> "HenrikB" == Henrik Bengtsson <[EMAIL PROTECTED]>
> on Thu, 16 Mar 2006 07:48:49 +0100 writes:
HenrikB> It is even better/more generic(!) to have:
HenrikB> extract <- function(...) UseMethod("extract")
HenrikB> "Specifying the object argument or the method
HenrikB>
It is even better/more generic(!) to have:
extract <- function(...) UseMethod("extract")
"Specifying the object argument or the method arguments of a generic
function will restrict any other methods with the same name to have
the same argument. By also excluding the object argument, default
fun
I would like to know if there is a way of directly calculating the
F-ratio of a random effect using the "aov" function. I have 2 factors
in my model, "population" which is random and "length" which is the
length of female fish within each population. The dependent variable is
"diam" which is th
Hi,
Are there any multidimenstional versions of runmed() and smooth.spline()
functions?
I need to fit surface into quite noisy 3D data.
Below is an example (2D) of kind of fittings I do.
Thank you,
Vlad
#=generating complex x,y dataset with gaussian & uniform noise==
x <- seq(1:1)
x2 <- rep(
On 3/15/06, Thomas Lumley <[EMAIL PROTECTED]> wrote:
> On Wed, 15 Mar 2006, Philippe Grosjean wrote:
>
> > the for() loop is very slow in S-PLUS. This is probably one of the
> > motivation of developing the apply() family of functions (as well as the
> > ugly For() loop) under this system.
> >
> >
Is there any way to store the value of warnings but avoid printing them?
A simplified example:
> out <- c(0.2,0.3,0.4)
> var <- c(2,3,4)
> outcome <- glm(out ~ var, family=binomial())
Warning message:
non-integer #successes in a binomial glm! in: eval(expr, envir, enclos)
I don't like the warning
Thanks for the tip. I've attempted the alternative
approach of using write.table, which allows you to see
the file and import it directly. Would the carriage
returns be visible with notepad or with the edit
command? Because I don't see the /r in the text file,
but still get it when I query the data
Hi,
I am reading a "|" delimited text file into R using read.table(). I am
using colClasses= to specify some variables as factors. Some of these
variables include missing values coded as "NA". Unfortunately the R code
I am using (pasted bellow) includes "NA" as one of the factor levels. Is
it
context grey wrote:
> That's approximately right, but the individual
> scatterplots
> are slightly stretched horizontally.
>
> Is there not any way to have the plots have true 1:1
> aspect ratio
> (given that the range of the data is the same on both
> axes)
> and still get a bounding box?.
W
I'm having trouble getting fun="cloglog" to work with plot on
a survfit object. Here are the data I used for the commands
that follow.
days status
2 0
2 0
5 1
9 0
14 1
16 0
16 0
17 0
29 1
30 0
37 1
37 0
39 1
44 0
44 0
58 0
60 1
67 1
68 1
82 1
82 1
86 0
86 0
89 1
93 0
97 1
100 0
100 0
100 0
> l
Patrick Baker sci.monash.edu.au> writes:
What I'd like to get some advice or insight on is whether
> there is an appropriate way to rescale the AIC values to permit
> comparisons across these models. Any suggestions would be very welcome.
> Cheers, Patrick Baker
>
>
Not a complete solutio
On 3/15/06, Sebastian Luque <[EMAIL PROTECTED]> wrote:
> Sebastian Luque <[EMAIL PROTECTED]> wrote:
>
> "Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
>
> >> Is this ok or is it what you are trying to avoid:
>
> >> factor(unlist(lapply(cutYield, as.character)))
>
> > Thank you Gabor. The problem
Liaw, Andy wrote:
> Could it be that you have the environment variable R_HOME (or something like
> that) defined somewhere? Just a wild guess...
No, R_HOME is not defined in the shell I was compiling in, but R_LIBS
was! Brilliant "wild guess."
Thanks,
Kevin
> Andy
>
> From: Kevin E. Thorpe
>
Could it be that you have the environment variable R_HOME (or something like
that) defined somewhere? Just a wild guess...
