机器学习之简单线性回归(梯度下降法)

zyt温柔发 · 发表于 2020-3-15 10:30

0.引入依赖
import numpy as np
import matplotlib.pyplot as plt
1.导入数据
points = np.genfromtxt("data.csv",delimiter=',')
#提取points中的两列数据，分别作为x，y
x = points[:,0]
y = points[:,1]
# 用plt画出散点图
plt.scatter(x,y)
plt.show()
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2.定义损失函数
# 损失函数是系数的函数，另外还要传入数据的x，y
def computer_cost(w,b,points):
total_cost = 0
M = len(points)
# 逐点计算平方损失误差，计算平均数
for i in range(M):
      x = points[i,0]
      y = points[i,1]
      total_cost+= (y-w*x-b)**2
return total_cost/M
3.定义模型超参数
alpha = 0.0001
initial_w = 0
initial_b = 0
num_iter = 10
4.定义核心梯度下降算法函数
def grad_desc(points,initial_w,initial_b,alpha,num_iter):
w = initial_w
b = initial_b
#定义一个列表list保存所有的损失函数值，用来显示下降的过程
cost_list = []
for i in range(num_iter):
      cost_list.append(computer_cost(w,b,points))
      w,b = step_grad_desc(w,b,alpha,points)
return [w,b,cost_list]
def step_grad_desc(current_w,current_b,alpha,points):
sum_grad_w = 0
sum_grad_b = 0
M = len(points)
for i in range(M):
      x = points[i,0]
      y = points[i,1]
      sum_grad_w+=(current_w*x+current_b-y)*x
      sum_grad_b+=current_w*x+current_b-y
# 用公式求当前梯度
grad_w = 2/M *sum_grad_w
grad_b = 2/M*sum_grad_b
# 梯度下降更新当前的w和b
updated_w = current_w - alpha* grad_w
updated_b = current_b - alpha* grad_b
return updated_w,updated_b
5.测试运行梯度下降算法，计算最优w和b
w,b,cost_list = grad_desc(points,initial_w,initial_b,alpha,num_iter)
print('w is:',w)
print('b is:',b)
cost=computer_cost(w,b,points)
print("cost is:",cost)
plt.plot(cost_list)
plt.show()
6.结果：
w is: 1.4774173755483797
b is: 0.02963934787473238
cost is: 112.65585181499748
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