使用Rcpp和openMP从截断正态分布快速采样

Inf*_*tor 7 r openmp rcpp

更新:

我试图实施德克的建议.评论?我现在正忙于JSM,但我想在为画廊编织Rmd之前得到一些反馈.我从犰狳换回正常的Rcpp,因为它没有增加任何价值.带有R ::的标量版本非常好.如果将mean/sd作为标量输入,而不是作为所需输出长度的向量,我应该在参数n中输入绘制数量.


有许多MCMC应用程序需要从截断的Normal分布中绘制样本.我建立在TN的现有实现上并添加了并行计算.

问题:

  1. 有没有人看到进一步的速度提升?在基准测试的最后一种情况下,rtruncnorm有时会更快.Rcpp实现总是比现有的包更快,但是可以进一步改进吗?
  2. 我在一个我无法分享的复杂模型中运行它,我的R会话崩溃了.但是,我不能系统地重现它,所以它可能是代码的另一部分.如果有人在TN工作,请测试并告诉我.更新:我没有更新代码的问题,但请告诉我.

我如何把事情放在一起:据我所知,最快的实现不在CRAN上,但源代码可以下载OSU stat.在我的基准测试中,msmtrunco​​rm中的竞争实现较慢.诀窍是有效地调整提案分布,其中指数很好地适用于截断的Normal的尾部.所以我拿了Chris的代码,"Rcpp'ed"它并添加了一些openMP香料.动态调度在这里是最佳的,因为取样可以根据边界花费更多或更少的时间.我发现一件令人讨厌的事情:当我想使用双打时,许多统计分布基于NumericVector类型.我只是编写了我的方式.

继承人Rcpp代码:

#include <Rcpp.h>
#include <omp.h>


// norm_rs(a, b)
// generates a sample from a N(0,1) RV restricted to be in the interval
// (a,b) via rejection sampling.
// ======================================================================

// [[Rcpp::export]]

double norm_rs(double a, double b)
{
   double  x;
   x = Rf_rnorm(0.0, 1.0);
   while( (x < a) || (x > b) ) x = norm_rand();
   return x;
}

// half_norm_rs(a, b)
// generates a sample from a N(0,1) RV restricted to the interval
// (a,b) (with a > 0) using half normal rejection sampling.
// ======================================================================

// [[Rcpp::export]]

double half_norm_rs(double a, double b)
{
   double   x;
   x = fabs(norm_rand());
   while( (x<a) || (x>b) ) x = fabs(norm_rand());
   return x;
}

// unif_rs(a, b)
// generates a sample from a N(0,1) RV restricted to the interval
// (a,b) using uniform rejection sampling. 
// ======================================================================

// [[Rcpp::export]]

double unif_rs(double a, double b)
{
   double xstar, logphixstar, x, logu;

   // Find the argmax (b is always >= 0)
   // This works because we want to sample from N(0,1)
   if(a <= 0.0) xstar = 0.0;
   else xstar = a;
   logphixstar = R::dnorm(xstar, 0.0, 1.0, 1.0);

   x = R::runif(a, b);
   logu = log(R::runif(0.0, 1.0));
   while( logu > (R::dnorm(x, 0.0, 1.0,1.0) - logphixstar))
   {
      x = R::runif(a, b);
      logu = log(R::runif(0.0, 1.0));
   }
   return x;
}

// exp_rs(a, b)
// generates a sample from a N(0,1) RV restricted to the interval
// (a,b) using exponential rejection sampling.
// ======================================================================

// [[Rcpp::export]]

double exp_rs(double a, double b)
{
  double  z, u, rate;

//  Rprintf("in exp_rs");
  rate = 1/a;
//1/a

   // Generate a proposal on (0, b-a)
   z = R::rexp(rate);
   while(z > (b-a)) z = R::rexp(rate);
   u = R::runif(0.0, 1.0);

   while( log(u) > (-0.5*z*z))
   {
      z = R::rexp(rate);
      while(z > (b-a)) z = R::rexp(rate);
      u = R::runif(0.0,1.0);
   }
   return(z+a);
}




// rnorm_trunc( mu, sigma, lower, upper)
//
// generates one random normal RVs with mean 'mu' and standard
// deviation 'sigma', truncated to the interval (lower,upper), where
// lower can be -Inf and upper can be Inf.
//======================================================================

// [[Rcpp::export]]
double rnorm_trunc (double mu, double sigma, double lower, double upper)
{
int change;
 double a, b;
 double logt1 = log(0.150), logt2 = log(2.18), t3 = 0.725;
 double z, tmp, lograt;

 change = 0;
 a = (lower - mu)/sigma;
 b = (upper - mu)/sigma;

 // First scenario
 if( (a == R_NegInf) || (b == R_PosInf))
   {
     if(a == R_NegInf)
       {
     change = 1;
     a = -b;
     b = R_PosInf;
       }

     // The two possibilities for this scenario
     if(a <= 0.45) z = norm_rs(a, b);
     else z = exp_rs(a, b);
     if(change) z = -z;
   }
 // Second scenario
 else if((a * b) <= 0.0)
   {
     // The two possibilities for this scenario
     if((R::dnorm(a, 0.0, 1.0,1.0) <= logt1) || (R::dnorm(b, 0.0, 1.0, 1.0) <= logt1))
       {
     z = norm_rs(a, b);
       }
     else z = unif_rs(a,b);
   }
 // Third scenario
 else
   {
     if(b < 0)
       {
     tmp = b; b = -a; a = -tmp; change = 1;
       }

     lograt = R::dnorm(a, 0.0, 1.0, 1.0) - R::dnorm(b, 0.0, 1.0, 1.0);
     if(lograt <= logt2) z = unif_rs(a,b);
     else if((lograt > logt1) && (a < t3)) z = half_norm_rs(a,b);
     else z = exp_rs(a,b);
     if(change) z = -z;
   }
   double output;
   output = sigma*z + mu;
 return (output);
}


// rtnm( mu, sigma, lower, upper, cores)
//
// generates one random normal RVs with mean 'mu' and standard
// deviation 'sigma', truncated to the interval (lower,upper), where
// lower can be -Inf and upper can be Inf.
// mu, sigma, lower, upper are vectors, and vectorized calls of this function
// speed up computation
// cores is an intege, representing the number of cores to be used in parallel
//======================================================================


// [[Rcpp::export]]

Rcpp::NumericVector rtnm(Rcpp::NumericVector mus, Rcpp::NumericVector sigmas, Rcpp::NumericVector lower, Rcpp::NumericVector upper, int cores){
  omp_set_num_threads(cores);
  int nobs = mus.size();
  Rcpp::NumericVector out(nobs);
  double logt1 = log(0.150), logt2 = log(2.18), t3 = 0.725;
    double a,b, z, tmp, lograt;

     int  change;

  #pragma omp parallel for schedule(dynamic)   
  for(int i=0;i<nobs;i++) {  

     a = (lower(i) - mus(i))/sigmas(i);
     b = (upper(i) - mus(i))/sigmas(i);
     change=0;
     // First scenario
     if( (a == R_NegInf) || (b == R_PosInf))
       {
         if(a == R_NegInf)
           {
              change = 1;
              a = -b;
              b = R_PosInf;
           }

         // The two possibilities for this scenario
         if(a <= 0.45) z = norm_rs(a, b);
         else z = exp_rs(a, b);
         if(change) z = -z;
       }
     // Second scenario
     else if((a * b) <= 0.0)
       {
         // The two possibilities for this scenario
         if((R::dnorm(a, 0.0, 1.0,1.0) <= logt1) || (R::dnorm(b, 0.0, 1.0, 1.0) <= logt1))
           {
                z = norm_rs(a, b);
           }
         else z = unif_rs(a,b);
       }

     // Third scenario
     else
       {
         if(b < 0)
           {
                tmp = b; b = -a; a = -tmp; change = 1;
           }

         lograt = R::dnorm(a, 0.0, 1.0, 1.0) - R::dnorm(b, 0.0, 1.0, 1.0);
         if(lograt <= logt2) z = unif_rs(a,b);
         else if((lograt > logt1) && (a < t3)) z = half_norm_rs(a,b);
         else z = exp_rs(a,b);
         if(change) z = -z;
       }
    out(i)=sigmas(i)*z + mus(i);          
  }

return(out);
}
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以下是基准:

libs=c("truncnorm","msm","inline","Rcpp","RcppArmadillo","rbenchmark")
if( sum(!(libs %in% .packages(all.available = TRUE)))>0){ install.packages(libs[!(libs %in% .packages(all.available = TRUE))])}
for(i in 1:length(libs)) {library(libs[i],character.only = TRUE,quietly=TRUE)}


#needed for openMP parallel
Sys.setenv("PKG_CXXFLAGS"="-fopenmp")
Sys.setenv("PKG_LIBS"="-fopenmp")

#no of cores for openMP version
cores = 4

#surce code from same dir
Rcpp::sourceCpp('truncnorm.cpp')


#sample size
nn=1000000


bb= 100
aa=-100
benchmark( rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),cores), rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),1),rtnorm(nn,rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn)),rtruncnorm(nn, a=aa, b=100, mean = 0, sd = 1) , order="relative", replications=3    )[,1:4]

aa=0 
benchmark( rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),cores), rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),1),rtnorm(nn,rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn)),rtruncnorm(nn, a=aa, b=100, mean = 0, sd = 1) , order="relative", replications=3    )[,1:4]

aa=2
benchmark( rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),cores), rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),1),rtnorm(nn,rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn)),rtruncnorm(nn, a=aa, b=100, mean = 0, sd = 1) , order="relative", replications=3    )[,1:4]

aa=50
benchmark( rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),cores), rtnm(rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn),1),rtnorm(nn,rep(0,nn),rep(1,nn),rep(aa,nn),rep(100,nn)),rtruncnorm(nn, a=aa, b=100, mean = 0, sd = 1) , order="relative", replications=3    )[,1:4]
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由于速度取决于上/下边界,因此需要进行几次基准测试.对于不同的情况,算法的不同部分启动.

Dir*_*tel 3

非常快速的评论:

  1. 如果您包含,RcppArmadillo.h则不需要包含Rcpp.h- 事实上,您不应该包含,我们甚至测试了这一点

  2. rep(oneDraw, n)拨打 n 次电话。我会编写一个函数,调用一次即可返回 n 次绘制——它会更快,因为您可以节省 n-1 次函数调用开销

  3. 您对许多统计分布的评论都是基于类型的NumericVector,当我想使用双打时,可能会揭示一些误解:NumericVector是我们内部 R 类型的方便代理类:没有副本。您可以自由使用std::vector<double>或选择您喜欢的任何形式。

  4. 我对截断法线知之甚少,所以我无法评论你的算法的细节。

  5. 完成后,请考虑在Rcpp Gallery上发表帖子。