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  • 【原】Coursera—Andrew Ng机器学习—编程作业 Programming Exercise 3—多分类逻辑回归和神经网络

    作业说明

    Exercise 3,Week 4,使用Octave实现图片中手写数字 0-9 的识别,采用两种方式(1)多分类逻辑回归(2)多分类神经网络。对比结果

    (1)多分类逻辑回归实现 lrCostFunction 计算代价和梯度。实现 OneVsAll 使用 fmincg 函数进行训练。使用 OneVsAll 里训练好的 theta 对 X 的数据类型进行预测,得到平均准确率。

    (2)多分类神经网络:两层 theta 权重值在 ex3weights 里已提供。参数不需要调,只需要在 predict 里进行矩阵计算,即可得到分类结果。

    数据集 :ex3data1.mat 。手写数字图片数据,5000个样例每张图片20px * 20px,也就是一共400个特征。数据集X维度为5000 * 400

         ex3weights.mat。神经网络每一层的权重。

     

    文件清单

    ex3.m - Octave/MATLAB script that steps you through part 1
    ex3 nn.m - Octave/MATLAB script that steps you through part 2
    ex3data1.mat - Training set of hand-written digits
    ex3weights.mat - Initial weights for the neural network exercise
    submit.m - Submission script that sends your solutions to our servers
    displayData.m - Function to help visualize the dataset
    fmincg.m - Function minimization routine (similar to fminunc)
    sigmoid.m - Sigmoid function
    [*] lrCostFunction.m - Logistic regression cost function
    [*] oneVsAll.m - Train a one-vs-all multi-class classi er
    [*] predictOneVsAll.m - Predict using a one-vs-all multi-class classi er
    [*] predict.m - Neural network prediction function

      * 为必须要完成的

    结论

    因为Octave里数组下标从1开始。所以这里将分类结果0用10替代。预测结果中的1-10代表图片数字为1,2,3,4,5,6,7,8,9,0

    矩阵运算 tricky 的地方在于维度对应,哪里需要转置很关键。

    多分类逻辑回归模型

    一、在数据集X里随机选择100个数字,绘制

    displayData.m:

    function [h, display_array] = displayData(X, example_width)
    %DISPLAYDATA Display 2D data in a nice grid
    %   [h, display_array] = DISPLAYDATA(X, example_width) displays 2D data
    %   stored in X in a nice grid. It returns the figure handle h and the 
    %   displayed array if requested.
    
    % Set example_width automatically if not passed in
    if ~exist('example_width', 'var') || isempty(example_width) 
        example_width = round(sqrt(size(X, 2)));
    end
    
    % Gray Image
    colormap(gray);
    
    % Compute rows, cols
    [m n] = size(X);
    example_height = (n / example_width);
    
    % Compute number of items to display
    display_rows = floor(sqrt(m));
    display_cols = ceil(m / display_rows);
    
    % Between images padding
    pad = 1;
    
    % Setup blank display
    display_array = - ones(pad + display_rows * (example_height + pad), ...
                           pad + display_cols * (example_width + pad));
    
    % Copy each example into a patch on the display array
    curr_ex = 1;
    for j = 1:display_rows
        for i = 1:display_cols
            if curr_ex > m, 
                break; 
            end
            % Copy the patch
            
            % Get the max value of the patch
            max_val = max(abs(X(curr_ex, :)));
            display_array(pad + (j - 1) * (example_height + pad) + (1:example_height), ...
                          pad + (i - 1) * (example_width + pad) + (1:example_width)) = ...
                            reshape(X(curr_ex, :), example_height, example_width) / max_val;
            curr_ex = curr_ex + 1;
        end
        if curr_ex > m, 
            break; 
        end
    end
    
    % Display Image
    h = imagesc(display_array, [-1 1]);
    
    % Do not show axis
    axis image off
    
    drawnow;
    
    end

    ex3.m里的调用

    load('ex3data1.mat'); % training data stored in arrays X, y
    m = size(X, 1);
    
    % Randomly select 100 data points to display
    rand_indices = randperm(m);
    sel = X(rand_indices(1:100), :);
    
    displayData(sel);

    运行效果如下:

     

    二、代价函数

    注意:θ0不参与正则化。

    正则化逻辑回归的代价函数如下,分为三项:

    梯度下降算法如下:

     

    lrCostFunction.m:

    function [J, grad] = lrCostFunction(theta, X, y, lambda)
    
    m = length(y); % number of training examples
    
    temp = theta; temp(1) = 0;   % because we don't add anything for j = 0  
    
    % 第一项
    part1 = -y' * log(sigmoid(X * theta));
    % 第二项
    part2 = (1 - y)' * log(1 - sigmoid(X * theta));
    
    % 正则项
    regTerm = lambda / 2 / m * temp' * temp;
    J = 1 / m * (part1 - part2) + regTerm; 
    
    % 梯度
    grad = 1 / m * X' *((sigmoid(X * theta) - y)) + lambda / m * temp;
    
    grad = grad(:);
    
    end

    正则化之后的代价函数同时返回代价 J 和 grad 梯度。这里和上个星期实现的二分类逻辑回归代码是一样的。

     三、训练oneVsAll多分类模型

    week2逻辑回归分类的目标是区分两种类型。写好代价函数之后,调用一次 fminunc 函数,得到 theta、就可以画出 boundary。

    但是多分类中需要训练 K 个分类器,所以多了一个 oneVsAll.m。其实就是循环 K 次,得到一个 theta矩阵(row 为 K,column为特征值个数)。

    function [all_theta] = oneVsAll(X, y, num_labels, lambda)
    
    % Some useful variables
    m = size(X, 1);
    n = size(X, 2);
    
    % You need to return the following variables correctly 
    all_theta = zeros(num_labels, n + 1);
    
    % Add ones to the X data matrix
    X = [ones(m, 1) X];
    
    initial_theta = zeros(n + 1, 1);
    
    options = optimset('GradObj', 'on', 'MaxIter', 50);
    
    % Run fmincg to obtain the optimal theta
    % This function will return theta and the cost 
    for i = 1 : num_labels,
      [theta] = ...
        fmincg(@(t)(lrCostFunction(t, X, (y == i), lambda)), ...
             initial_theta, options);
      all_theta(i,:) = theta';
    end;
    end

    多分类使用的 Octave 函数是 fmincg 不是 fminunc,fmincg更适合参数多的情况。

    注意这里用到了 y == i,这个操作将 y 中每个数据和 i 进行比较。得到的矩阵(这里是向量)相同的为1,不同的为0。

    ex3.m 里的调用

    lambda = 0.1;
    [all_theta] = oneVsAll(X, y, num_labels, lambda);

    四、使用OneVsAll分类器进行预测

    predictOneVsAll.m

    function p = predictOneVsAll(all_theta, X)
    
    m = size(X, 1);
    num_labels = size(all_theta, 1);
    
    % You need to return the following variables correctly 
    p = zeros(size(X, 1), 1);
    
    % Add ones to the X data matrix
    X = [ones(m, 1) X];  
       
    temp = X * all_theta';
    [x, p] = max(temp,[],2);
    
    end

    将 X 与 theta 相乘得到预测结果,得到一个 5000*10 的矩阵,每行对应一个分类结果(只有一个1,其余为0)。

    题目要求返回的矩阵 p 是一个 5000*1 的矩阵。每行是 1-K 的数字。使用 max 函数,得到两个返回值。第一个 x 是一个全1的向量,没用。第二个是 1 所在的 index,也就是对应的类别。

    ex3.m 中的调用:

    pred = predictOneVsAll(all_theta, X);
    fprintf('
    Training Set Accuracy: %f
    ', mean(double(pred == y)) * 100);

    三层神经网络

    这里g(z) 使用 sigmoid 函数。

    神经网络中,从上到下的每个原点是 feature 特征 x0, x1, x2...,不是实例。计算过程其实就是 feature 一层一层映射的过程。一层转换之后,feature可能变多、也可能变少。下一层 i+1层 feature 的个数是通过权重矩阵里当前 θ(i) 的 row 行数来控制。

    两层权重 θ 已经在 ex3weights.mat 里给出。从a1映射到a2权重矩阵 θ1为 25 * 401,从a2映射到a3权重矩阵 θ2为10 * 26。因为最后有10个分类。(这意味着运算的时候要注意转置)

    一、使用OneVsAll分类器进行预测

    权重已经在文件中给出了,只需要矩阵相乘,预测就好了

    predict.m :

    function p = predict(Theta1, Theta2, X)
    %PREDICT Predict the label of an input given a trained neural network
    %   p = PREDICT(Theta1, Theta2, X) outputs the predicted label ofxgiven the
    %   trained weights of a neural network (Theta1, Theta2)
    
    % Useful values
    m = size(X, 1);
    num_labels = size(Theta2, 1);
    
    % You need to return the following variables correctly 
    p = zeros(size(X, 1), 1);
    
    % ====================== YOUR CODE HERE ======================
    % Instructions: Complete the following code to make predictions using
    %               your learned neural network. You should set p to a 
    %               vector containing labels between 1 to num_labels.
    %
    % Hint: The max function might come in useful. In particular, the max
    %       function can also return the index of the max element, for more
    %       information see 'help max'. If your examples are in rows, then, you
    %       can use max(A, [], 2) to obtain the max for each row.
    %0
    
    % Add ones to thexdata matrix
    a1 = [ones(m, 1) X];    %5000x401
    a2 = sigmoid(a1 * Theta1');  %5000x401乘以401x25得到5000x25。即把401个feature映射到25
    
    a2 = [ones(m, 1) a2];    %5000x26
    a3 = sigmoid(a2 * Theta2');  %5000x26乘以26x10得到5000x10。即把26个feature映射到10
    
    [x,p] = max(a3,[],2);  %和上面逻辑回归多分类一样,最后使用 max 函数获得分类结果。
    % =========================================================================
    
    end

    二、预测结果

    em3_nn.m 里的调用

    %% Setup the parameters you will use for this exercise
    input_layer_size  = 400;  % 20x20 Input Images of Digits
    hidden_layer_size = 25;   % 25 hidden units
    num_labels = 10;          % 10 labels, from 1 to 10   
                              % (note that we have mapped "0" to label 10)
    
    load('ex3weights.mat');
    
    pred = predict(Theta1, Theta2, X);
    
    fprintf('
    Training Set Accuracy: %f
    ', mean(double(pred == y)) * 100);
    
    %  To give you an idea of the network's output, you can also run
    %  through the examples one at the a time to see what it is predicting.
    
    %  Randomly permute examples    随机获取数字,显示预测结果
    rp = randperm(m);
    
    for i = 1:m
        % Display 
        fprintf('
    Displaying Example Image
    ');
        displayData(X(rp(i), :));
    
        pred = predict(Theta1, Theta2, X(rp(i),:));
        fprintf('
    Neural Network Prediction: %d (digit %d)
    ', pred, mod(pred, 10));
        
        % Pause with quit option
        s = input('Paused - press enter to continue, q to exit:','s');
        if s == 'q'
          break
        end
    end

    运行结果:

    准确率 Training Set Accuracy: 97.520000

    这里我比较疑惑的一点是。代码中没对 0 进行特殊处理,为什么能预测成10。后来一想 整个计算过程都是权重矩阵控制的,给的 weights 文件里已经是调整好的结果了,0 就是会被分类成10。

    所以神经网络最主要的还是调参啊,调 θ 矩阵。

    三、关于矩阵相乘

    矩阵这里转置总是出问题。仔细思考一下

    (1)假设只有一个实例 x,而且它像图中一样是一个列向量。那根据神经网络的公式,最相符的应该是  θ(i) * x。最后的得到的是 K 维列向量。

    (2)但实际情况中,训练数据通常一行一个实例。如果严格按图和公式去计算,(θ(i) * XT)T 就需要将 X 先转置过来,最后再转置回去。应该是下面的代码,但是明显能看出比上面的方法麻烦:

    a1 = [ones(1, m); X'];      %401x5000
    a2 = sigmoid(Theta1 * a1);  %25x401 乘以 401x5000   得到25x5000
    
    a2 = [ones(1, m); a2];      %26x5000
    a3 = sigmoid(Theta2 * a2);  %10x26 乘以 26x5000    得到10x5000
    
    a3 = a3'; % 5000x10

    (3)所以用 X * θ(i)替换 (θ(i) * XT)T

    a1 = [ones(m, 1) X];    %5000x401
    a2 = sigmoid(a1 * Theta1');  %5000x401乘以401x25得到5000x25。即把401个feature映射到25
    
    a2 = [ones(m, 1) a2];    %5000x26
    a3 = sigmoid(a2 * Theta2');  %5000x26乘以26x10得到5000x10。即把26个feature映射到10

    完整代码

    https://github.com/madoubao/coursera_machine_learning/tree/master/homework/machine-learning-ex3/ex3

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  • 原文地址:https://www.cnblogs.com/maxiaodoubao/p/9939812.html
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