Question

我希望在split-apply-combine中实现一个简单的Rcpp例程，其中数据集（矩阵）被分成组，然后返回groupwise列的总和。这是一个在R中很容易实现的过程，但通常需要很长时间。我设法实现了Rcpp解决方案，该解决方案胜过R的性能，但我想知道我是否可以进一步改进它。为了说明这里的一些代码，首先使用R：

n <- 50000
k <- 50
set.seed(42)
X <- matrix(rnorm(n*k), nrow=n)
g=rep(1:8,length.out=n )

use.for <- function(mat, ind){
  sums <- matrix(NA, nrow=length(unique(ind)), ncol=ncol(mat))
  for(i in seq_along(unique(ind))){
    sums[i,] <- colSums(mat[ind==i,])
  }
  return(sums)
}

use.apply <- function(mat, ind){
  apply(mat,2, function(x) tapply(x, ind, sum))
}

use.dt <- function(mat, ind){ # based on Roland's answer
   dt <- as.data.table(mat)
   dt[, cvar := ind]
   dt2 <- dt[,lapply(.SD, sum), by=cvar]
   as.matrix(dt2[,cvar:=NULL])
}

事实证明，for - 循环实际上非常快，并且对Rcpp实施最简单（对我而言）。它的工作原理是为每个组创建一个子矩阵，然后在矩阵上调用colSums。这是使用RcppArmadillo：

实现的

#include <RcppArmadillo.h>
// [[Rcpp::depends(RcppArmadillo)]]
using namespace Rcpp;
using namespace arma;

// [[Rcpp::export]]
arma::mat use_arma(arma::mat X, arma::colvec G){

  arma::colvec gr = arma::unique(G);
  int gr_n = gr.n_rows;
  int ncol = X.n_cols;

  arma::mat out = zeros(gr_n, ncol); 

  for(int g=0; g<gr_n; g++){
   int g_id = gr(g);
   arma::uvec subvec = find(G==g_id);
   arma::mat submat = X.rows(subvec);
   arma::rowvec res = sum(submat,0);
   out.row(g) = res;     
  }
 return out;
}

但是，基于answers to this question，我了解到在C++中创建副本的成本很高（就像在R中一样），但是这些循环并不像R那样糟糕。 1}}。由于arma - 解决方案依赖于为每个组创建矩阵（代码中为submat），我的猜测是避免这种情况会进一步加快进程。因此，这里仅基于Rcpp的第二种实现仅使用循环：

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
NumericMatrix use_Rcpp(NumericMatrix X, IntegerVector G){

  IntegerVector gr = unique(G);
  std::sort(gr.begin(), gr.end());
  int gr_n = gr.size();
  int nrow = X.nrow(), ncol = X.ncol();

  NumericMatrix out(gr_n, ncol);

  for(int g=0; g<gr_n; g++){
     int g_id = gr(g);

      for (int j = 0; j < ncol; j++) {
      double total = 0;
        for (int i = 0; i < nrow; i++) {

          if (G(i) != g_id) continue;    // not sure how else to do this
          total += X(i, j);
        }
        out(g,j) = total;
      }
  }
      return out;
}

对这些解决方案进行基准测试，包括@Roland提供的use_dt版本（我之前的版本不公平地歧视data.table），以及@beginneR建议的dplyr解决方案，收益率以下内容：

 library(rbenchmark)
 benchmark(use.for(X,g), use.apply(X,g), use.dt(X,g), use.dplyr(X,g), use_arma(X,g), use_Rcpp(X,g), 
+           columns = c("test", "replications", "elapsed", "relative"), order = "relative", replications = 1000)
             test replications elapsed relative
# 5  use_arma(X, g)         1000   29.65    1.000
# 4 use.dplyr(X, g)         1000   42.05    1.418
# 3    use.dt(X, g)         1000   56.94    1.920
# 1   use.for(X, g)         1000   60.97    2.056
# 6  use_Rcpp(X, g)         1000  113.96    3.844
# 2 use.apply(X, g)         1000  301.14   10.156

