[I repost this message to the list since I used the wrong e-mail address...]
Dear Julia,
You can probably simplify Tyler's approach (who is totally right about
the highlighted problems) to the following:
TB <- as.data.frame(read.table("GPS_2009_2010R.csv", header = TRUE, sep
= ','))
TB$Date <- as.POSIXct(strptime(TB$DateTime, "%d.%m.%Y %H:%M:%S", tz =
"GMT"))
TB2 <- TB[!duplicated(paste(TB$Date, TB$Bird)), ]
## You can also store it directly into TB, but that way you can still
## check the original data that remains unchanged
tr <- as.ltraj(TB2[, c("LON","LAT")], TB2$Date, TB2$Bird)
How does that work?
Sincerely,
Mathieu.
Le 14/08/2010 10:31, Tyler Dean Rudolph a écrit :
Hi Julia,
You have two issues here: one is why do you have so many duplicated
observations? This may be in part due to having different birds sampled
at the same moment in time....
*> TB$gmt = TB$gmt[-dupz]*
Error in `$<-.data.frame`(`*tmp*`, "gmt", value = c(1265908114,
1265908416, :
replacement has 22079 rows, data has 22274
The reason you are getting this error is that you cannot reduce the
length of one column alone in a data.frame and keep the others at their
original length. You can, however, remove all associated observations
period:
TB = TB[-dupz,]
This works because every column has the same number of elements, and
since you know which elements of TB$gmt are duplicated you can use that
as an index of which rows you no longer want in the data frame.
But you are probably more interested in identifying duplicated date/time
values within the set of observed values for each unique bird, is that
not correct? In this case there is an extra step, albeit an important one:
# First you split your data frame into a list by unique bird
splitdata = split(TB, TB$Bird, drop=TRUE)
Note the drop=TRUE argument is in case you have levels of a factor with
no observations - which can happen - therefore you choose to drop those
levels (although you can still keep them if you have some compelling
reason, but it may complicate the results).
# Next you identify duplicates within each unique bird data set.
splitdupz = lapply(splitdata, function(birdup)
which(duplicated(birdup$gmt)))
str(splitdupz)
# If you've had a look and you've decided to remove them you can do it
this way
splitdata.nodupz = lapply(splitdata, function(birdup) birdup =
birdup[-which(duplicated(birdup$gmt))),])
# Now you can recompile your data frame with the duplicates from your
unique Bird data sets removed.
TB2 = do.call(rbind, splitdata.nodupz)
I don't have any of your sample data to work with so I can't verify this
has been debugged, but it should give you a pretty good head start.
When a create a ltraj for only one individual (eg. bird 18) it works
fine: tr18<-as.ltraj(xy[TB$Bird=="18",], date=TB$gmt[TB$Bird=="18"],
id="18")
One last thing to note is I'm not sure your id argument in the above
call should be "18"; it may be designed to work that way but I would
prefer to go with TB$Bird[TB$Bird=="18"], which will give you a vector
of values == "18" the length of nrow(TB).
Tyler
On 14/08/2010, at 3:04 , Tyler Dean Rudolph wrote:
I presume your date/time values are of class POSIXct and do include
times (if they don't adehabitat will think multiple observations on a
given day are duplicates)? If that's all good find out which POSIXct
values are repeated:
dupz = which(duplicated(mydates))
mydates = mydates[-dupz]
Tyler
On 2010-08-14, at 8:49, Julia Sommerfeld <
<mailto:[email protected]>
<mailto:[email protected]>[email protected]
<mailto:[email protected]>> wrote:
Hello,
Thank you for yoir fast reply. I have indeed overlapping dates, i.e.
several birds have the same start and end date. But I'm not sure
how to overcome this problem? Couldn't find anything in adehabitat.
Cheers
Julia
On 14/08/2010, at 2:38 , Tyler Dean Rudolph wrote:
your problem seems to be that you have duplicate (i.e. more than
one identical) dates for a given burst. I believe there are
arguments in adehabitat to check for this; if not you will need to
verify this first.
On 2010-08-14, at 8:16, Julia Sommerfeld <
<mailto:[email protected]> <mailto:[email protected]>
<mailto:[email protected]>[email protected]
<mailto:[email protected]>> wrote:
Dear All,
I would like to create an object of class "ltraj" for all GPS
positions (id=Bird) with time, but I always get the following
error message:
*Error in as.ltraj(xy, TB$gmt, id, burst, typeII = TRUE, slsp =
c("remove", : *
* non unique dates for a given burst*
Do I need to create a burst? Could anyone tell me what the exact
difference between "id" and "burst" in as.ltraj is?
When a create a ltraj for only one individual (eg. bird 18) it
works fine: tr18<-as.ltraj(xy[TB$Bird=="18",],
date=TB$gmt[TB$Bird=="18"], id="18")
But it doesn't work for all animals...
My script so far:
library(adehabitat)
library(ade4)
library(gpclib)
library(trip)
library(maps)
library(mapdata)
library(fields)
TB<-as.data.frame(read.table("GPS_2009_2010R.csv",header=T,sep=','))
xy<-TB[,c("LON","LAT")]
id<-TB[,c("Bird")]
### Conversion of the date to the format POSIX
gmt<-(TB$DateTime)
TB$gmt<-as.POSIXct(strptime(gmt, "%d.%m.%Y %H:%M:%S", tz="GMT"))
### Creation of the object of class "ltraj" for GPS positions with
time
tr<-as.ltraj(xy, TB$gmt, id, typeII=TRUE, slsp = c("remove",
"missing"))
My csv.file contains the following columns:
*Bird *(ID)
*DateTime* (date and time in one column)
*LON* (longitude)
*LAT* (latitude
Thank you very much.
Best regards,
Julia
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Julia Sommerfeld - PhD Candidate
Institute for Marine and Antarctic Studies
University of Tasmania
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TAS 7001
Phone: +61 458 247 348
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Julia Sommerfeld - PhD Candidate
Institute for Marine and Antarctic Studies
University of Tasmania
Private Bag 129, Hobart
TAS 7001
Phone: +61 458 247 348
Email: <mailto:[email protected]>[email protected]
<mailto:[email protected]>
<mailto:[email protected]>[email protected]
<mailto:[email protected]>
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