coursera 机器学习 linear regression 线性回归的小项目

  1 %% Machine Learning Online Class
  2 %  Exercise 1: Linear regression with multiple variables
  3 %
  4 %  Instructions
  5 %  ------------
  6 % 
  7 %  This file contains code that helps you get started on the
  8 %  linear regression exercise. 
  9 %
 10 %  You will need to complete the following functions in this 
 11 %  exericse:
 12 %
 13 %     warmUpExercise.m
 14 %     plotData.m
 15 %     gradientDescent.m
 16 %     computeCost.m
 17 %     gradientDescentMulti.m
 18 %     computeCostMulti.m
 19 %     featureNormalize.m
 20 %     normalEqn.m
 21 %
 22 %  For this part of the exercise, you will need to change some
 23 %  parts of the code below for various experiments (e.g., changing
 24 %  learning rates).
 25 %
 26 
 27 %% Initialization
 28 
 29 %% ================ Part 1: Feature Normalization ================
 30 
 31 %% Clear and Close Figures
 32 clear ; close all; clc
 33 
 34 fprintf(‘Loading data ...\n‘);
 35 
 36 %% Load Data
 37 data = load(‘ex1data2.txt‘);
 38 X = data(:, 1:2);
 39 y = data(:, 3);
 40 m = length(y);
 41 
 42 % Print out some data points
 43 fprintf(‘First 10 examples from the dataset: \n‘);
 44 fprintf(‘ x = [%.0f %.0f], y = %.0f \n‘, [X(1:10,:) y(1:10,:)]‘);
 45 
 46 fprintf(‘Program paused. Press enter to continue.\n‘);
 47 pause;
 48 
 49 % Scale features and set them to zero mean
 50 fprintf(‘Normalizing Features ...\n‘);
 51 
 52 [X mu sigma] = featureNormalize(X);
 53 
 54 % Add intercept term to X
 55 X = [ones(m, 1) X];
 56 
 57 
 58 %% ================ Part 2: Gradient Descent ================
 59 
 60 % ====================== YOUR CODE HERE ======================
 61 % Instructions: We have provided you with the following starter
 62 %               code that runs gradient descent with a particular
 63 %               learning rate (alpha). 
 64 %
 65 %               Your task is to first make sure that your functions - 
 66 %               computeCost and gradientDescent already work with 
 67 %               this starter code and support multiple variables.
 68 %
 69 %               After that, try running gradient descent with 
 70 %               different values of alpha and see which one gives
 71 %               you the best result.
 72 %
 73 %               Finally, you should complete the code at the end
 74 %               to predict the price of a 1650 sq-ft, 3 br house.
 75 %
 76 % Hint: By using the ‘hold on‘ command, you can plot multiple
 77 %       graphs on the same figure.
 78 %
 79 % Hint: At prediction, make sure you do the same feature normalization.
 80 %
 81 
 82 fprintf(‘Running gradient descent ...\n‘);
 83 
 84 % Choose some alpha value
 85 alpha = 0.01;
 86 num_iters = 400;
 87 
 88 % Init Theta and Run Gradient Descent 
 89 theta = zeros(3, 1);
 90 [theta, J_history] = gradientDescentMulti(X, y, theta, alpha, num_iters);
 91 
 92 % Plot the convergence graph
 93 figure;
 94 plot(1:numel(J_history), J_history, ‘-b‘, ‘LineWidth‘, 2);
 95 xlabel(‘Number of iterations‘);
 96 ylabel(‘Cost J‘);
 97 
 98 % Display gradient descent‘s result
 99 fprintf(‘Theta computed from gradient descent: \n‘);
100 fprintf(‘ %f \n‘, theta);
101 fprintf(‘\n‘);
102 
103 % Estimate the price of a 1650 sq-ft, 3 br house
104 % ====================== YOUR CODE HERE ======================
105 % Recall that the first column of X is all-ones. Thus, it does
106 % not need to be normalized.
107 price = 0; % You should change this
108 X_1 = [1 1650 3] ; 
109 price  = X_1 * theta ;
110 % ============================================================
111 
112 fprintf([‘Predicted price of a 1650 sq-ft, 3 br house ‘ ...
113          ‘(using gradient descent):\n $%f\n‘], price);
114 
115 fprintf(‘Program paused. Press enter to continue.\n‘);
116 pause;
117 
118 %% ================ Part 3: Normal Equations ================
119 
120 fprintf(‘Solving with normal equations...\n‘);
121 
122 % ====================== YOUR CODE HERE ======================
123 % Instructions: The following code computes the closed form 
124 %               solution for linear regression using the normal
125 %               equations. You should complete the code in 
126 %               normalEqn.m
127 %
128 %               After doing so, you should complete this code 
129 %               to predict the price of a 1650 sq-ft, 3 br house.
130 %
131 
132 %% Load Data
133 data = csvread(‘ex1data2.txt‘);
134 X = data(:, 1:2);
135 y = data(:, 3);
136 m = length(y);
137 
138 % Add intercept term to X
139 X = [ones(m, 1) X];
140 
141 % Calculate the parameters from the normal equation
142 theta = normalEqn(X, y);
143 
144 % Display normal equation‘s result
145 fprintf(‘Theta computed from the normal equations: \n‘);
146 fprintf(‘ %f \n‘, theta);
147 fprintf(‘\n‘);
148 
149 
150 % Estimate the price of a 1650 sq-ft, 3 br house
151 % ====================== YOUR CODE HERE ======================
152 price = 0; % You should change this
153 X_2 = [ 1 1650 3] ; 
154 price  = X_2 * theta ;
155 % ============================================================
156 
157 fprintf([‘Predicted price of a 1650 sq-ft, 3 br house ‘ ...
158          ‘(using normal equations):\n $%f\n‘], price);

