Convert start time and total duration to elapsed time per hour - r

Question

I have data on start time ('startTime', a date-time variable, POSIXct) and duration in minutes ('duration_minutes'):
df <- data.frame(id = c(1, 2, 3),
startTime = as.POSIXct(c("2018-01-01 12:15:31",
"2018-01-02 23:43:00",
"2018-01-03 11:00:11")),
duration_minutes = c(315, 120, 45))
I want to convert the start time and duration to elapsed time per hour, for each hour, from the hour of the start time to the last hour at the end of the duration:
df_result <- data.frame(id = c(1, 1, 1, 1, 1, 1, 2, 2, 2, 3),
startTime = c("2018-01-01 12:15:31","2018-01-01 13:00:00",
"2018-01-01 14:00:00","2018-01-01 15:00:00",
"2018-01-01 16:00:00","2018-01-01 17:00:00",
"2018-01-02 23:43:00","2018-01-03 00:00:00",
"2018-01-03 01:00:00",
"2018-01-03 11:00:11"),
duration_minutes = c(44.48, 60, 60, 60, 60, 30.5, 17, 60, 43, 45))
Please, advice with the possible solution.

Answer 1

Another possibility:
library(data.table)
library(lubridate)
setDT(df)
df[ , ceil_start := ceiling_date(start, "hour", change_on_boundary = TRUE)]
df[ , {
if(difftime(ceil_start, start, units = "min") > dur) {
.SD[ , .(start, dur)]
} else {
end <- start + dur * 60
time <- c(start,
seq(from = ceil_start,
to = floor_date(end, "hour"),
by = "hour"),
end)
.(start = head(time, -1), dur = `units<-`(diff(time), "mins"))
}
},
by = id]
# id start dur
# 1: 1 2018-01-01 12:15:31 44.48333 mins
# 2: 1 2018-01-01 13:00:00 60.00000 mins
# 3: 1 2018-01-01 14:00:00 60.00000 mins
# 4: 1 2018-01-01 15:00:00 60.00000 mins
# 5: 1 2018-01-01 16:00:00 60.00000 mins
# 6: 1 2018-01-01 17:00:00 30.51667 mins
# 7: 2 2018-01-02 23:43:00 17.00000 mins
# 8: 2 2018-01-03 00:00:00 60.00000 mins
# 9: 2 2018-01-03 01:00:00 43.00000 mins
# 10: 3 2018-01-03 11:00:11 45.00000 mins
# 11: 4 2018-01-03 11:35:00 25.00000 mins
# 12: 4 2018-01-03 12:00:00 10.00000 mins
# 13: 5 2018-01-03 00:00:00 60.00000 mins
# 14: 5 2018-01-03 01:00:00 0.00000 mins
Explanation
Convert data.frame to data.table (setDT). Round up start times to nearest hour (ceiling_date(start, "hour", ...). Use change_on_boundary = TRUE for easier handling of times without minutes and seconds (not in the data, but tested).
To handle cases when the end time (start + duration) is in the same hour as the start time (e.g. id = 3), check if difference between rounded time and start time is larger than duration (if(difftime(ceil_start, start, units = "min") > dur))). If so, just select the start and duration columns (.SD[ , .(start, dur)).
For other cases (else), calculate end time: end <- start + dur * 60. Create a sequence from the up-rounded start time ('ceil_start'), to the down-rounded end time, with an hourly increment (seq(from = ceil_start, to = floor_date(end, "hour"), by = "hour")). Concatenate with 'start' and 'end' times. Return all times except the last (head(time, -1) and calculate difference between time steps in minutes (`units<-`(diff(time), "mins")).
For times with H:M:S = 00:00:00 and duration is a multiple of 60 min, like id = 5, the current solution gives a row with a duration of 0 minutes for the last hour. While waiting for a more elegant solution, a quick and dirty way is just to delete such rows with duration = 0.
Data
Please note that I have added a case not included in original data, id = 4 (see also my comment above) and id = 5.
df <- data.frame(id = 1:5,
start = as.POSIXct(c("2018-01-01 12:15:31",
"2018-01-02 23:43:00",
"2018-01-03 11:00:11",
"2018-01-03 11:35:00",
"2018-01-03 00:00:00")),
dur = c(315, 120, 45, 35, 60))

Answer 2

Try this:
library(data.table)
library(lubridate)
library(magrittr)
df <-
setDT(df)[, start_ceiling := ceiling_date(startTime, "hour", change_on_boundary = TRUE)] %>%
.[, `:=` (
reps = ifelse(
startTime + (duration_minutes * 60) <= start_ceiling, 1, pmax(2, floor(duration_minutes / 60) + 1)
),
initial_diff = as.numeric(difftime(start_ceiling[1], startTime[1], units = "mins"))
), by = id] %>%
.[, df[df[, rep(.I, reps)]]] %>%
.[, startTime := pmax(startTime, floor_date(startTime, "hour") + hours(0:(.N - 1))), by = id] %>%
.[reps > 1, duration_minutes := c(initial_diff[.N],
rep(60, reps[.N] - 2),
(duration_minutes[.N] - initial_diff[.N]) %% 60), by = id] %>%
.[!(duration_minutes == 0 & reps > 1), ] %>%
.[, c("reps", "start_ceiling", "initial_diff") := NULL]
I've tested this with all the scenarios we've gathered so far, and this is the output:
id startTime duration_minutes
1: 1 2018-01-01 12:15:31 44.48333
2: 1 2018-01-01 13:00:00 60.00000
3: 1 2018-01-01 14:00:00 60.00000
4: 1 2018-01-01 15:00:00 60.00000
5: 1 2018-01-01 16:00:00 60.00000
6: 1 2018-01-01 17:00:00 30.51667
7: 2 2018-01-02 23:43:00 17.00000
8: 2 2018-01-03 00:00:00 60.00000
9: 2 2018-01-03 01:00:00 43.00000
10: 3 2018-01-03 11:00:11 45.00000
11: 4 2018-01-04 10:00:00 60.00000
12: 4 2018-01-04 11:00:00 5.00000
13: 5 2018-01-05 00:00:00 60.00000
14: 6 2018-01-06 11:35:00 25.00000
15: 6 2018-01-06 12:00:00 10.00000
16: 7 2018-01-07 00:00:00 60.00000
17: 7 2018-01-07 01:00:00 60.00000
Data used:
df <- data.frame(
id = c(1, 2, 3, 4, 5, 6, 7),
startTime = as.POSIXct(
c(
"2018-01-01 12:15:31",
"2018-01-02 23:43:00",
"2018-01-03 11:00:11",
"2018-01-04 10:00:00",
"2018-01-05 00:00:00",
"2018-01-06 11:35:00",
"2018-01-07 00:00:00"
)
),
duration_minutes = c(315, 120, 45, 65, 60, 35, 120)
)
df
id startTime duration_minutes
1 1 2018-01-01 12:15:31 315
2 2 2018-01-02 23:43:00 120
3 3 2018-01-03 11:00:11 45
4 4 2018-01-04 10:00:00 65
5 5 2018-01-05 00:00:00 60
6 6 2018-01-06 11:35:00 35
7 7 2018-01-07 00:00:00 120