How to fix gradient descent code?
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I am a novice trying to do a gradient descent with one variable, but cannot figure out how to fix my code (below). Not sure if my for-part is correct. This is the error message: "In an assignment A(:) = B, the number of elements in A and B must be the same." Please help?
data = load('data.txt' );
X = data(:, 1); y = data(:, 2);
m = length(y);
X = [ones(m, 1), data(:,1)]; % Add a column of ones to x
theta = zeros(2, 1); % initialize fitting parameters
num_iters = 1500;
alpha = 0.01;
J = computeCost(X, y, theta)
m = length(y);
J = sum(( X * theta - y ) .^2 )/( 2 * m );
[theta J_history] = gradientDescent(X, y, theta, alpha, num_iters)
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
h=(theta(1)+ theta(2)*X)';
theta(1) = theta(1) - alpha * (1/m) * h * X(:, 1);
theta(2) = theta(2) - alpha * (1/m) * h * X(:, 2);
% Save the cost J in every iteration
J_history(num_iters) = computeCost(X, y, theta);
end
2 Kommentare
Walter Roberson
am 30 Mär. 2016
Please show the complete error message, everything in red.
Jackwhale
am 30 Mär. 2016
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Weitere Antworten (2)
I don't know why you use such a complicated approach.
Just execute
data = load('data.txt' );
A = [ones(length(data(:,1)),1), data(:,1)];
b = data(:,2);
theta = A \ b
to get your optimum theta values.
Best wishes
Torsten.
14 Kommentare
Torsten
am 30 Mär. 2016
Your objective function is
f(theta)=(X*theta-y)'*(X*theta-y)/(2*m)
with
X = [ones(m, 1), data(:,1)] and y = data(:, 2).
The gradient descend method reads
theta(n+1)=theta(n)-alpha*grad(f(theta(n)))
Now determine grad(f(theta(n))) and iterate.
Note that the update formula for theta in your code from above is incorrect.
Best wishes
Torsten.
Jackwhale
am 30 Mär. 2016
Torsten
am 31 Mär. 2016
Please check whether this is correct:
data = load('data.txt' );
y = data(:,2);
m = length(y);
X = [ones(m,1), data(:,1)]; % Add a column of ones to x
num_iters = 1500;
alpha = 0.01;
J_history = zeros(num_iters, 1);
theta = zeros(2, 1); % initialize fitting parameters
J_history(1) = (X*theta-y)'*(X*theta-y)/(2*m);
for iter = 2:num_iters
theta = theta-alpha*X'*(X*theta-y)/m;
J_history(iter)=(X*theta-y)'*(X*theta-y)/(2*m);
end
Best wishes
Torsten.
Jackwhale
am 31 Mär. 2016
Torsten
am 31 Mär. 2016
And what does the above code give as J_history(1000) and theta for the first four iterations ?
Are the values different ?
Best wishes
Torsten.
Torsten
am 31 Mär. 2016
Can't be true.
For theta=0, the expression
(X*theta-y)'*(X*theta-y)/(2*m)
evaluates to
[1 6 4 2]*[1 6 4 2]'/(2*4) = 57/8 = 7.125,
but not 4.5161.
Best wishes
Torsten.
Jackwhale
am 31 Mär. 2016
Please run this code and output J_history(1000) and theta at the end.
y=[1; 6; 4; 2];
m = 4;
X = [1 5; 1 2; 1 4; 1 5];
num_iters = 1000;
alpha = 0.01;
J_history = zeros(num_iters, 1);
theta = zeros(2, 1); % initialize fitting parameters
J_history(1) = (X*theta-y)'*(X*theta-y)/(2*m);
for iter = 2:num_iters
theta = theta-alpha*X'*(X*theta-y)/m;
J_history(iter)=(X*theta-y)'*(X*theta-y)/(2*m);
end
Best wishes
Torsten.
Jackwhale
am 31 Mär. 2016
Jackwhale
am 31 Mär. 2016
Torsten
am 31 Mär. 2016
You seem to have a strange MATLAB version.
If I set
num_iters=1001,
I get
theta =
5.2147549
- 0.5733459
J_history(1001)
ans =
0.8554026
thus the results expected.
Best wishes
Torsten.
Torsten
am 31 Mär. 2016
I only need to supply the updates to theta within each iteration.
If you can't read from the code I supplied how theta is updated every iteration, then you should really start with MATLAB principles.
Agbakoba Chukwunoso
am 6 Dez. 2020
0 Stimmen
Pls help me out.. I'm trying to find gradientdescent with this code but when I run it, it returns gradientdescents to me not the value . 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); end end
2 Kommentare
Agbakoba Chukwunoso
am 6 Dez. 2020
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);
end
end
sivarao K
am 10 Nov. 2021
here 'y' not defined but it excuting how?
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