AIC is only defined up to an additive constant (as is log-likelihood).
It should not surprise you that the values for AIC differ between packages.
The real question is whether the change in AIC when going form one model to
anoth is the same. If not, one is wrong (at least).
-Original
Hi list,
I have a 1710x244 matrix of numerical values and I would like to calculate the
mean of every group of three consecutive values per column to obtain a new
matrix of 570x244. I could get it done using a for loop but how can I do that
using apply functions?
In addition to this, do I
On 15.10.2010 09:19, Santosh Srinivas wrote:
I'm still getting familiar with lapply
I have this date sequence
x- seq(as.Date(01-Jan-2010,format=%d-%b-%Y), Sys.Date(), by=1) #to
generate series of dates
I want to apply the function for all values of x . so I use lapply (Still a
newbie!)
I
Hi:
Look into the rollmean() function in package zoo.
HTH,
Dennis
On Fri, Oct 15, 2010 at 12:34 AM, David A. dasol...@hotmail.com wrote:
Hi list,
I have a 1710x244 matrix of numerical values and I would like to calculate
the mean of every group of three consecutive values per column to
one efficient way to do this, avoiding loops, is using the rowsum()
function, e.g.,
mat - matrix(rnorm(1710*244), 1710, 244)
id - rep(seq_len(570), each = 3)
means - rowsum(mat, id, FALSE) / 3
Regarding the second part of your question, indeed a gold rule of
efficient R programming when it
Hello Dennis,
That's a very good suggestion. I've attached a template here as a .png file,
I hope you can view it. This is what I've managed to achieve in S-Plus (we
use S-Plus at work but I also use R because there's some very good R
packages for PK data that I want to take advantage of that is
Dear Sir/Madam;
I'm not sure whether this is the correct contact for help.
I've been recently working with R on my project, unfortunately It sudenly
crashes!
It gives me the following message:
FATAL ERROR: unable to restore saved data in .RDATA
I decided to uninstall the copy (a R2.11.0) and
Thx.
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: 15 October 2010 13:11
To: Santosh Srinivas
Cc: r-help@r-project.org
Subject: Re: [R] Downloading file with lapply
On 15.10.2010 09:19, Santosh Srinivas wrote:
I'm still getting familiar with
orduek wrote:
Is there an option to import .sta files to R?
I know you can import SPSS ones.
thank you.
Hi:
I have the same problem.
Did you manage in the end to import .sta files to R?
thank you
--
View this message in context:
Hi
I want to set a variable to either 1 or 0 depending on an value of a
dataframe and then add this as a colum to the dataframe.
This could be done with a loop but as we are able to do questions on a
complete row or colum without a loop it would be sweet if it could be done.
for example:
On Fri, Oct 15, 2010 at 12:23 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi,
You might look at Reduce(). It seems faster. I converted the matrix
to a list in an incredibly sloppy way (which you should not emulate)
because I cannot think of the simple way.
Dennis provided the answer:
Dear colleagues,
I would like to ask you how to estimate
biweight M-estimator of Tukey with known
scale example.
I know how to estimate biweight M-estimator
if estimated scale is used using function
rml in package MASS.
library(MASS)
x-rnorm(1000)
rlm(x~1,psi=psi.bisquare)
But I would like to
Dear Joel,
On Fri, Oct 15, 2010 at 1:16 AM, Joel joda2...@student.uu.se wrote:
Hi
I want to set a variable to either 1 or 0 depending on an value of a
dataframe and then add this as a colum to the dataframe.
This could be done with a loop but as we are able to do questions on a
complete
try this:
table$VoteRight - as.numeric(table$age 18)
Best,
Dimitris
On 10/15/2010 10:16 AM, Joel wrote:
Hi
I want to set a variable to either 1 or 0 depending on an value of a
dataframe and then add this as a colum to the dataframe.
This could be done with a loop but as we are able to
Indeed I was close :)
Thx for the fast respond!
Have a good day
//Joel
--
View this message in context:
http://r.789695.n4.nabble.com/Set-value-if-else-tp2996667p2996682.html
Sent from the R help mailing list archive at Nabble.com.
