2017-05-04 115 views
0

我有大規模的QP問題,所以我在R中使用Gurobi優化器。但是,我希望使用廣義順序最小優化算法,但我無法在R軟件包中找到它。所以我試圖調用這個C函數,但失敗了。從R調用這個C函數(libqp_gsmo.c)的方法是什麼?

如何從R調用這個C函數?

#include <math.h> 
#include <stdlib.h> 
#include <stdio.h> 
#include <string.h> 
#include <stdint.h> 
#include <limits.h> 

#include "libqp.h" 

libqp_state_T libqp_gsmo_solver(const double* (*get_col)(uint32_t), 
     double *diag_H, 
     double *f, 
     double *a, 
     double b, 
     double *LB, 
     double *UB, 
     double *x, 
     uint32_t n, 
     uint32_t MaxIter, 
     double TolKKT, 
     void (*print_state)(libqp_state_T state)) 
{ 
    double *col_u; 
    double *col_v; 
    double *Nabla; 
    double minF_up; 
    double maxF_low; 
    double tau; 
    double F_i; 
    double tau_ub, tau_lb; 
    double Q_P; 
    uint32_t i, j; 
    uint32_t u, v; 
    libqp_state_T state; 

    Nabla = NULL; 

    /* ------------------------------------------------------------ */ 
    /* Initialization            */ 
    /* ------------------------------------------------------------ */ 

    /* Nabla = H*x + f is gradient*/ 
    Nabla = (double*)LIBQP_CALLOC(n, sizeof(double)); 
    if(Nabla == NULL) 
    { 
     state.exitflag=-1; 
     goto cleanup; 
    } 

    /* compute gradient */ 
    for(i=0; i < n; i++) 
    { 
    Nabla[i] += f[i]; 
    if(x[i] != 0) { 
     col_u = (double*)get_col(i);  
     for(j=0; j < n; j++) { 
      Nabla[j] += col_u[j]*x[i]; 
     } 
    } 
    } 

    if(print_state != NULL) 
    { 
    state.QP = 0; 
    for(i = 0; i < n; i++) 
     state.QP += 0.5*(x[i]*Nabla[i]+x[i]*f[i]); 

    print_state(state); 
    } 


    /* ------------------------------------------------------------ */ 
    /* Main optimization loop          */ 
    /* ------------------------------------------------------------ */ 

    state.nIter = 0; 
    state.exitflag = 100; 
    while(state.exitflag == 100) 
    { 
    state.nIter ++;  

    /* find the most violating pair of variables */ 
    minF_up = LIBQP_PLUS_INF; 
    maxF_low = -LIBQP_PLUS_INF; 
    for(i = 0; i < n; i++) 
    { 

     F_i = Nabla[i]/a[i]; 

     if(LB[i] < x[i] && x[i] < UB[i]) 
     { /* i is from I_0 */ 
     if(minF_up > F_i) { minF_up = F_i; u = i; } 
     if(maxF_low < F_i) { maxF_low = F_i; v = i; } 
     } 
     else if((a[i] > 0 && x[i] == LB[i]) || (a[i] < 0 && x[i] == UB[i])) 
     { /* i is from I_1 or I_2 */ 
     if(minF_up > F_i) { minF_up = F_i; u = i; } 
     } 
     else if((a[i] > 0 && x[i] == UB[i]) || (a[i] < 0 && x[i] == LB[i])) 
     { /* i is from I_3 or I_4 */ 
     if(maxF_low < F_i) { maxF_low = F_i; v = i; } 
     } 
    } 

    /* check KKT conditions */ 
    if(maxF_low - minF_up <= TolKKT) 
     state.exitflag = 4; 
    else 
    { 
     /* SMO update of the most violating pair */ 
     col_u = (double*)get_col(u); 
     col_v = (double*)get_col(v); 

     if(a[u] > 0) 
     { tau_lb = (LB[u]-x[u])*a[u]; tau_ub = (UB[u]-x[u])*a[u]; } 
     else 
     { tau_ub = (LB[u]-x[u])*a[u]; tau_lb = (UB[u]-x[u])*a[u]; } 

     if(a[v] > 0) 
     { tau_lb = LIBQP_MAX(tau_lb,(x[v]-UB[v])*a[v]); tau_ub = 
LIBQP_MIN(tau_ub,(x[v]-LB[v])*a[v]); } 
     else 
     { tau_lb = LIBQP_MAX(tau_lb,(x[v]-LB[v])*a[v]); tau_ub = 
LIBQP_MIN(tau_ub,(x[v]-UB[v])*a[v]); } 

     tau = (Nabla[v]/a[v]-Nabla[u]/a[u])/ 
      (diag_H[u]/(a[u]*a[u]) + diag_H[v]/(a[v]*a[v]) - 
2*col_u[v]/(a[u]*a[v])); 

     tau = LIBQP_MIN(LIBQP_MAX(tau,tau_lb),tau_ub); 

     x[u] += tau/a[u]; 
     x[v] -= tau/a[v]; 

     /* update Nabla */ 
     for(i = 0; i < n; i++) 
     Nabla[i] += col_u[i]*tau/a[u] - col_v[i]*tau/a[v]; 

    } 

    if(state.nIter >= MaxIter) 
     state.exitflag = 0; 

    if(print_state != NULL) 
    { 
     state.QP = 0; 
     for(i = 0; i < n; i++) 
     state.QP += 0.5*(x[i]*Nabla[i]+x[i]*f[i]); 

     print_state(state); 
    } 

    } 

    /* compute primal objective value */ 
    state.QP = 0; 
    for(i = 0; i < n; i++) 
    state.QP += 0.5*(x[i]*Nabla[i]+x[i]*f[i]); 

cleanup: 

    LIBQP_FREE(Nabla); 

    return(state); 
} 
+0

我最喜歡的工具,這是[RCPP(http://www.rcpp.org/)。 –

回答

0

說不上來,如果你可以直接打電話從R程序的C函數,但你可以嘗試從一個.so文件加載符號。

您編譯C文件作爲一個動態庫(的.so),並用 「的dlopen」 和 「dlsym進行」 也許它會工作

人的dlopen(C):https://linux.die.net/man/3/dlopen

人的dlsym( C):https://linux.die.net/man/3/dlsym

我知道你可以調用C函數從C++二進制這樣

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