Finding a vectorized way to execute for-loops with cross-line calculation

Question

I am trying to find a vectorial procedure that can replace the following code (which takes a long time to run):

for (i in 2:nrow(z)) { if (z$customerID[i]==z$customerID[i-1]) {z$timeDelta[i]<-(z$time[i]-z$time[i-1])} else {z$timeDelta[i]<- NA} } 

I tried looking for different application fragments, but did not find anything useful.

Here are some sample data:

 customerID time 1 2013-04-17 15:30:00 IDT 1 2013-05-19 11:32:00 IDT 1 2013-05-20 10:14:00 IDT 2 2013-03-14 18:41:00 IST 2 2013-04-24 09:52:00 IDT 2 2013-04-24 17:08:00 IDT 

And I want to get the following output:

 customerID time timeDelta* 1 2013-04-17 15:30:00 IDT NA 1 2013-05-19 11:32:00 IDT 31.83 1 2013-05-20 10:14:00 IDT 0.94 2 2013-03-14 18:41:00 IST NA 2 2013-04-24 09:52:00 IDT 40.59 2 2013-04-24 17:08:00 IDT 0.3 *I prefer the time will be in days 

Answer 1

 z$timeDelta <- NA z$timeDelta[-1] <- ifelse(tail(z$customerID,-1) == head(z$customerID,-1), diff(z$time)/24, NA) 

or shorter version

 z$timeDelta <- NA z$timeDelta[-1] <- ifelse(!diff(z$customerID), diff(z$time)/24, NA) 

Answer 2

This should work for you:

 do.call(rbind,lapply(split(mydf,mydf$customerID), function(df) within(df,timeDelta<-c(NA,diff(time)/24)))) 

Result:

  customerID time timeDelta 1.1 1 2013-04-17 15:30:00 NA 1.2 1 2013-05-19 11:32:00 31.8347222 1.3 1 2013-05-20 10:14:00 0.9458333 2.4 2 2013-03-14 18:41:00 NA 2.5 2 2013-04-24 09:52:00 40.5909722 2.6 2 2013-04-24 17:08:00 0.3027778 

Answer 3

It works:

 ## z <- read.table(text="customerID time ## 1 2013-04-17.15:30:00.IDT ## 1 2013-05-19.11:32:00.IDT ## 1 2013-05-20.10:14:00.IDT ## 2 2013-03-14.18:41:00.IST ## 2 2013-04-24.09:52:00.IDT ## 2 2013-04-24.17:08:00.IDT", header=TRUE) ## ## mydf$time <- z$time <- as.POSIXlt(gsub("\\.", " ", z$time)) do.call(rbind, lapply(split(z, z$customerID), function(x) { x$timeDelta <- c(NA, round(as.numeric(diff(x$time), units = "days"), 2)) x })) ## customerID time timeDelta ## 1.1 1 2013-04-17 15:30:00 NA ## 1.2 1 2013-05-19 11:32:00 31.83 ## 1.3 1 2013-05-20 10:14:00 0.95 ## 2.4 2 2013-03-14 18:41:00 NA ## 2.5 2 2013-04-24 09:52:00 40.63 ## 2.6 2 2013-04-24 17:08:00 0.30 

Answer 4

Using firstobs from doBy package:

 z$timeDelta <- c(NA, diff(z$time)) z$timeDelta[firstobs(z$customerID)] <- NA