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")

结果：

2.995391527138711 1.9801931425932864

import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegressionX = 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)*0.8plt.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 + breturn self.w*x + self.bdef 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，bgradient_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，bself.w -= gradient_w*learning_rateself.b -= gradient_b*learning_rate# 训练fitdef fit(self,X,y):count = 0 #算法执行优化了3000次，退出tol = 0.0001last_w = self.w + 0.1last_b = self.b + 0.1length = 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):breakcost = 0gradient_w = 0gradient_b = 0for i in range(length):cost_,gradient_w_,gradient_b_ = self.loss(X[i,0],y[i,0])cost += cost_/lengthgradient_w += gradient_w_/lengthgradient_b += gradient_b_/lengthprint('---------------------执行次数：%d。损失值是：%0.2f'%(count,cost))last_w = self.wlast_b = self.b# 更新截距和斜率self.gradient_descent(gradient_w,gradient_b,0.01)count+=1def 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")

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) + bplt.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')

 

#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

 

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 = 0for j in range(batch):y_ = 0 #预测的值 预测方程 y_ = f(x) = w1*x1 + w2*x2 + bfor 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")

 

 继续优化

1. x的num变大，从50变成500
2. 学习率调小