Reputation: 2710
Possible to avoid a FOR loop in this very simple R code?
The answers below are very helpful. But I oversimplified my original question. I figured I learn more if I oversimplify and then adapt to my actual need, but now I am stuck. There are other factors that drive the amortization. See more complete code here. I like the response using "amort$end_bal <- begin_bal * (1 - mpr)^amort$period" and "amort$pmt <- c(0, diff(amort$end_bal))* -1", but in addition npr increases the ending balances and ch_off decreases ending balances. Here´s the more complete code:
n_periods <- 8
begin_bal <- 10000
yld <- .20
npr <- .09
mpr <- .10
co <- .10
period = seq(0,n_periods,1)
fin = 0
pur = 0
pmt = 0
ch_off = 0
end_bal = begin_bal
for(i in 1:n_periods){
{fin[i+1] = end_bal[i]*yld/12}
{pur[i+1] = end_bal[i]*npr}
{pmt[i+1] = end_bal[i]*mpr}
{ch_off[i+1] = end_bal[i]*co/12}
end_bal[i+1] = end_bal[i]+pur[i+1]-pmt[i+1]-ch_off[i+1]}
amort <- data.frame(period,fin,pur,pmt,ch_off,end_bal)
Which gives the below correct output:
print(amort,row.names=FALSE)
period fin pur pmt ch_off end_bal
0 0.0000 0.0000 0.0000 0.00000 10000.000
1 166.6667 900.0000 1000.0000 83.33333 9816.667
2 163.6111 883.5000 981.6667 81.80556 9636.694
3 160.6116 867.3025 963.6694 80.30579 9460.022
4 157.6670 851.4020 946.0022 78.83351 9286.588
5 154.7765 835.7929 928.6588 77.38823 9116.334
6 151.9389 820.4700 911.6334 75.96945 8949.201
7 149.1534 805.4281 894.9201 74.57668 8785.132
8 146.4189 790.6619 878.5132 73.20944 8624.072
I´m new to R, and I understand one of its features is matrix/vector manipulation. In the below example I amortize an asset over 8 months, where each payment ("pmt") is 10% ("mpr") of the prior period balance ("end_bal"). The below works fine. I used a FOR loop. I understand FOR loops can be slow in large models and a better solution is use of R´s abundant vector/matrix functions. But I didn´t know how to do this in my example since each monthly payment is calculated by referencing the prior period ending balance.
So my questions are:
- Is there a more efficient way to do the below?
- How do I replace the 0 for pmt in period 0, with an empty space?
R code:
n_periods <- 8
begin_bal <- 100
mpr <- .10
# Example loan amortization
pmt = 0
end_bal = begin_bal
for(i in 1:n_periods){
{pmt[i+1] = end_bal[i]*mpr}
end_bal[i+1] = end_bal[i]-pmt[i+1]}
amort <- data.frame(period = 0:n_periods,pmt,end_bal)
amort
Results, which are correct:
> amort
period pmt end_bal
1 0 0.000000 100.00000
2 1 10.000000 90.00000
3 2 9.000000 81.00000
4 3 8.100000 72.90000
5 4 7.290000 65.61000
6 5 6.561000 59.04900
7 6 5.904900 53.14410
8 7 5.314410 47.82969
9 8 4.782969 43.04672
Upvotes: 0
Views: 77
Answers (2)
Reputation: 26218
Use R's vectorised calculations
n_periods <- 8
begin_bal <- 100
mpr <- .10
amort <- data.frame(period = seq(0, n_periods, 1))
amort$end_bal <- begin_bal * (1 - mpr)^amort$period
