week02.md

Multiple features

• Multiple features(variables)

  Notation:

  • n = number of features
  • $x^{(i)}$ = input (features) of $i^{th}$ training example
  • $x^{(i)}_j$ = value of feature $j$ in $i^{th}$ training example

  Hypotheses for Multivariate linear regression $$ h_\theta(x) = \theta_0 + \theta_1 x_1 + \theta_2 x_2 + ... + \theta_n x_n $$

  $$ --> h_\theta(x) = \theta^T x $$

  ​ Remark: for convenience of notation, define $x_0$ = 1

  ​ $x$ is (n+1) x 1 vector, $\theta$ is also (n+1) x 1 vector

• Gradient descent for multiple variables

  

• Gradient descent in practice

  Feature Scaling

  Idea: Make sure features are on a similar scale

  Target: Get every feature into approximately a $-1\leq x_i \leq 1$ range

  Mean normalization

  Remark: do not apply to $x_0$ = 1

  Learning rate

  • If $\alpha$ is too small; slow convergence.
  • If $\alpha$ too large; $J(\theta)$ may not decrease on every iteration; may not converge; slow convergence also possible.

• Features and polynomial regression

  $h_\theta(x) = \theta_0 + \theta_1(size) + \theta_x(size)^2$

• Normal Equation

  method to solve for $\theta$ analytically

  

  $m$ examples $(x^{(1)}, y^{(1)}),...(x^{(m)}, y^{(m)})$; $n$ features

  $\theta = (X^TX)^{-1}X^Ty$

  Octave: $pinv (X'*X)*X'*y$

​

• Normal equation and non-invertibility(option)

  non-invertible usually caused by:

  • Redundant features (linearly dependent)
  • Too many features($e.g. m\leq n$)
    • delete some features, or use regularization

Octave tutorial

• Basic operations/ Moving data around

  1~=2 % not equal
  1 && 0 %AND
  1 || 0 %or
  xor(0,1) %
  a = 3
  v = [1 2 3] %row vector
  v = [1;2;3] %column vector
  v = 1:0.1:2 %create a column vector, [1;1.1;1.2;...2]

  help ones %display the help text for ones function
  ones(2,3) %create a 2x3 matrixs, elements equals to 1
  rand(3,3) %create a random 3x3 matrixs, elements less than 1
  w = -6 + sqrt(10)*(rand(1,10))
  hist(w)
  hist(w, 5) %plot a histogram of the values in 5 bins
  eye(4) %return a identity matrix
  zeros(10,1) %return a 10x1 matrix whose elements are all 0
  
  size(A) %return the number of rows and columns of A
  size(X, 1) %return row of X
  size(x, 2) %return column of X
  
  [nr, nc] = size(A);
  length(A) %for empty object:0, for scalars:1, for vectors: elements qty
  		%for matrix: rows or columns qty, depends on which is larger
  V + ones(length(V), 1);
  
  data = load('featuresX.dat ');
  X = data(:, 1:3);
  y = data(:, 4);
  m = length(y); 
  
  A(2,:) %return 2nd row of matrix A
  A(:,2) %return 2nd column of matrix A
  A(:) %put all elements of A into a single vector
  A(:, 3:5) = B(:,1:3) %replace columns 3 to 5 of A with columns 1 to 3 of B
  A(:, 2: end) %return elements from 2nd to last column of A
  
  C = [A A] %column doubled
  C = [A,A] %column doubled
  C = [A;A] %row doubled
  A.*B %multiply corresponding elements of A,B
  abs(A) %absolute elements
  
  %max
  max(A) %return max element of matrix A
  [val, ind] = max(A, [], 1) %obtain max value of each column and its index, row vector format
  [val, ind] = max(A, [], 2) %obtain max value of each row and its index, column vector format
  
  pinv(A) % calculate inverse of matric 
  A' %calculate transpose of matrix A
  
  %Logical arrays
  a = 1:10; b = 3; % create a vector a, and a scalar b
  a == b; %return a vector same size of a, with ones at positions wjere the elements of a are equal to b, and zeros where they ar different

• Computing on data

  A = magic(3) %sum of elements in each row or column are equal
  sum(A) %summarize each elements
  prod(A) %multiply each elements
  floor(A) %取整
  max(A, [], 1) %obtain max value of each row
  max(A, [], 2) %obtain max value of each column
  max(max(A)) %obtain the max value of all elements
  sum(A, 1) %obtain sum of column, get a row vector
  sum(A, 2) %obtain sum of row, get a colum vector
  flipud(eye(10)) %inner argument matrix should be a nxn matrix

