如何在python中执行三次样条插值？
How to perform cubic spline interpolation in python?
我有两个列表来描述函数 y(x):
x = [0,1,2,3,4,5]
y = [12,14,22,39,58,77]
我想执行三次样条插值，以便在 x 的域中给定一些值 u，例如
u = 1.25
我能找到 y(u)。
简答：
from scipy import interpolate
def f(x):
x_points = [ 0, 1, 2, 3, 4, 5]
y_points = [12,14,22,39,58,77]
tck = interpolate.splrep(x_points, y_points)
return interpolate.splev(x, tck)
print(f(1.25))
长答案：
scipy 将样条插值中涉及的步骤分成两个操作，很可能是为了提高计算效率。
计算描述样条曲线的系数，
使用 splrep()。 splrep returns 一个元组数组，包含
系数.
这些系数被传入splev()来实际
在所需点 x 评估样条曲线（在本例中为 1.25）。
x 也可以是数组。调用 f([1.0, 1.25, 1.5]) returns
分别在 1、1.25 和 1,5 处插入点。
这种方法对于单次求值来说确实不方便，但由于最常见的用例是从少数几个函数求值点开始，然后反复使用样条曲线求插值，在实践中通常非常有用.
最小python3代码：
from scipy import interpolate
if __name__ == '__main__':
x = [ 0, 1, 2, 3, 4, 5]
y = [12,14,22,39,58,77]
# tck : tuple (t,c,k) a tuple containing the vector of knots,
# the B-spline coefficients, and the degree of the spline.
tck = interpolate.splrep(x, y)
print(interpolate.splev(1.25, tck)) # Prints 15.203125000000002
print(interpolate.splev(...other_value_here..., tck))
基于 cwhy 的评论和 youngmit 的回答
万一scipy没有安装：
import numpy as np
from math import sqrt
def cubic_interp1d(x0, x, y):
"""
Interpolate a 1-D function using cubic splines.
x0 : a float or an 1d-array
x : (N,) array_like
A 1-D array of real/complex values.
y : (N,) array_like
A 1-D array of real values. The length of y along the
interpolation axis must be equal to the length of x.
Implement a trick to generate at first step the cholesky matrice L of
the tridiagonal matrice A (thus L is a bidiagonal matrice that
can be solved in two distinct loops).
additional ref: www.math.uh.edu/~jingqiu/math4364/spline.pdf
"""
x = np.asfarray(x)
y = np.asfarray(y)
# remove non finite values
# indexes = np.isfinite(x)
# x = x[indexes]
# y = y[indexes]
# check if sorted
if np.any(np.diff(x) < 0):
indexes = np.argsort(x)
x = x[indexes]
y = y[indexes]
size = len(x)
xdiff = np.diff(x)
ydiff = np.diff(y)
# allocate buffer matrices
Li = np.empty(size)
Li_1 = np.empty(size-1)
z = np.empty(size)
# fill diagonals Li and Li-1 and solve [L][y] = [B]
Li[0] = sqrt(2*xdiff[0])
Li_1[0] = 0.0
B0 = 0.0 # natural boundary
z[0] = B0 / Li[0]
for i in range(1, size-1, 1):
Li_1[i] = xdiff[i-1] / Li[i-1]
Li[i] = sqrt(2*(xdiff[i-1]+xdiff[i]) - Li_1[i-1] * Li_1[i-1])
Bi = 6*(ydiff[i]/xdiff[i] - ydiff[i-1]/xdiff[i-1])
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
i = size - 1
Li_1[i-1] = xdiff[-1] / Li[i-1]
Li[i] = sqrt(2*xdiff[-1] - Li_1[i-1] * Li_1[i-1])
Bi = 0.0 # natural boundary
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
# solve [L.T][x] = [y]
i = size-1
z[i] = z[i] / Li[i]
for i in range(size-2, -1, -1):
z[i] = (z[i] - Li_1[i-1]*z[i+1])/Li[i]
# find index
index = x.searchsorted(x0)
np.clip(index, 1, size-1, index)
xi1, xi0 = x[index], x[index-1]
yi1, yi0 = y[index], y[index-1]
zi1, zi0 = z[index], z[index-1]
hi1 = xi1 - xi0
# calculate cubic
f0 = zi0/(6*hi1)*(xi1-x0)**3 + \
zi1/(6*hi1)*(x0-xi0)**3 + \
(yi1/hi1 - zi1*hi1/6)*(x0-xi0) + \
(yi0/hi1 - zi0*hi1/6)*(xi1-x0)
return f0
if __name__ == '__main__':
import matplotlib.pyplot as plt
x = np.linspace(0, 10, 11)
y = np.sin(x)
plt.scatter(x, y)
x_new = np.linspace(0, 10, 201)
plt.plot(x_new, cubic_interp1d(x_new, x, y))
plt.show()
如果您安装了 scipy 版本 >= 0.18.0，您可以使用 scipy.interpolate 中的 CubicSpline 函数进行三次样条插值。
您可以通过 运行 遵循 python 中的命令检查 scipy 版本：
#!/usr/bin/env python3
import scipy
scipy.version.version
如果您的 scipy 版本 >= 0.18.0，您可以 运行 下面的三次样条插值示例代码：
#!/usr/bin/env python3
import numpy as np
from scipy.interpolate import CubicSpline
# calculate 5 natural cubic spline polynomials for 6 points
# (x,y) = (0,12) (1,14) (2,22) (3,39) (4,58) (5,77)
x = np.array([0, 1, 2, 3, 4, 5])
y = np.array([12,14,22,39,58,77])
# calculate natural cubic spline polynomials
cs = CubicSpline(x,y,bc_type='natural')
# show values of interpolation function at x=1.25
print('S(1.25) = ', cs(1.25))