amort$pmt <- c(0, diff(amort$end_bal))* -1
amort
#> period end_bal pmt
#> 1 0 100.00000 0.000000
#> 2 1 90.00000 10.000000
#> 3 2 81.00000 9.000000
#> 4 3 72.90000 8.100000
#> 5 4 65.61000 7.290000
#> 6 5 59.04900 6.561000
#> 7 6 53.14410 5.904900
#> 8 7 47.82969 5.314410
#> 9 8 43.04672 4.782969
Created on 2021-05-12 by the reprex package (v2.0.0)
dplyr way for a different case (say)
n_periods <- 15
begin_bal <- 1000
mpr <- .07
library(dplyr)
seq(0, n_periods, 1) %>% as.data.frame() %>%
setNames('period') %>%
mutate(end_bal = begin_bal * (1 - mpr)^period,
pmt = -1 * c(0, diff(end_bal)))
#> period end_bal pmt
#> 1 0 1000.0000 0.00000
#> 2 1 930.0000 70.00000
#> 3 2 864.9000 65.10000
#> 4 3 804.3570 60.54300
#> 5 4 748.0520 56.30499
#> 6 5 695.6884 52.36364
#> 7 6 646.9902 48.69819
#> 8 7 601.7009 45.28931
#> 9 8 559.5818 42.11906
#> 10 9 520.4111 39.17073
#> 11 10 483.9823 36.42878
#> 12 11 450.1035 33.87876
#> 13 12 418.5963 31.50725
#> 14 13 389.2946 29.30174
#> 15 14 362.0439 27.25062
#> 16 15 336.7009 25.34308
Created on 2021-05-12 by the reprex package (v2.0.0)
Though OP has put another question in edited scenario, here's the approach suggested (for future reference)
n_periods <- 8
begin_bal <- 10000
yld <- .20
npr <- .09
mpr <- .10
co <- .10
library(dplyr)
seq(0, n_periods, 1) %>% as.data.frame() %>%
setNames('period') %>%
mutate(end_bal = begin_bal * (1 - (mpr + co/12 - npr))^period,
fin = c(0, (end_bal * yld/12)[-nrow(.)]),
pur = c(0, (end_bal * npr)[-nrow(.)]),
pmt = c(0, (end_bal * mpr)[-nrow(.)]),
ch_off = c(0, (end_bal * co/12)[-nrow(.)]))
#> period end_bal fin pur pmt ch_off
#> 1 0 10000.000 0.0000 0.0000 0.0000 0.00000
#> 2 1 9816.667 166.6667 900.0000 1000.0000 83.33333
#> 3 2 9636.694 163.6111 883.5000 981.6667 81.80556
#> 4 3 9460.022 160.6116 867.3025 963.6694 80.30579
#> 5 4 9286.588 157.6670 851.4020 946.0022 78.83351
#> 6 5 9116.334 154.7765 835.7929 928.6588 77.38823
#> 7 6 8949.201 151.9389 820.4700 911.6334 75.96945
#> 8 7 8785.132 149.1534 805.4281 894.9201 74.57668
#> 9 8 8624.072 146.4189 790.6619 878.5132 73.20944
Created on 2021-05-13 by the reprex package (v2.0.0)
Upvotes: 2
Reputation: 101034
If you are "lazy" (don't want to formulate the general expression of pmt and end_bal), you can define a recursive function f like blow
f <- function(k) {
if (k == 1) {
return(data.frame(pmt = 100 * mpr, end_bal = 100))
}
u <- f(k - 1)
end_bal <- with(tail(u, 1), end_bal - pmt)
pmt <- mpr * end_bal
rbind(u, data.frame(pmt, end_bal))
}
n_periods <- 8
res <- transform(
cbind(period = 0:n_periods, f(n_periods + 1)),
pmt = c(0, head(pmt, -1))
)
and you will see
> res
period pmt end_bal
1 0 0.000000 100.00000
2 1 10.000000 90.00000
3 2 9.000000 81.00000
4 3 8.100000 72.90000
5 4 7.290000 65.61000
6 5 6.561000 59.04900
7 6 5.904900 53.14410
8 7 5.314410 47.82969
9 8 4.782969 43.04672
Upvotes: 1
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