Calculating cumulative time in R

Question

I have a dataframe that looks like this:

     POI   LOCAL.DATETIME
1    1     2017-07-11 15:02:13
2    1     2017-07-11 15:20:28
3    2     2017-07-11 15:20:31
4    2     2017-07-11 15:21:13
5    3     2017-07-11 15:21:18
6    3     2017-07-11 15:21:21
7    2     2017-07-11 15:21:25
8    2     2017-07-11 15:21:59
9    1     2017-07-11 15:22:02
10   1     2017-07-11 15:22:05

I want to be able to calculate (probably with lubridate) the cumulative time spent at each POI and combine them into a table that looks something like this:

     POI   TOTAL.TIME
1    1     00:18:18
2    2     00:01:11
3    3     00:00:03

Also, I am not sure how to deal with the time between POIs, like the 3 seconds between rows 2 and 3. I think maybe I need to calculate the time from row 1 to row 3 instead of row 1 to row 2.

Pierre Lapointe · Accepted Answer · 2017-07-13T20:24:21.363

To get the total time in each group's periods, you first need to create a group index. I'm using rleid from data.table You can then, calculate the total time spent in each of these groups, and then summarise by the initial POI using sum.

df <- read.table(text="     POI   LOCAL.DATETIME
1     '2017-07-11 15:02:13'
1     '2017-07-11 15:20:28'
2     '2017-07-11 15:20:31'
2     '2017-07-11 15:21:13'
3     '2017-07-11 15:21:18'
3     '2017-07-11 15:21:21'
2     '2017-07-11 15:21:25'
2     '2017-07-11 15:21:59'
1     '2017-07-11 15:22:02'
1     '2017-07-11 15:22:05'",
                 header=TRUE,stringsAsFactors=FALSE)
df$LOCAL.DATETIME <- as.POSIXct(df$LOCAL.DATETIME)

library(dplyr)
df%>%
  mutate(grp=data.table::rleid(POI))%>%
  group_by(grp)%>%
  summarise(POI=max(POI),TOTAL.TIME=difftime(max(LOCAL.DATETIME),
                                     min(LOCAL.DATETIME),units="secs"))%>%
  group_by(POI)%>%
  summarise(TOTAL.TIME=sum(TOTAL.TIME))

# A tibble: 3 × 2
    POI TOTAL.TIME
  <int>     <time>
1     1  1098 secs
2     2    76 secs
3     3     3 secs

To get minute and seconds, you can use as.period from lubridate:

library(lubridate)
df%>%
  mutate(grp=data.table::rleid(POI))%>%
  group_by(grp)%>%
  summarise(POI=max(POI),TOTAL.TIME=difftime(max(LOCAL.DATETIME),
                                    min(LOCAL.DATETIME),units="secs"))%>%
  group_by(POI)%>%
  summarise(TOTAL.TIME=sum(TOTAL.TIME))%>%
  mutate(TOTAL.TIME =as.period((TOTAL.TIME), unit = "sec"))

    POI   TOTAL.TIME
  <int> <S4: Period>
1     1      18M 18S
2     2       1M 16S
3     3           3S

Isn't this method going to take the row 10 - row 1, row 8 - row 3, and row 6 - row5? I need to be able to do (row 2 - row 1) + (row 10 - row 9), etc. — Sam, Jul 13 '17 at 19:58
@P Lapointe Thank you so much, this is exactly what I was looking for. — Sam, Jul 14 '17 at 12:16

score 1 · Answer 2 · answered Jul 13 '17 at 20:29

Another data.table option is to create groupings of 2 rows for each POI, take the time difference between them, and finally sum it up by POI:

library(data.table)

dt <- as.data.table(df)
dt[, grp2 := (seq_len(.N)+1) %/% 2, by = POI]
dt[, time_diff := difftime(LOCAL.DATETIME, shift(LOCAL.DATETIME), unit = "min"), by = .(POI, grp2)]
dt[ , .(TOTAL.TIME = sum(time_diff, na.rm = T)), by = POI]

#   POI     TOTAL.TIME
#1:   1 18.300000 mins
#2:   2  1.266667 mins
#3:   3  0.050000 mins

Calculating cumulative time in R

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