斯坦福机器学习课程 Exercise 习题三

Exercise 3: Multivariate Linear Regression

预处理数据
Preprocessing the inputs will significantly increase gradient descent’s efficiency

Matlab代码

x=load('L:\\MachineLearning2016\\ex3x.dat');
y=load('L:\\MachineLearning2016\\ex3y.dat');
m = length(x(:,1));
x = [ones(m, 1), x];
sigma = std(x);
mu = mean(x);
x(:,2) = (x(:,2) - mu(2))./ sigma(2);
x(:,3) = (x(:,3) - mu(3))./ sigma(3);%%规整化输入数据
theta = zeros( size( x(1,:) ) )';
alpha= 0.18;
J = zeros(50, 1); %%这里只迭代50次
for num_iterations = 1:50
    J(num_iterations) =  (x*theta - y)' * (x * theta -y) /m/2; %% Calculate your cost function here %%
    theta = theta -( (x*theta -y)' * x)' * alpha /m /2; %% Result of gradient descent update %%
end

% now plot J
% technically, the first J starts at the zero-eth iteration
% but Matlab/Octave doesn't have a zero index
figure;
plot(0:49, J(1:50), '-')
xlabel('Number of iterations')
ylabel('Cost J')

%Prediction
realx =[1,1650,3];
realx(2) = (realx(2) - mu(2))./ sigma(2);
realx(3) = (realx(3) - mu(3))./ sigma(3);
realx*theta

Normal equations

不对数据进行预处理

x=load('L:\\MachineLearning2016\\ex3x.dat');
y=load('L:\\MachineLearning2016\\ex3y.dat');
m = length(x(:,1));
x = [ones(m, 1), x];
theta = (x'*x) \(x'*y);
J3=  (x*theta - y)' * (x * theta -y)/m/2;
realx =[1,1650,3];
realx*theta;

TIPS：Normal equations 好处是不用对数据进行规整化。缺点是矩阵运算比较占用计算机资源。

这里有个疑问，对要预测的数据的处理，[1,1650,3]，规整化是否是直接使用样本数据的均值和标准差。
注：上面的 scale data 只迭代了50次，与Normal equations的结果有较大误差。我就懒得去验证了。