Removing trailing zeros and decimal point in R - r

Question

I have the following data frame:
val
4.20
4.00
I would like to remove the trailing zeros and the decimal point if there are no additional decimals. Thus, my desired end result is
val
4.2
4
I know that I can use str_remove to get rid of the trailing zeros, but I am left with the decimal point on the 4. How can I update this code to drop that decimal point?
library(tidyverse)
data.frame(val = c("4.20", "4.00")) %>%
mutate(val = str_remove(val, "0+$"))
Edit: The numbers must be stored as a character.
Edit 2: Figured out solution below. Since data is stored as a character, solution needs to be robust to other instances in which you might deal with regular old character strings etc. This function deals with all contingencies:
decimal_func <- function(x) {
decimalVal_check <- function(y) {
case_when(str_count(y, "\\.") <= 1 & str_detect(str_replace(y, "\\.", ""), "^[:digit:]+$") == T ~ "Valid", TRUE ~ "Invalid")
}
if(decimalVal_check(x) == "Valid") {
if(str_count(x, "\\.") == 0) {
x
} else {
str_remove(x, "0+$") %>%
ifelse(substr(., nchar(.), nchar(.)) == ".", str_replace(., "\\.", ""), .)
}
} else {
x
}
}

Answer 1

One way:
data.frame(val = c("4.20", "4.00")) %>%
type.convert(as.is =TRUE) %>%
as_tibble()%>%
mutate(val = as.character(val))
# A tibble: 2 x 1
val
<chr>
1 4.2
2 4
Using str_remove:
data.frame(val = c("4.20", "4.00")) %>%
mutate(val = str_remove(val, '\\.?0+$'))
val
1 4.2
2 4
Any of the following can work:
formatC(c(1,2.40,5.06), zero.print = "")
[1] "1" "2.4" "5.06"
prettyNum(c(1,2.40,5.06), zero.print = "")
[1] "1" "2.4" "5.06"
prettyNum(c(1,2.40,5.06), drop0trailing = TRUE)
[1] "1" "2.4" "5.06"
formatC(c(1,2.40,5.06), drop0trailing = TRUE)
[1] "1" "2.4" "5.06"