## Aditional - find polynomial coefficients for different x regions
# if you want to print polynomial coefficients in form
# S0(0<=x<=1) = a0 + b0(x-x0) + c0(x-x0)^2 + d0(x-x0)^3
# S1(1< x<=2) = a1 + b1(x-x1) + c1(x-x1)^2 + d1(x-x1)^3
# ...
# S4(4< x<=5) = a4 + b4(x-x4) + c5(x-x4)^2 + d5(x-x4)^3
# x0 = 0; x1 = 1; x4 = 4; (start of x region interval)
# show values of a0, b0, c0, d0, a1, b1, c1, d1 ...
cs.c
# Polynomial coefficients for 0 <= x <= 1
a0 = cs.c.item(3,0)
b0 = cs.c.item(2,0)
c0 = cs.c.item(1,0)
d0 = cs.c.item(0,0)
# Polynomial coefficients for 1 < x <= 2
a1 = cs.c.item(3,1)
b1 = cs.c.item(2,1)
c1 = cs.c.item(1,1)
d1 = cs.c.item(0,1)
# ...
# Polynomial coefficients for 4 < x <= 5
a4 = cs.c.item(3,4)
b4 = cs.c.item(2,4)
c4 = cs.c.item(1,4)
d4 = cs.c.item(0,4)
# Print polynomial equations for different x regions
print('S0(0<=x<=1) = ', a0, ' + ', b0, '(x-0) + ', c0, '(x-0)^2 + ', d0, '(x-0)^3')
print('S1(1< x<=2) = ', a1, ' + ', b1, '(x-1) + ', c1, '(x-1)^2 + ', d1, '(x-1)^3')
print('...')
print('S5(4< x<=5) = ', a4, ' + ', b4, '(x-4) + ', c4, '(x-4)^2 + ', d4, '(x-4)^3')
# So we can calculate S(1.25) by using equation S1(1< x<=2)
print('S(1.25) = ', a1 + b1*0.25 + c1*(0.25**2) + d1*(0.25**3))
# Cubic spline interpolation calculus example
# https://www.youtube.com/watch?v=gT7F3TWihvk
如果你想要一个无依赖性的解决方案，就把它放在这里。
代码取自上面的答案：
def my_cubic_interp1d(x0, x, y):
"""
Interpolate a 1-D function using cubic splines.
x0 : a 1d-array of floats to interpolate at
x : a 1-D array of floats sorted in increasing order
y : A 1-D array of floats. The length of y along the
interpolation axis must be equal to the length of x.
Implement a trick to generate at first step the cholesky matrice L of
the tridiagonal matrice A (thus L is a bidiagonal matrice that
can be solved in two distinct loops).
