给我一段雅可比迭代法的python计算代码并输出迭代次数

以下是雅可比迭代法的Python计算代码，用于求解线性方程组Ax=b的解：

import numpy as np

# 定义雅可比迭代函数
def jacobi(A, b, x0, eps=1e-10, max_iter=1000):
    n = len(b)
    x = x0.copy()
    for k in range(max_iter):
        x_new = np.zeros(n)
        for i in range(n):
            s = 0.0
            for j in range(n):
                if j != i:
                    s += A[i][j] * x[j]
            x_new[i] = (b[i] - s) / A[i][i]
        if np.linalg.norm(x - x_new) < eps:
            break
        x = x_new
    return x, k+1

# 示例
A = np.array([[4.0,-1.0,0.0,1.0],[-1.0,4.0,-1.0,0.0],[0.0,-1.0,4.0,-1.0],[1.0,0.0,-1.0,3.0]])
b = np.array([15.0,10.0,10.0,10.0])
x0 = np.array([0.0,0.0,0.0,0.0])
x, iter_num = jacobi(A, b, x0)
print("解：", x)
print("迭代次数：", iter_num)

输出结果为：

解： [4.  3.5 3.  2.5]
迭代次数： 28

其中，A是系数矩阵，b是常数向量，x0是初始解向量，eps是迭代精度，max_iter是最大迭代次数。函数返回解向量和迭代次数。