I have sorted out how to do this - by much trial an error (a comment
from Deepayan in a post some years ago pointed which way to go, though
it took a lot of searching to find even where to start with it:
I've bound the two sets of data together, making the relevant part an
ordered factor. It's pro
How big is "big" ?
I am a fan of the "reshape" package. But It depends on your "big" factor.
Tal
--
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com/ (
"Unexpected symbol" often means missing punctuation.
In this case, it looks like
panel=function(){panel.xyplot(Kalibrierung$Spannung,Kalibrierung
$Magnetfeld)panel.abline(reg=test)}
this argument is missing a semi-colon ";" before "panel.abline".
panel=function(){panel.xyplot(Kalibrierung$Spann
am using ubuntu 9.10 and I am getting and x11 lib/headers not found.
I have installed libx11-dev. I thought this would solve the problem,
but it is not. Any help would be appreciated.
kind regards,
--
Stephen Sefick
Let's not spend our time and resources thinking about things that are
so litt
Dear R users,
I am running R 2.10.0 on OS X 10.5.8.
I had been running 2.10 successfully for about a week (and have used
previous R versions for 2+ years on the same computer) until 2 days ago it
failed to start up for me. Now when I try to start R, the application tries
to initiate for several
Hello,
after a long night I don't find any mistake anymore in my xyplot and it
doesn't work. I want to make a scatterplot with regression line.
Each of it alone is possible but both arguments together are not
working:
"unexpected symbol...
test<-lm(Magnetfeld~Spannung,data=Kalibrierung)
kalib<-
I want to draw 6 plots in ggobi, but do not know how to layout those plots in
one big screen, currently I can only draw in 6 screens which are scattered
every where. I have tried layout(),split.screen and seems they can only deal
with default device window in R.
Thanks a lot!
--
View this messa
TO be specific, here is how I graphed
plot(sigma, delta1, ylim=range(-0.5, 2), xlab='sigma', ylab='delta1--square
delta2--circle', pch=22, type='o')
par(new=TRUE)
plot(sigma, delta2, ylim=range(-0.5, 2), xlab='sigma', ylab='delta1--square
delta2--circle', col='red', axes=FALSE, type='o')
Thank
a graph question. Thanks a lot in advance.
I made two scatterplots on one graph (sigma vs. delta1, sigma vs. delta2)
(20 observations of delta1, delta2 and corresponding sigma) the x-axis is
sigma, the y-axis is either delta1 or delta2. I connected both scatterplots.
To seperate them, one curves
The loess.demo function in the TeachingDemos package helps visualize what is
happening with the loess algorithm, it may answer some of your questions.
Simply, the span can be greater than 1, a span of 3/4 mean use the 3/4 of the
points closest to where you are trying to predict, a span of 1 mean
At 7:45 PM -0800 11/19/09, Stropharia wrote:
Dear R users,
I am running R 2.10.0 on OS X 10.5.8.
I had been running 2.10 successfully for about a week (and have used
previous R versions for 2+ years on the same computer) until 2 days ago it
failed to start up for me. Now when I try to start R,
G'Day Marta,
Just found your post regarding Syntax hilighting for R in Crimson
Editor. I use Crimson for C++ and I find it to be very useful and
now I am starting to work with R.
Would you be kind enough to send me a copy? Thank you so much.
--
Best regards,
Ross
I AM REALLY LOST .
ychu066 wrote:
>
> I want to change the colour of the histogram in each panel by the levels
> of the conoditonal factor.
>
> I have 3 levels in the factor
>
> levels(data[,2])[1]
> levels(data[,2])[2]
> levels(data[,2])[3]
>
> what can i do , in order to c
Just a couple of suggestions. One would be to send your question to
R-sig-mac. The other would be to try starting R from the command line
instead of the gui, and see if the problem occurs in that context.
That might provide a little bit of debugging information, and might
also let you continue
Well, there is a difference between finding a root, and finding a
minimum or a maximum. So you would use one or the other depending on
which you need to do.
-Don
At 7:00 PM -0800 11/19/09, wrote:
I looked at the descriptions for uniroot and optimize and they are
somewhat different but the b
Dear R users,
I am running R 2.10.0 on OS X 10.5.8.
I had been running 2.10 successfully for about a week (and have used
previous R versions for 2+ years on the same computer) until 2 days ago it
failed to start up for me. Now when I try to start R, the application tries
to initiate for several
I appreciate the frank comments; I did try the examples on the help
page, but as someone new to R programming (it still makes my head hurt)
I missed the barplot example (I noticed it but assumed - now I know
wrongly - that barplot was providing only numerical data for plotCI).
I am sorry for my mi
On Nov 19, 2009, at 9:28 PM, Yihui Xie wrote:
Hi all,
Suppose I have a formula: a = log(y) ~ x1 + I(x2^2)
How can I extract the original variable names 'y', 'x1', 'x2' from
this formula? Thanks a lot!
Regards,
Yihui
> all.vars(a)
[1] "y" "x1" "x2"
See ?all.vars
HTH,
Marc Schwartz
Hi all,
Suppose I have a formula: a = log(y) ~ x1 + I(x2^2)
How can I extract the original variable names 'y', 'x1', 'x2' from
this formula? Thanks a lot!
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-6609 Web: http://yihui.name
Department of Statistics, Iowa State University
3211 Snedecor Hall, A
On Nov 19, 2009, at 7:38 PM, Dan Kortschak wrote:
Thank you for that, you have helped amazingly.Are you also able to
point
me to a barplot equivalent?
It is a great puzzle to me that people do not work through the
examples on a help page. Can you offer me a sensible explanation for
that
I looked at the descriptions for uniroot and optimize and they are somewhat
different but the book reference is the same and I am wondering if there are
reasons to pick one over the other?
Thank you.
Kevin
__
R-help@r-project.org mailing list
https:/
Yes, as Charlie said you can write a small C wrapper that calls R. The C
wrapper can then be called from Fortran.
For some examples:
see the package minpack.lm
(http://cran.r-project.org/web/packages/minpack.lm/index.html); the C
wrappers are the functions fcn_lmdif and fcn_lmder.c.
Maybe more
On 19 November 2009 at 20:57, David Scherrer wrote:
| I don't bring my head around how to link an external C++ library with some R
| code using R CMD SHLIB and/or R CMD INSTALL .
|
| So does anybody can tell me how I could easily link an external library that
| is e.g. in
| /usr/local/lib/mMyLibr
On 19 November 2009 at 20:29, Hao Cen wrote:
| Hi,
|
| I am writing a function in C that is meant to be called by R. In the C
| function, I used a gsl function gsl_stats_mean. The code is as simple as
| below
|
| void gsl(double *m, int *dim){
| int r, c;
| r = dim[0];
| c = di
I have been using lattice xyplot and am quite pleased, and I can use the
type=c("b","g") to have it print gridlines into the page, yet if I want to
have a line plot with points on it, how do I get the xYplot to print
gridlines (I use Hmisc xYplot because of its bands method which allows
plotting of
Dear all,
I don't bring my head around how to link an external C++ library with some R
code using R CMD SHLIB and/or R CMD INSTALL .
So does anybody can tell me how I could easily link an external library that
is e.g. in
/usr/local/lib/mMyLibrary ?
Many thanks,
David
[[alternative HTML
> chromosomes
id refseq namelength
1 0 NC_01.9 Homo sapiens chromosome 1 247249719
2 1 NC_02.10 Homo sapiens chromosome 2 242951149
3 2 NC_03.10 Homo sapiens chromosome 3 199501827
4 3 NC_04.10 Homo sapiens chromosome 4 191273063
5 4
Hi,
I am writing a function in C that is meant to be called by R. In the C
function, I used a gsl function gsl_stats_mean. The code is as simple as
below
void gsl(double *m, int *dim){
int r, c;
r = dim[0];
c = dim[1];
double mean = gsl_stats_mean(&m[0], 1, r);
oops, get a little copy-paste happy there. Just need
x <- "mia. SzaĚmitaĚĂł"
x.out<-gsub("Ě","I",x)
x.out<-gsub("Ăł","A",x.out)
x.out
-Ista
On Thu, Nov 19, 2009 at 7:51 PM, Ista Zahn wrote:
> It works the same for me as it does for non-ASCII characters:
>
> x <- "mia. SzaĚmitaĚĂł"
> x.out<-gsub
It works the same for me as it does for non-ASCII characters:
x <- "mia. SzaĚmitaĚĂł"
x.out<-gsub("Ăł","A",x)
x.out<-gsub("Ăł","A",x.out)
x.out<-gsub("Ě","I",x)
x.out<-gsub("Ăł","A",x.out)
x.out
[1]"mia.SzaImitaIA"
-Ista
On Thu, Nov 19, 2009 at 7:03 PM, Steven Kang wrote:
> Hi guys,
>
>
>
>
>
>
Thank you for that, you have helped amazingly.Are you also able to point
me to a barplot equivalent?
Dan
On Thu, 2009-11-19 at 18:53 -0500, David Winsemius wrote:
> > install.packages("plotrix")
> > library(plotrix)
> > ?plotCI
>
__
R-help@r-project
Hi guys,
Are there any feasible methods in searching & finding non-ASCII characters
in R?
For example, from the following object,
x <- mia. SzaÌmitaÌó
The desired output is,
x.out <- mia. SzaImitaIA
Your help in resolving this would be greatly appreciated.
[[alternative
On Nov 19, 2009, at 6:33 PM, Dan Kortschak wrote:
Hi,
I am trying to plot a set of means+/-SD calculated by an external
program (an RDBMS) since the data set is too large to happily fit in R
(740M x 100 values - which are summarised to 100 means/SD by the
DB). I
want to have a mean with whi
Hi,
I am trying to plot a set of means+/-SD calculated by an external
program (an RDBMS) since the data set is too large to happily fit in R
(740M x 100 values - which are summarised to 100 means/SD by the DB). I
want to have a mean with whiskers at +/-1SD.
Can anyone suggest a way to do this?
t
On Nov 19, 2009, at 3:10 PM, ychu066 wrote:
I want to change the colour of the histogram in each panel by the
levels of
the conoditonal factor.
I have 3 levels in the factor
levels(data[,2])[1]
levels(data[,2])[2]
levels(data[,2])[3]
what can i do , in order to chnage the colour ??
Why
Google for sql join and see the examples in Example 4 on the sqldf home page:
http://code.google.com/p/sqldf/#Example_4._Join
On Thu, Nov 19, 2009 at 2:30 PM, JoK LoQ wrote:
>
> Hello,
>
> I would like some help with sqldf syntax.
>
> Suppose I have table 1 and table 2.
> What do I have to
I am doing a project which involve reshaping a large dataset, can any of you
please sugguest me some good reading/websites/ examples can be in R and
SAS
Thanks everyone !!!
--
View this message in context:
http://old.nabble.com/REshaping-large-dataset-tp26421513p26421513.html
Sent from
I want to change the colour of the histogram in each panel by the levels of
the conoditonal factor.
I have 3 levels in the factor
levels(data[,2])[1]
levels(data[,2])[2]
levels(data[,2])[3]
what can i do , in order to chnage the colour ??
ychu066 wrote:
>
> Fore example my code is
>
Fore example my code is
histogram(~data[,8]|data[,2], ylab = "Frequency", xlab = "Score", xlim =
c(1,5), ylim = c(0,100),layout = c(3,1),col=data[,2] )
and i want the colour to be depended on the level of the factor in data[,2].
how do i do that ?
ychu066 wrote:
>
> what is that ?
>
>
>
As of last night I have this exact same problem, Enter, Backspace and the
arrow keys do not work for me in TinnR! I am using windows vista prof on a
64bit laptop. I have uninstalled, re-installed many times and even cleaned
the registry etc. Does anyone know of a reason and a fix for this problem?
To have a different colour for each histogram. Wasn't that your question?
Sorry if I misunderstood.
Did you try it?
ychu066 wrote:
>
> what is that ?
>
>
>
> Hrishi Mittal wrote:
>>
>> Add col=i in the histogram call.
>>
>
>
--
View this message in context:
http://old.nabble.com/How
Hello,
I would like some help with sqldf syntax.
Suppose I have table 1 and table 2.
