一元一次拟合f(x)=wx+b

使用已有的线性回归拟合函数

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')

lr = LinearRegression()
lr.fit(X, y)

w = lr.coef_[0, 0]
b = lr.intercept_[0]
print(w, b)
# 4.984999512157303 -3.7974579753883653

plt.scatter(X, y)

x = np.linspace(1, 11, 50)

plt.plot(x, w*x + b, color='green')

 

手写实现线性回归（简版）

# 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

 使用手写线性回归拟合函数

lm = LinearModel()

lm.fit(X,y)

w_,b_ = lm.result()

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

plt.plot(x,4.9850*x - 3.7974,color = 'green')

plt.plot(x,w*x + b,color = 'blue')

plt.title('自定义的算法拟合曲线',fontproperties = 'KaiTi')

 一元二次拟合

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

# 一元二次
# f(x) = w1*x**2 + w2*x + b

# 二元一次
# f(x1,x2) = w1*x1 + w2*x2 + b

X = np.linspace(0,10,num = 500).reshape(-1,1)

X = np.concatenate([X**2,X],axis = 1)
# X.shape
# (500, 2)

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,color = 'r')

plt.title('w1:%d.w2:%d.b:%d'%(w[0],w[1],b[0]))

 

使用sklearn自带的算法进行预测

lr = LinearRegression()

lr.fit(X,y)

print(lr.coef_,lr.intercept_)
# [1. 1.] 3.999999999999993

plt.scatter(X[:,1],y,marker = '*')

x = np.linspace(-2,12,100)

plt.plot(x,1*x**2 + 6*x + 1,color = 'green')

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 = [0.0001,0.0001]
w = np.random.randn(2)
b = np.random.randn(1)[0]
w_,b_ = gradient_descent(X,y,lr,5000,w,b)
print(w_,b_)
# [1.00325157 1.22027686] 2.1550745865631895