This is to announce the first release of the tikzDevice package on CRAN.
The package should be migrating to your mirror soon.
The tikzDevice package provides a new graphics device for R which enables
direct output of graphics in a LaTeX-friendly way. Please see the original
posting on R-Forge for
Hi all,
I have got 27 graphs to export (not a lot...I know!). How can I fit all of
them into a single file like PNG without adjusting the size of the graphs?
What's in my mind is like pasting graphs into Word, in which I can just
scroll down to view the graphs.
Thanks for your attention. Much ap
Still need assistance with this topic, so if anybody knows, please reply!
J_Laberga wrote:
>
> Hello,
>
> I'm trying to tackle a problem that would require the implementation of a
> recurrent NN. However, even though the CRAN is very big, I can’t seem to
> find a package for this. Does anybody
Hi All,
I have recently *re*-installed R-2.9.1 in my Linux machine. Since then,
I am unable to plot using the usual interactive device.
> plot(1:10)
This plots in a pdf file "Rplots.pdf" in my working directory.
> sessionInfo()
R version 2.9.1 (2009-06-26)
i686-pc-linux-gnu
locale:
LC_CTYPE=e
It still doesn't work. Now, R is asking me for the SSLEAY32.dll. If I download
it from the internet, and then put it in the file
\R-2.8.1\library\RPostgreSQL\libs, now they ask me for the VSINIT.dll file !
it's like an infinite loop... I've tried to intall again the RPostgreSQL
library but not
>
> # I tried defining a function like this
> myplot <- function(...)plot(..., pch=19, col=c("blue","red")[treatment])
>
> # So i can call it like this:
> with(mydfr, myplot(Xmeas, Ymeas))
>
> # but:
> Error in plot.xy(xy, type, ...) : object 'treatment' not found
>
basically that is something like
> I have got 27 graphs to export (not a lot...I know!). How can I fit all of
> them into a single file like PNG without adjusting the size of the graphs?
> What's in my mind is like pasting graphs into Word, in which I can just
> scroll down to view the graphs.
Pretty sure PNG can only cope with s
Hi All,
I am using "rpart" function to model my data:
library(rpart)
set.seed(1)
x1 = sample(c("a","b"),100,T)
x2 = runif(100)
y = ifelse((x1=="a" & x2<=0.5)|(x1=="b" & x2>0.5), 0, 1)
data = data.frame(x1,x2,y)
fit = rpart(y~x1+x2)
plot(fit); text(fit, use.n=TRUE)
Now I want to use variable "x1
Hi,
It's probably easiest with the pdf (or postscript) device,
pdf("all.pdf")
for(ii in 1:27)
plot(rnorm(10), main=paste("plot", ii))
dev.off()
Bitmap-based devices can generate sequential filenames (Rplot1.png,
Rplot2.png, ...) that you could combine in a single document using external
tools
Hi,
Using ggplot2, you could do something like this,
library(ggplot2)
myplot <- function(x, y, data, geom="point"){
ggplot(data=data, map=aes_string(x=x, y=y, colour = "treatment") ) +
layer(geom=geom) +
scale_colour_manual(values=c("red", "blue"))
}
d = data.frame(Xmeas=rnorm(10), Ymeas=rn
Thank you
- Original Message
From: spencerg
To: "Liaw, Andy"
Cc: Rolf Turner ; FMH ;
r-help@r-project.org
Sent: Wednesday, September 9, 2009 3:08:43 PM
Subject: Re: [R] Derivative of nonparametric curve
This may be overkill for your application, but you might be interested in
t
Thank you
From: Henrique Dallazuanna
Cc: r-help@r-project.org
Sent: Sunday, September 6, 2009 4:30:14 PM
Subject: Re: [R] Color index in image function
Try this to see which colors the codes are:
Color <- function(color){
z <- matrix(1:length(color), nco
Thank you
- Original Message
From: jim holtman
To: FMH
Cc: Schalk Heunis ; r-help@r-project.org
Sent: Tuesday, September 8, 2009 3:45:09 PM
Subject: Re: [R] Data in Array
Not sure what you mean by 'store', but you can use a list:
> a <- matrix(1, nrow = 5, ncol = 1)
> b <- matrix(2,
Hi list,
I've been producing autocorrelation functions of time series using the
acf function, and have found a series or two for which correlations of >
1 are given, which I think shouldn't happen.
Attached is the time series I'm using, and below is the R code (version
2.9.1) that I'm entering:
> col=c("blue","red")mydfr$[treatment]
Yes, but I would like to use the function for lots of other dataframes
as well, so embedding 'mydfr' in the function is not the ideal
solution...
remko
-
Remko Duursma
Post-Doctoral Fellow
Centre for Plants
You can wite them out as single files (png('file-%02d.png',...)) and
then import them to Word.
On Mon, Sep 14, 2009 at 1:55 AM, Chris Li wrote:
>
> Hi all,
>
> I have got 27 graphs to export (not a lot...I know!). How can I fit all of
> them into a single file like PNG without adjusting the size
On 14/09/2009 6:40 AM, Steve Jones wrote:
Hi list,
I've been producing autocorrelation functions of time series using the
acf function, and have found a series or two for which correlations of >
1 are given, which I think shouldn't happen.
Attached is the time series I'm using, and below is the
utkarshsinghal wrote:
Hi All,
I have recently *re*-installed R-2.9.1 in my Linux machine.
Hi,
Did you use the package manager of you linux distro, or did you compile
from source.
cheers,
Paul
Since then,
I am unable to plot using the usual interactive device.
> plot(1:10)
This plots in
Remko Duursma wrote:
col=c("blue","red")mydfr$[treatment]
Yes, but I would like to use the function for lots of other dataframes
as well, so embedding 'mydfr' in the function is not the ideal
solution...
The problem is that the info in 'treatment' is non-constant, and you
need to eithe
The example is reproducible! Did you see the first post?
remko
-
Remko Duursma
Post-Doctoral Fellow
Centre for Plants and the Environment
University of Western Sydney
Hawkesbury Campus
Richmond NSW 2753
Dept of Biological Science
Macquarie Unive
Apologies for the missing data. It can be downloaded from here (22Kb):
http://www.squaregoldfish.co.uk/sekrett/series.csv
Steve.
Duncan Murdoch wrote:
> On 14/09/2009 6:40 AM, Steve Jones wrote:
>> Hi list,
>>
>> I've been producing autocorrelation functions of time series using the
>> acf functi
>> col=c("blue","red")mydfr$[treatment]
>
> Yes, but I would like to use the function for lots of other dataframes
> as well, so embedding 'mydfr' in the function is not the ideal
> solution...
In that case I'd try something like:
myplot <- function(..., tmnt) {
plot(...,
pch=19,
On Sun, 2009-09-13 at 14:46 +0200, Erich Neuwirth wrote:
> months<-c("jan","feb","mar","apr","may","jun",
> "jul","aug","sep","oct","nov","dec")
>
> sortorder<-order(as.numeric(date2[,2]),match(date2[,1],months))
To save some key strokes, the above can be done via:
sortorder <- orde
I offer my sincere apologies for not reading the e-mail carefully, your
example is indeed reproducible. When you stop using the 'with' function,
this is I think what you would like:
myplot2 = function(formula, data, ...) {
plot(formula, data = data, ..., pch = 19, col =
c("blue","red")[data
G'day Steve,
On Mon, 14 Sep 2009 13:44:56 +0200
Steve Jones wrote:
> Apologies for the missing data. It can be downloaded from here (22Kb):
> http://www.squaregoldfish.co.uk/sekrett/series.csv
Well, the Details section of acf's help page states:
By default, no missing values are allowed.
Try this:
library(RDCOMClient)
w <- COMCreate('Word.Application')
w[["Visible"]] <- TRUE
w[["DisplayAlerts"]] <- FALSE
doc <- w$Documents()$Add(Template="Normal", NewTemplate=FALSE,
DocumentType=0)
fooPasteGraphs <- function(index, ...){
win.metafile()
plot(rnorm(100))
dev.off()
Hi
When I run a Chi squared test in R I am automatically given the chi squared
value and the degrees of freedom. How do I find these values when i've used
Fisher's exact test?
Cheers,
Katie
[[alternative HTML version deleted]]
__
R-help@r-p
Hello everybody,
I have been trying for some weeks to state the correct design of my
experiment as a GLM formula, and have not been able to find something
appropriate in Pinheiro & Bates, so I am posting it here and hope
somebody can help me.
