Extend data frame column with inflation in R - r

Question

I'm trying to extend some code to be able to:
1) read in a vector of prices
2) left join that vector of prices to a data frame of years (or years and months)
3) append/fill the prices for missing years with interpolated data based on the last year of available prices plus a specified inflation rate. Consider an example like this one:
prices <- data.frame(year=2018:2022,
wti=c(75,80,90,NA,NA),
brent=c(80,85,94,93,NA))
What I need is something that will fill the missing rows of each column with the last price plus inflation (suppose 2%). I can do this in a pretty brute force way as:
i_rate<-0.02
for(i in c(1:nrow(prices))){
if(is.na(prices$wti[i]))
prices$wti[i]<-prices$wti[i-1]*(1+i_rate)
if(is.na(prices$brent[i]))
prices$brent[i]<-prices$brent[i-1]*(1+i_rate)
}
It seems to me there should be a way to do this using some combination of apply() and/or fill() but I can't seem to make it work.
Any help would be much appreciated.

Answer 1

As noted by #camille, the problem with dplyr::lag is that it doesn't work here with consecutive NAs because it uses the "original" ith element of a vector instead of the "revised" ith element. We'd have to first create a version of lag that will do this by creating a new function:
impute_inflation <- function(x, rate) {
output <- x
y <- rep(NA, length = length(x)) #Creating an empty vector to fill in with the loop. This makes R faster to run for vectors with a large number of elements.
for (i in seq_len(length(output))) {
if (i == 1) {
y[i] <- output[i] #To avoid an error attempting to use the 0th element.
} else {
y[i] <- output[i - 1]
}
if (is.na(output[i])) {
output[i] <- y[i] * (1 + rate)
} else {
output[i]
}
}
output
}
Then it's a pinch to apply this across a bunch of variables with dplyr::mutate_at():
library(dplyr)
mutate_at(prices, vars(wti, brent), impute_inflation, 0.02)
year wti brent
1 2018 75.000 80.00
2 2019 80.000 85.00
3 2020 90.000 94.00
4 2021 91.800 93.00
5 2022 93.636 94.86

Answer 2

You can use dplyr::lag to get the previous value in a given column. Your lagged values look like this:
library(dplyr)
inflation_factor <- 1.02
prices <- data_frame(year=2018:2022,
wti=c(75,80,90,NA,NA),
brent=c(80,85,94,93,NA)) %>%
mutate_at(vars(wti, brent), as.numeric)
prices %>%
mutate(prev_wti = lag(wti))
#> # A tibble: 5 x 4
#> year wti brent prev_wti
#> <int> <dbl> <dbl> <dbl>
#> 1 2018 75 80 NA
#> 2 2019 80 85 75
#> 3 2020 90 94 80
#> 4 2021 NA 93 90
#> 5 2022 NA NA NA
When a value is NA, multiply the lagged value by the inflation factor. As you can see, that doesn't handle consecutive NAs, however.
prices %>%
mutate(wti = ifelse(is.na(wti), lag(wti) * inflation_factor, wti),
brent = ifelse(is.na(brent), lag(brent) * inflation_factor, brent))
#> # A tibble: 5 x 3
#> year wti brent
#> <int> <dbl> <dbl>
#> 1 2018 75 80
#> 2 2019 80 85
#> 3 2020 90 94
#> 4 2021 91.8 93
#> 5 2022 NA 94.9
Or to scale this and avoid doing the same multiplication over and over, gather the data into a long format, get lags within each group (wti, brent, or any others you may have), and adjust values as needed. Then you can spread back to the original shape:
prices %>%
tidyr::gather(key = key, value = value, wti, brent) %>%
group_by(key) %>%
mutate(value = ifelse(is.na(value), lag(value) * inflation_factor, value)) %>%
tidyr::spread(key = key, value = value)
#> # A tibble: 5 x 3
#> year brent wti
#> <int> <dbl> <dbl>
#> 1 2018 80 75
#> 2 2019 85 80
#> 3 2020 94 90
#> 4 2021 93 91.8
#> 5 2022 94.9 NA
Created on 2018-07-12 by the reprex package (v0.2.0).