机器学习Sklearn实战——手写线性回归

手写线性回归

import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
X = np.linspace(2,10,20).reshape(-1,1)
y = np.random.randint(1,6,size = 1)*X + np.random.randint(-5,5,size = 1)
#噪声 加盐
y += np.random.randn(20,1)*0.8
plt.scatter(X,y,color = "red")

w = lr.coef_[0,0]
b = lr.intercept_[0]
print(w,b)
plt.scatter(X,y)
x = np.linspace(1,11,50)
plt.plot(x,w*x + b,color= "green")

机器学习Sklearn实战——手写线性回归

结果:

2.995391527138711 1.9801931425932864
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
 
X = np.linspace(2, 10, 20).reshape(-1,1)
# f(x) = wx + b
y = np.random.randint(1, 6, size=1)*X + np.random.randint(-5, 5, size=1)
# 噪声,加盐
y += np.random.randn(20, 1)*0.8

plt.scatter(X, y, color = 'red')

w = lr.coef_[0, 0]
b = lr.intercept_[0]

x = np.linspace(1, 11, 50)
plt.plot(x, w*x + b, color='green')


# 使用梯度下降解决一元一次的线性问题:w,b
class LinearModel(object):
    def __init__(self):
        self.w = np.random.randn(1)[0]
        self.b = np.random.randn(1)[0]
    # 数学建模:将数据X和目标值关系用数学公式表达
    def model(self,x):#model 模型,f(x) = wx + b
        return self.w*x + self.b
    def loss(self,x,y):#最小二乘
        cost = (y - self.model(x))**2
    # 梯度就是偏导数,求解两个未知数:w,b
        gradient_w = 2*(y - self.model(x))*(-x)
        gradient_b = 2*(y - self.model(x))*(-1)
        return cost,gradient_w,gradient_b
    # 梯度下降
    def gradient_descent(self,gradient_w,gradient_b,learning_rate = 0.1):
    # 更新w,b
        self.w -= gradient_w*learning_rate
        self.b -= gradient_b*learning_rate
    # 训练fit
    def fit(self,X,y):
        count = 0 #算法执行优化了3000次,退出
        tol = 0.0001
        last_w = self.w + 0.1
        last_b = self.b + 0.1
        length = len(X)
        while True:
            if count > 3000:#执行的次数到了
                break
    # 求解的斜率和截距的精确度达到要求
            if (abs(last_w - self.w) < tol) and (abs(last_b - self.b) < tol):
                break
            cost = 0
            gradient_w = 0
            gradient_b = 0
            for i in range(length):
                cost_,gradient_w_,gradient_b_ = self.loss(X[i,0],y[i,0])
                cost += cost_/length
                gradient_w += gradient_w_/length
                gradient_b += gradient_b_/length
            print('---------------------执行次数:%d。损失值是:%0.2f'%(count,cost))
            last_w = self.w
            last_b = self.b
            # 更新截距和斜率
            self.gradient_descent(gradient_w,gradient_b,0.01)
            count+=1
    def result(self):
        return self.w,self.b
lm = LinearModel()
lm.fit(X,y)
lm.result()

结果:

(2.9489680632625297, 2.2698211503362224)
plt.scatter(X,y,color = "red")
plt.plot(x,2.94896*x + 2.2698211,color= "green")
plt.rcParams['font.sans-serif'] = ['Arial Unicode MS']
plt.title("自定义的算法拟合曲线")

机器学习Sklearn实战——手写线性回归

二元一次拟合

import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression

#f(x)w = w1*x**2 + w2*x + b
#一元二次

#f(x1,x2) = w1*x1 + w2*x2 +b
#二元一次

X = np.linspace(0,10,num = 50).reshape(-1,1)
X = np.concatenate([X**2,X],axis = 1)
X.shape
w = np.random.randint(1,10,size = 2)
b = np.random.randint(-5,5,size = 1)

#矩阵乘法
y = X.dot(w) + b

plt.plot(X[:,1],y,c="r")
plt.title("w1:%d.w2:%d.w3:%d"%(w[0],w[1],b[0]))

结果: 

Text(0.5, 1.0, 'w1:7.w2:1.w3:-5')

机器学习Sklearn实战——手写线性回归 

#使用sklearn自带的算法进行预测
from sklearn.linear_model import LinearRegression
lr = LinearRegression()
lr.fit(X,y)
print(lr.coef_,lr.intercept_)
plt.scatter(X[:,1],y,marker = "*")

x = np.linspace(-2,12,100)
plt.plot(x,7*x**2 + 1*x + -4.99,c="g")

 结果:

[7. 1.] -4.999999999999972

机器学习Sklearn实战——手写线性回归 

手写线性回归,拟合多属性,多元方程 

# epoch 训练的次数,梯度下降训练多少
def gradient_descent(X,y,lr,epoch,w,b):
    # 一批量多少,长度
    batch = len(X)
    for i in range(epoch):
    # d_lost:是损失的梯度
        d_loss = 0
    # 梯度,斜率梯度
        dw = [0 for _ in range(len(w))]
    # 截距梯度
        db = 0
        for j in range(batch):
            y_ = 0 #预测的值 预测方程 y_ = f(x) = w1*x1 + w2*x2 + b
            for n in range(len(w)):
                y_ += X[j][n]*w[n]
            y_ += b
    # (y - y_)**2 -----> 2*(y - y_)*(-1)
    # (y_- y)**2  -----> 2*(y_ - y)*(1)
            d_loss = -(y[j] - y_)
            for n in range(len(w)):
                dw[n] += X[j][n]*d_loss/float(batch)
            db += 1*d_loss/float(batch)
    # 更新一下系数和截距,梯度下降
        for n in range(len(w)):
            w[n] -= dw[n]*lr[n]
        b -= db*lr[0]
    return w,b
lr = [0.0001,0.001]
w = np.random.randn(2)
b = np.random.randn(1)[0]
gradient_descent(X,y,lr,500,w,b)

 结果:

(array([ 7.18689265, -1.25846592]), 0.6693960269813103)
plt.scatter(X[:,1],y,marker = "*")
plt.plot(x,7.1868*x**2 - 1.2584*x + 0.6694,c="g")

 机器学习Sklearn实战——手写线性回归

 继续优化

  1. x的num变大,从50变成500
  2. 学习率调小
上一篇:梯段下降算法


下一篇:6个城市街道电影效果lr预设