In this article, you will find **Coursera machine learning week 6 assignment answers – Andrew Ng. ** Use “Ctrl+F” To Find Any Questions or Answers. For Mobile Users, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Options to Get Any Random Questions Answer.

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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.

### Coursera machine learning week 6 assignment answers

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

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

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

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

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### FAQs

**Is Andrew Ng’s Machine Learning course good?**

It is the Best Course for Supervised Machine Learning! Andrew Ng Sir has been like always has such important & difficult concepts of Supervised ML with such ease and great examples, Just amazing!

**How do I get answers to coursera assignment?**

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**How long does it take to finish coursera Machine Learning?**

this specialization requires approximately **3 months** with 75 hours of materials to complete, and I finished it in 3 weeks and spent an additional 1 week reviewing the whole course.

**How do you submit assignments on Coursera Machine Learning?**

Submit a programming assignment **Open the assignment page for the assignment you want to submit**. Read the assignment instructions and download any starter files. Finish the coding tasks in your local coding environment. Check the starter files and instructions when you need to. Reference