additional ref: www.math.uh.edu/~jingqiu/math4364/spline.pdf
# original function code at:
This function is licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/
Original Author raphael valentin
Date 3 Jan 2018
Modifications made to remove numpy dependencies:
-all sub-functions by MR
This function, and all sub-functions, are licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
Mod author: Matthew Rowles
Date 3 May 2021
"""
def diff(lst):
"""
numpy.diff with default settings
"""
size = len(lst)-1
r = [0]*size
for i in range(size):
r[i] = lst[i+1] - lst[i]
return r
def list_searchsorted(listToInsert, insertInto):
"""
numpy.searchsorted with default settings
"""
def float_searchsorted(floatToInsert, insertInto):
for i in range(len(insertInto)):
if floatToInsert <= insertInto[i]:
return i
return len(insertInto)
return [float_searchsorted(i, insertInto) for i in listToInsert]
def clip(lst, min_val, max_val, inPlace = False):
"""
numpy.clip
"""
if not inPlace:
lst = lst[:]
for i in range(len(lst)):
if lst[i] < min_val:
lst[i] = min_val
elif lst[i] > max_val:
lst[i] = max_val
return lst
def subtract(a,b):
"""
returns a - b
"""
return a - b
size = len(x)
xdiff = diff(x)
ydiff = diff(y)
# allocate buffer matrices
Li = [0]*size
Li_1 = [0]*(size-1)
z = [0]*(size)
# fill diagonals Li and Li-1 and solve [L][y] = [B]
Li[0] = sqrt(2*xdiff[0])
Li_1[0] = 0.0
B0 = 0.0 # natural boundary
z[0] = B0 / Li[0]
for i in range(1, size-1, 1):
Li_1[i] = xdiff[i-1] / Li[i-1]
Li[i] = sqrt(2*(xdiff[i-1]+xdiff[i]) - Li_1[i-1] * Li_1[i-1])
Bi = 6*(ydiff[i]/xdiff[i] - ydiff[i-1]/xdiff[i-1])
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
i = size - 1
Li_1[i-1] = xdiff[-1] / Li[i-1]
Li[i] = sqrt(2*xdiff[-1] - Li_1[i-1] * Li_1[i-1])
Bi = 0.0 # natural boundary
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
# solve [L.T][x] = [y]
i = size-1
z[i] = z[i] / Li[i]
for i in range(size-2, -1, -1):
z[i] = (z[i] - Li_1[i-1]*z[i+1])/Li[i]
# find index
index = list_searchsorted(x0,x)
index = clip(index, 1, size-1)
xi1 = [x[num] for num in index]
xi0 = [x[num-1] for num in index]
yi1 = [y[num] for num in index]
yi0 = [y[num-1] for num in index]
zi1 = [z[num] for num in index]
zi0 = [z[num-1] for num in index]
hi1 = list( map(subtract, xi1, xi0) )
# calculate cubic - all element-wise multiplication
f0 = [0]*len(hi1)
for j in range(len(f0)):
f0[j] = zi0[j]/(6*hi1[j])*(xi1[j]-x0[j])**3 + \
zi1[j]/(6*hi1[j])*(x0[j]-xi0[j])**3 + \
(yi1[j]/hi1[j] - zi1[j]*hi1[j]/6)*(x0[j]-xi0[j]) + \
(yi0[j]/hi1[j] - zi0[j]*hi1[j]/6)*(xi1[j]-x0[j])
return f0
如果要获取值
from scipy.interpolate import CubicSpline
import numpy as np
x = [-5,-4.19,-3.54,-3.31,-2.56,-2.31,-1.66,-0.96,-0.22,0.62,1.21,3]
y = [-0.01,0.01,0.03,0.04,0.07,0.09,0.16,0.28,0.45,0.65,0.77,1]
value = 2
#ascending order
if np.any(np.diff(x) < 0):
indexes = np.argsort(x).astype(int)
x = np.array(x)[indexes]
y = np.array(y)[indexes]
f = CubicSpline(x, y, bc_type='natural')
specificVal = f(value).item(0) #f(value) is numpy.ndarray!!
print(specificVal)
如果要绘制插值函数。
np.linspace第三个参数增加“准确度”。
from scipy.interpolate import CubicSpline
import numpy as np
import matplotlib.pyplot as plt
x = [-5,-4.19,-3.54,-3.31,-2.56,-2.31,-1.66,-0.96,-0.22,0.62,1.21,3]
y = [-0.01,0.01,0.03,0.04,0.07,0.09,0.16,0.28,0.45,0.65,0.77,1]
#ascending order
if np.any(np.diff(x) < 0):
indexes = np.argsort(x).astype(int)
x = np.array(x)[indexes]
y = np.array(y)[indexes]
f = CubicSpline(x, y, bc_type='natural')
x_new = np.linspace(min(x), max(x), 100)
y_new = f(x_new)
plt.plot(x_new, y_new)
plt.scatter(x, y)
plt.title('Cubic Spline Interpolation')
plt.show()
输出：
在我之前的post中，我写了一个基于Cholesky开发的代码来求解三次算法生成的矩阵。不幸的是，由于平方根函数，它可能在某些点集（通常是 non-uniform 点集）上表现不佳。
