What is wrong with my IF/ELSE, or FOR loop, in below very simple R code?

Question

In response to some comments received, here´s what I am trying to do:

1. Replicate an XLS formula that works fine. The below is the beginning of what that XLS formula does
2. The complete XLS formula basically interpolates input variables along a time horizon (say 120 months). For example if you ONLY input mo 1 (along vector_1) = .25 (along vector_2), mo 2 (along vector_1) = .50 (along vector_2), XLS outputs mo 1 = .25, mo 2 = .50, mos 3-120 = 0. If you ONLY input mo 1 = 2 and month 3 = 4, then XLS interpolates and output is mo 1 = 2, mo 2 = 3, mo 3 = 4, mo 5-120 = 0. If you ONLY input mo 3 = 2, then output is mo 1 = 2, mo 2 = 2, mo 3 = 2, mo 4-120= 0. Etc. This R snippet is only the beginning of this formula
3. So all I am trying to do below, for now, is assign vector_2 variables along the mos 1-120 time scale, and if the current period in time horizon (vector_1) > period in the 1-120 scale, then a 0 value is returned.

I was worried about posting this, this is difficult to express clearly. And I want to learn enough to be able to do the rest of this on my own. Best way to learn!

I´m new to R and come from Excel/VBA world. I´m slowly converting an Excel formula to R equivalent. Will eventually change this to a function and explore using SEQ instead of this FOR loop. But for now, what am I doing wrong with the below? I don´t understand why I get NA in the 5th position of the output vector and 9 output elements instead of 10.

Code:

n_periods = 10
vector_1 <- c(1,2,3,4)
vector_2 <- c(0.20,0.10,0.05,0.3)
    
a=ifelse(vector_1[1] < 1, 0, vector_2[1])
  for(i in 1:n_periods){
    if(period[i+1] > max(vector_1)){a[i+1]=0} else {a[i+1] = vector_2[i+1]}
  }
a  

The output is below, whereby (i) the 5th position should show a 0 and not NA, (ii) there should be 10 output elements instead of 9 and (iii) I do not understand why I get that error message (output I´m looking for is 0.20 0.10 0.05 0.30 0.00 0.00 0.00 0.00 0.00 0.00):

 Error in if (period[i + 1] > max(vector_1)) { : 
  missing value where TRUE/FALSE needed
> a  
[1] 0.20 0.10 0.05 0.30   NA 0.00 0.00 0.00 0.00 

Answer 1

In your for-loop I am not getting what is period[i + 1] as there is no object defined period? However, as vectorised operations in base r you can do this achieve your output. One line of the code has been built upon this answer.

n_periods = 10

#ex-1
vector_1 <- c(1,2,3,4)
vector_2 <- c(0.20,0.10,0.05,0.3)

a <- rep(NA, n_periods)
a[vector_1] <- vector_2
a
#>  [1] 0.20 0.10 0.05 0.30   NA   NA   NA   NA   NA   NA

#ex-2
vector_1 <- c(2,3,4)
vector_2 <- c(0.10,0.05,0.3)

a <- rep(NA, n_periods)
a[vector_1] <- vector_2
a[seq_len(min(vector_1) -1)] <- a[min(vector_1)]
a[seq(max(vector_1)+1, n_periods, 1)] <- 0
a
#>  [1] 0.10 0.10 0.05 0.30 0.00 0.00 0.00 0.00 0.00 0.00

#ex-3
vector_1 <- c(3, 5)
vector_2 <- c(10, 30)

a <- rep(NA, n_periods)
a[vector_1] <- vector_2
a[seq_len(min(vector_1) -1)] <- a[min(vector_1)]
a[seq(max(vector_1)+1, n_periods, 1)] <- 0
a <- approx(seq_along(a)[!is.na(a)], a[!is.na(a)], seq_along(a))$y
a
#>  [1] 10 10 10 20 30  0  0  0  0  0

#complete solution check for more typical case
n_periods <- 20
vector_1 <- c(3, 5, 9)
vector_2 <- c(10, 30, 40)

a <- rep(NA, n_periods)
a[vector_1] <- vector_2
a[seq_len(min(vector_1) -1)] <- a[min(vector_1)]
a[seq(max(vector_1)+1, n_periods, 1)] <- 0
a <- approx(seq_along(a)[!is.na(a)], a[!is.na(a)], seq_along(a))$y
a
#>  [1] 10.0 10.0 10.0 20.0 30.0 32.5 35.0 37.5 40.0  0.0  0.0  0.0  0.0  0.0  0.0
#> [16]  0.0  0.0  0.0  0.0  0.0

^{Created on 2021-05-13 by the reprex package (v2.0.0)}