## Issue

Sometimes, thinking about the best algorithm is tricky. Well I have to work with some ranges of the form "letter/number/number", for example X86-Z95 would be a range. Actually I need to see if a given code is in the code range. I thought I’d generate an array containing the ranges and check if the code is in the array. But as I had a lot of work to implement the algorithm, I think there is an easier way to implement it. I would like to see some more efficient or elegant solution. I leave my code below.

Thanks in advance!

```
library(stringr)
generateInterval = function(str_interval) {
limits = strsplit(str_interval, '-')[[1]]
letters_limits = substr(limits, 1, 1)
numbers_limits = as.numeric(substr(limits, 2, 3))
if (letters_limits[1] == letters_limits[2]) {
interval = paste0(letters_limits[1], numbers_limits[1]:numbers_limits[2])
} else {
pos = which(LETTERS %in% letters_limits)
interval_letters = LETTERS[pos[1]:pos[2]]
numbers = as.numeric(substr(limits, 2, 3))
interval = c()
for (i in 1:length(interval_letters)) {
if (i == 1) {
interval = c(interval, paste0(interval_letters[i], numbers[1]:99))
} else if (i == length(interval_letters)) {
interval = c(interval, paste0(interval_letters[i], 0:numbers[2]))
} else {
interval = c(interval, paste0(interval_letters[i], 0:99))
}
}
}
interval = ifelse(str_length(interval) == 2,
paste0(substr(interval, 1, 1), 0, substr(interval, 2, 2)),
interval)
return(interval)
}
generateInterval('A72-B10')
```

## Solution

Following the OP’s comment here is a way in base R. The only thing that needs to be done is to realize that R can compare strings. And the one-line function will check if the codes are within range.

```
within_range <- function(x, lower, upper) lower <= x & x <= upper
within_range(c("X92", "X84", "A10", "Y30"), lower = "X86", upper = "Z95")
#> [1] TRUE FALSE FALSE TRUE
```

^{Created on 2022-02-11 by the reprex package (v2.0.1)}

Answered By – Rui Barradas

Answer Checked By – Marilyn (BugsFixing Volunteer)