本着与之前相同的精神，还有另一种使用 Thomas 算法 (TDMA) 的想法（参见 https://en.wikipedia.org/wiki/Tridiagonal_matrix_algorithm) to solve partially the tridiagonal matrix during its definition loop. However, the condition to use TDMA is that it requires at least that the matrix shall be diagonally dominant. However, in our case, it shall be true since |bi| > |ai| + |ci| with ai = h[i], bi = 2*(h[i]+h[i+1]), ci = h[i+1], with h[i] unconditionally positive. (see https://www.cfd-online.com/Wiki/Tridiagonal_matrix_algorithm_-TDMA(Thomas_algorithm)
我又参考了jingqiu的文档（见我之前的post，可惜link坏了，不过在网络缓存中还是可以找到的）。
TDMA求解器的优化版本可以描述如下：
def TDMAsolver(a,b,c,d):
""" This function is licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/
Author raphael valentin
Date 25 Mar 2022
ref. https://www.cfd-online.com/Wiki/Tridiagonal_matrix_algorithm_-_TDMA_(Thomas_algorithm)
"""
n = len(d)
w = np.empty(n-1,float)
g = np.empty(n, float)
w[0] = c[0]/b[0]
g[0] = d[0]/b[0]
for i in range(1, n-1):
m = b[i] - a[i-1]*w[i-1]
w[i] = c[i] / m
g[i] = (d[i] - a[i-1]*g[i-1]) / m
g[n-1] = (d[n-1] - a[n-2]*g[n-2]) / (b[n-1] - a[n-2]*w[n-2])
for i in range(n-2, -1, -1):
g[i] = g[i] - w[i]*g[i+1]
return g
当可以得到ai、bi、ci、di的每个个体时，组合自然三次样条的定义就变得容易了这 2 个单循环中的插值器函数。
def cubic_interpolate(x0, x, y):
""" Natural cubic spline interpolate function
This function is licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/
Author raphael valentin
Date 25 Mar 2022
"""
xdiff = np.diff(x)
dydx = np.diff(y)
dydx /= xdiff
n = size = len(x)
w = np.empty(n-1, float)
z = np.empty(n, float)
w[0] = 0.
z[0] = 0.
for i in range(1, n-1):
m = xdiff[i-1] * (2 - w[i-1]) + 2 * xdiff[i]
w[i] = xdiff[i] / m
z[i] = (6*(dydx[i] - dydx[i-1]) - xdiff[i-1]*z[i-1]) / m
z[-1] = 0.
for i in range(n-2, -1, -1):
z[i] = z[i] - w[i]*z[i+1]
# find index (it requires x0 is already sorted)
index = x.searchsorted(x0)
np.clip(index, 1, size-1, index)
xi1, xi0 = x[index], x[index-1]
yi1, yi0 = y[index], y[index-1]
zi1, zi0 = z[index], z[index-1]
hi1 = xi1 - xi0
# calculate cubic
f0 = zi0/(6*hi1)*(xi1-x0)**3 + \
zi1/(6*hi1)*(x0-xi0)**3 + \
(yi1/hi1 - zi1*hi1/6)*(x0-xi0) + \
(yi0/hi1 - zi0*hi1/6)*(xi1-x0)
return f0
此函数给出的结果与 scipy.interpolate 中的 function/class CubicSpline 相同，我们可以在下图中看到。
也可以实现可以这样描述的一阶和二阶解析导数：
f1p = -zi0/(2*hi1)*(xi1-x0)**2 + zi1/(2*hi1)*(x0-xi0)**2 + (yi1/hi1 - zi1*hi1/6) + (yi0/hi1 - zi0*hi1/6)
f2p = zi0/hi1 * (xi1-x0) + zi1/hi1 * (x0-xi0)
然后，很容易验证f2p[0]和f2p[-1]等于0，然后插值函数产生自然样条。
关于自然样条的附加参考：
https://faculty.ksu.edu.sa/sites/default/files/numerical_analysis_9th.pdf#page=167
使用示例：
import matplotlib.pyplot as plt
import numpy as np
x = [-8,-4.19,-3.54,-3.31,-2.56,-2.31,-1.66,-0.96,-0.22,0.62,1.21,3]
y = [-0.01,0.01,0.03,0.04,0.07,0.09,0.16,0.28,0.45,0.65,0.77,1]
x = np.asfarray(x)
y = np.asfarray(y)
plt.scatter(x, y)
x_new= np.linspace(min(x), max(x), 10000)
y_new = cubic_interpolate(x_new, x, y)
plt.plot(x_new, y_new)
from scipy.interpolate import CubicSpline
f = CubicSpline(x, y, bc_type='natural')
plt.plot(x_new, f(x_new), label='ref')
plt.legend()
plt.show()
总而言之，这个更新的算法应该比以前的代码（O（n））更稳定和更快地执行插值。与 numba 或 cython 相关联，它甚至会非常快。最后，它完全独立于Scipy。
重要的是，请注意，与大多数算法一样，有时对数据进行归一化（例如针对大数值或小数值）以获得最佳结果很有用。同样，在此代码中，我不检查 nan 值或有序数据。
无论如何，这次更新对我来说是一个很好的教训，我希望它能帮助到别人。如果您发现任何奇怪的事情，请告诉我。
我有两个列表来描述函数 y(x):
x = [0,1,2,3,4,5]
y = [12,14,22,39,58,77]
我想执行三次样条插值，以便在 x 的域中给定一些值 u，例如
u = 1.25
我能找到 y(u)。
简答：
from scipy import interpolate
def f(x):
x_points = [ 0, 1, 2, 3, 4, 5]
y_points = [12,14,22,39,58,77]
tck = interpolate.splrep(x_points, y_points)
return interpolate.splev(x, tck)
print(f(1.25))
长答案：
scipy 将样条插值中涉及的步骤分成两个操作，很可能是为了提高计算效率。
计算描述样条曲线的系数， 使用 splrep()。 splrep returns 一个元组数组，包含 系数.
