Coursera-Machine Learning-ex5

最新推荐文章于 2020-07-18 22:18:15 发布

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最新推荐文章于 2020-07-18 22:18:15 发布

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分类专栏： Deep Learning 文章标签： MachineLearning

本文链接：https://blog.youkuaiyun.com/language_zcx/article/details/85758453

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本文详细介绍了线性回归的成本函数实现与正则化处理，通过learningCurve和validationCurve函数，展示了如何评估模型在不同数据集上的表现，并选择最佳的正则化参数λ。

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Some Points:

The Results:

linearRegCostFunction.m

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

% 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));

diff = X * theta - y;
regu = lambda * theta(2:end)' * theta(2:end);
J = (diff' * diff + regu) / (2 * m);

grad = (X' * diff + lambda * [0; theta(2:end)]) / m;
grad = grad(:);

end

learningCurve.m

function [error_train, error_val] = ...
    learningCurve(X, y, Xval, yval, lambda)


% 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);


for i = 1:m
	theta = trainLinearReg(X(1:i, :), y(1:i), lambda);
	error_train(i) = linearRegCostFunction(X(1:i, :), y(1:i), theta, 0);
	% for the cross validation error,you should compute it over the entire cross validation set.
	error_val(i) = linearRegCostFunction(Xval, yval, theta, 0);
end

end

validationCurve.m

function [lambda_vec, error_train, error_val] = ...
    validationCurve(X, y, 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);


for i =1:length(lambda_vec)
	lambda = lambda_vec(i);
	theta = trainLinearReg(X, y, lambda);
	error_train(i) = linearRegCostFunction(X, y, theta, 0);
	error_val(i) = linearRegCostFunction(Xval, yval, theta, 0);
end


end