Problem while implementing "Gradient Descent Algorithm" in Matlab
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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
3 个评论
Racz Robert
2019-1-6
编辑：Racz Robert
2019-1-6
brackets mate, should work!
Cheers
Nancy Irisarri
2019-5-13
Calculation of k can be outside the for loop. Improves performance!
Ashok Saini
2022-7-4
hey have u found answer of your question
采纳的回答
更多回答（11 个）
Jayan Joshi
2019-10-15
编辑：Jayan Joshi
2019-10-15
predictions =X*theta;
theta=theta-(alpha/m*sum((predictions-y).*X))';
Margo Khokhlova
2015-10-19
编辑：Walter Roberson
2015-10-19
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);
1 个评论
Nancy Irisarri
2019-5-13
Calculation of k can be outside the for loop. Improves performance!
Shekhar Raj
2019-9-19
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 个评论
Jayan Joshi
2019-10-15
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
2021-1-24
How did you manage to vectorize it that much? I don't understand how to translate the formula to code, seems confusing
Sesha Sai Anudeep Karnam
2019-8-7
编辑：Sesha Sai Anudeep Karnam
2019-8-7
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 个评论
Shekhar Raj
2019-9-19
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
2021-8-15
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
ICHEN WU
2015-11-8
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 个评论
Walter Roberson
2015-11-8
I think we need the context of the rest of your code. Also, are you getting an error message?
Hi, Thank you for response. The other parts of the code are exactly the same. there is no error message. just the final result it generates is different.
I only replace this part
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);
to
theta(1)=theta(1)-(alpha/m)*sum( (X*theta)-y);
theta(2)=theta(2)-(alpha/m)*sum( ((X*theta)-y)'*X(:,2));
because I was thinking that I can use matrix for this instead of doing individual summation by 1:m. But the result of final theta(1,2) are different from the correct answer by a little bit. my answer: Theta found by gradient descent: -3.636063 1.166989 correct answer: Theta found by gradient descent: -3.630291 1.166362
Austin Lindquist
2016-3-7
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
2017-2-20
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
2017-4-6
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.
Ali Dezfooli
2016-6-17
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
2018-3-17
0 个投票
Looking at the problem, I also think that you cannot initiate Theta as Zero.
Rajeswari G
2021-1-2
0 个投票
error = (X * theta) - y;
theta = theta - ((alpha/m) * X'*error);
In this equation why we take x'?
1 个评论
Bee Ling TAN
2021-8-15
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.
Wamin Thammanusati
2021-2-21
编辑：Wamin Thammanusati
2021-2-21
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 个评论
Amber Hall
2021-8-15
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?
Chong Lu
2021-11-16
编辑：Walter Roberson
2021-11-27
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;
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