Problem while implementing "Gradient Descent Algorithm" in Matlab

19 Ansichten (letzte 30 Tage)
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
  3 Kommentare
Nancy Irisarri
Nancy Irisarri am 13 Mai 2019
Calculation of k can be outside the for loop. Improves performance!
Ashok Saini
Ashok Saini am 4 Jul. 2022
hey have u found answer of your question

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Matt J
Matt J am 11 Apr. 2015
j2 is not a scalar, but you are trying to assign it to a scalar location theta(2).
Did you intend for this line
k=1:m;
to be a for-loop
for k=1:m
  2 Kommentare
Atinesh S
Atinesh S am 11 Apr. 2015
Why j2 is not scalar, the expression
(1/m)*sum((theta(1)+theta(2).*X(k,2))-y(k))
is producing scalar result which can be multiplied by
X(k,2)
to produce scalar result. But on the matlab, I've also seen the result that is going to be stored in j2 is a vector. But Why ??
Matt J
Matt J am 12 Apr. 2015
k is not a scalar. You defined it to be the vector 1:m. Therefore X(k,2) is also a vector.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (11)

Jayan Joshi
Jayan Joshi am 15 Okt. 2019
Bearbeitet: Jayan Joshi am 15 Okt. 2019
predictions =X*theta;
theta=theta-(alpha/m*sum((predictions-y).*X))';

Margo Khokhlova
Margo Khokhlova am 19 Okt. 2015
Bearbeitet: Walter Roberson am 19 Okt. 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);

Shekhar Raj
Shekhar Raj am 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;
  2 Kommentare
Jayan Joshi
Jayan Joshi am 15 Okt. 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))';
Lomg Ma
Lomg Ma am 24 Jan. 2021
How did you manage to vectorize it that much? I don't understand how to translate the formula to code, seems confusing

Melden Sie sich an, um zu kommentieren.


Sesha Sai Anudeep Karnam
Sesha Sai Anudeep Karnam am 7 Aug. 2019
Bearbeitet: Sesha Sai Anudeep Karnam am 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??
  2 Kommentare
Shekhar Raj
Shekhar Raj am 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;
% ============================================================
Amber Hall
Amber Hall am 15 Aug. 2021
i've tried this code but still get error due to not enough input arguments for m = length(y) ? do you know what may be the cause as it appears i have coded correctly

Melden Sie sich an, um zu kommentieren.


ICHEN WU
ICHEN WU am 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));
  5 Kommentare
pavan B
pavan B am 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
Leon Cai am 6 Apr. 2017
yea I tried h = X*theta and it didn't work too, I'm thinking that when we use the variable h, as we update theta, the value of h will remain unchanged.

Melden Sie sich an, um zu kommentieren.


Ali Dezfooli
Ali Dezfooli am 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
Utkarsh Anand am 17 Mär. 2018
Looking at the problem, I also think that you cannot initiate Theta as Zero.

Rajeswari G
Rajeswari G am 2 Jan. 2021
error = (X * theta) - y;
theta = theta - ((alpha/m) * X'*error);
In this equation why we take x'?
  1 Kommentar
Bee Ling TAN
Bee Ling TAN am 15 Aug. 2021
This is because X is a 97x2 matrix. To perform dot products, only X' (2x97)will make the answer valid to be 2x1 vectors, entrys are theta(1)&theta(2) respectively.

Melden Sie sich an, um zu kommentieren.


Wamin Thammanusati
Wamin Thammanusati am 21 Feb. 2021
Bearbeitet: Wamin Thammanusati am 21 Feb. 2021
The code below works for this case (one variable) and also multiple variables -
for iter = 1:num_iters
Hypothesis = X * theta;
for i=1:size(X,2)
theta(i) = theta(i) - alpha/m * sum((Hypothesis-y) .* X(:,i));
end
end
  1 Kommentar
Amber Hall
Amber Hall am 15 Aug. 2021
having tried the same code i am struggling to understand what i am doing wrong - i receive error due to not enough jnput arguments for m = length(y) line. do you have any ideas?

Melden Sie sich an, um zu kommentieren.


Chong Lu
Chong Lu am 16 Nov. 2021
Bearbeitet: Walter Roberson am 27 Nov. 2021
temp1 = theta(1) - alpha*(sum(X*theta - y)/m);
temp2 = theta(2) - alpha*(sum((X*theta - y).*X(:,2))/m);
theta(1) = temp1;
theta(2) = temp2;

muhammad zohaib
muhammad zohaib am 27 Nov. 2021

Kategorien

Mehr zu MATLAB finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by