Hello everybody,
I am trying to do a spatial logit model with the spatially
autoregressive error structure - SAE. Right now, I did an
implementation and i would like to check the results.
Somebody knows with there is a package in R that estimates parameters
to spatial logit model with SAE structur
On 05/28/2010 05:31 PM, Dylan Beaudette wrote:
On Friday 28 May 2010, Frank E Harrell Jr wrote:
On 05/28/2010 03:49 PM, Dylan Beaudette wrote:
Hi,
I have fit a model using the rms package with the Gls() function.
Is there a way to get the model estimates, std errors, and p-values (i.e.
what y
?'::'
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of thmsfuller...@gmail.com
> Sent: Friday, May 28, 2010 12
Ron Burns wrote:
Hi all-
I have a lot of small (xml) data files that are saved by classification
in directories named in accordance with the classification. I would like
to zip up these directories and include the zipped file as part of the
data in a package (which I know how to do.)
Are th
Ron Burns wrote:
Hi all-
Is there an R function that returns the directory in which a package has
been installed?
Thanks
Ron
?system.file
For example:
> system.file(package="DistributionUtils")
[1] "C:/Users/dsco036/R/win-library/2.10/DistributionUtils"
David Scott
--
_
Sorry by the delay. You could do:
> my.data <- expand.grid(A=factor(1:4), B=factor(1:4), rep=1:4)
> my.data$y <- rbinom(my.data$A, 10, 0.5)
>
> model <- glm(cbind(y, 10-y)~A*B, family=binomial, data=my.data)
> anova(model, test="Chisq" )
Analysis of Deviance Table
Model: binomial, link: logit
Ah OK, I didn't get your question then.
a dist-object is actually a vector of numbers with a couple of attributes.
You can't just cut out values like that. The hclust function needs a perfect
distance matrix to use the calculations.
shortcut is easy : just do f <- f/2*max(f), and all values are b
Dear Tom,
You can use getAnywhere(); for example:
> mean <- 1
> getAnywhere("mean")
2 differing objects matching 'mean' were found
in the following places
.GlobalEnv
package:base
namespace:base
Use [] to view one of them
> getAnywhere("mean")[1]
[1] 1
> getAnywhere("mean")[2]
function (x
Your notes are bordering on harassment. Do you expect that everyone who
reads this list will reply "I do not have anything that will help you"
if they don't? By my count this is your 4th note asking for this help.
That being said I hope that you do find help somewhere or implement it
yoursel
On Friday 28 May 2010, Frank E Harrell Jr wrote:
> On 05/28/2010 03:49 PM, Dylan Beaudette wrote:
> > Hi,
> >
> > I have fit a model using the rms package with the Gls() function.
> >
> > Is there a way to get the model estimates, std errors, and p-values (i.e.
> > what you get with print(fit)) int
I can't run your code.
Please, just give me whatever comes on your screen when you run:
dput(q)
On Fri, May 28, 2010 at 10:57 PM, Ayesha Khan
wrote:
> I assume my matrix should look something like this?..
>
> >round(distance, 4)
>P00A P00B M02A M02B P04A P04B M06A M06B P0
I assume my matrix should look something like this?..
>round(distance, 4)
P00A P00B M02A M02B P04A P04B M06A M06B P08A
P08B M10A
P00B 0.9678
M02A 1.0054 1.0349
M02B 1.0258 1.0052 1.2106
P04A 1.0247 0.9928 1.0145 0.9260
P04B 0.9898 0.9769 0.9875 0.9855 0.6075
M06A 1.0159 0.
v <- dput(x,"sampledata.txt")
dim(v)
q <- v[1:10,1:10]
f =as.matrix(dist(t(q)))
distB=NULL
for(k in 1:(nrow(f)-1)) for( m in (k+1):ncol(f)) {
if(f[k,m] <2) distB=rbind(distB,c(k,m,f[k,m]))
}
#now distB looks like this
> distB
[,1] [,2] [,3]
[1,]12 1.6275568
[2,]13 0
Yes Joris. I did try that and it does produce the results. I am now
wondering why I wanted a matrix like structure in the first place. However,
I do want 'f' to contain values less than 2 only. but when i try to get rid
of values greater than 2 by doing N <- (f[f<2], f strcuture disrupts and
hclust
Hello:
I am working on getting some statistics related to clinical trials and
stuff. I have to work with ICD9 codes.
Is anyone aware of any R method that deals with ICD9 codes
verification and manipulation.
Thanks
Vishwanath
__
R-help@r-project.org ma
Hello,
Normally, if I type a function name, it shows the function definition.
When the function is masked by a variable with the same name, it
doesn't show the function definition any more. Can anyone please tell
me a way how to retrieve the function definition even if it is masked
by a variable?
errr, forget about the output of dput(q), but keep it in mind for next time.
f = dist(t(q))
hclust(f,method="single")
it's as simple as that.
Cheers
Joris
On Fri, May 28, 2010 at 10:39 PM, Ayesha Khan
wrote:
> v <- dput(x,"sampledata.txt")
> dim(v)
> q <- v[1:10,1:10]
> f =as.matrix(dist(t(q)))
Tal,
Wow, i cant believe how many different manipulations i went through trying
to coerce it into the format i wanted. The below works nearly perfectly, i had
to change the "mean" call to "sum". Im curious why you used mean? Other than
that thank you very much, i feel a little foolish ab
On May 28, 2010, at 5:58 PM, Christian Schoder wrote:
Dear R-users,
I use firm-level data in panel structure. I would like to drop all
firms that have less than x observations over the time scale in any
of the variables considered. I would appreciate any help that (a)
indicates relevant
Linux problem solved! (For me at any rate). Thanks to some hints
from my Linux contacts it transpires that the problem with
sort << EOT
"ABCD"
"A CD"
EOT
# "ABCD"
# "A CD"
sort << EOT
"ADCD"
"A CD"
EOT
# "A CD"
# "ADCD"
arises because, by default, the " " is ignored in sorting. Therefore
in the
Dear R-users,
I use firm-level data in panel structure. I would like to drop all firms that
have less than x observations over the time scale in any of the variables
considered. I would appreciate any help that (a) indicates relevant literature
or websites or (b) indicates the code that could s
see ?cv.glm under the heading "Value". The help files tell you what comes
out.
On Fri, May 28, 2010 at 10:19 PM, azam jaafari wrote:
> Hi
>
>
> Finally, I did leave-one-out cross validation in R for prediction error of
> logistic regression by cv.glm. But I don't know what are the produced
> dat
Hi all-
I have a lot of small (xml) data files that are saved by classification
in directories named in accordance with the classification. I would like
to zip up these directories and include the zipped file as part of the
data in a package (which I know how to do.)
Are there R functions fo
On 05/28/2010 03:49 PM, Dylan Beaudette wrote:
Hi,
I have fit a model using the rms package with the Gls() function.
Is there a way to get the model estimates, std errors, and p-values (i.e. what
you get with print(fit)) into latex format?
I have tried:
f<- Gls(...)
latex(f, file='')
... but
Hi all-
Is there an R function that returns the directory in which a package has
been installed?
Thanks
Ron
--
R. R. Burns
Physicist (Retired)
Oceanside, CA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Ted Harding
> Sent: Friday, May 28, 2010 1:15 PM
> To: r-help@r-project.org
> Cc: carslaw
> Subject: Re: [R] difference in sort order linux/Windows (R.2.11.0)
>
> On 28-May-10 14:
Hi,
I have fit a model using the rms package with the Gls() function.
Is there a way to get the model estimates, std errors, and p-values (i.e. what
you get with print(fit)) into latex format?
I have tried:
f <- Gls(...)
latex(f, file='')
... but I get the following error
Error in replace.su
-Messaggio originale-
Da: Steve Lianoglou [mailto:mailinglist.honey...@gmail.com]
Inviato: ven 28/05/2010 17.06
A: mau...@alice.it
Cc: r-h...@stat.math.ethz.ch
Oggetto: Re: [R] why biomaRt cannot extract 3UTR sequences for 1941 ENSGx ?
