Question

我在尝试编写一组模拟时遇到一些问题，其中我的变量的起始值是要在不同时间而不是单个观察值进行评估的向量。可以通过改变起点逐个运行模拟，然后将所有结果重新组合成另一个矩阵;但是，我处理大约2000个案例，因此需要更优雅，更快捷的方式。

这不是我使用的真实代码，而是同一问题的一个例子，让我们说：

time<-c(65,130,195,260)  #in days

simulation<-matrix(numeric(10*4),10,4)  

  #vessel matrix containing the desired number of simulations for 1 initial case

results<-NULL

initial<-.5   

   #This would be one case and I've got a vector with approx.2000

   #loop

for(i in 1:10){
  simulation[i,] = (initial*exp(-a*(time/260)) + 
                   b*(1-exp(-a*(time/260)))    +
                   c*(j/260))*rnorm(1)
}

results<-colMeans(simulation)


  # With this I end up with a row vector of 4 entries containing the average of 
  # the simulations for the first case at four dates.

如何用向量替换initial，让我们说initial=seq(from=0, to=10, by=.5)，其中我最终得到一个20X4的矩阵，其中每行仍然包含每个行的平均10个模拟日期？

Answer 1

因此，您可以使用replicate代替for循环，将其包装在函数中，然后您就完成了。

time<-c(65,130,195,260) 
initial<-.5   

# I assume a,b,c,j are constants
a<-b<-c<-j<-1

# Use replicate instead of a four loop to generate your matrix directly:
mat<-replicate(10,(initial*exp(-a*(time/260)) + b*(1-exp(-a*(time/260))) + c*(j/260))*rnorm(1))
# Get the means for all four variables (note matrix is transposed)
rowMeans(mat)

# Wrap this in a function:
f<-function(initial) {
  mat<-replicate(10,(initial*exp(-a*(time/260)) + b*(1-exp(-a*(time/260))) + c*(j/260))*rnorm(1))
  rowMeans(mat)
}

# Now repeat for a variety of "initial"
sapply(seq(from=0, to=10, by=.5),f)

# [,1]       [,2]      [,3]      [,4]       [,5]      [,6]        [,7]       [,8]
# [1,] 0.04131179 -0.1703167 0.2165282 0.4518704 -0.4305232 1.0390553 -0.09262142 -0.3767832
# [2,] 0.07293559 -0.1941923 0.2165282 0.4239343 -0.3889186 0.9154405 -0.08016291 -0.3217918
# [3,] 0.09756422 -0.2127866 0.2165282 0.4021776 -0.3565170 0.8191691 -0.07046022 -0.2789645
# [4,] 0.11674502 -0.2272679 0.2165282 0.3852335 -0.3312825 0.7441930 -0.06290376 -0.2456105

Answer 2

time<-c(65,130,195,260)  #in days

simulation<-matrix(numeric(10*4),10,4)  
#vessel matrix containing the desired number of simulations for 1 initial case

results<-NULL

#initial<-.5   

initial=seq(from=0.5, to=10, by=.5) # Changed This to be vector


#This would be one case and I've got a vector with approx.2000

#loop
## Writing to a function to help in reusing the code
MM <- function(initial){

for(i in 1:10){
  simulation[i,] = (initial*exp(-10*(time/260)) + 
                      20*(1-exp(-10*(time/260)))    +
                      20*(1000/260))*rnorm(1)
}

results<-colMeans(simulation)
return(results)
}

sapply(initial, FUN=MM) # This prints the colmeans for each inputted element in the vector initial

#Output:
#                 [,1]            [,2]            [,3]           [,4]
#[1,] 0.04560954756720 -7.358191844904 -8.098682864513 3.665639204753
#[2,] 0.04631255769454 -7.468652842648 -8.217009769192 3.717726620815
#[3,] 0.04637026427987 -7.477720033496 -8.226722633000 3.722002216291
#[4,] 0.04637500112485 -7.478464313845 -8.227519913413 3.722353178540
#               [,5]           [,6]           [,7]           [,8]
#[1,] 18.98057953410 21.03382334801 11.96959043580 63.54595548535
#[2,] 19.24268217595 21.31585924907 12.12529951449 64.34721302134
#[3,] 19.26419687094 21.33901016563 12.13808089400 64.41298424507
#[4,] 19.26596290465 21.34091050858 12.13913005352 64.41838307588
#                [,9]          [,10]          [,11]           [,12]
#[1,] -20.01917307564 12.56144497430 27.31607000721 -17.70877647896
#[2,] -20.26360373127 12.70980687528 27.62780928081 -17.90382189201
#[3,] -20.28366782130 12.72198516172 27.65339839866 -17.91983219447
#[4,] -20.28531478210 12.72298481635 27.65549888136 -17.92114640013
#               [,13]           [,14]          [,15]           [,16]
#[1,] -1.163127428493 -13.07282052205 20.91363938663 -15.43763356270
#[2,] -1.175475358083 -13.20640615074 21.11903955938 -15.58312496123
#[3,] -1.176488937866 -13.21737152689 21.13589983227 -15.59506762248
#[4,] -1.176572137562 -13.21827161978 21.13728380775 -15.59604793581
#              [,17]           [,18]           [,19]           [,20]
#[1,] 4.129982905656 -32.29365824471 -26.80599719664 -13.92279757936
#[2,] 4.167268037791 -32.57240758567 -27.02676696018 -14.03195652202
#[3,] 4.170328587811 -32.59528872494 -27.04488884592 -14.04091683368
#[4,] 4.170579813055 -32.59716692322 -27.04637638088 -14.04165234085

如果要在向量或列表上使用代码，则应使用apply系列函数。希望这有帮助！

使用R中的for循环模拟不同点处的向量

2 个答案: