Round off periods to nearest minutes using lubridate

Question

I was wondering if we can round off period to nearest minutes using lubridate. For ex - the value of time_diff is "2H 13M 9S". How can I get round off time_diff to nearest minute to get "2H 13M" (remove seconds)

library(lubridate)

time_1 <- ymd_hms("2021-01-01 12:12:36")
time_2 <- ymd_hms("2021-01-01 14:25:45")

time_diff <- seconds_to_period(difftime(time_2, time_1, units = 'sec'))

Related post: https://stackoverflow.com/q/17108515/680068 – zx8754 Sep 05 '21 at 21:05 — zx8754, Sep 05 '21 at 21:05

score 3 · Answer 1 · answered Sep 05 '21 at 21:05

3

Convert to integer, remove seconds, then convert to period:

x <- as.integer(difftime(time_2, time_1, units = "sec"))
seconds_to_period(x - (x %% 60))
# [1] "2H 13M 0S"

answered Sep 05 '21 at 21:05

zx8754

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Thanks, is it possible to remove the second component i.e. "2H 13M" – SiH Sep 06 '21 at 13:36

score 2 · Answer 2 · answered Sep 05 '21 at 20:49

2

We can convert to POSIXct and then use round_date

library(lubridate)
time_diff1 <- sprintf('%02d:%02d:%02d', time_diff@hour,
     time_diff@minute, second(time_diff))
 format(round_date(as.POSIXct(time_diff1, format = '%H:%M:%S'),
       unit = "minute"), '%H:%M')
[1] "02:13"

answered Sep 05 '21 at 20:49

akrun

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Thanks, is it possible to obtain in this format - "2H 13M", but not as character, I would like to further add/subtract it – SiH Sep 06 '21 at 13:38

Round off periods to nearest minutes using lubridate

2 Answers2

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