我的直觉（use_Rcpp优于use_arma）并没有成功。话虽如此，我想我if (G(i) != g_id) continue;函数中的行use_Rcpp会减慢一切。我很高兴了解有关替代方案的建议。

我很高兴我在R完成这项工作所花费的时间的一半时间内完成了相同的任务，但也许有几个Rcpp is much faster than R - 例子与我的期望相混淆，我想知道是否我可以加快速度。有没有人有想法？我也欢迎任何编程/编码评论，因为我对Rcpp和C++相对较新。

Answer 1

不，它不是你需要击败的for循环：

library(data.table)
#it doesn't seem fair to include calls to library in benchmarks
#you need to do that only once in your session after all

use.dt2 <- function(mat, ind){
  dt <- as.data.table(mat)
  dt[, cvar := ind]
  dt2 <- dt[,lapply(.SD, sum), by=cvar]
  as.matrix(dt2[,cvar:=NULL])
}

all.equal(use.dt(X,g), use.dt2(X,g))
#TRUE

benchmark(use.for(X,g), use.apply(X,g), use.dt(X,g), use.dt2(X,g),
          columns = c("test", "replications", "elapsed", "relative"), 
          order = "relative", replications = 50)

#             test replications elapsed relative
#4   use.dt2(X, g)           50    3.12    1.000
#1   use.for(X, g)           50    4.67    1.497
#3    use.dt(X, g)           50    7.53    2.413
#2 use.apply(X, g)           50   17.46    5.596

Answer 2

这是我对您的Rcpp解决方案的内联评论的批评。

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
NumericMatrix use_Rcpp(NumericMatrix X, IntegerVector G){

  // Rcpp has a sort_unique() function, which combines the
  // sort and unique steps into one, and is often faster than
  // performing the operations separately. Try `sort_unique(G)`
  IntegerVector gr = unique(G);
  std::sort(gr.begin(), gr.end());
  int gr_n = gr.size();
  int nrow = X.nrow(), ncol = X.ncol();

  // This constructor zero-initializes memory (kind of like
  // making a copy). You should use:
  // 
  //     NumericMatrix out = no_init(gr_n, ncol)
  //
  // to ensure the memory is allocated, but not zeroed.
  // 
  // EDIT: We don't have no_init for matrices right now, but you can hack
  // around that with:
  // 
  //     NumericMatrix out(Rf_allocMatrix(REALSXP, gr_n, ncol));
  NumericMatrix out(gr_n, ncol);

  for(int g=0; g<gr_n; g++){

     // subsetting with operator[] is cheaper, so use gr[g] when
     // you can be sure bounds checks are not necessary
     int g_id = gr(g);

      for (int j = 0; j < ncol; j++) {
      double total = 0;
        for (int i = 0; i < nrow; i++) {

          // similarily here
          if (G(i) != g_id) continue;    // not sure how else to do this
          total += X(i, j);
        }
        // IIUC, you are filling the matrice row-wise. This is slower as
        // R matrices are stored in column-major format, and so filling
        // matrices column-wise will be faster.
        out(g,j) = total;
      }
  }
      return out;
}

Answer 3

也许你正在寻找（奇怪的名字）rowsum

library(microbenchmark)
use.rowsum = rowsum

和

> all.equal(use.for(X, g), use.rowsum(X, g), check.attributes=FALSE)
[1] TRUE
> microbenchmark(use.for(X, g), use.rowsum(X, g), times=5)
Unit: milliseconds
             expr       min        lq    median        uq       max neval
    use.for(X, g) 126.92876 127.19027 127.51403 127.64082 128.06579     5
 use.rowsum(X, g)  10.56727  10.93942  11.01106  11.38697  11.38918     5

如何加快这个Rcpp功能？

3 个答案: