斯坦福机器学习课程 Exercise 习题三

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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 indexfigure;plot(0:49, J(1:50), '-')xlabel('Number of iterations')ylabel('Cost J')%Predictionrealx =[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。

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