Use sprintf() to add trailing zeros

Question

There has to be a simple way of doing this and I am overlooking it. But if I have a series of id and want to add trailing zeros where the character limit is not reached. I saw this solution on another post but can't seem to find it to link for reference.

df$id <- c(2331,29623,311,29623)

Doing this gets the leading zero:

df$id_new <- sprintf("%05s", df$id)

But doing this does not get a trailing zero:

df$id_new <- sprintf("%-05s", df$id)

Answer

Thanks to Richard below, I pulled the stringr package and used the following to test:

df$id_test <- str_pad(df$id, width=5, side="right", pad="0")

Produced:

id_test  
23310  
29623  
31100  
29623

Answer 1

I don't think sprintf does that, but you can use formatC to do that

x <- 123    
formatC(as.numeric(x), format = 'f', flag='0', digits = 2)
[1] "123.00"

Answer 2

As noted in the comments, sprintf() won't do this. But the stringi package has padding functions that are quick and easy.

id <- c(2331, 29623, 311, 29623)

library(stringi)
stri_pad_left(id, 5, 0)
# [1] "02331" "29623" "00311" "29623"
stri_pad_right(id, 5, 0)
# [1] "23310" "29623" "31100" "29623"

These functions dispatch directly to C code, so they should be sufficiently efficient.

Answer 3

If we insist - we can fool sprintf() into doing what it doesn't want to do:

# Helper function from: stackoverflow.com/a/13613183/4552295
strReverse <- function(x) sapply(lapply(strsplit(x, NULL), rev), paste, collapse="")

so_0padr <- function(x, width = 5) {
  strReverse(
    sprintf(
      "%0*d",
      width,
      as.integer(
        strReverse(
          as.character(x)
        )
      )
    )
  )
}

Resulting in

so_0padr(c(2331,29623,311,29623))
[1] "23310" "29623" "31100" "29623"