Reputation: 7846
I have a character containing both time and date and I would like to split it into date and time. I tried:
datetime <- as.character("12:00:00 03-May-2019")
date1 <- as.Date(datetime ,"%d/%m/%Y")
date1
time1 <- as.Date(datetime ,%H:%M:%S)
time1
but I just get NA returned.
Thank you for your help.
Upvotes: 0
Views: 176
Reputation: 886938
The format
should match the ones in the initial vector
.
as.Date(datetime, "%H:%M:%S %d-%b-%Y")
#[1] "2019-05-03" "2018-05-03" "2017-05-03"
If the intention it to split by space " "
strsplit(datetime, "\\s+")
Or with read.table
df1 <- read.table(text = datetime, header = FALSE, col.names = c("time", "date"))
df1$date <- as.Date(df1$date, "%d-%b-%Y")
df1
# time date
#1 12:00:00 2019-05-03
#2 02:01:00 2018-05-03
#3 11:00:22 2017-05-03
Or with tidyverse
library(tidyverse)
tibble(datetime) %>%
separate(datetime, into = c('date', 'time'), sep=" ") %>%
mutate(time = dmy(time))
# A tibble: 3 x 2
# date time
# <chr> <date>
#1 12:00:00 2019-05-03
#2 02:01:00 2018-05-03
#3 11:00:22 2017-05-03
datetime <- c("12:00:00 03-May-2019",
"02:01:00 03-May-2018",
"11:00:22 03-May-2017")
Upvotes: 1
Reputation: 2105
Here is a function that will extract the date (as a Date
object) and the time (as a character string). See this post and answers for some motivation for keeping the time as a character string.
extract_date_time <- function(dt){
dt_split <- strsplit(dt, split=" ")
lapply(dt_split, function(dts){
list(date=as.Date(dts[2], '%d-%b-%Y'), time=dts[1])
})
}
Example usage:
datetime <- as.character("12:00:00 03-May-2019")
extract_date_time(datetime)
You can also apply this function to a longer vector of strings like the one you originally have, like this:
datetimes <- c("12:00:00 03-May-2019",
"02:01:00 03-May-2018",
"11:00:22 03-May-2017")
extract_date_time(datetimes)
Upvotes: 1