Dear R Help,
I believe the glmer() function in lme4 automatically fits an
unstrucruted covariance matirx for the random effects.
Is that true?If so, do I have an option to somehow ask for a
diagonal structured covariance matrix?
Thank you,
Daniel Jeske
Department of Statistics
University of
Definitely out of sequence - it should be
[,1] [,2]
[1,]4 21
[2,]5 22
[3,]6 23
[4,]7 24
[5,]8 25
[6,]9 26
[7,] 10 27
[8,] 11 28
[9,] 12 29
[10,] 13 30
[11,] 14 31
On 16 November 2010 20:12, David Winsemius wrote:
>
> On No
Have you tried filter()?
filter(a, rep(1,7)/7)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ray Brownrigg
> Sent: Tuesday, November 16, 2010 12:05 PM
> To: r-help
Thanks
On 11/16/2010 12:41 PM, Seth Falcon wrote:
> Hi Abhijit,
>
> [I've cc'd R-help to keep the discussion on the list]
>
> On Tue, Nov 16, 2010 at 8:06 AM, Abhijit Dasgupta
> wrote:
>
>> Seth,
>>
>> I was looking for something like this too. I've a question. If
>> you're reading the data f
Can you try it with version 7.16 on R-Forge? Use
install.packages("odfWeave", repos="http://R-Forge.R-project.org";)
to get it.
Thanks,
Max
On Tue, Nov 16, 2010 at 8:26 AM, Søren Højsgaard
wrote:
> Dear Mike,
>
> Good point - thanks. The lines that caused the error mentioned above are
> sim
I am cc:ing the r-sig-mixed-mod...@r-project.org mailing list on this
reply as such questions are often answered more quickly on that list.
On Tue, Nov 16, 2010 at 2:00 PM, Daniel Jeske wrote:
> Dear R Help,
> I believe the glmer() function in lme4 automatically fits an
> unstrucruted covariance
Dear r-help,
I want to use tapply to calculate means for a variable. But there were several
infinite values in the observations.
How can I calculate means not considering these infinite values?
Thanks in advance.
Regards,
Lei
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Hi Lei,
Here is one option relying on is.finite()
## Messy data for means
dat <- data.frame(values = c(rnorm(7), 1:7, c(1, 2, 3, NA, 4, 5, 6),
c(1, 2, Inf, 4, 100, -Inf, NaN)), group = rep(letters[1:4], 7))
## use is.finite() to select for only finite numbers
tapply(dat$values, dat$group, fun
I found the problem.
For some reason, when I converted the list object with the data in it to
numeric, the values changed. This resulted in different clustering
results. Once that was fixed, the clustering was the same.
Thanks for the responses!
On Mon, Nov 15, 2010 at 2:37 PM, Peter Langfeld
Hi dear all,
i have a data (data.frame) which contain y and x coloumn(i.e.
y x
1 0.58545723 0.15113102
2 0.02769361 -0.02172165
3 1.00927527 -1.80072610
4 0.56504053 -1.12236685
5 0.58332337 -1.24263981
6 -1.70257274 0.46238255
7 -0.88501561 0.89484429
8
Hi all,
Say I fit a linear model, and saved it as 'test.lm'
Then if I use plot(test.lm)
it gives me 4 graphs
How do I ask for a 'subset' of it??
say just want the 1st graph,
the residual vs fitted values,
or the 1,3,4th graph?
I think I can use plot(test.lm[c(1,3,4)]) before,
but now, it's
You could try something like this:
Loop through your bootstrapped samples and store which ones have the
outlier you are looking for using code like:
count = c(count, outlier.value %in% boot.sample$outlier.variable)
Then subtract the count variable from the total number of samples to
get th
On Nov 16, 2010, at 5:01 PM, casperyc wrote:
Hi all,
Say I fit a linear model, and saved it as 'test.lm'
Then if I use plot(test.lm)
it gives me 4 graphs
How do I ask for a 'subset' of it??
?plot.lm # The answer is in the first sentence.
say just want the 1st graph,
the residual vs
There may be an easier way to do this, but you could always just do it
the long way.
Ex.
plot(residuals(test.lm)~fitted.values(test.lm))
Andrew Miles
On Nov 16, 2010, at 5:01 PM, casperyc wrote:
Hi all,
Say I fit a linear model, and saved it as 'test.lm'
Then if I use plot(test.lm)
i
Hi,
I did this exact thing for my masters, with intertidal fish, I just used a PCA?
have you tried that?
