Question

我有一些C ++和R的经验，但是是Rcpp的新手。最近，我在以前的一些项目中使用Rcpp取得了巨大的成功，因此决定将其应用于新项目。我惊讶于我的Rcpp代码可能比相应的R函数慢得多。我试图简化我的R函数以找出原因，但找不到任何线索。非常欢迎您的帮助和评论！

用于比较R和Rcpp实现的主要R函数：

main <- function(){

  n <- 50000
  Delta <- exp(rnorm(n))
  delta <- exp(matrix(rnorm(n * 5), nrow = n))
  rx <- matrix(rnorm(n * 20), nrow = n)
  print(microbenchmark(c1 <- test(Delta, delta, rx), times = 500))
  print(microbenchmark(c2 <- rcpp_test(Delta, delta, rx), times = 500))

  identical(c1, c2)
  list(c1 = c1, c2 = c2)
}

R实现：

test <- function(Delta, delta, rx){

  const <- list()
  for(i in 1:ncol(delta)){
    const[[i]] <- rx * (Delta / (1 + delta[, i]))
  }

  const

}

Rcpp实现：

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
List rcpp_test(NumericVector Delta, 
               NumericMatrix delta, 
               NumericMatrix rx) {

  int n = Delta.length();
  int m = rx.ncol();

  List c; 
  NumericMatrix c1;
  for(int i = 0; i < delta.ncol(); ++i){
    c1 = NumericMatrix(n, m);
    for(int k = 0; k < n; ++k){
      double tmp = Delta[k] / (1 + delta(k, i));
      for(int j = 0; j < c1.ncol(); ++j){
        c1(k, j) = rx(k, j) * tmp; 
      }
    }
    c.push_back(c1);
  }

  return c;

}

我了解使用Rcpp不能保证提高效率，但是鉴于这里显示的简单示例，我不明白为什么Rcpp代码运行得如此缓慢。

Unit: milliseconds
                         expr      min       lq     mean   median       uq      max neval
 c1 <- test(Delta, delta, rx) 13.16935 14.19951 44.08641 30.43126 73.78581 115.9645   500
Unit: milliseconds
                              expr      min       lq     mean  median       uq      max neval
 c2 <- rcpp_test(Delta, delta, rx) 143.1917 158.7481 171.6116 163.413 173.7677 247.5495   500

理想情况下，rx是我的项目中的矩阵列表。 for循环中的变量i将用于选择要计算的元素。在一开始，我怀疑将List传递给Rcpp可能会有很高的开销，因此在此示例中，我假设rx是用于所有i的固定矩阵。看来这不是缓慢的原因。

Answer 1

您的R代码似乎或多或少是最优的，即所有实际工作都是在编译后的代码中完成的。对于C ++代码，我发现的主要问题是在紧密循环中调用c1.ncol()。如果将其替换为m，则C ++解决方案的速度几乎与R相同。如果将RcppArmadillo添加到混合中，则语法将非常紧凑，但不会比纯Rcpp代码快。对我来说，这表明很难击败编写良好的R代码：

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

// [[Rcpp::export]]
List arma_test(const arma::vec& Delta,
           const arma::mat& delta,
           const arma::mat& rx) {
  int l = delta.n_cols;
  List c(l);

  for (int i = 0; i < l; ++i) {
    c(i) = rx.each_col() % (Delta / (1 + delta.col(i)));
  }

  return c;  
}

// [[Rcpp::export]]
List rcpp_test(NumericVector Delta, 
               NumericMatrix delta, 
               NumericMatrix rx) {

  int n = Delta.length();
  int m = rx.ncol();

  List c(delta.ncol()); 
  NumericMatrix c1;
  for(int i = 0; i < delta.ncol(); ++i){
    c1 = NumericMatrix(n, m);
    for(int k = 0; k < n; ++k){
      double tmp = Delta[k] / (1 + delta(k, i));
      for(int j = 0; j < m; ++j){
        c1(k, j) = rx(k, j) * tmp; 
      }
    }
    c(i) = c1;
  }

  return c;

