Logistic Algorithm分类算法的Octave仿真

本次Octave仿真解决的问题是，根据两门入学考试的成绩来决定学生是否被录取，我们学习的训练集是包含100名学生成绩及其录取结果的数据，需要设计算法来学习该数据集，并且对新给出的学生成绩进行录取结果预测。

 

首先，我们读取并绘制training set数据集：

%% Initialization
clear ; close all; clc

%% Load Data
%  The first two columns contains the exam scores and the third column
%  contains the label.

data = load('ex2data1.txt');
X = data(:, [1, 2]); y = data(:, 3);

%% ==================== Part 1: Plotting ====================
%  We start the exercise by first plotting the data to understand the 
%  the problem we are working with.

fprintf(['Plotting data with + indicating (y = 1) examples and o ' ...
         'indicating (y = 0) examples.\n']);

plotData(X, y);

% Put some labels 
hold on;
% Labels and Legend
xlabel('Exam 1 score')
ylabel('Exam 2 score')

% Specified in plot order
legend('Admitted', 'Not admitted')
hold off;

 

然后，我们来学习训练集，直接使用我们逻辑回归算法原理分析中梯度下降算法的结果：

function [theta, J_history] = gredientDescent(X,y,alpha,iteration);
 
  [m,n]=size(X);
  theta = zeros(n,1);
  for(i= 1:iteration)
     [J,grad] = costFunction(theta,X,y);
     J_history(i) = J;
     theta = theta-X'*(sigmoid(X*theta)-y)*alpha/m;
  endfor 
  
endfunction


function [J, grad] = costFunction(theta, X, y)
m = length(y); 
J = 0;
grad = zeros(size(theta));

tmp=ones(m,1);
h = sigmoid(X*theta);
h1=log(h);
h2=log(tmp-h);

y2=tmp-y;

J=(y'*h1+y2'*h2)/(-m);

grad=(X'*(h-y))/m;
end

 计算后得出的theta值为：

绘出的decision boundary几*完美，但唯一的问题是，貌似梯度下降算法的收敛速度相当之慢，我选择了参数alpha=0.5,iteration=500000，才收敛到此程度。

而对于内建函数fminunc,迭代4000次已可以达到相*的水*。