Sent from my iPhone
On 16 Nov 2010, at 17:01, Mike Gibson wrote:
>
> My objective is to look at differences in two species of fish from
> morphometric measurements. My morphometric me
Hi All -
Doug, thanks for your reply. The context I'm looking at is a Poisson
GLMM with random (intercept,slope) for each subject. The
variance-covariance matrix is 2x2. By unstructured, I meant a 3
parameter matrix (sig1^2,sig2^2,sig12), as compared to a (reduced)
alternative diagonal structur
Dennis and all,
Thank you for the help as I try to get this method for importing and batch
processing files organized. I currently have this set-up to import data from
two files in my working directory. "Var1" specifies data from file 1 and
file 2.
filenames=list.files()
library(plyr)
import
I have a growth curve, which is essentially an ECDF: Statistically,
it's F(x)...
> GrowthCurve
[1] 0.06919932 0.24154761 0.42206402 0.61412408 0.72228295 0.79727292
0.86605315 0.91271120 0.98258397 1.
I'd like to fit a Weibull Curve (then a LogLogistic) to this ECDF, and
have no clue ho
I am trying to understand my population abundance data and am looking into
analyses of change point to try and determine, at approximately what point
do populations begin to change (either decline or increasing).
Can anyone offer suggestions on ways to go about this?
I have looked into bcp and st
thank you very much for your idea,
if i write code as;
my data name is data.
samples<-function(data,num){
resamples<-lapply(1:num,function(i) sample(data,n,replace=TRUE))
list(resamples=resamples)}
>n=10
data<-rnorm(n=10,mean=5,sd=2)
data[1]=100
obj<-samples(data,1000)
i generate 1000 sample, i
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 11/16/2010 03:08 PM, Columbine Caroline Waring wrote:
> Officially I tried:
**A**
>> glm(count~md+ms+rf+sg+offset(log(Eff)), family=poisson,data=DepthHabGen)
>> glm(count~md+ms+rf+sg, offset=(log(Eff)), family=poisson,data=DepthHabGen)
> (which of
First, I would rename your function "samples" to "mksample"
to avoid confusion with the R function "sample".
Next, I would modify the function so that you are returning
a list of samples, instead of a list containing a list of
samples:
mksamples = function(data,num)
lapply(1:num,f
Thanks very much for your help. This was a case of some weird network card IP
reset hardware error that did something to permissions or something during the
install. Not exactly sure if this was part of the problem. But getting the
hardware corrected and reinstalling R 2.12.0 solved the prob
Your outlier has row.names "1". If this is selected in the bootstrap sample
once, it will also have row.names "1". If it is selected more than once the
row.names of the successive entries will begin with "1."
Here is a possibility you may wish to consider.
> txt <- textConnection("
+
Hello there
I have a list, Y, and each component of that list is a real-valued function
(that is, Y[[i]](u) returns a number).
I was wishing to build the mean function and the first thing I thought of
was
Ybar<-function(u){
mean(Y[[1:n]](u))
}
but obviously this doesn't work, since Y[[1:n]]
Try this:
u <- 1:10
mean(sapply(Y, function(f)match.fun(f)(u)))
On Tue, Nov 16, 2010 at 9:00 PM, Eduardo de Oliveira Horta <
eduardo.oliveiraho...@gmail.com> wrote:
> Hello there
>
> I have a list, Y, and each component of that list is a real-valued function
> (that is, Y[[i]](u) returns a numbe
Fencl, Martin eawag.ch> writes:
>
> Helllo,
> I am having trouble with running the library Playwith in the R-2.12.0. running
under 32bit Windows XP.
> After calling the library the error message "The procedure entry point
gdk_cairo_reset_clip could not
> be located in the dynamic library libgdk-
Eduardo -
I'd guess that
Ybar = function(u)mean(sapply(Y,function(fun)fun(u)))
will do what you want, but without a reproducible example,
it's hard to tell.
- Phil Spector
Statistical Computing Facility
It's implemented in the metafor package.
Using the example from the pdf that Marc pointed out:
library(metafor)
ai <- c(53, 121, 95, 103, 64, 7, 0)
bi <- c(2, 3, 14, 27, 51, 29, 13)
ci <- c(61, 152, 114, 66, 81, 28, 0)
di <- c(1, 5, 7, 12
Thank you so much
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Thank you so much
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Thank you both.
casper
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https://stat.ethz.ch/
The documentation for agrep says it uses the Levenshtein edit distance,
but it seems to get this wrong in certain cases when there is a
combination of deletions and substitutions. For example:
> agrep("abcd", "abcxyz", max.distance=1)
[1] 1
That should've been a no-match. The edit distance bet
On Nov 16, 2010, at 6:18 PM, Phil Spector wrote:
Eduardo -
I'd guess that
Ybar = function(u)mean(sapply(Y,function(fun)fun(u)))
I had an example to which this was offered and all it did was make
Ybar a function;
# Polymorphic Fn-object Version 1
Fs <- list(mode="language")
> Fs
$mode
[
Elijah DePalma ucr.edu> writes:
>
> Greetings,
>
> May you please suggest a package or function to use for fitting a GLMM
> (generalized linear mixed model) with spatially correlated random effects?
