Thank you Shige,I got the Zelig package to work for me. For anyone who cares,
here's the code:
library(Zelig)order(df$WEIGHT)z.out<- zelig(WEIGHT~DISTANCE, model =
"normal.gee",id = "MOTHER", data = df,corstr =
"exchangeable")summary(z.out)date.range<-0:51 # sequence of values over the
range of DIS
On Oct 17, 2011, at 9:45 PM, David Wolfskill wrote:
Sorry about the odd terminology, but I suspect that my intent might be
completely missed had I used "aggregate" or "classify" (each of which
appears to have some rather special meanings in statistical analysis
and
modeling).
I have some da
Hi Hannah,
You are correct that this is not the right place for LaTeX questions.
There are many resources online for TeX users. Here is one page that
may help you: http://en.wikibooks.org/wiki/LaTeX/Floats,_Figures_and_Captions
stackoverflow may be a better place to post questions, there are als
Dear all,
This may not be the right place for latex questions, but I
do not know where else to ask the question.
I have two subfigures. I want the first subfigure to have a caption on
the side of it and the second subfigure
to have a caption below it.
Thank you.
Hannah
[[
Sorry about the odd terminology, but I suspect that my intent might be
completely missed had I used "aggregate" or "classify" (each of which
appears to have some rather special meanings in statistical analysis and
modeling).
I have some data about software builds; one of the characteristics of
eac
I don't think you said how you packaged this stuff up
and that is the critical part of your question.
Here is an example that does what I think you want to do.
It uses save() and load() for the serialization and I
serialize an environent full of data and functions.
In one session of R make some da
Thank you David.
The plot() you sugested worked nicely.
You and Weyland both think ts class objects are difficult to use. I had not
considered doing differently, but I have just gotten many tutorials on package
zoo and how to use it in econometrics. Looks promising, will read everything
now.
On Mon, Oct 17, 2011 at 5:44 PM, Peter Langfelder <
peter.langfel...@gmail.com> wrote:
> On Mon, Oct 17, 2011 at 5:28 PM, Tyler Pirtle wrote:
> > Hi there,
> >
> > I'm trying to do something like a migration of an R program. I've got a
> > function and some variables
> > in an interactive-session
On Mon, Oct 17, 2011 at 5:28 PM, Tyler Pirtle wrote:
> Hi there,
>
> I'm trying to do something like a migration of an R program. I've got a
> function and some variables
> in an interactive-session:
>
> r <- .5
> ## estimatePi references r
> estimatePi <- function(seed) {
> set.seed(seed)
>
Hi there,
I'm trying to do something like a migration of an R program. I've got a
function and some variables
in an interactive-session:
r <- .5
## estimatePi references r
estimatePi <- function(seed) {
set.seed(seed)
numDraws <- 1e+05
x <- runif(numDraws, min = -r, max = r)
y <-
Oh please!
Try reading the relevant docs -- like an Intro to R (The chapter on Reading
data from files) or the R Data Import/Export manual.
You will find plenty of help on this list, but you owe us an honest effort
to help yourself first.
-- Bert
On Mon, Oct 17, 2011 at 11:56 AM, Sally Ann Sims
On Oct 17, 2011, at 5:23 PM, Iara Faria wrote:
Dear R helpers,
I am a beginner at R so please be gentle :)
I have already read manuals and FAQs, with no help.
I have a monthly time series data on public debt with 40 variables,
it starts on January 1994 and ends on June 2011.
I agree with W
I have about 27,000 survey responses from across about 150 Bus Routes, each
with potentially 100 stops. I've recorded the total Ons and Offs for each stop
on each bus run, as well as the stop pair each survey response corresponds to.
I wish to create weights based on the On and Off stop for eac
That worked perfectly, thanks!!
I'd assumed that it wasn't necessary for me to type out "...based on a MV
normal distribution that I made into 1-7 integers by blah blah blah"; the
distribution is normal-ish after I transform it around for the purposes I need
for the simulations. I'll be sure to
I'm not a huge fan of using R's ts class directly, much preferring to
use zoo or xts (both of which are packages) to enhance. That said, I
think your problem comes from identifying the data/variable name
incorrectly. You've only "ts"ed dados, but you try to plot divliq.pib;
try
plot(dados[,"divliq
Thank you David, I'm moving forward again. I was not aware of that website,
I was recently at
http://www.r-project.org/
and clicked on the 'mailing list' link and didn't see it there...with 8k+
posts, maybe its worth adding?
--
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On 18/10/11 10:35, Michael Parent wrote:
Hi, all,
I'm running a monte carlo simulation with missing data. The data are arranged
such that there are k columns and n rows over a set number of simulations (set
to 10 right now so it runs fast while I set everything up). The data are
integers, num
Dear List Members,
I am working on a below piece of code: Initially have created pie
charts with Enlish labels to present the data for Czech Republic, now
however I need to print the label with original Czech fonts.
When I copy paste from Word, certain fonts get 'simplified' distorting
original s
Dear R helpers,
I am a beginner at R so please be gentle :)
I have already read manuals and FAQs, with no help.
I have a monthly time series data on public debt with 40 variables, it starts
on January 1994 and ends on June 2011.
I am loading the data into R using read.csv and the data looks ok
On Oct 17, 2011, at 5:59 PM, emorway wrote:
Hi Carlos,
Thanks for the response. The plot you suggested was not in line
with what
I'm trying to produce in R. The last plot found at:
http://www.advsofteng.com/gallery_contour.html
is more along the lines of what I'm looking for. Notice th
Hi Carlos,
Thanks for the response. The plot you suggested was not in line with what
I'm trying to produce in R. The last plot found at:
http://www.advsofteng.com/gallery_contour.html
is more along the lines of what I'm looking for. Notice the scatter points
(shown by "x") are not laid o
Hello,
One alternative is to create some kind of contourplot/levelplot which will
enhance and show clearly the spikes and will smooth the rest of the
information. That perhaps will be slow in generating the plot but not with
the postscript file.
Regards,
Carlos Ortega
www.qualityexcellence.es
20
On Oct 17, 2011, at 5:11 PM, xhan wrote:
Hi,
I have a list of lists that I would like to convert into a dataframe
such
that the name(?) of the individual lists is replicated as rows with
each
item in the list listed in another column.
I have provided a partial from my list: (let's say th
Hi,
Please check if this representation helps you.
library(lattice)
library(latticeExtra)
p1.gr <- cloud( ts181 ~ x * z, data=FD, panel.3d.cloud = panel.3dbars,
scales = list(arrows = FALSE, just = "right"))
p2.gr <- cloud( ts1825 ~ x * z, data=FD, panel.3d.cloud = panel.3dbars,
scales = list(arr
Hi, all,
I'm running a monte carlo simulation with missing data. The data are arranged
such that there are k columns and n rows over a set number of simulations (set
to 10 right now so it runs fast while I set everything up). The data are
integers, numbers 1-7 only (normal distribution). The si
In the absence of a reproducible example (your example is not
reproducible as is), try this:
names(help.me) <- as.character(2:4)
library('plyr')
newDF <- ldply(help.me, rbind)
newDF[['.id']] <- as.numeric(newDF[['.id']])
ldply will create a new column named .id that contains the name of the
list
Yes, there seem to be a lot of elements in the mlogit model object and I don't
have some of that information now since the models have been estimated in
Biogeme. I had been trying some of the steps that you suggested (after I posted
my question) and as you mentioned, realized that this is danger
Hi,
I have a list of lists that I would like to convert into a dataframe such
that the name(?) of the individual lists is replicated as rows with each
item in the list listed in another column.
I have provided a partial from my list: (let's say the list is called
"help.me")
[[2]]
[1] 18 27 11
Hello
Trying to apply a model to each level of a factor
For example, i have three levels of a variable I call 'Code'...I want to
model the data under each level of code differently...I've attached a sample
data set... http://r.789695.n4.nabble.com/file/n3913431/data.txt data.txt
I.E for cod
The instructions *are* in the manual, and they do work: I checked for
R 2.14.0 beta only this morning.
On Mon, 17 Oct 2011, Nynese Tinsley wrote:
Hello All,
Has anyone had success building 64-bit R on Solaris 10 Sparc with the Oracle
Studio Compiler suite? Could anyone start give me a clue.
I'm not at all familiar with mlogit or Biogene so I can't really give
a definitive answer, but it seems that this would be a very dangerous
thing to try.
Looking at the mlogit manual, I see that, like most model objects in
R, mlogit has a variety of elements including many of those you
specificall
On Mon, Oct 17, 2011 at 4:31 PM, R. Michael Weylandt
wrote:
> I have no idea what's going on in the code you provided, particularly
> why there are no ratios given your subject line or why you are looping
> over an unused variable, but how about this:
>
> n <- colnames(d)
> n <- n[-length(n)] # Th
Dear Mattune,
I guess that you're using the effect() function in the effects package.
You've requested the effect for the covariate fixing the values of the
factors, rather than vice-versa. Try effect("a*b*c", Anocova.Model) or, for
more detail in the output, summary(effect("a*b*c", Anocova.Model)
Hello All,
Has anyone had success building 64-bit R on Solaris 10 Sparc with the Oracle
Studio Compiler suite? Could anyone start give me a clue. I have tried to no
avail.
Thanks,
Nynese
Nynese Tinsley, BSEE, MSCIS
UNIX Systems Analyst
Harvard School of Public Health
Center for Bio
Yes, the qplot works great, but do you know how to allow for multiple
plots? I want one variable to be plotted say from blue to red and
another say from yellow to green but in the same graph, each having
there own separate legends. I've tried print() and arrange() but no
luck.
Thanks again,
kb
On
Hi,
Looking at the fastICA library. I want to test separating out different sounds
from a recorded wav file.
But, I only have a SINGLE stream of data (one channel wav.) It appears as if
the fastICA won't let me separate more sounds than I have columns in my data
matrix. is this correct? If
Hello,
I am trying to run a small example with foreach, but I am having some
problems. Here is the code:
*library(doMC)
registerDoMC()
zappa = list()
frank = list()
foreach (i = 1:4) %dopar% {
zappa[[i]] = kmeans (iris[-5],4)
frank[[i]] = warnings()
}*
The code runs without error. However th
Here's my loadable data in case it helps. It creates 2 separate plots
which I'd like to be in the same graph with 2 separate legends.
library(ggplot2)
#Here's the 1st plot
x<-rnorm(100)
y<-rnorm(100)
z<-rnorm(100)
d <- data.frame(x,y,z)
dg<-qplot(x,y,colour=z,data=d)
dg + scale_colour_gradient(lo
I'm running the following ANCOVA model in R: X=a*b*d + d where d is the
covariate. I want to produce adjusted means for X for all possible
treatment combinations. a has 6 levels, b has 2 levels, c has 2 levels, so
this is a 6x2x2 = 24 possible treatment combinations. When I run:
effect("d", Anco
data.frames are quite efficient when you use
a column at a time, but not when used a row
at a time.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: Nathan Piekielek [mailto:npiekie...@gmail.com]
> Sent: Monday, October 17, 2011 12:37 PM
> To: William D
On Sun, Oct 16, 2011 at 7:20 PM, David Winsemius wrote:
>
> On Oct 16, 2011, at 1:17 AM, Richard M. Heiberger wrote:
>
>> How can I right justify the right-axis tick values? They appear in the
>> example below as left-justified.
>>
>> I have tried several different ways and all fail in different
Your original code works far faster when the input
is a matrix that when it is a data.frame. Selecting
a row from a data.frame is a very slow operation,
selecting a row from a matrix is quick. Modifying
a row or a single element in a data.frame is even
worse compared to do it on a matrix. I comp
You aren't reading the file in at all: notice that,
> summary("MooGoesTheCow",header=TRUE,as.is=TRUE)
Length Class Mode
1 character character
This is because you are asking for a summary of the string containing
the file name, not the file itself.
