Calculate time difference (difftime) between columns of different rows

问题

I have data on 'Start' and 'End' time for different jobs, grouped by 'owner':

Data <- data.frame(
  job = c(1, 2, 3, 4, 5),
  owner = c("name1", "name2", "name1", "name1", "name2"),
  Start = as.POSIXct(c("2015-01-01 15:00:00", "2015-01-01 15:01:00", "2015-01-01 15:13:00", "2015-01-01 15:20:00", "2015-01-01 15:39:02"), format="%Y-%m-%d %H:%M:%S"),
  End =   as.POSIXct(c("2015-01-01 15:11:11", "2015-01-01 15:17:21", "2015-01-01 15:17:00", "2015-01-01 15:31:21", "2015-01-01 15:40:11"), format="%Y-%m-%d %H:%M:%S")
)

For each owner, I want to calculate the idle time between the jobs for each owner, i.e. the difference between the 'End' time of one job and the 'Start' time of the next job.

How do I use difftime() to calculate this time difference between specific rows and times in different columns?

The result should look something like this:

job, owner, idletime
1, name1, NA
2, name2, NA
3, name1, 1.816667  # End of row 1 minus Start of row 3
4, name1, 3.0       # End of row 3 minus Start of row 4
...

回答1:

Here's a possible solution using data.table

library(data.table) # v 1.9.5+
setDT(Data)[, idletime := difftime(Start, shift(End), units = "mins"), by = owner]
#    job owner               Start                 End       idletime
# 1:   1 name1 2015-01-01 15:00:00 2015-01-01 15:11:11        NA mins
# 2:   2 name2 2015-01-01 15:01:00 2015-01-01 15:17:21        NA mins
# 3:   3 name1 2015-01-01 15:13:00 2015-01-01 15:17:00  1.816667 mins
# 4:   4 name1 2015-01-01 15:20:00 2015-01-01 15:31:21  3.000000 mins
# 5:   5 name2 2015-01-01 15:39:02 2015-01-01 15:40:11 21.683333 mins

Or using dplyr

library(dplyr)
Data %>%
  group_by(owner) %>%
  mutate(idletime = difftime(Start, lag(End), units = "mins"))

# Source: local data frame [5 x 5]
# Groups: owner
# 
#   job owner               Start                 End       idletime
# 1   1 name1 2015-01-01 15:00:00 2015-01-01 15:11:11        NA mins
# 2   2 name2 2015-01-01 15:01:00 2015-01-01 15:17:21        NA mins
# 3   3 name1 2015-01-01 15:13:00 2015-01-01 15:17:00  1.816667 mins
# 4   4 name1 2015-01-01 15:20:00 2015-01-01 15:31:21  3.000000 mins
# 5   5 name2 2015-01-01 15:39:02 2015-01-01 15:40:11 21.683333 mins

回答2:

If we are using base R, ave would be one option. We get the lag of the 'End' grouped by 'owner' using ave, use that as the second argument in difftime to create the 'idtime'.

Data$idtime <- with(Data, difftime(Start, ave(End, owner,FUN=lag), units='mins'))

Data
#  job owner               Start                 End         idtime
#1   1 name1 2015-01-01 15:00:00 2015-01-01 15:11:11        NA mins
#2   2 name2 2015-01-01 15:01:00 2015-01-01 15:17:21        NA mins
#3   3 name1 2015-01-01 15:13:00 2015-01-01 15:17:00  1.816667 mins
#4   4 name1 2015-01-01 15:20:00 2015-01-01 15:31:21  3.000000 mins
#5   5 name2 2015-01-01 15:39:02 2015-01-01 15:40:11 21.683333 mins

NOTE: I named the column name as 'idtime' to keep the code in a single line :-)

回答3:

library(dplyr)


Data <- data.frame(
  job = c(1, 2, 3, 4, 5),
  owner = c("name1", "name2", "name1", "name1", "name2"),
  Start = as.POSIXct(c("2015-01-01 15:00:00", "2015-01-01 15:01:00", "2015-01-01 15:13:00", "2015-01-01 15:20:00", "2015-01-01 15:39:02"), format="%Y-%m-%d %H:%M:%S"),
  End =   as.POSIXct(c("2015-01-01 15:11:11", "2015-01-01 15:17:21", "2015-01-01 15:17:00", "2015-01-01 15:31:21", "2015-01-01 15:40:11"), format="%Y-%m-%d %H:%M:%S")
)


Data %>% 
  group_by(owner) %>% 
  arrange(Start) %>% 
  mutate(lagEnd = lag(End),
         idletime = difftime(Start,lagEnd, units="mins")) %>%
  ungroup %>%
  arrange(job) %>%
  select(job,owner,idletime)

#   job owner        idletime
# 1   1 name1         NA mins
# 2   2 name2         NA mins
# 3   3 name1   1.816667 mins
# 4   4 name1   3.000000 mins
# 5   5 name2  21.683333 mins

来源：https://stackoverflow.com/questions/32199444/calculate-time-difference-difftime-between-columns-of-different-rows

标签

datetime

posixct

difftime