手寫二元線性迴歸梯度下降—Python代碼

代碼思路

1. 導入所需庫
2. 導入所需數據
3. 數據轉換—提取X,Y
4. 設置初始值
   事先指定係數值、學習率、迭代次數
5. 定義損失函數(二元線性迴歸)
   $J(\theta_i)=\frac{1}{2m}\sum_{i=1}^{m}[h_\theta(x^{(i)})-y^{(i)}]^2$其中$h_\theta(x^{(i)})=\theta_0+\theta_1x^{(1)}+\theta_2x^{(2)}$
6. 求解梯度下降
   $\frac{\partial J(\theta_i)}{\partial\theta_i}=\frac{1}{m}\sum_{i=1}^{m}(h_\theta(x^{(i)})-y^{(i)})*x^{(i)}=\frac{1}{m}\sum_{i=1}^{m}(\theta_0+\theta_1x^{(1)}+\theta_2x^{(2)}-y^{(i)})*x^{(i)}$

代碼實現

1. 導入所需庫

import numpy as np 

2. 導入所需數據

data = np.genfromtxt("線性迴歸/data/Delivery.csv",delimiter=',')

3. 數據轉換——提取X,Y

X = data[:,:-1]
Y = data[:,-1]

4. 設置初始值

theta0 = 0
theta1 = 0
theta2 = 0
epochs = 100
alpha = 0.001

5. 定義損失函數（根據上述公式編寫）

def cost(X,Y,theta0,theta1,theta2):
    loss = 0
    m = len(Y)
    for i in range(m):
        loss += (theta0+theta1*X[i,0]+theta2*X[i,1]-Y[i])**2
    loss = loss/(2*m)
    return loss

6. 定義梯度下降（根據上述公式編寫）

def grad_des(X,Y,theta0,theta1,theta2,alpha,epochs):
    m = len(Y)
    for z in range(epochs):
        theta0_grad = 0
        theta1_grad = 0
        theta2_grad = 0
        for i in range(m):
            theta0_grad = (theta0+theta1*X[i,0]+theta2*X[i,1]-Y[i])
            theta1_grad = (theta0+theta1*X[i,0]+theta2*X[i,1]-Y[i])*X[i,0]
            theta2_grad = (theta0+theta1*X[i,0]+theta2*X[i,1]-Y[i])*X[i,1]
        theta0_grad = theta0_grad/m
        theta1_grad = theta1_grad/m
        theta2_grad = theta2_grad/m
    theta0 -=alpha*theta0_grad
    theta1 -=alpha*theta1_grad
    theta2 -=alpha*theta2_grad
    return theta0,theta1,theta2

7. 輸出結果（係數值和損失值）

print('begin...theta0={},theta1={},theta2={},loss={}'.format(theta0,theta1,theta2,cost(X,Y,theta0,theta1,theta2)))
print('running...')
theta0,theta1,theta2 = grad_des(X,Y,theta0,theta1,theta2,alpha,epochs)
print('end...theta0={},theta1={},theta2={},loss={}'.format(theta0,theta1,theta2,cost(X,Y,theta0,theta1,theta2)))

數據資料

X1	X2	Y
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