you will have to check what function pargev() returns as a result. I
would guess that is probably a list. In any case, you could use
something like the following:
estIID50 <- lapply(IID50, function (m) pargev(lmom.ub(m)))
I hope it helps.
Best,
Dimitris
On 6/27/2012 1:31 PM, Al Ehan wrote:
Hi R-users,
I'm trying to repeat the same procedure to 1000 data set. I know this is
very easy, but I got stuck finding the right and fastest way in running it.
IID50=Riidf[1:50,1:1000] #where IID50 is a dataframe consist of 1000 time
series(as column) and 50 time scales (row).
#what I tried to do:
estIID50=rep(NA,1000)
for (i in 1:1000)
estIID50[i]=pargev(lmom.ub(IID50[1:50,i]))
#warning message
In estIID50[i] = pargev(lmom.ub(IID50[1:50, i])) :
number of items to replace is not a multiple of replacement length
#pargev is a function from lmomco package. I would like to apply it to the
1000 set of time series that I have in the IID50, without having to do it
manually.
#I dont understand what is the warning warns about
Can somebody help me?
[[alternative HTML version deleted]]
______________________________________________
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
and provide commented, minimal, self-contained, reproducible code.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/
______________________________________________
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
and provide commented, minimal, self-contained, reproducible code.
