python 用雅可比方法求解稀疏矩阵，完全不会写，求助a

import numpy as np
def Jacobi(A, b, iter_n, initial_guess=0):
n = len(A)

D = np.diag(A)
R = A - np.diag(D)
x_i = initial_guess * np.ones(n)
for i in range(iter_n):
    print('x_',i,'=',x_i)
    x_i = (b - R.dot(x_i)) / D

return x_i

def A_ij(n):
A = np.empty((n, n))
for i in range(n):
A[i, i] = 2
for i in range(n-1):
A[i,i+1]=A[i+1,i]=1
return A
def b_i(n):
b=np.empty(n)
b[0]=1
b[n-1]=-1
return b
def x0(n):
return np.zeros(n)
print(Jacobi(A,b,100,x0))

# -*- coding: utf-8 -*-

#Jacobi迭代法 输入系数矩阵mx、值矩阵mr、迭代次数n、误差c(以list模拟矩阵 行优先)

def Jacobi(mx,mr,n=100,c=0.0001):
    if len(mx) == len(mr):  #若mx和mr长度相等则开始迭代 否则方程无解
        x = [] #迭代初值 初始化为单行全0矩阵
        for i in range(len(mr)):
            x.append([0])
        count = 0 #迭代次数计数
        while count < n:
            nx = [] #保存单次迭代后的值的集合
            for i in range(len(x)):
                nxi = mr[i][0]
                for j in range(len(mx[i])):
                    if j!=i:
                        nxi = nxi+(-mx[i][j])*x[j][0]
                nxi = nxi/mx[i][i]
                nx.append([nxi]) #迭代计算得到的下一个xi值
            lc = [] #存储两次迭代结果之间的误差的集合
            for i in range(len(x)):
                lc.append(abs(x[i][0]-nx[i][0]))
            if max(lc) < c:
                return nx #当误差满足要求时 返回计算结果
            x = nx
            count = count + 1
        return False #若达到设定的迭代结果仍不满足精度要求 则方程无解
    else:
        return False

#调用 Jacobi(mx,mr,n=100,c=0.001) 示例
mx = [[8,-3,2],[4,11,-1],[6,3,12]]

mr = [[20],[33],[36]]
print(Jacobi(mx,mr,100,0.00001))

https://blog.csdn.net/cswfqxs_/article/details/84067711