我有简单的异或问题,我想学习使用libsvm中的RBF内核。当我使用XOR问题,像训练了Java LIBSVM:为什么svm_predict和svm_predict_probability在java libsvm中为xor问题提供了不同的结果?
x y 0,0 -1 0,1 1 1,0 1 1,1 -1
结果我得到一个测试向量(0,0)分类为-1,如果我使用svm.svm_predict,但+1如果我使用SVM .svm_predict_probability。即使返回的概率也是相反的。我使用的代码和结果如下。任何人都可以告诉我我在做什么错在这里?
public static void main(String[] args) {
svm_problem sp = new svm_problem();
svm_node[][] x = new svm_node[4][2];
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 2; j++) {
x[i][j] = new svm_node();
}
}
x[0][0].value = 0;
x[0][1].value = 0;
x[1][0].value = 1;
x[1][1].value = 1;
x[2][0].value = 0;
x[2][1].value = 1;
x[3][0].value = 1;
x[3][1].value = 0;
double[] labels = new double[]{-1,-1,1,1};
sp.x = x;
sp.y = labels;
sp.l = 4;
svm_parameter prm = new svm_parameter();
prm.svm_type = svm_parameter.C_SVC;
prm.kernel_type = svm_parameter.RBF;
prm.C = 1000;
prm.eps = 0.0000001;
prm.gamma = 10;
prm.probability = 1;
prm.cache_size=1024;
System.out.println("Param Check " + svm.svm_check_parameter(sp, prm));
svm_model model = svm.svm_train(sp, prm);
System.out.println(" PA "+ model.probA[0]);
System.out.println(" PB " + model.probB[0]);
System.out.println(model.sv_coef[0][0]);
System.out.println(model.sv_coef[0][1]);
System.out.println(model.sv_coef[0][2]);
System.out.println(model.sv_coef[0][3]);
System.out.println(model.SV[0][0].value + "\t" + model.SV[0][1].value);
System.out.println(model.SV[1][0].value + "\t" + model.SV[1][1].value);
System.out.println(model.SV[2][0].value + "\t" + model.SV[2][1].value);
System.out.println(model.SV[3][0].value + "\t" + model.SV[3][1].value);
System.out.println(model.label[0]);
System.out.println(model.label[1]);
svm_node[] test = new svm_node[]{new svm_node(), new svm_node()};
test[0].value = 0;
test[1].value = 0;
double[] l = new double[2];
double result_prob = svm.svm_predict_probability(model, test,l);
double result_normal = svm.svm_predict(model, test);
System.out.println("Result with prob " + result_prob);
System.out.println("Result normal " + result_normal);
System.out.println("Probability " + l[0] + "\t" + l[1]);
}
---------结果-------------
Param Check null
*
.
.
optimization finished, #iter = 3
nu = 0.0010000908050150552
obj = -2.000181612091545, rho = 0.0
nSV = 4, nBSV = 0
Total nSV = 4
PA 3.2950351477129125
PB -2.970957107176531E-12
1.0000908039844314
1.0000908060456788
-1.0000908039844314
-1.0000908060456788
0.0 0.0
1.0 1.0
0.0 1.0
1.0 0.0
-1
1
Result with prob 1.0
Result normal -1.0
Probability 0.03571492727188865 0.9642850727281113
显然,结果是完全相反的。这似乎发生在我选择作为测试的任何示例上。
有人可以对此有所了解吗? 在此先感谢
嗨,感谢您的回复,但我试图互换标签,但这似乎并不重要。它总是给错误的类别分配更高的概率,就好像有人忘记了撤销标志或什么的。 – Karthik 2011-05-17 10:37:05