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loop causes enlargement of array

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fima v
fima v on 13 Apr 2020
Commented: Star Strider on 14 Apr 2020
Hello, my theta vector 2X1 two values needs to be updated each itteration. instead i get theta to be size 1000 (as the number of the itterations)
Why my theta updating line causes theta to inflate?
Thanks
n=1000;
t=1.4;
sigma_R = t*0.001;
min_value_t = t-sigma_R;
max_value_t = t+sigma_R;
y_data = min_value_t + (max_value_t - min_value_t) * rand(n,1);
x_data=[1:1000];
L=0.0001; %learning rate
%plot(x_data,y_data);
itter=1000;
theta_0=0;
theta_1=0;
theta=[theta_0;theta_1];
itter=1000;
for i=1:itter
onss=ones(1,1000);
x_mat=[onss;x_data]';
pred=x_mat*theta;
residuals = (pred-y_data)';
theta_0=theta_0-((x_data.*residuals)*(L/n));
theta_1=theta_1-((x_data.*residuals)*(L/n));
theta=[theta_0,theta_1];
end

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Accepted Answer

Star Strider
Star Strider on 13 Apr 2020
The ‘x_data’ and ‘residuals’ variables are both (1x1000) vectors.
Subscripting them is likely the solution, although you need to decide that:
theta_0=theta_0-((x_data(i).*residuals(i))*(L/n));
theta_1=theta_1-((x_data(i).*residuals(i))*(L/n));
It may also be necessary to do this with other variables in the loop. You need to decide that as well.
.

  2 Comments

fima v
fima v on 14 Apr 2020
Hello Star Strider,in the big code bellow i have those two lines which are in the itter loop.
as you can see the temp0 is a scalar its not a vector.on the right side we create a scalar,after that we subtract two number and update the theta.so why i get a vector sizee 1000 for theta when i put it in a loop?
Thanks.
temp0=theta(1) - Learning_step_a * (1/m)* sum((hypothesis-y).* x);
temp1=theta(2) - Learning_step_a * (1/m) *sum(hypothesis-y);
% Machine Learning : Linear Regression
clear all; close all; clc;
%% ======================= Plotting Training Data =======================
t=1.2
n=1000;
sigma_R = t*0.001;
min_value_t = t-sigma_R;
max_value_t = t+sigma_R;
y = min_value_t + (max_value_t - min_value_t) * rand(n,1);
x = [1:1000];
% 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 of
Star Strider
Star Strider on 14 Apr 2020
One problem is that ‘x’ is (1x1000). Another problem is that ‘(hypothesis-y)’ is a (1000x1000) matrix, so the sum is going to be a (1x1000) row vector:
temp0=theta(1) - Learning_step_a * (1/m)* sum((hypothesis-y).* x);
The same soirt of problem occurs in the next line:
temp1=theta(2) - Learning_step_a * (1/m) *sum(hypothesis-y);
so when you try to assign them to a scalar:
theta(1)=temp0;
theta(2)=temp1;
your code throws the error.
If you want to assign them to a scalar, it would be best to calculate:
sum((hypothesis-y)
in a separate step, then refer to the individual element of it in the ‘theta’ assignments.
Either that, or make ‘theta’ row vectors:
theta(1,:)=temp0;
theta(2,:)=temp1;
It all depends on what you want your code to do.
.

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