Greetings R community,
I am making my first package and have run into the need to use other packages.
I pass all the checks in the command prompt running Rcmd check package.name.
In the Description file I have included:
Depends: R (>= 2.13), plotrix
Repository: CRAN
Now I create the zip
Hi:
Using Michael's example data, here's another approach:
x = data.frame(x = 1:5)
y = data.frame(y = 1:10)
z = data.frame(z = 1:3)
# Generate a vector to name the list components
nms <- c('x', 'y', 'z')
# Combine data frames into a list
L <- list(x, y, z)
# name them
names(L) <- nms
# Use lappl
Hello,
I got an error message saying
Error in lognet(x, is.sparse, ix, jx, y, weights, offset, alpha, nobs, :
NA/NaN/Inf in foreign function call (arg 5)
when I try to analysis a binary trait using glmnet(R) by running the
following code
library(glmnet)
Xori <- read.table("c:\\SNP.txt", s
Hello,
I am a new R user. I am trying to use the arima command, but I have a
question on intermediate lags. I want to run in R the equivalent Stata
command of ARIMA d.yyy, AR(5) MA(5 7). This would tell the program I am
interested in AR lag 5, MA lag 5, and MA lag 7, all while skipping the
inte
Hey everybody,
i am using the rugarch-package and its great!
I have a pretty easy problem, but i just dont get it, so thanks if you can
help me.
Normally i use:
/
data(DATANAME)
spec = ugarchspec()
fit = ugarchfit(data = x[,1], spec = spec)
fit
slotNames(fit)
names(fit@fit)
coef(fit)
infocri
Hi,
I am new to R.
I have a matrix that I have assigned to the object “colon”.
> colon<-read.table("c:\\alon.txt",header=T,row.names=1)
attach(colon)
names(colon)
The dimenstions are 2000 62.
Each of the 62 columns (titled norm1, norm2, norm3, etc) has 2000
different numbers (‘continuous’ va
hi friend i m david if you are having any type of problem in Statistics then
do visit http://www.transtutors.com/statistics-homework-help/ it has helped
me a lot. thanks. david
http://www.transtutors.com/statistics-homework-help/ Transtutors
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Also I have this problem that many times the method converges to a local
maximum...if I run it a few times, I often get the expected answer.
Is there any way to fix this problem by changing the convergence conditions?
I have tried changing the nls.control {stats} but hasn't helped much. Any
sugges
On Sep 20, 2011, at 7:15 PM, Rich Shepard wrote:
I'm having difficulty finding the syntax to use to specify the
beginning
and ending dates for the x-axis while plotting a zoo object. I
thought that
I had seen a message on this list that used start=as.Date("...")
end=as.Date("..."), but I c
Dear R-help,
I have a raster map which has a measure of profitability of land by
parcel over several regions of geographic aggregation (think of
counties). This data is stored in an ASCII file. If I type
plot(profit)
I get a raster map of profit per parcel for all regions. I want to
plot sepa
Actually, you appear to have re-assigned your object “colon” (from
"c:\\alon.txt") with a character vector of intended row.names.
so use
row.names(colon) <-paste("g",c(1:nrow(colon)),sep="")
B77S wrote:
>
> I don't have access to your "alon.txt" file (see ?dput for future posts),
> but...
>
Hi
I recall running across a function a while back which would return
information about running processes (such as their cpu and memory
usage), but I cannot seem to locate it. Wondering if someone would be
kind enough to refresh my memory. I vaguely recall it was parsing the
output of the 'ps
On Sep 20, 2011, at 7:37 PM, justin jarvis wrote:
> That still discards the other data columns. For example, in the
> data frame
>
> V1 V2 V3 V4
> 1 1 1 NA 1
> 2 1 NA 1 1
> 3 1 NA 1 1
> 4 1 11 1
> 5 1 11 1
>
> Suppose I was running a regression using V1 and V2. R will r
If different labs tested different tissue.types, I am not sure you can
effectively partition variance between labs(batches) and tissue.types.
