Separate hour and minutes in R - r

Question

I have a column for time, but it hasn't been separated by : or any thing. It looks like this:
person time
1 356
1 931
1 2017
1 2103
2 256
2 1031
2 1517
2 2206
How do I separate them?

Answer 1

There are different ways of approaching the issue. Which method you choose depends on your desired output.
For example, you could use stringr::str_split to split time into a list vector of hours and minutes using a positive look-ahead
library(tidyverse)
df %>% mutate(time = str_split(time, "(?=\\d{2}$)"))
# person time
#1 1 3, 56
#2 1 9, 31
#3 1 20, 17
#4 1 2, 13
#5 2 2, 56
#6 2 10, 31
#7 2 15, 17
#8 2 2, 26
Or we can use tidyr::separate to create two new columns hours and minutes
df %>% separate(time, c("hours", "minutes"), sep = "(?=\\d{2}$)")
# person hours minutes
#1 1 3 56
#2 1 9 31
#3 1 20 17
#4 1 2 13
#5 2 2 56
#6 2 10 31
#7 2 15 17
#8 2 2 26
In response to your comment you could use stringr::str_replace
df %>% mutate(time = str_replace(time, "(?=\\d{2}$)", ":"))
# person time
#1 1 3:56
#2 1 9:31
#3 1 20:17
#4 1 2:13
#5 2 2:56
#6 2 10:31
#7 2 15:17
#8 2 2:26
And the same in base R using sub
transform(df, time = sub("(?=\\d{2}$)", ":", time, perl = TRUE))
giving the same result.
Sample data
df <- read.table(text = "
person time
1 356
1 931
1 2017
1 213
2 256
2 1031
2 1517
2 226", header = T)

Answer 2

We can use strptime with sprintf in base R
df[c("hour", "min")] <- unclass(strptime(sprintf("%04d00", df$time),
"%H%M%S"))[c('hour', 'min')]
df
# person time hour min
#1 1 356 3 56
#2 1 931 9 31
#3 1 2017 20 17
#4 1 213 2 13
#5 2 256 2 56
#6 2 1031 10 31
#7 2 1517 15 17
#8 2 226 2 26
Or if it needs to only create a delimiter
tmp <- sub('(\\d{2})$', ':\\1', df$time)
tmp
#[1] "3:56" "9:31" "20:17" "2:13" "2:56" "10:31" "15:17" "2:26"
and then it can be separated in to two column with read.table
read.table(text = tmp, sep=":", header = FALSE, col.names = c('hour', 'min'))
data
df <- structure(list(person = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), time = c(356L,
931L, 2017L, 213L, 256L, 1031L, 1517L, 226L)),
class = "data.frame", row.names = c(NA,
-8L))

Answer 3

Another possibility:
res<-strsplit(gsub("(\\d+(?=\\d{2,}))(\\d{1,})",
"\\1:\\2",df$time,perl = T),":")
df$Minutes <- sapply(res,"[[",2)
df$Hr <- sapply(res,"[[",1)
df
Result:
person time Minutes Hr
1 1 356 56 3
2 1 931 31 9
3 1 2017 17 20
4 1 2103 03 21
5 2 256 56 2
6 2 1031 31 10
7 2 1517 17 15
8 2 2206 06 22
Data:
df <-structure(list(person = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), time = c(356L,
931L, 2017L, 2103L, 256L, 1031L, 1517L, 2206L)), row.names = c(NA,
-8L), class = "data.frame")

Answer 4

If you want to show time in HH:MM format, probably we can use sprintf with sub to enter semicolon (:) in between
sub("(\\d{2})(\\d{2})", "\\1:\\2",sprintf("%04d", df$time))
#[1] "03:56" "09:31" "20:17" "21:03" "02:56" "10:31" "15:17" "22:06"