Random Forest 随机森林

袋外错误率 Oob Error Rate

无需交叉认证或测试集，在训练过程中，每个样本被选中的概率为 1 - 1/e ≈ 63.2%。袋外样本的预测结果不会被用于训练，因此可以用来评估模型的泛化能力。

$f_6(x)=\sum_{i=1}^n([x_i+0.5])^2$

$f_7(x)=\sum_{l=1}^nix_i^4+random[0,1)$

$F_8(x)=\sum_{i=1}^n-x\sin(\sqrt{|x_i|})$

$F_9(x)=\sum_{i=1}^n[x_i^2-10\cos(2\pi x_i)+10]$

$F_{10}(x)=-20exp(-0.2\sqrt{\frac{1}{n}\sum_{i=1}^n\cos(2\pi x_i)+20+e})$

$F_{11}(x)=\frac{1}{4000}\sum_{i=1}^nx_i^2-\prod_{i=1}^n\cos(\frac{x_i}{\sqrt i})+1$

$$F_{12}(x)=\frac{\pi}{n}\{10\sin(\pi y_1)+\sum_{i=1}^n(y_i-1)^2[1+10\sin^2(\pi y_{i+1})]+(y_n-1)^2\}+\sum_{i=1}^nu(x_i,10,100,4)\\y_i=1+\frac{x_i+1}{4}\\u(x_i,a,k,m)=$$

$F_{13}(x)=0.1\{\sin(3\pi x_1)+\sum_{i=1}^n(x_i-1)^2[1+\sin^2(3\pi x_i+1)]+(x_n-1)^2[1+\sin^2(2\pi x_n)]\}+\sum_{i=1}^nu(x_i,5,100,4)$

$F_{14}(x)=(\frac{1}{500}+\sum_{j=1}^{25}\frac{1}{j+\sum_{i=1}^2(x_i-a_{ij})^6})^{-1}$

$F_{15}(x)=\sum_{i=1}^{11}[a_i-\frac{x_1(b_i^2+b_ix_2)}{b_i^2+b_ix_3+x_4}]^2$

$F_{16}(x)=4x_1^2-2.1x_1^4+\frac{1}{3}x_1^6+x_1x_2-4x_2^2+4x_2^4$

$F_{17}(x)=(x_2-\frac{5.1}{4\pi^2}x_1^2+\frac{5}{\pi}x_1-6)^2+10(1-\frac{1}{8\pi})\cos x_1+10$

$F_{18}(x)=[1+(x_1+x_2+1)^2(19-14x_1+3x_1^2-14_2+6x_1x_2+3x_2^2)]\times[30+(2x_1-3x_2)^2\times(18-32x_1+12x_1^2+48x_2-36x_1x_2+27x_2^2)]$