dear list
I am student M.S. statistics in department statistics . I am working in the
function "nls" in the [R 2.3.1] with 246 data and want to fit the "exp" model
to vectors( v and u ) but I have
a problem to use it
u
5.00e-13 2.179057e+03 6.537171e+03 1.089529e+04 1.525340e+04
1.9
If I understand your question, then you should just study some
examples/other people's code that initialize new objects as the return
values of functions. There are two small examples below. On the other
hand, if you really need a _global_ object, see help(assign).
## Example 1: last line is ret
Hello
If you want the multivariate t-distribution,
use rmvt() of the mvtnorm package.
If you want the Wishart distribution, one
can write a little nonce function:
library(mvtnorm)
rwis <- function(n,sigma){
crossprod(rmvnorm(n,sigma=sigma))
}
I'm not sure what you mean by the multivariate gam
Dear all,
As a response to the comment of Mr. Romain Francois,
I wrote a small set of REST web APIs for R Graphical Manuals.
It has become possible to get image URLs for a given package/function
name via the web API.
http://cged.genes.nig.ac.jp/RGM2/webAPI.php
Any comment would be gratefully app
Hi, in order to define multivariate distribution functions with known
marginals, you can also use the copula package, e.g.,
library(copula)
x <- mvdc(normalCopula(0.75), c("gamma", "gamma"),
list(list(shape = 2, rate = 3), list(shape = 3, rate = 2)))
x.samp <- rmvdc(x, 100)
dmvdc(x, x.samp)
You need better starting values:
f <- function(x) { c0 <- x[1]; ce <- x[2]; ae <- x[3]
sum((v - c0+(ce*(1-exp((-u)/ae^2)
}
g <- 4^(0:8)
g <- c(-g, g)
g <- expand.grid(c0 = g, ce = g, ae = g)
start <- g[which.min(apply(g, 1, f)), ]
nls(v ~ c0+(ce*(1-exp((-u)/ae))), data = data1, start = star
Dear all,
I have a problem with R CMD CHECK (R 2.6.0 on Windows XP)
Even if I use R CMD CHECK on the windlgs library I get:
* using log directory 'C:/PROGRA~1/R/RProgram/src/library/windlgs.Rcheck'
* using ARGUMENT '
' __ignored__ R version 2.6.0 (2007-10-03)
* checking for file 'windl
Hi
System: Linux Open SuSE 10.2
I upgraded to R 2.6.0 and wanted to upgrade my packages as well. I tried
to use the checkBuild=TRUE option in update.packages, it asked me if I
wanted to update the package and I answered with yes, and finally ig
tave me the error message unknown parameter "chec
Hi,
May I ask how I can save the coefficients and the p values into a table?
thanks.
jiong
The email message (and any attachments) is for the sole...{{dropped:11}}
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
Dear Olivier,
I had the same problem the other day.
It is important that you set the PATH in the right order.
I use the Rtools set from http://www.murdoch-sutherland.com/Rtools/ and created
a little batch file to set the PATH in the right order, please see below:
8<-
azadeh sadeghian yahoo.com> writes:
>
> I am student M.S. statistics in department statistics . I am working in the
function "nls" in the [R 2.3.1]
You should update to a more recent version of R (2.6.0 is current)
> with 246 data and want to fit the "exp" model to vectors( v and u ) but I
Great idea, thanks!
小笠原理 wrote:
>
> Dear all,
>
> As a response to the comment of Mr. Romain Francois,
> I wrote a small set of REST web APIs for R Graphical Manuals.
> It has become possible to get image URLs for a given package/function
> name via the web API.
>
> http://cged.genes.nig.ac
Van Campenhout Bjorn wrote:
>> I upgraded to R 2.6.0 and wanted to upgrade my packages as
>> well. I tried to use the checkBuild=TRUE option in
>> update.packages, it asked me if I wanted to update the
>> package and I answered with yes, and finally ig tave me the
>> error message unknown param
Ollivier TARAMASCO wrote:
> Dear all,
>
>
>
> I have a problem with R CMD CHECK (R 2.6.0 on Windows XP)
>
> Even if I use R CMD CHECK on the windlgs library I get:
So how are you calling it precisely?
Have you installed the recent tools for building R-2.6.0 and *packages*?
Uwe Ligges
>
This question teeters on the brink of general statistics rather than an
R-specific issue, but I'm hoping it boils down to how I coded V&R's EM
algorithm for finite mixture modeling and not something worse (i.e. that
my plan is impossible).
Briefly, I'm trying to simulate an experiment where th
Dear all,
I would like to be able to subset a data.frame in a special way. I will put
here an example:
Score Name
88 19_0070
88 19_0070
87 19_0070
79 002127_0658
79 002127_0658
77 002127_0658
So, for the abo
Thanks, Steve. I have read that document a bit, seems I need read it more.
I have 6 variables, all factors. 3^2 and 4^4
the following is the code
require(AlgDesign)
set.seed(1)
levels = c(v1=2,v2=2, v3=4,v4=4,v5=4,v6=4)
dat<-gen.factorial(levels,center=FALSE,varNames=names(level
Hi,
I am trying to get a title on a plot that contains both some formatting
and prints the value of an object. What I have been using to get the
italics is:
title(sub=expression(paste("Log-rank test ",italic("p"),"-value = ",p.val)))
But this prints "p.val" rather than the object value. I have
Hi all,
Suppose I have the following data.frame, with an id column and two
variables columns :
idX Y
0001 NA 21
0002 NA 13
0003 000145
0004 NA 71
0005 000320
What I would like to do is to create a new variable Z whose values are
the Y
Alberto Monteiro wrote:
> Prof Brian Ripley wrote:
>
>>Read ?par and the descriptiuon in 'An Introduction to R'. "din",
>> "fin", "mai", "omi", "pin" and "usr" are relevant.
>>
>
> Is there any hope that, instead of fin, din, pin, etc someday
> we will have fmm, dmm, pmm?
>
Why worry about that
?model.matrix
On Tue, 9 Oct 2007, Ajay Shah wrote:
> I am using the clever formula notation of R to first do an OLS. E.g. I
> say
>
> m <- lm(y ~ x + f)
>
> where f is a factor, and R automatically constructs the dummy
> variables. Very nice.
>
> I need to then go on to do some other ML estimati
On Wed, 10 Oct 2007, Rainer M Krug wrote:
> Hi
>
> System: Linux Open SuSE 10.2
>
> I upgraded to R 2.6.0 and wanted to upgrade my packages as well. I tried
> to use the checkBuild=TRUE option in update.packages, it asked me if I
> wanted to update the package and I answered with yes, and finally
Rob,
It might be worth upgrading at least mgcv and MASS to the current versions
(latest mgcv is 1.3-28, just gone to CRAN). Way back I vaguely remember that
there was an issue with glmmPQL (called by gamm) not picking up correlation
structure variables from the dataframe, but I can't now find
On Tue, 2007-10-09 at 20:04 +0530, Ajay Shah wrote:
> I am using the clever formula notation of R to first do an OLS. E.g. I
> say
>
> m <- lm(y ~ x + f)
>
> where f is a factor, and R automatically constructs the dummy
> variables. Very nice.
>
> I need to then go on to do some other ML estim
I am using the clever formula notation of R to first do an OLS. E.g. I
say
m <- lm(y ~ x + f)
where f is a factor, and R automatically constructs the dummy
variables. Very nice.
I need to then go on to do some other ML estimation using the same
design matrix that's used for the OLS. I could, o
Hi,
I was looking for a way to make Sweave output not only EPS and PDF,
but also PNG. Searching through the R mailing list archives, I found
the following post from Thibaut Jombard:
https://stat.ethz.ch/pipermail/r-help/2006-March/102122.html
and
http://tolstoy.newcastle.edu.au/R/devel/06/0
I have 400 cycles of a noisy waveform. I would like to find the average
(over the 400 cycles) of one cycle. Please tell me to do it in R. Any
suggestions welcome!
Thanks very much for any help.
Bill
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https://stat.ethz
Hi
[EMAIL PROTECTED] napsal dne 10.10.2007 12:10:29:
> Hi all,
>
> Suppose I have the following data.frame, with an id column and two
> variables columns :
>
> idX Y
> 0001 NA 21
> 0002 NA 13
> 0003 000145
> 0004 NA 71
> 0005 000320
Hello,
I want to use the quantile function so I read the doc but I don't understand
with this
> qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
[1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27
63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14 6
We do recommend
Rcmd build mpkg
to build a source package, followed by
Rcmd INSTALL --build mypkg_ver.tar.gz
to install it and zip up a version for distribution.
It is 'check' and 'build', and I believe R should refuse to use CHECK or
BUILD (as in does on all other platforms and at one time
On 10/9/07, Christos Hatzis <[EMAIL PROTECTED]> wrote:
> There are at least a couple of versions of such plots.
> Search for forest plots:
>
> help.search("forest plot")
>
Hi Christos
Could you be more specific? On my system I found nothing:
> help.search("forest plot")
No help files found match
>
> I'm trying to plot two distance matrices against each other
> (74x74,phylogenetic distance and phenotypic distances). However R
> gives an error message:"Error in plot.new() : figure margins too
> large".
>
> Is it because I have too many points to plot?
No. It *sounds* like you don't have
elyakhlifi mustapha wrote:
>
> Hello,
> I want to use the quantile function so I read the doc but I don't
> understand with this
>
>> qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
> [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39
> 63160.27 63205.65 63250.33 63295.04 63
elyakhlifi mustapha wrote:
> Hello,
> I want to use the quantile function so I read the doc but I don't understand
> with this
>
>
>> qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
>>
> [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27
> 63205.65 63250.33 632
Bill Simpson wrote:
> I have 400 cycles of a noisy waveform. I would like to find the average
> (over the 400 cycles) of one cycle. Please tell me to do it in R. Any
> suggestions welcome!
>
Hi Bill,
I recently had to do something like this for an "approach to target"
curve. The steps were:
1) C
Elyakhlifi Mustapha wrote:
>
> I want to use the quantile function so I read the doc but I don't
> understand with this
>
> > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
> [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39
> 63160.27 63205.65 63250.33 63295.04 63340.48 63
On 10/10/2007 6:41 AM, Prof Brian Ripley wrote:
> We do recommend
>
> Rcmd build mpkg
>
> to build a source package, followed by
>
> Rcmd INSTALL --build mypkg_ver.tar.gz
>
> to install it and zip up a version for distribution.
>
> It is 'check' and 'build', and I believe R should refuse to us
Oops... Check...
http://en.wikipedia.org/wiki/Chi-square_distribution
The Chi-square distribution for _any_ df has
mean = df and variance = 2df, so for large df, it's
approximately a normal distribution with mean = df and
sd = sqrt(2 df) [and not sqrt(df) as I wrote in a previous
message that, pr
Try this:
transform(d, z = y[match(x, id)])
On 10/10/07, Julien Barnier <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> Suppose I have the following data.frame, with an id column and two
> variables columns :
>
> idX Y
> 0001 NA 21
> 0002 NA 13
> 0003 000145
>
Thanks Jim for the help.
I forgot to say that one of my main snags is figuring how how to chop up
the one long (44100 samples) vector into pieces (400 of them).
This is my C-inspired way of thinking but there must be a better way in
R.
cyclelength=44100/400
for i= 1 to 400
for j= 1 to cycleleng
It really depends on what you want to do with the values. If you are
computing these in a loop, then I would suggest that you store the
output of 'summary' in a list and you can then extract the values
later. Can you provide an idea of what you want to do with them and
how you are computing them.
On 10-Oct-07 10:59:43, elyakhlifi mustapha wrote:
> Hello,
> I want to use the quantile function so I read the doc but I don't
> understand with this
>
>> qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
> [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39
> 63160.27 63205.65 632
Hi,
mods <- lapply(lapply(df[,which(sapply(df, is.factor))],
function(reg)lm(df$value~reg)), summary)
res <- lapply(mods, "[", c(4,10))
And for each model adjusted:
1-pf(res[[2]][[2]][1], res[[2]][[2]][2], res[[2]][[2]][3])
1-pf(res[[1]][[2]][1], res[[1]][[2]][2], res[[1]][[2]][3])
On 09/10/2
Henrique Dallazuanna wrote:
> Hi,
>
> mods <- lapply(lapply(df[,which(sapply(df, is.factor))],
> function(reg)lm(df$value~reg)), summary)
> res <- lapply(mods, "[", c(4,10))
>
> And for each model adjusted:
> 1-pf(res[[2]][[2]][1], res[[2]][[2]][2], res[[2]][[2]][3])
> 1-pf(res[[1]][[2]][1], res[
Hi Petr,
> d$z<-NA
> d$z[d$x %in% d$id] <- d$y[d$id %in% d$x]
>
> works in this particular case but it means you do not have multiple same
> ids and X
Thanks for the idea. But the problem is that I can have multiple
ids...
In fact in the meantime I found a solution by using row names :
R> d
try:
y <- matrix(x, ncol=400, byrow=TRUE)
On 10/10/07, Bill Simpson <[EMAIL PROTECTED]> wrote:
> Thanks Jim for the help.
>
> I forgot to say that one of my main snags is figuring how how to chop up
> the one long (44100 samples) vector into pieces (400 of them).
>
> This is my C-inspired way of
Dear R users,
I'm using the "reldist" add-on package to calculate relative distribution in R
as part of my research project. The subject is a general mental health score
ranging from 0 to 12 (integer values only) with 0 indicating no mental health
problem and positive values meaning some or sev
Is this what you want?
> x <- read.table(textConnection("Score Name
+ 88 19_0070
+ 88 19_0070
+ 87 19_0070
+ 79 002127_0658
+ 79 002127_0658
+ 77 002127_0658"), header=TRUE)
> # return best scores
> best <- by(x, x$Name, f
On 10/10/07, Jim Lemon <[EMAIL PROTECTED]> wrote:
> Alberto Monteiro wrote:
> > Prof Brian Ripley wrote:
> >
> >>Read ?par and the descriptiuon in 'An Introduction to R'. "din",
> >> "fin", "mai", "omi", "pin" and "usr" are relevant.
> >>
> >
> > Is there any hope that, instead of fin, din, pin, e
Try bquote as in:
http://tolstoy.newcastle.edu.au/R/e2/help/07/09/26353.html
On 10/10/07, Daniel Brewer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am trying to get a title on a plot that contains both some formatting
> and prints the value of an object. What I have been using to get the
> italics is
[EMAIL PROTECTED] napsal dne 10.10.2007 13:55:32:
> Hi Petr,
>
> > d$z<-NA
> > d$z[d$x %in% d$id] <- d$y[d$id %in% d$x]
> >
> > works in this particular case but it means you do not have multiple
same
> > ids and X
>
> Thanks for the idea. But the problem is that I can have multiple
> ids...
>
Hadley Wickham wrote:
>
>>> Is there any hope that, instead of fin, din, pin, etc someday
>>> we will have fmm, dmm, pmm?
>>
>> Why worry about that, Alberto, when you can use my ammazing function:
>>
>> mm2in(x) return(x/25.4)
>>
>> thus:
>>
>> par(pin=mm2in(126))
>
> But be sure to use it consis
Hi all,
I seem to be having a difficult time using the 'ape' package in R
when it comes to rooting trees. Here's a short screenshot:
> nirK.tree
Phylogenetic tree with 23 tips and 21 internal nodes.
Tip labels:
Burkholder, Burkholde3, Burkholde1, Burkholde4, Burkholde5,
Ralstonia2
Thanks. That works great if I do this:
title(sub=bquote(italic(p)-value == .(p.val)))
But if I add text to the beginning e.g.
title(sub=bquote(Log rank test italic(p)-value == .(p.val)))
I get an error message saying,
Error: syntax error, unexpected SYMBOL, expecting ',' in
"title(sub=bquote(Log
Petr PIKAL/CTCAP napsal dne 10.10.2007 15:03:28:
> [EMAIL PROTECTED] napsal dne 10.10.2007 13:55:32:
>
> > Hi Petr,
> >
> > > d$z<-NA
> > > d$z[d$x %in% d$id] <- d$y[d$id %in% d$x]
> > >
> > > works in this particular case but it means you do not have multiple
same
> > > ids and X
> >
> > Tha
Hello,
is there a possibility to set default options to qplot?
I need to draw a lot of graphs and would like to have all of them as point
plot but with a greater size and a fixed color for all dots.
Thanks for help.
Christoph
__
R-help@r-project.org
Hi All,
I entered a R statement, e.g. 1:20 = x or log("a") on an HTML
form and passed it to a R-CGI script. Obviously, neither of both
is a correct R statement or expression. However, my R-CGI script
could not return and report the error message to the Web site even
though it received the st
Hi,
I have some problems when using AlgDesign->optFederov() generating
designs.
I have 6 variables, all factors. 3^2 and 4^4, I want to have a design that
can take care of main effects and two interactions within 2 pair of
variables v3-v4 and v5-v6, the following is the code
##
Armin,
RSiteSearch("forest plot meta analysis")
will point you to the metaplot function in the rmeta package.
There is also a forestplot function in the same package that offers more
flexibility.
These plots are useful in meta analysis in general (e.g. showing odds ratios
from different studies).
The expression in plotmath must be valid R syntax.
plot(1)
p.val <- .1
title(sub=bquote(Log ~ rank ~ test ~ (italic(p)-value == .(p.val
On 10/10/07, Daniel Brewer <[EMAIL PROTECTED]> wrote:
> Thanks. That works great if I do this:
> title(sub=bquote(italic(p)-value == .(p.val)))
>
> But if
On 10/10/07, Christoph Krammer <[EMAIL PROTECTED]> wrote:
> Hello,
>
> is there a possibility to set default options to qplot?
>
> I need to draw a lot of graphs and would like to have all of them as point
> plot but with a greater size and a fixed color for all dots.
Sure. It's a bit of hack, bu
You need to follow the posting guide and provide commented, minimal,
self-contained, reproducible code.
I can only guess at what your code looks like, but the following
catches the error:
> z <- try(eval(parse(text="1:20 <- x")))
Error in eval(expr, envir, enclos) : object "x" not found
> str(z)
Hello all,
I read the following variable
> x
x
1 1_A1_ML1_a.DLL
2 11_B1_ML2_a.DLL
3 4_A1_ML3_a.DLL
4 55_C1_ML4_a.DLL
5 14_C1_ML5_a.DLL
I would like to disperse it in three variable such as
> x1
[1] 1 11 4 55 14
> x2
[1] "A1" "B1" "A1" "C1" "C1"
> x3
[1] "ML1" "ML
I forgot to attach my script. I work under pressure and make a lot of
these mistakes. Sorry.
I have attached it right now. I'm lso pasting the messages appearing
on R GUI in the following.
I see the average becomes NA which makes me think that for some reason
R interpreter is reading out of the loa
Hello and sorry that I still haven´t found a solution for my problem.
I need to extract the lower and upper triangle from a square matrix
including the diagonal. This diagonal is not zero in that special case.
I tried with as.dist
w<-as.dist(w, diag = TRUE)
> w
1 2 3 4 5
1 0
2 2 0
3
On Wed, 2007-10-10 at 10:13 -0400, Tom Sgouros wrote:
> Hello all:
>
> I'm sure this is a trivial request, but I'm still a beginner at this,
> and haven't been able to find it. I need to create simulated data based
> on some empirical distributions of a single variable. I've found R
> functions
Howard,
> though it received the statement. I tried to use some R built-in
> functions of try, tryCatch, eval, expression, as.expression,
> parse, deparse, etc. None of them worked.
How did they not work? What did you attempt? Did you use silent = TRUE
in try?
Please be more specific about t
Hello all:
I'm sure this is a trivial request, but I'm still a beginner at this,
and haven't been able to find it. I need to create simulated data based
on some empirical distributions of a single variable. I've found R
functions to help me simulate data based on analytical distributions, or
to
Dear R users,
I am trying to ´detect´ the trend in an artificial time series created
by the simple function
x=seq(pi,10*pi,0.1)
my.ts=0.1*x+sin(x)
my.ts=ts(my.ts,start=1800)
plot(my.ts)
I have tried stl(my.ts), but because I don´t have ´replications´ at
every time point, this somehow doesn´t
Hi all,
I'm trying to download a file from the net, and the download.file
command leaves it corrupted, i.e excel cannot open it. It seems as if
it's going ok, however.
Does anyone have an idea of what's going on?
regards,
Gustaf Rydevik
>download.file(url="http://www.who.int/entity/whosis/
This is a text part of the message.
It is shown for the users of old-style e-mail clients
__
This email has been scanned by the MessageLabs Email Security System.
For more information please visit http://www.messagelabs.com/email
Dear UseRs,
I wrote following function in order to solve Data Envelopment Analysis.
Reason for posting is that the function is slow when nrow(dat) is large.
I wonder if other functions could substitute the for() loop in the
code, such as mapply().
Can anybody help to rewrite the dea() function
On Wed, 2007-10-10 at 16:42 +0200, Birgit Lemcke wrote:
> Hello and sorry that I still haven´t found a solution for my problem.
>
> I need to extract the lower and upper triangle from a square matrix
> including the diagonal. This diagonal is not zero in that special case.
How can a site have a
jim holtman wrote:
> Is this what you want?
>
>
>> x <- read.table(textConnection("Score Name
>>
> + 88 19_0070
> + 88 19_0070
> + 87 19_0070
> + 79 002127_0658
> + 79 002127_0658
> + 77 002127_0658"), header=TRUE)
If all you're doing is fitting different responses to the same X data, then
you don't need model.matrix. See ?update, ?update.formula.
Bert Gunter
Genentech Nonclinical Statistic
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Ajay Shah
Sent: Tuesday, O
On 10/10/07, Birgit Lemcke <[EMAIL PROTECTED]> wrote:
> Hello and sorry that I still haven´t found a solution for my problem.
>
> I need to extract the lower and upper triangle from a square matrix
> including the diagonal. This diagonal is not zero in that special case.
> I tried with as.dist
>
>
On Wed, 10 Oct 2007, Gustaf Rydevik wrote:
> Hi all,
>
> I'm trying to download a file from the net, and the download.file
> command leaves it corrupted, i.e excel cannot open it. It seems as if
> it's going ok, however.
> Does anyone have an idea of what's going on?
That is a binary file. You n
On 10/10/07, Gustaf Rydevik <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I'm trying to download a file from the net, and the download.file
> command leaves it corrupted, i.e excel cannot open it. It seems as if
> it's going ok, however.
> Does anyone have an idea of what's going on?
>
> regards,
>
> Gu
This are distances but between male and female plants of the same
species.
species A B C D E --> male
A
B
C
D
|
female
I would like to do a Manteltest.
Birgit
Am 10.10.2007 um 16:59 schrieb Gustaf Rydevik:
Tom Sgouros as220.org> writes:
> What I have is twelve bins of data, and the population
> in each bin. The top bin is open-ended, and the whole distribution is
> more or less poisson-ish.
>
> I can think of a couple of ways to fake this ok, but is there a real R
> way to do it?
>
hist(rpois(
Hi Jim,
What you told me works well.
I have tried using 'eval(parse(text="1:20 = x"))' as well as
'try(parse(text="1:20 = x"))' before but not using eval anfd try
functions together. I just added try function following your
suggestion, e.g. 'try(eval(parse(text="1:20 = x")))'and then it
w
I am using gee to fit a log-linear marginal model to grouped data. I
get sensible results unfortunately followed by this warning
Warning message:
Note: Cgee had an error (code= 104 ). Results suspect. in:
gee(formula = event ~ exgr + agegr + indiv + offset(loginterval),
Does anyone knows what
On Wed, 10 Oct 2007, Michel Chavance wrote:
> I am using gee to fit a log-linear marginal model to grouped data. I
> get sensible results unfortunately followed by this warning
>
> Warning message:
> Note: Cgee had an error (code= 104 ). Results suspect. in:
> gee(formula = event ~ exgr + agegr +
Dear R graphics experts,
Do you know of any way of plotting semi-transparent points in R ?
I face this problem any time I want to e.g. plot thousands of points
that belong to two or three classes which I color code.
With the default opaque colors, the plotting order matters and
basically the po
Hello,
I problem is in the format of the date, my time series is like this:
2006070100 1244 6162
2006070101 1221 6060
2006070102 1214 6060
2006070103 1194 5959
2006070104 1182 5858
2006070105 1178
See ?rgb and its alpha argument, for example.
Uwe Ligges
Markus Loecher wrote:
> Dear R graphics experts,
> Do you know of any way of plotting semi-transparent points in R ?
> I face this problem any time I want to e.g. plot thousands of points
> that belong to two or three classes which I co
You may be better off using a complete distance matrix and a model
matrix describing plant sex, as explained in Legendre and Fortin 1989
and elsewhere.
Otherwise you will have to write your own Mantel code - all the examples
I know of (vegan and ecodist, primarily) assume that diagonals are
zero w
See ?rgb and the article by Paul Murrell in R-news 2004-2. (The details
of where it is supported have changed in the last three years, though.)
On Wed, 10 Oct 2007, Markus Loecher wrote:
> Dear R graphics experts,
> Do you know of any way of plotting semi-transparent points in R ?
> I face this
Rob,
Thanks for sending the example...
>
> m1=gamm(present~s(week,bs="cc")+s(week,bs="cc",by=y1),random=list(garden=~1
>) ,
> correlation=corAR1(form=~1|garden),family=binomial,data=count.data2)
>
> (ie removing the week term). This model proceeds - to a point...
>
> Maximum number of PQL itera
It can be as simple as:
> plot( rnorm(1000), rnorm(1000), col="#ff22", pch=16,cex=3)
Where the color is "#RRGGBBAA" and the AA portion is the
opacity/trasparency. Of course this only works if you are using a
graphics device that supports transparency. The windows device under R
2.6.0 does (
Hi,
I tried to install the caTools package manually
from the main R Project website and I get the
following error message when typing library('caTools')
Error in library("caTools") :
'caTools' is not a valid package -- installed <
2.0.0?
What am I doing wrong? Also, why is this package
Chris Jones-35 wrote:
>
> Hi all,
>
> I seem to be having a difficult time using the 'ape' package in R
> when it comes to rooting trees. Here's a short screenshot:
>
> [snip]
>
>
Can you post your dataset, or a subset of it that replicates
the problem, somewhere? (On the list if sho
One of the things that you should do is to use Rprof to see where time
is being spent. I would guess that is the not the 'for' loop, but
instead what is being done inside it. My guess it that most of the
time is being spent in the number of times that 'lp' is being called.
So the real problem mig
On 10/8/07, Sergio Correia <[EMAIL PROTECTED]> wrote:
> (Before starting: I'm a total R noob so please bear with me in case of
> any error or faux pas).
>
> Hi,
>
> For a small project, I'm writing a few simple R functions and calling
> them from python (using RPy). I'm sharing the code with a coup
You haven't told us your OS or R version (as we did ask).
But until last night there was not a version of caTools that worked
correctly in the current R (>= 2.6.0), so you may need to be patient.
On Wed, 10 Oct 2007, Judith Flores wrote:
> Hi,
>
> I tried to install the caTools package manual
Jim Holtman wrote:
>
> One of the things that you should do is to use Rprof
> to see where time is being spent.
>
Rprof is great! I didn't know such functionality existed.
But I am a very destructive user; I think I found a way
to break Rprof: if I do
f <- my.slow.function
then call
f(...)
Joao Santos-7 wrote:
>
> Hello,
>
> I problem is in the format of the date, my time series is like this:
>
> [snip]
>
> When I attempt to format the time like this:
>
> A <- read.table("file", sep="\t", col.names=c("date", "my1", "my2",
> "my3"))
> temp <- as.Date(A$date, format="%Y%m%d%H")
Hello,
I have a question about splitting label in regression tree.
for example, by using the following command
plot(m.regression.tree.pr,uniform=T,margin=0.05,branch=0.7,compress=F)
text(m.regression.tree.pr,pretty=0)
I get a tree with one label in some splitting node like this
bcu<0.004969
Now
There are a couple of outputs from Rprof. One shows time spent in the
function (and everything that it calls) and the other in the function
itself.
What I do is to post process the output from Rprof into a format that
is easier to understand. In this case, the Rprof run took 12.2
seconds of whic
Here you go
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=114
Also have a look @
http://www.bmj.com/cgi/reprint/322/7300/1479.pdf
HTH
Francois
- Original Message -
From: "Armin Goralczyk" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc:
Sent: Wednesday, October 10, 2007
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