R中的资产负债表建模

Question

I need to model a basic balance sheet, in which 1) final balance is initial balance plus cash flow, and 2) initial balance is equal to previous final balance. The following code works fine:

init_balance <- c(0,0,0,0)
cash_flow <- 1:4
final_balance <- c(0,0,0,0)

n <- length (final_balance)
for (i in 1:n) {
  final_balance[i] <- init_balance[i] + cash_flow[i]
  if(i < n) {
    init_balance[i+1] <- final_balance[i]
    }
}

> init_balance
[1]  0  1  3  6
> cash_flow
[1]  1  2  3  4
> final_balance
[1]  1  3  6 10

However, this implementation uses for loops and does not sound R-ish to me. All financial packages I have found are related to financial analysis, not financial modelling.

Would someone suggest another approach or package for this kind of modelling?

Thank you.

Answer 1

No need for loops and all you really need is cash_flow to start:

library(tidyverse)

df <- tibble(
  cash_flow = 1:4
)
df
#> # A tibble: 4 x 1
#>   cash_flow
#>       <int>
#> 1         1
#> 2         2
#> 3         3
#> 4         4

df %>%
  mutate(
    final_balance = cumsum(cash_flow),
    init_balance = lag(final_balance, 1, default = 0)
  )
#> # A tibble: 4 x 3
#>   cash_flow final_balance init_balance
#>       <int>         <int>        <dbl>
#> 1         1             1            0
#> 2         2             3            1
#> 3         3             6            3
#> 4         4            10            6

^{Created on 2019-03-21 by the reprex package (v0.2.1)}

Your final_balance is just the cumulative cash flows. For init_balance we are using lag of the final_balance by 1 period (think 1 period prior); we are also setting its default amount to zero instead of NA .

Answer 2

A base R-ish way could look like this.

finance <- data.frame(cash.flow=1:4)

within(finance, {
  final.balance <- cumsum(cash.flow)
  init.balance <- final.balance - cash.flow
})
#   cash.flow init.balance final.balance
# 1         1            0             1
# 2         2            1             3
# 3         3            3             6
# 4         4            6            10

The difference is that this solution (and that of the other answer) calculates with the whole vector at the same time and not with every single value, which could be termed "R-ish".