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机器学习(coursera 斯坦福)第六周编程作业

Preface

本文是机器学习第六周部分课程内容,以及编程作业答案记录,完整题目以及代码见github


作业

linearRegCostFunction.m

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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:end).^ 2);

grad = 1 / m * (X' * (X * theta - y));
grad(2:end) = grad(2:end) + lambda / m * theta(2:end);

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

grad = grad(:);

end

learningCurve.m

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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);
end

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

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

end

polyFeatures.m

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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 j = 1:p
X_poly(:,j) = X.^j;
end
% =========================================================================
end

validationCurve.m

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

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

end

课程部分内容记录

Evaluating a Hypothesis

Once we have done some trouble shooting for errors in our predictions by:

1.Getting more training examples
2.Trying smaller sets of features
3.Trying additional features
4.Trying polynomial features
5.Increasing or decreasing λ
6.We can move on to evaluate our new hypothesis.

A hypothesis may have a low error for the training examples but still be inaccurate (because of overfitting). Thus, to evaluate a hypothesis, given a dataset of training examples, we can split up the data into two sets: a training set and a test set. Typically, the training set consists of 70 % of your data and the test set is the remaining 30 %.


Diagnosing Bias vs. Variance

High bias (underfitting): both $J_{train}(\Theta)$ and $J_{CV}(\Theta)$ will be high. Also, $J_{CV}(\Theta) \approx J_{train}(\Theta)$
High variance (overfitting): $J_{train}(\Theta)$ will be low and $J_{CV}(\Theta)$ will be much greater than $J_{train}(\Theta)$
The is summarized in the figure below:

1.png


Learning Curves

2.png


Deciding What to Do Next Revisited

Our decision process can be broken down as follows:

  1. Getting more training examples: Fixes high variance
  2. Trying smaller sets of features: Fixes high variance
  3. Adding features: Fixes high bias
  4. Adding polynomial features: Fixes high bias
  5. Decreasing λ: Fixes high bias
  6. Increasing λ: Fixes high variance.

3.png


参考

Coursera吴恩达机器学习课程 总结笔记及作业代码——第6周有关机器学习的小建议
matlab函数 bsxfun浅谈(转载)
Octave学习随笔(更新至6.16)【bsxfun】
ldivide, .\

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