梯度下降法-3.实现线性回归中的梯度下降法

谁借莪１个温暖的怀抱￠ 2023-08-17 16:44 82阅读 0赞

# 实现线性回归中的梯度下降法 #

构造数据集

import numpy
    import matplotlib.pyplot as plt
    
    # 设置随机数种子
    numpy.random.seed(666)  
    x = 2 * numpy.random.random(size=100)
    y = x * 3. + 4. + numpy.random.normal(size=100)  #random.normal 为产生正太分布的噪点
    
    #假设此样本有一个特征值，把100个数转化为100行，1列
    X = x.reshape(-1,1)

![1465169-20190716204752466-771770990.png][]

绘制此数据集：

plt.scatter(x,y)
    plt.show()

![1465169-20190716204808344-139691616.png][]

### 使用梯度下降法训练 ###

目标：使  
\\\[ J = \\frac\{1\}\{m\}\\sum\_\{i=1\}^\{m\}(y^\{(i)\} - \\hat\{y\}^\{(i)\})^2 \\\]尽可能的小  
\\\[\\Lambda J = \\begin\{bmatrix\} \\frac\{\\partial J\}\{\\partial \\theta \_0\} \\\\ \\frac\{\\partial J\}\{\\partial \\theta \_1\} \\\\ \\frac\{\\partial J\}\{\\partial \\theta \_2\} \\\\ ... \\\\ \\frac\{\\partial J\}\{\\partial \\theta \_n\} \\end\{bmatrix\} = \\frac\{2\}\{m\}\\begin\{bmatrix\} \\sum(X^\{(i)\}\_b\\theta - y^\{(i)\}) \\\\ \\sum(X^\{(i)\}\_b\\theta - y^\{(i)\}) \\cdot X\_1^\{(i)\}\\\\ \\sum(X^\{(i)\}\_b\\theta - y^\{(i)\}) \\cdot X\_2^\{(i)\}\\\\ ...\\\\ \\sum(X^\{(i)\}\_b\\theta - y^\{(i)\}) \\cdot X\_3^\{(i)\} \\end\{bmatrix\}\\\]  
定义函数和导数的表达式

def J(theta,X_b,y):   #损失函数的表达式
        try:
            return numpy.sum((y - X_b.dot(theta))**2)/len(X_b)
        except:
            return float('inf') #返回float的最大值
       
    def dJ(theta,X_b,y):  #求导
        #要返回的导数矩阵
        res = numpy.empty(len(theta))    
        
        res[0] = numpy.sum(X_b.dot(theta)-y)
        for i in range(1,len(theta)):
            res[i] = ((X_b.dot(theta)-y).dot(X_b[:,i]))
            
        return res*2/len(X_b)

定义梯度下降的算法过程

def gradient_descent(X_b,y,init_theta,eta,n_iters=1e4,espilon=1e-8):
        
        theta = init_theta
        i_iters = 0
        
        #n_iters 表示梯度下降的次数，超过这个值，有可能算法不收敛，退出
        while n_iters>i_iters: 
            gradient = dJ(theta,X_b,y)  #偏导数
            last_theta = theta
            theta = theta - eta * gradient  #梯度下降，向极值移动   
            if abs(J(theta,X_b,y) - J(last_theta,X_b,y)) < espilon:
                break
            i_iters += 1 
        # 返回求出的theta值
        return theta

构造初始参数，并进行梯度下降的过程：

X_b = numpy.hstack([numpy.ones((len(X),1)),X])
    init_theta = numpy.zeros(X_b.shape[1])
    eta = 0.01
    theta = gradient_descent(X_b,y,init_theta,eta)

求得的theta值  
![1465169-20190716204841105-1635206704.png][]  
*创建数据集时，截距为4，斜率为3，由此可以看出，此梯度下降法成功的训练了此模型*

### 算法的封装 ###

def fit_gd(self,X_train,y_train,eta=0.01,n_iters=1e4):
    
            assert X_train.shape[0] == y_train.shape[0],\
                "size of x_train must be equal to the size of y_train"
    
            def J(theta,X_b,y):   
                try:
                    return numpy.sum((y - X_b.dot(theta))**2)/len(X_b)
                except:
                    return float('inf')  # 返回float的最大值
    
            def dJ(theta,X_b,y):  
                #要返回的导数矩阵
                res = numpy.empty(len(theta))    
        
                res[0] = numpy.sum(X_b.dot(theta)-y)
                for i in range(1,len(theta)):
                    res[i] = ((X_b.dot(theta)-y).dot(X_b[:,i]))
                return res*2/len(X_b)
    
            def gradient_descent(X_b,y,init_theta,eta,n_iters,espilon=1e-8):
                theta = init_theta
                i_iters = 0
                while n_iters>i_iters:
                    gradient = dJ(theta,X_b,y)  #偏导数
                    last_theta = theta
                    theta = theta - eta * gradient  #梯度下降，向极值移动
        
                    if abs(J(theta,X_b,y) - J(last_theta,X_b,y)) < espilon:
                        break
                    i_iters += 1
                return theta
    
            X_b = numpy.hstack([numpy.ones((len(X_train),1)),X_train])
    
            init_theta = numpy.zeros(X_b.shape[1])
            self._theta = gradient_descent(X_b,y_train,init_theta,eta,n_iters)
    
            self.coef_ = self._theta[1:]    #系数
            self.interception_ = self._theta[0] #截距
            return self

封装后调用：

from mylib.LineRegression import LineRegression
    lin_reg = LineRegression()
    # 用梯度下降法训练
    lin_reg.fit_gd(X,y)

![1465169-20190716214158359-1786492824.png][]

转载于:https://www.cnblogs.com/shuai-long/p/11197402.html

[1465169-20190716204752466-771770990.png]: /images/20230809/de147e977d1649ec929bf53deb1cc7ca.png
[1465169-20190716204808344-139691616.png]: /images/20230809/7484c4c5f5344ab986edab635dcbb985.png
[1465169-20190716204841105-1635206704.png]: /images/20230809/aa0ef79c9f9c454cad5f7c841cdd251d.png
[1465169-20190716214158359-1786492824.png]: /images/20230809/e4d582087f1942c38bf0df4691053119.png