In each experimental condition, described by
1) g
On 14/09/2009 7:44 AM, Steve Jones wrote:
Apologies for the missing data. It can be downloaded from here (22Kb):
http://www.squaregoldfish.co.uk/sekrett/series.csv
The dataset is 97% missing values. I suspect what you're seeing is what
the man page describes as follows:
"If the na.action fu
Dear List,
I tried installing the Rstem package from omegahat.org using my old
command
install.packages("Rstem", repos = "http://www.omegahat.org/R";, type =
"source")
but it seems omegahat.org is down. Any word on when it will be on
again (if ever)?
---
Sascha A. Wolfer, M.A.
Centre
On Sep 14, 2009, at 8:00 AM, Paul Hiemstra wrote:
I offer my sincere apologies for not reading the e-mail carefully,
your example is indeed reproducible. When you stop using the 'with'
function, this is I think what you would like:
myplot2 = function(formula, data, ...) {
plot(formula, d
On Sep 14, 2009, at 6:02 AM, Breach, Katherine wrote:
Hi
When I run a Chi squared test in R I am automatically given the chi
squared value and the degrees of freedom. How do I find these values
when i've used Fisher's exact test?
The function fisher.test uses hypergeometric distributio
inspired by useR! 2009, yeroon.net is a new project developing statistical
web applications to make popular R packages easier available. A first
version of the lme4 web interface is available at http://yeroon.net/lme4/
http://yeroon.net/lme4/ . It implements most lme4 features and some more.
A s
Wondering if anyone is aware of resources appropriate for home
schooled pre-teen/teen relatives?
Before anyone suggests, a lengthy google search was unsuccessful.
TIA, V.
--
Vince Fulco, CFA, CAIA
612.424.5477 (universal)
vful...@gmail.com
A posse ad esse non valet consequentia
“the possibi
R-help,
I 'm trying to optimize a model to data using log-likelihoods
but I encounter the following error message:
> l= c(49.4, 57.7,64.8,70.9,78.7,86.6,88.3,91.6,99,115)
> t=3:12
> fn <- function(params, l=l, t=t) {
Linf <- params[1]
k <- params[2]
Good day all,
I'm trying to plot a figure and ensure that the minimum and maximum
values are always displayed. However, the code below does not display
the minimum value, no matter what I try. Could someone please help?
Thanking you in advance,
George.
Code below for reproduction (apologies if
Hi,
is it possible to put a summary of an rpart-Object into a tktext-window?
Here is what I'm trying to do:
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
tt <- tktoplevel()
tex <- tktext(tt)
tkpack(tex)
tkinsert(tex, "end", summary(fit))
But since the summary of an object is a l
Readers,
I have been reading the r book (Crawley) and tried to use the
influence measures function for linear regression, as described. I
have one datum that I wish to show in the graph but exclude from the
regression and ab line.
x y
0 5
10 9
20 10
30 19
40 4
Wit
Try this:
tkinsert(tex, "end", paste(capture.output(summary(fit)), collapse = "\n"))
On Mon, Sep 14, 2009 at 10:43 AM, Anne Skoeries wrote:
> Hi,
>
> is it possible to put a summary of an rpart-Object into a tktext-window?
>
> Here is what I'm trying to do:
>
> fit <- rpart(Kyphosis ~ Age + Numb
I misunderstood what the help page was saying - I thought that the
missing values were what constituted the invalid function. Clearly I was
mistaken.
Looks like I'll have to compute the acf myself. Shame, but such is life.
Thanks for the input, everyone!
Steve.
Berwin A Turlach wrote:
> G'day S
Thanks! I just sent him an email. If anyone noticed any behavior change with
this function version 3.3.1 please report any finding.
Thanks in advance,
Max
Uwe Ligges-3 wrote:
>
>
>
> madmax1425 wrote:
>> Hi there,
>> I started getting a new error with the latest mclust package version
>> 3.
Hi,
On Sep 14, 2009, at 9:47 AM, e-letter wrote:
Readers,
I have been reading the r book (Crawley) and tried to use the
influence measures function for linear regression, as described. I
have one datum that I wish to show in the graph but exclude from the
regression and ab line.
x y
0
Estaré ausente de la oficina desde el 14/09/2009 y no volveré hasta el
24/09/2009.
Responderé a su mensaje cuando regrese.
[[alternative HTML version deleted]]
__
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Thanks! That's it! Perfect!
--
Anne Skoeries
Olgastr. 54
74072 Heilbronn
Phone: +49 (0)7131 - 390 33 33
Mobil: +49 (0)176 - 212 37 770
Mail: h...@anne-skoeries.de
Am 14.09.2009 um 15:50 schrieb Henrique Dallazuanna:
tkinsert(tex, "end", paste(capture.output(summary(fit)), collapse =
"\n"))
Hi, ALL,
Is there any R/bioconductor package to do SSVS (stochastic search variable
selection) for glm?
