神经网络入门-回归问题(梯度下降法)-python实现

程序实现的功能

给定一些点,拟合出回归直线,数据在百度云链接
在这里插入图片描述
1.以numpy格式读取csv文件

 points = np.genfromtxt("data.csv", delimiter=",")
 print(points)

打印一下point看一下numpy格式

在这里插入图片描述
2.初始化直线的参数 w,b,直线的形式如图所示,初始化w和b都为0
在这里插入代码片在这里插入图片描述

 initial_b = 0 # initial y-intercept guess
 initial_w = 0 # initial slope guess

3.计算损失函数(loss)

def compute_error_for_line_given_points(b, w, points):
    totalError = 0
    for i in range(0, len(points)):
        x = points[i, 0]
        y = points[i, 1]
        # computer mean-squared-error
        totalError += (y - (w * x + b)) ** 2
    # average loss for each point
    return totalError / float(len(points))

其中,这两句是numpy的调用格式

   x = points[i, 0]#相当于points[i][0],表示第i个点的第x坐标
   y = points[i, 1]#相当于points[i][1],表示第i个点的y坐标

这是我们构建的损失函数的形式,也就是损失平方和,再除以N

totalError += (y - (w * x + b)) ** 2

4.更新 b,w的值,采用梯度下降的方法,如图所示,w‘代表新的w值
在这里插入图片描述

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

loss函数对b,对w求偏导

 b_gradient += (2/N) * ((w_current * x + b_current) - y)
 w_gradient += (2/N) * x * ((w_current * x + b_current) - y)
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]
        # grad_b = 2(wx+b-y)
        b_gradient += (2/N) * ((w_current * x + b_current) - y)
        # grad_w = 2(wx+b-y)*x
        w_gradient += (2/N) * x * ((w_current * x + b_current) - y)
    # update w'
    new_b = b_current - (learningRate * b_gradient)
    new_w = w_current - (learningRate * w_gradient)
    return [new_b, new_w]
 new_b = b_current - (learningRate * b_gradient)
 new_w = w_current - (learningRate * w_gradient)

完整代码和数据打包到百度云:
链接:https://pan.baidu.com/s/1vWlsyR9BykcXVM1BYIQJuw
提取码:pj2z

发布了23 篇原创文章 · 获赞 9 · 访问量 4207

猜你喜欢

转载自blog.csdn.net/qq1225598165/article/details/102989069