小编给大家分享一下如何使用Python实现Newton插值法，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！
1. n阶差商实现
def diff(xi,yi,n): """ param xi:插值节点xi param yi:插值节点yi param n: 求几阶差商 return: n阶差商 """ if len(xi) != len(yi): #xi和yi必须保证长度一致 return else: diff_quot = [[] for i in range(n)] for j in range(1,n+1): if j == 1: for i in range(n+1-j): diff_quot[j-1].append((yi[i]-yi[i+1]) / (xi[i] - xi[i + 1])) else: for i in range(n+1-j): diff_quot[j-1].append((diff_quot[j-2][i]-diff_quot[j-2][i+1]) / (xi[i] - xi[i + j])) return diff_quot测试一下：
xi = [1.615,1.634,1.702,1.828]yi = [2.41450,2.46259,2.65271,3.03035]n = 3print(diff(xi,yi,n))返回的差商结果为：
[[2.53105263157897, 2.7958823529411716, 2.997142857142854], [3.0440197857724347, 1.0374252793901158], [-9.420631485362996]]
2. 牛顿插值实现
def Newton(x): f = yi[0] v = [] r = 1 for i in range(n): r *= (x - xi[i]) v.append(r) f += diff_quot[i][0] * v[i] return f测试一下：
x = 1.682print(Newton(x))结果为：
5944760289639732
完整Python代码
def Newton(xi,yi,n,x): """ param xi:插值节点xi param yi:插值节点yi param n: 求几阶差商 param x: 代求近似值 return: n阶差商 """ if len(xi) != len(yi): #xi和yi必须保证长度一致 return else: diff_quot = [[] for i in range(n)] for j in range(1,n+1): if j == 1: for i in range(n+1-j): diff_quot[j-1].append((yi[i]-yi[i+1]) / (xi[i] - xi[i + 1])) else: for i in range(n+1-j): diff_quot[j-1].append((diff_quot[j-2][i]-diff_quot[j-2][i+1]) / (xi[i] - xi[i + j])) print(diff_quot) f = yi[0] v = [] r = 1 for i in range(n): r *= (x - xi[i]) v.append(r) f += diff_quot[i][0] * v[i] return f以上是“如何使用Python实现Newton插值法”这篇文章的所有内容，感谢各位的阅读！相信大家都有了一定的了解，希望分享的内容对大家有所帮助，如果还想学习更多知识，欢迎关注编程网行业资讯频道！
