LibLinear(SVM包)使用说明之(三)实践
http://blog.csdn.net/zouxy09
我们在UFLDL的教程中,Exercise: Convolution and Pooling这一章节,已经得到了cnnPooledFeatures.mat特征。在该练习中,我们使用的是softmax分类器来分类的。在这里我们修改为用SVM来替代softmax分类器。SVM由Liblinear软件包来提供。这里是四分类问题,所以Liblinear会根据我们传入的训练样本训练四个二分类器,以实现四分类。以前由softmax分类器得到的准确率是80.406%。在这里换成Liblinear后,准确率变为80.75%。在这里差别不是很大。
在本文的例子中,我们增加了scale和Cross Validation,Cross Validation是用来选择一个最好的参数C的(不知道自己这两个步骤有没有正确,如有错误,还望大家提醒,谢谢)。
具体的代码如下:
%// Classification by LibLinear %// LibLinear: http://www.csie.ntu.edu.tw/~cjlin/liblinear/ %// Author : zouxy %// Date : 2013-9-2 %// HomePage : http://blog.csdn.net/zouxy09 %// Email : [email protected] clear; clc; %%% step1: load data fprintf(1,'step1: Load data...\n'); % pooledFeaturesTrain大小为400*2000*3*3 % pooledFeaturesTest大小为400*3200*3*3 % 第一维是特征个数,也就是特征图个数,第二维是样本个数,第三维是特征图的宽, % 第四维是特征图的高 load cnnPooledFeatures.mat; load stlTrainSubset.mat % loads numTrainImages, trainImages, trainLabels load stlTestSubset.mat % loads numTestImages, testImages, testLabels % B = permute(A,order) 按照向量order指定的顺序重排A的各维 train_X = permute(pooledFeaturesTrain, [1 3 4 2]); % 将每个样本的特征拉成一个列向量,每个样本一个列,矩阵大小为3600*2000 train_X = reshape(train_X, numel(pooledFeaturesTrain) / numTrainImages, numTrainImages); train_Y = trainLabels; % 2000*1 test_X = permute(pooledFeaturesTest, [1 3 4 2]); test_X = reshape(test_X, numel(pooledFeaturesTest) / numTestImages, numTestImages); test_Y = testLabels; % release some memory clear trainImages testImages pooledFeaturesTrain pooledFeaturesTest; %%% step2: scale the data fprintf(1,'step2: Scale data...\n'); % Using the same scaling factors for training and testing sets, % we obtain much better accuracy. Note: scale each attribute(feature), not sample % scale to [0 1] % when a is a vector, b = (a - min(a)) .* (upper - lower) ./ (max(a)-min(a)) + lower lower = 0; upper = 1.0; train_X = train_X'; X_max = max(train_X); X_min = min(train_X); train_X = (train_X - repmat(X_min, size(train_X, 1), 1)) .* (upper - lower) ... ./ repmat((X_max - X_min), size(train_X, 1), 1) + lower; test_X = test_X'; test_X = (test_X - repmat(X_min, size(test_X, 1), 1)) .* (upper - lower) ... ./ repmat((X_max - X_min), size(test_X, 1), 1) + lower; % Note: before scale the accuracy is 80.4688%, after scale it turns to 80.1875%, % and took more time. So is that my scale operation wrong or other reasons? % After adding bias, Accuracy = 80.75% (2584/3200) %%% step3: Cross Validation for choosing parameter fprintf(1,'step3: Cross Validation for choosing parameter c...\n'); % the larger c is, more time should be costed c = [2^-6 2^-5 2^-4 2^-3 2^-2 2^-1 2^0 2^1 2^2 2^3]; max_acc = 0; tic; for i = 1 : size(c, 2) option = ['-B 1 -c ' num2str(c(i)) ' -v 5 -q']; fprintf(1,'Stage: %d/%d: c = %d, ', i, size(c, 2), c(i)); accuracy = train(train_Y, sparse(train_X), option); if accuracy > max_acc max_acc = accuracy; best_c = i; end end fprintf(1,'The best c is c = %d.\n', c(best_c)); toc; %%% step4: train the model fprintf(1,'step4: Training...\n'); tic; option = ['-c ' num2str(c(best_c)) ' -B 1 -e 0.001']; model = train(train_Y, sparse(train_X), option); toc; %%% step5: test the model fprintf(1,'step5: Testing...\n'); tic; [predict_label, accuracy, dec_values] = predict(test_Y, sparse(test_X), model); toc;