如何在非单调曲线的预定 x 位置重新采样样条线

Question

这个出色的示例提供了为非单调曲线创建样条线的选项：非单调数据的三次样条线（不是一维函数）

但是，我现在需要另一个步骤 - 获取给定

y

值的

x

值。我无法弄清楚如何在没有循环的情况下扩展样条线的使用。我想我可以寻找一些

r

的解决方案，从而提供一些

x_i

，但这将是一个非常漫长的过程。

假设我需要

y

处的

x = 4.5

的值。但最接近的是 'spline(0.1)' 给出 array([4.54115695, 0.40519822])

Answer 1

找到了这个解决方案，首先使用上采样，然后使用径向基函数进行插值（由于非单调，唯一有效的解决方案

x

）

import numpy
import scipy.interpolate
import matplotlib.pyplot as plt

# very sparse path, non-monotonic

path_x = numpy.asarray((4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0),dtype=float)
path_y = numpy.asarray((0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668),dtype=float)


# Upsample using spline interpolation
# then downsample using griddata nearest interpolation


# defining arbitrary parameter to parameterize the curve
path_t = numpy.linspace(0,1,path_x.size)

# this is the position vector with
# x coord (1st row) given by path_x, and
# y coord (2nd row) given by path_y
r = numpy.vstack((path_x.reshape((1,path_x.size)),path_y.reshape((1,path_y.size))))

# creating the spline object
spline = scipy.interpolate.interp1d(path_t,r,kind='cubic')

# defining values of the arbitrary parameter over which
# you want to interpolate x and y
# it MUST be within 0 and 1, since you defined
# the spline between path_t=0 and path_t=1
t = numpy.linspace(numpy.min(path_t),numpy.max(path_t),1000)

# interpolating along t
# r[0,:] -> interpolated x coordinates
# r[1,:] -> interpolated y coordinates
rr = spline(t)
xi = rr[0,:]
yi = rr[1,:]

plt.plot(path_x,path_y,'ro')
plt.plot(xi,yi,'-')
plt.xlabel('x')
plt.ylabel('y')

from scipy.interpolate import Rbf

# xii = resample(xi,20) # resample at 20 points along x
# yii = scipy.interpolate.griddata(xi,yi,path_x,method='nearest')
rbfi = Rbf(t, xi, yi, function='linear')
yii = rbfi(path_t, path_x)
# plt.plot(path_x,path_y,'or')
# plt.plot(xi,yi,'x-k')
plt.plot(path_x, yii, 'gs')

plt.show()