文章目录
数据
数据集
( x ( i ) , y ( i ) ) , i = 1 , 2 , ⋯ , m \left(x^{(i)} , y^{(i)}\right) , i = 1 , 2 , \cdots , m (x(i),y(i)),i=1,2,⋯,m
实际值
y ( i ) y^{(i)} y(i)
估计值
h θ ( x ( i ) ) = θ 0 + θ 1 x ( i ) h_{\theta}\left(x^{(i)}\right) = \theta_{0} + \theta_{1} x^{(i)} hθ(x(i))=θ0+θ1x(i)
梯度下降算法
估计误差
h θ ( x ( i ) ) − y ( i ) h_{\theta}\left(x^{(i)}\right) - y^{(i)} hθ(x(i))−y(i)
代价函数
J ( θ ) = J ( θ 0 , θ 1 ) = 1 2 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 2 = 1 2 m ∑ i = 1 m ( θ 0 + θ 1 x ( i ) − y ( i ) ) 2 J(\theta) = J(\theta_{0} , \theta_{1}) = \cfrac{1}{2m} \displaystyle\sum\limits_{i = 1}^{m}{\left(h_{\theta}\left(x^{(i)}\right) - y^{(i)}\right)^{2}} = \cfrac{1}{2m} \displaystyle\sum\limits_{i = 1}^{m}{\left(\theta_{0} + \theta_{1} x^{(i)} - y^{(i)}\right)^{2}} J(θ)=J(θ0,θ1)=2m1i=1∑m(hθ(x(i))−y(i))2=2m1i=1∑m(θ0+θ1x(i)−y(i))2
学习率
- α \alpha α是学习率，一个大于 0 0 0的很小的经验值，决定代价函数下降的程度
参数更新
Δ θ j = ∂ ∂ θ j J ( θ 0 , θ 1 ) \Delta{\theta_{j}} = \cfrac{\partial}{\partial{\theta_{j}}} J(\theta_{0} , \theta_{1}) Δθj=∂θj∂J(θ0,θ1)
θ j : = θ j − α Δ θ j = θ j − α ∂ ∂ θ j J ( θ 0 , θ 1 ) \theta_{j} := \theta_{j} - \alpha \Delta{\theta_{j}} = \theta_{j} - \alpha \cfrac{\partial}{\partial{\theta_{j}}} J(\theta_{0} , \theta_{1}) θj:=θj−αΔθj=θj−α∂θj∂J(θ0,θ1)
$$
\left[
\begin{matrix}
\theta_{0} \
\theta_{1}
\end{matrix}
\right] :=
\left[
\begin{matrix}
\theta_{0} \
\theta_{1}
\end{matrix}
\right] -
\alpha
\left[
\begin{matrix}
\cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{0}}} \
\cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{1}}}
\end{matrix}
\right]
$$
[ ∂ J ( θ 0 , θ 1 ) ∂ θ 0 ∂ J ( θ 0 , θ 1 ) ∂ θ 1 ] = [ 1 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) 1 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) x ( i ) ] = [ 1 m ∑ i = 1 m e ( i ) 1 m ∑ i = 1 m e ( i ) x ( i ) ] e ( i ) = h θ ( x ( i ) ) − y ( i ) \left[ \begin{matrix} \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{0}}} \\ \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{1}}} \end{matrix} \right] = \left[ \begin{matrix} \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{\left(h_{\theta}\left(x^{(i)}\right) - y^{(i)}\right)} \\ \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{\left(h_{\theta}\left(x^{(i)}\right) - y^{(i)}\right) x^{(i)}} \end{matrix} \right] = \left[ \begin{matrix} \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)}} \\ \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)} x^{(i)}} \end{matrix} \right] \kern{2em} e^{(i)} = h_{\theta}\left(x^{(i)}\right) - y^{(i)} ∂θ0∂J(θ0,θ1)∂θ1∂J(θ0,θ1) = m1i=1∑m(hθ(x(i))−y(i))m1i=1∑m(hθ(x(i))−y(i))x(i) = m1i=1∑me(i)m1i=1∑me(i)x(i) e(i)=hθ(x(i))−y(i)