Weidong Gu
On Tue, Sep 20, 2011 at 2:14 PM, karena wrote:
> Hi,
>
> I am doing an analysis to see if these is tissue specific effects on the
> gene expres
Look at the help file for do.call. It is
most useful when you don't know how many
arguments will be given to the function you
are calling. E.g., if you know that the list
x is always 3 long then you can do
rbind(x[[1]], x[[2]], x[[3]])
to make a matrix out of the components, just as
do.call
I don't have access to your "alon.txt" file (see ?dput for future posts),
but...
I'm pretty sure info you want isn't in row.names(colon[1:2])
it should just be
text(x,y, label = colon[1:20])
??
HTH
baumeist wrote:
>
> Hi,
> I am new to R.
>
> I have a matrix that I have assigned to the o
You could use the na.action function on the fitted
object to see which observations were omitted. E.g.,
let's make a data.frame that we can actually do some
regressions with and try na.action():
> d <- data.frame(V1=11:15, V2=log(c(1,NA,NA,4,5)), V3=sqrt((-1):3),
V4=sin(1:5))
Warning message
Pardon my ignorance, but why is the do.call necessary? why not just execute
the rbind function? What's the advantage in putting it in a do.call "wrapper"?
On Sep 20, 2011, at 2:44 PM, William Dunlap wrote:
> In S+ do.call's first argument must be a character string
> that gives the name of
That still discards the other data columns. For example, in the data frame
V1 V2 V3 V4
1 1 1 NA 1
2 1 NA 1 1
3 1 NA 1 1
4 1 11 1
5 1 11 1
Suppose I was running a regression using V1 and V2. R will remove rows 2
and 3 due to the "NA." I would like a way to look at only th
Greetings!
2011/9/20 David Winsemius :
There is probably an environment/scoping problem as you stated. I am
using ESS and org-mode.
I started a new R code block and the problem disappeared.
Thanks for your kind feedback.
Henri-Paul
--
Henri-Paul Indiogine
Curriculum & Instruction
Texas A&
Diviya Smith gmail.com> writes:
>
> I dont think *r* is related to the problem. I am not trying to
> estimate *r* and
> so basically I am giving the model the correct value of *r* and so log(1-r)
> should not go to infinity.
>
> For test data, I generate data from the same model and add noise
I'm having difficulty finding the syntax to use to specify the beginning
and ending dates for the x-axis while plotting a zoo object. I thought that
I had seen a message on this list that used start=as.Date("...")
end=as.Date("..."), but I cannot find that message. I've tried ?plot,
?plotxy, ?pl
On Sep 20, 2011, at 6:49 PM, Henri-Paul Indiogine wrote:
My objective is to define a pattern for a grep() statement.
Thanks,
Henri-Paul
You are not illustrating the problem with enough detail to determine
the source of your errors. The errors do not appear when tested in
what appears to
Can you perhaps provide us a minimal working script in which this assignment
doesn't happen? If it really is just the line
pattern <- "[Aa]ccountability"
in the main body of the script, I can't for the life of me guess why it
wouldn't work. General keywords that come to mind are things like
"envi
My objective is to define a pattern for a grep() statement.
Thanks,
Henri-Paul
--
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Curriculum & Instruction
Texas A&M University
TutorFind Learning Centre
Email: hindiog...@gmail.com
Skype: hindiogine
Website: http://people.cehd.tamu.edu/~sindiogine
_
Hi Sarah!
2011/9/20 Sarah Goslee :
>> What works? Assigning a string to a function, thus replacing the
>> function in your search
>> path?
Bad choice for character variable name on my part. How about this?
pattern <- "[Aa]ccountability"
If I enter this at the CLI and then type "pattern", R ret
My apologies: not the name of a function, despite ?regexp having a return value.
But the rest of my questions are still relevant.
On Tue, Sep 20, 2011 at 6:32 PM, Sarah Goslee wrote:
> What works? Assigning a string to a function, thus replacing the
> function in your search
> path?
>
> Or is th
What works? Assigning a string to a function, thus replacing the
function in your search
path?
Or is that what doesn't work?
What do you expect? What are you trying to do? Using what commands?
Sarah
On Tue, Sep 20, 2011 at 6:19 PM, Henri-Paul Indiogine
wrote:
> Greegings!
>
> Any idea why this
Further to the plot suggested below, the plot
plot(log(A),log(B/A),pch="+",col="blue")
reveals an interesting structure to the data. Distinct curved
sequences are clearly visible. While their curved form is a
consequence of the fact that, for large A, A/B is close to 1
and so they tend to appro
On Sep 20, 2011, at 4:02 PM, Darrel Barbato wrote:
> Hi,
> I'm having some difficulty installing R on our server which is running
> RHEL (Red Hat Enterprise Linux) 5.2. I'm using RPM (Red Hat Package
> Manager) and by default it installs R in /usr/bin. Unfortunately due
> to a very small partition
Greegings!
Any idea why this works from the command line, but not from a source
file? This is driving me (more) insane.
regexp <- "[Aa]ccountability"
Thanks!
--
Henri-Paul Indiogine
Curriculum & Instruction
Texas A&M University
TutorFind Learning Centre
Email: hindiog...@gmail.com
Skype: h
I dont think *r* is related to the problem. I am not trying to estimate *r* and
so basically I am giving the model the correct value of *r* and so log(1-r)
should not go to infinity.
For test data, I generate data from the same model and add noise (using *r
norm*), with the following parameters -
We are contemplating the installation of R on a Windows server with Hyper-V but
we can't find any information on whether it can be done.
This transmission is intended solely for the person or organization to whom it
is addressed and it may contain privile
Hi,
I'm having some difficulty installing R on our server which is running
RHEL (Red Hat Enterprise Linux) 5.2. I'm using RPM (Red Hat Package
Manager) and by default it installs R in /usr/bin. Unfortunately due
to a very small partition size, I have run out of space in this
directory. I was hoping
Hi there,
I have the following problem on my macbook air with mac os x lion on it.
when any program tries to load the tcltk library the R GUI or R command line
freezes. also happens if I just use library(tcltk)
changing versions of tcltk and/or R doesn't help. currently I have the
following:
R ve
Might be of interest to list subscribers:
EMBL Advanced Course
R Programming and Development
EMBL Heidelberg, Germany
Monday 28 November - Tuesday 29 November 2011
The course will focus on two aspects of R programming and development.
In the first part, we will introduce object-oriented programmi
Hi,
I am doing an analysis to see if these is tissue specific effects on the
gene expression data .
Our data were collected from 6 different labs (batch effects). lab 1 has
tissue type 1 and tissue type 2, lab 2 has tissue 3, 4,5,6. The other labs
has one tissue type each. The 'sample' data is as
Thank you for the reply, it looks like the second option (te) will
work perfectly!
Max
On Tue, Sep 20, 2011 at 2:39 PM, Max Farrell wrote:
> One possibility is
>
> library(mgcv)
>
> ## isotropic thin plate spline smoother
> b <- gam(Y~s(X[,1],X[,2]))
> predict(b,newdata=list(X=W))
>
> ## ten
The following untested code might also be of some help to you:
x = data.frame(x = 1:5)
y = data.frame(y = 1:10)
z = data.frame(z = 1:3)
a = rnorm(50)
b = function(x) x^2
makeRowFrame <- function(){
n <- ls(envir = .GlobalEnv)
tempFunc <- function(VARIABLE){
VARIABLEM <- get(VARIAB
In S+ do.call's first argument must be a character string
that gives the name of the function, so replace
do.call(rbind, ...)
with
do.call("rbind", ...)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: Santosh [mailto:santosh2...@gmail.com]
Sent: Tuesday, September 20, 2011 2:55 AM
Diviya Smith wrote on 09/20/2011 01:03:22 PM:
>
> Hello there,
>
> I am using NLS for fitting a complex model to some data to estimate a
couple
> of the missing parameters. The model is -
> y ~
(C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r)
> where A, B and C are unknown
Hi Kelly,
c() creates a vector. You need data.frame() instead.
dfIntron <- c(mC_Explant=nrow(mC_Explant), mC_Callus=nrow(mC_Callus),
mC_RegenPlant=nrow(mC_RegenPlant)) # set colnames simultaneously.
Sarah
On Tue, Sep 20, 2011 at 2:23 PM, Vining, Kelly
wrote:
> Hello R users,
> I have a set of
Please keep the list in the loop for the archives:
The data provided looks like this:
V = structure(list(V1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L), .Label = "02/22/2011", class = "factor"), V2 = structure(c(1L,
1L, 2L,
Hello R users,
I have a set of data frames for which I am tallying row numbers, as shown
below.
> nrow(mC_Explant)
[1] 14480
> nrow(mC_Callus)
[1] 23320
> nrow(mC_RegenPlant)
[1] 8108
etc.
I want to create a new data frame which has the variable names as column
headings, and then a single row
As can be seen by plotting as follows:
plot(A,B,pch="+",col="blue") ## The raw data
plot(A,B-A,pch="+",col="blue") ## The differences versus A
lines(c(0,0.7),c(0,0))
Ted.
On 20-Sep-11 17:54:15, Timothy Bates wrote:
> Yes, in over 3/4s of the data points A is > B
which suggests the A
>
Pedro Mardones wrote on 09/20/2011 12:46:54 PM:
>
> Dear all;
>
> A very basic question. I have the following data:
>
>
>
> A <- 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163,
> 62,112,334,249,4
Hello there,
I am using NLS for fitting a complex model to some data to estimate a couple
of the missing parameters. The model is -
y ~ (C+((log(1-r))*exp(-A*d)*(-1+r+exp(d*(A-B)))/(r*(-A*d+d*B+log(1-r)
where A, B and C are unknown.
In order to test the model, I generate data by setting value
On Sep 20, 2011, at 12:46 PM, Pedro Mardones wrote:
> Dear all;
>
> A very basic question. I have the following data:
>
>
>
> A <- 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163,
> 62,112,334,249,
Yes, in over 3/4s of the data points A is > B… which suggests the A measure is
reading higher than the B measuring system.
length(A[A>B])/length(A)
On 20 Sep 2011, at 6:46 PM, Pedro Mardones wrote:
> Dear all;
>
> A very basic question. I have the following data:
>
>
Hi
I am having a problem using varImpPlot in randomForest. I get the error
message "Error in plot.window(xlim = xlim, ylim = ylim, log = "") : need
finite 'xlim' values"
When print $importance, several variables have NaN under %IncMSE. There
are no NaNs in the original data. Can someone hel
Hello,
I am using the kernlab package to do regression.
I have a data frame called Data6 which looks like this:
head(Data6)
WA PO ZA ZB ZC ZD KL
1 2.955447 6.378324 14.10622 0.134343 0.247120 0.734810 4.05988
2 2.939718 6.344122 14.03528 0.127512 0.
Dear all;
A very basic question. I have the following data:
A <- 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163,
62,112,334,249,45,244,211,175,174,26,375,346,153,32,
89,32,358,202,123,131,88,36,30,
Tarmo Remmel wrote on 09/20/2011 09:51:45 AM:
>
> Hello list members,
>
> I am working with simulated data for landscape pattern analysis. I have
> 1000 replicates of binary (2 colour) gridded landscapes at each
combination
> of 9 levels of class proportion and 11 levels of spatial
autocorrela
something like this should work:
> oldIDs <- c(1234, 1235, 1236)
> newIDs <- c(5000, 1234, 7000, 1236)
> # really new ones -- don't match the old ones
> (reallyNew <- setdiff(newIDs, oldIDs))
[1] 5000 7000
> # assign these back to the oldIDs for the next month
> (oldIDs <- c(oldIDs, reallyNew))
[1
A reproducible example would be useful, as yours isn't entirely clear.
But maybe something like this?
id.info <- data.frame(foreign=sort(unique(foreign.id)),
local=1:length(unique(foreign.id)))
and then use merge() to combine that with your actual data.
Sarah
On Tue, Sep 20, 2011 at 11:50 AM,
El Gorgonzola hotmail.com> writes:
> My question: how do you document methods of S4-classes in a package?
> It works fine for the method "show" but it does not work for "showall",
> for which I had to define the generic function first. What can I do to
> fix this, or is it better to just define
First off, let me apologize for the elementary question. I'm obviously a
novice.
Here's a stripped version of my problem.
March
foreign id = 1234, my id = 1
foreign id = 1235, my id = 2
foreign id = 1236, my id = 3
So we are adding new people for April, and things don't necessarily come in
order
Thank you Sarah and Petit bleu for your help.
I solved my problem with the code of Sarah.
Komine
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Rstudio and Rcmdr are very popular and for good reason, find a good book
while the latter is installing though.
IF you are using linux, rkward is fantastic to woRk with.
Ken Hutchison
On Tue, Sep 20, 2011 at 11:21 AM, Steve Lianoglou <
mailinglist.honey...@gmail.com> wrote:
> I just w
You did ask for a loop, but if you are willing to consider sapply(), the
result you want can be obtained with a single (admittedly long) command:
> slope<-sapply(1:100, function(x) lm(c(rnorm(1, mean=.01, sd=.001),
rnorm(1, mean=.1, sd=.01))~c(10,400))$coefficients[2])
> head(slope)
c(10, 400)
Hi
>
> I have a list of data frames like the following:
>
> set.seed(123)
> a<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
> b<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
> L<- list(a,b)
>
> All data frames in the list have the same dimensions. I need to
calcu
One possibility is
library(mgcv)
## isotropic thin plate spline smoother
b <- gam(Y~s(X[,1],X[,2]))
predict(b,newdata=list(X=W))
## tensor product smoother
b <- gam(Y~te(X[,1],X[,2]))
predict(b,newdata=list(X=W))
## variant tensor product smoother
b <- gam(Y~t2(X[,1],X[,2]))
predict(b,newd
I just wanted to second Rstudio:
http://rstudio.org/
Honestly, try this first -- it's an easy install, it just works, and
there's no real learning curve to the editor itself. Very n00b
friendly.
-steve
On Tue, Sep 20, 2011 at 9:28 AM, R. Michael Weylandt
wrote:
> RStudio
>
>
> Hope this help
You're passing a vector of length 3 to the legend argument of legend(), so R is
assuming that you have three things you want to display.
There are two alternatives:
1. Provide legend() a legend that's a single character vector.
2. Specify that you don't want symbols for the last two of the three l
Maybe you can use Rcmdr
http://socserv.mcmaster.ca/jfox/Misc/Rcmdr/
It is a very easy program.
Da: r-help-boun...@r-project.org [r-help-boun...@r-project.org] per conto di
Jack Siegrist [jack...@eden.rutgers.edu]
Inviato: martedì 20 settembre 2011 16.07
A:
Dear all,
Thanks a lot for the help. It worked very well in the end.
Best regards,
Marianne
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Maybe by adding
col=c("grey","white","white"), bg="white
but not sure ...
Have a nice end of day,
Ptit Bleu.
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Hello,
I am trying to estimate a multivariate regression of Y on X with
regression splines. Y is (nx1), and X is (nxd), with d>1. I assume the
data is generated by some unknown regression function f(X), as in Y =
f(X) + u, where u is some well-behaved regression error. I want to
estimate f(X) via
When sampling from a multi-variate truncated normal using rtmvnorm from
the tmvnorm-package, I experience extreme performance differences between
two of my computers. On my laptop computer, draws take ~5s, on my desktop
~30s. I need to run MCMCs with repeat calls to rtmvnorm on my desktop. The
I am struggling to get GAM model predictions from the top models calculated
using model.avg in the package "MuMIn".
My model looks something like the following:
gamp <- gam(log10(y)~s(x1,bs="tp",k=3)+s(x2,bs="tp",k=3)+
s(x3,bs="tp",k=3)+s(x4,bs="tp",k=3)+s(x5,bs="tp",k=3)+
s(x6
I have a list of data frames like the following:
set.seed(123)
a<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
b<- data.frame(x=runif(10), y = runif(10), sample = seq(1,10))
L<- list(a,b)
All data frames in the list have the same dimensions. I need to calculate
the sample means fo
HI,
This code is part of a code I used to do a linear regression:
>points(var1~var2,data=Regress,pch=21,bg="grey")
>reg11<-lm(var1~var2,data=Regress)
>abline(lm(var1~var2,data=Regress),lty=2,lwd=2,col="grey")
>legend("topleft",legend=
>c("NDII from composite",
>"y= 0.0007x - 0.1156",expression(p
Hi everybody. I'm trying to fit a weibull survival model with a spline
basis for the predictor, using the survival library. I've noticed that it
doesn't seem to be possible to use the aic method to choose the degrees of
freedom for the spline basis in a parametric regression (although it's
fine w
Marion Wenty wrote:
>
> ...Could anyone recommend an editor that is suitable for beginners?...
>
I had trouble with this for a long time. I tried several different programs,
but couldn't get any to work properly—they were either too complicated or
broken. I recently started using Notepad++ with
Hello list members,
I am working with simulated data for landscape pattern analysis. I have
1000 replicates of binary (2 colour) gridded landscapes at each combination
of 9 levels of class proportion and 11 levels of spatial autocorrelation.
The results are stored in an array as follows:
> dim(s
Hi Samir,
I do vaguely understand your intention. However I'm not sure taking
the mean of acf's makes any sense.
I do not know what your final goal is (what do you want to do with the
data?) but here is a suggestion. Instead of trying to aggregate acfs
computed on different series, why don't you:
1
Hi Jean and Jim!
Thanks for your suggestions.
Best,
Henri-Paul
--
Henri-Paul Indiogine
Curriculum & Instruction
Texas A&M University
TutorFind Learning Centre
Email: hindiog...@gmail.com
Skype: hindiogine
Website: http://people.cehd.tamu.edu/~sindiogine
_
Marion:
You should always first search CRAN for such queries. Had you done so, you
would have been led to:
http://www.sciviews.org/_rgui/
There is an extensive list of editors/IDE's there.
-- Bert
On Tue, Sep 20, 2011 at 6:26 AM, Marion Wenty wrote:
> Hello all,
>
> I am looking for an editor
On Tue, 20 Sep 2011, Marion Wenty wrote:
I am looking for an editor for R which has got functions beyond the normal
R editor that is included in the program.
I had a look at VIM but I think it's difficult if you are just starting
programming.
Marion,
VIM (Vi) is a line-oriented editor. T
On Sep 20, 2011, at 4:50 AM, mael wrote:
Hi,
I am trying to add a function in a linear quantile regresion to find a
breakpoint. The function I want to add is:
y=(k+ax)(xB)
How do I write it in the rq() function? Do I need to define the
parameters
in any way and how do
Hi everyone
I have got a quick question:
I the "seqinr" package:
*dist.alignment(x,"identity")*
This is calculating the square root of pairwise distances. Does anyone know
whether/how gaps are counted in this function?
Thank you.
Best wishes,
Bettina
[[alternative HTML version delet
RStudio
Hope this helps,
Michael Weylandt
On Tue, Sep 20, 2011 at 9:26 AM, Marion Wenty wrote:
> Hello all,
>
> I am looking for an editor for R which has got functions beyond the normal
> R
> editor that is included in the program.
>
> I had a look at VIM but I think it's difficult if you are
Hello all,
I am looking for an editor for R which has got functions beyond the normal R
editor that is included in the program.
I had a look at VIM but I think it's difficult if you are just starting
programming.
Could anyone recommend an editor that is suitable for beginners?
Thanks for your h
Good afternoon/morning readers. This is the first time I am trying to run
some Bayesian computation in R, and am experiencing a few problems.
I am working on a Poisson model for cancer rates which has a conjugate Gamma
prior.
1) The first question is precisely how I work out the parameters.
#Sup
hi eik, mark and sarah,
thank you for your help!
i was looking for the t-command but in the end the apply-command did the
trick.
marion
2011/9/20 Sarah Goslee
> It isn't entirely clear to me what you want, but here are all the
> possibilities
> I could think of. I hope one of them does what y
Jim Lemon wrote on 09/20/2011 04:15:46 AM:
>
> On 09/19/2011 04:46 PM, Henri-Paul Indiogine wrote:
> > Greetings!
> >
> > I am using the R library RQDA to assign certain codes to paragraphs of
> > documents in a collection. Several paragraphs are assigned more than
> > 1 code. E.g. often the co
On Sep 19, 2011, at 8:49 PM, justin jarvis wrote:
I have a data set with many missing observations. When I run a
regression, R of course discards the observations (the whole row) that
have "NA". I want to tabulate some baseline characteristics (column
means) but only for the observations that
Hi,
could anyone tell me how predict() predicts the meanError or
standardDerivation of a garchFit(1,1)-model,
knowing the coefficients mu, omega, alpha1, beta1 and of course all
datapoints?
Thanks and sorry for my poor english.
--
View this message in context:
http://r.789695.n4.nabble.com/pred
On Tue, Sep 20, 2011 at 4:31 PM, Duncan Murdoch wrote:
> On 11-09-20 3:08 AM, Ashim Kapoor wrote:
>
>> Dear R-helpers,
>>
>> Please look at the following minimal code.
>>
>> \documentclass[a4paper]{**article}
>> \begin{document}
>> <<>>==
>> library(zoo)
>> a<-zoo(1:4,order.by=Sys.time()**+1:4)
>>
On 11-09-20 7:25 AM, _Luc_ wrote:
ok great, as.numeric(as.character(data$x)) works perfectly.
I am sorry that i did not find this before making this post.
@ peter
what did not work was:
data$x2<- as.character(data$x)
data$x3<- as.numeric(data$x2)
That should have worked. Can you put togethe
ok great, as.numeric(as.character(data$x)) works perfectly.
I am sorry that i did not find this before making this post.
@ peter
what did not work was:
data$x2 <- as.character(data$x)
data$x3 <- as.numeric(data$x2)
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-
On 11-09-20 3:08 AM, Ashim Kapoor wrote:
Dear R-helpers,
Please look at the following minimal code.
\documentclass[a4paper]{article}
\begin{document}
<<>>==
library(zoo)
a<-zoo(1:4,order.by=Sys.time()+1:4)
str(a)
@
\end{document}
When I do R CMD Sweave,followed by pdflatex ,and view the final
Szeptember 12-től 26-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 12 till 26 September with no access to my
emails.
In urgent cases please contact
On 09/20/2011 10:57 AM, Achim Zeileis wrote:
On Tue, 20 Sep 2011, Tal Galili wrote:
I am not sure who to send this to - so I am writing this here.
It seems that the "index.html" pages on CRAN are gone.
For example:
http://cran.r-project.org/web/packages/ggplot2/index.html
Is gone. But the dir
It isn't entirely clear to me what you want, but here are all the possibilities
I could think of. I hope one of them does what you want.
> testmat <- matrix(1:8, ncol=2)
> testmat
[,1] [,2]
[1,]15
[2,]26
[3,]37
[4,]48
> paste(testmat, collapse=" ")
[1] "1 2 3 4
Hi,
I am trying to add a function in a linear quantile regresion to find a
breakpoint. The function I want to add is:
y=(k+ax)(xB)
How do I write it in the rq() function? Do I need to define the parameters
in any way and how do I do that? I'm a biologist new to R.
Thanks!
Hi marion,
just transpose the matrix:
mm<-matrix(LETTERS[1:20],nrow=5)
paste(t(mm),collapse="")
or - if you want the result seperated by rows
apply(mm,1,paste,collapse="")
cheers
Am 20.09.2011 11:55, schrieb Marion Wenty:
> I have another question concerning the paste command:
>
> now instead o
I have another question concerning the paste command:
now instead of a vector I would like to paste the elements of a matrix
together, which works in the same:
Mypastedmatrix <- paste(Mymatrix,collapse="")
My problem now is that the program does this BY COLUMN, but I would like to
have the eleme
Dear R- Splus experts,
In R, I have frequently used do.call with strsplit. and I have a hard time
with Splus.. any suggestions?
for example, the R code below:
do.call(rbind,strsplit(paste(letters[1:10],c(1:10))," "))
Thanks so much,
Santosh
On Fri, Dec 5, 2008 at 8:51 AM, William Dunlap wrote:
hello,
thank you for your answer!
yes, now it is working!
marion
2011/9/19 Duncan Murdoch
> On 11-09-19 7:30 AM, Marion Wenty wrote:
>
>> Hello,
>>
>> could someone help me with this problem?:
>>
>> I would like to create a latex-script inside of a character vector in
>> order
>> to being able t
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