多元线性回归
一、回归模型
- 多元线性回归的一般表达式
[f(oldsymbol x_i)=oldsymbol {w^Tx_i}+b
=
egin{bmatrix}
{w_{1}}&{w_{2}}&{cdots}&{w_{d}}\
end{bmatrix}
egin{bmatrix}
{x_{11}}\
{x_{21}}\
{vdots}\
{x_{d1}}\
end{bmatrix}+b
]
- 其中将数据D表示为一个矩阵X，将参数b包含在举证中方便后面的运算
[egin{align}
X&= left[
egin{matrix}
x_{11} & x_{12} & cdots & x_{1d} & 1 \
x_{21} & x_{22} & cdots & x_{2d} & 1 \
vdots & vdots & ddots & vdots & vdots \
x_{m1} & x_{m2} & cdots & x_{md} & 1 \
end{matrix}
ight]
=
left[
egin{matrix}
x^T_1 & 1 \
x^T_2 & 1 \
vdots & vdots \
x^T_m & 1 \
end{matrix}
ight]
=
left[
egin{matrix}
样本1 & 1 \
样本2 & 1 \
vdots & vdots \
样本m & 1 \
end{matrix}
ight]
\
w^*&=
left[
egin{matrix}
w_1 \
w_2 \
vdots \
w_d \
b \
end{matrix}
ight]
\
y&=Xw^*
end{align}
]
-
方法类似于单变量线性回归，先写出误差项
[e_i=y_i-f(x_i) ] -
误差累计和
[egin{align}
E
&= sum^m_{i=1}e^2_i \
&= sum^m_{i=1}(y_i -
left[
egin{matrix}
x_i & 1
end{matrix}
ight]
w^*)^2\
&= ||y-Xw^*||^2_2 \
&= (y-Xw^*)^T(y-Xw^*) \
&= yy^T - (w^*)^TX^Ty -y^TXw^* + (w^*)^TX^TXw^*
end{align}
]
- 误差对w求导
[egin{align}
frac{partial E}{partial w^*} &=-X^Ty-X^Ty+2X^TXw^*=2X^T(Xw^*-y)=0 \
w^*&=(X^TX)^{-1}X^Ty
end{align}
]
- 即可得线性回归模型为
[f(x_i)= left[
egin{matrix}
x_i & 1
end{matrix}
ight]
w^*
= left[
egin{matrix}
x_i & 1
end{matrix}
ight]
(X^TX)^{-1}X^Ty
=(x^*_i)^T(X^TX)^{-1}X^Ty
]
二、实例代码
实例分析
已知数据集
| 特征1 | 特征2 | 输出值 |
|---|---|---|
| 1 | 0.5 | 20.52 |
| 2 | 2.6 | 27.53 |
| 3 | 3.5 | 29.84 |
| 4 | 6.7 | 36.92 |
| 5 | 8.9 | 38.11 |
| 6 | 9.2 | 37.21 |
| 7 | 11.5 | 40.96 |
| 8 | 15.8 | 46.37 |
| 9 | 19.6 | 49.78 |
| 10 | 20.1 | 50.22 |
求解如下问题
- 利用最小二乘求解多远线性模型参数，并计算训练的均方误差
- 在得到模型的基础上，假如输入下面新数据，进行预测，并计算均方误差
| 特征1 | 特征2 | 输出值 |
|---|---|---|
| 11 | 0.85 | 52.38 |
| 12 | 16.54 | 55.66 |
| 13 | 3.37 | 58.31 |
代码：Matlab
x1 = [1 2 3 4 5 6 7 8 9 10]';
x2 = [0.5 2.6 3.5 6.7 8.9 9.2 11.5 15.8 19.6 20.1]';
y = [20.52 27.53 29.84 36.92 38.11 37.21 40.96 46.37 49.78 50.22]';
X = [x1 x2 ones(10,1)]
% 训练
w_ = pinv(X'*X)*X'*y
y_ = X*w_;
% 预测
x1_=[11 12 13]';
x2_=[0.85 16.54 3.37]';
y_new = [52.38 55.66 58.31]';
y_new_ = [x1_ x2_ ones(3,1)]*w_
%% 误差
E = sum((y_-y).^2)/10
E_t= sum((y_new_-y_new).^2)/3
plot3(x1, x2, y_, 'r-', x1, x2, y, 'bo'); hold on
legend('predict', 'real')
运行结果
w_ =
1.6144
0.6742
22.2324
y_new_ =
40.5635
52.7568
45.4914
E =
3.4949
E_t =
104.1249
代码：Python
1.数据构造
[egin{align}
X &= left[
egin{matrix}
x_{11} & x_{12} & cdots & x_{1d} & 1 \
x_{21} & x_{22} & cdots & x_{2d} & 1 \
vdots & vdots & ddots & vdots & vdots \
x_{m1} & x_{m2} & cdots & x_{md} & 1 \
end{matrix}
ight]
=
left[
egin{matrix}
x^T_1 & 1 \
x^T_2 & 1 \
vdots & vdots \
x^T_m & 1 \
end{matrix}
ight]
=
left[
egin{matrix}
样本1 & 1 \
样本2 & 1 \
vdots & vdots \
样本m & 1 \
end{matrix}
ight]\
end{align}
]
import numpy as np
import math
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
%matplotlib inline
x1 = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]).reshape(10,1)
x2 = np.array([0.5, 2.6, 3.5, 6.7, 8.9, 9.2, 11.5, 15.8, 19.6, 20.1]).reshape(10,1)
y = np.array([20.52, 27.53, 29.84, 36.92, 38.11, 37.21, 40.96, 46.37, 49.78, 50.22]).reshape(10,1)
X = np.hstack((x1, x2, np.ones((10,1),dtype=int)))
print('X:
',X)
结果：
X:
[[ 1. 0.5 1. ]
[ 2. 2.6 1. ]
[ 3. 3.5 1. ]
[ 4. 6.7 1. ]
[ 5. 8.9 1. ]
[ 6. 9.2 1. ]
[ 7. 11.5 1. ]
[ 8. 15.8 1. ]
[ 9. 19.6 1. ]
[10. 20.1 1. ]]
2.参数训练
[w^*=
left[
egin{matrix}
w_1 \
w_2 \
vdots \
w_m \
b \
end{matrix}
ight]
=(X^TX)^{-1}X^Ty
]
w_ = np.dot(np.dot(np.linalg.pinv(np.dot(np.transpose(X),X)), np.transpose(X)) ,y)
print('w:
',w_)
结果：
w:
[[ 1.61436503]
[ 0.67424588]
[22.23241285]]
3.预测
[y=Xw^*
]
y_ = np.dot(X, w_)
print('y:
',y_)
结果：
y:
[[24.18390082]
[27.2141822 ]
[29.43536853]
[33.20732038]
[36.30502636]
[38.12166515]
[41.28679571]
[45.80041804]
[49.97691742]
[51.92840539]]
4.计算均方误差
[E = frac{1}{m}sum^m_{i=1}(y_i-f(x_i))^2
]
E = np.sum(pow((y - y_), 2))/len(x1)
print('E:', E)
5.绘图
x1_T = x1.ravel()
x2_T = x2.ravel()
y_T = y.ravel()
y__T = y_.ravel()
fig = plt.figure()
ax = Axes3D(fig)
ax.scatter(x1_T, x2_T, y_T, label='real', alpha=1)
ax.plot(x1_T, x2_T, y__T, c='r', label='predict')
ax.legend()
plt.show()
结果：
