calculate mean of period in lubridate

Question

How to calculate mean of period in lubridate ?

library(lubridate)

# tibble
tbl <- tibble(time_1 = ymd_hms(c("2021-01-01 12:00:00"), ("2021-01-01 12:12:36"), ("2021-01-01 14:25:45")),
              time_2 = ymd_hms(c("2021-01-01 12:12:36"), ("2021-01-01 14:25:45"), ("2021-01-01 15:35:45")),
              time_period = seconds_to_period(difftime(time_2, time_1, units = 'sec')))

# calculate mean of period (WRONG ANSWER)
tbl %>% summarise(mean_val = mean(time_period))

Martin Gal · Accepted Answer · 2021-09-06T16:25:31.263

1

The period object doesn't seem to work well with mean. You could use

tbl %>%
  summarise(mean = seconds_to_period(mean(period_to_seconds(time_period))))

to get

# A tibble: 1 x 1
  mean      
  <Period>  
1 1H 11M 55S

To get another representation, you could try

tbl %>%
  summarise(
    mean = seconds_to_period(mean(period_to_seconds(time_period))),
    mean_val = mean(period_to_seconds(time_period)),
    days     = mean_val %/% (60*60*24),
    hours    = mean_val %/% (60*60),
    minutes  = mean_val %/% 60 %% 60 + round((mean_val %% 60)/60),
    value    = paste0(days, "d ", hours, "H ", minutes, "M")
    )

to get

# A tibble: 1 x 6
  mean       mean_val  days hours minutes value    
  <Period>      <dbl> <dbl> <dbl>   <dbl> <chr>    
1 1H 11M 55S     4315     0     1      12 0d 1H 12M

I'm pretty sure, lubridate offers better tools for formating a Period object but I don't have much experience here.

edited Sep 06 '21 at 16:25

answered Sep 06 '21 at 15:43

Martin Gal

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Thanks @Martin Gal, is it possible to round off and remove the seconds i.e. `1H 12M` not `1H 11M 55S` – SiH Sep 06 '21 at 15:51
Do you want it just displayed as `1H 12M` (as a string) or do you need a `Period` object you can work with? – Martin Gal Sep 06 '21 at 16:08
Preferably `period` but string will also work – SiH Sep 06 '21 at 16:10
1

Edited the anwer. – Martin Gal Sep 06 '21 at 16:25

calculate mean of period in lubridate

1 Answers1