这些系数被传入splev()来实际 在所需点
x评估样条曲线（在本例中为 1.25）。x也可以是数组。调用f([1.0, 1.25, 1.5])returns 分别在1、1.25和1,5处插入点。
这种方法对于单次求值来说确实不方便，但由于最常见的用例是从少数几个函数求值点开始，然后反复使用样条曲线求插值，在实践中通常非常有用.
最小python3代码：
from scipy import interpolate
if __name__ == '__main__':
x = [ 0, 1, 2, 3, 4, 5]
y = [12,14,22,39,58,77]
# tck : tuple (t,c,k) a tuple containing the vector of knots,
# the B-spline coefficients, and the degree of the spline.
tck = interpolate.splrep(x, y)
print(interpolate.splev(1.25, tck)) # Prints 15.203125000000002
print(interpolate.splev(...other_value_here..., tck))
基于 cwhy 的评论和 youngmit 的回答
万一scipy没有安装：
import numpy as np
from math import sqrt
def cubic_interp1d(x0, x, y):
"""
Interpolate a 1-D function using cubic splines.
x0 : a float or an 1d-array
x : (N,) array_like
A 1-D array of real/complex values.
y : (N,) array_like
A 1-D array of real values. The length of y along the
interpolation axis must be equal to the length of x.
Implement a trick to generate at first step the cholesky matrice L of
the tridiagonal matrice A (thus L is a bidiagonal matrice that
can be solved in two distinct loops).
additional ref: www.math.uh.edu/~jingqiu/math4364/spline.pdf
"""
x = np.asfarray(x)
y = np.asfarray(y)
# remove non finite values
# indexes = np.isfinite(x)
# x = x[indexes]
# y = y[indexes]
# check if sorted
if np.any(np.diff(x) < 0):
indexes = np.argsort(x)
x = x[indexes]
y = y[indexes]
size = len(x)
xdiff = np.diff(x)
ydiff = np.diff(y)
# allocate buffer matrices
Li = np.empty(size)
Li_1 = np.empty(size-1)
z = np.empty(size)
# fill diagonals Li and Li-1 and solve [L][y] = [B]
Li[0] = sqrt(2*xdiff[0])
Li_1[0] = 0.0
B0 = 0.0 # natural boundary
z[0] = B0 / Li[0]
for i in range(1, size-1, 1):
Li_1[i] = xdiff[i-1] / Li[i-1]
Li[i] = sqrt(2*(xdiff[i-1]+xdiff[i]) - Li_1[i-1] * Li_1[i-1])
Bi = 6*(ydiff[i]/xdiff[i] - ydiff[i-1]/xdiff[i-1])
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
i = size - 1
Li_1[i-1] = xdiff[-1] / Li[i-1]
Li[i] = sqrt(2*xdiff[-1] - Li_1[i-1] * Li_1[i-1])
Bi = 0.0 # natural boundary
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
# solve [L.T][x] = [y]
i = size-1
z[i] = z[i] / Li[i]
for i in range(size-2, -1, -1):
z[i] = (z[i] - Li_1[i-1]*z[i+1])/Li[i]
# find index
index = x.searchsorted(x0)
np.clip(index, 1, size-1, index)
xi1, xi0 = x[index], x[index-1]
yi1, yi0 = y[index], y[index-1]
zi1, zi0 = z[index], z[index-1]
hi1 = xi1 - xi0
# calculate cubic
f0 = zi0/(6*hi1)*(xi1-x0)**3 + \
zi1/(6*hi1)*(x0-xi0)**3 + \
(yi1/hi1 - zi1*hi1/6)*(x0-xi0) + \
(yi0/hi1 - zi0*hi1/6)*(xi1-x0)
return f0
if __name__ == '__main__':
import matplotlib.pyplot as plt
x = np.linspace(0, 10, 11)
y = np.sin(x)
plt.scatter(x, y)
x_new = np.linspace(0, 10, 201)
plt.plot(x_new, cubic_interp1d(x_new, x, y))
plt.show()
如果您安装了 scipy 版本 >= 0.18.0，您可以使用 scipy.interpolate 中的 CubicSpline 函数进行三次样条插值。
您可以通过 运行 遵循 python 中的命令检查 scipy 版本：
#!/usr/bin/env python3
import scipy
scipy.version.version
如果您的 scipy 版本 >= 0.18.0，您可以 运行 下面的三次样条插值示例代码：
#!/usr/bin/env python3
import numpy as np
from scipy.interpolate import CubicSpline
# calculate 5 natural cubic spline polynomials for 6 points
# (x,y) = (0,12) (1,14) (2,22) (3,39) (4,58) (5,77)
x = np.array([0, 1, 2, 3, 4, 5])
y = np.array([12,14,22,39,58,77])
# calculate natural cubic spline polynomials
cs = CubicSpline(x,y,bc_type='natural')
# show values of interpolation function at x=1.25
print('S(1.25) = ', cs(1.25))
