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Gradient Descent Matlab implementation

I have gone through many codes in stack overflow and made my own on same line. there is some problem with this code I am unable to understand. I am storing the value theta1 and theta 2 and also the cost function for analysis purpose. The data for x and Y can be downloaded from this Openclassroom page. It has x and Y data in 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 am getting the value of

theta as 0.0373 and 0.1900 it should be 0.0745 and 0.3800

this value is approximately double that I am expecting.

like image 872
Incpetor Avatar asked Nov 28 '22 02:11

Incpetor


2 Answers

I have been trying to implement the iterative step with matrices and vectors (i.e not update each parameter of theta). Here is what I came up with (only the gradient step is here):

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

The hard part to grasp is that the "sum" in the gradient step of the previous examples is actually performed by the matrix multiplication X'*err. You can also write it as (err'*X)'

like image 131
cmantas Avatar answered Nov 29 '22 17:11

cmantas


I managed to create an algorithm that uses more of the vectorized properties that Matlab support. My algorithm is a little different from yours but does the gradient descent process as you ask. After the execution and validation (using polyfit function) that i made, i think that the values in openclassroom (exercise 2) that are expected in variables theta(0) = 0.0745 and theta(1) = 0.3800 are wrong after 1500 iterations with step 0.07 (i do not take response of that). This is the reason that i plotted my results with the data in one plot and your required results with the data in another plot and i saw a big difference in data fitting procedure.

First of all have a look at the code :

% 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 

Ok the results are as follows :

With theta(0) and theta(1) that produced from my algorithm as a result the line fits the data.

Gradient descent - theta0=0.063883, theta1=0.750150

With theta(0) and theta(1) as fixed values as a result the line do not fit the data.

Gradient descent - theta0=0.0745, theta1=0.3800

like image 26
Konstantinos Monachopoulos Avatar answered Nov 29 '22 17:11

Konstantinos Monachopoulos