三次样条插值法
使用的是公式法迭代，没有用牛顿，我认为更加精准，牛顿只是方便手算误差自然大。
import time
import numpy as np
import sympy
from sympy import symbols, plot_implicit, Eq
from fractions import Fraction
import matplotlib.pyplot as plt
'''
程序名称：三次样条插值算法程序
程序功能：解决三种三次样条插值问题
程序作者：Yaung
'''
# 四舍五入函数
def round_up(n, m):
n = str(n)
if len(n) - n.index(".") - 1 == m + 1:
n += "01"
n = float(n)
return np.round(n, m)
while True:
# 界面展示
print("\t**********第一类固定边界(输入:1)")
print("\t\tS'(x0)=f0'\tS'(xn)=fn'")
print("\t**********第二类自由边界(输入:2)")
print("\t\tS''(x0)=f0''\tS''(xn)=fn''")
print("\t**********第三类非节点边界(输入:3)")
print("\t\tlimSp(x0+)=limSp(xn-)\tp=0,1,2")
print("\t**********退出程序(输入:4)")
# 选项输入
choice = eval(input('请输入你的选项数字:'))
if choice == 4:
exit() # 退出程序
# 输入数据的个数
N = eval(input('请输入数据的个数:'))
arr = input('请输入xk的所有值(每个值用空格隔开):')
X = np.array([float(i) for i in arr.split()])
arr = input('请输入每个xk所对应的函数值f(xk)(每个值用空格隔开):')
Y = np.array([float(i) for i in arr.split()])
C = np.array([0, 0])
if choice != 3:
arr = input('请输入两个边界条件(每个值用空格隔开):')
C = np.array([float(i) for i in arr.split()])
'''
测试
第二类
>>
2
4
1 2 4 5
1 3 4 2
0 0
3
<<
4.25
'''
# 基础公式
# 计算h
H = np.array([])
for i in range(0, N - 1):
H = np.r_[H, X[i + 1] - X[i]]
# 计算U
U = np.array([np.max])
for i in range(1, N - 1):
U = np.r_[U, round_up(H[i - 1] / (H[i] + H[i - 1]), 6)]
# 计算R
R = np.array([np.max])
for i in range(1, N - 1):
R = np.r_[R, round_up(H[i] / (H[i] + H[i - 1]), 6)]
# 计算G
G = np.array([3 * (Y[1] - Y[0]) / H[0] - H[0] / 2 * C[0]]) # 一开始第一个先按照第二类初始化
for i in range(1, N - 1):
# print(3*(U[0,i]*(Y[i+1]-Y[i])+R[0,i]*(Y[i]-Y[i-1])))
G = np.r_[G, 3 * (U[i] * (Y[i + 1] - Y[i]) / H[i] + R[i] * (Y[i] - Y[i - 1]) / H[i - 1])]
# 边界类型判断
if choice == 1:
# 第一类固定边界条件
# 求解方程组
A1 = np.array([[]])
for i in range(1, N - 1):
Ai = np.array([])
Ai = np.r_[Ai, [0 for j in range(i - 2)]]
if i > 1:
Ai = np.r_[Ai, R[i]]
Ai = np.r_[Ai, 2]
if i != N - 2:
Ai = np.r_[Ai, U[i]]
Ai = np.r_[Ai, [0 for j in range(N - 2 - Ai.size)]]
if i == 1:
A1 = np.c_[A1, [Ai]]
else:
A1 = np.r_[A1, [Ai]]
b1 = np.array([G[1] - R[1] * C[0]])
b1 = np.r_[b1, [G[i] for i in range(2, N - 2)]]
b1 = np.r_[b1, G[N - 2] - U[N - 2] * C[1]]
M = np.array([C[0]])
M = np.r_[M, np.linalg.solve(A1, b1)]
M = np.r_[M, C[1]]
elif choice == 2:
# 第二类自由边界条件
# 补充最后一个G
G = np.r_[G, 3 * (Y[N - 1] - Y[N - 2]) / H[N - 2] + H[N - 2] / 2 * C[1]]
# 解方程组求M
A2 = np.array([[2, 1]])
A2 = np.c_[A2, [[0 for i in range(N - 2)]]]
for i in range(1, N - 1):
Ai = np.array([])
Ai = np.r_[Ai, [0 for j in range(i - 1)]]
Ai = np.r_[Ai, [R[i], 2, U[i]]]
Ai = np.r_[Ai, [0 for j in range(N - Ai.size)]]
A2 = np.r_[A2, [Ai]]
# A2 = np.r_[A2,[0 for i in range(N-2)]]
A2 = np.r_[A2, [np.r_[[0 for i in range(N - 2)], [1, 2]]]]
b2 = np.array([G[i] for i in range(N)])
M = np.array(np.linalg.solve(A2, b2))