## Aditional - find polynomial coefficients for different x regions
# if you want to print polynomial coefficients in form
# S0(0<=x<=1) = a0 + b0(x-x0) + c0(x-x0)^2 + d0(x-x0)^3
# S1(1< x<=2) = a1 + b1(x-x1) + c1(x-x1)^2 + d1(x-x1)^3
# ...
# S4(4< x<=5) = a4 + b4(x-x4) + c5(x-x4)^2 + d5(x-x4)^3
# x0 = 0; x1 = 1; x4 = 4; (start of x region interval)
# show values of a0, b0, c0, d0, a1, b1, c1, d1 ...
cs.c
# Polynomial coefficients for 0 <= x <= 1
a0 = cs.c.item(3,0)
b0 = cs.c.item(2,0)
c0 = cs.c.item(1,0)
d0 = cs.c.item(0,0)
# Polynomial coefficients for 1 < x <= 2
a1 = cs.c.item(3,1)
b1 = cs.c.item(2,1)
c1 = cs.c.item(1,1)
d1 = cs.c.item(0,1)
# ...
# Polynomial coefficients for 4 < x <= 5
a4 = cs.c.item(3,4)
b4 = cs.c.item(2,4)
c4 = cs.c.item(1,4)
d4 = cs.c.item(0,4)
# Print polynomial equations for different x regions
print('S0(0<=x<=1) = ', a0, ' + ', b0, '(x-0) + ', c0, '(x-0)^2 + ', d0, '(x-0)^3')
print('S1(1< x<=2) = ', a1, ' + ', b1, '(x-1) + ', c1, '(x-1)^2 + ', d1, '(x-1)^3')
print('...')
print('S5(4< x<=5) = ', a4, ' + ', b4, '(x-4) + ', c4, '(x-4)^2 + ', d4, '(x-4)^3')
# So we can calculate S(1.25) by using equation S1(1< x<=2)
print('S(1.25) = ', a1 + b1*0.25 + c1*(0.25**2) + d1*(0.25**3))
# Cubic spline interpolation calculus example
# https://www.youtube.com/watch?v=gT7F3TWihvk
如果你想要一个无依赖性的解决方案，就把它放在这里。
代码取自上面的答案：
def my_cubic_interp1d(x0, x, y):
"""
Interpolate a 1-D function using cubic splines.
x0 : a 1d-array of floats to interpolate at
x : a 1-D array of floats sorted in increasing order
y : A 1-D array of floats. The length of y along the
interpolation axis must be equal to the length of x.
Implement a trick to generate at first step the cholesky matrice L of
the tridiagonal matrice A (thus L is a bidiagonal matrice that
can be solved in two distinct loops).
