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])
我想在曲线的离散点找到矢量的切线,g.g是曲线的起点和终点。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
有什么想法吗?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)
但是这些方法似乎都不符合我的要求。
答案 0 :(得分:5)
将der=1
赋予splev以获取样条曲线的导数:
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)
答案 1 :(得分:0)
以下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)
和切线(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)