Rather, use
X = read.csv("C
Thanks for the info. I am working with Recombinant Inbred Lines(RILs), not
pedigree, so most of what I could find does not apply to my work. I've found
a couple of tutorials explaining how to use analysis of variance components
with lme4, but I don't have multiple measurements per individual per
ex
I would assume that you would use 'read.csv'. I don't know where you
got the syntax for "summary";
x <- read.csv(("C:\\Documents\\R_dfiles\\H_N_T.csv",as.is=TRUE)
summary(x)
On Mon, Oct 17, 2011 at 2:56 PM, Sally Ann Sims wrote:
> Hello,
>
> I need some help getting started with data analysis.
Oh, I just saw your ggplot example, that works well too. However, it
seems much of the options are hidden for changing the range of colors
or the color types altogether. I'm currently looking through the
ggplot ref manuals.
Thanks,
kb
On Oct 2, 10:42 pm, Ben Bolker wrote:
> Duncan Murdoch gmail
Thanks, I consider all of those options and tried some, but the
z=color gradient seems the best option for my data.
kb
On Oct 2, 10:42 pm, Ben Bolker wrote:
> Duncan Murdoch gmail.com> writes:
>
>
>
>
>
>
>
>
>
>
>
> > On 11-10-02 1:11 PM, Kerry wrote:
> > > I have 3 columns of data and want to
> where the first column is the chromosome location and the second column is
> some value. What I'd like to do is have a histogram created for each chr
> location (i.e. a separate histogram for chr1, chr2, chr3, chr7, chr9, and
> chr22). I am just having a hard time getting everything to work out
Hello,
I need some help getting started with data analysis. Iâm having trouble
getting R to read my data file. Iâve referred to various R help
documentation, the website, and FAQs, but I donât see my situation listed.
I saved an Excel file (post-2007 Excel version) of data as a â.csvâ
Thanks. bb=replicate(2, rnorm(20)) was very useful for me.
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so here is the code I need help to fix the line for 'st' when it reaches
maximum, that's where i am getting stuck.
thanks
Sharad
set.seed(100)
d=data.frame(x=rnorm(20)+5,
x1=rnorm(20)+5,
x2=rnorm(20)+5,
x3=rnorm(20)+5,
x4=rnorm(20)+5,
x5=rnorm(20)+5,
x6=rnorm(20)+5,
x7=rnorm(20)+5,
x8=rno
On Sun, 16-Oct-2011 at 03:05PM -0700, Benjamin Cheah wrote:
|> Hi all,
[...]
|> It is particularly frustrating when you need to adjust axis title
|> positions separately for x and y axes - i.e. specify mgp twice -
|> once for each axis within the axis function. But this doesn't seem
|> to work -
Hi All,
I have some data from which I set four points to be breaks. Based on these
points, I cut the dataset into four groups and assign a number to it:
<=331.04 assign 0
>331.04 <=476.07 assign data-331.04/(476.07-331.04)
>476.07<=608.66 assign 1
>608.66 <=791.5 assign 791.5- data/(791.5-608.
Hello,
I'm almost positive R can do the following, I just haven't hit upon the
right package or search terms, however. Here's what I'm after: I've got
concentration output from two different models that I want to qualitatively
compare in a contour plot (or some variant of a contour plot). The p
Try vectorizing it a bit by looping over the columns.
E.g.,
f1 <- function (df)
{
# loop (backwards) over all columns in df whose
# names start with "D" to find the earliest one
# that is bigger than column "thold". I tested with
# df being a data.frame but a matrix sh
I think something like this should do it at a huge speed up, though
I'd advise you check it to make sure it does exactly what you want:
there's also nothing to guarantee that something beats the threshold,
so that might make the whole thing fall apart (though I don't think it
will)
# Sample data
d
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Seref Arikan
> Sent: Monday, October 17, 2011 9:11 AM
> To: r-help@r-project.org
> Subject: [R] Best practices for handling very small numbers?
