Dear Frank,
Thanks for your comments. But in my situation, I do not
have any future data and I want to calculate Mean Square Error for
prediction on future data. So, is it not it a good idea to go for LOO?
thanks
Alex
On Tue, Feb 24, 2009 at 7:15 PM, Frank E Harrell Jr <
f.har
Hi,
The flexmix package should provide what you need:
http://cran.r-project.org/web/packages/flexmix/index.html
The package has an extensive vignette.
HTH,
Tobias
Wen Gu wrote:
What's the best approach to running latent class analysis with R? I've
downloaded both randomLCA and poLCA packag
KmL, version 0.9.2
KmL is a new implematation of k-means for longitudinal data (or
trajectories). This algorithm is able to deal with missing value and
provides an easy way to re roll the algorithm several times, varying
the starting conditions and/or the number of clusters looked for. It
als
Argh! The second (concise) version should have |, not & !!!
-s
On 2/24/09, Stavros Macrakis wrote:
> "L'esprit de l'escalier" strikes again
>
> An even simpler statement of your original problem:
>
> Find the factors that A and B have in common.
>
> If A and B are fairly sma
Hi,
I am using rpart package to fit classification trees.
library(rpart)
fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
plot(fit,uniform=T)
text(fit, use.n=TRUE)
But I am unable to label the branches (not the nodes) of the tree. Can somebody
help me out in this?
T
Frank E Harrell Jr vanderbilt.edu> writes:
... Word and pdf
> It depends on how you copy. By all means use Insert ... Picture ...
> from file and directly insert pdf.
Please, tell me how you got this to work.
Dieter
__
R-help@r-project.org maili
What's the best approach to running latent class analysis with R? I've
downloaded both randomLCA and poLCA packages, but I am interesting in running a
standard LCA with individual records (not frequency table) as input data.
Wen Gu
John Jay College of Criminal Justice445 West 59 StreetNew York
Another way is:
> subset(df, select=c(var.b, var.c))
though I'd be willing to bet that using %in% is probably faster.[1]
Tony
[1] Unfortuantly I'm skint :-(
On 24 Feb, 20:10, Sean Zhang wrote:
> Dear R-helpers:
>
> I am an R novice and would appreciate answer to the following question.
>
> Wa
Hi list,
I don't know if somebody has spent a lot of time debugging strange
problems with if else positioning - the parser seems to recognize only
the syntax bellow - this is the only way of making these pieces of
code to work.
As far as i'm concerned, no examples were available (it would be so
a
I don't know if there is a direct, perl-like way to capture the matches, but
here is a solution:
> mdat <- gregexpr("[[:digit:]]{8}", txt)
> dates <- mapply(function(x, y) substr(txt, x, x + y - 1), mdat[[1]],
attr(mdat[[1]], "match.length"))
> dat
Hi Ira:
For your first question, under the hood of R, names<- is actually a
function so , when you do that, you need to say names(a)[2] rather
than names(a[2]). why this is is tricky and I wouldn't do it justice if
i tried to explain it. it's best if you do ?"names<-" at an R prompt and
read t
On Feb 25, 2009, at 12:12 AM, David Winsemius wrote:
On Feb 24, 2009, at 11:36 PM, Fuchs Ira wrote:
also unrelated: if I have two vectors and I want to combine them
to form a matrix ,is cbind (or rbind) the most direct way to do this?
e.g.
x=c(1,2,3)
y=c(3,4,5)
z=rbind(x,y)
Tha
On Feb 24, 2009, at 11:36 PM, Fuchs Ira wrote:
I was wondering why the following doesn't work:
> a=c(1,2)
> names(a)=c("one","two")
> a
one two
1 2
>
> names(a[2])
[1] "two"
>
> names(a[2])="too"
> names(a)
[1] "one" "two"
> a
one two
1 2
I must not be understanding some basic concept h
Try this:
library(gsubfn)
strapply("blah blah start=20080101 end=20090224", "start=(\\d{8})
end=(\\d{8})", c, perl = TRUE)[[1]]
or perhaps just:
strapply("blah blah start=20080101 end=20090224", "\\d{8}", perl = TRUE)[[1]]
On Tue, Feb 24, 2009
"L'esprit de l'escalier" strikes again
An even simpler statement of your original problem:
Find the factors that A and B have in common.
If A and B are fairly small (< 1e7, say), a very direct approach is:
which( ! (A %% 1:min(A,B)) & !(B %% 1:min(A,B)) )
Is this "brute forc
I was wondering why the following doesn't work:
> a=c(1,2)
> names(a)=c("one","two")
> a
one two
1 2
>
> names(a[2])
[1] "two"
>
> names(a[2])="too"
> names(a)
[1] "one" "two"
> a
one two
1 2
I must not be understanding some basic concept here.
Why doesn't the 2nd name change to "too"?
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
bbbnc [benwo...@gmail.com]
Sent: Wednesday, February 25, 2009 6:01 AM
To: r-help@r-project.org
Subject: [R] Simulating contingency table (Basic question, help please)
I'd like
> txt <- "blah blah start=20080101 end=20090224"
> nums <- sub(".*start=(\\d+).*end=(\\d+).*", "\\1 \\2", txt, perl=TRUE)
> nums <- strsplit(sub(".*start=(\\d+).*end=(\\d+).*", "\\1 \\2", txt,
> perl=TRUE), ' ')
I have written a short script to estimate the size of a test of non-constant
mean in an AR(1) time series. When I run the simulation on my PC (R version
2.7.1), I get the expected result: an empirical size much larger than the
nominal size. On the Red Hat machine (R version 2.7.2) in my departmen
Hello,
Newbie question: how do you capture groups in a regexp in R?
Let's say I have txt="blah blah start=20080101 end=20090224".
I'd like to get the two dates start and end.
In Perl, one would say:
my ($start,$end) = ($txt =~ /start=(\d{8}).*end=(\d{8})/);
I've
Michael Friendly wrote:
For the following model,
library(vcd)
arth.polr <- polr(Improved ~ Sex + Treatment + Age, data=Arthritis)
summary(arth.polr)
where Improved is an ordered, 3-level response I'm looking for a
*simple* way to test
the validity of the proportional odds assumption, typically
For the following model,
library(vcd)
arth.polr <- polr(Improved ~ Sex + Treatment + Age, data=Arthritis)
summary(arth.polr)
where Improved is an ordered, 3-level response I'm looking for a
*simple* way to test
the validity of the proportional odds assumption, typically done via a
score test
On Mon, Feb 23, 2009 at 9:52 PM, Fox, Gordon wrote:
> This is a seemingly simple problem - hopefully someone can help.
> Problem: we have two integers. We want (1) all the common factors, and
> (2) all the possible products of these factors. We know how to get (1),
> but can't figure out a general
Colleagues,
sorry for the simple question, but I'm novice in using R.
How can I print all needed statistics for models results while using
leaps package?
I want to see R squared and significance of each model and also B &
significance for coefficients.
Also I can't understand what data I can see wi
jimdare wrote:
Dear R Users,
I have three questions.
1) Is there a way to get the output tex file to include
\documentclass{report}
\begin{document}
and
\end{document}
so I can generate a PDF straight away. (I am trying to generate hundreds of
these and don't want to have to manually
OK, the one thing I figured out:
Is should be like:
biplot(test.pca, cex=c(2,1), col=c("red","green")...
to change size, colours etc separately. But I still don't know how
change lables of observations to symbols properly.
Tipps? Thanks again,
Axel
Dear R helpers,
When producing a PCA biplot,
Stavros Macrakis wrote:
> On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
>
>> ...Want to delete many variables in a dataframe
>> df<-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
>> df[,'var.a']<-NULL #this works for one single variable
>> df[,c('var.a','var.b')]<-NULL
Well, yes to all -- that's why I pointed this out. Sorting is (depending on
details) >= O(n log(n)) while the quantile algorithm uses R's partial sort
option, which I think would be O(n) for a single quantile. For n of a few
thousand or so, the difference shouldn't be noticeable. For large n, of
co
On Tue, Feb 24, 2009 at 3:01 PM, Bert Gunter wrote:
> Nothing wrong with prior suggestions, but strictly speaking, (fully) sorting
> the vector is unnecessary.
>
> y[y > quantile(y, 1- p/length(y))]
>
> will do it without the (complete) sort. (But sorting is so efficient anyway,
> I don't think yo
You're right. The help page is somewhat misleading at first read.
?`[<-.data.frame` states that (with added emphasis)
value A suitable replacement value: it will be repeated a whole number of
times if necessary and it may be coerced: see the Coercion section. *** If
NULL, deletes the column if
Dear R helpers,
When producing a PCA biplot, vectors of environmental variables (as red
arrows with labels) and scores of the observations (black labels
(observation names)) are plotted by default. How can I change the
graphical output? Let's say I would like that the scores are plottet
only
On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
> ...Want to delete many variables in a dataframe
> df<-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
> df[,'var.a']<-NULL #this works for one single variable
> df[,c('var.a','var.b')]<-NULL #does not work for multiple variab
Dear R Users,
I have three questions.
1) Is there a way to get the output tex file to include
\documentclass{report}
\begin{document}
and
\end{document}
so I can generate a PDF straight away. (I am trying to generate hundreds of
these and don't want to have to manually type this in ever
Manuel Morales wrote:
Hi list,
I'd like to install an archived version of lmer to compare results with
the current version. I guess one way to do this would be to download the
source, rename the package and then install it. Is there a better
alternative?
Yes: Install into a different librar
Hi list,
I'd like to install an archived version of lmer to compare results with
the current version. I guess one way to do this would be to download the
source, rename the package and then install it. Is there a better
alternative?
--
http://mutualism.williams.edu
signature.asc
Description: T
Setting to NULL works only if a single column is selected.
More generally,
df[, !(colnames(df) %in% c("var.b", "var.c")), drop=FALSE]
-Christos
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Sean Zhang
> Sent: Tuesday, Fe
One way is to use %in% like below:
df<-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
print(df)
df <- df[,!(names(df) %in% c('var.a','var.b')),drop=FALSE]
print(df)
On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
Dear R-helpers:
I am an R novice and would appreciate answer
Dear R-helpers:
I am an R novice and would appreciate answer to the following question.
Want to delete many variables in a dataframe.
Am able to delete one variable by assigning it as NULL
Have a large number of variables and would like to delete them without using
a for loop.
Is there a command
Nothing wrong with prior suggestions, but strictly speaking, (fully) sorting
the vector is unnecessary.
y[y > quantile(y, 1- p/length(y))]
will do it without the (complete) sort. (But sorting is so efficient anyway,
I don't think you could notice any difference).
-- Bert Gunter
Genentech
-
On Tue, 24 Feb 2009, bbbnc wrote:
I'd like to carry out a Monte Carlo simulation test where given data is a
contingency table. I think this is something to do with using rmultinonom(),
but I'm not sure how to code this, to simulate contingency tables. Could
anyone please help with how to use R t
Thanks for all
Peterko wrote:
>
> Hi, may be simle question, but a do not find it anywhere.
> Is there same function like max() ,but giving more results.
> max() give 1number-maximum
> I need funcion what give p bigest number.
> many thanks
>
--
View this message in context:
http://www.nabb
There's probably something built in to R but you can change the values
of the percentiles (p) below to get the value that corresponds to it.
rounding might be problematic also.
temp <- c(1,4,8,3,5)
p <- 0.8
temp[order(temp)][round(length(temp)*p)]
On Tue, Feb 24, 2009 at 2:36 PM, Peterko
Dear Peter,
Perhaps:
set.seed(1) # For reproducibility
p<-2 # Two biggest values of x
x<-rnorm(10)
tail(sort(x),p)
[1] 0.7383247 1.5952808
In a function mode,
foo<-function(x,p=2) tail(sort(x),p)
foo(x,p=3)
[1] 0.5757814 0.7383247 1.5952808
HTH,
Jorge
On Tue, Feb 24, 2009 at 2:36
On Tue, 24 Feb 2009 11:36:06 -0800 (PST) (SO, 55 Chs 3175 YOLD)
Peterko wrote:
>
> Hi, may be simle question, but a do not find it anywhere.
> Is there same function like max() ,but giving more results.
> max() give 1number-maximum
> I need funcion what give p bigest number.
> many thanks
How
Try ?relevel
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of choonhong ang
Sent: Tuesday, February 24, 2009 2:33 PM
To: r-help@r-project.org
Subject: Re: [R] Insurance data in library(MASS)
Hi,
In the result shown, the District
Hi, may be simle question, but a do not find it anywhere.
Is there same function like max() ,but giving more results.
max() give 1number-maximum
I need funcion what give p bigest number.
many thanks
--
View this message in context:
http://www.nabble.com/bigest-part-of-vector-tp22188901p22188901
Hi,
In the result shown, the District 1 is used as the base category. How to
change to make District 4 as a base category ?
On Mon, Feb 23, 2009 at 11:05 AM, choonhong ang wrote:
> I have used the insurance data from R library and I have 2 questions:
> I use the following:
> >library(MASS)
> >d
Thank you Greg! Yes this is the comparison I was looking for.
>>> Greg Snow 2009/02/24 06:07 PM >>>
The same as what? It is not clear what you are trying to compare.
Is this the comparison that you are looking for?:
> A <- scale(stackloss)
>
> fit1 <- lm( stack.loss ~ . - 1, data=as.data.fram
One more followup on this: I just added a function shapelist3d() to rgl
(so far only on R-forge, not CRAN) that automates a lot of this. To get
the plot as below, the following code works:
x <- rep(1:5, each=25)
y <- rep(rep(1:5, each=5), 5)
z <- rep(1:5, 25)
shapelist3d(cube3d(), x,y,z, size
Dear all,
Here is one more way to go though using rep() and then matrix():
> rows <- 1:3
> matrix(rep(rows,5),ncol=5)
[,1] [,2] [,3] [,4] [,5]
[1,]11111
[2,]22222
[3,]33333
HTH,
Jorge
On Tue, Feb 24, 2009 at 1:43 PM, Gabor Grothe
Suppose we want 3 rows and the ith row should have 5 columns of i.
Create a list whose ith component is the ith row and rbind them:
> rows <- 1:3
> do.call(rbind, lapply(rows, rep, 5))
[,1] [,2] [,3] [,4] [,5]
[1,]11111
[2,]22222
[3,]333
>
> > And, since my son asked me and I am basketball ignorant: Why are
> > basketball scores mostly much too close to equality? The arguments
> > (loose power when leading)
manchester.ac.uk> writes:
> Or: Once you are in the lead, become much more defensive against
> attacking play by the
On 24-Feb-09 17:25:53, Dieter Menne wrote:
> Tony Breyal googlemail.com> writes:
>
>>Cheers for that information; I've just registered for the useR meeting
>>in London and then about 10 minutes later got that same bit of spam
>>too which made me a wee bit suspicious.
>
> Welcome in the Fooled by
Hi Alex,
Give a look at:
http://search.r-project.org/cgi-bin/namazu.cgi?query=leave+one+out&max=20&result=normal&sort=score&idxname=Rhelp02a&idxname=functions&idxname=docs
Cheers
miltinho astronauta
brazil
On Tue, Feb 24, 2009 at 3:07 PM, Alex Roy wrote:
> Dear R user,
>
At 15:24 23/02/2009, Jessica L Hite/hitejl/O/VCU wrote:
THANKS so very much for your help (previous and future!). I have a two
follow-up questions.
1) You say that dispersion = 1 by definition dispersion changes from 1
to 13.5 when I go from binomial to quasibinomialdoes this suggest th
MarcioRibeiro wrote:
Hi Listers,
Is there a way that I can transpose an array...
Suppose I have the following array...
x<-array(c(1,2,3,4),dim=c(1,2,2))
, , 1
[,1] [,2]
[1,]12
, , 2
[,1] [,2]
[1,]34
And I would like to get the following result...
, , 1
[,1]
[1,]
Alex Roy wrote:
Dear R user,
I am working with LOO. Can any one who is working
with leave one out cross validation (LOO) could send me the code?
Thanks in advance
Alex
I don't think that LOO adequately penalizes for model uncertainty. I
recommend the bootstrap or 50
If you know the final size that your matrix will be, it is better to
preallocate the matrix, then insert the rows into the matrix:
mymat <- matrix( nrow=100, ncol=10 )
for( i in 1:100 ){
mymat[i, ] <- rnorm(10)
}
Even better than this is to use replicate or sapply if you can, they will t
Dear R user,
I am working with LOO. Can any one who is working
with leave one out cross validation (LOO) could send me the code?
Thanks in advance
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Try aperm(x, c(2,1,3))
--
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 MarcioRibeiro
> Sent: Tuesday, Februar
Here is a simple example (finding the parameters of a normal) that shows one
way to do it:
x <- rnorm(25, 100, 5)
tmpfun <- function(x) {
steps <- matrix( 0:1, nrow=1 )
myfn <- function(param) {
print(param)
I'm growing a large dataframe by composing new rows and then doing
row <- compute.new.row.somehow(...)
d <- rbind(d,row)
Is this a fast/preferred way?
Cheers,
Alexy
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
Dear all,
I'm using the samr-package to identify significantly differentially expressed
genes in microarray data.
So far, I had no problems, but when I used a large multiclass data set with 327
samples, I obtained the following error/warning message:
Warning message:
Inf
In factorial(length(y))
I'd like to carry out a Monte Carlo simulation test where given data is a
contingency table. I think this is something to do with using rmultinonom(),
but I'm not sure how to code this, to simulate contingency tables. Could
anyone please help with how to use R to simulate contingency tables like
t
Hi Listers,
Is there a way that I can transpose an array...
Suppose I have the following array...
x<-array(c(1,2,3,4),dim=c(1,2,2))
, , 1
[,1] [,2]
[1,]12
, , 2
[,1] [,2]
[1,]34
And I would like to get the following result...
, , 1
[,1]
[1,]1
[,2]2
, , 2
Hi again,
Looking more into test statistics i realized that maybe i can use the
power.prop.test to see if the difference between the 2 accuracies are zero or
not. Do you have any comments about that? Also, should i considered kappa
statistics also a kind of proportion and use the same test? I
Tony Breyal googlemail.com> writes:
>Cheers for that information; I've just registered for the useR meeting
>in London and then about 10 minutes later got that same bit of spam
>too which made me a wee bit suspicious.
Welcome in the Fooled by Randomness society. 2:0 is a bit away from
significa
Rob Steele suggested the same thing but I'm not sure I understand how to
implement this exactly. Is there any documentation that you could suggest?
This might be something that could be useful for the future.
Thanks,
Shimrit
On Tue, Feb 24, 2009 at 5:09 PM, Greg Snow wrote:
> It looks like you
I also received the spam, and have not registered for the conference.
I decided to do the noble experiment, and used their web interface to
unsubscribe to the newsletter to which they claim I had subscribed,
and for good measure added my name to their "do not contact" list.
albyn
On Tue, Feb 24,
The add1 function might be what you want, there is also addterm in the MASS
package and the leaps package can do some things along this line (plus more).
But before doing this, you may want to ask yourself what question you are
really trying to answer, then explore if this answers that question
Dear Gordon,
Try also,
> unique(apply(expand.grid(commfac,commfac),1,prod))
[1] 1 2 5 4 10 25
HTH,
Jorge
On Mon, Feb 23, 2009 at 9:52 PM, Fox, Gordon wrote:
> This is a seemingly simple problem - hopefully someone can help.
> Problem: we have two integers. We want (1) all the common fact
All
Mango Solutions are pleased to announce the first meeting of the London
useR Group.
DateTuesday 31st March
Time4pm to 7pm
Venue The Wall
45 Old Broad St
London
EC2N 1HU
Tel 020 7588 4845
S
Irina Ursachi itwm.fraunhofer.de> writes:
>
> Dear all,
>
> When using odfWeave, I get the following error:
>
> odfWeave(file.in,file.out)
> Copying Example2.odt
> Setting wd to /tmp/Rtmp8ekeDC/odfWeave24145003519
> Unzipping ODF file using unzip -o Example2.odt
> Archive: Example2.od
R-helpers,
A quick question regarding my wanting to run multiple regressions without
writing a loop.
Looking at a previous discussion :
http://tolstoy.newcastle.edu.au/R/e2/help/07/02/9740.html
my objective is to do the "opposite", i.e. instead of having the same
independent variable and testi
Fredrik Karlsson gmail.com> writes:
>
> Dear list,
>
> Sorry for bothering you with a pure odfSweave question, but I just ran
> into a problem that I cannot find the cause of.
> Anyonse seen this before? This file "used to work", but not anymore.
>
> Would apreciate all the help I could get.
>
It looks like you found a solution, but if you find yourself in this situation
again using optim, then one approach is to modify your function that you are
optimizing (or write a wrapper for it) to produce the tracing information for
you.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
On Tue, Feb 24, 2009 at 10:56 AM, Zheng, Xin (NIH) [C]
wrote:
> Hi there,
>
> I wanted to search some packages related with 'genetics'. I know I can do it
> on CRAN webpage. I'm just wondering if there's some function in R could do
> that. Debian apt offers one way. It would be perfect if R has
What did the data appear like when it was read in? Have you just
tried to read in the lines (readLines) to see if the decompression is
working? Does this compare to what you get if you decompress the file
outside of R? Not exactly sure what you mean by "force" since the
command it probably readi
Hi,
I'm looking for any information on calculating and using Shapley Values
in a TURF context. A paper was presented at an S-Plus conference about
this:
Conklin M., Lipovetsky S. Modern Marketing Research Combinatorial
Computations:
Shapley Value versus TURF Tools, Proceedings of 1998 Inter
> I am estimating the following coxph function with stratification and
> frailty, where each person had multiple events.
> m <- coxph(Surv(dtime1,status1) ~ gender +cage +uplf+ strata(enum)+
> frailty(id), xmodel)
>I want to predict the cumulative hazard for each person for their
Hi everyone,
I would like to test for the statistical significance(for what it worth ...) in
increasing classification accuracy and kappa statistics from different land
classifications. The classifications were done using other software (like
eCognition and See5), but the results were "sampled
Thanks Wacek. Sorry that I didn't clarify my question. I want to search
uninstalled packaged in repository. I just found RSitesearch could do that and
open the results in system browser. It's not flexible, although it's better
than none.
Xin
-Original Message-
From: Wacek Kusnierczyk [
Zheng, Xin (NIH) [C] wrote:
> Hi there,
>
> I wanted to search some packages related with 'genetics'. I know I can do it
> on CRAN webpage. I'm just wondering if there's some function in R could do
> that. Debian apt offers one way. It would be perfect if R has some builtin
> function like Debia
one way is:
mat <- matrix(rnorm(2500*12), 2500, 12)
ind <- rowSums(mat > 0) > 6
mat[ind, ]
I hope it helps.
Best,
Dimitris
A Ezhil wrote:
Dear All,
I have matrix of size 2500 x 12. I would like to select all the rows if 6 out 12 values are > 0. How can I do that in R?
Thanks in advance
The same as what? It is not clear what you are trying to compare.
Is this the comparison that you are looking for?:
> A <- scale(stackloss)
>
> fit1 <- lm( stack.loss ~ . - 1, data=as.data.frame(A))
> tmp1 <- fitted(fit1)
>
> fit2 <- svd(A[,-4])
> tmp2 <- fit2$u %*% t(fit2$u) %*% A[,4]
>
> al
Hi,
If you want to see any result by gene, you should apply the function by genes,
not by samples. As I understood, you are applying the test with the columns
Samplea and SamplVehicle, but what you need is a result per row (per gene).
Then you will see wich of the genes are significant.
Reg
Dear All,
I have matrix of size 2500 x 12. I would like to select all the rows if 6 out
12 values are > 0. How can I do that in R?
Thanks in advance.
Kind regards,
Ezhil
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https://stat.ethz.ch/mailman/listinfo/r-he
Hi there,
I wanted to search some packages related with 'genetics'. I know I can do it on
CRAN webpage. I'm just wondering if there's some function in R could do that.
Debian apt offers one way. It would be perfect if R has some builtin function
like Debian apt.
Appreciate any help.
Xin Zheng
Dear Tania,
Shouldn't that be varIdent(form = ~species.group) instead of
varIdent(form = ~1|species.group)?
Notice that this is untested as you did not provide a self-contained
example (as de posting guide asked you to do).
HTH,
Thierry
PS Next time try the mixed-models mailing list with quest
Kirk Wythers wrote:
How does one control the size and type of data symbols in pairs()? I am
trying to use the little dot (as in type=".") with absolutely no success.
I guess you mean pch="." and hence
pairs(data.frame, panel=panel.smooth, pch=".")
Uwe Ligges
Here is the pairs call I am
On Tue, Feb 24, 2009 at 9:58 AM, Martin Maechler
wrote:
>> "TL" == Thomas Lumley
>> on Tue, 24 Feb 2009 05:39:33 -0800 (PST) writes:
>
> TL> The same company caused a complaint about a year ago
> TL> https://stat.ethz.ch/pipermail/r-help/2008-March/157423.html
>
> thanks, Thomas
I am using R to access .mdb files (created in Microsoft access 2003) through
RODBC.
I am able to view the Tables and also list their attributes through the
sql. commands. However, when try to acces the primary keys (using
sqlPrimaryKeys /odbcPrimaryKeys) I get a value of '-1' on m
Cheers for that information; I've just registered for the useR meeting
in London and then about 10 minutes later got that same bit of spam
too which made me a wee bit suspicious.
On 24 Feb, 13:39, Thomas Lumley wrote:
> The same company caused a complaint about a year
> agohttps://stat.ethz.ch/p
Hi,
I am running R2.8.1 under Linux, and I am having trouble using the
variance functions in nlme
My basic model was something like:
model0 <- lme( log(growth) ~ light * species.group , data=data,
random=~light|species ) # with 20 odd species divided in 2 groups
Following the methods in Pinh
Yes, initially, it didn't work and thanks to one of the examples in the help
file, I found out that I need to set maximize = T...but thanks for your
suggestion anyway.
I mainly work with state space models and I'm currently dealing with a case
where the estimation time is halved (!!!) by spg().
Sh
How does one control the size and type of data symbols in pairs()? I
am trying to use the little dot (as in type=".") with absolutely no
success.
Here is the pairs call I am using:
pairs(data.frame, panel=function(x,y) {points(x,y); lines(lowess(x,y))})
or even simpler:
pairs(data.frame, pan
Hi I have managed to do a paired t-test with a data set
i have 5 colums of data im dealing with
GENE SampA SampB SampC SampVehicle
ctcc 859 na145 24
gtcg 45 5 54 69
and so on but they are much larger col
Hi Shimrit,
Make sure that you set maximize=TRUE in the control settings (since you have
fnscale = -1 in your optim() call).
A nice feature of spg() is that the entire code is in R, and can be readily
seen by just typing the function name at the R prompt. On smaller problems
(with only a few p
Hi Ravi,
Thanks for your great suggestion, it does exactly what I need as it provides
more insight into what is going on in the 'black box'. In addition, it's
much faster than optim(). I will use this function in the future.
Kind Regards,
Shimrit
On Tue, Feb 24, 2009 at 2:33 PM, Ravi Varadhan
Dear list,
Sorry for bothering you with a pure odfSweave question, but I just ran
into a problem that I cannot find the cause of.
Anyonse seen this before? This file "used to work", but not anymore.
Would apreciate all the help I could get.
/Fredrik
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