Andy
From: Kevin E. Thorpe
>
> I downloaded R-Patched today (to see if another problem I
> want to ask about is still present or if its just me - as per
> the posting gui
I downloaded R-Patched today (to see if another problem I want to
ask about is still present or if its just me - as per the posting
guide). I ran "tools/rsync-recommended" successfully. I then ran
"configure --enable-R-shlib" successfully. Then make stops with the
following error.
gcc -shared -
On Thu, 16 Mar 2006, Heinz Tuechler wrote:
>
> Thank you, Thomas. You are right, it works, but why then I find on the help
> page for Surv{survival} the following sentence:
> "To include a survival object inside a data frame, use the I() function.
> Surv objects are implemented as a matrix of 2 or
At 11:59 15.03.2006 -0600, Robert Baer wrote:
>This does work:
>coxph(survobj~group, data=df.test[[1]]) # this works like your original
>
>To get insight compare:
>str(survobj)
>str(df.test)
>str(df.test[[1]])
Thank you for your answer. It seems to me that your solution only works, as
long as the
Sebastian Luque <[EMAIL PROTECTED]> wrote:
"Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
>> Is this ok or is it what you are trying to avoid:
>> factor(unlist(lapply(cutYield, as.character)))
> Thank you Gabor. The problem with that is what if some levels do not
> appear in any member of cut
At 09:23 15.03.2006 -0800, Thomas Lumley wrote:
>On Wed, 15 Mar 2006, Heinz Tuechler wrote:
>
>> Dear All,
>>
>> a Surv object I put in a data frame behaves somehow unexpected (see
example).
>> If I do a Cox regression on the original Surv object it works. If I put it
>> in a data.frame and do the
Since all components of cutYield have the same levels, one
could do this to ensure that all levels are represented:
factor(unlist(lapply(cutYield, as.character)), levels = levels(cutYield[[1]]))
On 3/15/06, Sebastian Luque <[EMAIL PROTECTED]> wrote:
> "Gabor Grothendieck" <[EMAIL PROTECTED]> wro
"Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
> Is this ok or is it what you are trying to avoid:
> factor(unlist(lapply(cutYield, as.character)))
Thank you Gabor. The problem with that is what if some levels do not
appear in any member of cutYield? In that case, the factor created above
wou
Hi there,
Can R use "principal component analysis (PCA)" to do the clustering? Or
does PCA only be used to pick up the important variables?
Thank you!
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Is this ok or is it what you are trying to avoid:
factor(unlist(lapply(cutYield, as.character)))
On 3/15/06, Sebastian Luque <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've run into a ridiculous problem I can't find any solutions for in the
> archives or help pages:
>
> data(barley)
> cutYield <- wi
That's approximately right, but the individual
scatterplots
are slightly stretched horizontally.
Is there not any way to have the plots have true 1:1
aspect ratio
(given that the range of the data is the same on both
axes)
and still get a bounding box?. And, without getting
out
a ruler and m
Hi,
I've run into a ridiculous problem I can't find any solutions for in the
archives or help pages:
data(barley)
cutYield <- with(barley, by(yield, variety, cut, breaks = c(0, 30, 60, 90)))
As in this example, I'm using 'by' to return a factor for each level of
another factor. The problem is t
Hi, thanks for the help but I'm still having issues.
Basically, I have two matrices of equal dimension, one should produce
something similar to a heatmap.. The 2nd matrix should be the "heights"
for each value of the heatmap - producing a sort of surface plot.
Viewing this seems like a problem to
On Wed, 15 Mar 2006, Philippe Grosjean wrote:
> the for() loop is very slow in S-PLUS. This is probably one of the
> motivation of developing the apply() family of functions (as well as the
> ugly For() loop) under this system.
>
> Now, for() loops are much faster in R. Also, if you look at the R
In my opinion the main issue between using 'for' and
an apply function is the simplicity of the code. If it is
simpler and more understandable to use 'lapply' than
a 'for' loop in a situation, then use 'lapply'. If in a
different situation it is the 'for' loop that is simpler, then
use the 'for'
the for() loop is very slow in S-PLUS. This is probably one of the
motivation of developing the apply() family of functions (as well as the
ugly For() loop) under this system.
