R：格式化字符串并填充0

Question

Is there a function in R to get the following. Let's say i have some inputs like c("1456-3-13;11:56:98", "45:76:12", "981-56-54;11") . However, they all should be in in the form of the regex "/d{4}-/d{2}-/d{2};/d{2}:/d{2}:/d{2}"

Whenever a number is missing in the input it should be filled with zeros from left to right. So the output of processing the mentioned inputs should be c("1456-03-13;11:56:98", "0000-00-00;45:76:12", "0981-56-54;11:00:00").

Any idea how to do this in a clever way?:)

Answer 1

This is a bit messy, but it seems to work for this case. First I parse the string you have and convert to numeric, then I reformat all the values

tests<-c("1456-3-13;11:56:98", "45:76:12", "981-56-54;11", "1456-3-13")

matches<- stringr::str_match(tests, "(?:(\\d+(?=[-;]))(?:-(\\d+)(?:-(\\d+))?)?)?;?(\\d+)?(?::(\\d+)(?::(\\d+))?)?")
matches <- matches[,-1]
class(matches) <- "numeric"
matches[is.na(matches)]<-0

do.call("sprintf", c(list("%04d-%02d-%02d;%02d:%02d:%02d"), lapply(1:ncol(matches), function(i) matches[, i])))

# [1] "1456-03-13;11:56:98" "0000-00-00;45:76:12" "0981-56-54;11:00:00"
# [4] "1456-03-13;00:00:00"