Octave code for gradient descent using vectorization not updating cost function correctly
I have implemented following code for gradient descent using vectorization but it seems the cost function is not decrementing correctly.Instead the cost function is increasing with each iteration.
Assuming theta to be an n+1 vector, y to be a m vector and X to be design matrix m*(n+1)
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
n = length(theta); % number of features
J_history = zeros(num_iters, 1);
error = ((theta' * X')' - y)*(alpha/m);
descent = zeros(size(theta),1);
for iter = 1:num_iters
for i = 1:n
descent(i) = descent(i) + sum(error.* X(:,i));
i = i + 1;
end
theta = theta - descent;
J_history(iter) = computeCost(X, y, theta);
disp("the value of cost function is : "), disp(J_history(iter));
iter = iter + 1;
end
The compute cost function is :
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
for i = 1:m,
H = theta' * X(i,:)';
E = H - y(i);
SQE = E^2;
J = (J + SQE);
i = i+1;
end;
J = J / (2*m);
machine-learning octave vectorization gradient-descent
add a comment |
I have implemented following code for gradient descent using vectorization but it seems the cost function is not decrementing correctly.Instead the cost function is increasing with each iteration.
Assuming theta to be an n+1 vector, y to be a m vector and X to be design matrix m*(n+1)
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
n = length(theta); % number of features
J_history = zeros(num_iters, 1);
error = ((theta' * X')' - y)*(alpha/m);
descent = zeros(size(theta),1);
for iter = 1:num_iters
for i = 1:n
descent(i) = descent(i) + sum(error.* X(:,i));
i = i + 1;
end
theta = theta - descent;
J_history(iter) = computeCost(X, y, theta);
disp("the value of cost function is : "), disp(J_history(iter));
iter = iter + 1;
end
The compute cost function is :
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
for i = 1:m,
H = theta' * X(i,:)';
E = H - y(i);
SQE = E^2;
J = (J + SQE);
i = i+1;
end;
J = J / (2*m);
machine-learning octave vectorization gradient-descent
1
Shouldn'tfor i = 1:n
incrementi
for you? You're also doing it inside the loop. (Long time since I did any Octave...)
– Fred Foo
Oct 30 '14 at 15:15
yeah thats true
– Dcoder
Oct 30 '14 at 15:23
add a comment |
I have implemented following code for gradient descent using vectorization but it seems the cost function is not decrementing correctly.Instead the cost function is increasing with each iteration.
Assuming theta to be an n+1 vector, y to be a m vector and X to be design matrix m*(n+1)
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
n = length(theta); % number of features
J_history = zeros(num_iters, 1);
error = ((theta' * X')' - y)*(alpha/m);
descent = zeros(size(theta),1);
for iter = 1:num_iters
for i = 1:n
descent(i) = descent(i) + sum(error.* X(:,i));
i = i + 1;
end
theta = theta - descent;
J_history(iter) = computeCost(X, y, theta);
disp("the value of cost function is : "), disp(J_history(iter));
iter = iter + 1;
end
The compute cost function is :
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
for i = 1:m,
H = theta' * X(i,:)';
E = H - y(i);
SQE = E^2;
J = (J + SQE);
i = i+1;
end;
J = J / (2*m);
machine-learning octave vectorization gradient-descent
I have implemented following code for gradient descent using vectorization but it seems the cost function is not decrementing correctly.Instead the cost function is increasing with each iteration.
Assuming theta to be an n+1 vector, y to be a m vector and X to be design matrix m*(n+1)
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
n = length(theta); % number of features
J_history = zeros(num_iters, 1);
error = ((theta' * X')' - y)*(alpha/m);
descent = zeros(size(theta),1);
for iter = 1:num_iters
for i = 1:n
descent(i) = descent(i) + sum(error.* X(:,i));
i = i + 1;
end
theta = theta - descent;
J_history(iter) = computeCost(X, y, theta);
disp("the value of cost function is : "), disp(J_history(iter));
iter = iter + 1;
end
The compute cost function is :
function J = computeCost(X, y, theta)
m = length(y);
J = 0;
for i = 1:m,
H = theta' * X(i,:)';
E = H - y(i);
SQE = E^2;
J = (J + SQE);
i = i+1;
end;
J = J / (2*m);
machine-learning octave vectorization gradient-descent
machine-learning octave vectorization gradient-descent
edited Oct 30 '14 at 15:26
Dcoder
asked Oct 30 '14 at 15:09
DcoderDcoder
1342513
1342513
1
Shouldn'tfor i = 1:n
incrementi
for you? You're also doing it inside the loop. (Long time since I did any Octave...)
