• 使用Logistic Regression Algorithm进行多分类数字识别的Octave仿真


    所需解决的问题是,训练一个Logistic Regression系统,使之能够识别手写体数字1-10,每张图片为20px*20px的灰度图。训练样例的输入X是5000行400列的一个矩阵,每一行存储一张图片(20^2=400),共5000个训练样例,而y则为手写体所表示的数字1-10。

    利用Logistic Regression进行多分类应用,其基础是将问题本身化解为z个二分类问题,其中z为类别的个数。第一步,将向量m*1维y扩展为矩阵m*z维矩阵Y,向量n+1维向量theta扩展为矩阵z*(n+1)维矩阵Theta。其意义是将一维数据转换至二维,以0,1表示,从而使我们能够利用二分类来解决问题。如下图:

    第二步,利用内置函数fmincg来求解10组问题的最佳theta值,构建10*401维theta_all矩阵:

    function [all_theta] = oneVsAll(X, y, num_labels, lambda)
    m = size(X, 1);
    n = size(X, 2);
    
    all_theta = zeros(num_labels, n + 1);
    
    % Add ones to the X data matrix
    X = [ones(m, 1) X];
    
    % loop for every number, we train the theta of every number respectively.
    initial_theta = zeros(n+1,1);
    options = optimset('GradObj', 'on', 'MaxIter', 50);
    
    for(i=1:num_labels)
    
      y_b=(y==i);
    
      all_theta(i,:) = fmincg (@(t)(lrCostFunction(t, X,y_b, lambda)), ...
                       initial_theta, options);
    
    
    endfor
    

     其中用到的lrCostFunction函数如下:

    function [J, grad] = lrCostFunction(theta, X, y, lambda)
    %LRCOSTFUNCTION Compute cost and gradient for logistic regression with 
    %regularization
    %   J = LRCOSTFUNCTION(theta, X, y, lambda) computes the cost of using
    %   theta as the parameter for regularized logistic regression and the
    %   gradient of the cost w.r.t. to the parameters. 
    
    % Initialize some useful values
    m = length(y); % number of training examples
    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);
    
    theta(1)=0;
    
    J+=theta'*theta*lambda/(2*m);
    grad=((X'*(h-y))+lambda*theta)/m;
    
    grad = grad(:);
    
    end
    

    第三步,合并该问题,构建“可能性矩阵”,然后选择可能性最大的项作为系统的输出:

    function p = predictOneVsAll(all_theta, X)
    
    m = size(X, 1);
    num_labels = size(all_theta, 1);
    
    p = zeros(size(X, 1), 1);
    
    X = [ones(m, 1) X];
    
    probMatrix = X*all_theta';
    [pVector,p] = max(probMatrix,[],2);
    
    end
    

     

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  • 原文地址:https://www.cnblogs.com/rhyswang/p/6868865.html
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