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In the question "average time in a column in hr:min:sec format" the following example is given:

Col_Time = c('03:08:20','03:11:30','03:22:18','03:27:39')
library(chron)
mean(times(Col_Time))
[1] 03:17:27

How can I get hr:min:sec as result for the standard deviation? If I use the R function sd, the result looks like that:

sd(times(Col_Time))
[1] 0.006289466
Chris
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2 Answers2

3

sd is operating on the number internally representing the time (days for chron::times, seconds for hms and POSIXct, settable for difftime), which is fine. The only problem is that it is dropping the class from the result so it isn't printed nicely. The solution, then, is just to convert back to the time class afterwards:

x <- c('03:08:20','03:11:30','03:22:18','03:27:39')

chron::times(sd(chron::times(x)))
#> [1] 00:09:03

hms::as.hms(sd(hms::as.hms(x)))
#> 00:09:03.409836

as.POSIXct(sd(as.POSIXct(x, format = '%H:%M:%S')), 
           tz = 'UTC', origin = '1970-01-01')
#> [1] "1970-01-01 00:09:03 UTC"

as.difftime(sd(as.difftime(x, units = 'secs')), 
            units = 'secs')
#> Time difference of 543.4098 secs
alistaire
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1

You can use lubridate package. The hms function will convert time from characters to HMS format. Then use seconds to convert to seconds and calculate mean/sd. Finally, use seconds_to_period to get the result in HMS format.

library(lubridate)
Col_Time = c('03:08:20','03:11:30','03:22:18','03:27:39')

#Get the mean
seconds_to_period(mean(seconds(hms(Col_Time))))
# [1] "3H 17M 26.75S"

#Get the sd
seconds_to_period(sd(seconds(hms(Col_Time))))
#[1] "9M 3.40983612739285S"
MKR
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