– Fred Foo
Oct 30 '14 at 15:15
yeah thats true
– Dcoder
Oct 30 '14 at 15:23
add a comment |
1
Shouldn'tfor i = 1:n
incrementi
for you? You're also doing it inside the loop. (Long time since I did any Octave...)
– Fred Foo
Oct 30 '14 at 15:15
yeah thats true
– Dcoder
Oct 30 '14 at 15:23
1
1
Shouldn't
for i = 1:n
increment i
for you? You're also doing it inside the loop. (Long time since I did any Octave...)– Fred Foo
Oct 30 '14 at 15:15
Shouldn't
for i = 1:n
increment i
for you? You're also doing it inside the loop. (Long time since I did any Octave...)– Fred Foo
Oct 30 '14 at 15:15
yeah thats true
– Dcoder
Oct 30 '14 at 15:23
yeah thats true
– Dcoder
Oct 30 '14 at 15:23
add a comment |
2 Answers
2
active
oldest
votes
You can vectorise it even further:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
delta = (theta' * X'-y')*X;
theta = theta - alpha/m*delta';
J_history(iter) = computeCost(X, y, theta);
end
end
add a comment |
You can vectorize it better as follows
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
theta=theta-(alpha/m)*((X*theta-y)'*X)';
J_history(iter) = computeCost(X, y, theta);
end;
end;
The ComputeCost function can be written as
function J = computeCost(X, y, theta)
m = length(y);
J = 1/(2*m)*sum((X*theta-y)^2);
end;
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can vectorise it even further:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
delta = (theta' * X'-y')*X;
theta = theta - alpha/m*delta';
J_history(iter) = computeCost(X, y, theta);
end
end
add a comment |
You can vectorise it even further:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
delta = (theta' * X'-y')*X;
theta = theta - alpha/m*delta';
J_history(iter) = computeCost(X, y, theta);
end
end
add a comment |
You can vectorise it even further:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
delta = (theta' * X'-y')*X;
theta = theta - alpha/m*delta';
J_history(iter) = computeCost(X, y, theta);
end
end
You can vectorise it even further:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
delta = (theta' * X'-y')*X;
theta = theta - alpha/m*delta';
J_history(iter) = computeCost(X, y, theta);
end
end
answered Nov 17 '16 at 12:30
Rimma ShafikovaRimma Shafikova
10115
10115
add a comment |
add a comment |
You can vectorize it better as follows
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
theta=theta-(alpha/m)*((X*theta-y)'*X)';
J_history(iter) = computeCost(X, y, theta);
end;
end;
The ComputeCost function can be written as
function J = computeCost(X, y, theta)
m = length(y);
J = 1/(2*m)*sum((X*theta-y)^2);
end;
add a comment |
You can vectorize it better as follows
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
theta=theta-(alpha/m)*((X*theta-y)'*X)';
J_history(iter) = computeCost(X, y, theta);
end;
end;
The ComputeCost function can be written as
function J = computeCost(X, y, theta)
m = length(y);
J = 1/(2*m)*sum((X*theta-y)^2);
end;
add a comment |
You can vectorize it better as follows
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
theta=theta-(alpha/m)*((X*theta-y)'*X)';
J_history(iter) = computeCost(X, y, theta);
end;
end;
The ComputeCost function can be written as
function J = computeCost(X, y, theta)
m = length(y);
J = 1/(2*m)*sum((X*theta-y)^2);
end;
You can vectorize it better as follows
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
theta=theta-(alpha/m)*((X*theta-y)'*X)';
J_history(iter) = computeCost(X, y, theta);
end;
end;
The ComputeCost function can be written as
function J = computeCost(X, y, theta)
m = length(y);
J = 1/(2*m)*sum((X*theta-y)^2);
end;
answered Nov 13 '18 at 4:04
Vishnu PrasadVishnu Prasad
235
235
add a comment |
add a comment |
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1
Shouldn't
for i = 1:n
incrementi
for you? You're also doing it inside the loop. (Long time since I did any Octave...)– Fred Foo
Oct 30 '14 at 15:15
yeah thats true
– Dcoder
Oct 30 '14 at 15:23