>
> Greetings
> I have been experi
Hello R-community,
I am trying to populate a column (doy) in a large dataset with the first
column number that exceeds the value in another column (thold) using the
'apply' function.
Sample data:
pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177
D193 D209 D225 D241
Hi!
I have fitted a Negative Binomial model (glm.nb) and a Poisson model (glm
family=poisson) to some count data. Both have the same explanatory variables
& dataset
When I call sum(fitted(model.poisson)) for my GLM-Poisson model, I obtain
exactly the same number of counts as my data.
However,
Greetings
I have been experimenting with sampling from posterior distributions using
R. Assume that I have the following observations from a normal distribution,
with an unscaled joint likelihood function:
normsamples = rnorm(1000,8,3)
joint_likelihood = function(observations, mean, sigma){
r
I am trying to run an rlq analysis on some pretty cool data.
The error message I am getting states " Error in rlq(dudiR, dudiL, dudiQ,
scannf = TRUE, nf = 2) : Non equal row weights"
Problem for me is that the row weights are equal; that is if the set-up is
correct.
The contingency tables colum
Hi Everyone,
I am new to R Language and was wondering if someone could help me convert my
latitude-longitude coordinates to cartesian coordiantes using geoXY() from SoDA
package? I have been uploading my coordinates from a text file into R (they
originate as a dataframe) and them converti
On Oct 17, 2011, at 11:13 AM, Cem Girit wrote:
Hello,
I am having hard time obtaining a value from a function. "fit" is a
survival
function that produces some results, such as "median", "confidence
intervals" etc. But str() function does not list these values. How
can I
extract these to b
Often the summary() methods for the modelling functions
contain the kinds of details you are looking for. Sometimes
there are special extractor functions but often not.
You have to study the help file for summary.
and the vignettes for the package, and perhaps do a little
experimentation to figure
There are indeed many, many ways in R to do something like you've
described. If you are just a beginner, I'd recommend something simple:
suppose your data is called df and the columns named are "id", "x'',
and "y" respectively.
with(df, plot(x[id == 1], y[id==1], ylim = range(y), type="l")
# Call
Michael,
Thanks for the quick response. I currently have a few models that have been
estimated using Biogeme. I have attached a sample output file from the
estimation.
I am more concerned about creating a model object from scratch rather than the
reading part. Let us just say if we have a list
Bonjour
I have this data frame and I am newbie in R.
I want to ask if it is possible to draw 10 lines in a plot such that: a line
for every colomn, the x - axis is the second column and the y-axis is the
third one.
Thank you for any input
1 0 1094442
1 0.2 1163576.2
1 0.4 1238539.6
1
Assuming I'm reading your post correctly -- fit is the output object
and the lines following >fit are the console output, which implicitly
calls print -- it looks like the median isn't stored in fit but is
calculated on the fly in the print call. But this doesn't seem
consistent with your claim tha
You're going to have to say more about your text file if you want
meaningful help: specifically, what is in the file: data, output from
other software, etc?
There are probably two questions to deal with here: reading in
whatever is in the text file and mapping it to the correct R object
and settin
Hello,
I am having hard time obtaining a value from a function. "fit" is a survival
function that produces some results, such as "median", "confidence
intervals" etc. But str() function does not list these values. How can I
extract these to be able use them? For example, I need "median" value for
Hello all,
Has anyone tried to create an R mlogit model object from a text file? If yes,
what would be the best way to do it? I already have models that have been
estimated using other software and would like to use R to help me make
predictions for new data.
Thank you!
Bhargava Sana
thanks guys!! your great
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Hi Jason,
I would go for Zelig package to get simulated values and confidence
intervals. It can handle gee model.
Shige
On Mon, Oct 17, 2011 at 9:38 AM, JASON M. HILL wrote:
> Hello Fellow R
> Users,I have
> spent the last week trying to find a work around to this problem and I can't
> seem to
I would like to build a forest of regression trees to see how well some
covariates predict a response variable and to examine the importance of
the
covariates. I have a small number of covariates (8) and large number of
records (27368). The response and all of the covariates are continuous
vari
On Oct 17, 2011, at 8:04 AM, Sarah Goslee wrote:
Hi,
On Sun, Oct 16, 2011 at 6:24 PM, Bob Briggs
wrote:
Hello
I'm wanting to understand more about ptrend (a statistical
explanation via an internet website if possible) and also to know
what package in R would produce a ptrend.
Appr
Hello,
I have simulation results in the form of
Time V I
0.e+000 7.218354344368e-001 5.224478627497e-006
1.e-009 7.218354344368e-001 5.224477718002e-006
2.e-009 7.218354344368e-001 5.224477718
Like others have suggested, I think ggplot2 is probably the best way
to go about this, but if you'd rather use base graphics (and you never
indicated how you felt about ggplot2), you could do something like
this with tapply:
fcts <- letters[sample(9,1500,T)]
vals <- rnorm(1500)
df <- data.frame(ca
Got it. Thanks!
On Mon, Oct 17, 2011 at 9:40 AM, Prof Brian Ripley wrote:
> On Mon, 17 Oct 2011, Brian Smith wrote:
>
> Hi,
>>
>> I had a large file for which I require a subset of rows. Instead of
>> reading
>> it all into memory, I use the awk command to get the relevant rows.
>> However,
>> I
Hello Fellow R
Users,I have
spent the last week trying to find a work around to this problem and I can't
seem to solve it. I simply want to plot my GEE model result with 95% confidence
bands.
I am using the geepack package to run a basic GEE model involving
nestling weights, to a Gaussian distri
Paul Hiemstra knmi.nl> writes:
>
> Hi,
>
> When using ggplot, take a look at facet_wrap and geom_histogram.
>
> regards,
More specifically, try something along the lines of
d <- data.frame(f=factor(paste("chr",rep(c(1,2,3,7,9,22),each=50),sep="")),
v=runif(300))
library(ggplot2)
ggplo
Thanks a lot! Sorry for cross-posting, but I did it intentially
because I tend to believe Barry Rowlingson (Why R-help Must Die!), and
I will summarize the answers here later to StackOverflow.
Another user also told me this worked for 2.13.1, but not later versions.
Regards,
Yihui
--
Yihui Xie
P
Search: "mcmcglmm heritability" to see some discussions using the
mcmcglmm package. This package is discussed often
on the mixed model list. You can also use the kinship package. It will
take some time to get familiar with R. Work through
a few of the examples for variance component models, and the
On Sun, 2011-10-16 at 19:36 -0500, Yihui Xie wrote:
> Hi,
>
> I have spent a few hours on the R-exts manual and the documentation of
> parse_Rd() (as well as the PDF document in the references), but I
> still have not figured out what results=rd means. I thought I could
> use an R code fragment to
rugarch installs and works fine for me under windows 7 64-bit and R 2.13.1
64 bit.
On Mon, Oct 17, 2011 at 7:49 AM, user84 wrote:
> Hi,
>
> i am unable to install the rugarch package.
> More than that i do not even find this package in my list of possible
> packages.
> Its possible than the name
On Mon, 17 Oct 2011, Brian Smith wrote:
Hi,
I had a large file for which I require a subset of rows. Instead of reading
it all into memory, I use the awk command to get the relevant rows. However,
I'm doing it pretty inefficiently as I write the subset to disk, before
reading it into R. Is ther
On Mon, Oct 17, 2011 at 9:23 AM, Brian Smith wrote:
> Hi,
>
> I had a large file for which I require a subset of rows. Instead of reading
> it all into memory, I use the awk command to get the relevant rows. However,
> I'm doing it pretty inefficiently as I write the subset to disk, before
> readi
On Mon, 17 Oct 2011, user84 wrote:
Hi,
i am unable to install the rugarch package.
More than that i do not even find this package in my list of possible
packages.
Its possible than the name has changed, or the package is not longer
availiable?
