How to generate a sequence in R, that starts with 01 and ends with 99?

Question

I want: 01 02 03 ... 10, 11, 12 ... 97, 98, 99

I have tried using following:

seq("01":"19") # Doesn't help
paste0("0",seq("01":"19")) # Doesn't help

Please suggest something!

Tittle

Why not just 1:99? How are going to use this character vector? Seems like almost a dupe of this – Judicatory

I had a column having this format, a character vector of length 2. Wanted to map it there. – Tittle

Answer 1

We can use sprintf

sprintf("%02d", 1:99)
#[1] "01" "02" "03" "04" "05" "06" "07" "08" "09" "10" "11" "12" "13" "14" "15" "16" "17" "18" "19" "20" "21" "22" "23" "24" "25" "26" "27" "28" "29" "30" "31" "32"
#[33] "33" "34" "35" "36" "37" "38" "39" "40" "41" "42" "43" "44" "45" "46" "47" "48" "49" "50" "51" "52" "53" "54" "55" "56" "57" "58" "59" "60" "61" "62" "63" "64"
#[65] "65" "66" "67" "68" "69" "70" "71" "72" "73" "74" "75" "76" "77" "78" "79" "80" "81" "82" "83" "84" "85" "86" "87" "88" "89" "90" "91" "92" "93" "94" "95" "96"
#[97] "97" "98" "99"

---

Or a package option is str_pad

library(stringr)
str_pad(1:99, pad = 0,width = 2 , "left")

Farrow

Thanks a lot! actually i wanted c("01":"31","98","99") is there any in-function way to do it or i should use c(sprintf("%02d", 1:31),98,99) – Tittle

Answer 2

Another option is formatC. You can specifiy width = 2 to get 2 digit sequence.

formatC(seq(1, 99), width=2, flag=0)

#[1]"01" "02" "03" "04" "05" "06" "07" "08" "09" "10" "11" "12" "13" "14" "15" "16" "17"
#[18]"18" "19" "20" "21" "22" "23" "24" "25" "26" "27" "28" "29" "30" "31" "32" "33" "34"
#[35]"35" "36" "37" "38" "39" "40" "41" "42" "43" "44" "45" "46" "47" "48" "49" "50" "51"
#[52]"52" "53" "54" "55" "56" "57" "58" "59" "60" "61" "62" "63" "64" "65" "66" "67" "68"
#[69]"69" "70" "71" "72" "73" "74" "75" "76" "77" "78" "79" "80" "81" "82" "83" "84" "85"
#[86]"86" "87" "88" "89" "90" "91" "92" "93" "94" "95" "96" "97" "98" "99"

Deepsea