Reputation: 321
I apologize if this is a duplicate, I've searched through all of the "add leading zero" content I can find, and I'm struggling to find a solution I can work with. I have the following:
siteid<-c("1","11","111")
modifier<-c("44","22","11")
df<-data.frame(siteid,modifier)
and I want a modified siteid
that is always six (6) characters long with zeroes to fill the gaps. The Site ID can vary in nchar
from 1-3, the modifier is always a length of 2, and the number of zeroes can vary depending on the length of the site ID (so that 6 is always the final modified length).
I would like the following final output:
df
# siteid modifier mod.siteid
#1 1 44 440001
#2 11 22 220011
#3 111 11 110111
Thanks for any suggestions or direction. This could also be numeric, but it seems like character manipulation has more options...?
Upvotes: 0
Views: 49
Reputation: 27802
siteid<-c("1","11","111")
modifier<-c("44","22","11")
df<-data.frame(siteid,modifier, stringsAsFactors = FALSE)
df$mod.siteid = paste0( df$modifier,
formatC( as.numeric(df$siteid), width = 4, format = "d", flag="0") )
df
# siteid modifier mod.siteid
# 1 1 44 440001
# 2 11 22 220011
# 3 111 11 110111
Upvotes: 2
Reputation: 33603
The vocabulary here is left pad and paste here is one way using sprintf()
::
df$mod.siteid <- with(df, sprintf("%s%04d", modifier, as.integer(siteid)))
# Note:
# code simplified thanks to suggestion by Maurits.
Output:
siteid modifier mod.siteid
1 1 44 440001
2 11 22 220011
3 111 11 110111
Data:
df <- data.frame(
siteid = c("1", "11", "111"),
modifier = c("44", "22", "11"),
stringsAsFactors = FALSE
)
Extra: If you don't want to left pad with 0
, then using the stringi
package is one option: with(df, paste0(modifier, stringi::stri_pad_left(siteid, 4, "q")))
Upvotes: 2