What do I have to do to generate a table with columns 2,5,6 from table 1
(for example), and columns 3,4,5,9 from table 2, but only when values in
column 2 from table 1 are equal to values in column 5
what is that ?
Hrishi Mittal wrote:
>
> Add col=i in the histogram call.
>
--
View this message in context:
http://old.nabble.com/How-do-I-change-the-colour-and-format-for-the-trelli-plot---tp26418382p26421453.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi all,
I've finally started to use Rscript for my statistical scripting
needs, and find I'm being blocked by what must be a very simple
problem. Specifically, the command lines for my scripts usually
contain: (1) the script name, (2) one or more options and their
arguments, and finally, (3)
This was just answered by Deepayan earlier today:
This code did not produce the plot you have linked to. The answer to
your question depends on how you created the plot, so you have to tell
us that. Changing the color in all panels is easy:
histogram(rnorm(100), col = "goldenrod")
Different col
I have a problem using optim, so I am hoping someone can help me out with it:
Suppose I have the list:
list(D,R,P)
[[1]]
V1 V2 V3 V4
1 0 1 0 1
2 1 1 0 0
3 1 0 1 0
4 0 0 1 1
5 0 1 0 1
6 1 1 0 0
7 1 0 1 0
8 0 0 1 1
9 1 0 1 0
10 0 0 1 1
[[2]]
[1]
Hello,
In reading the loess description I see:
span: the parameter alpha which controls the degree of smoothing.
The default seems to be 0.75. Would it be possible to expand on this decription
so I can avoid trail and error? Can I increase this pass 'span' > 1?
Qualitatively to what degree ch
You could try
?example
and see what the help page says.
-Peter Ehlers
Peng Yu wrote:
Why I don't see 'par(ask=T)' in /usr/lib/R/library/graphics/R-ex/barplot.R?
Is 'par(ask=T)' implicitly called by example()?
On Mon, Sep 7, 2009 at 10:55 AM, RIOS,ALFREDO ARTURO wrote:
Hi Peng
I think t
On Nov 19, 2009, at 4:31 PM, Peng Yu wrote:
On Thu, Nov 19, 2009 at 4:27 PM, Marc Schwartz
wrote:
On Nov 20, 2009, at 10:21 AM, Peng Yu wrote:
There are a few version of apply() (e.g., lapply(), sapply()). I'm
wondering if there is one that does not return anything but just
silently apply a
Hi there,
I have created a single page, multi-panel, xyplot using lattice. Each panel
is a trial (total of 8) from an experiment with 3 variables on the x axis
and the observed value on the y axis. I have added the mean of all trials
so the entire plot looks like this (where M=mean panel and 1-
On Thu, Nov 19, 2009 at 5:31 PM, Peng Yu wrote:
> Is there a way to get the name of the list in the loop body?
This was just discussed! See this thread:
https://stat.ethz.ch/pipermail/r-help/2009-November/218919.html
__
R-help@r-project.org mailing
I want to use both the name and the content. Although, I could do the
following thing.
for(x in names(List)) {
do some thing with x
do some thing with List[[x]]
}
However, I'd prefer something like the following if R offers such
functionality. But it seems not.
for(x in List) {
do somethin
On Thu, Nov 19, 2009 at 4:27 PM, Marc Schwartz wrote:
> On Nov 20, 2009, at 10:21 AM, Peng Yu wrote:
>
>> There are a few version of apply() (e.g., lapply(), sapply()). I'm
>> wondering if there is one that does not return anything but just
>> silently apply a function to the list argument.
>>
>>
invisible(apply(...))
On Thu, Nov 19, 2009 at 5:21 PM, Peng Yu wrote:
> There are a few version of apply() (e.g., lapply(), sapply()). I'm
> wondering if there is one that does not return anything but just
> silently apply a function to the list argument.
>
> For example, the plot function is app
On Thu, Nov 19, 2009 at 2:21 PM, Peng Yu wrote:
> There are a few version of apply() (e.g., lapply(), sapply()). I'm
> wondering if there is one that does not return anything but just
> silently apply a function to the list argument.
>
> For example, the plot function is applied to each element in
On Nov 20, 2009, at 10:21 AM, Peng Yu wrote:
There are a few version of apply() (e.g., lapply(), sapply()). I'm
wondering if there is one that does not return anything but just
silently apply a function to the list argument.
For example, the plot function is applied to each element in 'alist'.
There are a few version of apply() (e.g., lapply(), sapply()). I'm
wondering if there is one that does not return anything but just
silently apply a function to the list argument.
For example, the plot function is applied to each element in 'alist'.
It is redundant to return anything from apply.
I've only recently started using R. One of the problems I come up
against is after having extracted a large dataset (>5M rows) out of
database, I realize I need another variable. In this case I have data
frame with dates. I want to find the minimum date for each value of x1
and add that minimum dat
Why I don't see 'par(ask=T)' in /usr/lib/R/library/graphics/R-ex/barplot.R?
Is 'par(ask=T)' implicitly called by example()?
On Mon, Sep 7, 2009 at 10:55 AM, RIOS,ALFREDO ARTURO wrote:
> Hi Peng
>
> I think this is what you are looking for
>
> par(ask=T)
>
> Alfredo
>
>
> On Sun Sep 06 12:52:31 E
On Thu, Nov 19, 2009 at 12:14 PM, Paul Warren Simonin
wrote:
>
>
> Hello,
>
> I am currently working on a modeling project using Fortran to run
> large repetitive loops (many DO loops). As part of this process I
> would like to use a model fit in R and currently stored as an R
> object. This
I'm currently using the mvtnorm package to model unobserved
heterogeneity in a structural model and using optim to estimate the
model. I have got good clues that convergence is not really a problem
but the hessian matrix estimate is very bad. To overcome this problem,
I'm constructing an OP
Hello,
I am currently working on a modeling project using Fortran to run
large repetitive loops (many DO loops). As part of this process I
would like to use a model fit in R and currently stored as an R
object. This is a rather complex model, a GAMM, and I am curious
whether ther
Hi
In my experiment I have 2 Groups (control patient) and 4 blocks of trials
per subject, n=12 for Control and n=11 for patient
The variances of the patients is very much higher than that of the Control.
1- I tried an ANOVA with Groups as between factor and Blocks as within
factor (repeated me
You probably want to put the map boundaries on top of the contour plot rather
than the other way around. Here is one example of doing that (find the correct
asp may be the hardest part):
library(maps)
x <- seq( -124.7, -67, length.out=25 )
y <- seq( 25, 49, length.out=25 )
z <- outer(x,y,'+')
You need to look at the assign function:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
(... and in addition to R FAQ 7.21 also perhaps read the rest of the R
FAQ.)
There are a ton of very similar questions in the r-help archives, so
you could als
On Fri, 20 Nov 2009, Matthias Demuzere wrote:
> Dear all,
>
> As a newbie in R I would like to do the following (simple?) thing:
>
> to plot a filled.contour plot over a map showing country boundaries (e.g.
> for Europe) What i do is:
> map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE,
Try something like:
gp <- lapply(
paste(("/Users/thomasjackson/Data/GEP&CO/GEP&CO",LETTERS[1:12],"/HPLC_",LETTERS[1:12],"12.csv",sep=''),
read.csv, header=TRUE, sep=',' )
names(gp) <- paste("GandP", LETTERS[1:12], sep='')
Now gp (or whatever you want to call it) will be a list w
Hi Thomas,
Here are two suggestions which storage all the files in a list:
# Parameters
route <- "/Users/thomasjackson/Data/GEP&CO/GEP&CO"
lts <- LETTERS[1:12]
toread <- paste(route,lts,"/HPLC_",lts,"12.csv",sep='')
# Reading in the data -- option 1
myfiles <- list()
for(i in 1:12) myfiles[i]
Dear Paul,
Thank you for your response. I am investigating a differentially expressed
gene detected method that cannot be applied on R 2.9.1 that currently I
used. The reason about why it cannot work due to the changes of new version
of R. And, recently, I study a lots of methods that developed b
Dear R Users,
I am trying to read in a series of csv files which vary by the letter on the
end of he file name. When I input what seems to be a logical for loop I get an
error message that doesn't make sense to me.
> for(i in 1:12){ paste("G&P", LETTERS[i],sep='')
> <-read.csv(paste("/Users
Hello,
I just started using R to do epidemiologic simulation research using the Cox
proportional hazard model. I have 2 covariates X1 and X2 which I want to
model as h(t,X)=h0(t)*exp(b1*X1+b2*X2). I assume independence of X from t.
After I simulate Time and Censor data vectors denoting the cens
Actually it may be that the documentation is at fault more than the
code. The help page for Plot.Fore says that the first argument is a
time series data set and although that seems to suggest that it should
be a ts object the code seems to be written assuming a plain numeric
vector; therefore, you
On Nov 19, 2009, at 12:23 PM, Richard M. Heiberger wrote:
Robert Terwilliger wrote:
Dear R experts,
I have a so-called person-level data frame that I need to transform
into a person-period data frame.
If the lingo is unclear, the data have one row for each subject, with
repeated measures dat
Awesome!
Thanks a ton!
I guess I had overlooked how it was really working.
I will still have to reflect on why it was working running it straight through,
but not being nested.
That is kind of a mystery. Oh well...
Thanks again.
- Original Message
From: baptiste auguie
To:
Can't be too hard to convert it in pixmap representation, but you need
to tell us hoe the representation looks like, probbaly with a short code
that makes it reproducible (e.g. generate some data that we can use in R).
Uwe Ligges
Rajarshi Guha wrote:
Hi, I have some code that uses rJava. One
Alla Bulashevska wrote:
Dear R users,
i try to use function dmvnorm(x, mean, sigma, log=FALSE)
from R package mvtnorm to calculate the probability of x
under the multivariate normal distribution with mean equal
to mean and covariance matrix sigma.
I become the following
Error in solve.default(
Dear all,
As a newbie in R I would like to do the following (simple?) thing:
to plot a filled.contour plot over a map showing country boundaries (e.g. for
Europe)
What i do is:
map('worldHires',xlim=c(-10,40),ylim=c(35,70),boundary = TRUE,border=0.1)
map.axes()
filled.contour(mslp, zlim=c(1000,1
Robert Terwilliger wrote:
Dear R experts,
I have a so-called person-level data frame that I need to transform
into a person-period data frame.
If the lingo is unclear, the data have one row for each subject, with
repeated measures data each in a separate column.
I need to transform these data
Ok,
Thanks all.
Rick.
--
From: "Achim Zeileis"
Sent: Thursday, November 19, 2009 3:06 PM
To: "Ricardo Gonçalves Silva"
Cc: "R-Help" ;
Subject: Re: [R] Problem with zoo and BootPR packages
On Thu, 19 Nov 2009, Ricardo Gonçalves Silva wrote:
H
On Thu, 19 Nov 2009, Ricardo Gonçalves Silva wrote:
Hi,
I'm trying to plot the forecasts I generated using the Plot.Fore function of
the BootPR package.
But I got an error from zoo:
My data:
Time Series:
Start = 1
End = 18
Frequency = 1
[1] 38731 38628 39117 92809 71984 31226 58613 7
Thanks for your reply! I just added some more details below.
Our code needs around 1GB of RAM and all machines and R configurations have
its default maximum above this number.
Our suspicion is that the windows server could run the code in half of its
current time (given the apparent factor of 2 b
Contact the BootPR maintainer regarding a bug in this line of Plot.Fore:
y1 <- zooreg(x, start, end, frequency)
where x is a ts object but that may not be used in that context.
as.zooreg is available for converting ts series (and certain other
objects) to zooreg objects.
2009/11/19 Ricardo Go
Dear R experts,
I have a so-called person-level data frame that I need to transform
into a person-period data frame.
If the lingo is unclear, the data have one row for each subject, with
repeated measures data each in a separate column.
I need to transform these data so that each subject has mul
Hi,
I'm trying to plot the forecasts I generated using the Plot.Fore function of
the BootPR package.
But I got an error from zoo:
My data:
Time Series:
Start = 1
End = 18
Frequency = 1
[1] 38731 38628 39117 92809 71984 31226 58613 72360 107956 92066
[11] 95208 99098 95848 120383
Dear List,
I thought it would be much easier to put a second query into a second mail.
I need to print 426*1 blocks of variance components data, where 426 is
the number of columns of data that have 1 permutations of variance
generated for each of them.
I have tried printing out a sma
Dear List,
I am trying to run a mixed model which, on the R console, prints output as
follows:
[1] "Marker"
[1] "perm no."
[1] NA
Linear mixed model fit by REML
Formula: peg.no.prm ~ 1 + (1 | family/f)
Data: modeldf
AIC BIC logLik deviance REMLdev
3119 3134 -1555 31123111
Random e
Hello everybody, I started to use Rexcel and I am getting an error with the
following code:
Sub AutoForma1_Clique()
Call RInterface.StartRServer
Call RInterface.RRun("setwd(""C:/Program Files/R/R-2.10.0/Working
Directory"")")
Call RInterface.RRun("getwd()")
End Sub
The error is the fo
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Stephan Dlugosz
> Sent: Thursday, November 19, 2009 7:03 AM
> To: r-help@r-project.org
> Subject: [R] Efficient cbind of elements from two lists
>
> Hi!
>
> I have a data.frame
On Nov 19, 2009, at 9:25 AM, Carlos Hernandez wrote:
Dear All,
I appreciate any advice or hints you could provide about the
following.
We are running R code in a server (running Windows XP and QuadCore
Xeon
processors, see details below) and we would like to use the server
efficiently. Ou
Dear Stephan,
Here is a suggestion using do.call():
res <- do.call(cbind, yourlist)
res
HTH,
Jorge
On Thu, Nov 19, 2009 at 10:03 AM, Stephan Dlugosz <> wrote:
> Hi!
>
> I have a data.frame "data" and splitted it.
>
> data <- split(data, data[,1])
>
> This is a quite slow procedure; and I do n
Hi!
I have a data.frame "data" and splitted it.
data <- split(data, data[,1])
This is a quite slow procedure; and I do not want to do it again. So,
any unsplit and "resplit" is no option for me.
But: I have to cbind "variables" to the splitted data from another list,
that contains of vectors
Dear All,
I appreciate any advice or hints you could provide about the following.
We are running R code in a server (running Windows XP and QuadCore Xeon
processors, see details below) and we would like to use the server
efficiently. Our code takes a bit more than 6 seconds per 25 iterations in
th
Hi,
I think your ddply call with a calculation inside ".( )" is the
problem. Are you sure you need to do this? Performing the cut outside
ddply seems to work fine,
determine_counts<-function()
{
min_range<-1
max_range<-30
bin_range_size<-5
Me_df<-data.frame(Data
Look at the ggobi program and the rggobi package for interactions between R and
ggobi.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>
On Nov 19, 2009, at 9:40 AM, Антон Морковин wrote:
Dear all,
I need to create n*n table with sums of all possible pair
combinations of numbers from n-row column. What function allows it?
?expand.grid
Best regards,
A.Morkovin
__
R-help@r-pro
Thanks to everybody who replied - I got three distinct, very useful
suggestions.
Bjarke Christensen
Romain Francois
Hello Sir
Thanx a lot, will try Pareto chart for my data
Regards
Our Thoughts have the Power to Change our Destiny.
Sunita
On Thu, Nov 19, 2009 at 12:29 PM, Petr PIKAL wrote:
> Well, from what you say it seems to me that you could also use Pareto
> charts together with some aggregation of da
Hello Sir
Thanx even I will try to work out on your suggestions, will keep you updated
on the progress. Thanx a lot
Regards
Our Thoughts have the Power to Change our Destiny.
Sunita
On Thu, Nov 19, 2009 at 3:50 PM, Jim Lemon wrote:
> On 11/19/2009 03:13 AM, Sunita Patil wrote:
>
>> Hello Sir
Dear all,
I need to create n*n table with sums of all possible pair combinations of
numbers from n-row column. What function allows it?
Best regards,
A.Morkovin
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
While putting my R code into functions, I've encountered a ddply function
nesting issue and need a bit of advice on the proper way to fix it. I've tried
several approahces, but neither worked and I need to have the ability to
include the "cut", "range", and "fullseq" methods within ddply. (For
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