Hi,
Is there any package in R which i can use for Coalescent Theory/Genelogical
Trees?
Is this possible to use R function in C language?
I would be very thankful to you.
Regards,
++
MUHAMMAD FAISAL Department of Statistics and Decion Support system,
Ted Harding; you are my HERO!
THANK YOU!
Ted.Harding-2 wrote:
>
> On 10-Jul-08 20:43:12, DaveFrisch wrote:
>>
>> Okay, so I'm fairly retarded, and asked a question about finding
>> the T-Value in the Fisher Exact method. I suppose what I'm truly
>> after can best be explained by the Biddle Co
Dear Sirs,
I would like to kindly ask for your assistance on the folllowing issue.
After uploading the most updated the versions of some libraries (namely:
survival and nlme) I became unable to open R. Every time I try to open R
a dialogue box pops up with the following message:
-
Geez guys,
I lost count of the replies to this one. It's probably just the "round
to the even digit on a 5" kludge mentioned in the Details of "round".
Jim
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PLEASE do read
This is great! Thank you very much Gabor.
rcoder
Gabor Grothendieck wrote:
>
> See ?lag and in zoo ?lag.zoo. Both pages have
> examples. Using lag.zoo here it is with your data:
>
> Lines <- "Date Apples Oranges Pears
> 1/7 2 35
> 2/7 1 4
Thank you a lot,
I am almost done, but unfortunately I have to manipulate values like
x
220a1
220ab1
220a12
to
y
220
220
220
Eventhough it is easy to macht a 3-digit number
[0-9]{3}
I habe no idea how to mach everything except a 3-digit number in order to
replace everything but the 3-digit
I have to test the hypothesis that a multivariate variable X = (X_1, X_2,
..., X_n) is a multivariate normal distribution with mean M = (M_1, ...,
M_n) and variance V = (V_1, ... , V_n) that I know. Obviously I have also
the sample-vector m and v.
I do not know the covariances of the population and
try something like this:
x<-c("220a1", "220ab1", "220a12", "a34dkaffdse223", "abc123")
sub(".*([[:digit:]]{3}).*", "\\1", x)
( ?regexp is also useful)
kk
On Fri, Jul 11, 2008 at 12:04 PM, Kunzler, Andreas <[EMAIL PROTECTED]>
wrote:
> Thank you a lot,
>
> I am almost done, but unfortunately I h
Dear R users,
I have been having a problem since I installed the new versions of TinnR and
R on my computer (1.19.4.7 and 2.6.0 respectively). I used to open R by the
command "start preferred RGUI", on the "R" tab of TinnR's menu. But now,
since I updated both softwares (which was necessary for s
Just one additional item. Since I posted, an optional
na.pad= argument has been added to lag.zooreg in the
devel version of zoo with which the code reduces to:
# next line grabs lag.zooreg from devel version of zoo
source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/R/w
Hi,
See
https://sourceforge.net/forum/forum.php?thread_id=1741502&forum_id=481901
This is a bug in the current version of Tinn-R (1.19.4.7). As far as I
know version 1.19.5.0 hasn't been released yet. The workarounds suggested by
JCFaria and Dean don't work for me.
I still have to reset the
I haven't used TinnR but in batchfiles
http://batchfiles.googlecode.com
the Rversions.hta script will bring up a dialog showing you
which versions of R you have, according to the registry,
and it will let you set the default version of R by setting the registry
appropriately. If TinnR uses the reg
strapply() in gsubfn is convenient for that since it matches by contents
rather than delimiters:
x <- factor(c("220", "220a", "221b", "B221", "220a1", "220ab1", "220a12"))
library(gsubfn)
strapply(as.character(x), "[0-9]{3}", simplify = c)
See
http://gsubfn.googlecode.com
On Fri, Jul 11, 2008 a
Does anyone know how to draw funnel plots in R with control limits?
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__
R-help@r-project.org mail
Re-installing R did the trick for me a while ago, don't know if this still
works with the current Tinn-R version.
Ptit_Bleu wrote:
>
> Hello Marcia,
>
> I had a similar problem when I updated Tinn-R.
> Under XP, in the file Document and Settings\... \Application
> Data\Tinn-R\ini\Tinn.ini,
>
Thank you for taking the time.
I get your point about presumptions.
If you issue the call my.plot( x, y, main = main.str ), it works fine. The
problem turns up only when you use a formula as an argument, hence the
suspicion that the behavior was unintended.
--- On Thu, 7/10/08, Bert Gunter <
Hello Marcia,
I had a similar problem when I updated Tinn-R.
Under XP, in the file Document and Settings\... \Application
Data\Tinn-R\ini\Tinn.ini,
I replaced
sPathRGui=\bin\Rgui.exe
by
rguiPreferred=C:\R\bin\Rgui.exe (to adjust to your path)
It worked for me. I hope it will work for you.
Have
How do you draw 1 and 3 sigma control limits on a funnel plot in R?
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Hello
I am a relatively new user of R and am struggling to use the 'network' package.
I have a correlation matrix (produced using 'cor'), and want to draw a network
where each item showing correlation above a threshold (say 0.5) is joined by a
green line, and each item showing correlation belo
A workaround is to use the title command instead:
my.plot(y ~ x, tdf)
title(main.str)
On Fri, Jul 11, 2008 at 8:43 AM, Amit Ganatra <[EMAIL PROTECTED]> wrote:
> Thank you for taking the time.
> I get your point about presumptions.
> If you issue the call my.plot( x, y, main = main.str ), it work
I'm sure this is possible with 'network', but i'm not very familiar
with that package. In case you don't get an answer on how
to do it with network, here is how to do it with the 'igraph' package:
library(igraph)
M <- matrix(runif(100)*2-1, 10, 10)
M[ lower.tri(M, diag=TRUE) ] <- 0
M[ abs(M) <
Hi All,
This didn't work for me :-(
[Tinn-R 1.19.5.7, R 2.7.0, Windows Server 2003, Standard x64 Edition]
When I edited Tinn.ini to have the right path it worked the first time I
startedd Tinn-R, but when I closed Tinn-R it replaced Tinn.ini with a new
file without my edits, so the next start o
Jonathan,
please stay on the list.
The first query i don't understand, can you send a pdf or explain
a bit what exactly happens.
As for the names, use them as row or column names in the matrix
(see ?colnames), and then do
V(g)$label <- V(g)$name
before the plotting.
G.
On Fri, Jul 11, 2
Hello,
I am wondering if a modified KS (Kolmogrov-Smirnov) test is availabe in R,
which can calculate the correct p value with ties for the data that have
discrete distribution.
Joshua
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Thnaks
Ignore the forst problem - seems to solve itself!
Trying 'V(g)$label <- V(g)$name' I recieve the following error:
Error in `V<-`(`*tmp*`, value = 0:9) : invalid indexing
The full code used (with matrix shown) is below - any help appreciated:
> data2
G1 G2 G3 G4 G5
Thank you. That does work for the title.
My intention for the program I am writing was to set some parameters with
default values that I wanted. Then each call would set other parameters as
required.
It all works fine apart from the calls where I pass a formula. It probably has
to do with the
Hmmm, it seems that the colum/row names are not added properly,
maybe this works only in the not-yet-released version.
You can do
V(gdata2)$label <- colnames(data2)
as a workaround.
G.
On Fri, Jul 11, 2008 at 03:32:15PM +0100, Dry, Jonathan R wrote:
> Thnaks
>
> Ignore the forst problem - s
You could replace the one line in my.plot with this:
eval.parent(substitute(plot( x, y, cex.axis=0.5, ...)))
On Fri, Jul 11, 2008 at 10:41 AM, Amit Ganatra <[EMAIL PROTECTED]> wrote:
>
> Thank you. That does work for the title.
> My intention for the program I am writing was to set some paramete
Hi,
I am I trying to learn mixture models of gamma distributions.
However sometime the calculation of the log-likelihood it gives is
positive instead of the usual negative, in particular when
the number of samples are small.
My questions are:
1. Is it possible that a loglikelihood is positive?
2.
Gundala Viswanath wrote:
Hi,
I am I trying to learn mixture models of gamma distributions.
However sometime the calculation of the log-likelihood it gives is
positive instead of the usual negative, in particular when
the number of samples are small.
My questions are:
1. Is it possible that a lo
Hi
Is it possible to subset an n-dimensional array by a vector of n dimensions?
E.g. assume I have
x <- array(1:24, dim=2:4)
x[1,1,2]
[1] 7
dims <- c(1,1,2)
I would like a function that I can supply x and dims as parameters to, and have
it return 7. Also, I would like to do something like
Hello,
After usefull suggestions by Paul Murrell, i have been trying to use
my.symbols to plot arrows of varying angles on my plot in order to create
a time series of wind direction... however,i have been unable to figure
out how the allignment of symbols works...
below i have included a simplifi
Dear R People:
I'm running Rmpi on a single machine and I have the following
statement from the command line:
mpirun -np 3 ./R --no-save < eek1.in >stuff4.out
The stuff4.out file only contains the third result. Is there a way to
fix this such that it shows all 3 sets, please
Thanks in advance
Hi everyone,
Is there a way to take an lm() model and strip it to a minimal form (or
convert it to another type of object) that can still used to predict the
dependent variable?
Background: I have a series of 6 lm() models, each of which are being
run on the same data frame of approximatel
Is there an easy way to compare complex numbers?
Here is a small example:
> (z1=polyroot(c(1,-.4,-.45)))
[1] 1.11-0i -2.00+0i
> (z2=polyroot(c(1,1,.25)))
[1] -2+0i -2+0i
> x=0
> if(any(identical(z1,z2))) x=99
>x
[1] 0
# real and imaginary parts:
>Re(z1); Im(z1)
[1]
Perfect, thank you.
--- On Fri, 7/11/08, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> From: Gabor Grothendieck <[EMAIL PROTECTED]>
> Subject: Re: [R] Ellipsis arguments for plot.formula
> To: [EMAIL PROTECTED]
> Cc: r-help@r-project.org
> Date: Friday, July 11, 2008, 9:45 AM
> You could rep
I think you are looking for subscripting with a
matrix:
x[cbind(1,1,2)]
See, for instance, the subscripting section of chapter 1
of S Poetry.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
Richard Pearson
I have a huge dataset for which I only want to read in a subset of it. Is it
possible to use read.table to read in only a subset of the data? For example,
something like read.table('~/data.txt', subset = chromosome=='1' )
If not, then why not? This seems to be a feature available in all other
s
on 07/11/2008 10:50 AM Woolner, Keith wrote:
Hi everyone,
Is there a way to take an lm() model and strip it to a minimal form (or
convert it to another type of object) that can still used to predict the
dependent variable?
Depending upon how much memory you need to conserve and what else
I am trying to fit a formula to my data, but I just can't find the right way
to do it.
My experiment consists of manipulating FRUITS and VEGETATION to two levels
each(intact or removed) on 12 experimental plots.
This leaves me with 4 treatment combinations
Fruit intact Vegetation removed
Fruit in
My understanding of matrix subscripting is that this can be used to access
arbitrary elements from an array and return them as a vector, but I don't
understand how that helps me here. I've now written a function that seems to do
what I originally wanted, but I've also realised I want to do assi
You are right Keith : it doesn't work. Sorry.
In fact it works for a previous version but not for the new ones. Bad luck.
Nice week-end anyway,
Ptit Bleu.
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On 7/11/2008 11:51 AM, David Stoffer wrote:
Is there an easy way to compare complex numbers?
Here is a small example:
(z1=polyroot(c(1,-.4,-.45)))
[1] 1.11-0i -2.00+0i
(z2=polyroot(c(1,1,.25)))
[1] -2+0i -2+0i
x=0
if(any(identical(z1,z2))) x=99
x
[1] 0
# real and im
I had written asking for a simple way to extract the
Index of the last value in a vector greater than some
cutoff, e.g., the index, 6, for a cutoff of 20 and this
example vector:
v <- c(20, 134, 45, 20, 24, 500, 20, 20, 20)
Thank you, Alain Guillet, for this simple solution sent
to me offlist:
Dear list,
I'm afraid this is a mundane question. Here's the background: I've
produced a function which allows me to sequentially measure angles and
distances from a specified reference point to a curve defined by
empirical data points while (i.e. using a while loop) the angle being
meas
If the data you want is contiguous, then just 'skip' the number of
records and then read the number you want.
If you want to select a random sample, then checkout
http://article.gmane.org/gmane.comp.lang.r.general/78318/match=random+read
In your case where you want to conditionally read based on
Hello,
I am trying to loop through a matrix column and find the first value <=0 and
store that matrix location in a variable called Start.
I have tried the following:
i <- 1
j <- 1
while (Matrix[i, j] > 0)
i = (i + 1)
#loop until matrix [i, j] value <= 0
#strore that row number in a variable
Sta
... for which you need ?connections and the "nrows" argument to read.table
and friends.
(also ?scan and its "nlines" argument)
-- Bert Gunter
Genentech
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of jim holtman
Sent: Friday, July 11, 2008 9:58 AM
To
It's not clear what you want to do with coalescent theory, but you may
find Jonathan Marchini's popgen package useful. If not, you may find
what you are looking for in one of the other packages listed in the
Genetics task view.
You can learn about the R-C interface in Section 5 of the "Writing R
E
If I understand correctly, you can try:
m <- matrix(sample(0:5, 24, rep = TRUE), nc = 3)
# A list with the index
lapply(apply(m == 0, 2, which), head, 1)
# A vector
unlist(lapply(apply(m == 0, 2, which), head, 1))
On Fri, Jul 11, 2008 at 2:06 PM, Rheannon <[EMAIL PROTECTED]> wrote:
>
> Hello,
Did you try 'RSiteSearch("unable to restore saved data in
.RData")'? This just produced 61 hits for me, the second of which
"http://finzi.psych.upenn.edu/R/Rhelp02a/archive/85451.html"; looks to me
like it might answer your question.
Hope this helps.
Spencer
Susana Est
Dieter,
Thanks for tip. Well, XML 1.93-2.2 is what's working for me in my
2.6.2 environment:
R2.6.2 environment has XML Version: 1.93-2.2
R2.7.1 environment has XML Version: 1.95-3
Max, I'll try rolling back the XML package in my 2.7.1 environment
this weekend to see if that makes a difference.
Hello R users,
I would apprecaite any help to my current problem. I wanted to plot the
following data:
3.3705997271301E-06
-0.000193767858719911
-4.30103236602221E-05
-1.22606947412076E-06
-2.14297322582901E-05
-5.66282817417529E-05
-2.56009760591276E-05
-2.39301719384927E-06
-0.00013440022135339
In case anyone's still interested, I now have (I think!) a complete solution
(thanks to a quick look at my new favourite document - S Poetry :-)
subsetArray <- function(x, subset) {
subsetString <- paste(subset, collapse=",")
subsetString <- gsub("NA","",subsetString)
evalString <- paste(e
Hello everybody!
I'm using R and I have a little problem about function "grep". I 've got to
make a new function in which "grep" is present. So the first argument of
"grep" is the string we want to find,ok..but in this case I define a
function "x" before , x receives an argument in a object "name
I think this is what you want
> table <- sample(LETTERS[1:5], 20,TRUE)
>
> name <- "A"
>
> myfun <- function(name) {
+ r <- grep (name, table )
+ return (r) }
>
> myfun(name)
[1] 4 7 14 18
> table
[1] "E" "B" "D" "A" "B" "B" "A" "B" "E" "B" "C" "C" "C" "A" "E" "D"
"D" "A" "D" "C"
>
On Fri, Ju
Can you provide the 'plot' command that you are using. Is this data
on the x or y axis, and then what is the other axis you are plotting
against? Have you tried using logs? Exactly what is it you would
like to see from the plot?
On Fri, Jul 11, 2008 at 2:25 PM, Neil Gupta <[EMAIL PROTECTED]> wr
> From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> Sent: Friday, July 11, 2008 12:14 PM
>
> on 07/11/2008 10:50 AM Woolner, Keith wrote:
> > Hi everyone,
> >
> >
> >
> > Is there a way to take an lm() model and strip it to a minimal form
(or
> > convert it to another type of object) that can still
Thanks- I was hoping I wouldn't have to loop, but using abs() is better than
comparing Re()s and Im()s, which was my original thought. I have to compare
all the roots, so I used something like this:
> for (i in 1:length(z1)) if(any(abs(z1[i]-z2[1:length(z2)])<1e-15))
> print("ouch")
... thank
hi,
I need to remove the list that does not have any values.Do suggest me the
way to do it.
result<- list()
> head(result)
[[1]]
[,1] [,2] [,3]
[[2]]
[,1] [,2] [,3]
[[3]]
[,1] [,2] [,3]
[[4]]
[,1] [,2] [,3]
[[5]]
[,1] [,2][,3]
[1,] "HIGD1C" "SNP_A
Perhaps something about like this:
result[which(unlist(lapply(result, length)) != 0)]
On Fri, Jul 11, 2008 at 4:57 PM, Rajasekaramya <[EMAIL PROTECTED]> wrote:
>
> hi,
>
> I need to remove the list that does not have any values.Do suggest me the
> way to do it.
>
> result<- list()
>
>> head(resul
Hello -
Andrew Rominger wrote:
Dear list,
I'm afraid this is a mundane question. Here's the background: I've
produced a function which allows me to sequentially measure angles and
distances from a specified reference point to a curve defined by
empirical data points while (i.e. using a whil
Thank u so much it did work
Rajasekaramya wrote:
>
> hi,
>
> I need to remove the list that does not have any values.Do suggest me the
> way to do it.
>
> result<- list()
>
>> head(result)
>
> [[1]]
> [,1] [,2] [,3]
>
> [[2]]
> [,1] [,2] [,3]
>
> [[3]]
> [,1] [,2] [,3]
>
Hi Ramya,
Perhaps
result[lapply(result, length) != 0]
does the work.
HTH,
Jorge
On Fri, Jul 11, 2008 at 3:57 PM, Rajasekaramya <[EMAIL PROTECTED]>
wrote:
>
> hi,
>
> I need to remove the list that does not have any values.Do suggest me the
> way to do it.
>
> result<- list()
>
> > head(res
on 07/11/2008 02:02 PM Woolner, Keith wrote:
From: Marc Schwartz [mailto:[EMAIL PROTECTED]
Sent: Friday, July 11, 2008 12:14 PM
on 07/11/2008 10:50 AM Woolner, Keith wrote:
Hi everyone,
Is there a way to take an lm() model and strip it to a minimal form
(or
convert it to another type of ob
Hi Richard,
what is wrong with Patrick's suggestion? I get
x <- array(1:24, dim=2:4)
x[rbind(c(1,1,2))]
## [1] 7
x[rbind(c(1,1,2))] <- 13
x[rbind(c(1,1,2))]
## [1] 13
And you could also do
do.call("[", list(x,1,1,2))
These should be a bit quicker than the eval/parse constructs - see also
What is "wrong" is, I think, that you can't write a substitute for x[1,2,]
this way. (Or, in general, empty indexes won't work.)
A related question - Is there a way to replace x[1,2,] with "["(1,2,
*something clever*) ?
Kenn
On Sat, Jul 12, 2008 at 12:48 AM, Wolfgang Huber <[EMAIL PROTECTED]>
Kenn Konstabel wrote:
What is "wrong" is, I think, that you can't write a substitute for
x[1,2,] this way. (Or, in general, empty indexes won't work.)
A related question - Is there a way to replace x[1,2,] with "["(1,2,
*something clever*) ?
Kenn
Dear Kenn,
I can:
> do.call("[", list(x
Hello,
I am trying to do some fairly straightforward data summarization, i.e., the
kind you would do with a pivot table in excel or by using SQL queires. I
have a moderately sized data set of ~70,000 records and I am trying to
compute some group averages and sum values within groups. the code exam
Hello,
I am trying to do some fairly straightforward data summarization, i.e., the
kind you would do with a pivot table in excel or by using SQL queires. I
have a moderately sized data set of ~70,000 records and I am trying to
compute some group averages and sum values within groups. the code exam
Hello,
Have you tried using the GUI Rattle from www.rattle.togaware.com . It
works pretty well for summarization.
Regards,
Ajay
www.decisionstats.com
On Sat, Jul 12, 2008 at 4:14 AM, sj <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I am trying to do some fairly straightforward data summarization, i
After installing a new version of R, start a console session, cd to
the bin subdirectory of the new R installation (C:\Program
Files\R\R-2.7.1\bin in my case) and run the program RSetReg.exe. This
will update your registry entries and Tinn=R should find the correct
version of R. If you have kept
On 7/10/08, Mike Lawrence <[EMAIL PROTECTED]> wrote:
> As I understand it, Duncan MacKay's solution involves simply pasting the
> factors together, as in:
>
> |_AX_|_AY_|_BX_|_BY_|
>
> Which isn't quite as aesthetically pleasing as what I I'm looking for:
>
> |___A___|___B___|
> |_X_|_Y_|_X_|_Y
The problem is that you do not really have categories. You draw 3 times
7 random normal variables and then try to subset one by the other.
Since, no of the values will perfectly coincide with another, your code
would create something like 7^3 categories. No wonder that you are
running out o
I am sorry. Upon inspection, you only tried to create 70,000 categories.
However, the calculations for creating the 140,000 subsetted values pti and
finc exhausted your memory or the memory allocated to/in R.
Best,
Daniel
-
cuncta stricte discussurus
--
On Fri, 11 Jul 2008, Woolner, Keith wrote:
From: Marc Schwartz [mailto:[EMAIL PROTECTED]
Sent: Friday, July 11, 2008 12:14 PM
on 07/11/2008 10:50 AM Woolner, Keith wrote:
Hi everyone,
Is there a way to take an lm() model and strip it to a minimal form
(or
convert it to another type of obj
Aha! With the help of Deepayan's ever-insightful hints, I was able to
do exactly what I wanted:
dotplot(
variety ~ yield | site * year
, data = barley
, subset = (site %in% c("Grand Rapids", "Duluth"))
, layout=c(4, 1)
, scales=list(
alter
I need to plot multiple sets of data, on one Ternary plot (from the VCD
package). Since they all need to be formatted differently, I can't just
coerce them into one matrix, I need three separate calls. Is there a way to
do this? Thanks
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I am running Ubuntu 8.04 through Wubi on a HP Pavilion dv4000 (my home
computer) and I recently installed R 2.7.1. I attempted to install the
randomForest package from within the R environment and from the Linux
terminal. Both attempts failed. I've tried several mirrors but with no
luck. Below
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