additional ref: www.math.uh.edu/~jingqiu/math4364/spline.pdf
# original function code at:
This function is licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/
Original Author raphael valentin
Date 3 Jan 2018
Modifications made to remove numpy dependencies:
-all sub-functions by MR
This function, and all sub-functions, are licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
Mod author: Matthew Rowles
Date 3 May 2021
"""
def diff(lst):
"""
numpy.diff with default settings
"""
size = len(lst)-1
r = [0]*size
for i in range(size):
r[i] = lst[i+1] - lst[i]
return r
def list_searchsorted(listToInsert, insertInto):
"""
numpy.searchsorted with default settings
"""
def float_searchsorted(floatToInsert, insertInto):
for i in range(len(insertInto)):
if floatToInsert <= insertInto[i]:
return i
return len(insertInto)
return [float_searchsorted(i, insertInto) for i in listToInsert]
def clip(lst, min_val, max_val, inPlace = False):
"""
numpy.clip
"""
if not inPlace:
lst = lst[:]
for i in range(len(lst)):
if lst[i] < min_val:
lst[i] = min_val
elif lst[i] > max_val:
lst[i] = max_val
return lst
def subtract(a,b):
"""
returns a - b
"""
return a - b
size = len(x)
xdiff = diff(x)
ydiff = diff(y)
# allocate buffer matrices
Li = [0]*size
Li_1 = [0]*(size-1)
z = [0]*(size)
# fill diagonals Li and Li-1 and solve [L][y] = [B]
Li[0] = sqrt(2*xdiff[0])
Li_1[0] = 0.0
B0 = 0.0 # natural boundary
z[0] = B0 / Li[0]
for i in range(1, size-1, 1):
Li_1[i] = xdiff[i-1] / Li[i-1]
Li[i] = sqrt(2*(xdiff[i-1]+xdiff[i]) - Li_1[i-1] * Li_1[i-1])
Bi = 6*(ydiff[i]/xdiff[i] - ydiff[i-1]/xdiff[i-1])
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
i = size - 1
Li_1[i-1] = xdiff[-1] / Li[i-1]
Li[i] = sqrt(2*xdiff[-1] - Li_1[i-1] * Li_1[i-1])
Bi = 0.0 # natural boundary
z[i] = (Bi - Li_1[i-1]*z[i-1])/Li[i]
# solve [L.T][x] = [y]
i = size-1
z[i] = z[i] / Li[i]
for i in range(size-2, -1, -1):
z[i] = (z[i] - Li_1[i-1]*z[i+1])/Li[i]
# find index
index = list_searchsorted(x0,x)
index = clip(index, 1, size-1)
xi1 = [x[num] for num in index]
xi0 = [x[num-1] for num in index]
yi1 = [y[num] for num in index]
yi0 = [y[num-1] for num in index]
zi1 = [z[num] for num in index]
zi0 = [z[num-1] for num in index]
hi1 = list( map(subtract, xi1, xi0) )
# calculate cubic - all element-wise multiplication
f0 = [0]*len(hi1)
for j in range(len(f0)):
f0[j] = zi0[j]/(6*hi1[j])*(xi1[j]-x0[j])**3 + \
zi1[j]/(6*hi1[j])*(x0[j]-xi0[j])**3 + \
(yi1[j]/hi1[j] - zi1[j]*hi1[j]/6)*(x0[j]-xi0[j]) + \
(yi0[j]/hi1[j] - zi0[j]*hi1[j]/6)*(xi1[j]-x0[j])
return f0
如果要获取值
from scipy.interpolate import CubicSpline
import numpy as np
x = [-5,-4.19,-3.54,-3.31,-2.56,-2.31,-1.66,-0.96,-0.22,0.62,1.21,3]
y = [-0.01,0.01,0.03,0.04,0.07,0.09,0.16,0.28,0.45,0.65,0.77,1]
value = 2
#ascending order
if np.any(np.diff(x) < 0):
indexes = np.argsort(x).astype(int)
x = np.array(x)[indexes]
y = np.array(y)[indexes]
f = CubicSpline(x, y, bc_type='natural')
specificVal = f(value).item(0) #f(value) is numpy.ndarray!!
print(specificVal)
如果要绘制插值函数。
np.linspace第三个参数增加“准确度”。
from scipy.interpolate import CubicSpline
import numpy as np
import matplotlib.pyplot as plt
x = [-5,-4.19,-3.54,-3.31,-2.56,-2.31,-1.66,-0.96,-0.22,0.62,1.21,3]
y = [-0.01,0.01,0.03,0.04,0.07,0.09,0.16,0.28,0.45,0.65,0.77,1]
#ascending order
if np.any(np.diff(x) < 0):
indexes = np.argsort(x).astype(int)
x = np.array(x)[indexes]
y = np.array(y)[indexes]
f = CubicSpline(x, y, bc_type='natural')
x_new = np.linspace(min(x), max(x), 100)