>
> Thank you,
> Elijah DePalma
Not easy, and I hope you have a lot of data.
Your choices
Hi all,
#
test=data.frame(x=1:26,y=-23.5+0.45*(1:26)+rnorm(26))
rownames(test)=LETTERS[1:26]
attach(test)
#test
test.lm=lm(y~x)
plot(test.lm,2)
identify(test.lm$res,,row.names(test))
# not working
plot(x,y)
identify(x,y,row.names(test))
# works fine
ident
Thanks, guys... but it seems these suggestions won't work.
Let me try to be more specific with a simple example:
Y<-list()
Y[[1]]<-function(u) sqrt(u)
Y[[2]]<-function(u) sin(u)
Y[[3]]<-function(u) 1/2*u
I wanted something equivalent to
Ybar<-function(u){
1/3*(Y[[1]](u) + Y[[2]](u) + Y[[3]](
HI, Dear R community,
I have used the following codes this morning, but this afternoon, I got the
following errors:
> x <- seq(0,10, by=1)
> y <- c(0.952, 0.947, 0.943, 0.941, 0.933, 0.932, 0.939, 0.932, 0.924,
0.918, 0.920) # missense
> z <- c(0.068, 0.082, 0.080, 0.099, 0.108, 0.107, 0.101, 0.1
Eduardo -
Thanks for the reproducible example!
Y<-list()
Y[[1]]<-function(u) sqrt(u)
Y[[2]]<-function(u) sin(u)
Y[[3]]<-function(u) 1/2*u
Ybar = function(u)mean(sapply(Y,function(fun)fun(u)))
Since integrate requires a function which accepts a vector
and returns a vector, we'd need to use Ve
Thank you very much! Works like a charm!
On Tue, Nov 16, 2010 at 10:24 PM, Phil Spector wrote:
> Eduardo -
> Thanks for the reproducible example!
>
>> Y<-list()
>> Y[[1]]<-function(u) sqrt(u)
>> Y[[2]]<-function(u) sin(u)
>> Y[[3]]<-function(u) 1/2*u
>> Ybar = function(u)mean(sapply(Y,function(
Another approach would be
> Y <- list(sqrt, sin, function(u) u/2)
> Ybar <- function(u) rowMeans(sapply(Y, function(fun) fun(u)))
>
> integrate(Ybar, 0, 1)
0.4587882 with absolute error < 5.6e-05
>
i.e. make the function vectorized directly.
Note, however, that if you had
Y[[4]] <- function(
G'day John,
On Tue, 16 Nov 2010 14:02:57 -0500
"Prof. John C Nash" wrote:
> Are the xxxPR routines now deprecated (particularly for 64 bit
> systems) or still OK to use?
They are still OK to use, and I use them occasionally.
> If OK, can anyone point to documentation and examples?
Section
Hi Diana,
Yes, this seems to be a little bug in the setparts function. The
following is a modified version which should work for any x > 0.
You'll see I've just changed a couple of lines...
setparts2 <- function (x)
{
if (length(x) == 1) {
if (x < 1)
stop("if single value, x
On Tue, Nov 16, 2010 at 1:49 PM, Peter Langfelder
wrote:
>
> It is easy to come up with examples where Cov(A, B) + Cov(B, A) is not
> positive definite. As an extreme example, consider a matrix A (say 10
> columns, 100 rows) such that the off-diagonal covariances are all zero
> and the columns are
> Peter,
>
> I see your point. As it turns out though, what I'm trying to
> calculate is heritability using a slightly modified version of an
> equation from multivariate quantitative genetics. Theoretically I
> suppose a heritability matrix could be non-positive definite, but in
> practice it al
Hi, R-folks:
I have been tryin many combination of parameter to make Matern variogram to
work, but I can't find the available one. I'm near to be crazy.
I tiped:
Año2003Selg.lf<-likfit(Año2003Selg,cov.model="matern",ini.cov.pars=c(1.5,14),kappa=2.5,fix.kappa=FALSE,nugget=0.08,lambda=0.008,fix.l
Hello there,
My name is Alireza. I am interested in utilizing R-program for measurement of
earnings management (accounting related issue). However, there are two source
programs which should be used simultaneously with the R-program to be able to
measure earnings management. Previously (i.e. in
I think the problem is with the (2,2) element in your hessian, unless
that is a typo.
