Implement Gradient Descent Matlab

Question

Implement Gradient Descent Matlab

I went through many stack overflow codes and made my own on the same line. There is some problem with this code that I cannot understand. I store the value of theta1 and theta 2, as well as a cost function for analysis purposes. Data for x and Y can be downloaded from this Openclassroom . It has x and Y data in the form of .dat files that you can open in notepad.

    %Single Variate Gradient Descent Algorithm%%
    clc
clear all
close all;
% Step 1 Load x series/ Input data and Output data* y series

x=load('D:\Office Docs_Jay\software\ex2x.dat');
y=load('D:\Office Docs_Jay\software\ex2y.dat');
%Plot the input vectors
plot(x,y,'o');
ylabel('Height in meters');
xlabel('Age in years');

% Step 2 Add an extra column of ones in input vector
[m n]=size(x);
X=[ones(m,1) x];%Concatenate the ones column with x;
% Step 3 Create Theta vector
theta=zeros(n+1,1);%theta 0,1
% Create temporary values for storing summation

temp1=0;
temp2=0;
% Define Learning Rate alpha and Max Iterations

alpha=0.07;
max_iterations=1;
      % Step 4 Iterate over loop
      for i=1:1:max_iterations

     %Calculate Hypothesis for all training example
     for k=1:1:m
        h(k)=theta(1,1)+theta(2,1)*X(k,2); %#ok<AGROW>
        temp1=temp1+(h(k)-y(k));
        temp2=temp2+(h(k)-y(k))*X(k,2);
     end
     % Simultaneous Update
      tmp1=theta(1,1)-(alpha*1/(2*m)*temp1);
      tmp2=theta(2,1)-(alpha*(1/(2*m))*temp2);
      theta(1,1)=tmp1;
      theta(2,1)=tmp2;
      theta1_history(i)=theta(2,1); %#ok<AGROW>
      theta0_history(i)=theta(1,1); %#ok<AGROW>
      % Step 5 Calculate cost function
      tmp3=0;
      tmp4=0;
      for p=1:m
        tmp3=tmp3+theta(1,1)+theta(2,1)*X(p,1);
        tmp4=tmp4+theta(1,1)+theta(2,1)*X(p,2);
      end
      J1_theta0(i)=tmp3*(1/(2*m)); %#ok<AGROW>
      J2_theta1(i)=tmp4*(1/(2*m)); %#ok<AGROW>


      end
      theta
      hold on;
      plot(X(:,2),theta(1,1)+theta(2,1)*X);

I get the value

theta as 0.0373 and 0.1900 it should be 0.0745 and 0.3800

this value roughly doubles what I expect.

+3

matlab machine-learning

Incpetor Feb 15 '14 at 15:18

source share

6 answers

cmantas · Answer 1 · 2015-10-19T13:05:25+0000

(.. theta). ( ):

h = X * theta;  # hypothesis
err = h - y;    # error
gradient = alpha * (1 / m) * (X' * err); # update the gradient
theta = theta - gradient;

, "" X'*err. (err'*X)'

Konstantinos Monachopoulos · Answer 2 · 2015-01-07T02:22:05+0000

, , Matlab. , , . ( polyfit), , , openclassroom ( 2), theta (0) = 0.0745 theta (1) = 0.3800, 1500 0.07 ( ). , .

, :

% Machine Learning : Linear Regression

clear all; close all; clc;

%% ======================= Plotting Training Data =======================
fprintf('Plotting Data ...\n')

x = load('ex2x.dat');
y = load('ex2y.dat');

% Plot Data
plot(x,y,'rx');
xlabel('X -> Input') % x-axis label
ylabel('Y -> Output') % y-axis label

%% =================== Initialize Linear regression parameters ===================
 m = length(y); % number of training examples

% initialize fitting parameters - all zeros
theta=zeros(2,1);%theta 0,1

% Some gradient descent settings
iterations = 1500;
Learning_step_a = 0.07; % step parameter

%% =================== Gradient descent ===================

fprintf('Running Gradient Descent ...\n')

%Compute Gradient descent

% Initialize Objective Function History
J_history = zeros(iterations, 1);

m = length(y); % number of training examples

% run gradient descent    
for iter = 1:iterations

   % In every iteration calculate hypothesis
   hypothesis=theta(1).*x+theta(2);

   % Update theta variables
   temp0=theta(1) - Learning_step_a * (1/m)* sum((hypothesis-y).* x);
   temp1=theta(2) - Learning_step_a * (1/m) *sum(hypothesis-y);

   theta(1)=temp0;
   theta(2)=temp1;

   % Save objective function 
   J_history(iter)=(1/2*m)*sum(( hypothesis-y ).^2);

end

% print theta to screen
fprintf('Theta found by gradient descent: %f %f\n',theta(1),  theta(2));
fprintf('Minimum of objective function is %f \n',J_history(iterations));

% Plot the linear fit
hold on; % keep previous plot visible 
plot(x, theta(1)*x+theta(2), '-')

% Validate with polyfit fnc
poly_theta = polyfit(x,y,1);
plot(x, poly_theta(1)*x+poly_theta(2), 'y--');
legend('Training data', 'Linear regression','Linear regression with polyfit')
hold off 

figure
% Plot Data
plot(x,y,'rx');
xlabel('X -> Input') % x-axis label
ylabel('Y -> Output') % y-axis label

hold on; % keep previous plot visible
% Validate with polyfit fnc
poly_theta = polyfit(x,y,1);
plot(x, poly_theta(1)*x+poly_theta(2), 'y--');

% for theta values that you are saying
theta(1)=0.0745;  theta(2)=0.3800;
plot(x, theta(1)*x+theta(2), 'g--')
legend('Training data', 'Linear regression with polyfit','Your thetas')
hold off

, :

(0) (1), , .

(0) (1) .

user1149913 · Answer 3 · 2014-02-15T16:52:46+0000

:

max_iterations 1. , , , , , .
1/(2 * m) . , .
. .
for-loops matlab. , res=X*theta-y; obj=.5/m*res'res; (res) (obj).

navid · Answer 4 · 2015-08-20T16:20:07+0000

temp1 = 0 temp2 = 0 ; , , , ..

Arumoy · Answer 5 · 2016-02-21T07:02:58+0000

Ɵ (theta) , .

, , 1/(2*m) 1/m . 2 , (h _Ɵ (x) - y) ², 2 * (h _Ɵ (x) - y). 2s .

:

J1_theta0(i)=tmp3*(1/(2*m)); %#ok<AGROW>
J2_theta1(i)=tmp4*(1/(2*m)); %#ok<AGROW>

J1_theta0(i)=tmp3*(1/m); %#ok<AGROW>
J2_theta1(i)=tmp4*(1/m); %#ok<AGROW>

, .

Tapojoy Das · Answer 6 · 2017-03-17T09:10:20+0000

37 38 :

tmp1=theta(1,1)-(alpha*(1/(2*m))*temp1);
tmp2=theta(2,1)-(alpha*(1/(2*m))*temp2);

. , :

  tmp1=theta(1,1)-(alpha*(1/(m))*temp1);
  tmp2=theta(2,1)-(alpha*(1/(m))*temp2);

..!

PS: , - .!

Implement Gradient Descent Matlab

More articles: