问题
I have a specific problem; I have data in the following format:
# USER_ID SUBMISSION_DATE CONTRACT_REF
1 1 20/6 1:00 W001
2 1 20/6 2:00 W002
3 1 20/6 3:30 W003
4 4 20/6 4:00 W004
5 5 20/6 5:00 W005
6 5 20/6 6:00 W006
7 7 20/6 7:00 W007
8 7 20/6 8:00 W008
9 7 20/6 9:00 W009
10 7 20/6 10:00 W0010
Now I need to somehow calculate the time difference between the different submissions (uniquely identifiable).
In other words: I have a table of submissions, in this table, there are all submissions for all users. I need to find a way how to calculate the time difference for each unique STUDENT-CONTRACT tuple between nth assignment and the (n-1)th assignment.
Also note that each new user has to has zero for the new assignment. So the output would look as follows:
# USER_ID SUBMISSION_DATE CONTRACT_REF TIME_DIFFRENCE
1 1 20/6 1:00 W001 0
2 1 20/6 2:00 W002 3600
3 1 20/6 3:30 W003 5400
4 4 20/6 4:00 W004 3600
5 5 20/6 5:00 W005 0
6 5 20/6 6:00 W006 3600
7 7 20/6 7:00 W007 0
8 7 20/6 8:00 W008 3600
9 7 20/6 9:00 W009 3600
10 7 20/6 10:00 W0010 3600
Note that the time may NOT be in seconds, but whatever is suitable.
My thoughts:
1) I presume this will require as.POSIXct somewhere so that R knows how to deal with the time
2) This may involve some package such as plyr, but I am so utterly lost in the documentation and examples are hard to find.
Thank you very much for all responses!
Best, Jakub
回答1:
Here's an attempt. Firstly, get the data:
dat <- read.csv(text="USER_ID,SUBMISSION_DATE,CONTRACT_REF
1,20/6 1:00,W001
1,20/6 2:00,W002
1,20/6 3:30,W003
4,20/6 4:00,W004
5,20/6 5:00,W005
5,20/6 6:00,W006
7,20/6 7:00,W007
7,20/6 8:00,W008
7,20/6 9:00,W009
7,20/6 10:00,W0010",header=TRUE)
Get the number from the contract ref and sort the data
dat$CR_NUM <- as.numeric(gsub("W","",dat$CONTRACT_REF))
dat <- with(dat,dat[order(USER_ID,CR_NUM),])
Convert the date to a POSIXct numeric representation
dat$SD_DATE <- as.numeric(with(dat,as.POSIXct(SUBMISSION_DATE,format="%d/%m %H:%M")))
Calculate a time difference with a 0 at the start using ave
dat$TIME_DIFF <- with(dat, ave(SD_DATE, USER_ID, FUN=function(x) c(0,diff(x)) ))
Result:
# not showing the calculated columns
dat[-c(4:5)]
USER_ID SUBMISSION_DATE CONTRACT_REF TIME_DIFF
1 1 20/6 1:00 W001 0
2 1 20/6 2:00 W002 3600
3 1 20/6 3:30 W003 5400
4 4 20/6 4:00 W004 0
5 5 20/6 5:00 W005 0
6 5 20/6 6:00 W006 3600
7 7 20/6 7:00 W007 0
8 7 20/6 8:00 W008 3600
9 7 20/6 9:00 W009 3600
10 7 20/6 10:00 W0010 3600
回答2:
Here's a slightly tighter version (with fewer "intermediate" columns). Note that using "difftime" rather than "diff" allows you to choose your time units (seconds, minutes, hours, etc.)
dat$DATE2 <- as.POSIXct(dat$SUBMISSION_DATE,format="%d/%m %H:%M")
getDtimes <- function(t) {
if(length(t)>0) c(0,difftime(t[-1], t[-length(t)], units="hours")) else(0)
}
dat$DTime <- unlist(with(dat, tapply(DATE2, USER_ID, getDtimes)))
The key (as above) is to convert times to POSIXt objects. tapply generates a list of the time difference vectors, which you then need to unlist.
来源：https://stackoverflow.com/questions/18092072/how-to-calculate-time-difference-between-datetimes-for-each-group-student-cont