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