Re: [R] R2 always increases as variables are added?
I know that "-1" indicates to remove the intercept term. But my question is why intercept term CAN NOT be treated as a variable term as we place a column consited of 1 in the predictor matrix. If I stick to make a comparison between a model with intercept and one without intercept on adjusted r2 term, now I think the strategy is always to use another definition of r-square or adjusted r-square, in which r-square=sum((y.hat)^2)/sum((y)^2). Am I in the right way? Thanks Li Junjie 2007/5/19, Paul Lynch <[EMAIL PROTECTED]>: > > In case you weren't aware, the meaning of the "-1" in y ~ x - 1 is to > remove the intercept term that would otherwise be implied. > --Paul > > On 5/17/07, Àî¿¡½Ü <[EMAIL PROTECTED]> wrote: > > Hi, everybody, > > > > 3 questions about R-square: > > -(1)--- Does R2 always increase as variables are added? > > -(2)--- Does R2 always greater than 1? > > -(3)--- How is R2 in summary(lm(y~x-1))$r.squared > > calculated? It is different from (r.square=sum((y.hat-mean > > (y))^2)/sum((y-mean(y))^2)) > > > > I will illustrate these problems by the following codes: > > -(1)--- R2 doesn't always increase as variables are > added > > > > > x=matrix(rnorm(20),ncol=2) > > > y=rnorm(10) > > > > > > lm=lm(y~1) > > > y.hat=rep(1*lm$coefficients,length(y)) > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 2.646815e-33 > > > > > > lm=lm(y~x-1) > > > y.hat=x%*%lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 0.4443356 > > > > > > This is the biggest model, but its R2 is not the > biggest, > > why? > > > lm=lm(y~x) > > > y.hat=cbind(rep(1,length(y)),x)%*%lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 0.2704789 > > > > > > -(2)--- R2 can greater than 1 > > > > > x=rnorm(10) > > > y=runif(10) > > > lm=lm(y~x-1) > > > y.hat=x*lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 3.513865 > > > > > > -(3)--- How is R2 in summary(lm(y~x-1))$r.squared > > calculated? It is different from (r.square=sum((y.hat-mean > > (y))^2)/sum((y-mean(y))^2)) > > > x=matrix(rnorm(20),ncol=2) > > > xx=cbind(rep(1,10),x) > > > y=x%*%c(1,2)+rnorm(10) > > > ### r2 calculated by lm(y~x) > > > lm=lm(y~x) > > > summary(lm)$r.squared > > [1] 0.9231062 > > > ### r2 calculated by lm(y~xx-1) > > > lm=lm(y~xx-1) > > > summary(lm)$r.squared > > [1] 0.9365253 > > > ### r2 calculated by me > > > y.hat=xx%*%lm$coefficients > > > (r.square=sum((y.hat-mean(y))^2)/sum((y-mean(y))^2)) > > [1] 0.9231062 > > > > > > Thanks a lot for any cue:) > > > > > > > > > > -- > > Junjie Li, [EMAIL PROTECTED] > > Undergranduate in DEP of Tsinghua University, > > > > [[alternative HTML version deleted]] > > > > __ > > R-help@stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > > -- > Paul Lynch > Aquilent, Inc. > National Library of Medicine (Contractor) > -- Junjie Li, [EMAIL PROTECTED] Undergranduate in DEP of Tsinghua University, [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] displaying intensity through opacity on an image (ONE SOLUTION)
Dear list,
I did not get any response yet, but after looking around R and other things, I
came up with something that works.
Basically, I use the rgb() function in R [though I could also use the hsv()
function] to help me with the colormap.
Anyway, doing a help on rgb gives:
This function creates "colors" corresponding to the given
intensities (between 0 and 'max') of the red, green and blue
primaries.
An alpha transparency value can also be specified (0 means fully
transparent and 'max' means opaque). If 'alpha' is not specified,
an opaque colour is generated.
The names argument may be used to provide names for the colors.
The values returned by these functions can be used with a 'col='
specification in graphics functions or in 'par'.
and later on.
Semi-transparent colors ('0 < alpha < 1') are supported only on a
few devices: at the time of writing only on the 'pdf' and (on
MacOS X) 'quartz' devices.
The hsv() function has a similar point on semi-transparent colors.
Ok, looks promising: I don't use a Mac, and my potential journal does not
accept .pdf, only .tiff or .eps, but we are not totally lost here.
So, I tried the following silly example in R:
> pdf()
> image( matrix(rep(1:5,5), nr = 5), col = gray(0:16/16))
> image( matrix(1:25, nr = 5), col = rgb(rep(1, 15), g=0, b=0, alpha =
rep(1:15)/15), add = T) # red with different opacities
> q()
(we are out of R).
And then look at the pdf file created: by default it is Rplots.pdf.
OK, now we can use gimp, simply to convert this to .eps. Alternatively on
linux, the command pdftops and then psto epsi on it would also work.
Yippee! Isn't R wonderful??
Hope this helps: though others may have known about this before, I certainly
did not know how to do this in R.
Best wishes,
Ranjan
On Thu, 17 May 2007 19:16:18 -0500 Ranjan Maitra <[EMAIL PROTECTED]> wrote:
> Dear colleagues,
>
> I have an image which I can display in the greyscale using image. On this
> image, for some pixels, which I know, I want to display their activity based
> on a third measure. One way to do that would be to color these differently,
> and use an opacity measure to display the third measure. An example of what I
> am trying to do is at:
>
> http://www.public.iastate.edu/~maitra/papers/mrm02.pdf
>
> page 26, for instance. There are two different kinds of voxels, given by
> greens and red. At the low end, there is transparency on the red scale and at
> the upper end there is opacity in the red and the green.
>
> A simpler example involving only one kind of voxels is on page 24 of the same
> paper. Either way, that figure was done using Matlab, but I was wondering how
> do i do this using R.
>
> Any suggestions, please?
>
> Many thanks and best wishes,
> Ranjan
>
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Re: [R] my ugly apply/sweep code needs help
Please include test data in your posts. We define
sweep.med to perform the sweep on an entire matrix. Then
we lapply f over group.sel where f(g) combines a column of all
g with sweep.med applied to the submatrix of data.mat whose
rows correspond to group.vec of g.
sweep.median2 <- function(data.mat, group.vec, group.sel) {
sweep.med <- function(x) sweep(x, 2, apply(x, 2, median))
f <- function(g) cbind(g+0, sweep.med(data.mat[group.vec == g,,drop
= FALSE ]))
do.call(rbind, lapply(group.sel, f))
}
# test
mat <- matrix(1:24, 6)
group.sel <- 1:2
group.vec <- rep(1:3, 2)
sweep.median(data.mat, group.vec, group.sel)
sweep.median2(data.mat, group.vec, group.sel)
On 5/18/07, Tyler Smith <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a matrix of data from from several groups. I need to center the
> data by group, subtracting the group median from each value, initially
> for two groups at a time. I have a working function to do this, but it
> looks quite inelegant. There must be a more straightforward way to do
> this, but I always get tangled up in apply/sweep/subset
> operations. Any suggestions welcome!
>
> Thanks,
>
> Tyler
>
> My code:
>
> Notes: data.mat is an nxm matrix of data. group.vec is a vector of
> length n with grouping factors. group.sel is a vector of length 2 of
> the groups to include in the analysis.
>
> sweep.median <- function (data.mat, group.vec, group.sel) {
>
> data.sub1 <- data.mat[group.vec %in% group.sel[1],]
> data.sub2 <- data.mat[group.vec %in% group.sel[2],]
>
> data.sub1.med <- apply(data.sub1, MAR=2, median)
> data.sub1.cent <- sweep(data.sub1, MARGIN=2, data.sub1.med)
>
> data.sub2.med <- apply(data.sub2, MAR=2, median)
> data.sub2.cent <- sweep(data.sub2, MARGIN=2, data.sub2.med)
>
> data.comb <- rbind(data.sub1.cent, data.sub2.cent)
> data.comb <- cbind(c(rep(group.sel[1],nrow(data.sub1.cent)),
> rep(group.sel[2],nrow(data.sub2.cent))),
> data.comb)
>
> return(data.comb)
> }
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] graphics on Ubunut
On 18 May 2007 at 20:25, Erin Hodgess wrote: | Dear R People: | | I'm working with R on the latest version of Ubuntu. | | However, I can't get graphics to appear, even with the | simplest plot commands. What does this show for you: > capabilities()["X11"] X11 TRUE > If you get 'FALSE', and by chance you built this yourself, then you probably omitted to install the X11 development packages, and overlooked the warning that configure gave you. You could consider installing the prebuilt Ubuntu binaries that are provided via CRAN and its mirrors; see the R FAQ. Hth, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] penalized maximum likelihood estimator
Dear r helper, are their any package in R in which i can find penalized maximum likelihood estimator with beta pdf from generalized extreme value distribution? i tried to find out the package from CRAN but I could not get that yet though i found some penalized functions but those are not my work related.if possible please help me to find out that package.thanks S.Murshed Looking for a deal? Find great prices on flights and hotels with Yahoo! FareChase. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] graphics on Ubunut
Dear R People: I'm working with R on the latest version of Ubuntu. However, I can't get graphics to appear, even with the simplest plot commands. Has anyone run into that, please? I'm using R-2.5.0 on Ubuntu Feisty Fawn. (Please don't puke. That's really the name) Thanks in advance, Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How can we add a legend to a set of graphs?
On 18/05/2007 7:33 PM, Judith Flores wrote:
> Hi,
>
>I have a set of 4 graphs and I need to add a legend
> that is shared by those 4 graphs. This is what I
> tried:
>
>> locator(1) # I placed the cursor in the center of the
> 4 graphs
> $x
> [1] 9.299001
>
> $y
> [1] 226.3201
>
>
>> legend(9.3,226.3,"and the rest of the legend
> arguments")# but the legend didn't show.
>
> The legend only appears when I place in inside any
> of the for plots. How can I place it outside these
> plots, in the center.
RSiteSearch("legend outside")
suggests using par(xpd=TRUE).
Duncan Murdoch
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[R] my ugly apply/sweep code needs help
Hi,
I have a matrix of data from from several groups. I need to center the
data by group, subtracting the group median from each value, initially
for two groups at a time. I have a working function to do this, but it
looks quite inelegant. There must be a more straightforward way to do
this, but I always get tangled up in apply/sweep/subset
operations. Any suggestions welcome!
Thanks,
Tyler
My code:
Notes: data.mat is an nxm matrix of data. group.vec is a vector of
length n with grouping factors. group.sel is a vector of length 2 of
the groups to include in the analysis.
sweep.median <- function (data.mat, group.vec, group.sel) {
data.sub1 <- data.mat[group.vec %in% group.sel[1],]
data.sub2 <- data.mat[group.vec %in% group.sel[2],]
data.sub1.med <- apply(data.sub1, MAR=2, median)
data.sub1.cent <- sweep(data.sub1, MARGIN=2, data.sub1.med)
data.sub2.med <- apply(data.sub2, MAR=2, median)
data.sub2.cent <- sweep(data.sub2, MARGIN=2, data.sub2.med)
data.comb <- rbind(data.sub1.cent, data.sub2.cent)
data.comb <- cbind(c(rep(group.sel[1],nrow(data.sub1.cent)),
rep(group.sel[2],nrow(data.sub2.cent))),
data.comb)
return(data.comb)
}
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[R] How can we add a legend to a set of graphs?
Hi, I have a set of 4 graphs and I need to add a legend that is shared by those 4 graphs. This is what I tried: >locator(1) # I placed the cursor in the center of the 4 graphs $x [1] 9.299001 $y [1] 226.3201 >legend(9.3,226.3,"and the rest of the legend arguments")# but the legend didn't show. The legend only appears when I place in inside any of the for plots. How can I place it outside these plots, in the center. Your help will be very much appreciated. Sincerely, Judith Get the Yahoo! toolbar and be alerted to new email wherever you're surfing. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] BATCH
it states that it does not find the object
donParEssai
which is not there. It is nor there since you have given the object the
name
donParCara
from what I see...
elyakhlifi mustapha wrote:
> hello,
> I tried to run programs in BATCH like you told me but to read the results
> it's a lil hard
> first to read the results it's important to write down an outfile
> but when I do this I've got stil the same answer
>
> R version 2.4.1 (2006-12-18)
> Copyright (C) 2006 The R Foundation for Statistical Computing
> ISBN 3-900051-07-0
> R est un logiciel libre livré sans AUCUNE GARANTIE.
> Vous pouvez le redistribuer sous certaines conditions.
> Tapez 'license()' ou 'licence()' pour plus de détails.
> R est un projet collaboratif avec de nombreux contributeurs.
> Tapez 'contributors()' pour plus d'information et
> 'citation()' pour la façon de le citer dans les publications.
> Tapez 'demo()' pour des démonstrations, 'help()' pour l'aide
> en ligne ou 'help.start()' pour obtenir l'aide au format HTML.
> Tapez 'q()' pour quitter R.
> [Sauvegarde de la session précédente restaurée]
>
>> #programme pour calculer les caractères
>>
>> donParCara <- read.table("C:/Documents and Settings/melyakhlifi/Mes
>> documents/feuilles
>> excel/copi_donnees6.csv",header=TRUE,sep=";",quote="",dec=",")
>> #print(subset(donParCara, select = c(Id_Cara,Form_C,X)))
>>
>>
>>
>> #valeurs observées pour les caractères observés
>>
>> C103 <- as.numeric(as.character(subset(donParEssai, Id_Essai == 1006961 &
>> Id_Cara == 103, c(Val_O,Surf_O))[,1]))
>>
> Erreur dans subset(donParEssai, Id_Essai == 1006961 & Id_Cara == 103,
> c(Val_O, :
> objet "donParEssai" non trouvé
> Exécution arrêtée
>
> I thnik that it's a problem in the options but I don't know how to do
> correct the errors.
>
>
>
> _
> Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo! Mail
> [[alternative HTML version deleted]]
>
>
>
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> No virus found in this incoming message.
> Checked by AVG Free Edition.
> Version: 7.5.467 / Virus Database: 269.7.3/809 - Release Date: 17.05.2007
> 17:18
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Re: [R] Trouble compiling XML package [Broadcast]
Oops - I see I was unclear. I'm using R-2.4.1 to install; 1.9.0 is the
version of XML I was trying to load. I'll try your change - I'm really
using this for reading XML.
Thanks,
Matt
-Original Message-
From: Duncan Temple Lang [mailto:[EMAIL PROTECTED]
Sent: Friday, May 18, 2007 4:18 PM
To: Wiener, Matthew
Cc: R-help
Subject: Re: [R] Trouble compiling XML package [Broadcast]
Wiener, Matthew wrote:
> Dear Prof. Lang -
>
> I am trying to install the XML library on a 64-bit SUSE linux system
> (version info below) running 2.4.1.
>
> I have gcc version 3.3.3, and libxml2 version 2.6.7. I know this is
not
> current, but I'm on a machine used and administered by others, and
> updating libxml2 would require updating libc, and things get pretty
> complicated from there.
>
> Trying to install through R (1.9-0),
Wow, that's really old!
> I eventually get the error:
> XMLTree.c: In function `xmlBufferWrite':
> XMLTree.c:729: error: void value not ignored as it ought to be
> make: *** [XMLTree.o] Error 1
>
You might try changing the routine xmlBufferWrite to
static int
xmlBufferWrite (void * context, const char * buffer, int len) {
xmlBufferAdd((xmlBufferPtr) context, (const xmlChar *) buffer,len);
return(len);
}
That will work if there are no errors when adding to the buffer.
But this is used when generating XML in a particular way, i.e. using
internal nodes. So let's just hope that you don't invoke
this code and if you do, that there are no errors.
> I manually downloaded version 1.8-0 and got the same problem. I took
a
> look at that part of the code, but do not understand enough to start
> tinkering with it. I was able to install an earlier version a couple
of
> years ago, and it was extremely useful (thanks!) but the relevant
> machine has been decommissioned.
>
> Can you make any suggestions about which component of my system this
> might indicate needs to be changed? I checked the mailing list
> archives, but didn't find anything. I'm hoping there's an alternative
> to changing libxml2, with all the cascading requirements that would
> bring (and no guarantee, with what I know now, that that's the
problem).
>
> Thanks,
>
> Matt Wiener
>
>> version
>_
> platform x86_64-unknown-linux-gnu
> arch x86_64
> os linux-gnu
> system x86_64, linux-gnu
> status
> major 2
> minor 4.1
> year 2006
> month 12
> day18
> svn rev40228
> language R
> version.string R version 2.4.1 (2006-12-18)
>
>
--
> Notice: This e-mail message, together with any
attachments,...{{dropped}}
>
> __
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Trouble compiling XML package
Wiener, Matthew wrote:
> Dear Prof. Lang -
>
> I am trying to install the XML library on a 64-bit SUSE linux system
> (version info below) running 2.4.1.
>
> I have gcc version 3.3.3, and libxml2 version 2.6.7. I know this is not
> current, but I'm on a machine used and administered by others, and
> updating libxml2 would require updating libc, and things get pretty
> complicated from there.
>
> Trying to install through R (1.9-0),
Wow, that's really old!
> I eventually get the error:
> XMLTree.c: In function `xmlBufferWrite':
> XMLTree.c:729: error: void value not ignored as it ought to be
> make: *** [XMLTree.o] Error 1
>
You might try changing the routine xmlBufferWrite to
static int
xmlBufferWrite (void * context, const char * buffer, int len) {
xmlBufferAdd((xmlBufferPtr) context, (const xmlChar *) buffer,len);
return(len);
}
That will work if there are no errors when adding to the buffer.
But this is used when generating XML in a particular way, i.e. using
internal nodes. So let's just hope that you don't invoke
this code and if you do, that there are no errors.
> I manually downloaded version 1.8-0 and got the same problem. I took a
> look at that part of the code, but do not understand enough to start
> tinkering with it. I was able to install an earlier version a couple of
> years ago, and it was extremely useful (thanks!) but the relevant
> machine has been decommissioned.
>
> Can you make any suggestions about which component of my system this
> might indicate needs to be changed? I checked the mailing list
> archives, but didn't find anything. I'm hoping there's an alternative
> to changing libxml2, with all the cascading requirements that would
> bring (and no guarantee, with what I know now, that that's the problem).
>
> Thanks,
>
> Matt Wiener
>
>> version
>_
> platform x86_64-unknown-linux-gnu
> arch x86_64
> os linux-gnu
> system x86_64, linux-gnu
> status
> major 2
> minor 4.1
> year 2006
> month 12
> day18
> svn rev40228
> language R
> version.string R version 2.4.1 (2006-12-18)
>
> --
> Notice: This e-mail message, together with any attachments,...{{dropped}}
>
> __
> R-help@stat.math.ethz.ch mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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[R] ayuda con macros
necesito que me manden lo mas pronto posible la macro que ejecuta la accion del
modelo lineal general 'glm', para una familia binomial, con funcion de enlace
la probit y la lo-log complementaria, bueno tengo la macro para lafuncion de
enlace logit es:
# macro logit
macro.logit <- function( y, m, X, b.0, mu, niter )
{
# Inicializaciones (BOPTIM = B0 o ) MU = y/m C5=g(y/m)=ETA
# y Observaciones de ocurrencias
# Considerar MU=PI, es decir Y/M seguido B(M,PI)/M
# Recordar V(y(NITER))=MUx(1-MU)/m
# X es la matriz de diseño
c1 <- y/m
mu <- c1
c5 <- log(c1/(1-c1)) # C5 = g(y/M) FUNCIÓN DE ENLACE
for (i in 1:niter)
{
c1 <- mu # C1 = MU
c2 <-((1/c1)*(1/(1-c1)))^2 # C2 = g'(C1)**2 (g'(MU)**2)
c3 <-c1*(1-c1)*c2/m # C3 = var(C1)*C2=(C1(1-C1)/M)*C2
=1/wkk) Depende
# de M y MU !!!
c4 <- 1/c3 # C4 = 1/C3 (=wkk) (Diagonal de W)
M6 <- as.matrix(diag(c4)) # M6 = W
c6 <- c5 # C6 = ETA (C5)
c7 <- (1/c1)*(1/(1-c1)) # C7 = g'(C1) (= g'(MU))
c8 <- y
c8 <- (c8/m)-c1# C8 = y/m-MU
c9 <- c6+c7*c8 # C9 = C6 + C7*C8 (z=ETA+g'(MU)*(Y/M-MU))
M11 <- t(X)%*%M6 %*% as.vector(c9) # M11 = XTWz
M12 <- t(X)%*%M6 %*% X # M12 = XTWX
M13 <- solve( M12 ) # M13 = (XTWX)-1
# browser() # > n # para ejecutar siguiente línea de comandos
b.fi <- as.vector( M13 %*% M11 ) # BOPTIM = (XTWX)-1 XTWz
c5 <- as.vector(X %*% b.fi) # C5 = ETA = X*BOPTIM (=Xb)
mu <- exp(c5)/(1+exp(c5)) # MU = LINK-1(C5)
}
list(estimadors=b.fi,prediccions=mu*m)
}
b.0 <- as.vector(c(0,0))
mu <- as.vector(rep(0.1,6))
X <- as.matrix(data.frame(dobson$uns,dobson$log.x))
niter<-30
exe1 <- macro.logit( dobson$y, dobson$m, X, b.0, mu, 30 )
exe1
# LA INSTRUCCION DIRECTA EN R ES:
rexe1 <- glm(cbind(y,m-y)~log.x,data=dobson, family=binomial)
rexe1
bueno esto como un ejemplo de lo que quiero. Estare infinitamente agradecido.
__
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Re: [R] partial correlation significance
On 18/05/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> among the many (5) methods that I found in the list to do partial
> correlation in the following two that I had a look I am getting different
> t-values. Does anyone have any clues on why is that? The source code is
> below. Thanks.
>
> pcor3 <- function (x, test = T, p = 0.05) {
> nvar <- ncol(x)
> ndata <- nrow(x)
> conc <- solve(cor(x))
> resid.sd <- 1/sqrt(diag(conc))
> pcc <- -sweep(sweep(conc, 1, resid.sd, "*"), 2, resid.sd, "*")
> #colnames(pcc) <- rownames(pcc) <- colnames(x)
> if (test) {
> t.df <- ndata - nvar
> t <- pcc/sqrt((1 - pcc^2)/t.df)
> print(t);
> pcc <- list(coefs = pcc, sig = t > qt(1 - (p/2), df = t.df))
> }
> return(pcc)
> }
>
>
> pcor4 <- function(x, y, z) {
> return(cor.test (lm(x~z)$resid,lm(y~z)$resid));
> }
>
>
Just to self-reply my question since I found the answer. The difference is
in the degrees of freedom. The variable t.df in pcor3 is smaller than the df
used for the test in pcor4, and how smaller it is depends on the number of
variables used in the partial correlation.
[[alternative HTML version deleted]]
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Re: [R] How to extract R codes that embedded in a HTML file
Works perfectly!
Thank you very much, Fritz!
Tao
From: Friedrich Leisch <[EMAIL PROTECTED]>
To: Tao Shi <[EMAIL PROTECTED]>, [EMAIL PROTECTED],
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
Subject: Re: [R] How to extract R codes that embedded in a HTML file
Date: Fri, 18 May 2007 15:14:48 +0200
> Original Message
> Subject: Re: [R] How to extract R codes that embedded in a HTML file
> usingStangle?
> Date: Thu, 17 May 2007 17:01:30 +
> From: Tao Shi <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED], [EMAIL PROTECTED]
> CC: r-help@stat.math.ethz.ch
> : <[EMAIL PROTECTED]>
> Hi Uwe,
> Thanks for the answer, but I still need a bit more clearification. I
> always
> thought that a .rnw or .snw file is a file mixing word processing
markup
> (e.g. tex or HTML) and R/S code using noweb syntax. Is the reason for
> 'Stangle' is not working with .rnw file with HTML due to there is no
proper
> driver available (like RweaveHTML driver for Sweave)? If yes, does
R2HTML
> package have plans to provide a such driver?
> Tao
The following does the trick for me:
R> mytangle <- function ()
list(setup = RtangleSetup, runcode = utils:::RtangleRuncode,
writedoc = RtangleWritedoc,
finish = utils:::RtangleFinish, checkopts = RweaveHTMLOptions)
R> Stangle("/PATH/TO/R/SITE-LIBRARY/R2HTML/samples/example1.snw",
driver=mytangle)
Writing to file example1.R
Best,
Fritz
_
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Live Hotmail.
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Re: [R] {10,20,30}>={25,30,15}
Van---
Perhaps I'm misunderstanding your question, but in a null hypothesis
framework, the only conclusion you can draw from failing to reject
the null hypothesis is that, based on your observed data, you were
unable to conclude that your null hypothesis was false. Put another
way, the correct conclusion for both of your hypothesis tests is
"inconclusive."
Kyle H. Ambert
Graduate Student, Dept. Behavioral Neuroscience
Oregon Health & Science University
[EMAIL PROTECTED]
On May 18, 2007, at 11:07 AM, [EMAIL PROTECTED] wrote:
> Hi There,
>
> Using t.test to test hypothesis about which one is greater, A or B?
> where A={10,20,30},B={25,30,15}.
>
> My question is which of the following conclusions is right?
>
> #hypothesis testing 1
>
> h0: A greater than or equal to B
> h1: A less than B
>
> below is splus code
> A=c(10,20,30)
> B=c(25,30,15)
> t.test(c(10,20,30),c(25,30,15),alternative="less")
>
> output:
> p-value=0.3359
>
> because p-value is not less than alpha (0.05), we
> cannot reject h0.
>
> so A greater than or equal to B.
>
>
> #hypothesis testing 2
>
> h0: A less than or equal to B
> h1: A greater than B
>
> below is splus code
>
> A=c(10,20,30)
> B=c(25,30,15)
> t.test(c(10,20,30),c(25,30,15),alternative="greater")
>
> output:
> p-value=0.6641
>
> because p-value is not less than alpha (0.05), we
> cannot reject h0.
>
> so A less than or euqal to B.
> #
>
> Thank you very much.
> Van
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-
> guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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Re: [R] {10,20,30}>={25,30,15}
At the alpha level you set, A is neither greater nor less than B. Supposing
you don't use paired t.test:
data: c(10, 20, 30) and c(25, 30, 15)
t = -0.4588, df = 3.741, p-value = 0.6717
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-24.06675 17.40008
sample estimates:
mean of x mean of y
20.0 23.3
it tells you what is the null and what is the alternative.
genomenet wrote:
>
> Hi There,
> Using t.test to test hypothesis about which one is greater, A or B?
> where A={10,20,30},B={25,30,15}.
> My question is which of the following conclusions is right?
> #hypothesis testing 1
> h0: A greater than or equal to B
> h1: A less than B
> below is splus code
> A=c(10,20,30)
> B=c(25,30,15)
> t.test(c(10,20,30),c(25,30,15),alternative="less")
> output:
> p-value=0.3359
> because p-value is not less than alpha (0.05), we
> cannot reject h0.
> so A greater than or equal to B.
> #hypothesis testing 2
> h0: A less than or equal to B
> h1: A greater than B
> below is splus code
> A=c(10,20,30)
> B=c(25,30,15)
> t.test(c(10,20,30),c(25,30,15),alternative="greater")
> output:
> p-value=0.6641
> because p-value is not less than alpha (0.05), we
> cannot reject h0.
> so A less than or euqal to B.
> #
> Thank you very much.
> Van
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] partial correlation significance
Hi,
among the many (5) methods that I found in the list to do partial
correlation in the following two that I had a look I am getting different
t-values. Does anyone have any clues on why is that? The source code is
below. Thanks.
pcor3 <- function (x, test = T, p = 0.05) {
nvar <- ncol(x)
ndata <- nrow(x)
conc <- solve(cor(x))
resid.sd <- 1/sqrt(diag(conc))
pcc <- -sweep(sweep(conc, 1, resid.sd, "*"), 2, resid.sd, "*")
#colnames(pcc) <- rownames(pcc) <- colnames(x)
if (test) {
t.df <- ndata - nvar
t <- pcc/sqrt((1 - pcc^2)/t.df)
print(t);
pcc <- list(coefs = pcc, sig = t > qt(1 - (p/2), df = t.df))
}
return(pcc)
}
pcor4 <- function(x, y, z) {
return(cor.test(lm(x~z)$resid,lm(y~z)$resid));
}
[[alternative HTML version deleted]]
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[R] Trouble compiling XML package
Dear Prof. Lang -
I am trying to install the XML library on a 64-bit SUSE linux system
(version info below) running 2.4.1.
I have gcc version 3.3.3, and libxml2 version 2.6.7. I know this is not
current, but I'm on a machine used and administered by others, and
updating libxml2 would require updating libc, and things get pretty
complicated from there.
Trying to install through R (1.9-0), I eventually get the error:
XMLTree.c: In function `xmlBufferWrite':
XMLTree.c:729: error: void value not ignored as it ought to be
make: *** [XMLTree.o] Error 1
I manually downloaded version 1.8-0 and got the same problem. I took a
look at that part of the code, but do not understand enough to start
tinkering with it. I was able to install an earlier version a couple of
years ago, and it was extremely useful (thanks!) but the relevant
machine has been decommissioned.
Can you make any suggestions about which component of my system this
might indicate needs to be changed? I checked the mailing list
archives, but didn't find anything. I'm hoping there's an alternative
to changing libxml2, with all the cascading requirements that would
bring (and no guarantee, with what I know now, that that's the problem).
Thanks,
Matt Wiener
> version
_
platform x86_64-unknown-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
major 2
minor 4.1
year 2006
month 12
day18
svn rev40228
language R
version.string R version 2.4.1 (2006-12-18)
--
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[R] {10,20,30}>={25,30,15}
Hi There,
Using t.test to test hypothesis about which one is greater, A or B?
where A={10,20,30},B={25,30,15}.
My question is which of the following conclusions is right?
#hypothesis testing 1
h0: A greater than or equal to B
h1: A less than B
below is splus code
A=c(10,20,30)
B=c(25,30,15)
t.test(c(10,20,30),c(25,30,15),alternative="less")
output:
p-value=0.3359
because p-value is not less than alpha (0.05), we
cannot reject h0.
so A greater than or equal to B.
#hypothesis testing 2
h0: A less than or equal to B
h1: A greater than B
below is splus code
A=c(10,20,30)
B=c(25,30,15)
t.test(c(10,20,30),c(25,30,15),alternative="greater")
output:
p-value=0.6641
because p-value is not less than alpha (0.05), we
cannot reject h0.
so A less than or euqal to B.
#
Thank you very much.
Van
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
On 5/18/07, jiho <[EMAIL PROTECTED]> wrote:
> On 2007-May-18 , at 18:21 , Gabor Grothendieck wrote:
> > In particular, we can use "[" directly instead of subset. This is the
> > same as your function except for the line marked ### :
> >
> > myfun2 <- function() {
> > foo = data.frame(1:10,10:1)
> > foos = list(foo)
> > fooCollumn=2
> > cFoo = lapply(foos, "[", fooCollumn) ###
> > return(cFoo)
> > }
> > myfun2() # test
> >
> > On 5/18/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> >> You need to study carefully what the semantics of 'subset' are. The
> >> function body of myfun is not in the evaluation environment. (The
> >> issue
> >> is 'subset', not 'lapply': select is an *expression* and not a
> >> value.)
> >>
> >> Hint: using subset() programmatically is almost always a mistake.
> >> R's
> >> subsetting function is '[': subset is a convenience wrapper.
>
> Thank you very much. Indeed it is much better this way. I got used to
> subset for data.frames because [ does not work with negative named
> arguments while select does. E.g.:
>x[,-c("name1","name2")]
> does not work while
>subset(x,select=-c("name1","name2"))
> works (it eliminates columns named name1 and name 2 from x). But I
> guess in most cases an other syntax can achieve the same thing with
> [, like:
>x[,-which(names(x)%in%c("name1","name2"))]
> it's just a little less clear.
which is not needed. Using builtin CO2:
CO2[ ! names(CO2) %in% c("Type", "conc" ]
or
CO2[ setdiff(names(CO2), c("Type", "conc")) ]
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Re: [R] How to repress the annoying complains from X window system
You might want to find out how R was installed and suggest the he/she installs the latest version of R. If from a pre-compiled binary, the update will likely be 2.5.0 and if from source, be sure that they install using the latest R 2.5.0 Patched source tarball from: ftp://ftp.stat.math.ethz.ch/Software/R/ In addition, have them run, as 'root', update.packages() from with an R session started using: R --vanilla to update any installed packages that might be causing a conflict. HTH, Marc On Fri, 2007-05-18 at 13:14 -0400, Hao Liu wrote: > Thanks for the input, however, I am using R 2.4.0. I don't know how the > SysAdmin installed or configured it though. > > I am not running Rcmdr, I am developing R GUI applications using Tcl/Tk > package, for some weird reason, those messages comes and goes... > > Thanks > Hao > > Marc Schwartz wrote: > > >On Fri, 2007-05-18 at 11:25 -0400, Hao Liu wrote: > > > > > >>Dear All: > >> > >>I am running some GUI functions in linux environment, they runs fine, > >>however I constantly get this kind of message in R console: > >> > >>Warning: X11 protocol error: BadWindow (invalid Window parameter) > >> > >>Is there a way to repress it? Or am I doing something wrong here.. it > >>does not interfere with the running of fucntion though. > >> > >>Thanks > >>Hao > >> > >> > > > >Upgrade your version of R. > > > >You have not provided sufficient details, but if I had to guess, you are > >either running RCmdr or using other tcl/tk based widgets. > > > >If correct, the error message that you are seeing was fixed back in R > >2.4.0: > > > >oThe X11() device no longer produces (apparently spurious) > > 'BadWindow (invalid Window parameter)' warnings when run from > > Rcmdr. > > > >HTH, > > > >Marc Schwartz > > > > > > > > > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to repress the annoying complains from X window system
Thanks for the input, however, I am using R 2.4.0. I don't know how the SysAdmin installed or configured it though. I am not running Rcmdr, I am developing R GUI applications using Tcl/Tk package, for some weird reason, those messages comes and goes... Thanks Hao Marc Schwartz wrote: >On Fri, 2007-05-18 at 11:25 -0400, Hao Liu wrote: > > >>Dear All: >> >>I am running some GUI functions in linux environment, they runs fine, >>however I constantly get this kind of message in R console: >> >>Warning: X11 protocol error: BadWindow (invalid Window parameter) >> >>Is there a way to repress it? Or am I doing something wrong here.. it >>does not interfere with the running of fucntion though. >> >>Thanks >>Hao >> >> > >Upgrade your version of R. > >You have not provided sufficient details, but if I had to guess, you are >either running RCmdr or using other tcl/tk based widgets. > >If correct, the error message that you are seeing was fixed back in R >2.4.0: > >o The X11() device no longer produces (apparently spurious) > 'BadWindow (invalid Window parameter)' warnings when run from > Rcmdr. > >HTH, > >Marc Schwartz > > > > [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
On 2007-May-18 , at 18:21 , Gabor Grothendieck wrote:
> In particular, we can use "[" directly instead of subset. This is the
> same as your function except for the line marked ### :
>
> myfun2 <- function() {
> foo = data.frame(1:10,10:1)
> foos = list(foo)
> fooCollumn=2
> cFoo = lapply(foos, "[", fooCollumn) ###
> return(cFoo)
> }
> myfun2() # test
>
> On 5/18/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
>> You need to study carefully what the semantics of 'subset' are. The
>> function body of myfun is not in the evaluation environment. (The
>> issue
>> is 'subset', not 'lapply': select is an *expression* and not a
>> value.)
>>
>> Hint: using subset() programmatically is almost always a mistake.
>> R's
>> subsetting function is '[': subset is a convenience wrapper.
Thank you very much. Indeed it is much better this way. I got used to
subset for data.frames because [ does not work with negative named
arguments while select does. E.g.:
x[,-c("name1","name2")]
does not work while
subset(x,select=-c("name1","name2"))
works (it eliminates columns named name1 and name 2 from x). But I
guess in most cases an other syntax can achieve the same thing with
[, like:
x[,-which(names(x)%in%c("name1","name2"))]
it's just a little less clear.
Thanks again.
JiHO
---
http://jo.irisson.free.fr/
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A programming question
On Fri, 2007-05-18 at 09:01 -0700, Anup Nandialath wrote:
> Dear Friends,
>
> My problem is related to how to measure probabilities from a probit
> model by changing one independent variable keeping the others
> constant.
>
> A simple toy example is like this
>
> Range for my variables is defined as follows
>
> y=0 or 1, x1 = -10 to 10, x2=-40 to 100, x3 = -5 to 5
>
> Model
>
> output <- glim(y ~ x1+x2+x3 -1, family=binomial(link="probit"))
> outcoef <- output$coef
> xbeta <- as.matrix(cbind(x1, x2, x3)
>
> predprob <- pnorm(xbeta%*%outcoef)
>
> now I have the predicted probabilities for y=1 as defined above. My
> problem is as follows
>
> Keep X2 at 20 and X3 at 2. Then compute the predicted probability
> (predprob) for the entire range of X1 ie from -10 to 10 with an
> increment of 1.
>
> Therefore i need the predicted probabilities when x1=-10,
> x1=-9,x1=9, x1=10 keeping the other constant.
>
> Can somebody give me some direction on how this can be programmed.
>
> Thanks in advance for your help
>
> Sincerely
>
> Anup
Anup,
What glim() function are you using?
Or is that a typo and should be glm()?
In either case, take a look at ?predict.glm which takes your fitted
glm() model and generates predicted values based upon specifying a data
frame ('newdata' argument) containing new values.
Be sure that your 'newdata' data frame contains the same columns AND
names as the data used to fit the model.
So you could do something like:
newdata <- data.frame(X2 = 20, X3 = 2, X1 = -10:10)
predict(model, newdata, type = "response")
BTW, if you also want the fitted values for the actual data used to
create the model, you can use fitted(model) rather than doing the matrix
multiplications directly. See ?fitted for more information.
HTH,
Marc Schwartz
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Re: [R] How to extract R codes that embedded in a HTML file
> Original Message
> Subject: Re: [R] How to extract R codes that embedded in a HTML file
> using Stangle?
> Date: Thu, 17 May 2007 17:01:30 +
> From: Tao Shi <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED], [EMAIL PROTECTED]
> CC: r-help@stat.math.ethz.ch
> : <[EMAIL PROTECTED]>
> Hi Uwe,
> Thanks for the answer, but I still need a bit more clearification. I
> always
> thought that a .rnw or .snw file is a file mixing word processing markup
> (e.g. tex or HTML) and R/S code using noweb syntax. Is the reason for
> 'Stangle' is not working with .rnw file with HTML due to there is no proper
> driver available (like RweaveHTML driver for Sweave)? If yes, does R2HTML
> package have plans to provide a such driver?
> Tao
The following does the trick for me:
R> mytangle <- function ()
list(setup = RtangleSetup, runcode = utils:::RtangleRuncode,
writedoc = RtangleWritedoc,
finish = utils:::RtangleFinish, checkopts = RweaveHTMLOptions)
R> Stangle("/PATH/TO/R/SITE-LIBRARY/R2HTML/samples/example1.snw",
driver=mytangle)
Writing to file example1.R
Best,
Fritz
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[R] A programming question
Dear Friends, My problem is related to how to measure probabilities from a probit model by changing one independent variable keeping the others constant. A simple toy example is like this Range for my variables is defined as follows y=0 or 1, x1 = -10 to 10, x2=-40 to 100, x3 = -5 to 5 Model output <- glim(y ~ x1+x2+x3 -1, family=binomial(link="probit")) outcoef <- output$coef xbeta <- as.matrix(cbind(x1, x2, x3) predprob <- pnorm(xbeta%*%outcoef) now I have the predicted probabilities for y=1 as defined above. My problem is as follows Keep X2 at 20 and X3 at 2. Then compute the predicted probability (predprob) for the entire range of X1 ie from -10 to 10 with an increment of 1. Therefore i need the predicted probabilities when x1=-10, x1=-9,x1=9, x1=10 keeping the other constant. Can somebody give me some direction on how this can be programmed. Thanks in advance for your help Sincerely Anup - Got a little couch potato? Check out fun summer activities for kids. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
In particular, we can use "[" directly instead of subset. This is the
same as your function except for the line marked ### :
myfun2 <- function() {
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos, "[", fooCollumn) ###
return(cFoo)
}
myfun2() # test
On 5/18/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> You need to study carefully what the semantics of 'subset' are. The
> function body of myfun is not in the evaluation environment. (The issue
> is 'subset', not 'lapply': select is an *expression* and not a value.)
>
> Hint: using subset() programmatically is almost always a mistake. R's
> subsetting function is '[': subset is a convenience wrapper.
>
> On Fri, 18 May 2007, jiho wrote:
>
> > Hello,
> >
> > I am facing a problem with lapply which I think''' may be a bug.
> > This is the most basic function in which I can reproduce it:
> >
> > myfun <- function()
> > {
> > foo = data.frame(1:10,10:1)
> > foos = list(foo)
> > fooCollumn=2
> > cFoo = lapply(foos,subset,select=fooCollumn)
> > return(cFoo)
> > }
> >
> > I am building a list of dataframes, in each of which I want to keep
> > only column 2 (obviously I would not do it this way in real life but
> > that's just to demonstrate the bug).
> > If I execute the commands inline it works but if I clean my
> > environment, then define the function and then execute:
> > > myfun()
> > I get this error:
> > Error in eval(expr, envir, enclos) : object "fooCollumn" not found
> > while fooCollumn is defined, in the function, right before lapply. In
> > addition, if I define it outside the function and then execute the
> > function:
> > > fooCollumn=1
> > > myfun()
> > it works but uses the value defined in the general environment and
> > not the one defined in the function.
> > This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
> > What did I do wrong? Is this indeed a bug? An intended behavior?
>
> It is a bug, in your function.
>
> --
> 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
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] Simple programming question
?cut
This would recode to a factor with numeric labels for its levels.
as.numeric(as.character(...))would then convert the labels to numeric values
that you can manipulate. This presumes that the variable you are coding is
numeric and you want to recode by binning the values into ordered bins.
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Lauri Nikkinen
Sent: Friday, May 18, 2007 8:02 AM
To: Gabor Grothendieck
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Simple programming question
Thank you all for your answers. Actually Gabor's first post was right in
that sense that I wanted to have "low" to all cases which are lower than
second highest. But how about if I want to convert/recode those "high",
"mid" and "low" to numeric to make some calculations, e.g. 3, 1, 0
respectively. How do I have to modify your solutions? I would also like to
apply this solution to many kinds of recoding situations.
-Lauri
2007/5/18, Gabor Grothendieck <[EMAIL PROTECTED]>:
>
> There was a problem in the first line in the case that the highest number
> is not unique within a category. In this example its not apparent since
> that never occurs. At any rate, it should be:
>
> f <- function(x) 4 - pmin(3, match(x, sort(unique(x), decreasing = TRUE)))
> factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
>
> Also note that the factor labels were arranged so that
> "low", "mid" and "high" correspond to levels 1, 2 and 3
> respectively.
>
> On 5/18/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> > Try this. f assigns 1, 2 and 3 to the highest, second highest and third
> highest
> > within a category. ave applies f to each category. Finally we convert
> it to a
> > factor.
> >
> > f <- function(x) 4 - pmin(3, match(x, sort(x, decreasing = TRUE)))
> > factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
> >
> >
> >
> > On 5/18/07, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> > > Hi R-users,
> > >
> > > I have a simple question for R heavy users. If I have a data frame
> like this
> > >
> > >
> > > dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> > > var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> > > dfr <- dfr[order(dfr$categ),]
> > >
> > > and I want to score values or points in variable named "var3"
> following this
> > > kind of logic:
> > >
> > > 1. the highest value of var3 within category (variable named "categ")
> ->
> > > "high"
> > > 2. the second highest value -> "mid"
> > > 3. lowest value -> "low"
> > >
> > > This would be the output of this reasoning:
> > >
> > > dfr$score <-
> > >
>
factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low
","low","high","mid","low","low"))
> > > dfr
> > >
> > > The question is how I do this programmatically in R (i.e. if I have
> 2000
> > > rows in my dfr)?
> > >
> > > I appreciate your help!
> > >
> > > Cheers,
> > > Lauri
> > >
> > >[[alternative HTML version deleted]]
> > >
> > > __
> > > R-help@stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> > >
> >
>
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Re: [R] How to repress the annoying complains from X window system
On Fri, 2007-05-18 at 11:25 -0400, Hao Liu wrote: > Dear All: > > I am running some GUI functions in linux environment, they runs fine, > however I constantly get this kind of message in R console: > > Warning: X11 protocol error: BadWindow (invalid Window parameter) > > Is there a way to repress it? Or am I doing something wrong here.. it > does not interfere with the running of fucntion though. > > Thanks > Hao Upgrade your version of R. You have not provided sufficient details, but if I had to guess, you are either running RCmdr or using other tcl/tk based widgets. If correct, the error message that you are seeing was fixed back in R 2.4.0: o The X11() device no longer produces (apparently spurious) 'BadWindow (invalid Window parameter)' warnings when run from Rcmdr. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
You need to study carefully what the semantics of 'subset' are. The
function body of myfun is not in the evaluation environment. (The issue
is 'subset', not 'lapply': select is an *expression* and not a value.)
Hint: using subset() programmatically is almost always a mistake. R's
subsetting function is '[': subset is a convenience wrapper.
On Fri, 18 May 2007, jiho wrote:
> Hello,
>
> I am facing a problem with lapply which I think''' may be a bug.
> This is the most basic function in which I can reproduce it:
>
> myfun <- function()
> {
> foo = data.frame(1:10,10:1)
> foos = list(foo)
> fooCollumn=2
> cFoo = lapply(foos,subset,select=fooCollumn)
> return(cFoo)
> }
>
> I am building a list of dataframes, in each of which I want to keep
> only column 2 (obviously I would not do it this way in real life but
> that's just to demonstrate the bug).
> If I execute the commands inline it works but if I clean my
> environment, then define the function and then execute:
> > myfun()
> I get this error:
> Error in eval(expr, envir, enclos) : object "fooCollumn" not found
> while fooCollumn is defined, in the function, right before lapply. In
> addition, if I define it outside the function and then execute the
> function:
> > fooCollumn=1
> > myfun()
> it works but uses the value defined in the general environment and
> not the one defined in the function.
> This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
> What did I do wrong? Is this indeed a bug? An intended behavior?
It is a bug, in your function.
--
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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
On 2007-May-18 , at 17:09 , Thomas Lumley wrote:
> On Fri, 18 May 2007, jiho wrote:
>> I am facing a problem with lapply which I think''' may be a bug.
>> This is the most basic function in which I can reproduce it:
>>
>> myfun <- function()
>> {
>> foo = data.frame(1:10,10:1)
>> foos = list(foo)
>> fooCollumn=2
>> cFoo = lapply(foos,subset,select=fooCollumn)
>> return(cFoo)
>> }
>>
>
>> I get this error:
>> Error in eval(expr, envir, enclos) : object "fooCollumn" not found
>> while fooCollumn is defined, in the function, right before lapply.
>
>> This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
>> What did I do wrong? Is this indeed a bug? An intended behavior?
>
> The problem is that subset() evaluates its "select" argument in an
> unusual way. Usually the argument would be evaluated inside myfun()
> and the value passed to lapply(), and everything would work as you
> expect.
> subset() bypasses the normal evaluation and explicitly evaluates
> the "select" argument in the calling frame, ie, inside lapply(),
> where fooCollumn is not visible.
> You could do
> lapply(foos, function(foo) subset(foo, select=fooCollum))
> capturing fooCollum by lexical scope. In R this is often a better
> option than passing extra arguments to lapply (or other functions
> that take function arguments).
Thank you very much, this works well indeed. I agree it is a bit
confusing, to say the least. The point is that supplying other
arguments in the ... of lapply worked for all other functions I tried
before (mean, sd, summary and even spline) so it is really a problem
with subset. Anyway, R is great even with such little flaws here and
there and as long as the community is there to support it, it will rule.
Cheers,
JiHO
---
http://jo.irisson.free.fr/
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Re: [R] Execute expression ( R -e) without close
You cannot. '-e' is modelled on batch tools, and is really back-end support for Rscript -e. See ?Startup for other ways to achieve what you seem to be trying to do. On Fri, 18 May 2007, marcelll wrote: > > I need to find a way how to execute an expression with the new command line > parameter in R (R -e "AnExpression") without R get closed after the > expression has been evaluated. > For example if i want to start R with a plot or start R with a *.RData file. > Any help or suggestions are appreciated ! > > -- 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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ordering in list.files
Try this:
library(gtools)
mixedsort(x)
On 5/18/07, Shubha Vishwanath Karanth <[EMAIL PROTECTED]> wrote:
> Hi R,
>
>
>
> My csv files are stored in the order, '1abc.csv', '2def.csv',
> '3ghi.csv', '10files.csv' in a folder. When I read this into R from
> list.files (R command: x=list.files("Z:/CSV/fold",full.names=F), I don't
> get the same order, instead I get the order as "10files.csv" "1abc.csv"
> "2def.csv""3ghi.csv". But I don't want this ordering. So, how do I
> maintain the oder which I have in my physical folder?
>
>
>
> Thanks in advance
>
> Shubha
>
>
>[[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] ordering in list.files
On Fri, 2007-05-18 at 20:16 +0530, Shubha Vishwanath Karanth wrote:
> Hi R,
> My csv files are stored in the order, '1abc.csv', '2def.csv',
> '3ghi.csv', '10files.csv' in a folder. When I read this into R from
> list.files (R command: x=list.files("Z:/CSV/fold",full.names=F), I don't
> get the same order, instead I get the order as "10files.csv" "1abc.csv"
> "2def.csv""3ghi.csv". But I don't want this ordering. So, how do I
> maintain the oder which I have in my physical folder?
> Thanks in advance
>
> Shubha
>From ?list.files in the Value section:
"The files are sorted in alphabetical order, on the full path if
full.names = TRUE."
Presumably you are on Windows and you have the folder view set to sort
the files in some order, possibly by the date/time of creation? Check
the folder settings to see how you have this set.
In R the list of files is sorted in alpha order and in this case, the
numbers are sorted based upon the order of the ASCII values of the
numeric characters, not in numeric value order.
You can try this approach using a regex and sub() to get the numeric
value parts of the file names, get the ordered indices and then pass
them back to the vector of file names:
> Files
[1] "10files.csv" "1abc.csv""2def.csv""3ghi.csv"
> Files[order(as.numeric(sub("([0-9]*).*", "\\1", Files)))]
[1] "1abc.csv""2def.csv""3ghi.csv""10files.csv"
See ?sub, ?regex and ?order for more information.
HTH,
Marc Schwartz
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[R] How to repress the annoying complains from X window system
Dear All: I am running some GUI functions in linux environment, they runs fine, however I constantly get this kind of message in R console: Warning: X11 protocol error: BadWindow (invalid Window parameter) Is there a way to repress it? Or am I doing something wrong here.. it does not interfere with the running of fucntion though. Thanks Hao __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
On Fri, 18 May 2007, jiho wrote:
> Hello,
>
> I am facing a problem with lapply which I think''' may be a bug.
> This is the most basic function in which I can reproduce it:
>
> myfun <- function()
> {
> foo = data.frame(1:10,10:1)
> foos = list(foo)
> fooCollumn=2
> cFoo = lapply(foos,subset,select=fooCollumn)
> return(cFoo)
> }
>
> I get this error:
> Error in eval(expr, envir, enclos) : object "fooCollumn" not found
> while fooCollumn is defined, in the function, right before lapply.
> This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
> What did I do wrong? Is this indeed a bug? An intended behavior?
No, it isn't a bug (though it may be confusing).
The problem is that subset() evaluates its "select" argument in an unusual
way. Usually the argument would be evaluated inside myfun() and the value
passed to lapply(), and everything would work as you expect.
subset() bypasses the normal evaluation and explicitly evaluates the
"select" argument in the calling frame, ie, inside lapply(), where
fooCollumn is not visible.
You could do
lapply(foos, function(foo) subset(foo, select=fooCollum))
capturing fooCollum by lexical scope. In R this is often a better option
than passing extra arguments to lapply (or other functions that take
function arguments).
-thomas
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[R] BATCH
hello,
I tried to run programs in BATCH like you told me but to read the results it's
a lil hard
first to read the results it's important to write down an outfile
but when I do this I've got stil the same answer
R version 2.4.1 (2006-12-18)
Copyright (C) 2006 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
R est un logiciel libre livré sans AUCUNE GARANTIE.
Vous pouvez le redistribuer sous certaines conditions.
Tapez 'license()' ou 'licence()' pour plus de détails.
R est un projet collaboratif avec de nombreux contributeurs.
Tapez 'contributors()' pour plus d'information et
'citation()' pour la façon de le citer dans les publications.
Tapez 'demo()' pour des démonstrations, 'help()' pour l'aide
en ligne ou 'help.start()' pour obtenir l'aide au format HTML.
Tapez 'q()' pour quitter R.
[Sauvegarde de la session précédente restaurée]
> #programme pour calculer les caractères
>
> donParCara <- read.table("C:/Documents and Settings/melyakhlifi/Mes
> documents/feuilles
> excel/copi_donnees6.csv",header=TRUE,sep=";",quote="",dec=",")
> #print(subset(donParCara, select = c(Id_Cara,Form_C,X)))
>
>
>
> #valeurs observées pour les caractères observés
>
> C103 <- as.numeric(as.character(subset(donParEssai, Id_Essai == 1006961 &
> Id_Cara == 103, c(Val_O,Surf_O))[,1]))
Erreur dans subset(donParEssai, Id_Essai == 1006961 & Id_Cara == 103, c(Val_O,
:
objet "donParEssai" non trouvé
Exécution arrêtée
I thnik that it's a problem in the options but I don't know how to do correct
the errors.
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Re: [R] length, mean, na.rm, na.omit...
On 5/18/2007 10:32 AM, Muenchen, Robert A (Bob) wrote: > Hi All, > > Can anyone tell me why the length function does not use na.rm? I know > how to work around it, I'm just curious to know why such a useful option > was left out. length() is used very frequently in other functions, so it is encoded as a primitive for speed. Adding an optional argument to it would slow it down. > I'm also interested in the logic of setting na.rm=TRUE as the default on > mean, sd, etc. This is the opposite of the many other stat packages I > have used, so I assume it provides some programming benefit that is not > obvious to me. That's also the opposite of what R does. Did you mean to ask why na.rm=FALSE is the default? I think it follows from thinking of NA as meaning "not known", rather than "missing at random". If you don't know why values are missing, you may get biased results by calculating the mean of the others: and R would rather not give you biased results. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple programming question
Thank you all for your answers. Actually Gabor's first post was right in
that sense that I wanted to have "low" to all cases which are lower than
second highest. But how about if I want to convert/recode those "high",
"mid" and "low" to numeric to make some calculations, e.g. 3, 1, 0
respectively. How do I have to modify your solutions? I would also like to
apply this solution to many kinds of recoding situations.
-Lauri
2007/5/18, Gabor Grothendieck <[EMAIL PROTECTED]>:
>
> There was a problem in the first line in the case that the highest number
> is not unique within a category. In this example its not apparent since
> that never occurs. At any rate, it should be:
>
> f <- function(x) 4 - pmin(3, match(x, sort(unique(x), decreasing = TRUE)))
> factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
>
> Also note that the factor labels were arranged so that
> "low", "mid" and "high" correspond to levels 1, 2 and 3
> respectively.
>
> On 5/18/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> > Try this. f assigns 1, 2 and 3 to the highest, second highest and third
> highest
> > within a category. ave applies f to each category. Finally we convert
> it to a
> > factor.
> >
> > f <- function(x) 4 - pmin(3, match(x, sort(x, decreasing = TRUE)))
> > factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
> >
> >
> >
> > On 5/18/07, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> > > Hi R-users,
> > >
> > > I have a simple question for R heavy users. If I have a data frame
> like this
> > >
> > >
> > > dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> > > var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> > > dfr <- dfr[order(dfr$categ),]
> > >
> > > and I want to score values or points in variable named "var3"
> following this
> > > kind of logic:
> > >
> > > 1. the highest value of var3 within category (variable named "categ")
> ->
> > > "high"
> > > 2. the second highest value -> "mid"
> > > 3. lowest value -> "low"
> > >
> > > This would be the output of this reasoning:
> > >
> > > dfr$score <-
> > >
> factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
> > > dfr
> > >
> > > The question is how I do this programmatically in R (i.e. if I have
> 2000
> > > rows in my dfr)?
> > >
> > > I appreciate your help!
> > >
> > > Cheers,
> > > Lauri
> > >
> > >[[alternative HTML version deleted]]
> > >
> > > __
> > > R-help@stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> > >
> >
>
[[alternative HTML version deleted]]
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Re: [R] subset arg in (modified) evalq
Try this:
e <- quote(summary(y + z))
all.vars(e)
On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> Sorry, I didn't explain myself clear enough. I knew about the select arg in
> subset(). My question was, given the expression expression(summary(x+y)),
> how to extract all names that will be looked up during its evaluation.
>
> As to checking performance assumptions, you are right, in most cases the
> overhead is negligible, but sometimes I work with really big data sets.
>
> Thanks a lot for your help,
> Vadim
>
>
> - Original Message -
> From: "Gabor Grothendieck" <[EMAIL PROTECTED]>
> To: "Vadim Ogranovich" <[EMAIL PROTECTED]>
> Cc: r-help@stat.math.ethz.ch
> Sent: Friday, May 18, 2007 9:53:26 AM (GMT-0600) America/Chicago
> Subject: Re: [R] subset arg in (modified) evalq
>
> I would check your performance assumption with an actual test before
> concluding such but at any rate subset does have a select argument. See
> ?subset
>
> On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> > Thanks Gabor! This does exactly what I wanted.
> >
> > One follow-up question, how to extract the var names, in this case y, z,
> > from the expression? The subset function creates a new object and this may
> > be expensive when the data has a lot of irrelevant collumns. So I thougth
> > that I could reduce this to the columns I actually need.
> >
> > Thanks,
> > Vadim
> >
> >
> >
> > - Original Message -
> > From: "Gabor Grothendieck" <[EMAIL PROTECTED]>
> > To: "Vadim Ogranovich" <[EMAIL PROTECTED]>
> > Cc: r-help@stat.math.ethz.ch
> > Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
> > Subject: Re: [R] subset arg in (modified) evalq
> >
> > Try this:
> >
> >with(subset(data, x > 0), summary(y + z))
> >
> >
> > On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> > > Hi,
> > >
> > > When using evalq to evaluate expressions within a say data.frame context
> I
> > often wish there was a 'subset' argument, much like in lm() or any ather
> > advanced regression model. I would be grateful for a tip how to do this.
> > >
> > > Here is an illustration of what I want:
> > >
> > > n <- 100
> > > data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
> > >
> > > # this works
> > > evalq({ i <- 0 > >
> > > # I want to do the above w/o explicit subscripting, e.g.
> > > myevalq(summary(y + z), subset=0 > >
> > > Thanks,
> > > Vadim
> > >
> > >[[alternative HTML version deleted]]
> > >
> > > __
> > > R-help@stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> > >
> >
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] subset arg in (modified) evalq
I would check your performance assumption with an actual test before
concluding such but at any rate subset does have a select argument. See
?subset
On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> Thanks Gabor! This does exactly what I wanted.
>
> One follow-up question, how to extract the var names, in this case y, z,
> from the expression? The subset function creates a new object and this may
> be expensive when the data has a lot of irrelevant collumns. So I thougth
> that I could reduce this to the columns I actually need.
>
> Thanks,
> Vadim
>
>
>
> - Original Message -
> From: "Gabor Grothendieck" <[EMAIL PROTECTED]>
> To: "Vadim Ogranovich" <[EMAIL PROTECTED]>
> Cc: r-help@stat.math.ethz.ch
> Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
> Subject: Re: [R] subset arg in (modified) evalq
>
> Try this:
>
>with(subset(data, x > 0), summary(y + z))
>
>
> On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > When using evalq to evaluate expressions within a say data.frame context I
> often wish there was a 'subset' argument, much like in lm() or any ather
> advanced regression model. I would be grateful for a tip how to do this.
> >
> > Here is an illustration of what I want:
> >
> > n <- 100
> > data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
> >
> > # this works
> > evalq({ i <- 0 >
> > # I want to do the above w/o explicit subscripting, e.g.
> > myevalq(summary(y + z), subset=0 >
> > Thanks,
> > Vadim
> >
> >[[alternative HTML version deleted]]
> >
> > __
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] ordering in list.files
Rename them so that they all have the same number of zero filled leading
digits:
01abc.csv, 02def.csv, ...
'10' comes before '1a' in the stardard sorting sequence.
The other way is to get a list of all your files, parse off the numerics at
the beginning and then sort in numerical order.
On 5/18/07, Shubha Vishwanath Karanth <[EMAIL PROTECTED]> wrote:
>
> Hi R,
>
>
>
> My csv files are stored in the order, '1abc.csv', '2def.csv',
> '3ghi.csv', '10files.csv' in a folder. When I read this into R from
> list.files (R command: x=list.files("Z:/CSV/fold",full.names=F), I don't
> get the same order, instead I get the order as "10files.csv" "1abc.csv"
> "2def.csv""3ghi.csv". But I don't want this ordering. So, how do I
> maintain the oder which I have in my physical folder?
>
>
>
> Thanks in advance
>
> Shubha
>
>
>[[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
[[alternative HTML version deleted]]
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Re: [R] Simple programming question
The solution already calculates it as numeric and only after that
does it convert it to factor so just omit the conversion:
f <- function(x) 4 - pmin(3, match(x, sort(unique(x), decreasing = TRUE)))
score <- ave(dfr$var3, dfr$categ, FUN = f)
As mentioned, this assigns 1 to low (everything other than the highest
two numbers in a category), 2 to the second highest and 3 to the highest.
If you want some other assignment, e.g. 3 is low, 1 is mid and 0 is high
then try:
c(3, 1, 0)[score]
On 5/18/07, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> Thank you all for your answers. Actually Gabor's first post was right in
> that sense that I wanted to have "low" to all cases which are lower than
> second highest. But how about if I want to convert/recode those "high",
> "mid" and "low" to numeric to make some calculations, e.g. 3, 1, 0
> respectively. How do I have to modify your solutions? I would also like to
> apply this solution to many kinds of recoding situations.
>
> -Lauri
>
>
> 2007/5/18, Gabor Grothendieck <[EMAIL PROTECTED]>:
> > There was a problem in the first line in the case that the highest number
> > is not unique within a category. In this example its not apparent since
> > that never occurs. At any rate, it should be:
> >
> > f <- function(x) 4 - pmin(3, match(x, sort(unique(x), decreasing = TRUE)))
> > factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
> >
> > Also note that the factor labels were arranged so that
> > "low", "mid" and "high" correspond to levels 1, 2 and 3
> > respectively.
> >
> > On 5/18/07, Gabor Grothendieck < [EMAIL PROTECTED]> wrote:
> > > Try this. f assigns 1, 2 and 3 to the highest, second highest and third
> highest
> > > within a category. ave applies f to each category. Finally we convert
> it to a
> > > factor.
> > >
> > > f <- function(x) 4 - pmin(3, match(x, sort(x, decreasing = TRUE)))
> > > factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
> > >
> > >
> > >
> > > On 5/18/07, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> > > > Hi R-users,
> > > >
> > > > I have a simple question for R heavy users. If I have a data frame
> like this
> > > >
> > > >
> > > > dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> > > > var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> > > > dfr <- dfr[order(dfr$categ),]
> > > >
> > > > and I want to score values or points in variable named "var3"
> following this
> > > > kind of logic:
> > > >
> > > > 1. the highest value of var3 within category (variable named "categ")
> ->
> > > > "high"
> > > > 2. the second highest value -> "mid"
> > > > 3. lowest value -> "low"
> > > >
> > > > This would be the output of this reasoning:
> > > >
> > > > dfr$score <-
> > > >
> factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
> > > > dfr
> > > >
> > > > The question is how I do this programmatically in R (i.e. if I have
> 2000
> > > > rows in my dfr)?
> > > >
> > > > I appreciate your help!
> > > >
> > > > Cheers,
> > > > Lauri
> > > >
> > > >[[alternative HTML version deleted]]
> > > >
> > > > __
> > > > R-help@stat.math.ethz.ch mailing list
> > > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > > > and provide commented, minimal, self-contained, reproducible code.
> > > >
> > >
> >
>
>
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[R] R: Simple programming question
try also this
dfr$score<-factor(dfr$var3 %in% sort(unique(dfr$var3), decr=T)[1:2] * dfr$var3,
labels=c("low", "mid", "high"))
Hope this helps,
Stefano
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di Lauri Nikkinen
Inviato: venerdì 18 maggio 2007 15.15
A: r-help@stat.math.ethz.ch
Oggetto: [R] Simple programming question
Hi R-users,
I have a simple question for R heavy users. If I have a data frame like this
dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
dfr <- dfr[order(dfr$categ),]
and I want to score values or points in variable named "var3" following this
kind of logic:
1. the highest value of var3 within category (variable named "categ") ->
"high"
2. the second highest value -> "mid"
3. lowest value -> "low"
This would be the output of this reasoning:
dfr$score <-
factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
dfr
The question is how I do this programmatically in R (i.e. if I have 2000
rows in my dfr)?
I appreciate your help!
Cheers,
Lauri
[[alternative HTML version deleted]]
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Re: [R] subset arg in (modified) evalq
Sorry, I didn't explain myself clear enough. I knew about the select arg in
subset(). My question was, given the expression expression(summary(x+y)), how
to extract all names that will be looked up during its evaluation.
As to checking performance assumptions, you are right, in most cases the
overhead is negligible, but sometimes I work with really big data sets.
Thanks a lot for your help,
Vadim
- Original Message -
From: " Gabor Grothendieck " < ggrothendieck @ gmail .com>
To: " Vadim Ogranovich " < vogranovich @ jumptrading .com>
Cc: r-help @stat.math. ethz .ch
Sent: Friday, May 18, 2007 9:53:26 AM ( GMT-0600 ) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
I would check your performance assumption with an actual test before
concluding such but at any rate subset does have a select argument. See
?subset
On 5/18/07, Vadim Ogranovich < vogranovich @ jumptrading .com> wrote:
> Thanks Gabor ! This does exactly what I wanted.
>
> One follow-up question, how to extract the var names, in this case y, z,
> from the expression? The subset function creates a new object and this may
> be expensive when the data has a lot of irrelevant collumns . So I thougth
> that I could reduce this to the columns I actually need.
>
> Thanks,
> Vadim
>
>
>
> - Original Message -
> From: " Gabor Grothendieck " < ggrothendieck @ gmail .com>
> To: " Vadim Ogranovich " < vogranovich @ jumptrading .com>
> Cc: r-help @stat.math. ethz .ch
> Sent: Friday, May 18, 2007 9:19:49 AM ( GMT-0600 ) America/Chicago
> Subject: Re: [R] subset arg in (modified) evalq
>
> Try this:
>
> with(subset(data, x > 0), summary(y + z))
>
>
> On 5/18/07, Vadim Ogranovich < vogranovich @ jumptrading .com> wrote:
> > Hi,
> >
> > When using evalq to evaluate expressions within a say data.frame context I
> often wish there was a 'subset' argument, much like in lm () or any ather
> advanced regression model. I would be grateful for a tip how to do this.
> >
> > Here is an illustration of what I want:
> >
> > n <- 100
> > data <- data.frame(x= rnorm (n), y= rnorm (y), z= rnorm (z))
> >
> > # this works
> > evalq ({ i <- 0 >
> > # I want to do the above w/o explicit subscripting , e.g.
> > myevalq (summary(y + z), subset=0 >
> > Thanks,
> > Vadim
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help @stat.math. ethz .ch mailing list
> > https ://stat. ethz .ch/mailman/ listinfo / r-help
> > PLEASE do read the posting guide
> http :// www . R-project .org/ posting-guide . html
> > and provide commented, minimal, self-contained , reproducible code.
> >
>
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lapply not reading arguments from the correct environment
subset() was not defined inside myfun(); try this version instead:
myfun <- function () {
foo <- data.frame(1:10, 10:1)
foos <- list(foo)
fooCollumn <- 2
my.subset <- function(...) subset(...)
cFoo <- lapply(foos, my.subset, select = fooCollumn)
cFoo
}
myfun()
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: "jiho" <[EMAIL PROTECTED]>
To:
Sent: Friday, May 18, 2007 4:41 PM
Subject: [R] lapply not reading arguments from the correct environment
> Hello,
>
> I am facing a problem with lapply which I think''' may be a bug.
> This is the most basic function in which I can reproduce it:
>
> myfun <- function()
> {
> foo = data.frame(1:10,10:1)
> foos = list(foo)
> fooCollumn=2
> cFoo = lapply(foos,subset,select=fooCollumn)
> return(cFoo)
> }
>
> I am building a list of dataframes, in each of which I want to keep
> only column 2 (obviously I would not do it this way in real life but
> that's just to demonstrate the bug).
> If I execute the commands inline it works but if I clean my
> environment, then define the function and then execute:
> > myfun()
> I get this error:
> Error in eval(expr, envir, enclos) : object "fooCollumn" not found
> while fooCollumn is defined, in the function, right before lapply.
> In
> addition, if I define it outside the function and then execute the
> function:
> > fooCollumn=1
> > myfun()
> it works but uses the value defined in the general environment and
> not the one defined in the function.
> This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
> What did I do wrong? Is this indeed a bug? An intended behavior?
> Thanks in advance.
>
> JiHO
> ---
> http://jo.irisson.free.fr/
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] subset arg in (modified) evalq
Thanks Gabor! This does exactly what I wanted.
One follow-up question, how to extract the var names, in this case y, z, from
the expression? The subset function creates a new object and this may be
expensive when the data has a lot of irrelevant collumns. So I thougth that I
could reduce this to the columns I actually need.
Thanks,
Vadim
- Original Message -
From: "Gabor Grothendieck" <[EMAIL PROTECTED]>
To: "Vadim Ogranovich" <[EMAIL PROTECTED]>
Cc: r-help@stat.math.ethz.ch
Sent: Friday, May 18, 2007 9:19:49 AM (GMT-0600) America/Chicago
Subject: Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x > 0), summary(y + z))
On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> Hi,
>
> When using evalq to evaluate expressions within a say data.frame context I
> often wish there was a 'subset' argument, much like in lm() or any ather
> advanced regression model. I would be grateful for a tip how to do this.
>
> Here is an illustration of what I want:
>
> n <- 100
> data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
>
> # this works
> evalq({ i <- 0
> # I want to do the above w/o explicit subscripting, e.g.
> myevalq(summary(y + z), subset=0
> Thanks,
> Vadim
>
> [[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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and provide commented, minimal, self-contained, reproducible code.
[R] ordering in list.files
Hi R,
My csv files are stored in the order, '1abc.csv', '2def.csv',
'3ghi.csv', '10files.csv' in a folder. When I read this into R from
list.files (R command: x=list.files("Z:/CSV/fold",full.names=F), I don't
get the same order, instead I get the order as "10files.csv" "1abc.csv"
"2def.csv""3ghi.csv". But I don't want this ordering. So, how do I
maintain the oder which I have in my physical folder?
Thanks in advance
Shubha
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] lapply not reading arguments from the correct environment
Hello,
I am facing a problem with lapply which I think''' may be a bug.
This is the most basic function in which I can reproduce it:
myfun <- function()
{
foo = data.frame(1:10,10:1)
foos = list(foo)
fooCollumn=2
cFoo = lapply(foos,subset,select=fooCollumn)
return(cFoo)
}
I am building a list of dataframes, in each of which I want to keep
only column 2 (obviously I would not do it this way in real life but
that's just to demonstrate the bug).
If I execute the commands inline it works but if I clean my
environment, then define the function and then execute:
> myfun()
I get this error:
Error in eval(expr, envir, enclos) : object "fooCollumn" not found
while fooCollumn is defined, in the function, right before lapply. In
addition, if I define it outside the function and then execute the
function:
> fooCollumn=1
> myfun()
it works but uses the value defined in the general environment and
not the one defined in the function.
This is with R 2.5.0 on both OS X and Linux (Fedora Core 6)
What did I do wrong? Is this indeed a bug? An intended behavior?
Thanks in advance.
JiHO
---
http://jo.irisson.free.fr/
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[R] length, mean, na.rm, na.omit...
Hi All, Can anyone tell me why the length function does not use na.rm? I know how to work around it, I'm just curious to know why such a useful option was left out. I'm also interested in the logic of setting na.rm=TRUE as the default on mean, sd, etc. This is the opposite of the many other stat packages I have used, so I assume it provides some programming benefit that is not obvious to me. Thanks, Bob = Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting Center U of TN Office of Information Technology 200 Stokely Management Center, Knoxville, TN 37996-0520 Voice: (865) 974-5230 FAX: (865) 974-4810 Email: [EMAIL PROTECTED] Web: http://oit.utk.edu/scc, News: http://listserv.utk.edu/archives/statnews.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple programming question
There was a problem in the first line in the case that the highest number
is not unique within a category. In this example its not apparent since
that never occurs. At any rate, it should be:
f <- function(x) 4 - pmin(3, match(x, sort(unique(x), decreasing = TRUE)))
factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
Also note that the factor labels were arranged so that
"low", "mid" and "high" correspond to levels 1, 2 and 3
respectively.
On 5/18/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Try this. f assigns 1, 2 and 3 to the highest, second highest and third
> highest
> within a category. ave applies f to each category. Finally we convert it to
> a
> factor.
>
> f <- function(x) 4 - pmin(3, match(x, sort(x, decreasing = TRUE)))
> factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
>
>
>
> On 5/18/07, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> > Hi R-users,
> >
> > I have a simple question for R heavy users. If I have a data frame like this
> >
> >
> > dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> > var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> > dfr <- dfr[order(dfr$categ),]
> >
> > and I want to score values or points in variable named "var3" following this
> > kind of logic:
> >
> > 1. the highest value of var3 within category (variable named "categ") ->
> > "high"
> > 2. the second highest value -> "mid"
> > 3. lowest value -> "low"
> >
> > This would be the output of this reasoning:
> >
> > dfr$score <-
> > factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
> > dfr
> >
> > The question is how I do this programmatically in R (i.e. if I have 2000
> > rows in my dfr)?
> >
> > I appreciate your help!
> >
> > Cheers,
> > Lauri
> >
> >[[alternative HTML version deleted]]
> >
> > __
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
__
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Re: [R] Simple programming question
According to your post you are assuming that there are only 3 unique
values for var3 within each category. But category C and D have 4 unique
values for var3.
split(dfr, dfr$categ)
...
$C
id categ var3 score
3 3 C6 high
7 7 C5 mid
11 11 C3 low
15 15 C1 low
...
If you meant something different, then just change myfun() below
gmax <- function(x, rnk=1){
## generalized maximum with rnk=1 being the bigest value (i.e. max)
return( sort( unique(x), decreasing=T )[rnk] )
}
myfun <- function(x){ ifelse( x==gmax(x,1), "high",
ifelse( x==gmax(x,2), "med", "low" ) ) }
out <- lapply( split(dfr$var3, dfr$categ), myfun )
data.frame( dfr, my.score = unsplit(out, dfr$categ) )
Regards, Adai
Lauri Nikkinen wrote:
> Hi R-users,
>
> I have a simple question for R heavy users. If I have a data frame like this
>
>
> dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> dfr <- dfr[order(dfr$categ),]
>
> and I want to score values or points in variable named "var3" following this
> kind of logic:
>
> 1. the highest value of var3 within category (variable named "categ") ->
> "high"
> 2. the second highest value -> "mid"
> 3. lowest value -> "low"
>
> This would be the output of this reasoning:
>
> dfr$score <-
> factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
> dfr
>
> The question is how I do this programmatically in R (i.e. if I have 2000
> rows in my dfr)?
>
> I appreciate your help!
>
> Cheers,
> Lauri
>
> [[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
__
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Re: [R] subset arg in (modified) evalq
Try this:
with(subset(data, x > 0), summary(y + z))
On 5/18/07, Vadim Ogranovich <[EMAIL PROTECTED]> wrote:
> Hi,
>
> When using evalq to evaluate expressions within a say data.frame context I
> often wish there was a 'subset' argument, much like in lm() or any ather
> advanced regression model. I would be grateful for a tip how to do this.
>
> Here is an illustration of what I want:
>
> n <- 100
> data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
>
> # this works
> evalq({ i <- 0
> # I want to do the above w/o explicit subscripting, e.g.
> myevalq(summary(y + z), subset=0
> Thanks,
> Vadim
>
>[[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
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Re: [R] Simple programming question
Try this. f assigns 1, 2 and 3 to the highest, second highest and third highest
within a category. ave applies f to each category. Finally we convert it to a
factor.
f <- function(x) 4 - pmin(3, match(x, sort(x, decreasing = TRUE)))
factor(ave(dfr$var3, dfr$categ, FUN = f), lab = c("low", "mid", "high"))
On 5/18/07, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
> Hi R-users,
>
> I have a simple question for R heavy users. If I have a data frame like this
>
>
> dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> dfr <- dfr[order(dfr$categ),]
>
> and I want to score values or points in variable named "var3" following this
> kind of logic:
>
> 1. the highest value of var3 within category (variable named "categ") ->
> "high"
> 2. the second highest value -> "mid"
> 3. lowest value -> "low"
>
> This would be the output of this reasoning:
>
> dfr$score <-
> factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
> dfr
>
> The question is how I do this programmatically in R (i.e. if I have 2000
> rows in my dfr)?
>
> I appreciate your help!
>
> Cheers,
> Lauri
>
>[[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] time series
Jessica, > I am working with a data file which is the record of precipitation > measurement normaly done every 10 minutes. I would like to check if there > are missing times in my data file. > > Is there a function existing able to check for that in R ? I'd use max(diff(time))==min(diff(time)). HTH, Rogerio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple programming question
try this:
dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
dfr <- dfr[order(dfr$categ), ]
dfr$score <- unlist(tapply(dfr$var3, dfr$categ, function (x) {
sn <- sort(unique(x), decreasing = TRUE)
labs <- c("high", "mid", rep("low", length(sn) - 2))
labs[match(x, sn)]
}))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: "Lauri Nikkinen" <[EMAIL PROTECTED]>
To:
Sent: Friday, May 18, 2007 3:15 PM
Subject: [R] Simple programming question
> Hi R-users,
>
> I have a simple question for R heavy users. If I have a data frame
> like this
>
>
> dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
> var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
> dfr <- dfr[order(dfr$categ),]
>
> and I want to score values or points in variable named "var3"
> following this
> kind of logic:
>
> 1. the highest value of var3 within category (variable named
> "categ") ->
> "high"
> 2. the second highest value -> "mid"
> 3. lowest value -> "low"
>
> This would be the output of this reasoning:
>
> dfr$score <-
> factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
> dfr
>
> The question is how I do this programmatically in R (i.e. if I have
> 2000
> rows in my dfr)?
>
> I appreciate your help!
>
> Cheers,
> Lauri
>
> [[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] subset arg in (modified) evalq
Hi,
When using evalq to evaluate expressions within a say data.frame context I
often wish there was a 'subset' argument, much like in lm() or any ather
advanced regression model. I would be grateful for a tip how to do this.
Here is an illustration of what I want:
n <- 100
data <- data.frame(x=rnorm(n), y=rnorm(y), z=rnorm(z))
# this works
evalq({ i <- 0https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Execute expression ( R -e) without close
I need to find a way how to execute an expression with the new command line parameter in R (R -e "AnExpression") without R get closed after the expression has been evaluated. For example if i want to start R with a plot or start R with a *.RData file. Any help or suggestions are appreciated ! -- View this message in context: http://www.nabble.com/Execute-expression-%28-R--e%29-without-close-tf3777652.html#a10681974 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Simple programming question
Hi R-users,
I have a simple question for R heavy users. If I have a data frame like this
dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4),
var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1))
dfr <- dfr[order(dfr$categ),]
and I want to score values or points in variable named "var3" following this
kind of logic:
1. the highest value of var3 within category (variable named "categ") ->
"high"
2. the second highest value -> "mid"
3. lowest value -> "low"
This would be the output of this reasoning:
dfr$score <-
factor(c("high","mid","low","low","high","mid","mid","low","high","mid","low","low","high","mid","low","low"))
dfr
The question is how I do this programmatically in R (i.e. if I have 2000
rows in my dfr)?
I appreciate your help!
Cheers,
Lauri
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] svychisq
Dear All I am trying to use svychisq with a two-dimensional table 4 x 5. The command I am using is summary(svytable(~dietperception+dietstatus,dudleyls1rake,na.rm=TRUE),"C hisq") It is throwing up an error message as follows: Error in NCOL(y) : only 0's may be mixed with negative subscripts In addition: Warning messages: 1: is.na() applied to non-(list or vector) in: is.na(rowvar) 2: is.na() applied to non-(list or vector) in: is.na(colvar) The dietperception data set does have some NA's where as there are none in dietstatus. The table is svytable(~dietperception+dietstatus,dudleyls1rake) dietstatus dietperceptionGood OK Poor Very Poor Unclassified Perceive healthy 6669.15287 6306.38635 47563.49174 80030.97096 12340.28453 Neither agree not disagree250.68278 204.88193 6086.84308 35575.32925 2158.47668 Perceive unhealthy 0.0 171.49710 2075.8023026390.92946318.73213 Don't know 0.022.33107 334.44880 4402.99293562.91532 I wondered if you could give me any idea where I am gong wrong. Many thanks Angela Dr Angela Moss Public Health Information Analyst Dudley PCT St. John's House Union Street Dudley DY2 8PP Tel: 01384 366091 Fax: 01384 366485 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 9 Courses: Upcoming July 2007 R/S+ schedule by XLSolutions Corp / New Website!
__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] penalized maximum likelihood estimator
dear R-helper, I tried to find out a package in which i can have penalized maximum likelihood estimator applying on generalized extreme value distribution with beta function) but could not. would you please help me to know the name of the package. thanks for your help. S.Murshed --- [EMAIL PROTECTED] wrote: > Send R-help mailing list submissions to > r-help@stat.math.ethz.ch > > To subscribe or unsubscribe via the World Wide Web, > visit > https://stat.ethz.ch/mailman/listinfo/r-help > or, via email, send a message with subject or body > 'help' to > [EMAIL PROTECTED] > > You can reach the person managing the list at > [EMAIL PROTECTED] > > When replying, please edit your Subject line so it > is more specific > than "Re: Contents of R-help digest..." > > > Today's Topics: > >1. Re: use mathematics formula (hadley wickham) >2. creating different strata (raymond chiruka) >3. converting a data frame to ts objects (fatih > ozgul) >4. Running R function as a Batch process (d. > sarthi maheshwari) >5. Re: creating different strata (raymond > chiruka) >6. Re: Filled step-function? (Beate Kowalczyk) >7. Re: creating different strata (Petr > Klasterecky) >8. Re: use mathematics formula (John Kane) >9. Re: Running R function as a Batch process > (Vladimir Eremeev) > 10. more woes trying to convert a data.frame to a > numerical > matrix (Andrew Yee) > 11. Re: Filled step-function? (Petr Klasterecky) > 12. Re: more woes trying to convert a data.frame > to a numerical > matrix (ONKELINX, Thierry) > 13. Re: Filled step-function? (Jim Lemon) > 14. Re: more woes trying to convert a data.frame > to a numerical > matrix (Marc Schwartz) > 15. Re: more woes trying to convert a data.frame > to a numerical > matrix (Dimitris Rizopoulos) > 16. Re: more woes trying to convert a data.frame > to a numerical > matrix (Andrew Yee) > 17. Re: Running R function as a Batch process > (Hanke, Alex) > 18. Re: more woes trying to convert a data.frame > to a numerical > matrix (Marc Schwartz) > 19. Is it possible to pass a Tcl/Tk component as > argument to a > function (Hao Liu) > 20. Re: converting a data frame to ts objects > (Gabor Grothendieck) > 21. Re: more woes trying to convert a data.frame > to a numerical > matrix (Andrew Yee) > 22. Re: log rank test p value (Terry Therneau) > 23. effective df in local polinomial regression > (Simone Vantini) > 24. Re: more woes trying to convert a data.frame > to a numerical > matrix (Andrew Yee) > 25. Re: Is it possible to pass a Tcl/Tk component > as argument to > afunction (John Fox) > 26. Re: Is it possible to pass a Tcl/Tk component > as argument to > a function (Duncan Murdoch) > 27. Re: more woes trying to convert a data.frame > to a numerical > matrix (Marc Schwartz) > 28. partial least regression > (=?gb2312?Q?=D5=D4=D3=F1=D6=D2?=) > 29. Re: partial least regression (Gabor > Grothendieck) > 30. Installing SJava - problem (mister_bluesman) > 31. Re: Installing SJava - problem (Prof Brian > Ripley) > 32. Re: more woes trying to convert a data.frame > to a numerical > matrix (Liaw, Andy) > 33. Re: Installing SJava - problem > (mister_bluesman) > 34. Re: more woes trying to convert a data.frame > to a numerical > matrix (Marc Schwartz) > 35. Re: urca package - summary method - (Pfaff, > Bernhard Dr.) > 36. Re: Installing SJava - problem > (mister_bluesman) > 37. Re: how to reduce in a grid ? (Norma Leyva) > 38. Re: Installing SJava - problem (Prof Brian > Ripley) > 39. Re: Installing SJava - problem > (mister_bluesman) > 40. substitute "x" for "pattern" in a list, while > preservign list > "structure". lapply, gsub, list...? (new > ruser) > 41. Unable to compile "Matrix" package (Vittorio > De Martino) > 42. Re: substitute "x" for "pattern" in a list, > while preservign > list "structure". lapply, gsub, list...? (Marc > Schwartz) > 43. read.table opening a website incl Password > (Roland Rau) > 44. Re: lmer function (Douglas Bates) > 45. Re: Unable to compile "Matrix" package > (Douglas Bates) > 46. lmer error confusion (Rick DeShon) > 47. Re: read.table opening a website incl Password > (Chuck Cleland) > 48. Re: substitute "x" for "pattern" in a list, > while preservign > list "structure". lapply, gsub, list...? > (Gabor Grothendieck) > 49. Re: lmer error confusion (Douglas Bates) > 50. Re: read.table opening a website incl Password > (Prof Brian Ripley) > 51. Re: read.table opening a website incl Password > (Bos, Roger) > 52. Re: substitute "x" for "pattern" in a list, > while preservign > list "structure". lapply, gsub, list...? (Marc > Schwartz) > 53. Re: read.table opening a website incl Password > (Roland Rau) > 54. Re: substitute "x" for "pattern" in a list, > while preservign > list "structure". lapply, gsub,
[R] time series
Dear all, I am working with a data file which is the record of precipitation measurement normaly done every 10 minutes. I would like to check if there are missing times in my data file. Is there a function existing able to check for that in R ? Thanks by advance, Jessica __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Cross-validation for logistic regression with lasso2
Hello, I am trying to shrink the coefficients of a logistic regression for a
sparse dataset, I am using the lasso (lasso2) and I am trying to determine
the shrinkinage factor by cross-validation. I would like please some of the
experts here to tell me whether i'm doing it correctly or not. Below is my
dataset and the functions I use
w=
a b c d e P A
0 0 0 0 0 1 879
1 0 0 0 0 1 3
0 1 0 0 0 7 7
0 0 1 0 0 230 2
0 0 0 1 0 450 7
0 0 0 0 1 4
#The GLM output shows that the coefficients c and d are larger than 10:
resp=cbind(w$P,w$A)
summary(glm(resp~a+b+c+d+e,data=w,family=binomial))
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -6.779 1.001 -6.775 1.24e-11 ***
a 5.680 1.528 3.718 0.000201 ***
b 6.779 1.134 5.976 2.29e-09 ***
c 11.524 1.227 9.392 < 2e-16 ***
d 10.942 1.071 10.220 < 2e-16 ***
e 3.688 1.124 3.282 0.001031 **
#so I wrote this below using the lasso2 package to determine the best
shrinkage factor using the gcv cross-validation:
for (i in seq(1,40,1)) {
glmba=gl1ce(resp~a+b+c+d+e, data = w, family = binomial(),bound=i)
ecco=round(gcv(glmba,type="Tibshirani",gen.inverse.diag =1e11),digits=3)
print(ecco)
}
#and it gives me 21 with the lowest gcv.
