Create date column from datetime in R - r

Question

I am new to R and I am an avid SAS programmer and am just having a difficult time wrapping my head around R.
Within a data frame I have a date time column formatted as a POSIXct with the following the column appearing as "2013-01-01 00:53:00". I would like to create a date column using a function that extracts the date and a column to extract the hour. In an ideal world I would like to be able to extract the date, year, day, month, time and hour all within the data frame to create these additional columns within the data frame.

Answer 1

It is wise to always to be careful with as.Date(as.POSIXct(...)):
E.g., for me in Australia:
df <- data.frame(dt=as.POSIXct("2013-01-01 00:53:00"))
df
# dt
#1 2013-01-01 00:53:00
as.Date(df$dt)
#[1] "2012-12-31"
You'll see that this is problematic as the dates don't match. You'll hit problems if your POSIXct object is not in the UTC timezone as as.Date defaults to tz="UTC" for this class. See here for more info: as.Date(as.POSIXct()) gives the wrong date?
To be safe you probably need to match your timezones:
as.Date(df$dt,tz=Sys.timezone()) #assuming you've just created df in the same session:
#[1] "2013-01-01"
Or safer option #1:
df <- data.frame(dt=as.POSIXct("2013-01-01 00:53:00",tz="UTC"))
as.Date(df$dt)
#[1] "2013-01-01"
Or safer option #2:
as.Date(df$dt,tz=attr(df$dt,"tzone"))
#[1] "2013-01-01"
Or alternatively use format to extract parts of the POSIXct object:
as.Date(format(df$dt,"%Y-%m-%d"))
#[1] "2013-01-01"
as.numeric(format(df$dt,"%Y"))
#[1] 2013
as.numeric(format(df$dt,"%m"))
#[1] 1
as.numeric(format(df$dt,"%d"))
#[1] 1

Answer 2

Use the lubridate package. For example, if df is a data.frame with a column dt of type POSIXct, then you could:
df$date = as.Date(as.POSIXct(df$dt, tz="UTC"))
df$year = year(df$dt)
df$month = month(df$dt)
df$day = day(df$dt)
# and so on...
If your can store your data in a data.table, then this is even easier:
df[, `:=`(date = as.Date(as.POSIXct(dt, tz="UTC")), year = year(dt), ...)]