Question

我对梯度下降算法的简单实现有疑问- 我编写了以下代码，只是将GD的数学运算转换为matlab代码，但并没有收敛。由于此实现方式的简单性，我认为在这里缺少有关Matlab调试的非常关键的信息。

此代码的目标是估计线性回归系数，该系数使我们得到最小误差。因此，我加载了一些数据（'trees.data.txt'），然后将这些数据分成了训练集和测试集，我想使用训练和梯度下降来提取系数，但是不幸的是它不能收敛。

这是迭代方程- theta = theta-（alpha / m）*（（X * theta-y）'* X）';

任何人都可以解释一下是什么问题吗？

我的实现-

 new DropdownButton(
                            value: _current,
    items: dataMaker.map((item) {
                              return new DropdownMenuItem<String>(
                                child: new Text(item.fabricante,
                                    textAlign: TextAlign.center),
                                value: item.fabricante, //FAIL
                              );
                            }).toList(),
                            onChanged: (selected) => setState(() {
      _current = selected;
    });

这是前10次迭代的输出-

% Load trees data from file.
data = load('trees.data.txt');
data=data'; % put examples in columns

% Include a row of 1s as an additional intercept feature.
data = [ ones(1,size(data,2)); data ];

% Shuffle examples.
data = data(:, randperm(size(data,2)));

% Split into train and test sets
% The last row of 'data' is the median home price.
train.X = data(1:end-1,1:400);
train.y = data(end,1:400);

test.X = data(1:end-1,401:end);
c = data(end,401:end);

m=size(train.X,2);
n=size(train.X,1);

% Initialize the coefficient vector theta to random values.
theta = rand(n,1);
X = test.X;
y = test.y;

theta_=zeros(size(theta));
delta =0.01; % convergence tolerance
alpha = 0.001; % learning rate
shift = 1000; % big number
iter = 0;
formatSpec = 'iteration: %d, error: %2.4f\n';


while (shift > delta)

   iter = iter +1;
   grad =zeros(size(theta));

   for i = 1:m
       grad = grad + (train.X(:,i)'*theta - train.y(i)).*train.X(:,i);
   end
    %theta_= theta-(alpha*(1/m)*((theta'*X-y)*X')');
    theta_= theta - alpha*(1/m)*grad;
    shift = norm(theta_ - theta);

  fprintf(formatSpec, iter, shift);  
  theta = theta_;
  clear theta_;
end

梯度下降的实现（Matlab）

0 个答案: