机器学习之多项式回归

零、模型

1、公式

y = b0 + b1X1+B2X2^2 +BnXn^n

2.R平方

R^2 = 1- 剩余平方和/共平方和

R^2为值为0到1，越大拟合效果越好

3.广义R平方（增加惩罚作用）

R^2 = 1- 剩余平方和/共平方和 *（(n-1)/n-p-1）

n :数据个数 p:自变量个数

一、导入标准库

In [16]:

# Importing the libraries 导入库
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
# 使图像能够调整
%matplotlib notebook 
#中文字体显示  
plt.rc('font', family='SimHei', size=8)

二、导入数据

In [10]:

# Importing the dataset 导入数据
dataset = pd.read_csv('./Position_Salaries.csv')

X = dataset.iloc[:, 1:2].values  # 职位级别
y = dataset.iloc[:, 2].values    # 薪水
dataset

Out[10]:

	Position	Level	Salary
0	Business Analyst	1	45000
1	Junior Consultant	2	50000
2	Senior Consultant	3	60000
3	Manager	4	80000
4	Country Manager	5	110000
5	Region Manager	6	150000
6	Partner	7	200000
7	Senior Partner	8	300000
8	C-level	9	500000
9	CEO	10	1000000

In [11]:

X 

Out[11]:

array([[ 1],
       [ 2],
       [ 3],
       [ 4],
       [ 5],
       [ 6],
       [ 7],
       [ 8],
       [ 9],
       [10]], dtype=int64)

In [12]:

y

Out[12]:

array([  45000,   50000,   60000,   80000,  110000,  150000,  200000,
        300000,  500000, 1000000], dtype=int64)

三、创建回归模型

In [13]:

from sklearn.linear_model import LinearRegression
lin_reg = LinearRegression()
lin_reg.fit(X,y)

Out[13]:

LinearRegression(copy_X=True, fit_intercept=True, n_jobs=1, normalize=False)

四、创建多项式回归

In [29]:

from sklearn.preprocessing import PolynomialFeatures
poly_reg = PolynomialFeatures(degree = 4)
X_poly = poly_reg.fit_transform(X) # 自变量不同次数的矩阵
lin_reg_2 = LinearRegression()
lin_reg_2.fit(X_poly,y)
X_poly 

Out[29]:

array([[  1.00000000e+00,   1.00000000e+00,   1.00000000e+00,
          1.00000000e+00,   1.00000000e+00],
       [  1.00000000e+00,   2.00000000e+00,   4.00000000e+00,
          8.00000000e+00,   1.60000000e+01],
       [  1.00000000e+00,   3.00000000e+00,   9.00000000e+00,
          2.70000000e+01,   8.10000000e+01],
       [  1.00000000e+00,   4.00000000e+00,   1.60000000e+01,
          6.40000000e+01,   2.56000000e+02],
       [  1.00000000e+00,   5.00000000e+00,   2.50000000e+01,
          1.25000000e+02,   6.25000000e+02],
       [  1.00000000e+00,   6.00000000e+00,   3.60000000e+01,
          2.16000000e+02,   1.29600000e+03],
       [  1.00000000e+00,   7.00000000e+00,   4.90000000e+01,
          3.43000000e+02,   2.40100000e+03],
       [  1.00000000e+00,   8.00000000e+00,   6.40000000e+01,
          5.12000000e+02,   4.09600000e+03],
       [  1.00000000e+00,   9.00000000e+00,   8.10000000e+01,
          7.29000000e+02,   6.56100000e+03],
       [  1.00000000e+00,   1.00000000e+01,   1.00000000e+02,
          1.00000000e+03,   1.00000000e+04]])

五、画图比较

1.线性回归

In [24]:

plt.scatter(X,y,color="red")
plt.plot(X,lin_reg.predict(X),color ="blue")
plt.title(u"真话还是假话（线性模型）")
plt.xlabel(u"职位水平")
plt.ylabel(u"薪水")
plt.show()
print('r-squared', lin_reg.score(X,y))

('r-squared', 0.66904123319298947)

2.多项式回归

In [28]:

X_grid = np.arange(min(X),max(X),0.1)
X_grid = X_grid.reshape(len(X_grid),1)
plt.scatter(X,y,color="red")
plt.plot(X_grid,lin_reg_2.predict(poly_reg.fit_transform(X_grid)),color ="blue")
plt.title(u"真话还是假话（多项式模型）")
plt.xlabel(u"职位水平")
plt.ylabel(u"薪水")
plt.show()
print('r-squared', lin_reg_2.score(X_poly,y))

('r-squared', 0.99739228917066114)

结论：多项式回归结果明显优于简单线性回归结果（实际应用时应考虑过拟合的问题）

六、薪水预测

1.线性回归

In [19]:

lin_reg.predict(6.5)

Out[19]:

array([ 330378.78787879])

2.多项式回归预测

In [20]:

lin_reg_2.predict(poly_reg.fit_transform(6.5))

Out[20]:

array([ 158862.45265154])

七、项目地址

https://coding.net/u/RuoYun/p/Python-of-machine-learning/git/tree/master/00%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0/2.%E5%9B%9E%E5%BD%92%E7%AE%97%E6%B3%95/2.%E5%A4%9A%E5%85%83%E7%BA%BF%E6%80%A7%E5%9B%9E%E5%BD%92?public=true