[scikit-learn 机器学习] 2. 简单线性回归

1. 简单线性回归

代码语言：javascript复制

import numpy as np
import matplotlib.pyplot as plt

X = 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

代码语言：javascript复制

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

代码语言：javascript复制

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}

代码语言：javascript复制

# 方差
x_bar = X.mean() # 11.2
variance = ((X-x_bar)**2).sum()/(len(X)-1)
variance # 23.2

np.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. By
        default `ddof` is zero.

• 协方差

cov(x,y) = frac{sum(x_i-bar x)(y_i - bar y)}{n-1}

代码语言：javascript复制

# 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 # 22.65

np.cov(X.transpose(), y)

代码语言：javascript复制

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*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}

代码语言：javascript复制

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

线性回归 模型

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