 1 function [X_norm, mu, sigma] = featureNormalize(X)
 2 %FEATURENORMALIZE Normalizes the features in X 
 3 %   FEATURENORMALIZE(X) returns a normalized version of X where
 4 %   the mean value of each feature is 0 and the standard deviation
 5 %   is 1. This is often a good preprocessing step to do when
 6 %   working with learning algorithms.
 7 
 8 % You need to set these values correctly
 9 X_norm = X;
10 mu = zeros(1, size(X, 2));
11 sigma = zeros(1, size(X, 2));
12 
13 % ====================== YOUR CODE HERE ======================
14 % Instructions: First, for each feature dimension, compute the mean
15 %               of the feature and subtract it from the dataset,
16 %               storing the mean value in mu. Next, compute the 
17 %               standard deviation of each feature and divide
18 %               each feature by it‘s standard deviation, storing
19 %               the standard deviation in sigma. 
20 %
21 %               Note that X is a matrix where each column is a 
22 %               feature and each row is an example. You need 
23 %               to perform the normalization separately for 
24 %               each feature. 
25 %
26 % Hint: You might find the ‘mean‘ and ‘std‘ functions useful.
27 %       
28 mu = mean(X); % 1 * ( n + 1 )
29 sigma = std(X);
30 for i = 1 : size(X,1)
31     X_norm(i,:) = ( X_norm(i,:) - mu ) ./ sigma ; % 对每个行向量做减法
32 end
33 %  X_norm(1:10,:)
34 
35 % ============================================================
36 
37 end

 1 function [theta] = normalEqn(X, y)
 2 %NORMALEQN Computes the closed-form solution to linear regression 
 3 %   NORMALEQN(X,y) computes the closed-form solution to linear 
 4 %   regression using the normal equations.
 5 
 6 theta = zeros(size(X, 2), 1);
 7 
 8 % ====================== YOUR CODE HERE ======================
 9 % Instructions: Complete the code to compute the closed form solution
10 %               to linear regression and put the result in theta.
11 %
12 
13 % ---------------------- Sample Solution ----------------------
14 
15 theta = pinv(X‘*X)*X‘*y; 
16 
17 
18 % -------------------------------------------------------------
19 
20 
21 % ============================================================
22 
23 end