Thanks,
Cindy
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PLEAS
Hi,
On Sep 14, 2009, at 12:03 PM, ivo welch wrote:
Sorry, one more: on OSX, I deleted my old 2.9.2 R.app, and installed
the 64
bit version of 2.9.0. I then did an "install.packages("car")" under
my new
2.9.0. It seems to have worked, but alas, I still get an error that
package
'car' was
Hello *
I would like to reshape wide the following dataset:
> rl <- read.dta("intermedi/rapporti_lavoro.dta")
> [c("id_rl","prog","sil_pi","sil_cf","sil_dat_avv")]
> dim(rl)
[1] 12964 5
> object.size(rl)
1194728 bytes
> head(rl)
id_rl prog sil_pi sil_cf sil_dat_avv
1 638
http://jumpmath.org/program
Looks good but no personal experience of it.
--- On Mon, 9/14/09, Vince Fulco wrote:
> From: Vince Fulco
> Subject: [R] Teaching material for children...
> To: r-help@r-project.org
> Received: Monday, September 14, 2009, 9:17 AM
> Wondering if anyone is aware of
> r
I'm sure this is easy, but I'm having a hard time figuring out how to
recode some data in R.
I have a variable "numpeers" which is valued 1, 2, or 3. I also have
three other variables called "w9zd9_1," "w9zd9_2," and "w9zd9_3." I
want to use these variables to create a new item called "distot."
S
Dear All,
I have a problem which *should* be pretty straightforward to resolve - but I
can't work out how!
I have a list of 3 coefficient estimates for 4 different datasets:
Coefs<-list(c(1,0.6,0.5),c(0.98,0.65,0.4),c(1.05,0.55,0.45),c(0.99,0.50,0.47))
All I want to do is take the sum (or mean
hey,
On Mon, Sep 14, 2009 at 07:51:42AM -0700, John Kane wrote:
> p <- ggplot(bmm, aes(x="age", y="bm", colour="pp", group="pp"))
> p <- p + geom_line()
remove the quotes and it will work:
ggplot(bmm, aes(x=age, y=bm, colour=pp, group=pp))+geom_line()
regards,
Stefan
___
> "s" == smu
> on Mon, 14 Sep 2009 17:20:27 +0200 writes:
s> hey, I can not find a function for the following problem,
s> hopefully you can help me.
s> I have a vactor like this one
s> v = c(NA,NA,TRUE,TRUE,NA,TRUE,NA,TRUE,TRUE,TRUE)
s> and I would like to the
dear R wizards: I am looking for a binary package distribution of R 2.9.2
for OSX . Looking at http://r.research.att.com/ , there seems to be only a
binary for 2.9.0 . is the 2.9.2 version binary package available
somewhere? (at this point, would it make sense to elevate the 64-bit
version to
wesley mathew wrote:
Hello All
I have some problem for installing XML_2.6-0.tar . I am working in widows
and R version is R-2.9.1
>*install.packages("XML")*
After selecting a CRAN mirror
**
*Error :-* Warning: unable to access index for repository
http://cran.pt.r-project.org/bin/windows/
Try the tail() function
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Peng Yu
Sent: Monday, September 14, 2009 11:32 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] How to refer to the last a few rows?
Hi,
x=matrix(1:60,nr=6)
I can
Sorry, one more: on OSX, I deleted my old 2.9.2 R.app, and installed the 64
bit version of 2.9.0. I then did an "install.packages("car")" under my new
2.9.0. It seems to have worked, but alas, I still get an error that package
'car' was built under R version 2.9.2 . Where exactly does R under OS
Hi,
I'm running into an error message for the "anova"-function I never got
before with the Design (Version 2.2-0) package.
There seems to be a missing function "Varcov", please check my
function calls (it's in german but I think you get the error):
> library(Design)##
This is an error in the most recent Hmisc update. The authors promised
to deliver another update that fixes this issue.
Uwe Ligges
Sascha Wolfer wrote:
Hi,
I'm running into an error message for the "anova"-function I never got
before with the Design (Version 2.2-0) package.
There seems t
Hello
On Mon, Sep 14, 2009 at 2:53 PM, Steve Jones wrote:
> Looks like I'll have to compute the acf myself. Shame, but such is life.
