【题解】编程作业ex5: Regularized Linear Regression and Bias/Variance (Machine Learning)

吐槽：这个作业用了几个小时，，光是复习前面的就整了半天=。=不过问题不大，也就debug了好久，虽然最后我还是编出来了+_+以后一定要一鼓作气完成某个课程。。。

题目：

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linearRegCostFunction我的解法：

为什么想写个这个呢，虽然挺简单的，但是一开始复习的时候以为我编写过就直接抄以前ex2的代码了，结果发现没有sigmoid嘛，仔细看看注意到linear才反应过来。。就很蠢吧。。

function [J, grad] = linearRegCostFunction(X, y, theta, lambda)

%LINEARREGCOSTFUNCTION Compute cost and gradient for regularized linear 

%regression with multiple variables

%   [J, grad] = LINEARREGCOSTFUNCTION(X, y, theta, lambda) computes the 

%   cost of using theta as the parameter for linear regression to fit the 

%   data points in X and y. Returns the cost in J and the gradient in grad

% Initialize some useful values

m = length(y); % number of training examples

% You need to return the following variables correctly 

J = 0;

grad = zeros(size(theta));

% ====================== YOUR CODE HERE ======================

% Instructions: Compute the cost and gradient of regularized linear 

%               regression for a particular choice of theta.

%

%               You should set J to the cost and grad to the gradient.

%

h = X * theta;

J = 1/(2*m) * (sum((h - y).^2) + lambda * sum(theta(2:end,1).^2));

grad = 1/m * X' * (h - y) + lambda/m * theta; 

grad(1) = (1/m * X' * (h - y))(1);

% =========================================================================

grad = grad(:);

end

learningCurve我的题解：

这有点坑人的，我后面俩函数的参数一开始写的lambda不是0，不是那题目给的就是lambda为0的情况嘛，我一直调试觉得是没问题啊，结果submit每次没分，写到最后一个函数的时候参数改成0的时候发现这里改一下就通过了。。就很汗颜= =||以及最后那个validationCurve函数就是把1到m的循环改成1到lambda个数的循环，所以不用放了。。

function [error_train, error_val] = ...

    learningCurve(X, y, Xval, yval, lambda)

%LEARNINGCURVE Generates the train and cross validation set errors needed 

%to plot a learning curve

%   [error_train, error_val] = ...

%       LEARNINGCURVE(X, y, Xval, yval, lambda) returns the train and

%       cross validation set errors for a learning curve. In particular, 

%       it returns two vectors of the same length - error_train and 

%       error_val. Then, error_train(i) contains the training error for

%       i examples (and similarly for error_val(i)).

%

%   In this function, you will compute the train and test errors for

%   dataset sizes from 1 up to m. In practice, when working with larger

%   datasets, you might want to do this in larger intervals.

%

% Number of training examples

m = size(X, 1);

% You need to return these values correctly

error_train = zeros(m, 1);

error_val   = zeros(m, 1);

% ====================== YOUR CODE HERE ======================

% Instructions: Fill in this function to return training errors in 

%               error_train and the cross validation errors in error_val. 

%               i.e., error_train(i) and 

%               error_val(i) should give you the errors

%               obtained after training on i examples.

%

% Note: You should evaluate the training error on the first i training

%       examples (i.e., X(1:i, :) and y(1:i)).

%

%       For the cross-validation error, you should instead evaluate on

%       the _entire_ cross validation set (Xval and yval).

%

% Note: If you are using your cost function (linearRegCostFunction)

%       to compute the training and cross validation error, you should 

%       call the function with the lambda argument set to 0. 

%       Do note that you will still need to use lambda when running

%       the training to obtain the theta parameters.

%

% Hint: You can loop over the examples with the following:

%

%       for i = 1:m

%           % Compute train/cross validation errors using training examples 

%           % X(1:i, :) and y(1:i), storing the result in 

%           % error_train(i) and error_val(i)

%           ....

%           

%       end

%

% ---------------------- Sample Solution ----------------------

for i = 1:m,

  X_test = X(1:i, :);

  y_test = y(1:i);

  theta = trainLinearReg(X_test, y_test, lambda);

  error_train(i) = linearRegCostFunction(X_test, y_test, theta, 0);

  error_val(i) = linearRegCostFunction(Xval, yval, theta, 0);

endfor

% -------------------------------------------------------------

% =========================================================================

end