例子：预测电影评分
有如下信息
| Movie | Alice (1) | Bob (2) | Carol (3) | Dave (4) |
| Love at last | 5 | 5 | 0 | 0 |
| Romance forever | 5 | ? | ? | 0 |
| Cute puppies of love | ? | 4 | 0 | ? |
| Nonstop car chases | 0 | 0 | 5 | 4 |
| Swords vs. karate | 0 | 0 | 5 | ? |
定义
nu = 用户的数量
nm = 电影的数量
r(i, j) = 1 如果用户 j 给电影 i 打分
y(i, j) = 当 r(i, j) = 1 的情况下，用户 j 给电影 i 打的分数（0-5）
目标：预测 ？的值（未评分用户对电影的评分）
现假设每个电影有两个特征
| Movie | Alice (1) | Bob (2) | Carol (3) | Dave (4) | x1 (remoance) |
x2 (remoance) |
| Love at last | 5 | 5 | 0 | 0 | 0.9 | 0 |
| Romance forever | 5 | ? | ? | 0 | 1.0 | 0.01 |
| Cute puppies of love | ? | 4 | 0 | ? | 0.99 | 0 |
| Nonstop car chases | 0 | 0 | 5 | 4 | 0.1 | 1.0 |
| Swords vs. karate | 0 | 0 | 5 | ? | 0 | 0.9 |
这样就有了电影特征的训练集，比如对于电影 Love at least 的特征向量为
\[{x^{\left( 1 \right)}} = \left[ {\begin{array}{*{20}{c}}
{{x_0}}\\
{{x_1}}\\
{{x_2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
{0.9}\\
0
\end{array}} \right]\]
对于每个用户 j ，学习其对应的单数 θ(j) ∈ R3，然后用 (θ(1))Tx(i) 预测用户 j 对于电影 i 的评分。
用 m(j) = 代表用户 j 评分的电影的数量，则定义学习目标
\[\underbrace {\min }_{{\theta ^{\left( j \right)}}}\frac{1}{{2{m^{\left( j \right)}}}}\sum\limits_{i:{r^{\left( {i,j} \right)}} = 1} {{{\left( {{{\left( {{\theta ^{\left( j \right)}}} \right)}^T}\left( {{x^{\left( i \right)}}} \right) - {y^{\left( {i,j} \right)}}} \right)}^2}} + \frac{\lambda }{{2{m^{\left( j \right)}}}}\sum\limits_{k = 1}^n {{{\left( {\theta _k^{\left( j \right)}} \right)}^2}} \]
在推荐系统中会将 m(j) 去掉，因为他是常数且不会影响计算得到的 θ(j)。
\[\underbrace {\min }_{{\theta ^{\left( j \right)}}}\frac{1}{2}\sum\limits_{i:{r^{\left( {i,j} \right)}} = 1} {{{\left( {{{\left( {{\theta ^{\left( j \right)}}} \right)}^T}\left( {{x^{\left( i \right)}}} \right) - {y^{\left( {i,j} \right)}}} \right)}^2}} + \frac{\lambda }{2}\sum\limits_{k = 1}^n {{{\left( {\theta _k^{\left( j \right)}} \right)}^2}} \]
定义所有用户的学习目标
\[\underbrace {\min }_{{\theta ^{\left( 1 \right)}},...,{\theta ^{\left( {{n_u}} \right)}}}\frac{1}{2}\sum\limits_{j = 1}^{{n_u}} {\left[ {\sum\limits_{i:{r^{\left( {i,j} \right)}} = 1} {{{\left( {{{\left( {{\theta ^{\left( j \right)}}} \right)}^T}\left( {{x^{\left( i \right)}}} \right) - {y^{\left( {i,j} \right)}}} \right)}^2}} + \lambda \sum\limits_{k = 1}^n {{{\left( {\theta _k^{\left( j \right)}} \right)}^2}} } \right]} \]
然后运用梯度下降算法得到最优的 θ
\[\begin{array}{l}
\theta _k^{\left( j \right)}: = \theta _k^{\left( j \right)} - \alpha \sum\limits_{i:{r^{\left( {i,j} \right)}} = 1} {\left( {{{\left( {{\theta ^{\left( j \right)}}} \right)}^T}\left( {{x^{\left( i \right)}}} \right) - {y^{\left( {i,j} \right)}}} \right)x_k^{\left( i \right)}} ---for-k=0\\
\theta _k^{\left( j \right)}: = \theta _k^{\left( j \right)} - \alpha \left( {\sum\limits_{i:{r^{\left( {i,j} \right)}} = 1} {\left( {{{\left( {{\theta ^{\left( j \right)}}} \right)}^T}\left( {{x^{\left( i \right)}}} \right) - {y^{\left( {i,j} \right)}}} \right)x_k^{\left( i \right)} + + \lambda \theta _k^{\left( j \right)}} } \right)---for-k≠0
\end{array}\]