Date and Time in the same field, how to separate in R?

Question

In a table column, I have the values of DATE and TIME in the same field.

Example:

23:02:1989 14:22

I need to separate the Date and Time Value.

How should I do?

Thank you,

Answer 1

You can use the function strptime() and others more that I put in the example:

x <- "23:02:1989 14:22"
dataHora <-strptime(x, "%d:%m:%Y %H:%M") # converte para o formato adequado
# [1] "1989-02-23 14:22:00 GMT"

And if you want to separate in date and time:

data <- as.Date(dataHora, "%d:%m:%Y")
hora <- format(dataHora, "%H:%M")

Answer 2

Another alternative is to use the package lubridate, that is part of the tidyverse.

Then we would have:

library(lubridate)
x <- "23:02:1989 14:22"
data_hora <- dmy_hm(x)
class(data_hora)
[1] "POSIXct" "POSIXt" 

The function dmy_hm() parses dates that are in day, month, year, time and minute format.

The solution offered by @Willianvieira leaves the hours in a text format. In this case it can also be done with text manipulation. In tidyverse, this is done with the package stringr.

library(stringr)
separados <- str_split(x, " ")[[1]]
separados[1]
[1] "23:02:1989"
separados[2]
[1] "14:22"

After the separation of the texts it is possible to parse the date and time separately.

dt_data <- dmy(separados[1])
class(dt_data)
[1] "Date"
hm_data <- hm(separados[2])
class(hm_data)
[1] "Period"
attr(,"package")
[1] "lubridate"

Another widely used alternative with the lubridate, which has a tone of gambiarra, is to make all the elements of a date and time vector have the same day. So the differences in the hour are thrown to the foreground.

hora <- data_hora + duration(5:14,  "hours")
year(hora) <- 1989
month(hora) <- 1
day(hora) <- 1
hora
 [1] "1989-01-01 19:22:00 UTC" "1989-01-01 20:22:00 UTC"
 [3] "1989-01-01 21:22:00 UTC" "1989-01-01 22:22:00 UTC"
 [5] "1989-01-01 23:22:00 UTC" "1989-01-01 00:22:00 UTC"
 [7] "1989-01-01 01:22:00 UTC" "1989-01-01 02:22:00 UTC"
 [9] "1989-01-01 03:22:00 UTC" "1989-01-01 04:22:00 UTC"