Dear all,
when using .jinit() I get the message .jinit() : Cannot create Java
virtual machine (-1).
The details:
I am using the Dismo package.Dismo has a function 'maxent' that communi-cates
with this program(MaxEnt).MaxEnt is available as a stand-alone Java program.
It is normal when I exec
Hi,
I have a data set of 1 rows and ran a linear regression by using the
function lmres = lm(formula,data)
then got the fitted value by using the value fit=fitted(lmres).
but the number of rows in the fitted one is about 9548,
What could be the reason for reduction in the number of rows in
Dear all,
when using .jinit() I get the message .jinit() : Cannot create Java
virtual machine (-1).
sessionInfo() R version 2.14.0 (2011-10-31)Platform: i386-pc-mingw32/i386
(32-bit)locale:[1] LC_COLLATE=Chinese_People's Republic of China.936 [2]
LC_CTYPE=Chinese_People's Republic of China.9
Dear all,
I get the error when I use maxent.jar:
Error .jcall(mxe, "S", "fit", c("autorun", "-e", afn, "-o", dirout, :
java.lang.NoSuchMethodError:
density.Params.readFromArgs([Ljava/lang/String;)Ljava/lang/String;
sessionInfo() result:
R version 2.14.0 (2011-10-31)Platform: i386-pc-ming
On 11/14/2011 05:59 PM, Prasanth V P wrote:
require(plotrix)
xy.pop<- c(17,15,13,11,9,8,6,5,4,3,2,2,1,3)
xx.pop<- c(17,14,12,11,11,8,6,5,4,3,2,2,2,3)
agelabels<- c("0-4","5-9","10-14","15-19","20-24","25-29","30-34",
"35-39","40-44","45-49","50-54","55-59","60-64","65+")
xycol<-color.gr
Hi
>
> Hi,
>
> I have a data set of 1 rows and ran a linear regression by using the
> function lmres = lm(formula,data)
>
> then got the fitted value by using the value fit=fitted(lmres).
>
> but the number of rows in the fitted one is about 9548,
>
> What could be the reason for reductio
Dear all
I am again (as usual) lost in regular expression use for selection. Here
are my data:
> dput(mena)
c("138516_10g_50ml_50c_250utes1_m53.00-_s1.imp",
"138516_10g_50ml_50c_250utes1_m54.00_s1.imp",
"138516_10g_50ml_50c_250utes1_m55.00_s1.imp",
"138516_10g_50ml_50c_250utes1_m56.00_s1.imp"
Hi,
Try grepl instead of sub,
mena[grepl("m5.", mena)]
HTH,
baptiste
On 14 November 2011 21:45, Petr PIKAL wrote:
> Dear all
>
> I am again (as usual) lost in regular expression use for selection. Here
> are my data:
>
>> dput(mena)
> c("138516_10g_50ml_50c_250utes1_m53.00-_s1.imp",
> "138516
On 11/14/2011 07:45 PM, Petr PIKAL wrote:
Dear all
I am again (as usual) lost in regular expression use for selection. Here
are my data:
dput(mena)
c("138516_10g_50ml_50c_250utes1_m53.00-_s1.imp",
"138516_10g_50ml_50c_250utes1_m54.00_s1.imp",
"138516_10g_50ml_50c_250utes1_m55.00_s1.imp",
"138
Hi
>
> Hi,
>
> Try grepl instead of sub,
>
> mena[grepl("m5.", mena)]
It does not select those "m5?" strings from those character vectors. I
need as an output a vector
m53, m54, m55, m56, m57, m58, m59
Regards
Petr
>
> HTH,
>
> baptiste
>
> On 14 November 2011 21:45, Petr PIKAL wrote:
Good morning R list,
My apologies if this has *already* answered elsewhere, but I have not found
the answer that I am looking for.
I have a character string, i.e.
form<-c('~ A + B + C + C / D + E + E / F + G + H + I + J + K + L * M')
Now, my aim is to find the position of all those instances o
Hi
> On 11/14/2011 07:45 PM, Petr PIKAL wrote:
> > Dear all
> >
> > I am again (as usual) lost in regular expression use for selection.
Here
> > are my data:
> >
> >> dput(mena)
> > c("138516_10g_50ml_50c_250utes1_m53.00-_s1.imp",
> > "138516_10g_50ml_50c_250utes1_m54.00_s1.imp",
> > "138516_10g_
>From: Joshua Wiley
>dat$RTotal <- cumsum(dat$BAL)
Wow, that's really great. I'm starting to really enjoy using R. My statistical
needs are not that great, but I love the way that R handles tabular data.
__
R-help@r-project.org mailing list
https
On 14.11.2011 10:22, Petr PIKAL wrote:
Hi
On 11/14/2011 07:45 PM, Petr PIKAL wrote:
Dear all
I am again (as usual) lost in regular expression use for selection.
Here
are my data:
dput(mena)
c("138516_10g_50ml_50c_250utes1_m53.00-_s1.imp",
"138516_10g_50ml_50c_250utes1_m54.00_s1.imp",
"
Hi Jim,
It's working perfectly fine with the "rxcol" parameter. I am just
wondering how could I miss that..!!!
By the way, many thanks for pointing it out... Otherwise, I would have
been using the old version of R for just getting the required plot.
Much Appreciated,
Prasanth.
-Original Mess
I am performing document clustering on a set of documents using R. I
performed hierarchical clustering using hclust and have identified the
cluster corresponding to each data point. I would like to lablel each
cluster automatically in order to identify the top keywords associated with
each cluster.
Does
library( stringr )
str_extract( mena, "m5[0-9]" )
achieve what you are looking for?
Rgds,
Rainer
On Monday 14 November 2011 10:22:09 Petr PIKAL wrote:
> Hi
>
> > On 11/14/2011 07:45 PM, Petr PIKAL wrote:
> > > Dear all
> > >
> > > I am again (as usual) lost in regular expression use for
Dear R developers,
I want to draw an arrow in a figure with lty=2. The
lty argument also affects the edge of the arrow, which is
unfortunate. Feature or bug?
Is there some way to draw an arrow with intact edge, still
with lty=2?
Example code:
plot(1:10)
arrows(4, 5, 6, 7, lty=2)
Best wishes,
Hello Paul,
just a guess: different sample sizes! In your first call, the sample is shorter
than in your second. Hence, you can test this, if you curtail your data set in
your second call and then you should obtain the same result, i.e.:
> library(vars)
> data(Canada)
> test <- summary(CADFtest
Hi,
I have this same problem using read.xls in a table that I have NA values.
Look:
Using the importing from a csv file using read.table
> summary(dados)
TratV1V2
A:5 Min. :1.0 Min. :1.000
B:5 1st Qu.:1.0 1st Qu.:1.000
Median :1.5 Median :2.000
Hi
Thank you. It is a pure magic, something taught in Unseen University.
this is what I got as a help for selecting only letters from set of
character vector.
> vzor
[1] "61A" "62C/27" "65A/27" "66C/29" "69A/29" "70C/31"
"73A/31"
[8] "74C/33" "77A/33" "81A/35" "82C/37" "85A/37"
On 14.11.2011 11:27, Petr PIKAL wrote:
Hi
Thank you. It is a pure magic, something taught in Unseen University.
this is what I got as a help for selecting only letters from set of
character vector.
vzor
[1] "61A" "62C/27" "65A/27" "66C/29" "69A/29" "70C/31"
"73A/31"
[8] "74C/33
Hallo,
I am trying to define expectation as an integral
and use uniroot to find the distribution parameter
for a given expectation.
However I fail to understand how to define properly
the functions involved and pass the parameters correctly.
Can anyone help me out?
Thanks,
Gerrit Draisma.
Thi
On 14-Nov-11 09:57:52, Matthias Gondan wrote:
> Dear R developers,
>
> I want to draw an arrow in a figure with lty=2. The
> lty argument also affects the edge of the arrow, which is
> unfortunate. Feature or bug?
>
> Is there some way to draw an arrow with intact edge, still
> with lty=2?
>
>
Hi,
> Dear R developers,
>
> I want to draw an arrow in a figure with lty=2. The
> lty argument also affects the edge of the arrow, which is
> unfortunate. Feature or bug?
>
> Is there some way to draw an arrow with intact edge, still
> with lty=2?
AFAIK there is no such option in the arrow func
Hello.
Consider the following matrix:
mp <- matrix(c(0,1/4,1/4,3/4,0,1/4,1/4,3/4,1/2),3,3,byrow=T)
> mp
[,1] [,2] [,3]
[1,] 0.00 0.25 0.25
[2,] 0.75 0.00 0.25
[3,] 0.25 0.75 0.50
The eigenvectors of the previous matrix are 1, 0.25 and 0.25 and it is not a
diagonalizable matrix.
When you
The LaF package provides methods for fast access to large ASCII files.
Currently the following file formats are supported:
* comma separated format (csv) and other separated formats and
* fixed width format.
It is assumed that the files are too large to fit into memory, although
the package ca
> Ted Harding
> on Mon, 14 Nov 2011 10:39:35 + writes:
> On 14-Nov-11 09:57:52, Matthias Gondan wrote:
>> Dear R developers,
>>
>> I want to draw an arrow in a figure with lty=2. The lty
>> argument also affects the edge of the arrow, which is
>> unfortuna
Dear R users,
Thanks to a new contributor, Luigi Cerulo, amap provides 2 more metrics,
and another agglomeration linkage for distance computation, and hierarchical
clustering.
The new centroid linkage method implemented in the cluster3
software tool (http://bonsai.hgc.jp/~mdehoon/software/cluste
Hello,
I currently get anova results out of the aov function (see below) I use the
model.tables and I believe it gives me back the model parameters of the fit
(betas), however I don't see the intercept (beta_0) and don't understand what
the "rep" output means and there is no description in the
Hello David,
thanks for your answer.
I have done as you told me, however the fit is very poor, much worse than that
obtained from using the whole dataset (without upper bound).
Any idea?
Thanks,
Michele
On Nov 4, 2011, at 8:56 PM, David Winsemius wrote:
>
> On Nov 3, 2011, at 7:54 AM, Michele
Hello fellow R-users,
Iâve been mulling this problem over for some time now and have decided it is
something I have to deal with but canât, so here goes:
I have a dataset (called maindata, it is 271 columns *13890 rows so I wont post
the entire thing here, Iâll just explain the situation!)
> Consider the following matrix:
> > mp <- matrix(c(0,1/4,1/4,3/4,0,1/4,1/4,3/4,1/2),3,3,byrow=T)
> > mp
> [,1] [,2] [,3]
> [1,] 0.00 0.25 0.25
> [2,] 0.75 0.00 0.25
> [3,] 0.25 0.75 0.50
> The eigenvectors of the previous matrix are 1, 0.25 and 0.25 and it is not a
> diagonaliza
I found an old thread on R-Sig-Finance with the same problem and a possible
solution https://stat.ethz.ch/pipermail/r-sig-finance/2007q2/001362.html
I've used with success a few times but it seems a bit slow.
If anyone has a better way of modelling state-dependent volatility using
one of the avai
Hi,
On Mon, Nov 14, 2011 at 4:20 AM, Michael Griffiths
wrote:
> Good morning R list,
>
> My apologies if this has *already* answered elsewhere, but I have not found
> the answer that I am looking for.
>
> I have a character string, i.e.
>
>
> form<-c('~ A + B + C + C / D + E + E / F + G + H + I +
Inicio del mensaje reenviado:
> De: Arnau Mir
> Fecha: 14 de noviembre de 2011 13:24:31 GMT+01:00
> Para: Martin Maechler
> Asunto: Re: [R] How to compute eigenvectors and eigenvalues?
>
> Sorry, but I can't explain very well.
>
>
> The matrix 4*mp is:
>
> 4*mp
> [,1] [,2] [,3]
> [1,]
Dear Arnau,
In this and a subsequent message, you seem to incorrectly infer that the two
equal eigenvalues of the matrix imply that it's singular. The rank of the
matrix is equal to the number of *nonzero* eigenvalues, here 3, and so the
matrix is nonsingular. That two eigenvalues are equal simply
Yes there are few NA in the Data
--
View this message in context:
http://r.789695.n4.nabble.com/help-in-fitted-values-in-lm-function-tp4038642p4039207.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
ht
In order for the information criteria to be able to select the model, all the
models have to be estimated on the same sample. Therefore, all the models
are estimated and compared using the same sample used for the model
containing the largest number of lags. You can find this and other details
in
L
Hi,
I'm running R in supercomputer where clusters already are configurered, i.e.
number of procesors are predefined and I cannot change it. Can I still use
package like snow? I'm worrying that command like makeCluster won't be very
'healthy'. Or its ok to use it?
Thanks in advance, Tomas
--
View
David's answers were correct. You are looking deep into the code when
there is no reason to to so.
1. h(t|(X=x,Z=z)) = exp(Beta0 + XZBeta1)
Most statisticians will tell you that this is an unwise model. The
reason is that if you replace X with "X+1" the fit changes, which is
almost never desira
Try this:
EV <- function(lamb){
fnc <- function(x) x * dexp(x, lamb)
integrate(fnc, 0, Inf)$value
}
Your problem is that there's nothing to translate th to lambda in your
code for E.
Michael
On Mon, Nov 14, 2011 at 5:32 AM, Gerrit Draisma wrote:
> Hallo,
> I am trying to define expec
Thank you Sarah,
Your reply was very helpful. I have the added difficulty that I am not only
dealing with single A-Z characters, but quite often have the following
situation:
form<-c('~Sentence+LEGAL+Intro+Intro/Intro1+Intro*LEGAL+benefit+benefit/benefit1+product+action+mean+CTA*help')
and again
Hello.
I just installed R 2.14.0 and then did
> install.packages("splm", repos="http://R-Forge.R-project.org";)
in order to replicate your problem.
It worked "almost" fine. The only problem I had was with some missing
packages: being a fresh install, all suggested and required packages had
to b
You need to explicitly pass th to your function with the ... argument
of integrate.
E <- function(th){
integrate(function(x,th) x*g(x, th), 0, Inf, th)$value
}
Also, it's value, not Value, which might be producing errors of another sort.
Michael
On Mon, Nov 14, 2011 at 9:16 AM, Gerrit Drais
Glad it worked! The one "gotcha" is that it does not handle missing
values, so for example:
> cumsum(c(1, 2, 3, NA, 4))
[1] 1 3 6 NA NA
both the NA, and everything after it become NA (missing). If you find
yourself working with tables and frequencies and the like, you may
also like (if you h
Hi
r-help-boun...@r-project.org napsal dne 14.11.2011 14:54:05:
> Thank you Sarah,
>
> Your reply was very helpful. I have the added difficulty that I am not
only
> dealing with single A-Z characters, but quite often have the following
> situation:
>
> form<-c('~Sentence+LEGAL+Intro+Intro/Intr
Hello Kevin,
Thank you. I will delete the folder and run an application called
CCleaner (free). That will remove all the broken registry entries.
There should be a problem free update path for R installation. As
you rightfully mentioned the R- manual is not clear about package upd
Dear all,
I have the following dataset with results from an experiment with individual
bats that performed two tasks related to prey capture under different
conditions:
X variables:
indiv - 5 individual bats used in the experiment; all of which performed both
tasks
task - 2 tasks that each ind
Dear all,
I am working on a 64 bits Linux system.
I issue the following R commands:
> rm(list=ls()) # To remove all objects in the workspace.
> gc() # To free memory.
used (Mb) gc trigger (Mb) max used (Mb)
Ncells 124250 6.7 35 18.7 35 18.7
Vcells 124547 1.0 786432 6.0 476934 3.7
> gc(
Hi,
On Mon, Nov 14, 2011 at 8:54 AM, Michael Griffiths
wrote:
> Thank you Sarah,
>
> Your reply was very helpful. I have the added difficulty that I am not only
> dealing with single A-Z characters, but quite often have the following
> situation:
>
> form<-c('~Sentence+LEGAL+Intro+Intro/Intro1+In
If you want fitted() to return NAs in the positions where
there were NA's in data, use na.action=na.exclude in your
call to lm(). E.g.,
> z <- data.frame(y=1:5, x=c(1,NA,3,3,5))
> fitted(lm(y~x, data=z))
1345
1.25 3.25 3.25 5.25
> fitted(lm(y~x, data=z, na.action=na.exc
1. Post this on the R-sig-mixed-models list, not here.
2. (The following "advice" should be treated cautiously): Forget it! With
only 5 bats, you have too little information to estimate variance
components. Treat the bats as fixed effects and fit via lm (or glm if some
of your responses are not ga
This caught me learning R, and no doubt thousands of others.
When would one ever want the results of fitted() or residuals() to NOT match
the data frame rows which went into the model? Certainly making shrinking the
results the default is not what 99% of user will want if they need to access
On Nov 14, 2011, at 6:11 AM, Michele Mazzucco wrote:
Hello David,
thanks for your answer.
I have done as you told me, however the fit is very poor, much worse
than that obtained from using the whole dataset (without upper bound).
Any idea?
Counter questions in the absence of data:
??? Is
dear R-team
I need to find the min, max values for each patient from dataset and keep
the output of it as a dataframe with the following columns
- Patient nr
- Region (remains same per patient)
- Min score
- Max score
Patient Region Score Time
11 X19 28
21
I am also using statConn so I will let you know if I hear anything new.
-Original Message-
From: Cem Girit [mailto:gi...@comcast.net]
Sent: Monday, November 14, 2011 8:52 AM
To: 'Kevin Burton'
Cc: r-help@r-project.org
Subject: RE: [R] Upgrade R?
Hello Kevin,
Thank you. I will de
Hi Laura,
This looks suspiciously like homework. Nonetheless, you may wish to
check out ?cbind.
Sarah
On Mon, Nov 14, 2011 at 11:10 AM, B Laura wrote:
> dear R-team
>
> I need to find the min, max values for each patient from dataset and keep
> the output of it as a dataframe with the following
I am sorry to ask this group but the maintainer of this package did not
leave an email address.
Has anyone used or is using the 'rugarch' package with time-series data
(ts)? I try to fit a GARCH model to my data using the following:
> gf <- ugarchfit(data=l[["MEN"]]$series, spec=spec)
David,
here is the smallest dataset
# Bid Price SurvivalCensored
0.030.029 1 1
0.030.029 11 1
0.030.029 10 1
0.030.029 9 1
0.030.029 8 1
0.030.029 7 1
0.030.029 6 1
0.030.029 5 1
0.030.02
A non-helpful reply on a "language" issue.
"Truncated" data are quite different than "censored" data and require
different methodologies to analyze. You -- and many others in their
postings -- have appeared to confuse the two here.
-- Bert
On Mon, Nov 14, 2011 at 8:20 AM, Michele Mazzucco wro
Hi Laura,
You were close. Just use range() instead of min/max:
## your data (read in and then pasted the output of dput() to make it easy)
dat <- structure(list(Patient = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 6L,
I'm very new to R and am trying to create my first loop.
I have:
x <-c(0:200)
A <- dpois(x,exp(4.5355343))
B <- dpois(x,exp(4.5355343 + 0.0118638))
C <- dpois(x,exp(4.5355343 -0.0234615))
D <- dpois(x,exp(4.5355343 + 0.0316557))
E <- dpois(x,exp(4.5355343 + 0.0004716))
F <- dpois(x,exp(4.5355343
Hello,
I am using a grid.layout for combining multiple ggplot-plots.
So far I am doing it this way to get a pdf/eps:
pdf("/path/to/my/file.pdf") #or postscript("/path/to/my/file.eps")
grid.newpage()
pushViewport(viewport(layout = grid.layout(nrow=2, ncol=2,
widths = unit(c
Hi,
On Mon, Nov 14, 2011 at 10:59 AM, Davg wrote:
> I'm very new to R and am trying to create my first loop.
>
> I have:
>
> x <-c(0:200)
> A <- dpois(x,exp(4.5355343))
> B <- dpois(x,exp(4.5355343 + 0.0118638))
> C <- dpois(x,exp(4.5355343 -0.0234615))
> D <- dpois(x,exp(4.5355343 + 0.0316557))
x <-c(0:200)
A <- dpois(x,exp(4.5355343))
B <- dpois(x,exp(4.5355343 + 0.0118638))
C <- dpois(x,exp(4.5355343 -0.0234615))
D <- dpois(x,exp(4.5355343 + 0.0316557))
E <- dpois(x,exp(4.5355343 + 0.0004716))
F <- dpois(x,exp(4.5355343 + 0.056437))
G <- dpois(x,exp(4.5355343 + 0.1225822))
total <- A +
x <-c(0:200)
dat <- data.frame(
A = dpois(x,exp(4.5355343)),
B = dpois(x,exp(4.5355343 + 0.0118638)),
C = dpois(x,exp(4.5355343 -0.0234615)),
D = dpois(x,exp(4.5355343 + 0.0316557)),
E = dpois(x,exp(4.5355343 + 0.0004716)),
F = dpois(x,exp(4.5355343 + 0.056437)),
G = dpois(x,exp(4.53
But the awesome thing is you don't need a for loop at all thanks to
the magic of R's vectorization!
