Algorithm 2 Improved splicing method

1: Input: An individual A, i.e., $A=[f_1,f_2,\cdots ,f_p]$, $f_j\in \{0,1\}$, Sample $X\in n\times p$, $Y\in n\times 1$, threshold $\tau$; Maximal size of the subset of genes $S_{max}$;       2: The individual A is divided into active set $\mathcal{A}=\left\{f_i\right|f_i=1,i=1,\cdots ,p\}$ and inactive set $\mathcal{I}=\left\{f_i\right|f_i=0,i=1,\cdots ,p\}$;       3: if $\left|\mathcal{A}\right|>S_{max}$  then       4:     Calculate the score of each gene in active set $\mathcal{A}$ in terms of backward sacrifice.       5:     Delete $\left|\mathcal{A}\right|-S_{max}$ genes of the lowest score in active set $\mathcal{A}$ to obtain a new active set $\mathcal{A}$; Then deleted $\left|\mathcal{A}\right|-S_{max}$ genes are added into inactive set $\mathcal{I}$ to obtain a new inactive set $\mathcal{I}$.       6:     ${\beta }_{\mathcal{A}}={\left(X_{\mathcal{A}}^T{X}_{\mathcal{A}}\right)}^{-1}{X}_{\mathcal{A}}^{T}Y$, $d_{\mathcal{I}}=X_{\mathcal{I}}^T(Y-X_{\mathcal{A}}{\beta }_{\mathcal{A}})/n$, where $X_{\mathcal{A}}\in n\times \left|\mathcal{A}\right|$, $X_{\mathcal{I}}\in n\times \left|\mathcal{I}\right|$;       7: else       8:     ${\beta }_{\mathcal{A}}={\left(X_{\mathcal{A}}^T{X}_{\mathcal{A}}\right)}^{-1}{X}_{\mathcal{A}}^{T}Y$, $d_{\mathcal{I}}=X_{\mathcal{I}}^T(Y-X_{\mathcal{A}}{\beta }_{\mathcal{A}})/n$, where $X_{\mathcal{A}}\in n\times \left|\mathcal{A}\right|$, $X_{\mathcal{I}}\in n\times \left|\mathcal{I}\right|$;       9: end if       10: repeat       11:     Calculate Loss function $L=L_0=\frac{1}{2n}\left|\right|Y-X_{\mathcal{A}}{\beta }_{\mathcal{A}}{\left|\right|}_2^2$;       12:     Calculate the score ${\xi }_j$ of each gene in active set $\mathcal{A}$ in terms of backward sacrifice.       13:     Calculate the score of each gene ${\zeta }_j$ in inactive set $\mathcal{I}$ in terms of forward sacrifice.       14:     for  $k=1,2,\cdots ,min\left(\right|\mathcal{A}|,|\mathcal{I}\left|\right)$ do       15:        ${\mathcal{A}}_k=\{j\in \mathcal{A}:{\sum }_{i\in \mathcal{A}}I({\xi }_j\ge {\xi }_i)\le k\}$, ${\mathcal{I}}_k=\{j\in \mathcal{I}:{\sum }_{i\in \mathcal{I}}I({\zeta }_j\le {\zeta }_i)\le k\}$;       16:        $\tilde{\mathcal{A}}=(\mathcal{A}\setminus {\mathcal{A}}_k)\bigcup {\mathcal{I}}_k$, $\tilde{\mathcal{I}}=(\mathcal{I}\setminus {\mathcal{I}}_k)\bigcup {\mathcal{A}}_k$, ${\tilde{d}}_{\tilde{\mathcal{I}}}=X_{\tilde{\mathcal{I}}}^T(Y-X_{\tilde{\mathcal{A}}}{\beta }_{\tilde{\mathcal{A}}})$, ${\tilde{\beta }}_{\tilde{\mathcal{A}}}={\left(X_{\tilde{\mathcal{A}}}^T{X}_{\tilde{\mathcal{A}}}\right)}^{-1}{X}_{\tilde{\mathcal{A}}}^{T}Y$;       17:        Calculate Loss function $L_n=\frac{1}{2n}\left|\right|Y-X_{\tilde{\mathcal{A}}}{\tilde{\beta }}_{\tilde{\mathcal{A}}}{\left|\right|}_2^2$;       18:        if  $L_n<L$  then       19:           ${\beta }_{\mathcal{A}}={\tilde{\beta }}_{\tilde{\mathcal{A}}}$, $d_{\mathcal{I}}={\tilde{d}}_{\tilde{\mathcal{I}}}$, $\mathcal{A}=\tilde{\mathcal{A}}$, $\mathcal{I}=\tilde{\mathcal{I}}$, $L=L_n$;       20:        end if       21:     end for       22: until  $L_0-L<\tau$       23: Merge active set $\mathcal{A}$ with inactive set $\mathcal{I}$ to generate a new individual A.       24: Output: A new individual A.