numpy实现回归问题求解

loss函数的定义:

​​​为防止loss求和后过大,通常求和后除以样本数量,即取平均值​​

优化过程中新w,b的计算:

​​​lr为学习率​​

​​为防止梯度过大,一般求和后会除以样本数量,即取平均值​​

代码:

import numpy as np

# 计算均方误差(定义loss=(w * x_i + b - y_i)**2 | 遍历每一个[i]求和)
def points_avg2_error(b,w,points):
    totalError = 0
    for i in range(len(points)):
        x = points[i, 0]
        y = points[i, 1]

        totalError += (y - (w * x + b)) ** 2
    return totalError / len(points)

# 计算每一步的梯度
def step_graident(b,w,points,learning_rate):
    # 梯度
    b_gradient = 0
    w_gradient = 0
    # 样本数量
    N = len(points)
    # 定义loss=(w * x_i + b - y_i)**2 | 遍历每一个[i]求和
    # 偏loss/b-w求出来的导数都是求和的形式
    for i in range(N):
        x = points[i, 0]
        y = points[i, 1]
        # 偏loss/偏b ==> b梯度的表示
        b_gradient += (2/N)*((w * x + b) - y)
        # 偏loss/偏w ==> w梯度的表示
        w_gradient += (2/N)* x * ((w * x + b) - y)

    # 沿着梯度的反方向优化,使其达到局部最优点
    b = b - (learning_rate * b_gradient)
    w = w - (learning_rate * w_gradient)
    return b,w

# 梯度下降(优化w,b,使得loss最小)
def gradient_descent_runner(points,b,w,learning_rate,num_iterations):

    for i in range(num_iterations):
        b,w = step_graident(b,w,np.array(points),learning_rate)

    return b,w

def run():
    # 读取数据
    points = np.genfromtxt("data.csv",delimiter=",")
    # 学习率
    learning_rate = 0.0001
    # 初始化参数b,w
    init_b = 0
    init_w = 0
    # 学习次数
    num_iterations = 1000
    print("梯度下降开始前 b = {0}, w = {1}, error(均方误差) = {2}".format(init_b,init_w,points_avg2_error(init_b,init_w,points)))

    print("Running....")

    # 梯度下降优化后的b,w
    b,w = gradient_descent_runner(points,init_b,init_w,learning_rate,num_iterations)
    print("梯度下降开始后 b = {0}, w = {1}, error(均方误差) = {2}".format(b, w, points_avg2_error(b, w, points)))

if __name__=="__main__":
    run()