三次 Hermite 样条插值 python

我想计算一个三次多项式，该多项式由其函数值和指定点的导数定义。

https://en.wikipedia.org/wiki/Cubic_Hermite_spline https://en.wikipedia.org/wiki/Cubic_Hermite_spline

我知道 scipy 的插值方法。具体来说

splprep http://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.splprep.html#scipy.interpolate.splprep插值 N 维样条 和splev http://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.splev.html#scipy.interpolate.splev来评估其衍生品。

是否有一个接受函数值的Python例程f(x)及衍生品f'(x)对应值x并计算适合给定数据的样条表示。

举个例子：
我在空间中有两个由坐标定义的对象位置x,y,z我知道速度x',y',z'物体在这些位置的情况。我现在可以插入对象随时间 t 在两点之间经过的路径吗？考虑所有给定的参数。

延伸ev-br 的回答 https://stackoverflow.com/a/36655118/3388962，这里有一些示例代码，说明了用法BPoly.from_derivatives https://docs.scipy.org/doc/scipy-0.19.0/reference/generated/scipy.interpolate.BPoly.from_derivatives.html在点之间进行插值n具有指定导数的尺寸。

import numpy as np
from scipy import interpolate

def sampleCubicSplinesWithDerivative(points, tangents, resolution):
    '''
    Compute and sample the cubic splines for a set of input points with
    optional information about the tangent (direction AND magnitude). The 
    splines are parametrized along the traverse line (piecewise linear), with
    the resolution being the step size of the parametrization parameter.
    The resulting samples have NOT an equidistant spacing.

    Arguments:      points: a list of n-dimensional points
                    tangents: a list of tangents
                    resolution: parametrization step size
    Returns:        samples

    Notes: Lists points and tangents must have equal length. In case a tangent
           is not specified for a point, just pass None. For example:
                    points = [[0,0], [1,1], [2,0]]
                    tangents = [[1,1], None, [1,-1]]

    '''
    resolution = float(resolution)
    points = np.asarray(points)
    nPoints, dim = points.shape

    # Parametrization parameter s.
    dp = np.diff(points, axis=0)                 # difference between points
    dp = np.linalg.norm(dp, axis=1)              # distance between points
    d = np.cumsum(dp)                            # cumsum along the segments
    d = np.hstack([[0],d])                       # add distance from first point
    l = d[-1]                                    # length of point sequence
    nSamples = int(l/resolution)                 # number of samples
    s,r = np.linspace(0,l,nSamples,retstep=True) # sample parameter and step

    # Bring points and (optional) tangent information into correct format.
    assert(len(points) == len(tangents))
    data = np.empty([nPoints, dim], dtype=object)
    for i,p in enumerate(points):
        t = tangents[i]
        # Either tangent is None or has the same
        # number of dimensions as the point p.
        assert(t is None or len(t)==dim)
        fuse = list(zip(p,t) if t is not None else zip(p,))
        data[i,:] = fuse

    # Compute splines per dimension separately.
    samples = np.zeros([nSamples, dim])
    for i in range(dim):
        poly = interpolate.BPoly.from_derivatives(d, data[:,i])
        samples[:,i] = poly(s)
    return samples

为了演示此函数的使用，我们指定点和切线。该示例进一步演示了更改切线的“大小”时的效果。

# Input.
points = []
tangents = []
resolution = 0.2
points.append([0.,0.]); tangents.append([1,1])
points.append([3.,4.]); tangents.append([1,0])
points.append([5.,2.]); tangents.append([0,-1])
points.append([3.,0.]); tangents.append([-1,-1])
points = np.asarray(points)
tangents = np.asarray(tangents)

# Interpolate with different tangent lengths, but equal direction.
scale = 1.
tangents1 = np.dot(tangents, scale*np.eye(2))
samples1 = sampleCubicSplinesWithDerivative(points, tangents1, resolution)
scale = 2.
tangents2 = np.dot(tangents, scale*np.eye(2))
samples2 = sampleCubicSplinesWithDerivative(points, tangents2, resolution)
scale = 0.1
tangents3 = np.dot(tangents, scale*np.eye(2))
samples3 = sampleCubicSplinesWithDerivative(points, tangents3, resolution)

# Plot.
import matplotlib.pyplot as plt
plt.scatter(samples1[:,0], samples1[:,1], marker='o', label='samples1')
plt.scatter(samples2[:,0], samples2[:,1], marker='o', label='samples2')
plt.scatter(samples3[:,0], samples3[:,1], marker='o', label='samples3')
plt.scatter(points[:,0], points[:,1], s=100, c='k', label='input')
plt.axis('equal')
plt.title('Interpolation')
plt.legend()
plt.show()

This results in the following plot:

需要注意三点：

• 以下也可以应用于二维以上。
• 样品之间的间距不固定。实现等距采样的一种简单方法是在返回的结果之间进行线性插值samples，正如在例如中所讨论的那样这个帖子 https://stackoverflow.com/a/19122075/3388962.
• The specification of the tangents is optional, however BPoly.from_derivatives does not ensure smooth transitions between the splines at this position. If for example tangents[1] in the above sample is set to None, sampleCubicSplinesWithDerivative(points, tangents, resolution), the result will look like this: