I am trying to do something simple with ggplot. I wish to draw a density
plot split by group, and fill each group with a different colour (and
each with an alpha =0.25). I have tried a number of variations of the
following, but cannot find a way to define the colour of the fill, its
transparency an
Marc Schwartz wrote:
It's always dangerous to say always, but "in general", the use of:
Path/To/File
with *forward* slashes, will work on both Windows and Linux, including
the use of relative paths.
Also, notice the existence of file.path() and system.file(). (Other
separators than forw
Dear Arams,
I would suggest to use lme() instead of lmer(), and then to use a variance function to model the
heteroscedasticity in the within-group errors, such as:
model.new=update(model,weights=varPower(form=~primary.covariate))
where model and model.new are lme fits, and primary.covariate
Dear Jon,
The function stepAIC() in Venable´s & Ripley´s MASS library does the job.
Best wishes
Christoph
Jon Zadra schrieb:
Hello,
I make frequent use of the *step()* and, for plotting, *all.effects()
*functions for *lm()* objects. I am now doing more with *lme()* random
effects models,
And what should the matrix look like?
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
Daren Tan wrote:
How to convert the strings into matrix ?
(strings <- strsplit(c("1 2 3", "1 2", "1 3 4 5"), " "))[[1]
On Thu, 24 Jul 2008, Christoph Scherber wrote:
Dear Jon,
The function stepAIC() in Venable´s & Ripley´s MASS library does the job.
Actually, it is extractAIC method in the MASS package that does the job
for stepAIC() or step(), so if you attach MASS then step() should work.
As for 'all.eff
I want the matrix to look like this:
[,1] [,2] [,3] [,4]
[1,] "1" "2" "3"
[2,] "1" "2" [3,] "1" "2" "4" "5"
I tried to use do.call(rbind, strings) but failed due to unequal row lengths.
> Date: Thu, 24 Jul 2008 09:35:21 +0100> From: [EMAIL PROTECTED]> To: [EMAIL
> PROTECTED]> CC: [EMAIL
I frequently work with hourly data in GMT format, and ensure that any data I
read into R are expressed in GMT through as.POSIXct(mydate, tz = "GMT").
I am interested in processing air pollution data in GMT where where peaks in
emissions (say from road traffic sources) tend to occur at the same L
Daren Tan wrote:
> I want the matrix to look like this:
>
> [,1] [,2] [,3] [,4]
> [1,] "1" "2" "3"
> [2,] "1" "2" [3,] "1" "2" "4" "5"
>
> I tried to use do.call(rbind, strings) but failed due to unequal row lengths.
>
you can't {r,c}bind them because of different lengths.
one way to r
Hello All,
Could anybody point me out any port of "Global Optimization Test
Problems" OR "Constraint Satisfaction Test Problems" part of Coconut[%]
benchmark or other such similar benchmarks for R.
[%] Coconut benchmark:
http://www.mat.univie.ac.at/users/neum/public_html/glopt/coconu
On Thu, 24 Jul 2008, DCC wrote:
I frequently work with hourly data in GMT format, and ensure that any data I
read into R are expressed in GMT through as.POSIXct(mydate, tz = "GMT").
That makes no sense. POSIXct times are absolute times, not in any
timezone. You can display them in any time
 Dear R experts
Â
I generated rules using apriori method in arules package. Though I can access
rules using %in% function but I am unable to access a specific rule by its
unique identifier number. I want to use rule number for further analysis..
Â
Thanking you in advance.
Â
Daniel
Amsterdam
Â
Hi all,
Is there an equivalent to the general * operator in R, where * can
represent any character?
I have a dataset, one column being date, date ranging between
April-September, 97-06. I would like to be able to acquire the data for
a specific month, say September, so that I can take average t
hello all,
i'm running R-2.6.1 on solaris 10 (x86). I added the package aplpack
without problems, but when i try to use it, i got errors:
library(aplpack)
Loading required package: tcltk
Error in namespaceExport(ns, exports) :
undefined exports :addTclPath, as.tclObj, is.tclObj, is.tkwin,
tclf
Something like this might do what you want:
dates<-paste(c("2008"), c("Jan", "Feb", "Sep", "Sep", "Dec"), 1:3,
sep="")
temp<-rnorm(length(dates))*30
data1<-data.frame(Dates=dates, Temp=temp)
data.s1<-data1[grep("Sep", data1[["Dates"]]),]
> data.s1
Dates Temp
3 2008Sep3 22.263627
4 2
Williams, Robin wrote:
Hi all,
Is there an equivalent to the general * operator in R, where * can
represent any character?
I have a dataset, one column being date, date ranging between
April-September, 97-06. I would like to be able to acquire the data for
a specific month, say September, so
On Wed, 2008-07-23 at 12:46 -0700, jcarmichael wrote:
> Hello everyone. I am new to R, so please bear with me. I am trying to find
> an easy way to export descriptive statistics and other information about my
> data frame to a LaTeX format.
>
> I have found the describe function in PrettyR to be
Hi all,
Thanks for the help with my previous post.
I have just two more questions for the minute.
I think I said in a previous post that I like to use the terminal,
i.e. run rterm.exe. On exiting the terminal, I am asked if I want to
save the workspace. If I hit y (yes), the workspace is just
Dear R-Users,
I am trying to apply a bootstrap to a GAM in order to calculate the 95%
confidence intervals for a smooth curve obtained by the “plot.gam”
function of the mgcv package. Nonetheless, I am getting some
difficulties in transposing the results for the graphs.
I used the following c
Wacek Kusnierczyk wrote:
> Daren Tan wrote:
>
>> I want the matrix to look like this:
>>
>> [,1] [,2] [,3] [,4]
>> [1,] "1" "2" "3"
>> [2,] "1" "2" [3,] "1" "2" "4" "5"
>>
>> I tried to use do.call(rbind, strings) but failed due to unequal row lengths.
>>
>>
> you can't {r,c}bi
Guys,
I'm running LAM 7.1.4 and R/Rmpi on a SuSE 9.3 box.
When I do
>library(Rmpi)
The whole thing crashes with the following message:
-
No SSI RPI modules said that they were available to run.
This should only happe
Try this:
print(t(sapply(strings, '[', 1:max(sapply(strings, length, na=".")
On 7/24/08, Daren Tan <[EMAIL PROTECTED]> wrote:
>
> I want the matrix to look like this:
>
> [,1] [,2] [,3] [,4]
> [1,] "1" "2" "3"
> [2,] "1" "2" [3,] "1" "2" "4" "5"
>
> I tried to use do.call(rbind, strings
I have created following interactive plot :
library(TeachingDemos)
plott = function(x)
{
return(hist(rnorm(as.integer(1000, 10, x)), xlab=NA))
}
tkexamp(plott, list(x=list('slider',from=1,to=40, resolution=0.1, init=2)),
plotloc='top')
Here everything works fine, but the problaem is wh
lm(x~y+z)
this should work- I think
On Thu, Jul 24, 2008 at 7:01 AM, Williams, Robin <
[EMAIL PROTECTED]> wrote:
> Hi all,
> Thanks for the help with my previous post.
> I have just two more questions for the minute.
> I think I said in a previous post that I like to use the terminal,
> i.e. run
On 7/24/2008 7:34 AM, Megh Dal wrote:
I have created following interactive plot :
library(TeachingDemos)
plott = function(x)
{
return(hist(rnorm(as.integer(1000, 10, x)), xlab=NA))
}
tkexamp(plott, list(x=list('slider',from=1,to=40, resolution=0.1, init=2)),
plotloc='top')
Here every
I tried to run the following example from section 4.1.4 of the "Scatterplot3d
-
an R package for Visualizing Multivariate Data" vignette and got an error on
the part that plots the regression plane:
> library(scatterplot3d)
> data(trees)
> s3d <- scatterplot3d(trees, type = "h", color = "blue",
+
Hi Robin,
>> I ... can't get lm to work despite reading the help. I can get it to work
>> with a single
>> explanatory variable, EG model <- lm(data$admissions~data$maxitemp)
I'll answer just the second of your questions. Advice: don't just read the
help file, look at the examples and run them;
Dear List,
I ran into some trouble handling missing values.
Assume 2 vectors (numeric) including NAs
a <- c(rep(seq(1,4),4),NA,NA)
b <- c(sample(1:2,14,replace=T),NA,NA,1,2)
I want to replace the values of vector a that are smaller than 2 and
larger than 3 into NAs only in case vector b equals
On 7/24/2008 7:01 AM, Williams, Robin wrote:
Hi all,
Thanks for the help with my previous post.
I have just two more questions for the minute.
I think I said in a previous post that I like to use the terminal,
i.e. run rterm.exe. On exiting the terminal, I am asked if I want to
save the work
On 7/24/2008 8:02 AM, Tom La Bone wrote:
I tried to run the following example from section 4.1.4 of the "Scatterplot3d
-
an R package for Visualizing Multivariate Data" vignette and got an error on
the part that plots the regression plane:
library(scatterplot3d)
data(trees)
s3d <- scatterplot3d
Thanks for this. Can you please explain me what are those MDI and SDI and the
diff in laymans language?
--- On Thu, 7/24/08, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> From: Duncan Murdoch <[EMAIL PROTECTED]>
> Subject: Re: [R] Interactive plot
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
>
If you really want to work with character strings, you could do something like
data.s1 <- data$date[grep(".*-Sep-.*", date)]
(see ?regexp for more on pattern matching)
but there are better ways to handle dates. You should convert the
character strings into Date objects: see ?Date. And check out t
Q1:
Have a look at:
?save
?save.image
?load
Q2:
I am not sure this is right:
model<-lm(data$admissions~data$maxitemp)
Try and code a line that looks more like this with the variables the
names of the columns in your data:
model<-lm(admissions ~ maxitemp + minitemp, data=data)
Have a read
On 7/24/2008 8:24 AM, Megh Dal wrote:
Thanks for this. Can you please explain me what are those MDI and SDI and the
diff in laymans language?
Not really. Just try them both, the differences are obvious.
Duncan Murdoch
--- On Thu, 7/24/08, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
From:
Hi Robin,
to answer your question: I think you are looking for a regular expression, have
a look at regex().
In your particular case, where all the strings show the same structure, you can
also use substr() to cut out the substring of interest.
More generally, you could consider to convert your
> I ran into some trouble handling missing values.
>
> Assume 2 vectors (numeric) including NAs
>
> a <- c(rep(seq(1,4),4),NA,NA)
> b <- c(sample(1:2,14,replace=T),NA,NA,1,2)
>
> I want to replace the values of vector a that are smaller than 2 and
> larger than 3 into NAs only in case vector b e
You didn't mention your OS but if its Windows then see:
http://tolstoy.newcastle.edu.au/R/help/05/07/8950.html
and try adding this line to that -- actually it gives me an
error message (from Excel?) using Excel 2007 on Vista,
RDCOMClient 0.92-0 and "R version 2.7.1 RC (2008-06-16 r45949)"
but it
Hi
I have the process id,
How can I know whether it is in active state or not
Thanks
K.Ravichandra
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/lis
Dear List-serv members:
How can I structure a for loop so that it will increment the counter by two
using R? I am just looking for a simple example that shows where the
incrementing part should go. Thanks!
Jennifer
--
Jennifer Hayes Clark
Assistant Professor
Department of Political Science
Un
Dear R users,
Â
I have a dataframe with lot of duplicate cases and I want to delete duplicate
ones which have low rank and keep that case which has highest rank.
e.g
Â
> df1
 cno     rank
1Â 1342Â Â Â 0.23
2Â 1342Â Â Â 0.14
3Â 1342Â Â Â 0.56
4Â Â 2568Â Â Â 0.15
5Â 2568Â Â Â 0.89
Kurapati, Ravichandra (Ravichandra) wrote:
Hi
I have the process id,
How can I know whether it is in active state or not
This is not an R question. Find out how to do it in your operating
system (whatever that is), and do the same in R using system().
Duncan Murd
Dear Daniel,
Try this:
x=read.table(textConnection("cno rank
1 13420.23
2 13420.14
3 13420.56
4 25680.15
5 25680.89"),header=TRUE,sep="")
x[cumsum(tapply(x$rank,x$cno,which.max)),]
cno rank
3 1342 0.56
5 2568 0.89
HTH,
Jorge
On Thu, Jul 24, 2008 at 10:00 AM, D
Try this:
aggregate(x$rank, list(cno=x$cno), max)
On 7/24/08, Daniel Wagner <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> I have a dataframe with lot of duplicate cases and I want to delete duplicate
> ones which have low rank and keep that case which has highest rank.
> e.g
>
> > df1
> cno
On Thu, 2008-07-24 at 02:25 +0530, Arun Kumar Saha wrote:
> There is a R function to calculate weighted mean : weighted.mean() under
> stats package. Is there any direct R function for calculating weighted
> variance as well?
Here are two ways; weighted.var() is via the usual formula and
weighted.
Daniel -
First, use order() to arrange the data.frame into an appropriate format.
Then, use duplicated() with the negation operator to get rid of the
duplicated values.
Daniel Wagner wrote:
Dear R users,
Â
I have a dataframe with lot of duplicate cases and I want to delete duplicate ones
Hi, I'm trying to use fastICA package but I only get an error message
like following.
> library(fastICA)
Error in dyn.load(file, DLLpath = DLLpath, ...) :
unable to load shared library
'/usr/lib/R/library/fastICA/libs/fastICA.so':
/usr/lib/R/library/fastICA/libs/fastICA.so: undefined symbol:
Hi all,
I am getting the following error message.
Does somebody know what needs to happen here?
I have tried re-installing the RGtk2 package and also downloading a .dll
file and installing it in the RGtk2 file folder
Error in dyn.load(x, as.logical(local), as.logical(now)) :
unable to
Thank you Frank and all for your advices.
Here I attach the raw data from the Pawinski's paper. I have obtained
permission from the corresponding Author to post it here for everyone.
The only condition of use is that the Authors retain ownership of the
data, and any publication resulting from the
on 07/24/2008 09:00 AM Daniel Wagner wrote:
Dear R users,
I have a dataframe with lot of duplicate cases and I want to delete duplicate ones which have low rank and keep that case which has highest rank.
e.g
df1
cno rank
1 13420.23
2 13420.14
3 13420.56
4 25680.1
this works
cno = c(rep(1342,times=3),rep(2568,times=2))
rank= c(.23,.14,.56,.15,.89)
df1 = data.frame(cno,rank)[order(cno,rank),]
cnou= unique(cno)
ind = match(cno,cnou)
where = tapply(rank,ind,length)
where = cumsum(as.numeric(where))
df1[where,]
regards,
PF
2008/7
Hy Richie, thank you for the quick response.
Unfortunately my problems hold on.
Once again: 2 vectors (numeric) including NAs
My intention: I want to replace the values of vector a that are smaller
than 2 and larger than 3 into NAs only in case vector b equals 1
a <- c(rep(seq(1,4),4),NA,NA)
b
Hi,
I'm using which to find the position of a value in my data table, and it is
returning the correct position and the position of another number that differs
by an extra decimal place. For example, when I do:
where1<-which(Operons==3573.1,arr.ind=TRUE)
it returns the position of that number and
Greg and all,
Just another thought on bias and variability. As I tried to explain, I
perceive this problem as a very practical problem.
The equation, that is the goal of this work, is supposed to serve the
clinicians to estimate a pharmacokinetic parameter. It therefore must be
simple and al
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
[EMAIL PROTECTED] wrote:
Hi,
I'm using which to find the position of a value in my data table, and it is
returning the correct position and the position of another number that differs
by an ex
Duncan Murdoch wrote:
On 7/24/2008 8:02 AM, Tom La Bone wrote:
I tried to run the following example from section 4.1.4 of the
"Scatterplot3d
-
an R package for Visualizing Multivariate Data" vignette and got an
error on
the part that plots the regression plane:
library(scatterplot3d)
data
On Fri, 25 Jul 2008, paulandpen wrote:
Hi all,
I am getting the following error message.
Does somebody know what needs to happen here?
I have tried re-installing the RGtk2 package and also downloading a .dll file
and installing it in the RGtk2 file folder
Error in dyn.load(x, as.logical(lo
On Thu, Jul 24, 2008 at 2:23 AM, Williams Scott
<[EMAIL PROTECTED]> wrote:
> I am trying to do something simple with ggplot. I wish to draw a density
> plot split by group, and fill each group with a different colour (and
> each with an alpha =0.25). I have tried a number of variations of the
> fol
Dear all,
I am fitting a multivariate linear model with 7 response variables and 1
explanatory variable.
The following matrix P:
P <- cbind(
c(1,-1,0,0,0,0,0),
c(2,2,2,2,2,-5,-5),
c(1,0,0,-1,0,0,0),
c(-2,-2,0,-2,2,2,2),
c(-2,1,0,1,0,0,0),
c(0,-1,0,1,0,0,0))
should consist of orthogonal elemen
On Thu, Jul 24, 2008 at 04:45:14PM +0200, Kunzler, Andreas wrote:
> Hy Richie, thank you for the quick response.
>
> Unfortunately my problems hold on.
>
> Once again: 2 vectors (numeric) including NAs
> My intention: I want to replace the values of vector a that are smaller
> than 2 and larger
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Kunzler, Andreas
> Sent: Thursday, July 24, 2008 7:45 AM
> To: [EMAIL PROTECTED]
> Cc: r-help@r-project.org; [EMAIL PROTECTED]
> Subject: Re: [R] NAs - NAs are not allowed in subscripted assignments
>
On 24-Jul-08 15:30:57, Christoph Scherber wrote:
> Dear all,
> I am fitting a multivariate linear model with 7 response variables and
> 1 explanatory variable.
>
> The following matrix P:
>
> P <- cbind(
> c(1,-1,0,0,0,0,0),
> c(2,2,2,2,2,-5,-5),
> c(1,0,0,-1,0,0,0),
> c(-2,-2,0,-2,2,2,2),
> c(-2
Clark, Jennifer H wrote:
> Dear List-serv members:
> How can I structure a for loop so that it will increment the counter by two
> using R? I am just looking for a simple example that shows where the
> incrementing part should go. Thanks!
>
>
for (i in seq(, , by=2)) { ... }
vQ
___
Dear Ted,
Thanks for your help; is there any straightforward way to construct an orthogonal contrast matrix by
hand? Of course
contrasts(as.factor(letters[1:6]))
would also do the job, but user-defined contrasts would make more sense.
Ideally, I would like to have a function that "tests" for
Erik Iverson biostat.wisc.edu> writes:
>
> See
>
>
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
>
> naw3 duke.edu wrote:
> > I'm using which to find the position of a value in my data table
[snip]
> For example, when I do:
> >
> > where1<
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Gabor Csardi
> Sent: Thursday, July 24, 2008 8:39 AM
> To: Kunzler, Andreas
> Cc: r-help@r-project.org; [EMAIL PROTECTED];
> [EMAIL PROTECTED]
> Subject: Re: [R] NAs - NAs are not allowed in subscripte
On 24-Jul-08 15:35:45, Erik Iverson wrote:
> See
>
> http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-the
> se-numbers-are-equal_003f
That's not an applicable response: The difference between 3573.1
and 3573.15 is far too large for them to be confused by "==".
Thr confusion on
Ben Bolker wrote:
Erik Iverson biostat.wisc.edu> writes:
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
naw3 duke.edu wrote:
I'm using which to find the position of a value in my data table
[snip]
For example, when I do:
where
To quote (or as nearly so as I can) Einstein's famous remark:
"Make everything as simple as possible ... but no simpler"
Moreover, "as possible" here means "maintaining fidelity to scientific
validity," not "simple enough for me to understand." So I don't think a
physicist can explain relativist
On Thu, Jul 24, 2008 at 09:30:54AM -0700, Nordlund, Dan (DSHS/RDA) wrote:
[...]
> > > a <- c(rep(seq(1,4),4),NA,NA)
> > > b <- c(rep(seq(1,2),7),NA,NA,1,2)
> >
> > Andreas,
> >
> > what is wrong with
> >
> > a[ (a < 2 | a > 3) & b==1 ] <- NA
> >
> > ? Isn't this what you want?
> >
[...]
>
>
I'm trying to calculate the percent change for a time-series variable.
Basically the first several observations often look like this,
x <- c(100, 0, 0, 150, 130, 0, 0, 200, 0)
and then later in the life of the variable they're are generally no more
0's. So when I try to calculate the percent c
What are the arguments against fidelity of this concept to scientific
validity?
The concept of predictive performance was devised by one of you,
biostatisticians - not me! I accept the authoritative view of the person
that did it, especially because I do understand it.
When I think of it, ex
I'm trying to calculate the percent change for a time-series variable.
Basically the first several observations often look like this,
x <- c(100, 0, 0, 150, 130, 0, 0, 200, 0)
and then later in the life of the variable they're are generally no more
0's. So when I try to calculate the percent c
Does anybody have any suggestions regarding applying standard regression
packages lm(), hccm(), and others within a parallel environment? Most of
the packages I've found only deal with iterative processes (bootstrap) or
simple linear algebra. While the latter might help, I'd rather not program
th
Hi, I have a R package, which is a .tar.tar file. And it is said to be
source code for all platforms. And the author said it should install on any
system (but he didn't know about Windows). I am wondering if I can use this
package in Windows R.
Thank you,
Cindy
[[alternative HTML version d
> -Original Message-
> From: Gabor Csardi [mailto:[EMAIL PROTECTED]
> Sent: Thursday, July 24, 2008 9:59 AM
> To: Nordlund, Dan (DSHS/RDA)
> Cc: r-help@r-project.org
> Subject: Re: [R] NAs - NAs are not allowed in subscripted assignments
>
> On Thu, Jul 24, 2008 at 09:30:54AM -0700, Nordl
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
[EMAIL PROTECTED] wrote:
| Hi guys,
|
| Thank you all for looking at this, and I'm sorry - I was making a silly
| mistake.
| I was using a previous version of the data table in R, while looking at
| a new
| file in excel (its easier for me to view the
As always, 'str(rules)' will provide you with a view on the internal
structure of 'rules'. By the way, 'rules' is of class 'rules', see the
"rules-class" entry in the ARULES help pages, with more explanations on the
meaning of attributes.
For example, '[EMAIL PROTECTED]' will show support, confid
on 07/24/2008 07:59 AM Clark, Jennifer H wrote:
Dear List-serv members: How can I structure a for loop so that it
will increment the counter by two using R? I am just looking for a
simple example that shows where the incrementing part should go.
Thanks!
It would be helpful to have an idea of w
Dear Gabor,
I am really sorry about the file attachment. As you might have figured
out I am quite new to the forum interaction techniques. I will keep your
suggestion in mind. Thanks for taking the time to test the data I sent you. I
found out where exactly the problem is. The surprising p
Hi,
I want to test for independence in my GLS model fitp2, but when I try to use the
dwtest function in the lmtest library, I get the error message "Error in
terms.default(formula) : no terms component".
The model and data set are below. Any suggestions would be really helpful!
Thanks a lot in ad
Michal Figurski wrote:
Thank you all for your words of wisdom.
I start getting into what you mean by bootstrap. Not surprisingly, it
seems to be something else than I do. The bootstrap is a tool, and I
would rather compare it to a hammer than to a gun. People say that
hammer is for driving na
Michal Figurski wrote:
Thank you Frank and all for your advices.
Here I attach the raw data from the Pawinski's paper. I have obtained
permission from the corresponding Author to post it here for everyone.
The only condition of use is that the Authors retain ownership of the
data, and any public
I wonder what such a benchmark test would be good for.
Relevant and speaking tests in Global Optimization are formulated as models
in AMPL, GAMS, GMPK, MPS, etc. Rewriting a model in matrix form can be a
tedious manual task and I have not yet seen it.
Very few optimization packages in R accept s
Hi all,
I have a set of data which looks similar to this (only way bigger!)
linktyperegister
F D a
E D b
R D c
A T d
D T a
F D b
. . .
. . .
and would like to
On Thu, Jul 24, 2008 at 10:39:34AM -0700, Nordlund, Dan (DSHS/RDA) wrote:
[...]
> Yes, it does help. I was misunderstanding how logical values are
> used for indexing. I assumed incorrectly that a value would be
> returned only if the index expression evaluated as TRUE. It would
> seem that the
Hi!
Suppose I have a factor:
> fac1 <- factor(rep(c("dog","cat","tree"),c(2,3,4)))
> fac1
[1] dog dog cat cat cat tree tree tree tree
Levels: cat dog tree
> length(fac1)
[1] 9
Now I want to get rid of the dogs:
> fac1 <- fac1[3:9]
> fac1
[1] cat cat cat tree tree tree tree
Levels: cat do
Try:
fac1[3:9][drop=T]
On Thu, Jul 24, 2008 at 4:19 PM, Edna Bell <[EMAIL PROTECTED]> wrote:
> Hi!
>
> Suppose I have a factor:
>
>> fac1 <- factor(rep(c("dog","cat","tree"),c(2,3,4)))
>> fac1
> [1] dog dog cat cat cat tree tree tree tree
> Levels: cat dog tree
>> length(fac1)
> [1] 9
>
> No
Hi Edna,
Because I am "always" subsetting, I keep the following function handy
mydata[] <- lapply(mydata, function(x) if(is.factor(x)) x[,drop=T] else x)
This will strip out all factor levels that have been dropped by a previous
subsetting operation. For novice users of R (though I am not sugge
Hi yet again!
Thank you for being patient with me.
Is there a "how to" for ggplot2, please? I would like to look at it,
but have no idea where to start, please.
Thanks,
Edna Bell
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On 25/07/2008, at 5:24 AM, Robert Baer wrote:
I'm trying to calculate the percent change for a time-series
variable.
Basically the first several observations often look like this,
x <- c(100, 0, 0, 150, 130, 0, 0, 200, 0)
and then later in the life of the variable they're are generally
Hi again.
Sorry to be a pest.
What are Cairo graphics and how to find out about them, please?
Thanks yet again,
Edna
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PLEASE do read the posting guide http://www.R-projec
--- On Thu, 7/24/08, Michal Figurski <[EMAIL PROTECTED]> wrote:
> From: Michal Figurski <[EMAIL PROTECTED]>
> Subject: Re: [R] Coefficients of Logistic Regression from bootstrap - how to
> get them?
> To:
> Cc: "r-help@r-project.org"
> Received: Thursday, July 24, 2008, 11:02 AM
> Greg and a
Dear R Experts,
I am trying to create several subsets that I want to use as separate
datasets. After separate subsets are created for each country, I want to
do some calculations for each. I did the following:
data <- data.frame(na.omit(read.spss("trial.sav")))
attach (data)
Netherlands <-
Dear R users,
I try to analyse a self paced reading experiment where I have two fixed
variables (Relativiser, Attachment), two random variables (Participant,
ItemNbr), and one covariable (BiasValue). The dependant variable is RT, reading
time region by region.
My aim is to remove the variance i
Returning to the old question ...
> Is it possible to subset an n-dimensional array by a
> vector of n dimensions?
On Sat, Jul 12, 2008 at 1:34 AM, Wolfgang Huber <[EMAIL PROTECTED]> wrote:
> > do.call("[", list(x,1,2,TRUE))
> [1] 3 9 15 21
>
> See also http
Spencer -
Spencer wrote:
Dear R Experts,
I am trying to create several subsets that I want to use as separate
datasets. After separate subsets are created for each country, I want to
do some calculations for each. I did the following:
data <- data.frame(na.omit(read.spss("trial.sav")))
at
> What are Cairo graphics and how to find out about them, please?
http://www.google.de/search?hl=de&q=cairo+graphics
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
ht
A professor here at Vanderbilt sent me the following code. Each of the
text strings should right justify against the center vertical bar, but
because of font issues it doesn't. I understand that there are
workarounds, but I was just curious if this was consistent across all
platforms. On linux
Hi Cindy,
>> Hi, I have a R package, which is a .tar.tar file. And it is said to be
>> source code for all
>> platforms. ... I am wondering if I can use this package in Windows R.
If it is source code you would first need to compile it to binary format
before you can use it. Can you do that?
W
or sapply as I have recently found to be useful.
On Thu, Jul 24, 2008 at 4:33 PM, Erik Iverson <[EMAIL PROTECTED]>
wrote:
> Spencer -
>
> Spencer wrote:
>
>>
>> Dear R Experts,
>>
>> I am trying to create several subsets that I want to use as separate
>> datasets. After separate subsets are creat
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