Cheers
David Cross
d.cr...@tcu.edu
www.davidcross.us
On Nov 16, 2010, at 8:33 PM, Jimmy Martina wrote:
Hi, R-folks:
I have been tryin many combination of parameter to make Matern
variogram to work,
Fitting curves to an ECDF will result in a fit that has the same precision as
the ECDF if variances are calculated correctly. So why not stop with the
ECDF as your estimator?
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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R-helpers,
I have had difficulty installing the "pcvsuite" package on R version 2.12.0
(2010-10-15). The pcvsuite package is not available on CRAN, but is located
for download at the following website at the University of Washington:
Windows version
http://labs.fhcrc.org/pepe/dabs/pcvsuite_1.
Hi,
I hope this is a simple question. I am having trouble changing the scale of a
secondary y-axis on a barplot. When I run the code below the limits set for the
first axis are always applied to the second axis as well. I am using the latest
R version 2.12.0.
For example, if I have 3 vectors
Ben,
Thank you for your assistance.
Going back to basics and using the data set as you suggested has resulted in a
win.
Set A works!
using +offset(log(variable)) or ,offest=(log(Eff)) is the same as using
exposure(variable) program stata.
I went back and isloated a problem with code be
Hello,
I'm so confused why I can't run Jarque-Bera test on my data. I have 9968
observation and I want to run Jarque-Bera test on them, but no matter how
hard I am trying I can't get it work. please let me know what should I do.
Best,
Kiana
[[alternative HTML version deleted]]
__
Okay here is a solution that works in less than 60 minutes but i feel likes
its messy, if anyone has an alternative solution i would very much
appreciate your insights.
#Create test data
TNode<-c(1:20,21)
FNode<-c(rev(1:20),22)
Volume<-c(rep(100,20),200)
ClassCode=c(rep("Local",20),rep("Freeway
Hello.
First, I'm thankful about your wonderful project.
However, I have serious worries about the reliability of R. I found
the next bug which I consider important because in my job everytime We
work with datanames like next. Please see below:
b=data.frame(matrix(1:9,ncol=3))
names(b)=c("q99
> Date: Tue, 16 Nov 2010 17:39:57 -0800
> From: peter.langfel...@gmail.com
> To: jbass...@cs.gmu.edu
> CC: r-help@r-project.org
> Subject: Re: [R] Non-positive definite cross-covariance matrices
>
> > Peter,
> >
> > I see your point. As it turns out
2010/11/17 Kiana Basiri
> Hello,
> I'm so confused why I can't run Jarque-Bera test on my data. I have 9968
> observation and I want to run Jarque-Bera test on them, but no matter how
> hard I am trying I can't get it work. please let me know what should I do.
>
> Best,
> Kiana
>
>
Did you check
On Nov 16, 2010, at 7:02 PM, José Fernando Zea Castro wrote:
Hello.
First, I'm thankful about your wonderful project.
However, I have serious worries about the reliability of R. I found
the next bug which I consider important because in my job everytime We
work with datanames like next. Pleas
Hi:
Try this:
# Function to generate one sample from the data frame
sampler <- function(df) {
s1 <- sample(nrow(df), 1, replace = FALSE)
s2 <- sample(setdiff(1:nrow(df), s1), 2, replace = FALSE)
list(sample1 = df[s1, grep('^C', names(df))],
sample2 = df[s2, grep('^W', nam
As long as the names are unique, there is not a problem to shorten them.
El mié, 17-11-2010 a las 01:02 +0100, José Fernando Zea Castro escribió:
> Hello.
>
> First, I'm thankful about your wonderful project.
>
> However, I have serious worries about the reliability of R. I found
> the next bug
Here is a better approach that will keep the axis ticks as well on the two axes.
## define a lattice "axis function"
axis.L <-
function(side, ..., line.col)
{
if (side %in% c("bottom", "left")) {
col <- trellis.par.get("axis.text")$col
axis.default(side, ..., line.col = col
I figured it out myself.
Again, Michael and Petr, many thanks to both of you!!! :)
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Hi Nick,
I've used MCMC to fit change point regressions to a variety of
ecological data and prefer this approach to strucchange and similar
because I feel I have more control over the model, ie. I find it
easier to tailor the form of the model to biological / demographic
processes. I also find the
Hi:
The input to jarque.bera.test() is a numeric vector or time series. Try
running the function str() on your input object to see if it is of the
correct type. If you have a vector that is not numeric or a time series
object, you need to convert it to one with something like as.numeric(myvec).
Th
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