Coursera: Machine Learning (Week 6) [Assignment Solution] - Andrew NG

▸ Regularized linear regression to study models with different bias-variance properties.

I have recently completed the Machine Learning course from Coursera by Andrew NG.

While doing the course we have to go through various quizzes and assignments.

Here, I am sharing my solutions for the weekly assignments throughout the course.

These solutions are for reference only.

> It is recommended that you should solve the assignments by yourself honestly then only it makes sense to complete the course.

> But, In case you stuck in between, feel free to refer to the solutions provided by me.

NOTE:

Don't just copy-paste the code for the sake of completion. 

Even if you copy the code, make sure you understand the code first.

Click here to check out week-5 assignment solutions, Scroll down for the solutions for week-6 assignment.




In this exercise, you will implement regularized linear regression and use it to study models with different bias-variance properties. Before starting on the programming exercise, we strongly recommend watching the video lectures and completing the review questions for the associated topics.

Recommended Machine Learning Courses:

1. Coursera: Machine Learning  
2. Coursera: Deep Learning Specialization
3. Coursera: Machine Learning with Python
4. Coursera: Advanced Machine Learning Specialization
5. Udemy: Machine Learning
6. LinkedIn: Machine Learning
7. Eduonix: Machine Learning
8. edX: Machine Learning
9. Fast.ai: Introduction to Machine Learning for Coders

It consist of the following files:

• ex5.m - Octave/MATLAB script that steps you through the exercise
• ex5data1.mat - Dataset
• submit.m - Submission script that sends your solutions to our servers
• featureNormalize.m - Feature normalization function
• fmincg.m - Function minimization routine (similar to fminunc)
• plotFit.m - Plot a polynomial fit
• trainLinearReg.m - Trains linear regression using your cost function
• [*] linearRegCostFunction.m - Regularized linear regression cost function
• [*] learningCurve.m - Generates a learning curve
• [*] polyFeatures.m - Maps data into polynomial feature space
• [*] validationCurve.m - Generates a cross-validation curve
• Video - YouTube videos featuring Free IOT/ML tutorials

* indicates files you will need to complete

linearRegCostFunction.m :

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.
  %DIMENSIONS:
  %   X = 12x2 = m x 1
  %   y = 12x1 = m x 1
  %   theta = 2x1 = (n+1) x 1
  %   grad = 2x1 = (n+1) x 1
  
  h_x = X * theta; % 12x1
  J = (1/(2*m))*sum((h_x - y).^2) + (lambda/(2*m))*sum(theta(2:end).^2); % scalar
  
  % grad(1) = (1/m)*sum((h_x-y).*X(:,1)); % scalar == 1x1
  grad(1) = (1/m)*(X(:,1)'*(h_x-y)); % scalar == 1x1
  grad(2:end) = (1/m)*(X(:,2:end)'*(h_x-y)) + (lambda/m)*theta(2:end); % n x 1
  % =========================================================================
  
  grad = grad(:);

end

learningCurve.m :

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 ----------------------
  
  %DIMENSIONS:
  %   error_train = m x 1
  %   error_val   = m x 1 
  
  for i = 1:m
      Xtrain = X(1:i,:);
      ytrain = y(1:i);
      
      theta = trainLinearReg(Xtrain, ytrain, lambda);
      
      error_train(i) = linearRegCostFunction(Xtrain, ytrain, theta, 0); %for lambda = 0;
      error_val(i)   = linearRegCostFunction(Xval, yval, theta, 0); %for lambda = 0;
  end
  
  % -------------------------------------------------------------
  
  % =========================================================================

end

polyFeatures.m :

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); % m 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.
  %
  %
  % Here, X does not include X0 == 1 column
  
  %%%% WORKING: Using for loop %%%%%%
  % for i = 1:p
  %     X_poly(:,i) = X(:,1).^i;
  % end
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  X_poly(:,1:p) = X(:,1).^(1:p); % w/o for loop
  
  % =========================================================================
end

Check-out our free tutorials on IOT (Internet of Things):

validationCurve.m :

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
  %
  %
  
  % Here, X & Xval are already including x0 i.e 1's column in it
  
  m = size(X, 1);
  
  %% %%%%% WORKING: BUT UNNECESSARY for loop for i is inovolved %%%%%%%%%%%
  % for i = 1:m
  %     for j = 1:length(lambda_vec);
  %         lambda = lambda_vec(j);
  %         Xtrain = X(1:i,:);
  %         ytrain = y(1:i);
  %
  %         theta = trainLinearReg(Xtrain, ytrain, lambda);
  %
  %         error_train(j) = linearRegCostFunction(Xtrain, ytrain, theta, 0); % lambda = 0;
  %         error_val(j)   = linearRegCostFunction(Xval, yval, theta, 0); % lambda = 0;
  %     end
  % end
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  %% %%%%%%% WORKING: BUT UNNECESSARY for loop for i is inovolved %%%%%%%%%%%
  % for j = 1:length(lambda_vec)
  %     lambda = lambda_vec(j);
  %     for i = 1:m
  %         Xtrain = X(1:i,:);
  %         ytrain = y(1:i);
  %
  %         theta = trainLinearReg(Xtrain, ytrain, lambda);
  %
  %         error_train(j) = linearRegCostFunction(Xtrain, ytrain, theta, 0); % lambda = 0;
  %         error_val(j)   = linearRegCostFunction(Xval, yval, theta, 0); % lambda = 0;
  %     end
  % end
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  %% %%% NOT WORKING: BUT UNNECESSARY for loop inside learningCurve function is inovolved %%%%%%
  % for j = 1:length(lambda_vec)
  %     lambda = lambda_vec(j);
  %
  %     [error_train_temp, error_val_temp] = ...
  %     learningCurve(X, y, ...
  %                   Xval, yval, ...
  %                   lambda);
  %
  %     error_train(j) = error_train_temp(end);
  %     error_val(j) = error_val_temp(end);
  % end
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  
  %% %%%%% WORKING: OPTIMISED (Only 1 for loop) %%%%%%%%%%%
  for j = 1:length(lambda_vec)
      lambda = lambda_vec(j);
      
      theta = trainLinearReg(X, y, lambda);
      error_train(j) = linearRegCostFunction(X, y, theta, 0); % lambda = 0;
      error_val(j)   = linearRegCostFunction(Xval, yval, theta, 0); % lambda = 0
  end
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  
  % =========================================================================
  
end

I tried to provide optimized solutions like vectorized implementation for each assignment. If you think that more optimization can be done, then put suggest the corrections / improvements.

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Click here to see solutions for all Machine Learning Coursera Assignments.

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Feel free to ask doubts in the comment section. I will try my best to solve it.

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This is the simplest way to encourage me to keep doing such work.

Thanks and Regards,

-Akshay P. Daga