Get the time between consecutive dates stored in a single column - r

Question

I am trying to figure out how to get the time between consecutive events when events are stored as a column of dates in a dataframe.
sampledf=structure(list(cust = c(1L, 1L, 1L, 1L), date = structure(c(9862,
9879, 10075, 10207), class = "Date")), .Names = c("cust", "date"
), row.names = c(NA, -4L), class = "data.frame")
I can get an answer with
as.numeric(rev(rev(difftime(c(sampledf$date[-1],0),sampledf$date))[-1]))
# [1] 17 196 132
but it is really ugly. Among other things, I only know how to exclude the first item in a vector, but not the last so I have to rev() twice to drop the last value.
Is there a better way?
By the way, I will use ddply to do this to a larger set of data for each cust id, so the solution would need to work with ddply.
library(plyr)
ddply(sampledf,
c("cust"),
summarize,
daysBetween = as.numeric(rev(rev(difftime(c(date[-1],0),date))[-1]))
)
Thank you!

Answer 1

Are you looking for this?
as.numeric(diff(sampledf$date))
# [1] 17 196 132
To remove the last element, use head:
head(as.numeric(diff(sampledf$date)), -1)
# [1] 17 196
require(plyr)
ddply(sampledf, .(cust), summarise, daysBetween = as.numeric(diff(date)))
# cust daysBetween
# 1 1 17
# 2 1 196
# 3 1 132

Answer 2

You can just use diff.
as.numeric(diff(sampledf$date))
To leave off the last, element, you can do:
[-length(vec)] #where `vec` is your vector
In this case I don't think you need to leave anything off though, because diff is already one element shorter:
test <- ddply(sampledf,
c("cust"),
summarize,
daysBetween = as.numeric(diff(sampledf$date)
))
test
# cust daysBetween
#1 1 17
#2 1 196
#3 1 132