三次样条插值Python实现

函数

$$y=\frac{1}{1+x^2}$$

算法分析

• 在每个分段区间上是三次多项式（这就是三次样条中的三次的来源）
• 在整个区间（开区间）上二阶导数连续（当然啦，这里主要是强调在节点上的连续）
• 加上边界条件。边界条件只需要给出两个方程。构建一个方程组，就可以解出所有的参数。

这里话，根据第一类样条作为边界。（就是知道两端节点的导数数值，然后来做三次样条插值）

但是这里也分为两种情况，分别是这个数值是随便给的一个数，还是说根据函数的在对应点上数值给出。

情况一：两边导数数值给出

这里假设数值均为1。即 $f^{\prime }(x_0)=f^{\prime }(x_n)=1$的情况。

情况一图像

情况一代码

import numpy as np from sympy import * import matplotlib.pyplot as plt def f(x): return 1 / (1 + x  2) def cal(begin, end, i): by = f(begin) ey = f(end) I = Ms[i] * ((end - n)  3) / 6 + Ms[i + 1] * ((n - begin)  3) / 6 + (by - Ms[i] / 6) * (end - n) + ( ey - Ms[i + 1] / 6) * (n - begin) return I def ff(x): # f[x0, x1, ..., xk] ans = 0 for i in range(len(x)): temp = 1 for j in range(len(x)): if i != j: temp *= (x[i] - x[j]) ans += f(x[i]) / temp return ans def calM(): lam = [1] + [1 / 2] * 9 miu = [1 / 2] * 9 + [1] # Y = 1 / (1 + n 2) # df = diff(Y, n) x = np.array(range(11)) - 5 # ds = [6 * (ff(x[0:2]) - df.subs(n, x[0]))] ds = [6 * (ff(x[0:2]) - 1)] for i in range(9): ds.append(6 * ff(x[i: i + 3])) # ds.append(6 * (df.subs(n, x[10]) - ff(x[-2:]))) ds.append(6 * (1 - ff(x[-2:]))) Mat = np.eye(11, 11) * 2 for i in range(11): if i == 0: Mat[i][1] = lam[i] elif i == 10: Mat[i][9] = miu[i - 1] else: Mat[i][i - 1] = miu[i - 1] Mat[i][i + 1] = lam[i] ds = np.mat(ds) Mat = np.mat(Mat) Ms = ds * Mat.I return Ms.tolist()[0] def calnf(x): nf = [] for i in range(len(x) - 1): nf.append(cal(x[i], x[i + 1], i)) return nf def calf(f, x): y = [] for i in x: y.append(f.subs(n, i)) return y def nfSub(x, nf): tempx = np.array(range(11)) - 5 dx = [] for i in range(10): labelx = [] for j in range(len(x)): if x[j] >= tempx[i] and x[j] < tempx[i + 1]: labelx.append(x[j]) elif i == 9 and x[j] >= tempx[i] and x[j] <= tempx[i + 1]: labelx.append(x[j]) dx = dx + calf(nf[i], labelx) return np.array(dx) def draw(nf): plt.rcParams['font.sans-serif'] = ['SimHei'] plt.rcParams['axes.unicode_minus'] = False x = np.linspace(-5, 5, 101) y = f(x) Ly = nfSub(x, nf) plt.plot(x, y, label='原函数') plt.plot(x, Ly, label='三次样条插值函数') plt.xlabel('x') plt.ylabel('y') plt.legend() plt.savefig('1.png') plt.show() def lossCal(nf): x = np.linspace(-5, 5, 101) y = f(x) Ly = nfSub(x, nf) Ly = np.array(Ly) temp = Ly - y temp = abs(temp) print(temp.mean()) if __name__ == '__main__': x = np.array(range(11)) - 5 y = f(x) n, m = symbols('n m') init_printing(use_unicode=True) Ms = calM() nf = calnf(x) draw(nf) lossCal(nf) 

情况二：两边导数数值由函数本身算出

这里假设数值均为1。即 $f^{\prime }(x_i)=S^{\prime }(x_i)(i=0,n)$的情况。

情况二图像

情况二代码

import numpy as np from sympy import * import matplotlib.pyplot as plt def f(x): return 1 / (1 + x  2) def cal(begin, end, i): by = f(begin) ey = f(end) I = Ms[i] * ((end - n)  3) / 6 + Ms[i + 1] * ((n - begin)  3) / 6 + (by - Ms[i] / 6) * (end - n) + ( ey - Ms[i + 1] / 6) * (n - begin) return I def ff(x): # f[x0, x1, ..., xk] ans = 0 for i in range(len(x)): temp = 1 for j in range(len(x)): if i != j: temp *= (x[i] - x[j]) ans += f(x[i]) / temp return ans def calM(): lam = [1] + [1 / 2] * 9 miu = [1 / 2] * 9 + [1] Y = 1 / (1 + n  2) df = diff(Y, n) x = np.array(range(11)) - 5 ds = [6 * (ff(x[0:2]) - df.subs(n, x[0]))] # ds = [6 * (ff(x[0:2]) - 1)] for i in range(9): ds.append(6 * ff(x[i: i + 3])) ds.append(6 * (df.subs(n, x[10]) - ff(x[-2:]))) # ds.append(6 * (1 - ff(x[-2:]))) Mat = np.eye(11, 11) * 2 for i in range(11): if i == 0: Mat[i][1] = lam[i] elif i == 10: Mat[i][9] = miu[i - 1] else: Mat[i][i - 1] = miu[i - 1] Mat[i][i + 1] = lam[i] ds = np.mat(ds) Mat = np.mat(Mat) Ms = ds * Mat.I return Ms.tolist()[0] def calnf(x): nf = [] for i in range(len(x) - 1): nf.append(cal(x[i], x[i + 1], i)) return nf def calf(f, x): y = [] for i in x: y.append(f.subs(n, i)) return y def nfSub(x, nf): tempx = np.array(range(11)) - 5 dx = [] for i in range(10): labelx = [] for j in range(len(x)): if x[j] >= tempx[i] and x[j] < tempx[i + 1]: labelx.append(x[j]) elif i == 9 and x[j] >= tempx[i] and x[j] <= tempx[i + 1]: labelx.append(x[j]) dx = dx + calf(nf[i], labelx) return np.array(dx) def draw(nf): plt.rcParams['font.sans-serif'] = ['SimHei'] plt.rcParams['axes.unicode_minus'] = False x = np.linspace(-5, 5, 101) y = f(x) Ly = nfSub(x, nf) plt.plot(x, y, label='原函数') plt.plot(x, Ly, label='三次样条插值函数') plt.xlabel('x') plt.ylabel('y') plt.legend() plt.savefig('1.png') plt.show() def lossCal(nf): x = np.linspace(-5, 5, 101) y = f(x) Ly = nfSub(x, nf) Ly = np.array(Ly) temp = Ly - y temp = abs(temp) print(temp.mean()) if __name__ == '__main__': x = np.array(range(11)) - 5 y = f(x) n, m = symbols('n m') init_printing(use_unicode=True) Ms = calM() nf = calnf(x) draw(nf) lossCal(nf)