我自己做个网站怎么做,做ppt封面的网站,文具电子商务网站开发内容,wordpress主题虚拟会员下载文章目录 [toc]数据数据集实际值估计值 梯度下降算法估计误差代价函数学习率参数更新 Python实现导包数据预处理迭代过程结果可视化完整代码 结果可视化线性拟合结果代价变化 数据
数据集 ( x ( i ) , y ( i ) ) , i 1 , 2 , ⋯ , m \left(x^{(i)} , y^{(i)}\right) , i 1 ,… 文章目录 [toc]数据数据集实际值估计值 梯度下降算法估计误差代价函数学习率参数更新 Python实现导包数据预处理迭代过程结果可视化完整代码 结果可视化线性拟合结果代价变化
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)),i1,2,⋯,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))θ0θ1x(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)2m1i1∑m(hθ(x(i))−y(i))22m1i1∑m(θ0θ1x(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−αΔθjθj−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) m1i1∑m(hθ(x(i))−y(i))m1i1∑m(hθ(x(i))−y(i))x(i) m1i1∑me(i)m1i1∑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) m1i1∑me(i)m1i1∑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) m1XTem1XT(Xθ−y)
由上述推导得 Δ θ 1 m X T e \Delta{\theta} \cfrac{1}{m} X^{T} e Δθm1XTe 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−αΔθθ−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, cblue)
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 pltx 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, cblue)
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
