[人工智能]Pytorch--简单线性回归

梯度下降
在求解损失函数的最小值时，可以通过梯度下降法来一步步的迭代求解，得到最小化的损失函数和模型参数值。
𝑙𝑜𝑠𝑠 = 𝑥2 ? sin(𝑥)



线性方程
𝑦 = 𝑤 ? 𝑥 + 𝑏
例如
1.567 = w * 1 + b
3.043 = w * 2 + b
我们可以解的W = 1.477，B = 0.089
但是在现实生活中存在一定的噪声，所以我们任需要引入一个噪声来模拟
𝑦 = 𝑤 ? 𝑥 + 𝑏 + 𝜖
𝜖 ∽ 𝑁(0.01, 1)（符合正态分布）
所以损失函数𝑙𝑜𝑠𝑠 = 𝑊𝑋 + 𝑏 ? 𝑦 2
计算步骤
1、𝑙𝑜𝑠𝑠 = σ𝑖 𝑤 ? 𝑥𝑖 + 𝑏 ? 𝑦𝑖 2
2、Minimize 𝑙𝑜𝑠𝑠
3、𝑤′ ? 𝑥 + 𝑏′ → 𝑦

#𝒍𝒐𝒔𝒔 = 𝑾𝑿 + 𝒃 ? 𝒚 𝟐
#计算求和
def error_Sum(b,w,points):
    totalError=0
    for i in range(0,len(points)):
        x=points[i,0]
        y=points[i,1]
        totalError+=(y-(w*x+b))**2
    return totalError/float(len(points))

##计算导数（求损失函数极小值）
#𝒘′ = 𝒘 ? 𝒍𝒓 ?𝛁𝒍𝒐𝒔𝒔    𝛁𝒘 𝒍𝒐𝒔𝒔 = 𝑾𝑿 + 𝒃 ? 𝒚 𝟐

def step_gradient(b_current,w_current,points,learningRate):
    b_gradient=0
    w_gradient=0
    N=float(len(points))
    for i in range(0,len(points)):
        x=points[i,0]
        y = points[i, 1]
        b_gradient+=(2/N)*(((w_current*x)+b_current)-y)
        w_gradient+=(2/N)*x*(((w_current*x)+b_current)-y)
    new_b=b_current-learningRate*b_gradient
    new_w=w_current-learningRate*w_gradient
    return [new_b,new_w]

### 得到b,w
def gradient_run(points,starting_b,starting_w,learning_rate,num_iterations):
    b=starting_b
    w=starting_w
    for i in range(num_iterations):
        b,w=step_gradient(b,w,np.array(points),learning_rate)
    return [b,w]

#运行函数
def run():
    points = np.genfromtxt("data.csv", delimiter=",")
    learning_rate = 0.0001
    initial_b = 0  # initial y-intercept guess
    initial_w = 0  # initial slope guess
    num_iterations = 1000
    print("Starting gradient descent at b = {0}, w = {1}, error = {2}"
          .format(initial_b, initial_w,
                  error_Sum(initial_b, initial_w, points))
          )
    print("Running...")
    [b, w] = gradient_run(points, initial_b, initial_w, learning_rate, num_iterations)
    print("After {0} iterations b = {1}, w = {2}, error = {3}".
          format(num_iterations, b, w,
                 error_Sum(b, w, points))
          )

最终完整代码

import numpy as np
#𝒍𝒐𝒔𝒔 = 𝑾𝑿 + 𝒃 ? 𝒚 𝟐
#计算求和
def error_Sum(b,w,points):
    totalError=0
    for i in range(0,len(points)):
        x=points[i,0]
        y=points[i,1]
        totalError+=(y-(w*x+b))**2
    return totalError/float(len(points))

##计算导数（求损失函数极小值）
#𝒘′ = 𝒘 ? 𝒍𝒓 ?𝛁𝒍𝒐𝒔𝒔    𝛁𝒘 𝒍𝒐𝒔𝒔 = 𝑾𝑿 + 𝒃 ? 𝒚 𝟐

def step_gradient(b_current,w_current,points,learningRate):
    b_gradient=0
    w_gradient=0
    N=float(len(points))
    for i in range(0,len(points)):
        x=points[i,0]
        y = points[i, 1]
        b_gradient+=(2/N)*(((w_current*x)+b_current)-y)
        w_gradient+=(2/N)*x*(((w_current*x)+b_current)-y)
    new_b=b_current-learningRate*b_gradient
    new_w=w_current-learningRate*w_gradient
    return [new_b,new_w]

### 得到b,w
def gradient_run(points,starting_b,starting_w,learning_rate,num_iterations):
    b=starting_b
    w=starting_w
    for i in range(num_iterations):
        b,w=step_gradient(b,w,np.array(points),learning_rate)
    return [b,w]

def run():
    points = np.genfromtxt("data.csv", delimiter=",")
    learning_rate = 0.0001
    initial_b = 0  # initial y-intercept guess
    initial_w = 0  # initial slope guess
    num_iterations = 1000
    print("Starting gradient descent at b = {0}, w = {1}, error = {2}"
          .format(initial_b, initial_w,
                  error_Sum(initial_b, initial_w, points))
          )
    print("Running...")
    [b, w] = gradient_run(points, initial_b, initial_w, learning_rate, num_iterations)
    print("After {0} iterations b = {1}, w = {2}, error = {3}".
          format(num_iterations, b, w,
                 error_Sum(b, w, points))
          )
if __name__=='__main__':
    run()