本文主要實(shí)現(xiàn)手寫數(shù)字識(shí)別，利用多類邏輯回歸與神經(jīng)網(wǎng)絡(luò)兩種方法實(shí)現(xiàn)

Multi-class Classification

數(shù)據(jù)源
There are 5000 training examples in ex3data1.mat, where each training example is a 20 pixel by 20 pixel grayscale image of the digit. Each pixel is represented by a floating point number indicating the grayscale intensity at that location. The 20 by 20 grid of pixels is “unrolled” into a 400-dimensional vector. Each of these training examples becomes a single row in our data matrix X. This gives us a 5000 by 400 matrix X where every row is a training example for a handwritten digit image.

The second part of the training set is a 5000-dimensional vector y that contains labels for the training set. To make things more compatible with Octave/MATLAB indexing, where there is no zero index, we have mapped the digit zero to the value ten. Therefore, a “0” digit is labeled as “10”, while the digits “1” to “9” are labeled as “1” to “9” in their natural order.

數(shù)據(jù)可視化
通過displaydata函數(shù)可以察看數(shù)據(jù)

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_rowsfor i = 1:display_colsif curr_ex > m, break; end% Copy the patch% Get the max value of the patchmax_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;endif curr_ex > m, break; end end% Display Image h = imagesc(display_array, [-1 1]);% Do not show axis axis image offdrawnow;end

向量化邏輯回歸

由于數(shù)據(jù)較多，使用向量化可以使訓(xùn)練更加高效，提高速度。

代價(jià)函數(shù)的向量化
代價(jià)函數(shù)為

Let us define X and θ as

gradient向量化

小技巧

Debugging Tip: Vectorizing code can sometimes be tricky. One com- mon strategy for debugging is to print out the sizes of the matrices you are working with using the size function. For example, given a data ma- trix X of size 100 × 20 (100 examples, 20 features) and θ, a vector with dimensions 20×1, you can observe that Xθ is a valid multiplication oper- ation, while θX is not. Furthermore, if you have a non-vectorized version of your code, you can compare the output of your vectorized code and non-vectorized code to make sure that they produce the same outputs.

綜上，可以寫出lrCostFunction

function [J, grad] = lrCostFunction(theta, X, y, lambda) m = length(y); % number of training examples% You need to return the following variables correctly J = 0; grad = zeros(size(theta)); temp=[0;theta(2:end)]; % 先把theta(1)拿掉，不參與正則化 J= -1 * sum( y .* log( sigmoid(X*theta) ) + (1 - y ) .* log( (1 - sigmoid(X*theta)) ) ) / m + lambda/(2*m) * temp' * temp ; grad = ( X' * (sigmoid(X*theta) - y ) )/ m + lambda/m * temp ;grad = grad(:);end

一對(duì)多分類
在這個(gè)問題中，需要分成10份

小技巧

Octave/MATLAB Tip: Logical arrays in Octave/MATLAB are arrays which contain binary (0 or 1) elements. In Octave/MATLAB, evaluating the expression a == b for a vector a (of size m×1) and scalar b will return a vector of the same size as a with ones at positions where the elements of a are equal to b and zeroes where they are different. To see how this works for yourself, try the following code in Octave/MATLAB:a = 1:10; % Create a and bb = 3;a == b % You should try different values of b here

由上可得oneVsAll function

function [all_theta] = oneVsAll(X, y, num_labels, lambda) m = size(X, 1); n = size(X, 2); X = [ones(m, 1) X];options = optimset('GradObj', 'on', 'MaxIter', 50);initial_theta = zeros(n + 1, 1);for c = 1:num_labelsall_theta(c,:) = fmincg (@(t)(lrCostFunction(t, X, (y == c), lambda)), ...initial_theta, options); endend

One-vs-all Prediction
根據(jù)上面得到了訓(xùn)練后的θ ，根據(jù)訓(xùn)練的結(jié)果，得到各個(gè)分類的可能性，取最大的一個(gè)，即是結(jié)果

function p = predictOneVsAll(all_theta, X) [a,p] = max(sigmoid( X * all_theta'),[],2) ; % 返回每行最大值的索引位置，也就是預(yù)測的數(shù)字 end pred = predictOneVsAll(all_theta, X);fprintf('\nTraining Set Accuracy: %f\n', mean(double(pred == y)) * 100);

可得準(zhǔn)確率為95%

SourceCode:machine-learning-ex3

神經(jīng)網(wǎng)絡(luò)實(shí)現(xiàn)

神經(jīng)網(wǎng)絡(luò)實(shí)現(xiàn)主要步驟如下：

• 初始化與導(dǎo)入數(shù)據(jù)
• Compute Cost (Feedforward)
• Implement Regularization
• 隨機(jī)初始化參數(shù)
• Implement Backpropagation
• 梯度檢驗(yàn)
• 訓(xùn)練神經(jīng)網(wǎng)絡(luò)
• 預(yù)測

反向傳播算法的具體實(shí)現(xiàn)

function [J grad] = nnCostFunction(nn_params, ...input_layer_size, ...hidden_layer_size, ...num_labels, ...X, y, lambda)Theta1 = reshape(nn_params(1:hidden_layer_size * (input_layer_size + 1)), ...hidden_layer_size, (input_layer_size + 1));Theta2 = reshape(nn_params((1 + (hidden_layer_size * (input_layer_size + 1))):end), ...num_labels, (hidden_layer_size + 1));% Setup some useful variables m = size(X, 1);% You need to return the following variables correctly J = 0; Theta1_grad = zeros(size(Theta1)); Theta2_grad = zeros(size(Theta2));%% 對(duì)y進(jìn)行處理 Y(find(y==3))= [0 0 1 0 0 0 0 0 0 0]; 用于 Feedforward cost function 1和2 Y=[]; E = eye(num_labels); % 要滿足K可以是任意，則不能寫eye(10)！！ for i=1:num_labelsY0 = find(y==i); % 找到等于y=i的序列號(hào),替換向量Y(Y0,:) = repmat(E(i,:),size(Y0,1),1); end%% regularized Feedforward cost function lambda=1 % 計(jì)算前向傳輸 Add ones to the X data matrix -jin X = [ones(m, 1) X]; a2 = sigmoid(X * Theta1'); % 第二層激活函數(shù)輸出 a2 = [ones(m, 1) a2]; % 第二層加入b a3 = sigmoid(a2 * Theta2'); temp1 = [zeros(size(Theta1,1),1) Theta1(:,2:end)]; % 先把theta(1)拿掉，不參與正則化 temp2 = [zeros(size(Theta2,1),1) Theta2(:,2:end)]; temp1 = sum(temp1 .^2); % 計(jì)算每個(gè)參數(shù)的平方，再就求和 temp2 = sum(temp2 .^2);cost = Y .* log(a3) + (1 - Y ) .* log( (1 - a3)); % cost是m*K(5000*10)的結(jié)果矩陣 sum(cost(:))全部求和 J= -1 / m * sum(cost(:)) + lambda/(2*m) * ( sum(temp1(:))+ sum(temp2(:)) ); %% 計(jì)算 Gradient delta_1 = zeros(size(Theta1)); delta_2 = zeros(size(Theta2));for t = 1:m% step 1a_1 = X(t,:)'; % a_1 = [1 ; a_1];z_2 = Theta1 * a_1; a_2 = sigmoid(z_2); a_2 = [1 ; a_2];z_3 = Theta2 * a_2;a_3 = sigmoid(z_3);% step 2err_3 = zeros(num_labels,1);for k = 1:num_labels err_3(k) = a_3(k) - (y(t) == k);end% step 3err_2 = Theta2' * err_3; % err_2有26行！！！err_2 = err_2(2:end) .* sigmoidGradient(z_2); % 去掉第一個(gè)誤差值，減少為25. sigmoidGradient(z_2)只有25行！！！% step 4delta_2 = delta_2 + err_3 * a_2';delta_1 = delta_1 + err_2 * a_1'; end% step 5 Theta1_temp = [zeros(size(Theta1,1),1) Theta1(:,2:end)]; Theta2_temp = [zeros(size(Theta2,1),1) Theta2(:,2:end)]; Theta1_grad = 1 / m * delta_1 + lambda/m * Theta1_temp; Theta2_grad = 1 / m * delta_2 + lambda/m * Theta2_temp ;% Unroll gradients grad = [Theta1_grad(:) ; Theta2_grad(:)];end

訓(xùn)練神經(jīng)網(wǎng)絡(luò)代碼

options = optimset('MaxIter', 50);% You should also try different values of lambda lambda = 1;% Create "short hand" for the cost function to be minimized costFunction = @(p) nnCostFunction(p, ...input_layer_size, ...hidden_layer_size, ...num_labels, X, y, lambda);% Now, costFunction is a function that takes in only one argument (the % neural network parameters) [nn_params, cost] = fmincg(costFunction, initial_nn_params, options);

預(yù)測結(jié)果代碼

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 of X given 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);h1 = sigmoid([ones(m, 1) X] * Theta1'); h2 = sigmoid([ones(m, 1) h1] * Theta2'); [dummy, p] = max(h2, [], 2);% =========================================================================end

準(zhǔn)確率大概95%左右，根據(jù)隨機(jī)初始化的參數(shù)會(huì)有1%左右的誤差
SourceCode:machine-learning-ex4

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