How to extract ONLY time from strptime()?

Question

Part of the data looks like:

head(my_data[,c(1,2)], 3)
       Date Time 
1 16/12/2006 17:24:00 
2 16/12/2006 17:25:00 
3 16/12/2006 17:26:00

By the way, str() $Date, $Time are all chr now.

I want to keep them in two cols with correct format, so I use

x <- as.Date(my_data$Date, "%d/%m/%Y")

to get the 1st col in date format :

x[1:5]
[1] "2006-12-16" "2006-12-16" "2006-12-16" "2006-12-16" "2006-12-16"

But in the 2nd col, when I'm trying to use

y <- strptime(my_data$Time, "%H:%M:%S")

The output automatically add default date and timezone of my computer.

y[1:4]
[1] "2017-01-10 17:24:00 CST" "2017-01-10 17:25:00 CST" 
[2] "2017-01-10 17:26:00 CST" "2017-01-10 17:27:00 CST"

What should I do if I just want to keep the time, without date and timezone?
Is sub() with some regular expression the only way to achieve this?

If you want to have a pure time object, you need a package that defines such a class. You could use package `chron`. However, I would suggest to combine date and time into a `POSIXct` object. There are nuisances such as leap years and leap seconds that need to be taken into account. — Roland, Jan 10 '17 at 07:08

akrun · Accepted Answer · 2017-01-10T07:06:24.793

4

We can use sub to extract the 'time' component

sub(".*\\s+", "",  y)
#[1] "17:24:00" "17:25:00" "17:26:00"

and if we want a time class, use the times from chron

library(chron)
times(my_data$Time)
#[1] 17:24:00 17:25:00 17:26:00

edited Jan 10 '17 at 07:06

answered Jan 10 '17 at 07:00

akrun

Thank you so much! I think that's what I want to get. the `chron` package. – hepeng Jan 10 '17 at 07:16

Ronak Shah · Answer 2 · 2017-01-10T07:18:24.790

2

We can use format to extract the time component

format(strptime(y, "%H:%M:%S"),"%H:%M:%S")

Another alternative with lubridate package

library(lubridate)
hms(my_data$Time)

edited Jan 10 '17 at 07:18

answered Jan 10 '17 at 07:03

Ronak Shah

With `format` I can get only time listed, with `str()` get `chr`. But `format()` is very helpful for some other part of my code. Thanks! – hepeng Jan 10 '17 at 07:14
@hepeng Try the `hms` function from `lubridate`. – Ronak Shah Jan 10 '17 at 07:19
`hms(date())` doesn't work! I think this suggestion is wrong altogether! – agent18 Feb 27 '19 at 22:41
@ThejKiran works for me when I have `x <- c("17:24:00","17:25:00","17:26:00")` and then I do `lubridate::hms(x)`. – Ronak Shah Mar 05 '19 at 04:49

2 Answers2