机器学习——吴恩达 ——编程作业ex1——python

断断续续看了视频，一直想抽时间编程，找到作业后一通操作，先用Octave搞了下，又花了几个小时用python做了下吴恩达机器学习的ex1第一部分，有所收获，分享至此

代码如下;
main.py

import numpy as np
import matplotlib.pyplot as plt
import function
from mpl_toolkits.mplot3d import Axes3D

# 读取数据
filename = 'ex1data1.txt'
data = np.loadtxt(filename, delimiter=',')  # 由于文件用逗号分割数据，故指定分割数据的符号

# 把数据赋值给矩阵X和Y
X = data[:, 0]
Y = data[:, 1]
# print(np.shape(X))
# 获取训练数据长度
m = len(Y)

# 绘制原始数据的分布
plt.plot(X, Y, 'rx')
plt.xlabel('Population of City in 10,000s')
plt.ylabel('Profit in $10,000s')

# os.system('pause')

# 补充X的第一列为1
X = np.c_[np.ones([len(X), 1]), X]
print(X)

# 初始化参数
theta = np.zeros(2)
# theta = [-1, 2]
alpha = 0.01
iteras = 1500

# 计算代价函数
cost_result = function.compute_cost(X, Y, theta)
print(cost_result)

# 更新theta值
theta = function.update_grad(X, Y, theta, alpha, iteras)
print(theta)

# print(np.shape(theta))
# 绘制拟合曲线
plt.plot(X[:, 1], X @ theta, '-')

# 结果测试
predict1 = np.dot([1, 3.5], theta)
print('预测值为：', predict1 * 10000)

# 可视化理解
theta0_vals = np.linspace(-10, 10, 100)
theta1_vals = np.linspace(-1, 4, 100)

J_vals = np.zeros([len(theta0_vals), len(theta1_vals)])

for i in range(len(theta0_vals)):
    for j in range(len(theta1_vals)):
        t = [theta0_vals[i], theta1_vals[j]]
        J_vals[i, j] = function.compute_cost(X, Y, t)

J_vals = J_vals.T

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(theta0_vals, theta1_vals, J_vals, cmap='rainbow')
plt.title('3d图像', fontproperties='SimHei', fontsize=20)
plt.xlabel('theta0_vals')
plt.ylabel('theta1_vals')
plt.show()

# 绘制等值图
plt.figure()
plt.contourf(theta0_vals, theta1_vals, J_vals, 10)
plt.contour(theta0_vals, theta1_vals, J_vals, colors='black')
plt.plot(theta[0], theta[1], 'rx', markerSize=20, LineWidth=1)
plt.title('等值线图', fontproperties='SimHei', fontsize=20)
plt.xlabel('theta0')
plt.ylabel('theta1')
plt.show()

function.py

import numpy as np


def compute_cost(X, Y, theta):
    m = len(Y)
    H = np.dot(X, theta)
    J = np.sum((H - Y) ** 2) / (2 * m)
    return J


def update_grad(X, Y, theta, alpha,iteras):
    m = len(Y)
    J_history = np.zeros([iteras,1])
    for i in range(iteras):
        theta = theta - (alpha / m) * (X.T @ (X @ theta - Y))
        J_history[i] = compute_cost(X, Y, theta)
    return theta

效果：

图一：

图二：

图三：