try this:
sapply(estIID50, "[[", 'para')
Best,
Dimitris
On 6/27/2012 2:17 PM, Al Ehan wrote:
Sorry, one more simple question. how do I pick, from the generated
lapply, the $para for each X. say here I have 2 generated list from
previous function. how do I take only the $para for each X?
Tha
Sorry, one more simple question. how do I pick, from the generated lapply,
the $para for each X. say here I have 2 generated list from previous
function. how do I take only the $para for each X?
Thank you so much for your kindness.
$X1
$X1$type
[1] "gev"
$X1$para
xi alpha ka
It's working I am excited than ever!!
Thank you so much
On Wed, Jun 27, 2012 at 12:45 PM, Dimitris Rizopoulos <
d.rizopou...@erasmusmc.nl> wrote:
> 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 someth
Hello,
Try the following.
estIID50 <- apply( IID50, 2, function(x) pargev(lmom.ub(x)) )
And see ?apply
Hope this helps,
Rui Barradas
Em 27-06-2012 12:31, Al Ehan escreveu:
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 findin
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
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