本文主要是介绍等额本息和等额本金还款在R中的实现，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
等额本金还款
library(tidyverse)
Eqprincipal <- function(principal, r, Loanterm){n <- Loanterm * 12r <- r / 12Prepay <- rep(principal / n, n)Pint <- rep(NA, n)Pint[1] <- principal * rRepayPrint <- rep(NA, n)RepayPrint[1] <- Prepay[1] + Pint[1]Resipr <- rep(NA, n)Resipr[1] <- principal - Prepay[1]
for (i in c(2:n)){Pint[i] <- Resipr[i - 1] * rRepayPrint[i] <- Prepay[i] + Pint[i]Resipr[i] <- principal - i * Prepay[i]
}result <- tibble('每月偿还本金' = Prepay, '偿还利息' = Pint,'偿还本息' = RepayPrint,'剩余本金' = Resipr)return(result)
}
a <- Eqprincipal(1000000, 0.049, 20)
# 每月偿还本金 偿还利息 偿还本息 剩余本金
# <dbl> <dbl> <dbl> <dbl>
# 1 4167. 4083. 8250 995833.
# 2 4167. 4066. 8233. 991667.
# 3 4167. 4049. 8216. 987500
# 4 4167. 4032. 8199. 983333.
# 5 4167. 4015. 8182. 979167.
# 6 4167. 3998. 8165. 975000
# 7 4167. 3981. 8148. 970833.
# 8 4167. 3964. 8131. 966667.
# 9 4167. 3947. 8114. 962500
# 10 4167. 3930. 8097. 958333.
# # ... with 230 more rows
等额本息还款
Eqprint <- function(principal, r, Loanterm){n <- Loanterm * 12r <- r / 12Pint <- rep(principal * r, n)Pint[1] <- principal * rRepayPrint <- rep(principal * r * (1 + r)^n / ((1 + r)^n - 1), n)Prepay <- rep(NA, n)Prepay[1] <- RepayPrint[1] - Pint[1]Resipr <- rep(NA, n)Resipr[1] <- principal - Prepay[1]for (i in c(2:n)){Pint[i] <- Resipr[i - 1] * rPrepay[i] <- RepayPrint[i] - Pint[i]Resipr[i] <- Resipr[i - 1] - Prepay[i]}result <- tibble('每月偿还本金' = Prepay, '偿还利息' = Pint,'偿还本息' = RepayPrint,'剩余本金' = Resipr)return(result)
}
b <- Eqprint(1000000, 0.049, 20)
# 每月偿还本金 偿还利息 偿还本息 剩余本金
# <dbl> <dbl> <dbl> <dbl>
# 1 2461. 4083. 6544. 997539.
# 2 2471. 4073. 6544. 995068.
# 3 2481. 4063. 6544. 992586.
# 4 2491. 4053. 6544. 990095.
# 5 2502. 4043. 6544. 987594.
# 6 2512. 4033. 6544. 985082.
# 7 2522. 4022. 6544. 982560.
# 8 2532. 4012. 6544. 980027.
# 9 2543. 4002. 6544. 977485.
# 10 2553. 3991. 6544. 974932.
# # ... with 230 more rows
两种还款方式的利息差额计算
chae <- b$偿还利息 - a$偿还利息
chae1 <- tibble(chae)
# # A tibble: 240 x 1
# chae
# <dbl>
# 1 0
# 2 6.96
# 3 13.9
# 4 20.8
# 5 27.6
# 6 34.4
# 7 41.2
# 8 47.9
# 9 54.6
# 10 61.2
# # ... with 230 more rows
sum(chae)
# [1] 78624.05
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