R 计算所有时间点现金流的净现值

我有以下数据框包含几个项目的现金流。例如：

test <- data.frame(ID = c(rep("A",3), rep("B",4)), 
        time = c("y3","y2","y1","y4","y3","y2","y1"),
        Cfs= c(rep(1,3),rep(2,4)),
        interest = c(rep(0.1,3),rep(0.05,4)))

ID  time    CFs interest
A   y3      1   0.1
A   y2      1   0.1
A   y1      1   0.1
B   y4      2   0.05
B   y3      2   0.05
B   y2      2   0.05
B   y1      2   0.05

通过阅读一些旧帖子，我能够计算出每个项目总现金流的净现值，但我不确定在每个时间段如何计算。另外，由于实际的数据集相当大（300k+），所以我也尝试避免循环

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谢谢

您可能会发现其中一些帮助函数很有用

dcf <- function(x, r, t0=FALSE){
  # calculates discounted cash flows (DCF) given cash flow and discount rate
  #
  # x - cash flows vector
  # r - vector or discount rates, in decimals. Single values will be recycled
  # t0 - cash flow starts in year 0, default is FALSE, i.e. discount rate in first period is zero.
  if(length(r)==1){
    r <- rep(r, length(x))
    if(t0==TRUE){r[1]<-0}
  }
  x/cumprod(1+r)
}

npv <- function(x, r, t0=FALSE){
  # calculates net present value (NPV) given cash flow and discount rate
  #
  # x - cash flows vector
  # r - discount rate, in decimals
  # t0 - cash flow starts in year 0, default is FALSE
  sum(dcf(x, r, t0))
}

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我正在努力理解这个问题，为了计算净现值，您需要将所有

CFs

贴现回时间

0

。您希望如何在每个时段使用它？

dcf <- function(x, r, t0=FALSE){
  # calculates discounted cash flows (DCF) given cash flow and discount rate
  #
  # x - cash flows vector
  # r - vector or discount rates, in decimals. Single values will be recycled
  # t0 - cash flow starts in year 0, default is FALSE, i.e. discount rate in first period is zero.
  if(length(r)==1){
    r <- rep(r, length(x))
    if(t0==TRUE){r[1]<-0}
  }
  x/cumprod(1+r)
}

npv <- function(x, r, t0=FALSE){
  # calculates net present value (NPV) given cash flow and discount rate
  #
  # x - cash flows vector
  # r - discount rate, in decimals
  # t0 - cash flow starts in year 0, default is FALSE
  sum(dcf(x, r, t0))
}

library(dplyr)

test %>% mutate_if(is.factor, as.character) %>% 
  arrange(ID, time) %>% 
  group_by(ID) %>% 
  mutate(DCF=cumsum(dcf(x=Cfs, r=interest)))

#> # A tibble: 7 x 5
#> # Groups:   ID [2]
#>      ID  time   Cfs interest       DCF
#>   <chr> <chr> <dbl>    <dbl>     <dbl>
#> 1     A    y1     1     0.10 0.9090909
#> 2     A    y2     1     0.10 1.7355372
#> 3     A    y3     1     0.10 2.4868520
#> 4     B    y1     2     0.05 1.9047619
#> 5     B    y2     2     0.05 3.7188209
#> 6     B    y3     2     0.05 5.4464961
#> 7     B    y4     2     0.05 7.0919010