Reputation: 371
I'm attempting to implement gradient descent using code from :
Gradient Descent implementation in octave
I've amended code to following :
X = [1; 1; 1;]
y = [1; 0; 1;]
m = length(y);
X = [ones(m, 1), data(:,1)];
theta = zeros(2, 1);
iterations = 2000;
alpha = 0.001;
for iter = 1:iterations
theta = theta -((1/m) * ((X * theta) - y)' * X)' * alpha;
end
theta
Which gives following output :
X =
1
1
1
y =
1
0
1
theta =
0.32725
0.32725
theta is a 1x2 Matrix but should'nt it be 1x3 as the output (y) is 3x1 ?
So I should be able to multiply theta by the training example to make a prediction but cannot multiply x by theta as x is 1x3 and theta is 1x2?
Update :
%X = [1 1; 1 1; 1 1;]
%y = [1 1; 0 1; 1 1;]
X = [1 1 1; 1 1 1; 0 0 0;]
y = [1 1 1; 0 0 0; 1 1 1;]
m = length(y);
X = [ones(m, 1), X];
theta = zeros(4, 1);
theta
iterations = 2000;
alpha = 0.001;
for iter = 1:iterations
theta = theta -((1/m) * ((X * theta) - y)' * X)' * alpha;
end
%to make prediction
m = size(X, 1); % Number of training examples
p = zeros(m, 1);
htheta = sigmoid(X * theta);
p = htheta >= 0.5;
Upvotes: 0
Views: 121
Reputation: 66835
You are misinterpreting dimensions here. Your data consists of 3 points, each having a single dimension. Furthermore, you add a dummy dimension of 1s
X = [ones(m, 1), data(:,1)];
thus
octave:1> data = [1;2;3]
data =
1
2
3
octave:2> [ones(m, 1), data(:,1)]
ans =
1 1
1 2
1 3
and theta
is your parametrization, which you should be able to apply through (this is not a code, but math notation)
h(x) = x1 * theta1 + theta0
thus your theta should have two dimensions. One is a weight for your dummy dimension (so called bias) and one for actual X dimension. If your X has K dimensions, theta would have K+1. Thus, after adding a dummy dimension matrices have following shapes:
X is 3x2
y is 3x1
theta is 2x1
so
X * theta is 3x1
the same as y
Upvotes: 2