Hi,
Two things:
1. You mistakenly posted this
Hi
Finally, I did leave-one-out cross validation in R for prediction error of
logistic regression by cv.glm. But I don't know what are the produced
data(almost 700)? does delta show me error estimation?
cost<-function(a,b)mean(abs(a-b))
#SALIC=binary response
salic.lr<-glm(profilesample$SALI
On 28-May-10 14:37:39, Duncan Murdoch wrote:
> On 28/05/2010 9:24 AM, (Ted Harding) wrote:
>> An experiment:
>>
>> sort(c("AACD","A CD"))
>> # [1] "AACD" "A CD"
>>
>> sort(c("ABCD","A CD"))
>> # [1] "ABCD" "A CD"
>>
>> sort(c("ACCD","A CD"))
>> # [1] "ACCD" "A CD"
>>
>> sort(c("ADC
On 28/05/2010 1:58 PM, Paul wrote:
Hello
It seems that sweave always runs in the global environment.
By default it uses the RweaveEvalWithOpt function to evaluate
expressions, and they are evaluated in the global environment. It's
possible to change that. You need to make your own "drive
Hi Jon,
does the empirical cumulative distribution function do what you want?
dat$q.score <- ecdf(dat$score)(dat$score)
?ecdf
HTH
Stephan
Jonathan Beard schrieb:
Hello all,
Thanks in advance for you attention.
I would like to generate a third value that represents the quantile
value of a va
Hi all,
Sorry I have too many questions. I could not think of a way to fix
problem. Can anyone give some suggestions on fixing this?
Hannah
2010/5/28 li li
> Hi Sarah,
> Thanks for your kind help. I now know where the problem is.
> Hannah
>
>
Hi Ayesha,
I wish to help you, but without a simple self contained example that shows
your issue, I will not be able to help.
Try using the ?dput command to create some simple data, and let us see what
you are doing.
Best,
Tal
Contact
Details:---
Hi Sarah,
Thanks for your kind help. I now know where the problem is.
Hannah
2010/5/28 Sarah Goslee
> My initial guess appears to be right: you're working with something
> exceedingly sensitive to floating point precision. You may have to
> reconsider your methods.
>
Hello all,
Thanks in advance for you attention.
I would like to generate a third value that represents the quantile
value of a variable in a data frame.
# generating data
x <- as.matrix(seq(1:30))
y <- as.matrix(rnorm(30, 20, 7))
tmp1 <- cbind(x,y)
dat <- as.data.frame(tmp1)
colnames(dat) <- c(
On May 28, 2010, at 12:30 PM, UM wrote:
hi,
I have been trying to do this in R (have implemented it in Excel)
but I have
been using a very inefficent way (loops etc.). I have matrix A
(columns are
years and ages are rows) and matrix B (columns are birth yrs and
rows are
ages)
I would
Thanks Tal & Joris!
I created my distance matrix distA by using the dist() function in R
manipulating my output in order to get a matrix.
distA =as.matrix(dist(t(x2))) # x2 being my original dataset
as according to the documentaion on dist()
For the default method, a "dist" object, or a matrix (of
My initial guess appears to be right: you're working with something
exceedingly sensitive to floating point precision. You may have to
reconsider your methods.
Your problem is:
rho.f(rho = 0.3)
gives a different answer than
rho.f(seq(0, 1, by=.1)[4])
even though
all.equal(0.3, seq(0, 1, by=.1)[
Hello
It seems that sweave always runs in the global environment. I want to
run sweave from within a function, and pass a variable into sweave,
however when I do this, sweave doesn't see the variable.
Here's my example test_sweave.Rnw file
|%
\documentclass[a4paper]{article}
\usepackage[OT1]{
> **Disclaimer: I have no idea what your data represents or how
(in)appropriate any of these tests may be**
R can do the tests you mentioned (and many more).
Wilcoxon test:
wilcox.test(x=group1, y=group2, paired=FALSE)
see ?wilcox.test
I am not sure whether it is still valid but in case o
I modified my codes. However it looks like it still has the same problem.
Again, rho.f(0.3) gives the right answer. rho.f(corr[4])
gives wrong answer even though corr[4]==0.3.
The codes are attached.
Thank you very much!!!
> rho.f(0.3)
$est.1
[1] 0.000 0.000 0.000 0.000 0.8333
Am 28.05.2010 17:25, schrieb Cedrick W. Johnson:
I now have a bunch of pkgs that aren't loading due to the fact that
they were built before 2.10.0 -- There's some *ancient* packages, like
(sma) that I was able to figure out what we were using, and pull out
the relevant functions and just tempor
Hi,
Can you provide sample data? It seems like in both matrices, you have
ages in the rows. Do you just want to calculate birth years in matrix
B from ages and years in matrix A?
It may also help to give us some of the details of what you are doing
once you have transformed it prior to transfor
Provide a minimal example to start with. This sounds more like voodoo than
anything else.
Cheers
Joris
On Fri, May 28, 2010 at 6:30 PM, UM wrote:
>
> hi,
> I have been trying to do this in R (have implemented it in Excel) but I
> have
> been using a very inefficent way (loops etc.). I have matri
Josh,
Good point. And it actually accidentally solved my problem.
Previously I attached .Rdata files. After your email I exported this
dataset into a csv file and then re-loaded it to see if the problem
persists, and... the problem disappeared.
Nevertheless, if I load the original .Rdata fil
Linda,
There are different views about whether someone doing statistical
analysis should first take a certain number of statistics course. I
think for your issue some background information would certainly help.
You have not correctly interpreted the paper. The main point is that
for most c
On May 28, 2010, at 12:03 PM, Sarah Goslee wrote:
From your code:
mean <- c(rep(mu0, mzero), rep(mu1,m-mzero))
mean() is a function. If you overwrite it with data, you may mess
other things up -
any function you call that calls mean will now fail (at best).
Actually, it's bad but not quite
hi,
I have been trying to do this in R (have implemented it in Excel) but I have
been using a very inefficent way (loops etc.). I have matrix A (columns are
years and ages are rows) and matrix B (columns are birth yrs and rows are
ages)
I would like to first turn matrix A into matrix B
And th
Dear all,
using the hkb estimator obtained from lm.ridge in the below equation
(Formula 6 from this article:
http://www3.interscience.wiley.com/cgi-bin/fulltext/122484280/PDFSTART
beta^hat(k) = ((x'x + kI)^-1)x'y, where
x matrix of independent variables
y vector of dependent variables
k hkb esti
Hi,
It did not return the results you wanted because you tried to feed the
entire data frame to ifelse(). Does something like this do what you
want?
apply(tempr, 2, function(x) {ifelse(x < 0, 0, x)})
Josh
On Fri, May 28, 2010 at 8:37 AM, wrote:
> I have a data frame with both positive and n
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Prof. John C Nash
> Sent: Friday, May 28, 2010 7:46 AM
> Cc: r-help@r-project.org
> Subject: Re: [R] Wait for keystroke or timeout
>
> Matt's suggestion works in Linux (I use Ubu
You might want to ask on R-SIG-Debian
https://stat.ethz.ch/mailman/listinfo/r-sig-debian
Cedrick W. Johnson wrote:
Hi-
Looks like this morning, I did the ultimate in foobar to a main prod
box. I was using apt-get upgrade on the box and totally missed the fact
that my entire R installation we
temp2 = tempr
temp2[temp2<0] = 0
HTH
On Fri, May 28, 2010 at 8:37 AM, wrote:
> I have a data frame with both positive and negative values, and I want to
> make all the negative values equal zero, so i can eventually take an
> average.
> I've tried
> temp2 <- ifelse(tempr<0, 0, tempr)
> but it
This is really a user interface issue and the standard user interface is
different between platforms. Would tcltk (or RGTK or ...) be a possible
solution for you? tcltk is fairly consistent across platforms and does provide
for this type of thing (you can have a button to press to continue and
Do you mean replace values of a column?
> df <- data.frame("Jan" = 1:3,"Feb" = 11:13)
> df
Jan Feb
1 1 11
2 2 12
3 3 13
> df$Jan <- 21:23
> df
Jan Feb
1 21 11
2 22 12
3 23 13
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/Re-help-to-replace-
Does anyone use the R library GenABEL? I am using it to calculate SNP
interactions.
I have a list of 100 SNPs, I need to look at the interaction between each of
two SNPs among the list. my question is how to perform this in GenABEL. I
want to use the "lm" function, but don't know how to use the SN
Dear Terry,
Thanks so much for your help; I'm a bit of an R novice at the moment (as you
can probably tell from my failure to use the data argument!) so any help is
most welcome.
I'm hoping to use this model to generate transition probabilities for a
Markov model and, as such, I was wondering if
I have a data frame with both positive and negative values, and I want
to make all the negative values equal zero, so i can eventually take
an average.
I've tried
temp2 <- ifelse(tempr<0, 0, tempr)
but it doesn't seem to work.
Any suggestions?
Thanks!
___
Hello,
I can't have different data these data came from mice that have lived under
certain condition in the lab! I have just read the mentioned publication
"Should the median test be retired from general use?" It says in the
conclusion "If one felt that the data could not come from a Cauchy or sla
Bingo! Thx Gabor.
Thank you too Tal, I looked briefly at the package and it looks like a nice
interface. I keep it in mind for later.
Cheers
Joris
On Fri, May 28, 2010 at 2:25 PM, Gabor Grothendieck wrote:
> Try this:
>
> as.numeric(gsub("\\D", "", X))
>
> On Fri, May 28, 2010 at 8:21 AM, Jori
I think you tried to start JGR from the console, which will usually not work.
JGR has to be started using the launcher which is available at
http://jgr.markushelbig.org/Download.html
The irmb function will not only need the package installed, but also requires
you to use the JGR console for yo
You mean there is a function "mean" in R, so I should
avoid to use it, right?
so I do the following:
mean1 <- c(rep(mu0, mzero), rep(mu1,m-mzero))
var.f <- function(rho, m, J) {
(1-rho)*diag(m)+rho*J%*%t(J)
}
Thanks!
2010/5/28 Sarah Goslee
> From your code:
>
> mean <- c(rep(mu0, mzero),
On Fri, May 28, 2010 at 6:58 AM, linda Porz wrote:
> Hello,
>
> I can't have different data these data came from mice that have lived under
> certain condition in the lab! I have just read the mentioned publication
> "Should the median test be retired from general use?" It says in the
> conclusion
>From your code:
mean <- c(rep(mu0, mzero), rep(mu1,m-mzero))
mean() is a function. If you overwrite it with data, you may mess
other things up -
any function you call that calls mean will now fail (at best).
var.f <- function(rho) {
(1-rho)*diag(m)+rho*J%*%t(J)
}
var.f() is a complete funct
I am not sure about "overwrite mean() with data". My purpose was
to generate random numbers that are from a multivariate normal
distribution with the mean vector.
For the var.f function, since I already specify m and J, so the only
variable is really rho, so I wrote it as a function of rho only.
Hi-
Looks like this morning, I did the ultimate in foobar to a main prod
box. I was using apt-get upgrade on the box and totally missed the fact
that my entire R installation went from 2.10.0 to 2.11.0.
I now have a bunch of pkgs that aren't loading due to the fact that they
were built befor
Hi,
Two things:
1. You mistakenly posted this to the R-help list, when you should have
(and probably meant to) send to the bioconductor list. You might want
to repost there if you can't figure out the problem.
2. I just tried your query with 4 transcript IDs (one of them was a
duplicate) and it
I executed the following lines several times from a script as well as pasting
them in an R shell.
Systematically biomaRt is failing.
The problem is to extract the 3UTR sequences corresponding to a vector
containing 1941
Ensembl Transcript numbers (some are duplicated ... is this s problem ?)
Ple
On 28/05/2010 10:14 AM, Christopher David Desjardins wrote:
Perfect. Thanks.
Also using R 2.11.0 on Fedora I didn't get any warnings with my command.
That's a serious problem. Can you give more details (i.e. just plain R,
R under ESS, etc.)?
Duncan Murdoch
Chris
On 05/28/2010 09:09 AM
Matt's suggestion works in Linux (I use Ubuntu and Debian variants), but I
haven't yet
been able to get it to work in Windows. In a DOS terminal, I can run Cygwin's
blas.exe via
blas -c "read -t 1 -n 1"
and get the right functioning, but when embedded in R in various ways, I get
several err
On 28/05/2010 9:29 AM, Christopher David Desjardins wrote:
Hi,
I am trying to recreate the right graph on page 524 of Gelman's 2006
paper "Prior distributions for variance parameters in hierarchical
models" in Bayesian Analysis, 3, 515-533. I am only interested, however,
in recreating the port
There are a bunch of problems in your code:
you overwrite mean() with data, and that could screw things up.
you have a function var.f that isn't passed all the arguments it needs.
est.4 is defined several times, each overwriting the previous.
First you need to clean up these sorts of problems sinc
On 28/05/2010 9:24 AM, (Ted Harding) wrote:
An experiment:
sort(c("AACD","A CD"))
# [1] "AACD" "A CD"
sort(c("ABCD","A CD"))
# [1] "ABCD" "A CD"
sort(c("ACCD","A CD"))
# [1] "ACCD" "A CD"
sort(c("ADCD","A CD"))
# [1] "A CD" "ADCD"
sort(c("AECD","A CD"))
# [1] "A CD"
Thanks, this is a good solution
Dr. Iasonas Lamprianou
Assistant Professor (Educational Research and Evaluation)
Department of Education Sciences
European University-Cyprus
P.O. Box 22006
1516 Nicosia
Cyprus
Tel.: +357-22-713178
Fax: +357-22-590539
Honorary Research Fellow
Depar
Thanks very much for your reply. The function is a bit long. I attached it.
rho.f () is in second text document. It uses rada1.mnorm() function which
is contained in the first document.
Thank you very very much!!!
2010/5/28 Dennis Murphy
> Hi:
>
> The problem is that input arguments such as c
G'day Chris,
On Fri, 28 May 2010 08:29:30 -0500
Christopher David Desjardins wrote:
> Hi,
> I am trying to recreate the right graph on page 524 of Gelman's 2006
> paper "Prior distributions for variance parameters in hierarchical
> models" in Bayesian Analysis, 3, 515-533. I am only interested
Perfect. Thanks.
Also using R 2.11.0 on Fedora I didn't get any warnings with my command.
Chris
On 05/28/2010 09:09 AM, Berwin A Turlach wrote:
curve(2*dcauchy(x, location=0, scale=25), from=0, to=200)
__
R-help@r-project.org mailing list
https://st
Thanks that works. I am presuming that the density on the Y-axis would
be wrong in the case of a half-Cauchy distribution and in fact should be
doubled if it's folded at 0?
Chris
On 05/28/2010 09:02 AM, Uwe Ligges wrote:
Am 28.05.2010 15:29, schrieb Christopher David Desjardins:
Hi,
I am tr
Hi:
The problem is that input arguments such as corr[4] have to be evaluated
within the body of your function, and apparently you haven't written it to
do so. Unfortunately, I can't help further because my clairvoyance package
is still in the concept development stage. In the meantime, it would be
Hi Hannah,
No, we can't help because we have no idea what rho.f does - you didn't
provide the requested reproducible example. Without more information,
the only thing I can think of is that your function might be ridiculously
sensive to numeric precision (though that seems unlikely):
> corr <- se
Am 28.05.2010 15:29, schrieb Christopher David Desjardins:
Hi,
I am trying to recreate the right graph on page 524 of Gelman's 2006
paper "Prior distributions for variance parameters in hierarchical
models" in Bayesian Analysis, 3, 515-533. I am only interested, however,
in recreating the porti
Download and install
install.packages("RcmdrPlugin.HH")
library(RcmdrPlugin.HH)
Then there are two options.
The Rcmdr menu item
Statistics > Means > One-way ANOVA...
has a checkbox for pairwise comparison of means"
It uses glht in the multcomp package.
The second option, which I prefer, is to
Hi all,
I have a function rho.f which gives a list of estimators. I have the
following problems.
rho.f(0.3) gives me the right answer. However, if I use rho.f(corr[4]) give
me a different
answer, even though corr[4]==0.3.
This prevents me from using a for loop. Can someone give me some help?
Hi,
I am trying to recreate the right graph on page 524 of Gelman's 2006
paper "Prior distributions for variance parameters in hierarchical
models" in Bayesian Analysis, 3, 515-533. I am only interested, however,
in recreating the portion of the graph for the overlain prior density
for the hal
An experiment:
sort(c("AACD","A CD"))
# [1] "AACD" "A CD"
sort(c("ABCD","A CD"))
# [1] "ABCD" "A CD"
sort(c("ACCD","A CD"))
# [1] "ACCD" "A CD"
sort(c("ADCD","A CD"))
# [1] "A CD" "ADCD"
sort(c("AECD","A CD"))
# [1] "A CD" "AECD"
## (with results for "AFCD", ... "AZC
cobler_squad needs more basic help than doing lda. The data input just
doesn't make sense.
If vowel_feature is a data frame, than G <- vowel_feature[15] creates
another data frame containing the 15th variable in vowel_feature, so "G"
is the name of a data frame, not a variable in a data frame.
Could you provide us with data to test the code? use dput (and limit the
size!)
eg:
dput(vowel_features)
dput(mask_features)
Without this information, it's impossible to say what's going wrong. It
looks like you're doing something wrong in the selection. What should
vowel_features[15] return?
Before polluting your workspace with objects, look at how you might
use a 'list' to collect them all together, especially if you are going
to do processing on them later as a group, or if you want to easily
save/load them. You could do the following:
> myList <- list()
> for (i in 1:10) myList[[p
Hi all,
I was experiencing a similar problem with some code which uses the package
maCorrPlot (BioConductor)
http://www.bioconductor.org/packages/2.6/bioc/html/maCorrPlot.html
to compute the correlation between different variables. This code was
working
apparently fine under R 2.9 (but it was rai
Try this:
as.numeric(gsub("\\D", "", X))
On Fri, May 28, 2010 at 8:21 AM, Joris Meys wrote:
> Dear all,
>
> I have a vector of filenames which begins like this :
> X <- c("OrthoP1_DNA_str.aln", "OrthoP10_DNA_str.aln",
> "OrthoP100_DNA_str.aln",
> "OrthoP101_DNA_str.aln", "OrthoP102_DNA_str.aln",
Thanks Michael,
I don't have matlab.
How might I check this ?
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (
Although you asked for a loop this may be clearer if you just have to
do it once and only have 10 x and 10 y columns. Here we use the built
in anscombe data frame which has columns x1, x2, x3, x4, y1, y2, y3,
y4:
transform(anscombe, r1 = y1 / x1, r2 = y2 / x2, r3 = y3 / x3, r4 = y4 / x4)
In th
See :
http://www.statmethods.net/stats/anova.html
?TukeyHSD
Cheers
Joris
On Fri, May 28, 2010 at 2:11 PM, Iasonas Lamprianou wrote:
> Hi everybody
>
> does anyone know how I can run ANOVA post-hoc tests using R commander or R
> in general?
>
> Thank you
>
>
> Dr. Iasonas Lamprianou
>
>
>
>
> ___
Dear all,
I have a vector of filenames which begins like this :
X <- c("OrthoP1_DNA_str.aln", "OrthoP10_DNA_str.aln",
"OrthoP100_DNA_str.aln",
"OrthoP101_DNA_str.aln", "OrthoP102_DNA_str.aln", "OrthoP103_DNA_str.aln",
"OrthoP104_DNA_str.aln", "OrthoP105_DNA_str.aln", "OrthoP106_DNA_str.aln",
"Orth
On Fri, May 28, 2010 at 12:52 PM, Andre Easom wrote:
> Hi,
>
> I'm a novice R user, much more used to SAS. My problem is pretty simple -
> basically, in a data frame, I have variables named
> x1,,x10 and y1,...,y10; and I would like to create r1 = x1 / y1 etc
>
> Apologies if this is way too
Hi everybody
does anyone know how I can run ANOVA post-hoc tests using R commander or R in
general?
Thank you
Dr. Iasonas Lamprianou
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guid
Sorry, I should have written it an other way:
test$z <- test[[2]]/test[[3]]
Which is then really easy to fit into a for loop
Ivan
Le 5/28/2010 14:05, Ivan Calandra a écrit :
Hi,
Would this do:
test <- data.frame(a=LETTERS[1:10], x=1:10, y=seq(0.1,1,0.1)) #create
some data.frame
test$z <- te
Hi,
Would this do:
test <- data.frame(a=LETTERS[1:10], x=1:10, y=seq(0.1,1,0.1)) #create
some data.frame
test$z <- test$x/test$y #add a column
?
HTH,
Ivan
Le 5/28/2010 13:52, Andre Easom a écrit :
Hi,
I'm a novice R user, much more used to SAS. My problem is pretty simple -
basically, in
Hi Christofer,
I don't know what .Net is doing, but for R these globals are dependent on
your machine and platform.
?.Machine
?.Platform
Don't know if you can actually hack R into believing otherwise.
Did you consider the possibility that the underlying algorithms differ
between .Net and R?
Chee
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