#then I determine the shrunken coefficients:
>gl1ce( resp ~ a + b + c + d + e, data = w, family = binomial(), bound =
21)
Coefficients:
(Intercept) a b c d
e
-4.7498162.7762154.3426618.9565838.6615931.264660
Family:
Family: binomial
Link function: logit
The absolute L1 bound was : 21
The Lagrangian for the bound is : 1.843283
Thanks
--
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Inverse gamma
Patrick Wang wrote: > > assume I need to generate X from inverse gamma with parameter (k, > beta). > > should I generate from Y from gamma(-k, beta), > > then take X=1/Y? > Check the Borg of All Wisdom... http://en.wikipedia.org/wiki/Inverse-gamma_distribution Generate Y from gamma(k, 1/beta) (using... rgamma(n = number.of.points, shape = k, scale = 1/beta) ... or ... rgamma(n = number.of.points, shape = k, rate = beta) ) and take X = 1/Y (unless your beta is not the rate parameter...) Alberto Monteiro __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] unscuscribe
unsuscribe - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using lm() with variable formula [Broadcast]
One way to do it is by giving a data frame with the right variables to
lm() as the first argument each time. If lm() is given a data frame as
the first argument, it will treat the first variable as the LHS and the
rest as the RHS of the formula.
As examples, you can do:
lm(myData[c("height", "weight", "BP", "Cals")])
(The drawback to this is that the "formula" in the fitted model object
looks a bit strange...)
Andy
From: Chris Elsaesser
>
> New to R; please excuse me if this is a dumb question. I
> tried to RTFM;
> didn't help.
>
> I want to do a series of regressions over the columns in a data.frame,
> systematically varying the response variable and the the
> terms; and not
> necessarily including all the non-response columns. In my case, the
> columns are time series. I don't know if that makes a difference; it
> does mean I have to call lag() to offset non-response terms. I can not
> assume a specific number of columns in the data.frame; might
> be 3, might
> be 20.
>
> My central problem is that the formula given to lm() is different each
> time. For example, say a data.frame had columns with the following
> headings: height, weight, BP (blood pressure), and Cals
> (calorie intake
> per time frame). In that case, I'd need something like the following:
>
> lm(height ~ weight + BP + Cals)
> lm(height ~ weight + BP)
> lm(height ~ weight + Cals)
> lm(height ~ BP + Cals)
> lm(weight ~ height + BP)
> lm(weight ~ height + Cals)
> etc.
>
> In general, I'll have to read the header to get the argument labels.
>
> Do I have to write several functions, each taking a different
> number of
> arguments? I'd like to construct a string or list representing the
> varialbes in the formula and apply lm(), so to say [I'm mainly a Lisp
> programmer where that part would be very simple. Anyone have
> a Lisp API
> for R? :-}]
>
> Thanks,
> chris
>
> Chris Elsaesser, PhD
> Principal Scientist, Machine Learning
> SPADAC Inc.
> 7921 Jones Branch Dr. Suite 600
> McLean, VA 22102
>
> 703.371.7301 (m)
> 703.637.9421 (o)
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
--
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Re: [R] naive question about using an object as the name of another object
On 17/05/2007 10:11 PM, Andrew Yee wrote:
> This is a dumb question, but I'm having trouble finding the answer to this.
>
> I'd like to do the following:
>
> x<-"asdf"
>
> and then have
>
> the object x.y become automatically converted/represented as asdf.y (sort of
> akin to macro variables in SAS where you would do:
> %let x=asdf and do &x..y)
>
> What is the syntax for having x represented as "asdf" in x.y ?
You can use assign( gsub("x", x, "x.y"), x.y ), but this is not a normal
thing to do in R programs: R is not SAS. You should investigate using a
list and setting a member of it, e.g.
asdf$y <- value
or
f <- function(value) list(y=value)
asdf <- f(value)
depending on what you are trying to do.
Duncan Murdoch
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Re: [R] .jinit() problem
Please let me know if you require any more information. Thanks mister_bluesman wrote: > > Hello there. > > When I try to start the jvm using .jinit() after loading library(rJava) I > don't seem to be able to as I get the message: > > Error in .jinit() : Cannot create Java Virtual Machine > > What is going on here? I have java 1.6 installed on my XP machine > > Thanks again. > > -- View this message in context: http://www.nabble.com/.jinit%28%29-problem-tf3774265.html#a10679584 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] .jinit() problem
Please let me know if you needany other information mister_bluesman wrote: > > Hello there. > > When I try to start the jvm using .jinit() after loading library(rJava) I > don't seem to be able to as I get the message: > > Error in .jinit() : Cannot create Java Virtual Machine > > What is going on here? I have java 1.6 installed on my XP machine > > Thanks again. > > -- View this message in context: http://www.nabble.com/.jinit%28%29-problem-tf3774265.html#a10679578 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Anderson-Darling GoF (re-sent)
Hi,
I'm not a statistician so sorry for possible trivial questions ...
I want to perform a GoF test on sample data against several distribution
(like Extreme Value, Phase Type, Pareto, ...).
Since I suspect a long-tailed behaviour on data I want to use
Anderson-Darling (AD) GoF test because it's well known it's more
sensible to tail data.
Looking at R packages the only AD test is the AD normality test
("ad.test") in the "nortest" package. So I think this function is not
for me since long-tailed samples aren't normally distribuited (right?!)
I've found the Marsaglia article ("Evaluating the Anderson Darling
distribution") where it seems I can consider the ECDF (empirical CDF)
and the theoretical as a uniformly [0,1] distributed data and then
perform the test like I had to compare two uniform distribution. The
problem is the theoretical CDF ( i.e. the parameters of theoretical
distribution) has been estimated from the data against which I want to
make the test. I've read somewhere it's not a good technique to compare
the distribution with the above way because the resulting AD test might
be biased.
So, finally, I don't know how to proceed ...
Can anyone give me a help or any reference (please remember I'm not a
statistician so do not write too technically)??
Thanks a lot to everyone!!
-- Marco
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[R] Fwd: Re: Goodness-of-fit test for gamma distribution?
Thanks Petr. Comments below: At 03:40 PM 18/05/2007, Petr Klasterecky wrote: >Sean Connolly napsal(a): >>Hi all, >>I am wondering if anyone has written (or knows of) a function that >>will conduct a goodness-of-fit test for a gamma distribution. I am >>especially interested in test statistics have some asymptotic >>parametric distribution that is independent of sample size or >>values of fitted parameters (e.g., a chi-squared distribution with some Petr's reply: >The GOF test will always depend on the parameter values, since it >has to estimate them (if you don't provide them yourself). Anyway, >the gamma family is so versatile that you can fit *some* gamma >distribution to almost any nonnegative continuous data. Sean's reply to Petr: An example of what I'm looking for would be the "K-squared" statistic that tests for normality (D'Agostino and Pearson 1973, Biometrika 60: 613, also in Zar, 1996, Biostatistical Analysis, p89). The expected distribution of the test statistic is approximately chi-squared with 2df, regardless of values of estimated parameters or sample size (provided sample size is sufficiently large). Petr's reply: >Maybe it is easier and sufficient to use the Kolmogorov - Smirnov >test, that is implemented as ks.test() in R. However, I am not able >to check your reference, so my comment may not be what you want at all. Sean's reply to Petr: My understanding is that the K-S test requires that parameters be specified (i.e., not estimated from data), and that the test statistic depends on sample size. Am I missing something? Thanks again. Sean >>Sean R. Connolly, PhD >>Associate Professor >>ARC Centre of Excellence for Coral Reef Studies, and >>School of Marine and Tropical Biology >>James Cook University >>Townsville, QLD 4811 >>AUSTRALIA >>__ >>R-help@stat.math.ethz.ch mailing list >>https://stat.ethz.ch/mailman/listinfo/r-help >>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>and provide commented, minimal, self-contained, reproducible code. > >-- >Petr Klasterecky >Dept. of Probability and Statistics >Charles University in Prague >Czech Republic __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Anderson-Darling GoF
Hi,
I'm not a statistician so sorry for possible trivial questions ...
I want to perform a GoF test on sample data against several distribution
(like Extreme Value, Phase Type, Pareto, ...).
Since I suspect a long-tailed behaviour on data I want to use
Anderson-Darling (AD) GoF test because it's well known it's more sensible to
tail data.
Looking at R packages the only AD test is the AD normality test ("ad.test")
in the "nortest" package. So I think this function is not for me since
long-tailed samples aren't normally distribuited (right?!)
I've found the Marsaglia article ("Evaluating the Anderson Darling
distribution") where it seems I can consider the ECDF (empirical CDF) and
the theoretical as a uniformly [0,1] distributed data and then perform the
test like I had to compare two uniform distribution. The problem is the
theoretical CDF (i.e. the parameters of theoretical distribution) has been
estimated from the data against which I want to make the test. I've read
somewhere it's not a good technique to compare the distribution with the
above way because the resulting AD test might be biased.
So, finally, I don't know how to proceed ...
Can anyone give me a help or any reference (please remember I'm not a
statistician so do not write too technically)??
Thanks a lot to everyone!!
-- Marco
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Re: [R] add info
you could use attributes, e.g., dat <- data.frame(x = 1:3, y = letters[1:3]) attr(dat, "name") <- "my data.frame" attr(dat, "author") <- "John Smith" attr(dat, "date") <- "2007-05-18" ## dat attributes(dat) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: "XinMeng" <[EMAIL PROTECTED]> To: Sent: Friday, May 18, 2007 9:05 AM Subject: [R] add info > hi all: > If there's a dataframe: > > x y > 1 a > 2 b > 3 c > > The info of the data such as : > > name > date > author > > The result I want is: > > name > date > author > x y > 1 a > 2 b > 3 c > > In other words,I wanna add the info above the dataframe. > > How can I do it ? > > Thanks a lot! > > > My best > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