This will do it (and much faster than an R level loop would):
x = 0:200 # Note that you don't need a c() since you aren't
concatenating 0:200 with anything
A <- dpois(x,exp(4.5355343))
B <- dpois(x
I took a stab at this using ddply() from the plyr package. How's this
look to you?
x<- textConnection("Col Patient Region Score Time
11 X19 28
21 X20 126
31 X22 100
41 X25 191
52 Y121
62 Y
Hi,
I've just downloaded and installed R 2.14.0 using the windows binary on
a 64bit windows machine running windows 7.
Rterm / RGui work as expected, as does
R CMD --help
and
R CMD BATCH --help
however
R CMD check --help
returns no information and I seem to be unable to check a package.
Va
Hi Martyn,
My guess is that you need to add the directory where R is located to
your Windows PATH variable. It sounds like Windows just doesn't know
where to find R.
HTH,
Josh
On Mon, Nov 14, 2011 at 8:48 AM, Martyn Byng wrote:
> Hi,
>
> I've just downloaded and installed R 2.14.0 using the w
On Nov 14, 2011, at 4:20 AM, Michael Griffiths wrote:
Good morning R list,
My apologies if this has *already* answered elsewhere, but I have
not found
the answer that I am looking for.
I have a character string, i.e.
form<-c('~ A + B + C + C / D + E + E / F + G + H + I + J + K + L * M')
Hi Josh,
Thanks for that, which directory needs to be in the path?
There is a file called R.exe in
C:\Program Files\R\R-2.14.0\bin
C:\Program Files\R\R-2.14.0\bin\x64
and
C:\Program Files\R\R-2.14.0\bin\i386
I currently have
C:\Program Files\R\R-2.14.0\bin\x64
in the path (which is, I'm gues
Bert,
I think there is a misunderstanding here.
Some data is censored, but I want to fit the data with a distribution
in the interval [0,24] only. Also, please note that I have other
datasets having values larger than 1000.
Cheers,
Michele
On 14 Nov 2011, at 18:28, Bert Gunter wrote:
A n
Dear David,
You do not need a loop. The vectors are equaly sized, so sum them and then
plot the vector with the sums:
total <- A+B+C+D+E+F+G
plot (total, type="l")
Regards,
Emilio
2011/11/14 Davg
> I'm very new to R and am trying to create my first loop.
>
> I have:
>
> x <-c(0:200)
> A <- dpo
FYI,
you can use tools such as Path Manager 1.1.1 (GPL) on Windows:
http://www.softpedia.com/get/System/System-Miscellaneous/Path-Manager.shtml
to list, modify (add, remove, reorder, remove duplicates, normalize)
your PATH environment variable. For each directory it identifies in
PATH it will
See if
Rcmd check --help
works.
I've always been use that form (I'm on Win7 64-bit).
/Henrik
On Mon, Nov 14, 2011 at 9:06 AM, Martyn Byng wrote:
> Hi Josh,
>
> Thanks for that, which directory needs to be in the path?
>
> There is a file called R.exe in
>
> C:\Program Files\R\R-2.14.0\bin
>
Dear R users,
Is it possible to plot 2 different fields with image.plot().
For example at first a digitale elevation model and overlying another data
field, where NA values are transparent, so that the digitale elevation model
is still visble at this areas.
Maybe there is another plotting optio
Hi Martyn,
I would not expect you to need directories besides Rversion\bin\x64.
That is all I normally have in my path...perhaps because I do a
complete install?
I do not really know how or why it works, so I do not have any great
insight---just my experience. Perhaps someone else will chime in
On Nov 14, 2011, at 1:16 PM, Chris82 wrote:
Dear R users,
Is it possible to plot 2 different fields with image.plot().
There is an add parameter described on the helop page. Have you tried?
For example at first a digitale elevation model and overlying
another data
field, where NA values
Hi,
Thanks for the help.
I have now sorted out the issue.
Somehow the text "C:\Progra~1\R\R-2.14.0\bin\x64\R" had got added to the end of
an environment variable called "comspec".
Removing the extraneous text solved the problem - so the only mystery now is
how I managed to paste the text into
Thank you all!
It's working perfectly. I will have a look for an online guide.
--
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R-help@r
http://r.789695.n4.nabble.com/file/n4040429/model.jpg
Hi All
I need some help about construct MLE logit for Binary Autogressive Moving
Average model.
Please see the model in the PDF attach file.
This is what i did.
y<-c(0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1) # y
is
Hello,
I would like to get help on the instalation of R.
I have too few free space in my pc hard disk. So I wonder if it is possible
to install R on an external removable hard drive.
Can it be done? How should I proceed?
Thank you for your help.
best regards,
Francisca A. S. dos Santos Bronner
I can get an array of strings for the data that I want using 'paste()' as
follows:
paste('ma', 1:am$arma[2], '=', coef(am)[1:am$arma[2] + am$arma[1]], sep='')
This results in a vector of strings like:
[1] "ma1=1.17760133668255" "ma2=0.649795570407939" "ma3=0.329456750858276"
What I
I haven't personally used it, but I think this might work:
http://sourceforge.net/projects/rportable/
Michael
2011/11/14 Francisca Soares dos Santos :
> Hello,
>
> I would like to get help on the instalation of R.
> I have too few free space in my pc hard disk. So I wonder if it is possible
> to
The one included in the standard R installation -- which can be
accessed by typing help.start() at your prompt -- is quite good for
beginners (and very conveniently located). If you tell us a bit more
about yourself, we can help direct you to others as well:
specifically,
1) Prior programming expe
Do it in 2 steps:
z <- as.list( coef(am)[1:am$arma[2] + am$arma[1]])
names(z) <- paste("ma",seq_along(z), sep="")
-- Bert
On Mon, Nov 14, 2011 at 10:40 AM, Kevin Burton wrote:
> I can get an array of strings for the data that I want using 'paste()' as
> follows:
>
>
>
> paste('ma', 1:am$arma[2]
Is there a library that provides power calculation and sample size
estimation for nonlinear regression?
The task is easy for linear regression with the "pwr" package, but I
can't find a method for nonlinear regression (estimated with the "nls"
package).
-- -- -- -- -- -- -- -- -- -- -- -- -
See R Windows FAQ 2.6
On Monday, 14 November 2011, R. Michael Weylandt
wrote:
> I haven't personally used it, but I think this might work:
> http://sourceforge.net/projects/rportable/
>
> Michael
>
> 2011/11/14 Francisca Soares dos Santos :
>> Hello,
>>
>> I would like to get help on the instala
Path manager has a nice feature. Thanks. While I can see that rcmd is in
the path setting, it is still not recognized as an internal command. That
is something really confuses me.
Jixiang Wu
On Mon, Nov 14, 2011 at 11:30 AM, Henrik Bengtsson wrote:
> FYI,
>
> you can use tools such as Path Mana
Since installing R 2.14.0 on my Mac (a Mac Pro running 10.6.8) an issue has
arisen when using the vi editor in conjunction with the edit() command. More
specifically, commented lines disappear from edited functions when using
[functionname.R] <- edit().
That is, if you have created a function c
hello all R experts,
> how do I calculate the reliability between the two groups
> using the ICCs?
--
Sincerely,
Supreet kaur,
Biomedical research engineer,
Nationwide Childrens Hospital,
Columbus, OH
(614)355-3509
[[alternative HTML version deleted]]
__
If you don't want to go with the simple method mentioned by David and Ted, or
you just want some more theory, you can check out:
http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle and implement that.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s.
How to add units (e.g. "cm") to the color key of a lattice levelplot?
The plots looks fantastic, but it would be nice to indicate somewhere
near the end of the color key that the values associated with its colors
are in centimeters or some other physical units.
The only thing I find is the po
May I suggest you consult your local statistician. For reasons that (s)he
can answer, your request makes little sense.
Hint: Nonlinear regression is much different than linear regression: The
design matrix -- and hence the variance of estimators -- is a function of
the parameters being estimated.
Groupwise data summarization is a very common task, and it is worth
learning the various ways to do it in R. Josh showed you one way to
use aggregate() from the base package and Michael showed you one way
of using the plyr package to do the same; another way would be
ddply(df, .(Patient, Region),
There is a difference between parsed functions and .R files. What you see when
you type the name of a function alone on the R command line is a text
representation of the parsed function that is ready to run in RAM. That has
none of the comments or whitespace from the function as you wrote it.
I have had similar problems.
I have several installations of R and now I have no control over which one
opens when I try opening a RData file. The RGUI is registered more than
once, but they all have the exact same appearance in the "choose programs"
menu.
It's become particularly annoying now t
So far in 2011 JSS has published 4 (four !) special volumes. If you have
additional suggestions for special
volumes, let us know. Also, submit your JSS-adapted package vignettes.
If you like what you see, friend us at
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