机器学习-梯度下降法-多元线性回归

导入要用到的包

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

读入数据

data = genfromtxt(r"G:\work\python\jupyter_notebook_work\机器学习\回归\Delivery.csv",delimiter = ‘,’)
print(data)

[[100. 4. 9.3]
[ 50. 3. 4.8]
[100. 4. 8.9]
[100. 2. 6.5]
[ 50. 2. 4.2]
[ 80. 2. 6.2]
[ 75. 3. 7.4]
[ 65. 4. 6. ]
[ 90. 3. 7.6]
[ 90. 2. 6.1]]

切分数据

x_data = data[:,:-1]
y_data = data[:,-1]
print(x_data)
print(y_data)

[[100. 4.]
[ 50. 3.]
[100. 4.]
[100. 2.]
[ 50. 2.]
[ 80. 2.]
[ 75. 3.]
[ 65. 4.]
[ 90. 3.]
[ 90. 2.]]
[9.3 4.8 8.9 6.5 4.2 6.2 7.4 6. 7.6 6.1]

学习率 learning rate

lr = 0.0001

参数

theta0 = 0
theta1 = 0
theta2 = 0

最大迭代次数

epochs = 1000

最小二乘法

def compute_error(theta0, theta1, theta2, x_data, y_data):
totalError = 0
for i in range(0,len(x_data)):
totalError += (y_data[i] - (theta1 * x_data[i, 0] + theta2 * x_data[i, 1] + theta0)) ** 2
return totalError / float(len(x_data))

def gradient_descent_runner(x_data, y_data, theta0, theta1, theta2, lr, epochs):
# 计算总数据量
m = float(len(x_data))
# 循环epochs次
for i in range(epochs):
theta0_grad = 0
theta1_grad = 0
theta2_grad = 0
# 计算梯度的总和再求平均
for j in range(0, len(x_data)):
theta0_grad += (1/m) * ((theta1 * x_data[j, 0] + theta2 * x_data[j, 1] + theta0) - y_data[j])
theta1_grad += (1/m) * x_data[j, 0] * ((theta1 * x_data[j, 0] + theta2 * x_data[j, 1] + theta0) - y_data[j])
theta2_grad += (1/m) * x_data[j, 1] * ((theta1 * x_data[j, 0] + theta2 * x_data[j, 1] + theta0) - y_data[j])
# 更新b和k
theta0 = theta0 - (lr * theta0_grad)
theta1 = theta1 - (lr * theta1_grad)
theta2 = theta2 - (lr * theta2_grad)
return theta0, theta1, theta2

print(“Starting theta0 = {0}, theta1 = {1}, theta2 = {2}, error = {3}”.format(theta0, theta1, theta2, compute_error(theta0, theta1, theta2, x_data, y_data)))
print(“Running…”)
theta0, theta1, theta2 = gradient_descent_runner(x_data, y_data, theta0, theta1, theta2, lr, epochs)
print(“After {0} iterations theta0 = {1}, theta1 = {2}, theta2 = {3}, error = {4}”.format(epochs, theta0, theta1, theta2, compute_error(theta0, theta1, theta2, x_data, y_data)))

Starting theta0 = 0, theta1 = 0, theta2 = 0, error = 47.279999999999994
Running…
After 1000 iterations theta0 = 0.006971416196678632, theta1 = 0.08021042690771771, theta2 = 0.07611036240566814, error = 0.7731271432218118

ax = plt.figure().add_subplot(111, projection = ‘3d’)
ax.scatter(x_data[:,0], x_data[:, 1], y_data, c = ‘r’, marker = ‘o’, s = 100) # 点为红色三角形
x0 = x_data[:, 0]
x1 = x_data[:, 1]
#生成网格矩阵
x0, x1 = np.meshgrid(x0, x1)
z = theta0 + x0 * theta1 + x1 * theta2

画3D图

ax.plot_surface(x0, x1, z)
#设置坐标轴
ax.set_xlabel(‘Miles’)
ax.set_ylabel(‘Num of Deliveries’)
ax.set_zlabel(‘Time’)

显示图像

plt.show()

网格矩阵示例

x0 , x1 = np.meshgrid([1,2,3], [4,5,6])
‘’’
(1,4) (2,4) (3,4)
(1,5) (2,5) (3,5)
(1,6) (2,6) (3,6)
‘’’
x0

array([[1, 2, 3],
[1, 2, 3],
[1, 2, 3]])

x1

array([[4, 4, 4],
[5, 5, 5],
[6, 6, 6]])

plt.scatter(x0, x1)
plt.show()