>
Wouldn't na.omit be an option?
Liviu
__
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https://stat.ethz.ch/mailman/listinfo/r-hel
Hi,
try this,
rowMeans(as.data.frame(Coefs))
# or apply(as.data.frame(Coefs), 1, mean)
HTH,
baptiste
2009/9/14 Masca, N.
> Dear All,
>
> I have a problem which *should* be pretty straightforward to resolve - but
> I can't work out how!
>
> I have a list of 3 coefficient estimates for 4 diffe
Hi,
On Sep 14, 2009, at 11:48 AM, Martin Batholdy wrote:
Hi,
is there a way to make a call for an input at some point of a
process ..?
I don't know how to describe it well ...
like;
please enter your first name:
>
and then, what is typed in should be saved into a variable.
See: ?re
Hi,
tail(x,2)
or
x[seq(nrow(x)-1, nrow(x)), ]
HTH,
baptiste
2009/9/14 Peng Yu
> Hi,
>
> x=matrix(1:60,nr=6)
>
> I can refer the last 2 rows by
> x[5:6,]
>
> If I don't know the total number of rows is 6, is there a way to refer
> the last 2 rows?
>
> Regards,
> Peng
>
> _
Hello
On Mon, Sep 14, 2009 at 2:17 PM, Vince Fulco wrote:
> Wondering if anyone is aware of resources appropriate for home
> schooled pre-teen/teen relatives?
>
There is http://r-tutor.com/, but it is for college students/beginners
in stats. It would make sense to repost the question on
r-sig-tea
Hi,
is there a way to make a call for an input at some point of a
process ..?
I don't know how to describe it well ...
like;
please enter your first name:
>
and then, what is typed in should be saved into a variable.
__
R-help@r-project.org m
Please read r-help messages from 2 days ago that contain a simple
work-around.
Frank
Sascha Wolfer wrote:
Hi,
I'm running into an error message for the "anova"-function I never got
before with the Design (Version 2.2-0) package.
There seems to be a missing function "Varcov", please check m
Try this:
v <- c(NA,NA,TRUE,TRUE,NA,TRUE,NA,TRUE,TRUE,TRUE)
ick <-rle(v)
foo <- unlist(apply(matrix(ick$lengths),1,seq))
foo[is.na(v)] <- NA
foo
[1] NA NA 1 2 NA 1 NA 1 2 3
-Don
At 5:20 PM +0200 9/14/09, smu wrote:
hey,
I can not find a function for the following problem, hopefu
Thanks to both of you.
I knew it was something stupid. I just didn't realise it was that stupid.
--- On Mon, 9/14/09, baptiste auguie wrote:
> From: baptiste auguie
> Subject: Re: [R] ggplot2 graphing multiple lines of data
> To: "R R-help"
> Received: Monday, September 14, 2009, 11:34 AM
Hi, I have seen the answer to this sometime before but I just can't find it
again - pointers appreciated.
I have a package that contains some data.frames saved as .Rda files in the
data/ directory. When the package is loaded I would like to have them be
available in the workspace (without the user
The suggestions from others to use lattice's xyplot, or ggplot2 are good.
If you want an explicit loop you can do something like:
for ( nm in unique(mydat$Name) ) {
with( subset( mydf, Name==nm) ,
{
plot(Time, Value, title=nm)
readline('CR to continue ')
}
)
}
At 5:19
wesley mathew wrote:
Hello All
I have some problem for installing XML_2.6-0.tar . I am working in widows
and R version is R-2.9.1
>*install.packages("XML")*
After selecting a CRAN mirror
**
*Error :-* Warning: unable to access index for repository
http://cran.pt.r-project.org/bin/windows/co
Try this also:
with(rle(v), unlist(sapply(lengths, FUN = seq)) * v)
On Mon, Sep 14, 2009 at 12:20 PM, smu wrote:
> hey,
>
> I can not find a function for the following problem, hopefully you can
> help me.
>
> I have a vactor like this one
>
> v = c(NA,NA,TRUE,TRUE,NA,TRUE,NA,TRUE,TRUE,TRUE)
>
Hi Holger,
On Sep 14, 2009, at 10:57 AM, Hollix wrote:
Hi folks,
I created a subset of a dataframe (i.e., selected only men):
subdata <- subset(data,data$gender==1)
After a residual diagnostic of a regression analysis, I detected three
outliers:
linmod <- lm(y ~ x, data=subdata)
plot(linmo
Hello,
I have a very unusual situation with an SVM and wanted to get the
group's opinion.
We developed an experiment where we train the SVM with one set of data
(train data) and then test with a completely independent set of data
(test data). The results were VERY good.
I found and error
On Sep 14, 2009, at 11:59 AM, Steve Lianoglou wrote:
Hi,
On Sep 14, 2009, at 11:44 AM, ivo welch wrote:
dear R wizards: I am looking for a binary package distribution of
R 2.9.2
for OSX . Looking at http://r.research.att.com/ , there seems to
be only a
binary for 2.9.0 . is the 2.9.2
wesley mathew wrote:
>
> Hello All
>
> I have some problem for installing XML_2.6-0.tar . I am working in widows
> and R version is R-2.9.1
>
>
Unfortunately, I think there are some problems with CRAN being able to build
the XML package for Windows, at least the page:
http://cran.r-proje
Hi,
x=matrix(1:60,nr=6)
I can refer the last 2 rows by
x[5:6,]
If I don't know the total number of rows is 6, is there a way to refer
the last 2 rows?
Regards,
Peng
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PLE
Try this:
Assuming your vector is x:
ylimits<-c(min(x)- .05,max(x)+ .05)
plot(x,type="l",yaxt="n",ann=FALSE,ylim=ylimits)
By the way, if you are going to supply much data, it is better practice to use
dput() . Try dput(x) to see what I mean.
--- On Mon, 9/14/09, Jorgy
try the following:
v <- c(NA,NA,TRUE,TRUE,NA,TRUE,NA,TRUE,TRUE,TRUE)
l <- as.numeric(!is.na(v))
tmp <- cumsum(l)
tmp - cummax((!l) * tmp)
I hope it helps.
Best,
Dimitris
smu wrote:
hey,
I can not find a function for the following problem, hopefully you can
help me.
I have a vactor like t
linmod2 <- update(linmod, data = subdata[-c(11,22,33),])
Hollix wrote:
Hi folks,
I created a subset of a dataframe (i.e., selected only men):
subdata <- subset(data,data$gender==1)
After a residual diagnostic of a regression analysis, I detected three
outliers:
linmod <- lm(y ~ x, data=subda
At a quick glance, your code seems to be deleting columns not rows
try y[-c(11,22,33), ]
--- On Mon, 9/14/09, Hollix wrote:
> From: Hollix
> Subject: [R] Eliminate cases in a subset of a dataframe
> To: r-help@r-project.org
> Received: Monday, September 14, 2009, 10:57 AM
>
> Hi folks,
>
Masca, N. wrote:
>
> Dear All,
>
> I have a problem which *should* be pretty straightforward to resolve - but
> I can't work out how!
>
> I have a list of 3 coefficient estimates for 4 different datasets:
>
> Coefs<-list(c(1,0.6,0.5),c(0.98,0.65,0.4),c(1.05,0.55,0.45),c(0.99,0.50,0.47))
>
>
Hi folks,
I created a subset of a dataframe (i.e., selected only men):
subdata <- subset(data,data$gender==1)
After a residual diagnostic of a regression analysis, I detected three
outliers:
linmod <- lm(y ~ x, data=subdata)
plot(linmod)
Say, the cases 11,22, and 33 were outliers.
Here comes
Hello All
I have some problem for installing XML_2.6-0.tar . I am working in widows
and R version is R-2.9.1
>*install.packages("XML")*
After selecting a CRAN mirror
**
*Error :-* Warning: unable to access index for repository
http://cran.pt.r-project.org/bin/windows/contrib/2.9
Warning: unab
Hi,
On Sep 14, 2009, at 11:44 AM, ivo welch wrote:
dear R wizards: I am looking for a binary package distribution of R
2.9.2
for OSX . Looking at http://r.research.att.com/ , there seems to be
only a
binary for 2.9.0 . is the 2.9.2 version binary package available
somewhere? (at this p
Noah,
It may be just me - but how does "any" of your questions on prediction
modeling relate to R?
It seems to me that you have been getting a lot of "free" consulting from
this forum that is supposed to be a forum for help on R-related issues.
Ravi.
---
hey,
I can not find a function for the following problem, hopefully you can
help me.
I have a vactor like this one
v = c(NA,NA,TRUE,TRUE,NA,TRUE,NA,TRUE,TRUE,TRUE)
and I would like to the TRUE values by the their "local sequence
number".
This means, the result should look thike this:
c(NA,NA,
alternatively, use aes_string,
p <- ggplot(bmm, aes_string(x="age", y="bm", colour="pp", group="pp"))
p <- p + geom_line()
p
HTH,
baptiste
2009/9/14 smu
> hey,
>
> On Mon, Sep 14, 2009 at 07:51:42AM -0700, John Kane wrote:
> > p <- ggplot(bmm, aes(x="age", y="bm", colour="pp", group="pp"))
Of course if w9zd9_1, w9zd9_2, w9zd9_3 were elements of a data.frame (or
even a list), you could use indexing,
w9zd9 = data.frame(w9zd9_1 = 1:10, w9zd9_2 = 1:10, w9zd9_3 = -2*1:10)
average = function(numbers=1, w9zd9){
if(numbers == 1L) return(w9zd9[ ,1]) # vector case fails with rowMeans
rowM
I want to comput prime component of a graph.
please guide mi.
thanks alot
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-proje
Dear R colleagues-- Apologies for belabouring, but if it wasn't
obvious already, I was speaking specifically to R-Project resources
for young adults. As I was querying the R-help list, thought that was
crystal clear but I've received an offer or two of assistance of a
more generic nature.
Best, V
Hello fellow R's,
I´ve been learning to use the ggplot2 library, and after a full day of
work I still have a couple of basic questions.
Here is an example:
mydata=data.frame(x=runif(20),y=runif(20),n=runif(20))
mydata2=data.frame(x=c(0.4,0.6,0.5),y=c(0.4,0.4,0.6))
ggplot(mydata, aes(x, y)) +
I suppose there are a couple of ways...
if (numbers==1) {
distot <-
} else if (numbers ==2) {
distot <-
} else if (numbers==3) {
distot <-
} else distot <- NA
(letting you fill in the right hand side within each block)
There's also
distot <- switch(numbers,
w9zd9_1,
(w9zd9_1 + w9zd
On 2009-September-11 , at 13:55 , wrote:
Maybe:
do.call(rbind, lapply(with(xy <- rbind(x, y), split(xy, list(a, b),
drop = TRUE)), tail, 1))
On Fri, Sep 11, 2009 at 3:45 AM, jo wrote:
Thanks for the post-processing ideas. But is there any way to do that
in one step?
Thanks but by "in o
This is quite a nasty one! I've been having a look at it, and I think
it would be useful to go back to first principles.
Given a series u[1:N], the correlation coefficient between u[1:(N-k)]
and u[(k+1):N] is
c(k) = cov(u[1:(N-k)],u[(k+1):N])/sqrt(var(u[1:(N-k)])*var(u(k+1):N]))
in which, in par
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Henrique
> Dallazuanna
> Sent: Monday, September 14, 2009 9:59 AM
> To: smu
> Cc: r-help@r-project.org
> Subject: Re: [R] local sequence function
>
> Try this also:
>
> with(rl
Some day I may figure out how ggplot2 works.
I am trying to plot 5 columns of data on a graph (similar to a simple matplot)
===
library(ggplot2)
bmi <- structure(list(pct = 2:21, P10 = c(14.6, 14.5, 14.2, 13.9, 13.7,
13.7,
Hi,
On Sep 14, 2009, at 10:19 AM, asie soheili wrote:
I want to comput prime component of a graph.
please guide mi.
If you're working with graphs, you'll want to check out the igraph
package:
http://cran.r-project.org/web/packages/igraph/
http://igraph.sourceforge.net/
-steve
--
Steve L
> "iw" == ivo welch
> on Mon, 14 Sep 2009 12:03:59 -0400 writes:
iw> Sorry, one more: on OSX, I deleted my old 2.9.2 R.app, and installed
the 64
iw> bit version of 2.9.0. I then did an "install.packages("car")" under my
new
iw> 2.9.0. It seems to have worked, but alas,
Results can be slightly different when matrix algebra routines are called.
Here's your example again. When the prediction is computed directly using
matrix multiplication, the result is the same as 'predict' produces (at least
in this case.)
set.seed(1)
n <- 100
x <- rnorm(n)
y <- rnorm(n)
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Luis Ridao Cruz
> Sent: Monday, September 14, 2009 6:24 AM
> To: R-help
> Subject: [R] Error: C stack usage is too close to the limit
>
> R-help,
>
> I 'm trying to optimize a m
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