods, eg 我试图通过多种方法使用numpy和scipy找到解决方案,例如
tck, u= scipy.interpolate.splprep(data)
but none of the methods seem satisfy what I want. 但似乎没有一种方法可以满足我的要求。
Give der=1
to splev to get the derivative of the spline: 将der=1
赋予splev以得到样条的导数:
from scipy import interpolate
import numpy as np
t=np.linspace(0,1,200)
x=np.cos(5*t)
y=np.sin(7*t)
tck, u = interpolate.splprep([x,y])
ti = np.linspace(0, 1, 200)
dxdt, dydt = interpolate.splev(ti,tck,der=1)
ok, I found the solution which is a little modification of "pv" above (note that splev works only for 1D vectors) One problem I was having originally with "tck, u= scipy.interpolate.splprep(data)" is that it requires a min of 4 points to work (Matlab works with two points). 好的,我找到了上面“ pv”的一点修改的解决方案(请注意,splev仅适用于一维矢量)我最初使用“ tck,u = scipy.interpolate.splprep(data)”遇到的一个问题是至少需要4分才能工作(Matlab需要2分)。 I was using two points. 我当时使用了两点。 After increasing the data points, it works as i want. 增加数据点后,它可以按我的要求工作。
Here is the solution for completeness: 这是完整性的解决方案:
import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate
data = np.array([[-1452.18133319 , 3285.44737438, -7075.49516676],
[-1452.20175668 , 3285.29632734, -7075.49110863],
[-1452.32645025 , 3284.37412457, -7075.46633213],
[-1452.38226151 , 3283.96135828, -7075.45524248]])
distance=np.array([0., 0.15247556, 1.0834, 1.50007])
data = data.T
tck,u = interpolate.splprep(data, u=distance, s=0)
yderv = interpolate.splev(u,tck,der=1)
and the tangents are (which matches the Matlab results if the same data is used): 和切线是(如果使用相同的数据,则匹配Matlab结果):
(-0.13394599723751408, -0.99063114953803189, 0.026614957159932656)
(-0.13394598523149195, -0.99063115868512985, 0.026614950816003666)
(-0.13394595055068903, -0.99063117647357712, 0.026614941718878599)
(-0.13394595652952143, -0.9906311632471152, 0.026614954146007865)
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