y_new = f(x_new)
plt.plot(x_new, y_new)
plt.scatter(x, y)
plt.title('Cubic Spline Interpolation')
plt.show()
输出：
在我之前的post中，我写了一个基于Cholesky开发的代码来求解三次算法生成的矩阵。不幸的是，由于平方根函数，它可能在某些点集（通常是 non-uniform 点集）上表现不佳。
本着与之前相同的精神，还有另一种使用 Thomas 算法 (TDMA) 的想法（参见 https://en.wikipedia.org/wiki/Tridiagonal_matrix_algorithm) to solve partially the tridiagonal matrix during its definition loop. However, the condition to use TDMA is that it requires at least that the matrix shall be diagonally dominant. However, in our case, it shall be true since |bi| > |ai| + |ci| with ai = h[i], bi = 2*(h[i]+h[i+1]), ci = h[i+1], with h[i] unconditionally positive. (see https://www.cfd-online.com/Wiki/Tridiagonal_matrix_algorithm_-TDMA(Thomas_algorithm)
我又参考了jingqiu的文档（见我之前的post，可惜link坏了，不过在网络缓存中还是可以找到的）。
TDMA求解器的优化版本可以描述如下：
def TDMAsolver(a,b,c,d):
""" This function is licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/
Author raphael valentin
Date 25 Mar 2022
ref. https://www.cfd-online.com/Wiki/Tridiagonal_matrix_algorithm_-_TDMA_(Thomas_algorithm)
"""
n = len(d)
w = np.empty(n-1,float)
g = np.empty(n, float)
w[0] = c[0]/b[0]
g[0] = d[0]/b[0]
for i in range(1, n-1):
m = b[i] - a[i-1]*w[i-1]
w[i] = c[i] / m
g[i] = (d[i] - a[i-1]*g[i-1]) / m
g[n-1] = (d[n-1] - a[n-2]*g[n-2]) / (b[n-1] - a[n-2]*w[n-2])
for i in range(n-2, -1, -1):
g[i] = g[i] - w[i]*g[i+1]
return g
当可以得到ai、bi、ci、di的每个个体时，组合自然三次样条的定义就变得容易了这 2 个单循环中的插值器函数。
def cubic_interpolate(x0, x, y):
""" Natural cubic spline interpolate function
This function is licenced under: Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/
Author raphael valentin
Date 25 Mar 2022
"""
xdiff = np.diff(x)
dydx = np.diff(y)
dydx /= xdiff
n = size = len(x)
w = np.empty(n-1, float)
z = np.empty(n, float)
w[0] = 0.
z[0] = 0.
for i in range(1, n-1):
m = xdiff[i-1] * (2 - w[i-1]) + 2 * xdiff[i]
w[i] = xdiff[i] / m
z[i] = (6*(dydx[i] - dydx[i-1]) - xdiff[i-1]*z[i-1]) / m
z[-1] = 0.
for i in range(n-2, -1, -1):
z[i] = z[i] - w[i]*z[i+1]
# find index (it requires x0 is already sorted)
index = x.searchsorted(x0)
np.clip(index, 1, size-1, index)
xi1, xi0 = x[index], x[index-1]
yi1, yi0 = y[index], y[index-1]
zi1, zi0 = z[index], z[index-1]
hi1 = xi1 - xi0
# calculate cubic
f0 = zi0/(6*hi1)*(xi1-x0)**3 + \
zi1/(6*hi1)*(x0-xi0)**3 + \
(yi1/hi1 - zi1*hi1/6)*(x0-xi0) + \
(yi0/hi1 - zi0*hi1/6)*(xi1-x0)
return f0
此函数给出的结果与 scipy.interpolate 中的 function/class CubicSpline 相同，我们可以在下图中看到。
也可以实现可以这样描述的一阶和二阶解析导数：
f1p = -zi0/(2*hi1)*(xi1-x0)**2 + zi1/(2*hi1)*(x0-xi0)**2 + (yi1/hi1 - zi1*hi1/6) + (yi0/hi1 - zi0*hi1/6)
f2p = zi0/hi1 * (xi1-x0) + zi1/hi1 * (x0-xi0)
然后，很容易验证f2p[0]和f2p[-1]等于0，然后插值函数产生自然样条。
关于自然样条的附加参考： https://faculty.ksu.edu.sa/sites/default/files/numerical_analysis_9th.pdf#page=167
使用示例：
import matplotlib.pyplot as plt
import numpy as np
x = [-8,-4.19,-3.54,-3.31,-2.56,-2.31,-1.66,-0.96,-0.22,0.62,1.21,3]
y = [-0.01,0.01,0.03,0.04,0.07,0.09,0.16,0.28,0.45,0.65,0.77,1]
x = np.asfarray(x)
y = np.asfarray(y)
plt.scatter(x, y)
x_new= np.linspace(min(x), max(x), 10000)
y_new = cubic_interpolate(x_new, x, y)
plt.plot(x_new, y_new)
from scipy.interpolate import CubicSpline
f = CubicSpline(x, y, bc_type='natural')
plt.plot(x_new, f(x_new), label='ref')
plt.legend()
plt.show()
总而言之，这个更新的算法应该比以前的代码（O（n））更稳定和更快地执行插值。与 numba 或 cython 相关联，它甚至会非常快。最后，它完全独立于Scipy。
重要的是，请注意，与大多数算法一样，有时对数据进行归一化（例如针对大数值或小数值）以获得最佳结果很有用。同样，在此代码中，我不检查 nan 值或有序数据。
无论如何，这次更新对我来说是一个很好的教训，我希望它能帮助到别人。如果您发现任何奇怪的事情，请告诉我。