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Coursera 강의/Machine Learning

[Machine Learning] Exam 5 (Week 6)

by 별준 2020. 8. 19.
해당 내용은 Andrew Ng 교수님의 Machine Learning 강의(Coursera)를 정리한 내용입니다.

 

6주차 Programming Assignment는 다음과 같다.

linearRegCostFunction.m - Regularized linear regression의 Cost Function과 Grad J를 구하는 과제

learningCurve.m - Learning Curve를 생성하는 과제

polyFeatures.m - input X를 P차원의 방정식의 모델로 매핑하는 과제

validationCurve.m - CV Curve를 생성하는 과제

 

 

전체 코드는 GitHub 사이트에서 참조할 수 있다.

https://github.com/junstar92/Coursera/tree/master/MachineLearning/ex5

 

 

[linearRegCostFunction.m]

3주차의 Regularized Linear Regression에 대해서 Cost Function J(θ)J(\theta)θjJ(θ)\frac{\partial}{\partial\theta_j}J(\theta)를 구하는 코드를 작성해야 한다.

J(θ)J(\theta)θjJ(θ)\frac{\partial}{\partial\theta_j}J(\theta)의 식은 이전글을 참조하자.

2020/08/08 - [Machine Learning/Andrew Ng의 Machine Learning] - [Machine Learning] Regularization 정규화

 

코드는 아래와 같다.

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.
%
tempTheta = theta;
tempTheta(1) = 0;
h = X * theta;
error = h - y;
J = 1/(2*m) * (sum(sum(error.^2)) + (lambda * sum(sum(theta(2:end,:).^2))));
grad = 1/m * (X' * error + lambda * tempTheta);
% =========================================================================
grad = grad(:);
end

J(θ)J(\theta)θjJ(θ)\frac{\partial}{\partial\theta_j}J(\theta)의 식을 알고있다면, 코드를 이해하는데 큰 어려움을 없을 것이다.

부가적으로 설명하자면, 여기서 X는 x0x_0항이 이미 포함된 X이며, theta는 모두 1로 초기화된 값이다.

ex5.m

 

[learningCurve.m]

Jtrain(θ)J_{train}(\theta)JCV(θ)J_{CV}(\theta)를 구해서 training set size, m에 대한 그래프를 그리면 된다. 그래프는 ex5.m에서 그려주기 때문에 우리는 Jtrain(θ)J_{train}(\theta)JCV(θ)J_{CV}(\theta)의 값만 계산하면 된다.

다음의 함수를 구현하는 것이다.

주의해야 할 점은 m개수에 따른 Error를 모두 구해야 한다.

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

우선 Training Set(X, y)로 hθ(x)h_\theta(x)Jtrain(θ)J_{train}(\theta)를 구하고, CV Set(Xval, yval)로 JCV(θ)J_{CV}(\theta)를 구하면 된다. 마찬가지로 m개수에 따라 반복을 해주어야 한다.

코드는 다음과 같다.

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
theta = trainLinearReg(X(1:i, :), y(1:i, :), lambda);
error_train(i) = linearRegCostFunction(X(1:i, :), y(1:i, :), theta, 0);
error_val(i) = linearRegCostFunction(Xval, yval, theta, 0);
endfor
% -------------------------------------------------------------
% =========================================================================
end

 

 

[polyFeatures.m]

input X에 대해서 P차원으로 매핑하여 궁극적으로는 underfitting을 해결하도록 모델을 결정하는 코드를 작성하면 된다.

아래와 같이 P차원의 방정식으로 가설함수를 만든다.

ex5.m

위와 같이 1차원부터 P차원까지 구하면, 마지막에 θ0\theta_0에 대한 부분을 추가해주기 때문에, x,x2,...,xpx, x^2, ..., x^p만 구해주면 된다.

코드는 다음과 같다.

function [X_poly] = polyFeatures(X, p)
%POLYFEATURES Maps X (1D vector) into the p-th power
% [X_poly] = POLYFEATURES(X, p) takes a data matrix X (size m x 1) and
% maps each example into its polynomial features where
% X_poly(i, :) = [X(i) X(i).^2 X(i).^3 ... X(i).^p];
%
% You need to return the following variables correctly.
X_poly = zeros(numel(X), p);
% ====================== YOUR CODE HERE ======================
% Instructions: Given a vector X, return a matrix X_poly where the p-th
% column of X contains the values of X to the p-th power.
%
%
for i = 1:p
X_poly(:, i) = X.^i;
endfor
% =========================================================================
end

 

 

[validationCurve.m]

λ\lambda에 따른 Training Error와 CV Error를 구하는 함수를 구현해야 한다.

위와 같은 공식을 사용하여서, Jtrain(θ),Jcv(θ)J_{train}(\theta), J_{cv}(\theta)를 구한다.

 

코드는 아래와 같다.

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
%
%
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);
endfor
% =========================================================================
end

 

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