• Plotting data

  t = [0:0.01:0.98];
  y1 = sin(2*pi*4*t);
  plot(t, y1)
  hold on %displayed on a single graph
  plot(t, y1, 'r') %display in red line
  xlabel('time') %xlabel, ylabel, title, etc
  legend('sin', 'cos') %display a legend for the current axes using the specified strings as labels
  print -dpng 'myPlot.png' %save as picture
  close %close figure
  figure(1); plot(t, y1)
  
  subplot(2,10,1) %create a 2x10 plot, current 1st plot activate
  axis([0 1 1 2]) %change axis range, x from 0 to 1, y from 1 to 2
  clf %clear figures

• Control statement: for, while, if statements

  v = zeros(10,1);
  for i = 1: 10,
  	v(i) = 2^i;
  end 
  
  i = 1;
  while i<=5,
  	v(i) = 100;
  	i = i + 1;
  end 
  
  if i >= 3,
  	xxx
  else,
  	xxx
  end
  
  fprint('now you have almost finished basic octave learning. \n');
  pause; %suspend the execution for N seconds, e.g. pause (10)

• Vectorial implementation

  • can easily been solved in coding environments

Programming Exercise 1: Linear Regression

• Part i: linear regression with one variable

  ex1.m

  • load data

  data = load('ex1data1.txt')  
  X = data(:, 1); y = data(:, 2); %load data into variable X and y
  m = length(y); %number of training example

  • plot data

    '''
    function plotData(x, y)
    figure; 
    plot(x, y, 'rx', MarkerSize', 10);
    ylabel('Profit in $10,000s');
    xlabel('Population of City in 10,000s');
    end
    '''
  ployData(X, y) 

  • cost

  X = [ones(m, 1), data(:,1)]; %add a column of ones to x
  theta = zeros(2,1); %initialize fitting parameter
  iterations = 1500;
  alpha = 0.01;

  • computing the cost $J(\theta)$

    '''
    function J = computeCost(X, y, theta)
    m = length(y);
    J = 1/(2*m) * sum((X * theta-y).^2);
    end
    '''
  J = computeCost(X, y, theta)

  • Gradient descent

    '''
    function theta = gradientDescent(X, y, theta, alpha, num_iters)
    m = length(y);
    J_history = zeros(num_iters, 1);
    for iter = 1:num_iters
        a = 1/m*sum(X(:,1)'(X*theta-y));
    	  b = 1/m*sum(X(:,2)'*(X*theta-y));
        delta = [a;b];
        theta = theta - alpha*delta;
        J_history(iter) = computeCost(X, y, theta);
    end
    
    end 
    '''
  theta = gradientDescent(X, y, theta, alpha, iterations); %run gradient descent

  • Plot the linear fit

  hold on; % keep previous plot visible
  plot(X(:,2), X*theta, '-')
  legend('Training data', 'Linear regression')
  hold off % don't overlay any more plots on this figure

  • Predict values for population sizes of 35,000

  predict1 = [1, 3.5] *theta;
  fprintf('For population = 35,000, we predict a profit of %f\n',...
      predict1*10000);

  • Visualize $J(\theta)$

  % Grid over which we will calculate J
  theta0_vals = linspace(-10, 10, 100);
  theta1_vals = linspace(-1, 4, 100);
  
  % initialize J_vals to a matrix of 0's
  J_vals = zeros(length(theta0_vals), length(theta1_vals));
  
  % Fill out J_vals
  for i = 1:length(theta0_vals)
      for j = 1:length(theta1_vals)
  	  t = [theta0_vals(i); theta1_vals(j)];
  	  J_vals(i,j) = computeCost(X, y, t);
      end
  end
  
  % Because of the way meshgrids work in the surf command, we need to
  % transpose J_vals before calling surf, or else the axes will be flipped
  J_vals = J_vals';
  % Surface plot
  figure;
  surf(theta0_vals, theta1_vals, J_vals)
  xlabel('\theta_0'); ylabel('\theta_1');
  
  % Contour plot
  figure;
  % Plot J_vals as 15 contours spaced logarithmically between 0.01 and 100
  contour(theta0_vals, theta1_vals, J_vals, logspace(-2, 3, 20))
  xlabel('\theta_0'); ylabel('\theta_1');
  hold on;
  plot(theta(1), theta(2), 'rx', 'MarkerSize', 10, 'LineWidth', 2);

• Part ii: linear regression with multiple variables

  n.a.

  ​

Changelog

• 17.10.02 detail octave tutorial
• 17.09.10 init create @draachen

Reference

• Andrea Ng: Machine Learning
• Octave documentation pages