Now, for() loops are much faster in R. Also, if you look at the R code
in apply(), you will realize that there is a for
> From: Thomas Lumley
>>
>> On Tue, 14 Mar 2006, John McHenry wrote:
>>
>> > Thanks, Gabor & Thomas.
>> >
>> > Apologies, but I used an example that obfuscated the question that I
>> > wanted to ask.
>> >
>> > I really wanted to know how to have extra arguments in
>> functions that
>> > would al
?sample
You must use replace=FALSE to guarantee 1000 different rows
mymatrix[sample(12000,1000,replace=FALSE),]
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
"The business of the statistician is to catalyze the scientific learning
process." - George E. P. Box
>
Is something like this what your looking for:
x <- matrix(c(rnorm(100)),ncol=10)
sub <- sample(5, replace=TRUE) # For sampling with replacement
x[sub,]
Harold
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of mark salsburg
Sent: Wednesday, March 15, 2
Define extract like this:
extract <- function(e, n, ...) UseMethod("extract")
# test -- no warning
extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)
On 3/15/06, Philippe Grosjean <[EMAIL PROTECTED]> wrote:
> Hello,
> I just notice this:
> > x <- c(1:4,0:5, 4, 11)
> > library(pastecs)
> Load
Dear group,
I would like to generate a 1000 random rows from a MATRIX with dimensions
12,000 by 20 (i.e. to generate a 1000 by 20 matrix of random rows)
Does the function sample() work for this???
thank you in advance
[[alternative HTML version deleted]]
___
Hello,
I just notice this:
> x <- c(1:4,0:5, 4, 11)
> library(pastecs)
Loading required package: boot
> tp <- turnpoints(x)
> extract(tp, no.tp = FALSE, peak = TRUE, pit = FALSE)
[1] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE
FALSE
Warning message:
arguments after the
On Wed, 2006-03-15 at 21:45 +0100, Philippe Grosjean wrote:
> What Fred is looking for is local minima/maxima, also known as turning
> points, or pits/peaks in a series. You can look at ?turnpoints in
> pastecs library.
>
> > x <- c(1:4,0:5, 4, 11)
> > x
> [1] 1 2 3 4 0 1 2 3 4 5
What Fred is looking for is local minima/maxima, also known as turning
points, or pits/peaks in a series. You can look at ?turnpoints in
pastecs library.
> x <- c(1:4,0:5, 4, 11)
> x
[1] 1 2 3 4 0 1 2 3 4 5 4 11
> tp <- turnpoints(x)
> summary(tp)
Turning points for: x
nbr obs
What you want seems to be the valleys and peaks in the data. If so, try:
RSiteSearch("find peaks")
which points to a post by Philippe Grosjean, pointing to the pastesc
package:
> library(pastecs)
Loading required package: boot
> tp <- turnpoints(x)
> which(tp$peaks)
[1] 4 10
> which(tp$pits)
On Wed, 2006-03-15 at 11:32 -0800, Fred J. wrote:
> Hi
>
> Is there a function which determines the location, i.e., index of
> the all minimums or maximums of a numeric vector.
> Which.min(x) only finds the (first) of such.
>
> > x <- c(1:4,0:5, 4, 11)
> > x
>[1] 1 2 3 4
http://www.r-project.org/Rlogo.jpg
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erin Hodgess
Sent: Wednesday, March 15, 2006 2:39 PM
To: r-help@stat.math.ethz.ch
Subject: [R] R icon image file
Dear R People:
I would like to include a link to the R ho
Try
order(x, decreasing=TRUE/FALSE)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Fred J.
Sent: Wednesday, March 15, 2006 2:32 PM
To: r-help@stat.math.ethz.ch
Subject: [R] which.minimums not which.min
Hi
Is there a function which determines
\r is a carriage return character which some editors may use as a line
terminator when writing files. My guess is that RSQLite writes your
data frame to a temp file using \r as a line terminator and then runs
a script to have SQLite import the data (together with \r - this would
be the problem), b
Dear R People:
I would like to include a link to the R home page
on a web page for students.
I would like to have the "R" icon as part of the link.
Where is the image file please? (for the icon)
Thanks,
Sincerely,
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
Hi
Is there a function which determines the location, i.e., index of the all
minimums or maximums of a numeric vector.
Which.min(x) only finds the (first) of such.
> x <- c(1:4,0:5, 4, 11)
> x
[1] 1 2 3 4 0 1 2 3 4 5 4 11
> which.min(x)
[1] 5
> which.max(x)
[
I think you ca try
>identify()
-
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PLEASE do read the posting guide! ht
I meant there are more than 7 * 52 days in a year.
On 3/15/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Given that there are more than 7 * 52 days in a week you may
> need to think about what it is that you want.
>
> This will give you Jan 1 for Week 1, Jan 8 for Week 2, etc.
>
> as.Date(pa
Given that there are more than 7 * 52 days in a week you may
need to think about what it is that you want.
This will give you Jan 1 for Week 1, Jan 8 for Week 2, etc.
as.Date(paste(YEAR,1,1,sep="-")) + 7 * (WEEK - 1)
That or some variation of that might be suitable. See the help
desk article in
On 3/15/2006 1:38 PM, Vivek Satsangi wrote:
> Folks,
> I am documenting what I finally did, for the next person who comes along...
>
> Following Dr. Murdoch's suggestion, I looked at qqplot. The following
> approach might be helpful to get to the same information as given by
> qqplot.
> To summari
What am I doing wrong, or is the \r that I'm getting
in the example below a bug?
> a <- (1:10)
> b <- (LETTERS[1:10])
> df <- as.data.frame(cbind(a, b))
>
> df
a b
1 1 A
2 2 B
3 3 C
4 4 D
5 5 E
6 6 F
7 7 G
8 8 H
9 9 I
10 10 J
> library(RSQLite)
> drv <- dbDriver("SQLite")
>
Folks,
I am documenting what I finally did, for the next person who comes along...
Following Dr. Murdoch's suggestion, I looked at qqplot. The following
approach might be helpful to get to the same information as given by
qqplot.
To summarize the ask: given x, y, xw and yw, show (visually is okay)
On Tue, 14 Mar 2006, Dean Sonneborn wrote:
> Would anyone with experience with the map functions know how to divide
> Czechoslovakia into the Czech Republic and Slovakia. They have been two
> separate countries for some time now. I'm thinking about the worldhires
> map database in particular.
On Tue, 14 Mar 2006, Stefan Pohl wrote:
> Hi,
>
> is there an R package with which is it possible to display postal codes
> for Germany on a map?
If you know the geographical or projected coordinates of the postcodes,
then there are plenty of possibilities, but this kind of data is typically
o
Hi,
This one is quick: how to ask R to print "0.1" as ".1", i.e, what I want is
> 0.1
.1
Many thanks,
Dimitri
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PLEASE do read the posting guide! http://www.R-projec
On Sun, 12 Mar 2006, Ferran Carrascosa wrote:
> Hi r-users,
>
> I would like to know if R have any solution to the "Address standardization".
> The problem is to classify a database of addresses with the real
> addresses of a streets of Spain. Ideally, I would like to assign
> Postal code, census
This does work:
coxph(survobj~group, data=df.test[[1]]) # this works like your original
To get insight compare:
str(survobj)
str(df.test)
str(df.test[[1]])
Then note the 2nd sentence of the following from ?coxph
Arguments:
formula: a formula object, with the response on the left of a '~'
Thanks everyone.
Obvious when you think about it, and you check that both the matrices
your trying it with are actually matrices... instead of one being a
list.
On Wed, 2006-15-03 at 06:03 -0500, tom wright wrote:
> Can someone please give me a pointer here.
> I have two matrices
>
> matA
>
On Wed, 2006-03-15 at 17:54 +0100, jia ding wrote:
> Hi,
>
> I have a file named:
> test_R.txt
> aaa 2
> bbb 5
> ccc 7
> sss 3
> xxx 8
>
> I want to have a plot:
> test<-read.table("test_R.txt",col.name=c("Name","Score"))
> par(mfrow=c(1,2))
It's not clear what the purpose is here, at leas
Try something like:
xp <- barplot(test$Score, space=.5)
axis(1, at=xp, labels=as.character(test$Name))
See ?barplot more more detail.
Andy
From: jia ding
>
> Hi,
>
> I have a file named:
> test_R.txt
> aaa 2
> bbb 5
> ccc 7
> sss 3
> xxx 8
>
> I want to have a plot:
> test<-read.table("
On Wed, 15 Mar 2006, Heinz Tuechler wrote:
> Dear All,
>
> a Surv object I put in a data frame behaves somehow unexpected (see example).
> If I do a Cox regression on the original Surv object it works. If I put it
> in a data.frame and do the regression on the data frame it does not work.
> Seemin
Haleh,
This question would be better asked on the Bioconductor mailing list.
You haven't told us what version of R you are using. I suspect you
have a version mismatch.
With R 2.2.x you should be able to do the following to get MergeMaid
installed:
>From the R prompt do:
source("http://biocond
Mike White wrote:
> I am having difficulty setting different xlim values in the lattice
> histogram plot function.
> An example is shown below. I think I need to convert the limits data.frame
> to a list of paired values but don't know how. Any help would be
> appreciated.
>
> library(lattice)
Em Quarta 15 Março 2006 10:26, Gabor Grothendieck escreveu:
> as.Date(paste(YEAR, WEEK, 0), "%Y %U %w")
Hi,
it works, but it use a year with 53 weeks, I need to use with 52 weeks, how to
change this?
Thanks
Ronaldo
--
Errigal Mountains <--> Tailoring manure
-- anagrama
--
|>
Hi,
I have a file named:
test_R.txt
aaa 2
bbb 5
ccc 7
sss 3
xxx 8
I want to have a plot:
test<-read.table("test_R.txt",col.name=c("Name","Score"))
par(mfrow=c(1,2))
barplot(test$Score)
name<-test$Name
axis(1,at=1:length(test$Name),labels=paste(name))
Q1, if you try the script above,you will
You might also wish to read the relevant chapter of V&R's S PROGRAMMING.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
"The business of the statistician is to catalyze the scientific learning
process." - George E. P. Box
> -Original Message-
> From: [EMAI
Dear All,
a Surv object I put in a data frame behaves somehow unexpected (see example).
If I do a Cox regression on the original Surv object it works. If I put it
in a data.frame and do the regression on the data frame it does not work.
Seemingly it has to do with the class attribute, because if I
I am having difficulty setting different xlim values in the lattice
histogram plot function.
An example is shown below. I think I need to convert the limits data.frame
to a list of paired values but don't know how. Any help would be
appreciated.
library(lattice)
mat <- as.data.frame(matrix(abs(c(
Dear tom,
is the following what you are looking for?
> a=matrix(runif(9),3,3)
> a
[,1] [,2] [,3]
[1,] 0.9484247 0.9765431 0.6169739
[2,] 0.8423545 0.3137295 0.4031847
[3,] 0.6724235 0.1076373 0.2356923
> b<-matrix(sample(c(TRUE,FALSE),size=9,replace=TRUE),3,3)
> b
[
This is really elementary indexing in S language:
matA[matB]
Best,
Philippe Grosjean
tom wright wrote:
> Can someone please give me a pointer here.
> I have two matrices
>
> matA
> A B C
> 1 5 2 4
> 2 2 4 3
> 3 1 2 4
>
> matB
>
matA[matB] or matA[!matB]
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://www.med.kuleuven.be/biostat/
http:/
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of tom wright
> Sent: Wednesday, March 15, 2006 6:04 AM
> To: R-Stat Help
> Subject: [R] matrix indexing
>
>
> Can someone please give me a pointer here.
> I have two matrices
>
> matA
> A B
On Wed, 2006-03-15 at 06:03 -0500, tom wright wrote:
> Can someone please give me a pointer here.
> I have two matrices
>
> matA
> A B C
> 1 5 2 4
> 2 2 4 3
> 3 1 2 4
>
> matB
> A B C
> 1 TRUEFALSE TRUE
Try this:
matA[c(matB)]
In fact even this works for your example although in general it
couldbe problematic since a two column matrix index has
special meaning:
matA[matB]
On 3/15/06, tom wright <[EMAIL PROTECTED]> wrote:
> Can someone please give me a pointer here.
> I have two matrices
>
> m
Hello,
I wanted to install MergeMaid package in v 2.2.1. I could install it but
couldn't use without its dependant, Biobase. at biobase installation, I got the
following error message
In method for function "split": expanding the signature
to include omitted arguments in definition: drop = "mi
Can someone please give me a pointer here.
I have two matrices
matA
A B C
1 5 2 4
2 2 4 3
3 1 2 4
matB
A B C
1 TRUEFALSE TRUE
2 FALSE TRUETRUE
3 FALSE FALSE FALSE
how d
Hello,
I want a 2x1 multi-figure, with each plot 5" square.
Test code:
x<-rnorm(10,0,1)
y<-rnorm(10,0,1)
par(pty="s", mfrow=c(2,1), fin=c(5,5))
plot(x,y)
plot(y,x)
but this does not work (overplots the two figures). Substituting pin for fin
works, but is not what I want. Are mfrow and fin incomp
On 3/15/2006 8:31 AM, Vivek Satsangi wrote:
> Folks,
> Normally, in a data frame, one observation counts as one observation
> of the distribution. Thus one can easily produce a CDF and (in Splus
> atleast) use cdf.compare to compare the CDF (BTW: what is the R
> equivalent of the SPlus cdf.compare(
Ronny Hannemann wrote:
> Hi everyone,
>
> I'm very new to R and I like to learn a lot... actually I have a little
> problem concerning errorbars with xyplot.
>
> My data look like
> run target hemi x
> 1 1 Nichts Links 0.0007743240
> 2 2 Nichts Links -0.0008153365
> 3 1 T
Try this:
L <- list(letters, head(LETTERS))
L[[2]][[4]] # D
sapply(L, length) # 26 6
L <- c(L, 1:4)
L[[3]][[2]] # 2
L <- c(L, list(1:5))
On 3/15/06, Arnau Mir Torres <[EMAIL PROTECTED]> wrote:
> Hello.
>
> I want to create a list of vectors but each component of the list has a
> different length
Folks,
Normally, in a data frame, one observation counts as one observation
of the distribution. Thus one can easily produce a CDF and (in Splus
atleast) use cdf.compare to compare the CDF (BTW: what is the R
equivalent of the SPlus cdf.compare() function, if any?)
However, if each point should no
Hello all,
I'm trying to calculate the Maximum likelihood of individuals to get the
ancestry.
I mixd 3 populations 15 generations in proportion of 20% 20% 60% when each
population
sorce have diferent genome (0 1 and 2) with frequencies for each one.
So now i have individuals looks like 0 0 2 1 1 2
Try:
as.Date(paste(YEAR, WEEK, 0), "%Y %U %w")
On 3/15/06, Ronaldo Reis-Jr. <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have these vectors:
>
> > WEEK <- rep(c(1:52),2)
> > YEAR <- rep(c(2000,1999),c(52,52))
>
> How to make a vector of Date with weeks in years? I try as.date from survival
> package, b
Hi everyone,
I'm very new to R and I like to learn a lot... actually I have a little
problem concerning errorbars with xyplot.
My data look like
run target hemi x
1 1 Nichts Links 0.0007743240
2 2 Nichts Links -0.0008153365
3 1 Target Links -0.0015825950
4 2 Target Li
if you are interested in a solution using the tcltk package, then an
idea is to base a solution on the code for the demo "tkcanvas".
after installing the tcltk package, then
require(tcltk)
demo(tkcanvas)
Katharine Mullen
Department of Physics and Astronomy
Faculty of Sciences
Vrije Univers
Hi,
I have these vectors:
> WEEK <- rep(c(1:52),2)
> YEAR <- rep(c(2000,1999),c(52,52))
How to make a vector of Date with weeks in years? I try as.date from survival
package, but it dont work with weeks, just only with days, months etcs.
Thanks
Ronaldo
--
"Realmente minha cidade e muito facul
Hi,
I have these vectors:
> WEEK <- rep(c(1:52),2)
> YEAR <- rep(c(2000,1999),c(52,52))
How to make a vector of Date with weeks in years? I try as.date from survival
package, but it dont work with weeks, just only with days, months etcs.
Thanks
Ronaldo
--
"Realmente minha cidade e muito facul
On Tue, 2006-03-14 at 23:52 -0800, Michael wrote:
> Hi all,
>
> I am trying to use GAM to work on some data... Are there any resources
> providing hands-on tutorial/guide on how to do GAM on data in R?
> Specifically, I am not sure about which model to choose, and smooth models
> with which effect
Thanks I think you have both answered my question (reckon Ill go S3 on
that). As an adjunct to this do you know what might be the best
reference to the S4 methods current implementation.
I have ordered the Chambers book "Programming with Data", and I have a
short tutorial-- "S4 Classes in 15 pages
On 3/15/2006 3:21 AM, Fred J. wrote:
> Hi
>
>
> I have a list of 12000 rational numbers as inputs, running some of R
> functions will surly accumulate some round-off errors, Is there a way to
> have R do its calculations using rational numbers as input to minimize the
> round-off error?
R
You need to know the covariance of the Xs. The sum is just a linear
function of the Xs, so its variance is a function of a quadratic form
involving the covariance matrix of the Xs.
Andy
From: Antonio, Fabio Di Narzo
>
> Hi all.
> A statistical question. I have to estimate the variance of the su
Mi nueva dirección de correo es: [EMAIL PROTECTED]
New e-mail address: [EMAIL PROTECTED]
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On 3/15/06, Martin Maechler <[EMAIL PROTECTED]> wrote:
> > "Stephen" == Stephen Henderson <[EMAIL PROTECTED]>
> > on Tue, 14 Mar 2006 16:32:56 - writes:
>
> Stephen> Hello I've checked through previous postings but
> Stephen> don't see a fully equivalent problem-just a few
>
?cutree
>
> Hi all,
>
> Does hclust provide concrete clustered results? I could not
> see how to use it to make 6 clusters... and it does not give
> the 6 cluster labels...
>
> How to use the result of hclust?
>
> thanks a lot,
>
> Michael.
>
> [[alternative HTML version deleted]]
>
?cutree
?plot.hclust & ?identify.hclust
hc<- hclust(dist(tab, "manhattan"), "ward")
plot(hc, hang=-1)
(x <- identify(hc))
cutree(hc, 2)
Michael a écrit :
>Hi all,
>
>Does hclust provide concrete clustered results? I could not see how to use
>it to make 6 clusters... and it does not giv
Hi all.
A statistical question. I have to estimate the variance of the sum:
X(1) + X(2) + ...+ X(n)
from an observed sample, where X(i) are *correlated* and not necessarly
identically distributed. Someone can suggest a simple strategy
(I hope by exploiting some already present R package) for obta
Hi all,
Does hclust provide concrete clustered results? I could not see how to use
it to make 6 clusters... and it does not give the 6 cluster labels...
How to use the result of hclust?
thanks a lot,
Michael.
[[alternative HTML version deleted]]
___
> Now I want to select the best clustered class, how can I click on the data
> point, and the program returns the index of that cluster(its class number,
> or color number)?
Have a look at identify()
cu
Philipp
--
Dr. Philipp PagelTel. +49-8161-71 2131
Dept
Hello
I have big trouble getting R to work correctly with X11 fonts on Ubuntu
Breezy 5.10. I was hoping somebody could help me with this issue.
The first part of the problem is that I get the error "could not find
any X11 fonts" for any command with graphical ouput, for example
"demo(graphics)
Dear R-Users
I used the nlme library to fit a linear mixed model (lme). The random effect
standard errors and correlation reported are based on a Log-Cholesky
parametrization. Can anyone tell me how to get the Covariance matrix of the
random effects, given the above mentioned parameters ba
Dear Fred
You should change your code from
> x <- strptime(ts, "%m/%d/%y %I:%M:%S %p")
to
> x <- strptime(ts, "%m/%d/%Y %I:%M:%S %p")
"Y" instead of "y", since your year includes the century
(2006 and not 06)
Then it should work.
Regards,
Christoph
Hello,
I don't know if it is the most efficient way to do but my solution is:
x <- vector("list", 10) #creates a list with a length = 10
then in my loop (where i is iterated) :
x[[i]] <- my.vector
I hope this could help you
François Michonneau
Hello.
I want to create a list of vectors bu
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