Replace the first characters in a vector of strings in R

Question

So I have a vector of character strings (not formatted as time intentionally).

TIME
00:35 AM
08:34 AM
10:10 AM
10:00 PM
09:45 PM
02:00 PM
07:47 PM
12:22AM

I would like to make it so that the leading 0's are gone. If there are two leading 0's then it must be 12. Notice the extra space in the final entry, some data is formatted like that so the solution must take that into account.

Final outcome desired:

TIME
12:35 AM
 8:34 AM
10:10 AM
10:00 PM
 9:45 PM
 2:00 PM
 7:47 PM
12:22 AM

Tidyverse solution preferred but I attempted using str_remove to no avail.

that's also correct however if that gets in the way of solving the real problem then that can be ignored — Johnny Thomas, Jul 09 '20 at 15:34
@Nate had it backwards, there's no space when there should be — Johnny Thomas, Jul 09 '20 at 15:35
Does this answer your question? [Formatting a date in R without leading zeros](https://stackoverflow.com/questions/25387160/formatting-a-date-in-r-without-leading-zeros) — Ian Campbell, Jul 11 '20 at 20:43

score 1 · Accepted Answer · answered Jul 09 '20 at 15:35

library(magrittr) # Other packages import the %>% also nowadays
x %>% sub("^00", "12", .) %>% sub("^0", "", .)
# [1] "12:35 AM"  "8:34 AM"   "10:10 AM"  "10:00 PM"  "9:45 PM"   "2:00 PM"   "7:47 PM"   "12:22  AM"

Reproducible data:

x <- c("00:35 AM", "08:34 AM", "10:10 AM", "10:00 PM", "09:45 PM", "02:00 PM", "07:47 PM", "12:22  AM")

score 0 · Answer 2 · answered Jul 09 '20 at 15:38

use base sub and regex to substitute 00:00 for 12:00 at the start of the string:

sub("^00", "12", "00:35 AM")

will return ""12:35 AM". To remove leading 0s use sub afterwards:

sub("^0", "", "03:35 AM")"

will return "3:35 AM" Hope you get the idea

Best

score 0 · Answer 3 · answered Jul 09 '20 at 15:40

0

You can use this:

Time <- c("00:35 AM", "08:34 AM", "10:10 AM", "22:00 PM", "21:45 PM", "02:00 AM", "07:47 AM", "14:22  PM")

Time2 <- ifelse(substring(Time,1,1)=='0',substring(Time,2,nchar(Time)),Time)

answered Jul 09 '20 at 15:40

Duck

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Ric S · Answer 4 · 2020-07-09T16:01:58.780

This is a solution that takes into consideration also the cases where AM or PM is attached to the digit part of the variable TIME

data %>% 
  mutate(TIME = ifelse(!grepl(" ", TIME),
                       sub("(AM$)|(PM$)", " \\1\\2", sub("^0", "", sub("^00", "12", TIME))),
                       sub("^0", "", sub("^00", "12", TIME))))

The " \\1\\2" replacement means that it takes either the first or the second parenthesized subexpressions - (AM$) or (PM$) - of the regex pattern.

Output

#       TIME
# 1 12:35 AM
# 2  8:34 AM
# 3 10:10 AM
# 4 10:00 PM
# 5  9:45 PM
# 6  2:00 PM
# 7  7:47 PM
# 8 12:22 AM

Replace the first characters in a vector of strings in R

4 Answers4