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March 10, 2016 04:10
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Non-linear classification example
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% my training data. | |
% so if x > 3 || x < 7, y = 1, otherwise y = 0. | |
x = 1:100; | |
y = [0, 0, 0, 1, 1, 1, 1, zeros(1, 93)]; | |
% instead of theta' * x, I'm trying to create | |
% a non-linear decision boundary. | |
% So instead of y = theta_0 + theta_1 * x, I use: | |
function result = h(x, theta) | |
result = sigmoid(theta(1) + theta(2) * x + theta(3) * ((x - theta(4))^2)); | |
end | |
function result = sigmoid(z) | |
result = 1 / (1 + e ^ (-z)); | |
end | |
% cost function, works correctly | |
function distance = cost(theta) | |
distance = 0; | |
x = 1:100; | |
y = [0, 0, 0, 1, 1, 1, 1, zeros(1, 93)]; | |
for i = 1:length(x) % arrays in octave are indexed starting at 1 | |
if (y(i) == 1) | |
distance += -log(h(x(i), theta)); | |
else | |
distance += -log(1 - h(x(i), theta)); | |
end | |
end | |
% get how far off we were on average | |
distance = distance / length(x); | |
end | |
alpha = 1; | |
iters = 500; | |
m = length(x); | |
% initial values | |
theta = [0, 0, 0, 0]; | |
% I'm not using gradient descent. | |
% Instead I use Octave's built-in function | |
% fminunc to find the optimal values of theta for me. | |
opt = fminunc(@cost, theta); | |
disp(opt); | |
% for each number between 1 and 10, | |
% print out the probability that y(i) = 1 | |
for i = 1:100 | |
disp([i, h(i, opt) * 100]); | |
end |
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