找到离散数据点的切线向量

Question

我有一个在空间中最少有两个点的向量,例如:

A = np.array([-1452.18133319  3285.44737438 -7075.49516676])
B = np.array([-1452.20175668  3285.29632734 -7075.49110863])

我想在沿着曲线的离散点处找到矢量的切线,gg是曲线的开始和结束.我知道如何在Matlab中做到这一点,但我想用Python做.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    

有任何想法吗？我尝试使用numpy和scipy使用多种方法找到解决方案,例如

tck, u= scipy.interpolate.splprep(data)

但是没有一种方法能满足我的要求.

Answer 1

给der=1splev以获取样条的导数:

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)