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I'm solving a programming assignment in machine learning course. In which I've to implement "Gradient Descent Algorithm" like below

I'm using the following code

data = load('ex1data1.txt');

% text file conatins 2 values in each row separated by commas

X = [ones(m, 1), data(:,1)];

theta = zeros(2, 1);

iterations = 1500;

alpha = 0.01;

function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)

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

J_history = zeros(num_iters, 1);

for iter = 1:num_iters

k=1:m;

j1=(1/m)*sum((theta(1)+theta(2).*X(k,2))-y(k))

j2=((1/m)*sum((theta(1)+theta(2).*X(k,2))-y(k)))*(X(k,2))

theta(1)=theta(1)-alpha*(j1);

theta(2)=theta(2)-alpha*(j2);

J_history(iter) = computeCost(X, y, theta);

end

end

theta = gradientDescent(X, y, theta, alpha, iterations);

On running the above code I'm getting this error message

Jayan Joshi
on 15 Oct 2019

Edited: Jayan Joshi
on 15 Oct 2019

predictions =X*theta;

theta=theta-(alpha/m*sum((predictions-y).*X))';

Margo Khokhlova
on 19 Oct 2015

Edited: Walter Roberson
on 19 Oct 2015

Well, sort of super late, but you just made it wrong with the brackets... This one works for me:

k=1:m;

j1=(1/m)*sum((theta(1)+theta(2).*X(k,2))-y(k))

j2=(1/m)*sum(((theta(1)+theta(2).*X(k,2))-y(k)).*X(k,2))

theta(1)=theta(1)-alpha*(j1);

theta(2)=theta(2)-alpha*(j2);

Sesha Sai Anudeep Karnam
on 7 Aug 2019

Edited: Sesha Sai Anudeep Karnam
on 7 Aug 2019

temp0 = theta(1)-alpha*((1/m)*(theta(1)+theta(2).*X(k,2)-y(k)));

temp1 = theta(2)- alpha*((1/m)*(theta(1)+theta(2).*X(k,2)-y(k)).*X(k,2));

theta(1) = temp0;

theta(2) = temp1;

% this code gives approximate values but while submitting I'm getting 0points for this

% Theta found by gradient descent:

% -3.588389

% 1.123667

% Expected theta values (approx)

% -3.6303

% 1.1664

% How to overcome this??

Shekhar Raj
on 19 Sep 2019

Below code gave the exact value -

for iter = 1:num_iters

% ====================== YOUR CODE HERE ======================

% Instructions: Perform a single gradient step on the parameter vector

% theta.

%

% Hint: While debugging, it can be useful to print out the values

% of the cost function (computeCost) and gradient here.

%

Prediction = X * theta;

temp1 = alpha/m * sum((Prediction - y));

temp2 = alpha/m * sum((Prediction - y) .* X(:,2));

theta(1) = theta(1) - temp1;

theta(2) = theta(2) - temp2;

% ============================================================

Shekhar Raj
on 19 Sep 2019

Below Code works for me -

Prediction = X * theta;

temp1 = alpha/m * sum((Prediction - y));

temp2 = alpha/m * sum((Prediction - y) .* X(:,2));

theta(1) = theta(1) - temp1;

theta(2) = theta(2) - temp2;

Jayan Joshi
on 15 Oct 2019

Thank you this really helped. I tried more vectorized form of this and it worked.

predictions =X*theta;

theta=theta-(alpha/m*sum((predictions-y).*X))';

ICHEN WU
on 8 Nov 2015

Can you tell me why my answer is not correct? I felt they are the same.

theta(1)=theta(1)-(alpha/m)*sum( (X*theta)-y);

theta(2)=theta(2)-(alpha/m)*sum( ((X*theta)-y)'*X(:,2));

Austin Lindquist
on 7 Mar 2016

By assigning theta(1) before assigning theta(2), you've introduced a side effect.

One way of writing it:

temp1 = theta(1)-(alpha/m)*sum(X*theta-y);

theta(2) = theta(2)-(alpha/m)*sum((X*theta-y)'*X(:,2));

theta(1) = temp1;

pavan B
on 20 Feb 2017

above one works perfect .try below code of mine too

earlier i used h = X * theta; a0 = (1/m)*sum((h-y)); a1 = (1/m)*sum((h-y)'*x1); surprisingly it didn't work

working code: x1 = X(:,2); a0 = (1/m)*sum((X * theta-y)); a1 = (1/m)*sum((X * theta-y)'*x1); a = [a0;a1]; theta = theta- (alpha*a);

if anyone find out whats wrong with my earlier code it would be appreciated.

Leon Cai
on 6 Apr 2017

Ali Dezfooli
on 17 Jun 2016

In this line

X = [ones(m, 1), data(:,1)];

You add bias to your X, but in the formula of your picture (Ng's slides) when you want to compute theta(2) you should remove it.

Utkarsh Anand
on 17 Mar 2018

Looking at the problem, I also think that you cannot initiate Theta as Zero.

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