elif choice == 3:
# 第三类非节点边界条件9
# 新增U，R，G的最后一个值
U = np.r_[U, H[N - 2] / (H[0] + H[N - 2])]
R = np.r_[R, H[0] / (H[0] + H[N - 2])]
G = np.r_[G, 3 * (U[N - 1] * (Y[1] - Y[0]) / H[0] + R[N - 1] * (Y[N - 1] - Y[N - 2]) / H[N - 2])]
# 解方程组求M
A3 = np.array([[]])
for i in range(1, N):
Ai = np.array([])
if i == N - 1:
Ai = np.r_[Ai, U[N - 1]]
Ai = np.r_[Ai, [0 for j in range(i - 3)]]
else:
Ai = np.r_[Ai, [0 for j in range(i - 2)]]
if i > 1:
Ai = np.r_[Ai, R[i]]
Ai = np.r_[Ai, 2]
if i != N - 1:
Ai = np.r_[Ai, U[i]]
if i == 1:
Ai = np.r_[Ai, [0 for j in range(N - 2 - Ai.size)]]
Ai = np.r_[Ai, R[1]]
else:
Ai = np.r_[Ai, [0 for j in range(N - 1 - Ai.size)]]
if i == 1:
A3 = np.c_[A3, [Ai]]
else:
A3 = np.r_[A3, [Ai]]
b3 = np.array([G[i] for i in range(1, N)])
M = np.array(np.linalg.solve(A3, b3))
M = np.r_[M[N - 2], M]
# 求出全部表达式
x = sympy.symbols("x") # 申明未知数"x"
S = np.array([])
for i in range(X.size - 1):
S = np.r_[S, [(H[i] + 2 * (x - X[i])) / np.power(H[i], 3) * np.power(x - X[i + 1], 2) * Y[i] + (
H[i] - 2 * (x - X[i + 1])) / np.power(H[i], 3) * np.power(x - X[i], 2) * Y[i + 1] + (
x - X[i]) * np.power(x - X[i + 1], 2) / np.power(H[i], 2) * M[i] + (
x - X[i + 1]) * np.power(x - X[i], 2) / np.power(H[i], 2) * M[i + 1]]]
while True:
# 输入预测值
x1 = eval(input('请输入需要预测的值:'))
xl = 0
xlid = 0
xr = 0
xrid = 0
for i in range(X.size):
if X[i] > x1:
xr = X[i]
xrid = i
xl = X[i - 1]
xlid = i - 1
break
y = (H[xlid] + 2 * (x - X[xlid])) / np.power(H[xlid], 3) * np.power(x - X[xrid], 2) * Y[xlid] + (
H[xlid] - 2 * (x - X[xrid])) / np.power(H[xlid], 3) * np.power(x - X[xlid], 2) * Y[xrid] + (
x - X[xlid]) * np.power(x - X[xrid], 2) / np.power(H[xlid], 2) * M[xlid] + (
x - X[xrid]) * np.power(x - X[xlid], 2) / np.power(H[xlid], 2) * M[xrid]
y1 = y.evalf(subs={x: x1})
# 打印数据
print("方程组的解为：")
print(M)
print("三次样条插值的表达式为：")
print(S)
# 打印预测值
print("预测值为：")
print(y1)
# 画图
picture = plt.figure()
# plt.ion()
plt.scatter(X, Y, marker='.', c='b')
# plt.pause(0.01)
# 画出预测值
plt.scatter(x1, y1, marker='.', c='r')
# plt.pause(0.01)
# 画函数曲线
for i in range(S.size):
XX = np.arange(X[i], X[i + 1], 0.01)
XX = np.array(XX)
YY = np.array([])
for j in range(XX.size):
Z = S[i]
K = Z.evalf(subs={x: XX[j]})
YY = np.r_[YY, K]
plt.plot(XX, YY, color='k')
# plt.pause(0.01)
# plt.pause(0.01)
# plt.ioff() # 关闭interactive mode
plt.show(block=True)
tmpFlag = eval(input('输入\'1\'继续预测，输入\'2\'重新执行程序。'))
if tmpFlag != 1:
plt.close()
break
plt.close()
tmpFlag = eval(input('输入\'1\'继续程序，输入\'2\'退出程序。'))
if tmpFlag != 1:
break
测试样例
# 第二类
>>
2
4
1 2 4 5
1 3 4 2
0 0
3
<<
4.25
----以上为个人思考与见解，有误请指点，有想法也可联系交流！文章来源：https://www.toymoban.com/news/detail-721249.html
~~~~~~~~~~~~~~ 谢谢观看！文章来源地址https://www.toymoban.com/news/detail-721249.html
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