[ ∂ J ( θ 0 , θ 1 ) ∂ θ 0 ∂ J ( θ 0 , θ 1 ) ∂ θ 1 ] = [ 1 m ∑ i = 1 m e ( i ) 1 m ∑ i = 1 m e ( i ) x ( i ) ] = [ 1 m ( e ( 1 ) + e ( 2 ) + ⋯ + e ( m ) ) 1 m ( e ( 1 ) x ( 1 ) + e ( 2 ) x ( 2 ) + ⋯ + e ( m ) x ( m ) ) ] = 1 m [ 1 1 ⋯ 1 x ( 1 ) x ( 2 ) ⋯ x ( m ) ] [ e ( 1 ) e ( 2 ) ⋮ e ( m ) ] = 1 m X T e = 1 m X T ( X θ − y ) \begin{aligned} \left[ \begin{matrix} \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{0}}} \\ \cfrac{\partial{J(\theta_{0} , \theta_{1})}}{\partial{\theta_{1}}} \end{matrix} \right] &= \left[ \begin{matrix} \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)}} \\ \cfrac{1}{m} \displaystyle\sum\limits_{i = 1}^{m}{e^{(i)} x^{(i)}} \end{matrix} \right] = \left[ \begin{matrix} \cfrac{1}{m} \left(e^{(1)} + e^{(2)} + \cdots + e^{(m)}\right) \\ \cfrac{1}{m} \left(e^{(1)} x^{(1)} + e^{(2)} x^{(2)} + \cdots + e^{(m)} x^{(m)}\right) \end{matrix} \right] \\ &= \cfrac{1}{m} \left[ \begin{matrix} 1 & 1 & \cdots & 1 \\ x^{(1)} & x^{(2)} & \cdots & x^{(m)} \end{matrix} \right] \left[ \begin{matrix} e^{(1)} \\ e^{(2)} \\ \vdots \\ e^{(m)} \end{matrix} \right] = \cfrac{1}{m} X^{T} e = \cfrac{1}{m} X^{T} (X \theta - y) \end{aligned} ∂θ0∂J(θ0,θ1)∂θ1∂J(θ0,θ1) = m1i=1∑me(i)m1i=1∑me(i)x(i) = m1(e(1)+e(2)+⋯+e(m))m1(e(1)x(1)+e(2)x(2)+⋯+e(m)x(m)) =m1[1x(1)1x(2)⋯⋯1x(m)] e(1)e(2)⋮e(m) =m1XTe=m1XT(Xθ−y)
- 由上述推导得
Δ θ = 1 m X T e \Delta{\theta} = \cfrac{1}{m} X^{T} e Δθ=m1XTe
θ : = θ − α Δ θ = θ − α 1 m X T e \theta := \theta - \alpha \Delta{\theta} = \theta - \alpha \cfrac{1}{m} X^{T} e θ:=θ−αΔθ=θ−αm1XTe
Python实现
导包
import numpy as np
import matplotlib.pyplot as plt
数据预处理
x = np.array([4, 3, 3, 4, 2, 2, 0, 1, 2, 5, 1, 2, 5, 1, 3])
y = np.array([8, 6, 6, 7, 4, 4, 2, 4, 5, 9, 3, 4, 8, 3, 6])
m = len(x)
x = np.c_[np.ones((m, 1)), x]
y = y.reshape(m, 1)
迭代过程
alpha = 0.01 # 学习率
iter_cnt = 1000 # 迭代次数
cost = np.zeros(iter_cnt) # 代价数据
theta = np.zeros((2, 1))
for i in range(iter_cnt):
h = x.dot(theta) # 估计值
error = h - y # 误差值
cost[i] = 1 / (2 * m) * error.T.dot(error) # 代价值
# cost[i] = 1 / (2 * m) * np.sum(np.square(error)) # 代价值
# 更新参数
delta_theta = 1 / m * x.T.dot(error)
theta -= alpha * delta_theta
结果可视化
# 线性拟合结果
plt.scatter(x[:, 1], y, c='blue')
plt.plot(x[:, 1], h, 'r-')
plt.savefig('../pic/fit.png')
plt.show()
# 代价结果
plt.plot(cost)
plt.savefig('../pic/cost.png')
plt.show()
完整代码
import numpy as np
import matplotlib.pyplot as plt
x = np.array([4, 3, 3, 4, 2, 2, 0, 1, 2, 5, 1, 2, 5, 1, 3])
y = np.array([8, 6, 6, 7, 4, 4, 2, 4, 5, 9, 3, 4, 8, 3, 6])
m = len(x)
x = np.c_[np.ones((m, 1)), x]
y = y.reshape(m, 1)
alpha = 0.01 # 学习率
iter_cnt = 1000 # 迭代次数
cost = np.zeros(iter_cnt) # 代价数据
theta = np.zeros((2, 1))
for i in range(iter_cnt):
h = x.dot(theta) # 估计值
error = h - y # 误差值
cost[i] = 1 / (2 * m) * error.T.dot(error) # 代价值
# cost[i] = 1 / (2 * m) * np.sum(np.square(error)) # 代价值
# 更新参数
delta_theta = 1 / m * x.T.dot(error)
theta -= alpha * delta_theta
# 线性拟合结果
plt.scatter(x[:, 1], y, c='blue')
plt.plot(x[:, 1], h, 'r-')
plt.savefig('../pic/fit.png')
plt.show()
# 代价结果
plt.plot(cost)
plt.savefig('../pic/cost.png')
plt.show()
结果可视化
线性拟合结果
代价变化
标签：right,matrix,梯度,回归,np,cfrac,线性,theta,left From： https://blog.csdn.net/from__2024_04_11/article/details/139156319