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。

巴特西

斯坦福机器学习课程 Exercise 习题三

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