0

我有一个空间中至少有两个点的向量,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处找到向量的切线,gg 曲线的起点和终点。我知道如何在 Matlab 中做到这一点,但我想在 Python 中做到这一点。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?我尝试使用多种方法使用 numpy 和 scipy 找到解决方案,例如

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

但没有一种方法似乎满足我想要的。

4

2 回答 2

6

der=1splev 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)
于 2012-11-16T15:32:43.780 回答
0

好的,我找到了对上面的“pv”稍作修改的解决方案(请注意,splev 仅适用于一维向量)我最初遇到的一个问题是“tck, u= scipy.interpolate.splprep(data)”是它至少需要 4 个点才能工作(Matlab 需要两个点)。我用了两点。增加数据点后,它可以按我的意愿工作。

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)

并且切线是(如果使用相同的数据,则与 Matlab 结果匹配):

(-0.13394599723751408, -0.99063114953803189, 0.026614957159932656)
(-0.13394598523149195, -0.99063115868512985, 0.026614950816003666)
(-0.13394595055068903, -0.99063117647357712, 0.026614941718878599)
(-0.13394595652952143, -0.9906311632471152, 0.026614954146007865)
于 2012-11-17T05:01:36.030 回答