Does this do what you want?
library(lattice)
rand1 - rnorm(50)
rand2 - rnorm(50)
theplot - xyplot(rand1 ~ rand2, xlab=x axis,
ylab=y axis)
thefile - plotproblem.eps
trellis.device(postscript, file=thefile, color=F,
horizontal=FALSE, width=12, height=4, paper=special)
print(theplot,
Hi
I just started using RGui.exe under widnows.
I have a text file containing date arranged in columns and rows, each column
has the same format, each row with different formats. 3 of the columns are
something like this 1/12/2006 3:59:45 PM
I need to calculate the different in seconds
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?
thanks
In mva package, try ?prcomp.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of souvik banerjee
Sent: Wednesday, March 15, 2006 12:44 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Help on factanal.fit.mle
Hi
Can anybody please suggest me about the
Try this link also.
http://tolstoy.newcastle.edu.au/R/help/02b/0943.html
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of souvik banerjee
Sent: Wednesday, March 15, 2006 12:44 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Help on factanal.fit.mle
Hi
Can
Hello.
I want to create a list of vectors but each component of the list has a
different length.
For example:
Example=list()
Example=list(Example,c(1,2,3))
Example=list(Example,c(11,12,13,14,15))
If I want the first component of the Example list, I have to write:
Example[[1]][[2]]. R
Have you looked at:
An Introduction to R: Software for StatisticalModelling Computing by
Petra Kuhnert and Bill Venables
which is available at http://cran.r-project.org/other-docs.html
Hope this helps.
Best,
Matthias
Hi all,
I am trying to use GAM to work on some data... Are there any
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
Stephen hints. I have been trying to set a new method for
Stephen the
Hi all,
I am doing some clustering and the clustered results are presented in a
pairs plot showing 4 clusters...
I made the data points belong to 4 different clusters displaying different
colors.
Now I want to select the best clustered class, how can I click on the data
point, and the program
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 but
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
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
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):
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.
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]]
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
?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 give the 6
?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]]
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
Stephen hints. I
Mi nueva dirección de correo es: [EMAIL PROTECTED]
New e-mail address: [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
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 sum:
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 has no
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 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 effective
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
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
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
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
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, but it dont work
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
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
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.
For
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 Target
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()
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
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
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 =
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
matA
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
2
-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 C
1
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/
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
A
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
[,1]
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 -
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
You might also wish to read the relevant chapter of VR'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: [EMAIL
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$Name),labels=paste(name))
Q1, if you try the script above,you will get 5
Em Quarta 15 Março 2006 10:26, Gabor Grothendieck escreveu:
as.Date(paste(YEAR, WEEK, 0), %Y %U %w)
Hi,
it works, but it use a year with 53 weeks, I need to use with 52 weeks, how to
change this?
Thanks
Ronaldo
--
Errigal Mountains -- Tailoring manure
-- anagrama
--
| // |
Mike White wrote:
I am having difficulty setting different xlim values in the lattice
histogram plot function.
An example is shown below. I think I need to convert the limits data.frame
to a list of paired values but don't know how. Any help would be
appreciated.
library(lattice)
mat -
Haleh,
This question would be better asked on the Bioconductor mailing list.
You haven't told us what version of R you are using. I suspect you
have a version mismatch.
With R 2.2.x you should be able to do the following to get MergeMaid
installed:
From the R prompt do:
On Wed, 15 Mar 2006, Heinz Tuechler wrote:
Dear All,
a Surv object I put in a data frame behaves somehow unexpected (see example).
If I do a Cox regression on the original Surv object it works. If I put it
in a data.frame and do the regression on the data frame it does not work.
Seemingly
Try something like:
xp - barplot(test$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:
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 least in this example.
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
A
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 '~'
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 data
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!
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
only
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.
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
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)
con -
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 summarize the
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
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.
I think you ca try
identify()
-
[[alternative HTML version deleted]]
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PLEASE do read the posting guide!
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)
[1]
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
\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),
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
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
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 0 1 2 3
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)
[1]
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
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 4 11
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 first
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]]
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)
Loading required
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,
?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
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 allow, per the example code, 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
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'
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 code
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
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
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
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 -
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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R-help@stat.math.ethz.ch mailing list
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
would
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] wrote:
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 regression
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 cutYield?
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
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 3
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
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
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
I
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 with that is what if
Patrick Baker patrick.baker at 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
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
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