实现Python简单实现jacobi迭代

def Jacobi(mx,mr,n=100,c=0.0001):
    if len(mx) == len(mr): # mx为系数矩阵，mr
        x = []
        for i in range(len(mr)):
            x.append([0])  # x矩阵初始化
            count = 0  # 迭代次数的计算
        while count < n:
            nx = []  #保存单次迭代后的值的集合
            for i in range(len(x)):  # 迭代计算
                nxi = mr[i][0]
                for j in range(len(x)):
                     if j != i:
                        nxi += (-mx[i][j])*x[j][0]
                nxi = nxi/mx[i][i]
                nx.append([nxi])
            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

# 例子
mx = [[8, -3, 2], [4, 11, -1], [6, 3, 12]]

mr = [[20], [33], [36]]

print(Jacobi(mx, mr, 100, 0.00001))

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