__
On Oct 15, 2010, at 3:46 AM, Anh Nguyen wrote:
Hello Dennis,
That's a very good suggestion. I've attached a template here as
a .png file,
I hope you can view it. This is what I've managed to achieve in S-
Plus (we
use S-Plus at work but I also use R because there's some very good R
x - data.frame(x=1:10)
require(gtools)
x$y - ifelse(odd(x$x),0,1)
HTH
R.
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
-Mensaje
Hi,
For this example:
O - c(0 0 0 2 0 0 2 0)
I want to create an array every time O[i] 0. The array should be in the
form;
R[j] - array(-1, dim=c(2,O[i]))
i.e. if O[i] 0 4 times I want 4 R arrays.
Does anyone have any suggestions?
Thanks,
Doug
--
View this message in context:
Hi Dennis and Dimitris,
thanks for your answers. I am trying the rollmean() function and also the
rollapply() function because I also want to calculate CV for the values.
For this I created a co.var() function. I am having problems using them.
co.var-function(x)(
+sd(x)/mean(x)
+)
On Fri, Oct 15, 2010 at 9:55 AM, dpender d.pen...@civil.gla.ac.uk wrote:
Hi,
For this example:
O - c(0 0 0 2 0 0 2 0)
I want to create an array every time O[i] 0. The array should be in the
form;
R[j] - array(-1, dim=c(2,O[i]))
i.e. if O[i] 0 4 times I want 4 R arrays.
Does
Hi, Doug,
maybe
columns - c( 0, 3, 0, 2, 0, 1)
lapply( columns[ columns 0],
function( o) array( -1, dim = c( 2, o)))
does what you want?
Regards -- Gerrit
-
AOR Dr. Gerrit Eichner Mathematical
Dear List,
I am doing some simulation in R and need basic help!
I have a list of animal families for which i know the number of species in each
family.
I am working under the assumption that a species has a 7.48% chance of being at
risk.
I want to simulate the number of species expected to
Have a look at the package smoothmest.
Christian
On Fri, 15 Oct 2010, Ondrej Vozar wrote:
Dear colleagues,
I would like to ask you how to estimate
biweight M-estimator of Tukey with known
scale example.
I know how to estimate biweight M-estimator
if estimated scale is used using function
hello,
i was shortly asking the list for help with some interaction contrasts (see
below) for which
i had to change the reference level of the model on the fly (i read a post
that this is possible in
multcomp).
if someone has a clue how this is coded in multcomp; glht() - please point
me
Hi:
To get the plots precisely as you have given them in your png file, you're
most likely going to have to use base graphics, especially if you want a
separate legend in each panel. Packages ggplot2 and lattice have more
structured ways of constructing such graphs, so you give up a bit of
Barry, Gerrit,
That was what I am after but unfortunately only the starting point. I am
now trying to amend a function that inserts the R matrices into a dataset in
the correct places:
i.e.
H - c(0.88, 0.72, 0.89, 0.93, 1.23, 0.86, 0.98, 0.85, 1.23)
T - c(7.14, 7.14, 7.49, 8.14, 7.14, 7.32,
I've rolled up R-2.12.0.tar.gz a short while ago. This is a development
release which contains a number of new features.
Also, a number of mostly minor bugs have been fixed. See the full list
of changes below.
You can get it from
http://cran.r-project.org/src/base/R-2/R-2.12.0.tar.gz
or wait
Hi
r-help-boun...@r-project.org napsal dne 14.10.2010 10:34:12:
Thanks Dennis.
One more thing if you don't mind. How to I abstract the individual H
and T
“arrays” from f(m,o,l) so as I can combine them with a date/time array
and
write to a file?
Try to look at ?merge function
Thanks for the advice Gabor,
I was indeed not starting and finishing with sqldf(). Which was why it was
not working for me. Please forgive a blatantly obvious mistake.
I have tried what U suggested and unfortunately R is still having problems
doing the join. The problem seems to be one of
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