机器学习之线性回归

2018-12-09 本文已影响15人 MrWisdom

设线性回归的训练集为
$T = [X|Y]$
其中 $X \in R^{m \times p}$ , $Y\in R^{m \times 1}$ , m为训练集的样本个数，p为样本特征数。
作线性回归，设其回归模型为
$h_i(X) = [1,x_1^{(i)},x_2^{(i)},\cdots,x_p^{(i)}]\cdot[\theta_0,\theta_1,\theta_2,\cdots,\theta_p]^T = \mathbf{X^{(i)}}\mathbf{\theta}$
其中 $\mathbf{X, \theta}$ 均为p+1维向量。
最小化性能指标
$J(\theta) = \frac{1}{2m}\sum_{i=1}^{m}(H(X)-Y)^2$
梯度为
$\dot J(\theta) = \frac{1}{m}\sum_{i=1}^{m}X^T(H(X)-Y)$
其中, $H(X)= [h_1(x),h_2(x),\cdots,h_m(x)]^T$
Matlab代码实现为

clear ; close all; clc
data = load('ex1data2.txt');
X = data(:, 1:2);
y = data(:, 3);
m = length(y);

%normalization
mu = mean(X);       %  mean value 
sigma = std(X);     %  standard deviation
X_norm  = (X - repmat(mu,size(X,1),1)) ./  repmat(sigma,size(X,1),1);

X = [ones(m, 1) X];     % Add intercept term to X

% Choose some alpha value
alpha = 0.01;
num_iters = 8500;
theta = zeros(3, 1);
J_history = zeros(num_iters, 1);

for iter = 1:num_iters
    theta = theta - alpha / m * X' * (X * theta - y); 
    J_history(iter) = computeCostMulti(X, y, theta);
end

function J = computeCostMulti(X, y, theta)
m = length(y);
J = 0;
J = sum((X * theta - y).^2) / (2*m);
end

机器学习之线性回归

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