为了进行C++/Rcpp编程,我实现了一个(样本)标准偏差函数:
#include <Rcpp.h>
#include <vector>
#include <cmath>
#include <numeric>
// [[Rcpp::export]]
double cppSD(Rcpp::NumericVector rinVec)
{
std::vector<double> inVec(rinVec.begin(),rinVec.end());
int n = inVec.size();
double sum = std::accumulate(inVec.begin(), inVec.end(), 0.0);
double mean = sum / inVec.size();
for(std::vector<double>::iterator iter = inVec.begin();
iter != inVec.end(); ++iter){
double temp;
temp= (*iter - mean)*(*iter - mean);
*iter = temp;
}
double sd = std::accumulate(inVec.begin(), inVec.end(), 0.0);
return std::sqrt( sd / (n-1) );
}
考虑到可以在 vector 上调用它,我还决定从Armadillo库中测试
stddev
函数:#include <RcppArmadillo.h>
// [[Rcpp::depends(RcppArmadillo)]]
using namespace Rcpp;
// [[Rcpp::export]]
double armaSD(arma::colvec inVec)
{
return arma::stddev(inVec);
}
然后,我针对基本R函数
sd()
对这两个函数进行了基准测试,以获取一些大小不同的 vector :Rcpp::sourceCpp('G:/CPP/armaSD.cpp')
Rcpp::sourceCpp('G:/CPP/cppSD.cpp')
require(microbenchmark)
##
## sample size = 1,000: armaSD() < cppSD() < sd()
X <- rexp(1000)
microbenchmark(armaSD(X),sd(X), cppSD(X))
#Unit: microseconds
# expr min lq median uq max neval
# armaSD(X) 4.181 4.562 4.942 5.322 12.924 100
# sd(X) 17.865 19.766 20.526 21.287 86.285 100
# cppSD(X) 4.561 4.941 5.321 5.701 29.269 100
##
## sample size = 10,000: armaSD() < cppSD() < sd()
X <- rexp(10000)
microbenchmark(armaSD(X),sd(X), cppSD(X))
#Unit: microseconds
# expr min lq median uq max neval
# armaSD(X) 24.707 25.847 26.4175 29.6490 52.455 100
# sd(X) 51.315 54.356 55.8760 61.1980 100.730 100
# cppSD(X) 26.608 28.128 28.8885 31.7395 114.413 100
##
## sample size = 25,000: armaSD() < cppSD() < sd()
X <- rexp(25000)
microbenchmark(armaSD(X),sd(X), cppSD(X))
#Unit: microseconds
# expr min lq median uq max neval
# armaSD(X) 66.900 67.6600 68.040 76.403 155.845 100
# sd(X) 108.332 111.5625 122.016 125.817 169.910 100
# cppSD(X) 70.320 71.0805 74.692 80.203 102.250 100
##
## sample size = 50,000: cppSD() < sd() < armaSD()
X <- rexp(50000)
microbenchmark(armaSD(X),sd(X), cppSD(X))
#Unit: microseconds
# expr min lq median uq max neval
# armaSD(X) 249.733 267.4085 297.8175 337.729 642.388 100
# sd(X) 203.740 229.3975 240.2300 260.186 303.709 100
# cppSD(X) 162.308 185.1140 239.6600 256.575 290.405 100
##
## sample size = 75,000: sd() < cppSD() < armaSD()
X <- rexp(75000)
microbenchmark(armaSD(X),sd(X), cppSD(X))
#Unit: microseconds
# expr min lq median uq max neval
# armaSD(X) 445.110 479.8900 502.5070 520.5625 642.388 100
# sd(X) 310.931 334.8780 354.0735 379.7310 429.146 100
# cppSD(X) 346.661 380.8715 400.6370 424.0140 501.747 100
对于较小的样本,我的C++函数
cppSD()
比stats::sd()
快,但由于stats::sd()
是矢量化的,所以对于较大尺寸的 vector ,它的速度却慢得多,这让我并不感到惊讶。但是,我没想到arma::stddev()
函数会导致这种性能下降,因为它似乎也以 vector 方式运行。我使用arma::stdev()
的方式是否存在问题,或者仅仅是以可以更有效地处理较大 vector 的方式(以我假设的stats::sd()
编写)C
?任何输入将不胜感激。更新:
尽管我的问题最初是关于正确使用
arma::stddev
的问题,而不是关于尝试找到可能的最有效方法来计算样本标准差的问题,但有趣的是Rcpp::sd
sugar函数的性能如此出色。为了使事情变得更有趣,我将下面的arma::stddev
和Rcpp::sd
函数与我从JJ Allaire的Rcpp Gallery文章中的两篇改编的RcppParallel
版本进行了基准测试-here和here:library(microbenchmark)
set.seed(123)
x <- rnorm(5.5e06)
##
Res <- microbenchmark(
armaSD(x),
par_sd(x),
sd_sugar(x),
times=500L,
control=list(warmup=25))
##
R> print(Res)
Unit: milliseconds
expr min lq mean median uq max neval
armaSD(x) 24.486943 24.960966 26.994684 25.255584 25.874139 123.55804 500
par_sd(x) 8.130751 8.322682 9.136323 8.429887 8.624072 22.77712 500
sd_sugar(x) 13.713366 13.984638 14.628911 14.156142 14.401138 32.81684 500
这是在运行64位linux,i5-4200U CPU @ 1.60GHz处理器的笔记本电脑上;但是我猜想
par_sd
和sugar_sd
之间的差异在Windows机器上不会那么重要。以及
RcppParallel
版本的代码(该代码更长一些,并且需要C++ 11兼容的编译器来用于operator()
函数的重载InnerProduct
中使用的lambda表达式):#include <Rcpp.h>
#include <RcppParallel.h>
// [[Rcpp::depends(RcppParallel)]]
// [[Rcpp::plugins(cpp11)]]
/*
* based on: http://gallery.rcpp.org/articles/parallel-vector-sum/
*/
struct Sum : public RcppParallel::Worker {
const RcppParallel::RVector<double> input;
double value;
Sum(const Rcpp::NumericVector input)
: input(input), value(0) {}
Sum(const Sum& sum, RcppParallel::Split)
: input(sum.input), value(0) {}
void operator()(std::size_t begin, std::size_t end) {
value += std::accumulate(input.begin() + begin,
input.begin() + end,
0.0);
}
void join(const Sum& rhs) {
value += rhs.value;
}
};
/*
* based on: http://gallery.rcpp.org/articles/parallel-inner-product/
*/
struct InnerProduct : public RcppParallel::Worker {
const RcppParallel::RVector<double> x;
const RcppParallel::RVector<double> y;
double mean;
double product;
InnerProduct(const Rcpp::NumericVector x,
const Rcpp::NumericVector y,
const double mean)
: x(x), y(y), mean(mean), product(0) {}
InnerProduct(const InnerProduct& innerProduct,
RcppParallel::Split)
: x(innerProduct.x), y(innerProduct.y),
mean(innerProduct.mean), product(0) {}
void operator()(std::size_t begin, std::size_t end) {
product += std::inner_product(x.begin() + begin,
x.begin() + end,
y.begin() + begin,
0.0, std::plus<double>(),
[&](double lhs, double rhs)->double {
return ( (lhs-mean)*(rhs-mean) );
});
}
void join(const InnerProduct& rhs) {
product += rhs.product;
}
};
// [[Rcpp::export]]
double par_sd(const Rcpp::NumericVector& x_)
{
int N = x_.size();
Rcpp::NumericVector y_(x_);
Sum sum(x_);
RcppParallel::parallelReduce(0, x_.length(), sum);
double mean = sum.value / N;
InnerProduct innerProduct(x_, y_, mean);
RcppParallel::parallelReduce(0, x_.length(), innerProduct);
return std::sqrt( innerProduct.product / (N-1) );
}
最佳答案
您在实例化Armadillo对象方面犯了一个细微的错误-这导致复制并因此降低了性能。
改用const arma::colvec & invec
接口(interface),一切都很好:
R> sourceCpp("/tmp/sd.cpp")
R> library(microbenchmark)
R> X <- rexp(500)
R> microbenchmark(armaSD(X), armaSD2(X), sd(X), cppSD(X))
Unit: microseconds
expr min lq median uq max neval
armaSD(X) 3.745 4.0280 4.2055 4.5510 19.375 100
armaSD2(X) 3.305 3.4925 3.6400 3.9525 5.154 100
sd(X) 22.463 23.6985 25.1525 26.0055 52.457 100
cppSD(X) 3.640 3.9495 4.2030 4.8620 13.609 100
R> X <- rexp(5000)
R> microbenchmark(armaSD(X), armaSD2(X), sd(X), cppSD(X))
Unit: microseconds
expr min lq median uq max neval
armaSD(X) 18.627 18.9120 19.3245 20.2150 34.684 100
armaSD2(X) 14.583 14.9020 15.1675 15.5775 22.527 100
sd(X) 54.507 58.8315 59.8615 60.4250 84.857 100
cppSD(X) 18.585 19.0290 19.3970 20.5160 22.174 100
R> X <- rexp(50000)
R> microbenchmark(armaSD(X), armaSD2(X), sd(X), cppSD(X))
Unit: microseconds
expr min lq median uq max neval
armaSD(X) 186.307 187.180 188.575 191.825 405.775 100
armaSD2(X) 142.447 142.793 143.207 144.233 155.770 100
sd(X) 382.857 384.704 385.223 386.075 405.713 100
cppSD(X) 181.601 181.895 182.279 183.350 194.588 100
R>
这是基于我的代码版本的,其中所有内容都是一个文件,并且按照我的建议定义了
armaSD2
-从而赢得了成功。#include <RcppArmadillo.h>
// [[Rcpp::depends(RcppArmadillo)]]
#include <vector>
#include <cmath>
#include <numeric>
// [[Rcpp::export]]
double cppSD(Rcpp::NumericVector rinVec) {
std::vector<double> inVec(rinVec.begin(),rinVec.end());
int n = inVec.size();
double sum = std::accumulate(inVec.begin(), inVec.end(), 0.0);
double mean = sum / inVec.size();
for(std::vector<double>::iterator iter = inVec.begin();
iter != inVec.end();
++iter){
double temp = (*iter - mean)*(*iter - mean);
*iter = temp;
}
double sd = std::accumulate(inVec.begin(), inVec.end(), 0.0);
return std::sqrt( sd / (n-1) );
}
// [[Rcpp::export]]
double armaSD(arma::colvec inVec) {
return arma::stddev(inVec);
}
// [[Rcpp::export]]
double armaSD2(const arma::colvec & inVec) { return arma::stddev(inVec); }
/*** R
library(microbenchmark)
X <- rexp(500)
microbenchmark(armaSD(X), armaSD2(X), sd(X), cppSD(X))
X <- rexp(5000)
microbenchmark(armaSD(X), armaSD2(X), sd(X), cppSD(X))
X <- rexp(50000)
microbenchmark(armaSD(X), armaSD2(X), sd(X), cppSD(X))
*/
关于c++ - R stats::sd()与arma::stddev()与Rcpp实现的性能,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/24253228/