本文为 scikit-learn机器学习（第2版）学习笔记

1. 简单线性回归

import numpy as np
import matplotlib.pyplot as pltX = np.array([[6],[8],[10],[14],[18]])
y = np.array([7,9,13,17.5,18])
plt.title("pizza diameter vs price")
plt.xlabel('diameter')
plt.ylabel('price')
plt.plot(X,y,'r.') # r表示颜色红

from sklearn.linear_model import LinearRegression
model = LinearRegression()
model.fit(X,y)test_pizza = np.array([[12]])
pred_price = model.predict(test_pizza)
pred_price
# array([13.68103448])

• 误差 $\sum (y_i-f(x_i){)}^2$

print("误差平方和：%.2f" % np.mean((model.predict(X)-y)**2))
误差平方和：1.75

• 方差 $var(x)=\frac{\sum (x_i-\bar{x}{)}^2}{n-1}$

# 方差
x_bar = X.mean() # 11.2
variance = ((X-x_bar)**2).sum()/(len(X)-1)
variance # 23.2np.var(X, ddof=1) # np内置的方差，ddof为校正选项
###################
ddof : int, optional"Delta Degrees of Freedom": the divisor used in the calculation is``N - ddof``, where ``N`` represents the number of elements. Bydefault `ddof` is zero.

• 协方差 $cov(x,y)=\frac{\sum (x_i-\bar{x})(y_i-\bar{y})}{n-1}$

# 协方差，两个变量之间的相关性
y_bar = y.mean()
covariance = np.multiply((X-x_bar).transpose(), y-y_bar).sum()/(len(X)-1)
covariance # 22.65np.cov(X.transpose(), y)

array([[23.2 , 22.65],[22.65, 24.3 ]])

假设模型为 $y=a+bx$

$b=\frac{cov(x,y)}{var(x)}=22.65/23.2=0.98$

$a=\bar{y}-b\bar{x}=12.9-0.98\ast 11.2=1.92$

模型为 $y=1.92+0.98x$

2. 评价模型

$$R^2=1-\frac{\sum (y_i-f(x_i){)}^2}{\sum (y_i-\bar{y}{)}^2}$$

X_test = np.array([8,9,11,16,12]).reshape(-1,1)
y_test = [11,8.5,15,18,11]
r_squared = model.score(X_test, y_test)
r_squared # 0.6620052929422553