No: see http://cran.r-project.org/web/packages/ru
Hi,
I had a large file for which I require a subset of rows. Instead of reading
it all into memory, I use the awk command to get the relevant rows. However,
I'm doing it pretty inefficiently as I write the subset to disk, before
reading it into R. Is there a way that I can read it into an R object
Thank you very much for your answer.
It works just as I want!.
Kenneth
El sáb, 15-10-2011 a las 20:39 +0530, Deepayan Sarkar escribió:
> On Fri, Oct 14, 2011 at 10:02 AM, David Winsemius
> wrote:
> > Legends are built in columns. You need to find a graphics symbol to put in
> > the "points" col
Tinn-R also highlights .rd files and is free.
Me, I use TextPad with a latex syntax file.
S Ellison
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Bogaso Christofer
> Sent: 16 October 2011 13:57
> To: r-help@r-project.org
Hi,
i am unable to install the rugarch package.
More than that i do not even find this package in my list of possible
packages.
Its possible than the name has changed, or the package is not longer
availiable?
Is there a similar package avaliable for garch modelling except the fGarch
what i am usin
Adam Zeilinger umn.edu> writes:
> I have a log likelihood function that I was able to optimize using
> mle2. I have two years of the data used to fit the function and I would
> like to fit both years simultaneously to test if the model parameter
> estimates differ between years, using likelih
On Mon, 17 Oct 2011, Annette Prochnow wrote:
Hello,
I have the following problem:
I successfully installed R (Version 2.1.3.0 for Windows) on my Notebook
(Windows 7) in June. Now I used the same setup file for a PC (Windows XP)
and got a message from the anti virus software (Avira), that the s
Hi,
When using ggplot, take a look at facet_wrap and geom_histogram.
regards,
Paul
On 10/17/2011 12:14 PM, Sarah Goslee wrote:
> Hi,
>
> On Mon, Oct 17, 2011 at 8:07 AM, a217 wrote:
>> I have a dataframe in the general format:
>>
>> chr1 0.5
>> chr1 0
>> chr1 0.75
>> chr2 0
>> chr2 0
>> chr3 1
Couldn't you also just count the number of characters with nchar()?
Michael Weylandt
On Oct 17, 2011, at 1:57 AM, Henri-Paul Indiogine wrote:
> I think I have it and my apologies for spamming the list. I should
> not work this late on Sunday :-)
>
> I think that it should be
>
> as.integer(r
Dear R users,
I work with Matlab & R simultaneously. Usually I used to read successfully
mat files (cell, structure, etc.) into R, till Matlab 2011b release.
For reading mat files (or maybe I should say, converting it into R variable
or structure) Im using *readMat* (from R.matlab package)
>> how can i set the position of the first and last tick to the
>> borderline of a plot?
>>
>> The plot should look like this one made in Gnuplot [1].
>> Gnu-R adds some space between the ticks and the end of plot.
>
> do you mean like this?
>
> plot(rnorm(25),rnorm(25), xaxs ="i", yaxs="i", xlim=
Hello,
I have the following problem:
I successfully installed R (Version 2.1.3.0 for Windows) on my Notebook
(Windows 7) in June. Now I used the same setup file for a PC (Windows XP)
and got a message from the anti virus software (Avira), that the setup file
contains the Trojan "TR/ATRAPS.G
Hi,
On Mon, Oct 17, 2011 at 8:07 AM, a217 wrote:
> I have a dataframe in the general format:
>
> chr1 0.5
> chr1 0
> chr1 0.75
> chr2 0
> chr2 0
> chr3 1
> chr3 1
> chr3 0.5
> chr7 0.75
> chr9 1
> chr9 1
> chr22 0.5
> chr22 0.5
Using dput to give us some reproducible data would be even better.
On Sat, 2011-10-15 at 09:11 -0700, lincoln wrote:
> #Uwe:
> #Gavin:
>
> I have read carefully your thread but I am not sure to understand what you
> are suggesting (my gaps in statistics!). You say that it should be due to
> the /Hauck Donner/ effect and that it is not a quasi separation or
> sep
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