Coursera机器学习 week6 编程作业代码

这是Coursera上 Week4 的 “神经网络的表示” 的编程作业代码。经过测验，全部通过。

下面是 linearRegCostFunction.m 的代码：

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.
%

J = 1/(2*m) * sum((X*theta-y).^2) + lambda/(2*m) * sum(theta.^2); % 计算包含惩罚项的误差函数
J = J - lambda/(2*m) * theta(1)^2; % 减去theta0的平方，不需要惩罚theta0

grad(1) = 1/m * sum((X*theta-y)); % theta0的偏导
for i = 2:size(theta,1)
    grad(i) = 1/m * sum((X*theta-y).*X(:,i)) + lambda/m * theta(i); % theta1-thetan的偏导
end

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

grad = grad(:);

end

下面是 learningCurve.m  的代码：

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 ----------------------
n = size(Xval,1); % number of cross validation set
for i = 1:m
    theta = trainLinearReg(X(1:i,:), y(1:i,:), lambda);
    error_train(i) = 1/(2*i) * sum((X(1:i,:)*theta - y(1:i,:)).^2);  % training error
    error_val(i) = 1/(2*n) * sum((Xval*theta - yval).^2);            % cross validation error
end

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

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

end

下面是 validationCurve.m  的代码：

function [lambda_vec, error_train, error_val] = ...
    validationCurve(X, y, Xval, yval)
%VALIDATIONCURVE Generate the train and validation errors needed to
%plot a validation curve that we can use to select lambda
%   [lambda_vec, error_train, error_val] = ...
%       VALIDATIONCURVE(X, y, Xval, yval) returns the train
%       and validation errors (in error_train, error_val)
%       for different values of lambda. You are given the training set (X,
%       y) and validation set (Xval, yval).
%

% Selected values of lambda (you should not change this)
lambda_vec = [0 0.001 0.003 0.01 0.03 0.1 0.3 1 3 10]';

% You need to return these variables correctly.
error_train = zeros(length(lambda_vec), 1);
error_val = zeros(length(lambda_vec), 1);

% ====================== YOUR CODE HERE ======================
% Instructions: Fill in this function to return training errors in 
%               error_train and the validation errors in error_val. The 
%               vector lambda_vec contains the different lambda parameters 
%               to use for each calculation of the errors, i.e, 
%               error_train(i), and error_val(i) should give 
%               you the errors obtained after training with 
%               lambda = lambda_vec(i)
%
% Note: You can loop over lambda_vec with the following:
%
%       for i = 1:length(lambda_vec)
%           lambda = lambda_vec(i);
%           % Compute train / val errors when training linear 
%           % regression with regularization parameter lambda
%           % You should store the result in error_train(i)
%           % and error_val(i)
%           ....
%           
%       end
%
%
m = size(X, 1);
n = size(Xval, 1);
for i = 1:length(lambda_vec)
    lambda = lambda_vec(i);
    theta = trainLinearReg(X, y, lambda);
    error_train(i) = 1/(2*m) * sum((X*theta - y).^2);        % training error
    error_val(i) = 1/(2*n) * sum((Xval*theta - yval).^2);    % cross validation error
end

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

end