Question

在给定具有按行结果概率的矩阵的情况下，我需要对结果变量进行采样。

set.seed(1010) #reproducibility

#create a matrix of probabilities
#three possible outcomes, 10.000 cases
probabilities <- matrix(runif(10000*3),nrow=10000,ncol=3)
probabilities <- probabilities / Matrix::rowSums(probabilities)

我想出的最快方法是apply（）和sample（）的组合。

#row-wise sampling using these probabilities
classification <- apply(probabilities, 1, function(x) sample(1:3, 1, prob = x))

但是，在我正在执行的操作中，这是计算瓶颈。您是否知道如何加快此代码的速度/如何更有效地采样？

谢谢！

Answer 1

RLave关于Rcpp可能成为前进之路的评论（您还需要RcppArmadillo用于sample()）；我使用以下C ++代码创建了这样的功能：

// [[Rcpp::depends(RcppArmadillo)]]
#include <RcppArmadilloExtensions/sample.h>

using namespace Rcpp;

// [[Rcpp::export]]
IntegerVector sample_matrix(NumericMatrix x, IntegerVector choice_set) {
    int n = x.nrow();
    IntegerVector result(n);
    for ( int i = 0; i < n; ++i ) {
        result[i] = RcppArmadillo::sample(choice_set, 1, false, x(i, _))[0];
    }
    return result;
}

然后我通过以下方式在我的R会话中提供该功能

Rcpp::sourceCpp("sample_matrix.cpp")

现在，我们可以针对您的初始方法在R中对其进行测试，以及可以使用purrr::map()和lapply()的其他建议：

set.seed(1010) #reproducibility

#create a matrix of probabilities
#three possible outcomes, 10.000 cases
probabilities <- matrix(runif(10000*3),nrow=10000,ncol=3)
probabilities <- probabilities / Matrix::rowSums(probabilities)
probabilities_list <- split(probabilities, seq(nrow(probabilities)))

library(purrr)
library(microbenchmark)

microbenchmark(
    apply = apply(probabilities, 1, function(x) sample(1:3, 1, prob = x)),
    map = map(probabilities_list, function(x) sample(1:3, 1, prob = x)),
    lapply = lapply(probabilities_list, function(x) sample(1:3, 1, prob = x)),
    rcpp = sample_matrix(probabilities, 1:3),
    times = 100
)

Unit: milliseconds
   expr       min        lq      mean    median        uq       max neval
  apply 307.44702 321.30051 339.85403 342.36421 350.86090 434.56007   100
    map 254.69721 265.10187 282.85592 286.21680 295.48886 363.95898   100
 lapply 249.68224 259.70178 280.63066 279.87273 287.10062 691.21359   100
   rcpp  12.16787  12.55429  13.47837  13.81601  14.25198  16.84859   100
 cld
   c
  b 
  b 
 a

节省大量时间。

Answer 2

如果您愿意将probabilities放在list中，则purrr::map或lapply似乎要快一些：

probabilities <- matrix(runif(10000*3),nrow=10000,ncol=3)
probabilities <- probabilities / Matrix::rowSums(probabilities)
probabilities_list <- split(probabilities, seq(nrow(probabilities)))

library(purrr)
set.seed(1010)
classification_list <- map(probabilities_list, function(x) sample(1:3, 1, prob = x))

set.seed(1010)
classification_list <- lapply(probabilities_list, function(x) sample(1:3, 1, prob = x))

基准化：

microbenchmark::microbenchmark(
  apply = {classification = apply(probabilities, 1, function(x) sample(1:3, 1, prob = x))},
  map = {classification = map(probabilities_list, function(x) sample(1:3, 1, prob = x))},
  lapply = {classification = lapply(probabilities_list, function(x) sample(1:3, 1, prob = x))},
  times = 100
)
# Unit: milliseconds
#  expr      min       lq     mean   median       uq      max neval
# apply 39.92883 42.59249 48.39247 45.03080 47.86648 94.39828   100
#   map 35.54077 37.13866 42.19719 39.95046 41.56323 66.05167   100
#lapply 34.54861 36.48664 42.69512 39.20139 52.31494 59.29200   100

有100.000个案例

# Unit: milliseconds
# expr      min       lq     mean   median       uq      max neval
# apply 457.5310 520.4926 572.5974 552.1674 611.5640 957.3997   100
#   map 391.4751 457.7326 488.3286 482.1459 512.2054 899.1380   100
#lapply 386.2698 443.6732 491.9957 475.4160 507.3677 868.6725   100

Answer 3

您可以考虑

vapply和
并行化：parallel::parApply

使用您的probabilities矩阵：

set.seed(1010) #reproducibility

#create a matrix of probabilities
#three possible outcomes, 10.000 cases
probabilities <- matrix(runif(10000*3), nrow=10000,ncol=3)
probabilities <- probabilities / Matrix::rowSums(probabilities)
classification <- apply(probabilities, 1, function(x) sample(1:3, 1, prob = x))

vapply

通过指定FUN.VALUE的类，您也许可以使其变得快速。

classification2 <- vapply(split(probabilities, 1:nrow(probabilities)),
                          function(x) sample(1:3, 1, prob = x),
                          FUN.VALUE = integer(1), USE.NAMES = FALSE)
head(classification2)
#> [1] 1 3 3 1 2 3

平行包装

benchmarkme::get_cpu()
#> $vendor_id
#> [1] "GenuineIntel"
#> 
#> $model_name
#> [1] "Intel(R) Core(TM) i5-4288U CPU @ 2.60GHz"
#> 
#> $no_of_cores
#> [1] 4

在上述环境中，

cl <- parallel::makeCluster(4)
doParallel::registerDoParallel(cl, cores = 4)

parApply()可以完成apply()的工作。

classification3 <- parallel::parApply(cl, probabilities, 1, function(x) sample(1:3, 1, prob = x))
head(classification3)
#> [1] 2 2 2 2 3 3

比较包括apply()在内的三个解决方案，

microbenchmark::microbenchmark(
  question = { # yours
    apply(probabilities, 1, function(x) sample(1:3, 1, prob = x))
  },
  vapp = {
    vapply(split(probabilities, 1:nrow(probabilities)), function(x) sample(1:3, 1, prob = x), FUN.VALUE = integer(1), USE.NAMES = FALSE)
  },
  parr = {
    parallel::parApply(cl, probabilities, 1, function(x) sample(1:3, 1, prob = x))
  }
)
#> Unit: milliseconds
#>      expr      min       lq     mean   median       uq       max neval
#>  question 49.93853 58.39965 65.05360 62.98119 68.28044 182.03267   100
#>      vapp 44.19828 54.84294 59.47109 58.56739 62.05269 146.14792   100
#>      parr 43.33227 48.16840 53.26599 50.87995 54.17286  98.67692   100

parallel::stopCluster(cl)

在R中有效地应用sample（）

3 个答案:

vapply

平行包装