Append characters to strings in column which match expression R

August 8, 2023

I have a date/time column in an inconsistent format:

df <- data.frame(date.time=c("10-29-2022 09:46:40", "02-27-2023 22:53:53", "12-15-2022 02:54:03", "01-09-2023 14:44", "03-05-2023 14:58", "02-18-2023 19:46:35", "12-10-2022 16:50"))

I want to add ":00" to the time stamps that don’t include seconds (notice some strings are shorter than others). I can find the strings without seconds using gsubs(pattern=) but when I try to add on ":00" to the end I’m replacing the whole string:

df %>%
  mutate_at(.vars = "date.time", 
            .funs = gsub,
            pattern = "^\\d\\d-\\d\\d-\\d\\d\\d\\d \\d\\d:\\d\\d$",
            replacement = "\\d\\d-\\d\\d-\\d\\d\\d\\d \\d\\d:\\d\\d:00")

>Solution :

In this case, you can use a simple ifelse with nchar and paste0 with no need for regex:

df$date.time_new <- ifelse(nchar(df$date.time) < max(nchar(df$date.time)), 
                           paste0(df$date.time, ":00"), 
                           df$date.time)

Output:

            date.time       date.time_new
1 10-29-2022 09:46:40 10-29-2022 09:46:40
2 02-27-2023 22:53:53 02-27-2023 22:53:53
3 12-15-2022 02:54:03 12-15-2022 02:54:03
4    01-09-2023 14:44 01-09-2023 14:44:00
5    03-05-2023 14:58 03-05-2023 14:58:00
6 02-18-2023 19:46:35 02-18-2023 19:46:35
7    12-10-2022 16:50 12-10-2022 16:50:00