我正在模拟~30个相关的随机变量，让我们说500K次。然后我将所有随机变量求和以得到我想要的结果。我想找出这笔钱的99.5％百分位数，这很容易完成。另外，我想让所有相邻场景在随机变量级别上达到99.5％的实现。也就是说，对于99.499％-percentile，我想知道每个随机变量的不同实现是什么。下面是执行此操作的代码的程式化示例。然而，事实证明，＆＃34;顺序＆＃34;功能需要相当长的时间。有关如何最小化下面的代码运行时的任何建议？在这个特定的例子中，它似乎不是一个问题，但我需要做上述约500次。
require("MASS")
#Parameters to run
n=30
N_simulations=500000
Percentile=0.995
window_size=101
#Preparing multivariate normal parameters
cov_matrix<-matrix(0.5,n,n)
diag(cov_matrix)<-1
mu<-rep(0,30)
#running the simulation
simulation_matrix<-t(mvrnorm(n=N_simulations, mu=mu, Sigma= cov_matrix)) #RVs as rows and simulations as columns
losses<-colSums(simulation_matrix)
VaR<-quantile(losses,probs=0.995)
#Method 1 - using the order function to get adjacent scenarios
unsorted_losses<-cbind(seq(1:N_simulations),losses)
sorted_losses<-unsorted_losses[order(unsorted_losses[,2],decreasing=FALSE),]
sorted_scenario<-N_simulations*Percentile
window<-seq(-(window_size-1)/2,(window_size-1)/2,1)
window<-sorted_scenario+window
scenarios<-as.vector(sorted_losses[window,1]) # gives all the adjacent scenarios
#Method 2 - using partial sort function (does not work properly when duplicate values)
names(unsorted_losses)<-seq(1:N_simulations)
z=2*(1-Percentile)*N_simulations
a<-sort(-unsorted_losses,partial=1:z)
sorted_losses<- - a[1:z] #highest loss to smallest loss
u<-names(unsorted_losses[match(sorted_losses,unsorted_losses)]) #getting the rownames in unsorted losses
names(sorted_losses)<-u
sorted_scenario<-N_simulations-N_simulations*Percentile
window<-seq(-(window_size-1)/2,(window_size-1)/2,1)
window<-sorted_scenario+window #gives the r
scenarios<-as.numeric(names(sorted_losses[window])) # gives all the adjacent scenarios from the rownames
# code proposed by Stackoverflow users
window_size<-1001
percentile_step<-1/N_simulations
window<-seq(-(window_size-1)/2,(window_size-1)/2,1)
percentiles<-window*percentile_step*Percentile+Percentile
window_losses<-quantile(losses,probs=percentiles,type=3)
a<-losses[losses %in% window_losses] # returns only 997 TRUE, not 1001!!