}

/*** R
test <- function(Delta, delta, rx){

  const <- list()
  for(i in 1:ncol(delta)){
    const[[i]] <- rx * (Delta / (1 + delta[, i]))
  }

  const

}

n <- 50000
Delta <- exp(rnorm(n))
delta <- exp(matrix(rnorm(n * 5), nrow = n))
rx <- matrix(rnorm(n * 20), nrow = n)
bench::mark(test(Delta, delta, rx),
            arma_test(Delta, delta, rx),
            rcpp_test(Delta, delta, rx))
 */

输出：

# A tibble: 3 x 14
  expression     min    mean  median     max `itr/sec` mem_alloc  n_gc n_itr
  <chr>      <bch:t> <bch:t> <bch:t> <bch:t>     <dbl> <bch:byt> <dbl> <int>
1 test(Delt…  84.3ms  85.2ms  84.9ms  86.6ms     11.7     44.9MB     2     4
2 arma_test… 106.5ms 107.7ms 107.7ms 108.9ms      9.28    38.1MB     3     2
3 rcpp_test… 101.9ms 103.2ms 102.2ms 106.6ms      9.69    38.1MB     1     4
# … with 5 more variables: total_time <bch:tm>, result <list>, memory <list>,
#   time <list>, gc <list>

我还明确地将输出列表初始化为所需的大小，以避免使用push_back，但这并没有太大的区别。对于Rcpp中的矢量一样的数据结构，您绝对应该避免使用push_back，因为每次扩展矢量时都会创建一个副本。

Answer 2

I would like to add to @RalfStubner's excellent answer.

You will notice that we are making many allocations in the first for loop (i.e. c1 = NumerMatrix(n, m)). This can be expensive as we are initializing every element to 0 in addition to allocating memory. We can change this to the following for increased efficiency:

NumericMatrix c1 = no_init_matrix(n, m)

I also went ahead and added the keyword const where possible. It's debatable if doing this allows the compiler to optimize certain pieces of code, but I still add it where I can for code clarity (i.e. "I don't want this variable to change"). With that, we have:

// [[Rcpp::export]]
List rcpp_test_modified(const NumericVector Delta, 
                        const NumericMatrix delta, 
                        const NumericMatrix rx) {

    int n = Delta.length();
    int m = rx.ncol();
    int dCol = delta.ncol();

    List c(dCol);

    for(int i = 0; i < dCol; ++i) {
        NumericMatrix c1 = no_init_matrix(n, m);

        for(int k = 0; k < n; ++k) {
            const double tmp = Delta[k] / (1 + delta(k, i));

            for(int j = 0; j < m; ++j) {
                c1(k, j) = rx(k, j) * tmp; 
            }
        }

        c[i] = c1;
    }

    return c;

}

And here are some benchmarks (Armadillo solution left out):

bench::mark(test(Delta, delta, rx),
            rcpp_test_modified(Delta, delta, rx),
            rcpp_test(Delta, delta, rx))
# A tibble: 3 x 14
  expression     min   mean  median    max `itr/sec` mem_alloc  n_gc n_itr total_time result memory time 
  <chr>      <bch:t> <bch:> <bch:t> <bch:>     <dbl> <bch:byt> <dbl> <int>   <bch:tm> <list> <list> <lis>
1 test(Delt… 12.27ms 17.2ms 14.56ms 29.5ms      58.1    41.1MB    13     8      138ms <list… <Rpro… <bch…
2 rcpp_test…  7.55ms 11.4ms  8.46ms   26ms      87.8    38.1MB    16    21      239ms <list… <Rpro… <bch…
3 rcpp_test… 10.36ms 15.8ms 13.64ms 28.9ms      63.4    38.1MB    10    17      268ms <list… <Rpro… <bch…
# … with 1 more variable: gc <list>

And we see a 50% improvement with the Rcpp version.

为什么在我的示例中Rcpp实现比R函数要慢得多？

2 个答案: