[R] (no subject)
*** On Tue, 28 Sep 2004, YiYao_Jiang wrote: I can't setup a library for some functions. Can anybody kindly give me an example about setup a library. From build a package to library. *** Actually I have read " Writing R Extensions"， I tried many command: build, check, INSTALL, library , package.skeleton. But I still can't compile a package, Can anybody kindly give me an example about it? Thanks. Best Regards YiYao Jiang Product Division/ product Testing Department Semiconductor Manufacturing International Corporation 18 ZhangJiang Road, PuDong New Area, Shanghai ZIP: 201203 Tel:86-21-5080-2000 Ext. 15173 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] memory increase for large r simulation
Please read the appropriate FAQ (rw-FAQ), as the posting guide asks. ^ 2.7 There seems to be a limit on the memory it uses! lokks relevant, and is. (So was some more info given with that error message.) On Tue, 28 Sep 2004, shane stanton wrote: > Hi, > > I am running R from windows 95. I have a large > simulation, which R does not get very far through > before telling me: > > "error cannot allocate vector of size 871875 Kb" > > and warning message "reached total allocation of 127 > Mb" > > I have been trying to increase the memory allocation > to R from my computer, using various commands at the R > prompt, such as memory.limit(size=..) to which R > responds "NULL" or "cannot decrease memory limit" (no > matter how large I try to make the argument of > memory.limit. > > Does anybody have any ideas re how I can get this > simulation to run? > > Many thanks, > > Shane Stanton > > __ > [EMAIL PROTECTED] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > > -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] sapply behavior
On Mon, 27 Sep 2004, Liaw, Andy wrote:
> The problem is that temp2 is a data frame, and the function you are
> sapply()ing to returns a row from a data frame. A data frame is really a
> list, with each variable corresponding to a component. If you extract a row
> of a data frame, you get another data frame, not a vector, even if all
> variables are the same type. sapply() can really `simplify' the right way
> if it's given a vector (or matrix). Consider:
>
> > str(temp2)
> `data.frame': 6 obs. of 4 variables:
> $ A1: int 5 2 4 6 1 3
> $ A2: int 5 2 4 6 1 3
> $ A3: int 5 2 4 6 1 3
> $ A4: int 5 2 4 6 1 3
> > temp2 <- as.matrix(temp2)
> > str(temp2)
> int [1:6, 1:4] 5 2 4 6 1 3 5 2 4 6 ...
> - attr(*, "dimnames")=List of 2
> ..$ : chr [1:6] "1" "2" "3" "4" ...
> ..$ : chr [1:4] "A1" "A2" "A3" "A4"
> > str(sapply(1:6,function(x){xmat<-temp2[temp2[,1]==x,,drop=F]; xmat[1,]}))
> int [1:4, 1:6] 1 1 1 1 2 2 2 2 3 3 ...
> - attr(*, "dimnames")=List of 2
> ..$ : chr [1:4] "A1" "A2" "A3" "A4"
> ..$ : NULL
>
> (The is.matrix() function probably just check whether the dim attribute is a
> vector of length 2, and not a data frame (as it says in ?is.matrix). The
> newtemp2 object you get is a list with 24 components, each component is a
> vector of one integer, and has a dim attribute of c(4, 6). Not what I would
> call a matrix.)
That *is* a matrix, though, and is useful for lists of length greater than
one. A matrix in R is just a vector with a dim attribute (and length the
product of the dims's), so as well as any of the atomic vectors it can be
a generic vector aka list.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] memory increase for large r simulation
Hi, I am running R from windows 95. I have a large simulation, which R does not get very far through before telling me: "error cannot allocate vector of size 871875 Kb" and warning message "reached total allocation of 127 Mb" I have been trying to increase the memory allocation to R from my computer, using various commands at the R prompt, such as memory.limit(size=..) to which R responds "NULL" or "cannot decrease memory limit" (no matter how large I try to make the argument of memory.limit. Does anybody have any ideas re how I can get this simulation to run? Many thanks, Shane Stanton __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] (no subject)
Hi, On Tue, 28 Sep 2004, YiYao_Jiang wrote: > Hello everybody: > > I can't setup a library for some functions. > Can anybody kindly give me an example about setup a library. From build a package to > library. I am guessing you meant you want to compile a package from source? Have you looked at Writing R Extensions? HTH, Kevin Ko-Kang Kevin Wang PhD Student Centre for Mathematics and its Applications Building 27, Room 1004 Mathematical Sciences Institute (MSI) Australian National University Canberra, ACT 0200 Australia Homepage: http://wwwmaths.anu.edu.au/~wangk/ Ph (W): +61-2-6125-2431 Ph (H): +61-2-6125-7407 Ph (M): +61-40-451-8301 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[Hmisc] proposal for change in latex.summary.formula.reverse (was: [R] Sweave: superfluous newline (`\\') in tex file)
Hi, well, it wasn't a Sweave problem after all. Seems to have something to do with the function `latex.summary.formula.reverse' in the package Hmisc. Following change seems to help (at least for my purpose): --88--- *** /usr/local/lib/R/library/Hmisc/R/Hmisc 2004-09-28 07:37:26.0 +0200 --- /usr/local/lib/R/library/Hmisc/R/Hmisc.original 2004-09-28 07:38:28.0 +0200 *** *** 16156,16162 'Numbers after percents are frequencies.', sep="\n") if(length(testUsed)) ! legend <-paste(legend,'\n\n', if(length(testUsed)==1)'Test used:' else 'Tests used:', if(length(testUsed)==1) paste(testUsed,'test') else paste(paste('$^{',1:length(testUsed),'}$',testUsed, --- 16156,16162 'Numbers after percents are frequencies.', sep="\n") if(length(testUsed)) ! legend <-paste(legend,'\n','\n\n', if(length(testUsed)==1)'Test used:' else 'Tests used:', if(length(testUsed)==1) paste(testUsed,'test') else paste(paste('$^{',1:length(testUsed),'}$',testUsed, --8 8--- But since my knowledge of R is rather small I have no idea if these has any other drawbacks. Any comments? How would I go about changing this for my setup only? Is renaming Hmisc to something like pdHmisc ok? Patrick -- Do you remember when you only had to pay for windows when *you* broke them? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] (no subject)
Hello everybody: I can't setup a library for some functions. Can anybody kindly give me an example about setup a library. From build a package to library. Thanks! Best Regards YiYao Jiang __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Re: tseries Package for R
Wizon wrote: I am using Mac OSX. I am first starting to use R and have not installed any packages yet. I searched through the CRAN site for a Mac OSX version, but did not find one. I downloaded the tar.gz package. Will this work on the Mac? Is there a way to get a package that I don't have to compile? Thanks in advance for your help. Adam Dear Adam Sorry, I can't help you since I never used R under MacOS. However, I cc your mail to R-help. Maybe you get some feedback from there. Best regards Adrian __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] An index of all possible combinations of variables in a data fram e
Alan Simpson robertsresearch.com.au> writes: : Does anybody know of any way to create an index of all the possible : combinations of variables (factors) in a data frame? ie for 3 factors A, B & : C we have : : A : B : C : AB : AC : BC : ABC : : which equates to columns 1, 2, 3, 1:2, (1,3), 2:3 and 1:3. : : I realise that a function like model.matrix does this, but how to get the : seqence of the index? : Is this what you are looking for? R> attributes(terms(~ a * b * c))$term.labels [1] "a" "b" "c" "a:b" "a:c" "b:c" "a:b:c" __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] KS test
Hi, > I have a sample, and I want to test if its distribution is normal. So I > would like to perform a one sample KS test but I cannot find the "y" > argument representing the "character string naming a distribution > function" according to the ?ks.test help page. > It is impossible forme to have a list of these character strings naming a > distribution function. > > I think it is "pnorm" for a normal distribution, bu I would like to be sure... > >From examples in ks.test manual: # Does x come from a shifted gamma distribution with shape 3 and scale 2? ks.test(x+2, "pgamma", 3, 2) # two-sided ks.test(x+2, "pgamma", 3, 2, alternative = "gr") So you can be sure, that it will be "pnorm" for normal distribution. Vaidotas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] An index of all possible combinations of variables in a data fram e
Hello list Does anybody know of any way to create an index of all the possible combinations of variables (factors) in a data frame? ie for 3 factors A, B & C we have A B C AB AC BC ABC which equates to columns 1, 2, 3, 1:2, (1,3), 2:3 and 1:3. I realise that a function like model.matrix does this, but how to get the seqence of the index? Any help would be greatly appreciated. Regards Alan Simpson Roberts Research Group [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] optim error in arima
Hi, On Mon, 27 Sep 2004 09:02:39 -0700 (PDT), Nathaniel B. Derby <[EMAIL PROTECTED]> wrote: > Hello, > > I'm fitting a series of ARIMA models to a data set to compare fits. After > taking the logs of the data and then differencing them to induce stationarity, > I execute > > arima( y, order=c( p, 0, q ), seasonal=list( order=c( P, 0, Q ), period=7 ) ) > > for various values of p, q, P and Q. For one set of these values, I get > > Error in optim(init[mask], armafn, method = "BFGS", hessian = TRUE ... : > non-finite finite-difference value [0] > > which tells me that when computing derivatives of the objective function > (armafn) by finite differencing, one of the values is NA, +Inf or -Inf. Any > ideas? I would like to print some values of armafn, but how do I get that > from my data set, and what would I look for? This probably means, that for your values of p, q, P, Q the arima model is not valid arima model, i.e. it is not stationary. Vaidotas __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Course***Splus/R: Complementing and Extending Statistical Computing for SAS Users
XLSolutions Corporation (www.xlsolutions-corp.com) is pleased to announce a two-day course, "Splus/R: Complementing and Extending Statistical Computing for SAS Users" www.xlsolutions-corp.com/Rsas.htm Dates/City: October 28-29 Raleigh, NC November 4-5 Boston, MA This course is designed for SAS users who want to learn how to complement and extend statistical computing with Splus and/or R system. The course will give participants a strong foundation for becoming a versatile programmer. Course Description: This two-day course focuses on a broad spectrum of topics: *Data manipulations in S and R (data frame and matrix operations) and SAS (the data step) -- issues of importing, formatting, transformation, cataloging, exporting *Splus/R Functions vs macros in SAS for programming repetitive processes. *The iteration models of SAS vs whole-object modeling *Specific comparison: linear modeling, glms, gees, lmes. *etc Complete course description: www.xlsolutions-corp.com/Rsas.htm Earlybird ends October 10th. Ask for group discount! Registration: Email Sue Turner: [EMAIL PROTECTED] Phone: 206-686-1578 Share Your Thoughts: Are there any additional topics you would like for this course to address? Would you like for this course to be offered in another city? Please let us know by contributing to our recommendation list: [EMAIL PROTECTED] Elvis Miller, PhD Manager Training and Technical Support North American Division XLSolutions Corporation Email: [EMAIL PROTECTED] Phone: 206-686-1578 Web: www.xlsolutions-corp.com __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] smoothing noisy data with a twist
One thing I would try is to supply 1/sd(y) as the weights to smoothers that can accept them; e.g., loess(). Andy > From: Matt Gibbs > > Hi, > > I have a set of observations (x,y), derived from a previous > estimation. > For each observation I also have an estimated variance s(y) > derived from > the first stage. > > The problem is that I need to smooth the data (x,y) while taking into > account the fact that the y's have been estimated at a previous stage > and thus already come with a variance. So, if I smooth the data I > somehow need to take into account *two errors*, one from the > smoothing > and the other from the already noisy data that I start off with. > > Does anyone have any idea how to do this? > > thanks, matt. > > __ > [EMAIL PROTECTED] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > > __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] private on site R training solicited
Hi, R is catching up even with the hardest Excelists. At our company we are exploring the possibility of having some of the R&D staff introduced to R during a short on site class in San-Francisco, California. The number of people in the group will be about 5 - 7. If you are interested in delivering the class please e-mail me the info for consideration. Thanks, Vadim Ogranovich Evnine & Vaughan Assoc. [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] smoothing noisy data with a twist
Hi, I have a set of observations (x,y), derived from a previous estimation. For each observation I also have an estimated variance s(y) derived from the first stage. The problem is that I need to smooth the data (x,y) while taking into account the fact that the y's have been estimated at a previous stage and thus already come with a variance. So, if I smooth the data I somehow need to take into account *two errors*, one from the smoothing and the other from the already noisy data that I start off with. Does anyone have any idea how to do this? thanks, matt. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] passing formula arg to mgcv::gam
How about
environment(formula) <- environment()
Vadim Ogranovich evafunds.com> writes:
:
: This is a self-response .
:
: It was indeed a problem with environments. One way to get around is
to "reset" the environment, e.g. inside
: callGam do
: formula <- as.formula(unclass(formula))
:
: Not too aesthetic, but works. Is there a less kludgy way to do this?
:
: BTW, forgot to mention. This is R-1.9.1 on RH-7.3.
:
: Thanks,
: Vadim
:
: > -Original Message-
: > From: r-help-bounces stat.math.ethz.ch
: > [mailto:r-help-bounces stat.math.ethz.ch] On Behalf Of Vadim
: > Ogranovich
: > Sent: Monday, September 27, 2004 3:24 PM
: > To: r-help stat.math.ethz.ch
: > Subject: [R] passing formula arg to mgcv::gam
: >
: > Hi,
: >
: > I have a function, callGam, that fits a gam model to a subset
: > of a dataframe. The argument to callGam is a formula, the
: > subset is determined inside the function itself. My naÃve
: > approach generates and error, see below. I guess this is
: > because 'idx' is loocked up in the environment of 'formula',
: > but I am too ignorant about environments to be able to tell
: > for sure. Could someone please suggest a way around?
: >
: > Thanks,
: > Vadim
: >
: > > library("mgcv")
: > >
: > > callGam <- function(formula) {
: > + idx <- seq(10)
: > + gam(formula, data=data.frame(x=rnorm(100), y=rnorm(100)),
: > + subset=idx) }
: > >
: > > gam.fit <- callGam(y ~ x)
: > Error in eval(expr, envir, enclos) : Object "idx" not found
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Re: [R] sapply behavior
Elizabeth Purdom stanford.edu> writes:
:
: Hi,
:
: I use sapply very frequently, but I have recently noticed a behavior of
: sapply which I don't understand and have never seen before. Basically,
: sapply returns what looks like a matrix, says it a matrix, and appears to
: let me do matrix things (like transpose). But it is also a list and behaves
: like a list when I subset it, not a vector (so I can't sort a row for
: instance). I don't know where this is coming from so as to avoid it, nor
: how to handle the beast that sapply is returning. I double checked my old
: version of R and apparently this same thing happens in 1.8.0, though I
: never experienced it. I had a hard time reproducing it, and I don't know
: what's setting it off, but the code below seems to do it for me. (I'm using
: R on Windows XP, either 1.8.0 or 1.9.1)
:
: Thanks for any help,
: Elizabeth Purdom
:
:
: > temp2<-matrix(sample(1:6,6,replace=F),byrow=F,nrow=6,ncol=4)
: > colnames(temp2)<-paste("A",as.character(1:4),sep="")
: > temp2<-as.data.frame(temp2)
: >
: newtemp2<-sapply((1:6),function(x){xmat<-temp2[temp2[,1]==x,,drop=F];return
(xmat[1,])})
: > print(newtemp2) #looks like matrix
: [,1] [,2] [,3] [,4] [,5] [,6]
: A1 123456
: A2 123456
: A3 123456
: A4 123456
: > is.matrix(newtemp2) #says it's matrix
: [1] TRUE
: > class(newtemp2)
: [1] "matrix"
: > is.list(newtemp2) #but also list
: [1] TRUE
: > newtemp2[,1] #can't subset and get a vector back; same thing happens for
: rows.
: $A1
: [1] 1
:
: $A2
: [1] 1
:
: $A3
: [1] 1
:
: $A4
: [1] 1
: #other things about it:
: > names(newtemp2)
: NULL
: > dimnames(newtemp2)
: [[1]]
: [1] "A1" "A2" "A3" "A4"
:
: [[2]]
: NULL
: > dim(newtemp2)
: [1] 4 6
: > length(newtemp2)
: [1] 24
The problem is that your function is returning a one row data frame
and when sapply tries to simplify the resulting list of 6 data frames
that gives a list with dimensions rather what you were expecting
which is a vector with dimensions.
Let us call the original anonymous function in your post (i.e. the one
passed to sapply there), f. We can modify it to produce f2 which is like
f except that we wrap the return expression in c() to turn it into a
vector:
f2 <- function(x){xmat<-temp2[temp2[,1]==x,,drop=F];return(c(xmat[1,]))}
sapply(1:6, f2)
If you really do want to return a one row data frame then
use rbind to bind the data frames together rather than sapply:
do.call("rbind", lapply(1:6, f))
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Re: [R] sapply behavior
Elizabeth Purdom wrote:
I use sapply very frequently, but I have recently noticed a behavior of
sapply which I don't understand and have never seen before. Basically,
sapply returns what looks like a matrix, says it a matrix, and appears
to let me do matrix things (like transpose). But it is also a list and
behaves like a list when I subset it, not a vector (so I can't sort a
row for instance). I don't know where this is coming from so as to avoid
it, nor how to handle the beast that sapply is returning. I double
checked my old version of R and apparently this same thing happens in
1.8.0, though I never experienced it. I had a hard time reproducing it,
and I don't know what's setting it off, but the code below seems to do
it for me. (I'm using R on Windows XP, either 1.8.0 or 1.9.1)
Thanks for any help,
Elizabeth Purdom
> temp2<-matrix(sample(1:6,6,replace=F),byrow=F,nrow=6,ncol=4)
> colnames(temp2)<-paste("A",as.character(1:4),sep="")
> temp2<-as.data.frame(temp2)
It is this coercion to the data frame that is injecting a list-like
property into the result. Try your script without that line and it will
work as you expect.
> newtemp2<-sapply((1:6),function(x){xmat<-temp2[temp2[,1]==x,,drop=F];return(xmat[1,])})
> print(newtemp2) #looks like matrix
[,1] [,2] [,3] [,4] [,5] [,6]
A1 123456
A2 123456
A3 123456
A4 123456
The best thing to do in a situation like this is to use the str function
to see the details of the structure of the object.
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RE: [R] sapply behavior
The problem is that temp2 is a data frame, and the function you are
sapply()ing to returns a row from a data frame. A data frame is really a
list, with each variable corresponding to a component. If you extract a row
of a data frame, you get another data frame, not a vector, even if all
variables are the same type. sapply() can really `simplify' the right way
if it's given a vector (or matrix). Consider:
> str(temp2)
`data.frame': 6 obs. of 4 variables:
$ A1: int 5 2 4 6 1 3
$ A2: int 5 2 4 6 1 3
$ A3: int 5 2 4 6 1 3
$ A4: int 5 2 4 6 1 3
> temp2 <- as.matrix(temp2)
> str(temp2)
int [1:6, 1:4] 5 2 4 6 1 3 5 2 4 6 ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:6] "1" "2" "3" "4" ...
..$ : chr [1:4] "A1" "A2" "A3" "A4"
> str(sapply(1:6,function(x){xmat<-temp2[temp2[,1]==x,,drop=F]; xmat[1,]}))
int [1:4, 1:6] 1 1 1 1 2 2 2 2 3 3 ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:4] "A1" "A2" "A3" "A4"
..$ : NULL
(The is.matrix() function probably just check whether the dim attribute is a
vector of length 2, and not a data frame (as it says in ?is.matrix). The
newtemp2 object you get is a list with 24 components, each component is a
vector of one integer, and has a dim attribute of c(4, 6). Not what I would
call a matrix.)
HTH,
Andy
> From: Elizabeth Purdom
>
> Hi,
>
> I use sapply very frequently, but I have recently noticed a
> behavior of
> sapply which I don't understand and have never seen before.
> Basically,
> sapply returns what looks like a matrix, says it a matrix,
> and appears to
> let me do matrix things (like transpose). But it is also a
> list and behaves
> like a list when I subset it, not a vector (so I can't sort a row for
> instance). I don't know where this is coming from so as to
> avoid it, nor
> how to handle the beast that sapply is returning. I double
> checked my old
> version of R and apparently this same thing happens in 1.8.0,
> though I
> never experienced it. I had a hard time reproducing it, and I
> don't know
> what's setting it off, but the code below seems to do it for
> me. (I'm using
> R on Windows XP, either 1.8.0 or 1.9.1)
>
> Thanks for any help,
> Elizabeth Purdom
>
>
> > temp2<-matrix(sample(1:6,6,replace=F),byrow=F,nrow=6,ncol=4)
> > colnames(temp2)<-paste("A",as.character(1:4),sep="")
> > temp2<-as.data.frame(temp2)
> >
> newtemp2<-sapply((1:6),function(x){xmat<-temp2[temp2[,1]==x,,d
> rop=F];return(xmat[1,])})
> > print(newtemp2) #looks like matrix
> [,1] [,2] [,3] [,4] [,5] [,6]
> A1 123456
> A2 123456
> A3 123456
> A4 123456
> > is.matrix(newtemp2) #says it's matrix
> [1] TRUE
> > class(newtemp2)
> [1] "matrix"
> > is.list(newtemp2) #but also list
> [1] TRUE
> > newtemp2[,1] #can't subset and get a vector back; same
> thing happens for
> rows.
> $A1
> [1] 1
>
> $A2
> [1] 1
>
> $A3
> [1] 1
>
> $A4
> [1] 1
> #other things about it:
> > names(newtemp2)
> NULL
> > dimnames(newtemp2)
> [[1]]
> [1] "A1" "A2" "A3" "A4"
>
> [[2]]
> NULL
> > dim(newtemp2)
> [1] 4 6
> > length(newtemp2)
> [1] 24
>
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RE: [R] using tcltk in R under ESS/XEmacs on Windows
Tony:
It is indeed most amazing that you have been able to get ESS/Emacs
running under Windows. And much appreciated by those of us who have to
moil around
with MessySoft machines.
Thanks to all who have contributed to the effort.
Cheers,
Shawn Boles
-Original Message-
From: A.J. Rossini [mailto:[EMAIL PROTECTED]
Sent: Saturday, September 25, 2004 8:04 PM
To: Liaw, Andy
Cc: R-Help; [EMAIL PROTECTED]; ESS (Help list)
Subject: Re: [R] using tcltk in R under ESS/XEmacs on Windows
It most likely is Windows specific. It's most amazing that we
actually have ESS/(X)Emacs working under windows in the first place.
Unfortunately, I'm in transit for the next few weeks, but I'll
probably have a windows machine on my desk sometime after that. Argh.
best,
-tony
On Fri, 24 Sep 2004 17:48:18 -0400, Liaw, Andy <[EMAIL PROTECTED]>
wrote:
> > From: Marc Schwartz
> >
> > On Fri, 2004-09-24 at 15:02, Liaw, Andy wrote:
> > > Sorry for the cross-post. Not sure where the problem is...
> > >
> > > A while back I posted an R function to R-help:
> > >
> > > cd <- function (dir = tclvalue(tkchooseDirectory()),
> > saveOld = FALSE,
> > > loadNew = TRUE) {
> > > stopifnot(require(tcltk))
> > > if (saveOld)
> > > save.image(compress = TRUE)
> > > setwd(dir)
> > > rm(list = ls(all = TRUE, envir = .GlobalEnv), envir =
> > .GlobalEnv)
> > > if (loadNew && file.exists(".RData")) {
> > > loaded <- load(".RData", envir = .GlobalEnv)
> > > return(invisible(loaded))
> > > }
> > >
> > > where the default value for the `dir' argument is to run
> > the tcltk directory
> > > chooser and get the directory name chosen. (Thanks to
> > Prof. John Fox for
> > > the tcltk part!!) While this function works fine under
> > Rgui on Windows, it
> > > doesn't work when running R within ESS (5.2.3) and XEmacs
> > (21.4.13). The
> > > directory chooser never shows up, and dir just gets the
> > empty string. Does
> > > anyone have any idea what could be the problem? I'd very
> > much appreciate
> > > any pointers.
> > >
> > > Best,
> > > Andy
> >
> > Andy,
> >
> > This works under FC2 using ESS 5.2.3 with XEmacs version 21.4.15, so
> > presumably there is something specific to the Windows
implementation?
>
> Given Prof. Fox's follow-up and your obvservation, I guess the problem
_is_
> Windows-specific. 8-(
>
> > Also, two things:
> >
> > 1. You are missing a closing brace above, which I presume may be a
> > simple copy and paste issue.
>
> Yes. My apologies.
>
> > 2. If you successfully change the directory, the cd()
> > function itself is
> > deleted from the global environment via your rm(...), as you
currently
> > have it implemented. I am not sure if this is intentional or not.
>
> Well, sort of. I've placed it in a small package along with other
handy
> stuff, so that won't be a problem.
>
> Best,
> Andy
>
>
>
> > HTH,
> >
> > Marc
> >
> >
> >
> >
>
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>
--
A.J. Rossini
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[R] sapply behavior
Hi,
I use sapply very frequently, but I have recently noticed a behavior of
sapply which I don't understand and have never seen before. Basically,
sapply returns what looks like a matrix, says it a matrix, and appears to
let me do matrix things (like transpose). But it is also a list and behaves
like a list when I subset it, not a vector (so I can't sort a row for
instance). I don't know where this is coming from so as to avoid it, nor
how to handle the beast that sapply is returning. I double checked my old
version of R and apparently this same thing happens in 1.8.0, though I
never experienced it. I had a hard time reproducing it, and I don't know
what's setting it off, but the code below seems to do it for me. (I'm using
R on Windows XP, either 1.8.0 or 1.9.1)
Thanks for any help,
Elizabeth Purdom
> temp2<-matrix(sample(1:6,6,replace=F),byrow=F,nrow=6,ncol=4)
> colnames(temp2)<-paste("A",as.character(1:4),sep="")
> temp2<-as.data.frame(temp2)
>
newtemp2<-sapply((1:6),function(x){xmat<-temp2[temp2[,1]==x,,drop=F];return(xmat[1,])})
> print(newtemp2) #looks like matrix
[,1] [,2] [,3] [,4] [,5] [,6]
A1 123456
A2 123456
A3 123456
A4 123456
> is.matrix(newtemp2) #says it's matrix
[1] TRUE
> class(newtemp2)
[1] "matrix"
> is.list(newtemp2) #but also list
[1] TRUE
> newtemp2[,1] #can't subset and get a vector back; same thing happens for
rows.
$A1
[1] 1
$A2
[1] 1
$A3
[1] 1
$A4
[1] 1
#other things about it:
> names(newtemp2)
NULL
> dimnames(newtemp2)
[[1]]
[1] "A1" "A2" "A3" "A4"
[[2]]
NULL
> dim(newtemp2)
[1] 4 6
> length(newtemp2)
[1] 24
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RE: [R] passing formula arg to mgcv::gam
This is a self-response :-).
It was indeed a problem with environments. One way to get around is to "reset" the
environment, e.g. inside callGam do
formula <- as.formula(unclass(formula))
Not too aesthetic, but works. Is there a less kludgy way to do this?
BTW, forgot to mention. This is R-1.9.1 on RH-7.3.
Thanks,
Vadim
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Vadim
> Ogranovich
> Sent: Monday, September 27, 2004 3:24 PM
> To: [EMAIL PROTECTED]
> Subject: [R] passing formula arg to mgcv::gam
>
> Hi,
>
> I have a function, callGam, that fits a gam model to a subset
> of a dataframe. The argument to callGam is a formula, the
> subset is determined inside the function itself. My naïve
> approach generates and error, see below. I guess this is
> because 'idx' is loocked up in the environment of 'formula',
> but I am too ignorant about environments to be able to tell
> for sure. Could someone please suggest a way around?
>
> Thanks,
> Vadim
>
> > library("mgcv")
> >
> > callGam <- function(formula) {
> + idx <- seq(10)
> + gam(formula, data=data.frame(x=rnorm(100), y=rnorm(100)),
> + subset=idx) }
> >
> > gam.fit <- callGam(y ~ x)
> Error in eval(expr, envir, enclos) : Object "idx" not found
> >
>
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[R] passing formula arg to mgcv::gam
Hi,
I have a function, callGam, that fits a gam model to a subset of a dataframe. The
argument to callGam is a formula, the subset is determined inside the function itself.
My naïve approach generates and error, see below. I guess this is because 'idx' is
loocked up in the environment of 'formula', but I am too ignorant about environments
to be able to tell for sure. Could someone please suggest a way around?
Thanks,
Vadim
> library("mgcv")
>
> callGam <- function(formula) {
+ idx <- seq(10)
+ gam(formula, data=data.frame(x=rnorm(100), y=rnorm(100)), subset=idx)
+ }
>
> gam.fit <- callGam(y ~ x)
Error in eval(expr, envir, enclos) : Object "idx" not found
>
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Re: [R] Gridbase basic question
Hi
Sean Davis wrote:
All,
I have a simple plot(x,y) and I would like to then insert rectangles of
some length (in native coordinates) and height fixed to 0.5 in native
coordinates. I can't quite get the code right to do this. Can anyone
give me a quick example of how to do this? I looked the gridBase index
and the tutorial (from R-news?) but just haven't gotten it down yet.
> plot(1:10,1:10)
> par(new=T);vps <- baseViewports()
> pushViewport(vps$inner,vps$figure,vps$plot)
viewport[GRID.VP.28]
At this point you are within a viewport which has x- and y-scales
corresponding to the plot(1:10, 1:10).
> pushViewport(viewport(x=unit(1,"native"),y=unit(2,"native")))
viewport[GRID.VP.29]
You have just created a new viewport at location (1, 2) in the plot, but
the scales on this new viewport are the default (0, 1). i.e., you are
now in a completely different coordinate system. Also, this new
viewport is as wide and as high as the plot region -- for example, it
extends well beyong the left edge of the window/page.
grid.rect(height=unit(0.5,"native"),width=unit(1.5,"native"),just='botto
m')
This draws a rectangle half as high as the current viewport and 1.5
times as wide (the native scale in the current viewport is (0, 1) in
both dimensions). Importantly, the "native" coordinate systems you are
referring to no longer correspond to the scales on the plot.
This draws a very large rectangle going from 2 to 7 (y) and to 8 (x).
Three things:
(i) If drawing rectangles relative to the current "native" (or user)
coordinates is all you want to do then you could just use rect() and
ignore gridBase altogether. For example, ...
x <- sample(1:10, 10)
y <- 1:10
w <- runif(10)
h <- 0.5
plot(1:10,1:10)
rect(x - w/2, y - h/2, x + w/2, y + h/2)
(ii) Using grid and gridBase, the above example becomes ...
plot(1:10,1:10)
par(new=T);vps <- baseViewports()
pushViewport(vps$inner,vps$figure,vps$plot)
grid.rect(x=x, y=y, width=w, height=h, default.units="native")
popViewport(3)
... but as mentioned, this is like using a sledge hammer to kill a cat
or whatever the expression is.
(iii) There would be justification in using grid and gridBase if you
want to draw more than just a rectangle, especially if you want to use
coordinates other than native. Here's a trivial example (adds fixed
size "whiskers" to the corners of the rectangles) ...
plot(1:10,1:10)
par(new=T);vps <- baseViewports()
pushViewport(vps$inner,vps$figure,vps$plot)
for (i in 1:10) {
pushViewport(viewport(x=x[i], y=y[i], width=w[i], height=h,
default.units="native"))
grid.rect()
grid.segments(0, 0, unit(-1, "mm"), unit(-1, "mm"))
grid.segments(0, 1, unit(-1, "mm"),
unit(1, "npc") + unit(1, "mm"))
grid.segments(1, 1,
unit(1, "npc") + unit(1, "mm"),
unit(1, "npc") + unit(1, "mm"))
grid.segments(1, 0,
unit(1, "npc") + unit(1, "mm"),
unit(-1, "mm"))
popViewport()
}
... (but pushing a viewport per data point like this is a LOT slower).
Hope that helps
Paul
--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/
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Re: [R] Looking for .Call functions
Nathaniel B. Derby wrote:
Hi,
In my ongoing quest to track down the source of an error (see message
"[R] optim error in arima" above), I find in the cource code for arima0
the following:
arma0f <- function(p) {
par <- as.double(fixed)
par[mask] <- p
.Call("arma0fa", G, par, PACKAGE = "stats")
}
I would like to know what the function "arma0f" does. Does the above
mean that there is a function called "arma0fa" somewhere in R? Where is
it? I couldn't find anything in Rinternals.h.
It means that in one of the .c source files in
$RSRC/src/library/stats/src there will be a C function declared as
SEXP arma0fa(SEXP, SEXP);
(In fact it is declared in ts.h and defined in pacf.c in that directory)
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Re: [R] multinom object :way of plotting??
Dear Camille, You might be interested in a paper (available at http://socserv.socsci.mcmaster.ca/jfox/logit-effect-displays.pdf) that Bob Andersen and I wrote on this topic. The paper deals with graphing multinomial-logit and proportional-odds models. Regards, John On 27 Sep 2004 18:16:53 +0100 Camille Szmaragd <[EMAIL PROTECTED]> wrote: > Dear all, > > I'm fitting a multinom function to my dataset > (multinom(outcome~age+K+D)) and I need to present my results on a > poster. Does someone know a nice way of doing that? I think I saw in > an archive that you cannot plot a multinom.object, is it true? > > Thank you by advance for your help, > > Cheers > > Camille > > __ > [EMAIL PROTECTED] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html John Fox Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Looking for .Call functions
Hi,
In my ongoing quest to track down the source of an error (see message "[R] optim error in
arima" above), I find in the cource code for arima0 the following:
arma0f <- function(p) {
par <- as.double(fixed)
par[mask] <- p
.Call("arma0fa", G, par, PACKAGE = "stats")
}
I would like to know what the function "arma0f" does. Does the above mean that there is a
function called "arma0fa" somewhere in R? Where is it? I couldn't find anything in
Rinternals.h.
Thanks,
Nate
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Re: [R] Re: cannot assign dimnames
"Dan Bebber" <[EMAIL PROTECTED]> writes: > Dear list, > > Please ignore my earlier message on this topic. I was under the mistaken > impression that dimnames() named the dimensions themselves, rather than the > indices within the dimensions. The dimnames list can be named though: > matrix(1:4,2,dimnames=list(foo=1:2,bar=c(10,20))) bar foo 10 20 1 1 3 2 2 4 -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Re: cannot assign dimnames
Gabor Grothendieck myway.com> writes:
:
: Dan Bebber forestecology.co.uk> writes:
:
: : Please ignore my earlier message on this topic. I was under the mistaken
: : impression that dimnames() named the dimensions themselves, rather than the
: : indices within the dimensions.
:
: What you are referring to is done in R by referring to the names of
: the dimnames, as opposed to the dimnames, themselves. Here is an
: example:
:
: R> mat <- matrix(1:12,4, dimnames = list(letters[1:4],LETTERS[1:3]))
: R> names(dimnames(mat)) <- c("FirstDim", "SecondDim")
: R> mat
: SecondDim
: FirstDim A B C
:a 1 5 9
:b 2 6 10
:c 3 7 11
:d 4 8 12
Perhaps I should have also mentioned that the above could be done
in a single line like this:
R> mat <- matrix(1:12,4, dimnames = list(FirstDim = letters[1:4], SecondDim =
LETTERS[1:3]))
R> mat
SecondDim
FirstDim A B C
a 1 5 9
b 2 6 10
c 3 7 11
d 4 8 12
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Re: [R] Re: cannot assign dimnames
Dan Bebber forestecology.co.uk> writes:
: Please ignore my earlier message on this topic. I was under the mistaken
: impression that dimnames() named the dimensions themselves, rather than the
: indices within the dimensions.
What you are referring to is done in R by referring to the names of
the dimnames, as opposed to the dimnames, themselves. Here is an
example:
R> mat <- matrix(1:12,4, dimnames = list(letters[1:4],LETTERS[1:3]))
R> names(dimnames(mat)) <- c("FirstDim", "SecondDim")
R> mat
SecondDim
FirstDim A B C
a 1 5 9
b 2 6 10
c 3 7 11
d 4 8 12
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[R] [R-pkgs] new version of package "distr" available
We would like to announce the availability on CRAN of a new version (1.4) of our package "distr" . - Changes from 1.3 to 1.4 -To avoid name collisions with short forms for TRUE and FALSE: classes T and F (T- and F-distributions) renamed to Td and Fd -The package is now loaded as a binary => considerable speed gain -Using subsititute, the bodies of the r,d,p,q-function-slots distributions show the parameter values with which they were generated -Convolutions and applications of the math group may now be traced in r-slot of a distribution object, compare r(sin(Norm()) + cos(Unif() * 3 + 2)) -Parameters of a distribution (mean, sd, etc) are now tested on length 1 *we see the objects as implementations of univariate distributions, so vectors make no sense here; rather one could gather several objects with possibly different parameters to a vector of distributions. Of course, the original R-functions rnorm etc remain unchanged and still allow for vector-valued parameters. * -Classes "Parameter" , "Distribution" , "UnivariateDistribution" are no longer VIRTUAL -"AbscontParameter" and "DiscreteParameter" are replaced by "Parameter" -Type of slots d,p,q and param is changed to "OptionalFunction" and "OptionalParameter" respectively - Short Description of "distr": "distr" is to provide a conceptual treatment of random variables (r.v.'s) by means of S4--classes. A virtual mother class "Distribution" is introduced. All distributions of the "base" package are implemented as subclasses of either "AbscontDistribution" or "DiscreteDistribution". Using these classes, we also provide (default) methods to automatically generate the image distributions under unary mathematical operations as well as a general convolution algorithm. Additionally, we also provide classes for a standardized treatment of simulations (also under contaminations) and evaluations of statistical procedures on such simulations. - DESCRIPTION: Package: distr Version: 1.4 Date: 2004/09/23 Title: distr Authors: Peter Ruckdeschel <[EMAIL PROTECTED]>, Matthias Kohl <[EMAIL PROTECTED]>, Thomas Stabla <[EMAIL PROTECTED]>, Florian Camphausen <[EMAIL PROTECTED]> Maintainer: Peter Ruckdeschel <[EMAIL PROTECTED]> Description: S4 Classes for Distributions Depends: R (>= 1.9.0), (versions for <=1.8.x, on URL cited below), setRNG (>= 2004.3-1) License: GPL version 2 or later URL: http://www.uni-bayreuth.de/departments/math/org/mathe7/DISTR/ Reference: http://www.uni-bayreuth.de/departments/math/org/mathe7/DISTR/distr.pdf We look forward to receiving questions, comments and suggestions Peter Ruckdeschel Matthias Kohl Thomas Stabla Florian Camphausen ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] exponential correlation?
On Sat, Sep 25, 2004 at 01:21:34PM -0700, Spencer Graves wrote: > The "nlme" package has "corExp" for estimating the nuggett effect > n and the range d for an exponential spatial correlation structure = n + > (1-n)*exp(-r/d), where r = distance between two observations. See > library(nlme); ?corExp, plus Pinheiro and Bates (2000) Mixed-Effects > Models for S and S-Plus (Springer). > Is this what you want? > Thanks, for your answer. I'm afraid I was'nt very clear on what I needed. In fact, I have a set of paired quantitatve values. When I plot them, it looks like if the values where following a 1/exp(x) equation. I would like to make a regression on this dataset to find the equation and eventually find the x-value where the slope becomes more tough (I'm not sure it's the appropriate word, I hope you'll understand anyway). In fact I need to find the cutoff value. I'm not sure of what method to use to achieve that. Here is a subset of the dataset (Y are RR): X Y 82.0 1.00 72.0 2.45 53.0 3.88 45.0 8.76 71.0 1.00 60.0 2.00 40.0 4.76 20.0 12.05 8.0 11.62 87.0 1.00 78.0 1.03 65.0 1.21 41.0 2.78 75.0 1.00 67.5 1.53 45.0 1.81 30.0 3.76 75.0 1.00 62.0 1.90 42.0 4.10 17.0 5.40 86.0 1.00 so far I tried to do a non linear regression (nls): > fm1<-nls(rrm ~ a/(exp(clr*c)),start=list(a=1,c=0.033),trace=T) 437.4397 : 1.000 0.033 120.0999 : 7.78174860 0.01381373 70.89977 : 13.15418549 0.03034625 67.18951 : 14.63044976 0.03070568 67.18892 : 14.62649511 0.03066069 67.18892 : 14.6270777 0.0306623 >summary(fm1) Formula: rrm ~ a/(exp(clr * c)) Parameters: Estimate Std. Error t value Pr(>|t|) a 14.627078 2.217950 6.595 2.01e-06 *** c 0.030662 0.004876 6.288 3.87e-06 *** --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 1.833 on 20 degrees of freedom Correlation of Parameter Estimates: a c 0.8091 But I'm not sure I'm going the right way. If somebody could give me an advice on the way to proceed, it would be great. To find the cutoff value for x, should try to use the equation obtained by the regression or should I use the residuals between original data and the fitted values? Best Regards Blaise PS: I hope I've been more explicit this time and that my english is readable. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
