将预计算的 chi2 内核与 libsvm (matlab) 一起使用时结果不佳

Question

我正在尝试 libsvm，并按照该示例在软件随附的 heart_scale 数据上训练 svm。我想使用我自己预先计算的 chi2 内核。训练数据的分类率下降到 24%。我确定我正确计算了内核，但我想我一定是做错了什么。代码如下。你能看出有什么错误吗？非常感谢您的帮助。

%read in the data:
[heart_scale_label, heart_scale_inst] = libsvmread('heart_scale');
train_data = heart_scale_inst(1:150,:);
train_label = heart_scale_label(1:150,:);

%read somewhere that the kernel should not be sparse
ttrain = full(train_data)';
ttest = full(test_data)';

precKernel = chi2_custom(ttrain', ttrain');
model_precomputed = svmtrain2(train_label, [(1:150)', precKernel], '-t 4');

这是内核的预计算方式：

function res=chi2_custom(x,y)
a=size(x);
b=size(y);
res = zeros(a(1,1), b(1,1));
for i=1:a(1,1)
    for j=1:b(1,1)
        resHelper = chi2_ireneHelper(x(i,:), y(j,:));
        res(i,j) = resHelper;
    end
end
function resHelper = chi2_ireneHelper(x,y)
a=(x-y).^2;
b=(x+y);
resHelper = sum(a./(b + eps));

使用不同的 svm 实现 (vlfeat)，我获得了大约 90% 的训练数据分类率（是的，我在训练数据上进行了测试，只是为了看看发生了什么）。所以我很确定 libsvm 结果是错误的。

Answer 1

使用支持向量机时，将数据集归一化作为预处理步骤非常重要。 Normalization 将属性放在相同的尺度上，并防止具有较大值的属性对结果产生偏差。它还提高了数值稳定性（最大限度地减少了浮点表示引起的上溢和下溢的可能性）。

确切地说，您对卡方核的计算略有偏差。取而代之的是下面的定义，并为它使用这个更快的实现：

function D = chi2Kernel(X,Y)
    D = zeros(size(X,1),size(Y,1));
    for i=1:size(Y,1)
        d = bsxfun(@minus, X, Y(i,:));
        s = bsxfun(@plus, X, Y(i,:));
        D(:,i) = sum(d.^2 ./ (s/2+eps), 2);
    end
    D = 1 - D;
end

现在考虑使用与您相同的数据集的以下示例（代码改编自我的previous answer）：

%# read dataset
[label,data] = libsvmread('./heart_scale');
data = full(data);      %# sparse to full

%# normalize data to [0,1] range
mn = min(data,[],1); mx = max(data,[],1);
data = bsxfun(@rdivide, bsxfun(@minus, data, mn), mx-mn);

%# split into train/test datasets
trainData = data(1:150,:);    testData = data(151:270,:);
trainLabel = label(1:150,:);  testLabel = label(151:270,:);
numTrain = size(trainData,1); numTest = size(testData,1);

%# compute kernel matrices between every pairs of (train,train) and
%# (test,train) instances and include sample serial number as first column
K =  [ (1:numTrain)' , chi2Kernel(trainData,trainData) ];
KK = [ (1:numTest)'  , chi2Kernel(testData,trainData)  ];

%# view 'train vs. train' kernel matrix
figure, imagesc(K(:,2:end))
colormap(pink), colorbar

%# train model
model = svmtrain(trainLabel, K, '-t 4');

%# test on testing data
[predTestLabel, acc, decVals] = svmpredict(testLabel, KK, model);
cmTest = confusionmat(testLabel,predTestLabel)

%# test on training data
[predTrainLabel, acc, decVals] = svmpredict(trainLabel, K, model);
cmTrain = confusionmat(trainLabel,predTrainLabel)

测试数据上的结果：

Accuracy = 84.1667% (101/120) (classification)
cmTest =
    62     8
    11    39

在训练数据上，我们得到了大约 90% 的准确率，如您所料：

Accuracy = 92.6667% (139/150) (classification)
cmTrain =
    77     3
     8    62

Answer 2

问题出在下面一行：

resHelper = sum(a./(b + eps));

应该是：

resHelper = 1-sum(2*a./(b + eps));