I think this does what you want.  It uses 'findInterval' to determine where
a possible match is:

>
myvscan<-data.frame(c(1,NA,1.5),as.POSIXct(c("12:00:00","12:14:00","12:20:00"),
format="%H:%M:%S"))
> # convert to numeric
> names(myvscan)<-c("Latitude","DateTime")
> myvscan$tn <- as.numeric(myvscan$DateTime)  # numeric for findInterval
>
mygarmin<-data.frame(c(20,30,40),as.POSIXct(c("12:00:00","12:10:00","12:15:00"),
format="%H:%M:%S"))
> names(mygarmin)<-c("Latitude","DateTime")
> mygarmin$tn <- as.numeric(mygarmin$DateTime)
>
> # use 'findInterval'
> na.indx <- which(is.na(myvscan$Latitude))  # find NAs
> # replace with garmin latitude
> myvscan$Latitude[na.indx] <-
mygarmin$Latitude[findInterval(myvscan$tn[na.indx], mygarmin$tn)]
>
>
> myvscan
  Latitude            DateTime         tn
1      1.0 2009-05-22 12:00:00 1243008000
2     30.0 2009-05-22 12:14:00 1243008840
3      1.5 2009-05-22 12:20:00 1243009200
>


On Fri, May 22, 2009 at 12:45 AM, Tim Clark <mudiver1200@yahoo.com> wrote:

>
> Dear List,
>
> I need some help in coming up with a function that will take two data sets,
> determine if a value is missing in one, find a value in the second that was
> taken at about the same time, and substitute the second value in for where
> the first should have been.  My problem is from a fish tracking study.  We
> put acoustic tags in fish and track them for several days.  Location data is
> supposed to be automatically recorded every time we detect a "ping" from the
> fish.  Unfortunately the GPS had some problems and sometimes the fishes
> depth was recorded but not its location.  I fortunately had a back-up GPS
> that was taking location data every five minutes.  I would like to merge the
> two files, replacing the missing value in the vscan (automatic) file with
> the location from the garmin file.  Since we were getting vscan records
> every 1-2 seconds and garmin records every 5 minutes, I need to find the
> right place in the vscan file to place the garmin record - i.e. the
>  closest in time, but not greater than 5 minutes.  I have written a
> function that does this. However, it works with my test data but locks up my
> computer with my real data.  I have several million vscan records and
> several thousand garmin records.  Is there a better way to do this?
>
>
> My function and test data:
>
> myvscan<-data.frame(c(1,NA,1.5),times(c("12:00:00","12:14:00","12:20:00")))
> names(myvscan)<-c("Latitude","DateTime")
> mygarmin<-data.frame(c(20,30,40),times(("12:00:00","12:10:00","12:15:00")))
> names(mygarmin)<-c("Latitude","DateTime")
>
> minute.diff<-1/24/12   #Time diff is in days, so this is 5 minutes
> for (k in 1:nrow(myvscan))
> {
> if (is.na(myvscan$Latitude[k]))
> {
> if ((min(abs(mygarmin$DateTime-myvscan$DateTime[k]))) < minute.diff )
> {
> index.min.date<-which.min(abs(mygarmin$DateTime-myvscan$DateTime[k]))
> myvscan$Latitude[k]<-mygarmin$Latitude[index.min.date]
> }}}
>
> I appreciate your help and advice.
>
> Aloha,
>
> Tim
>
>
>
>
> Tim Clark
> Department of Zoology
> University of Hawaii
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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