### Re: [R] R Anova

```

Keizer_61 wrote:
I am really struggling with this question.

Three students take part of an experiment.

student  smoking  non-smokingcancer
110.5   7.5 6.5
2  9.5   6.5 8.4
3  8.5   7.2 5.5

the proper inferences is .05

we need to conduct Anova and have a inference of .05.

How do you enter this in R?

How do you calculate the F-test using R for this if we have smoking,
non-smoking and cancer all equal.

any suggestions?

Try to do the homework yourself and read the manuals?

Uwe Ligges

thanks

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```

### [R] How to solve difficult equations?

```
This below is not solvable with uniroot to find a:
fn=function(a){
b=(0.7/a)-a
(1/(a+b+1))-0.0025
}
uniroot(fn,c(-500,500))  gives
Error in uniroot(fn, c(-500, 500)) : f() values at end points not of
opposite sign

I read R-help posts and someone wrote a function:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html
but it is not very precise. Is there any 'standard function in R that can
solve this? thanks.
--
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```

### Re: [R] simulate values

```

Soare Marcian-Alin wrote:
Hello,

I want to simulate 100 values of the ARMA Process with this function:

x[i] = 0.5 * x[i-1] + 0.2 * x[i-2] + x[i] + 0.9 * x[i-1] + 0.2 * x[i-2] +
0.3 * x[i-3]

There is no kind of noise in your model, hence no need to do any
simulation so far ...

Uwe Ligges

which possibilities do I have?

Alin Soare

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### [R] How to solve difficult equations?

```I don't see the problem, except that you might want to think about
what the error message is telling you.

A little exploration of your function always helps, too.

ss - seq(-2, 2, len = 100)
plot(ss, fn(ss), type = l)
uniroot(fn, c(-1, 1))
Erreur dans uniroot(fn, c(-1, 1)) : f() values at end points not of
opposite sign

fn(-1)
[1] 3.330833
fn(0)
[1] -0.0025
fn(1)
[1] 0.5857353
uniroot(fn, c(-1, 0))
\$root
[1] -0.6999466

\$f.root
[1] -13118.83

\$iter
[1] 18

\$estim.prec
[1] 7.70751e-05

uniroot(fn, c(0, 1))
\$root
[1] 0.001760625

\$f.root
[1] 8.86832e-06

\$iter
[1] 3

\$estim.prec
[1] 6.103516e-05

This below is not solvable with uniroot to find a:
fn=function(a){
b=(0.7/a)-a
(1/(a+b+1))-0.0025
}
uniroot(fn,c(-500,500))  gives
Error in uniroot(fn, c(-500, 500)) : f() values at end points not of
opposite sign

I read R-help posts and someone wrote a function:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html
but it is not very precise. Is there any 'standard function in R
that can
solve this? thanks.

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```

### Re: [R] Analysis of Variance

```

CrazyJoe wrote:
Hello

Blind  toy_Car toy_truck   toy_boat
1 6.3 7.5 5.4
2 3.4 8.1 6.1
3 2.2 4.4 5.1

How do we calculate the F-statistic in R.

Any help is really appreciated.

You may want to try to do the homework yourself and read the manuals...
It also makes sense to specify the model rather than just providing some
data.

Uwe Ligges

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```

### Re: [R] How to solve difficult equations?

```plot(fn,-1,1)

On 25 Apr 2007, at 09:15, francogrex wrote:

This below is not solvable with uniroot to find a:
fn=function(a){
b=(0.7/a)-a
(1/(a+b+1))-0.0025
}
uniroot(fn,c(-500,500))  gives
Error in uniroot(fn, c(-500, 500)) : f() values at end points not of
opposite sign

I read R-help posts and someone wrote a function:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html
but it is not very precise. Is there any 'standard function in R
that can
solve this? thanks.
--
View this message in context: http://www.nabble.com/How-to-solve-
difficult-equations--tf3643595.html#a10175603
Sent from the R help mailing list archive at Nabble.com.

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guide.html
and provide commented, minimal, self-contained, reproducible code.

Ingmar Visser
Department of Psychology, University of Amsterdam
1018 WB Amsterdam
The Netherlands
t: +31-20-5256735

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```

### Re: [R] simulate values

```On Wed, 25 Apr 2007, Uwe Ligges wrote:

Soare Marcian-Alin wrote:
Hello,

I want to simulate 100 values of the ARMA Process with this function:

x[i] = 0.5 * x[i-1] + 0.2 * x[i-2] + x[i] + 0.9 * x[i-1] + 0.2 * x[i-2] +
0.3 * x[i-3]

There is no kind of noise in your model, hence no need to do any
simulation so far ...

My guess is that ... x[i] + 0.9 * x[i-1] + 0.2 * x[i-2] + 0.3 * x[i-3]
was meant to be for a noise series z.

In any case, to simulate an 'ARMA Process', see ?arima.sim.

Uwe Ligges

which possibilities do I have?

Alin Soare

[[alternative HTML version deleted]]

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--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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### [R] Unsubscription Confirmation

```Thank you for subscribing. You have now unsubscribed and no more messages will
be sent.

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```

### [R] for loops

```Hello everybody
I'm very new at using R so probably this is a very stupid question.
I have a matrix of p columns and I have to calculate for each of them the
two sample t-statistic and p-value and to save the results into two different
vectors.
I have divided my matrix into two submatrices: submatrix A containing the first
n1 rows (p columns) and submatrix B containing the remaining n2 (total
rows=n1+n2).
How can I do this with for loop construction?
Friendly regards
Silvia

--
Passa a Infostrada. ADSL e Telefono senza limiti e senza canone Telecom

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```

### Re: [R] How to solve difficult equations?

```On 25-Apr-07 07:15:55, francogrex wrote:

This below is not solvable with uniroot to find a:
fn=function(a){
b=(0.7/a)-a
(1/(a+b+1))-0.0025
}
uniroot(fn,c(-500,500))  gives
Error in uniroot(fn, c(-500, 500)) : f() values at end points
not of opposite sign

I read R-help posts and someone wrote a function:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html
but it is not very precise. Is there any 'standard function
in R that can solve this? thanks.

First, No:

Let alpha denote 0.0025, and beta 0.7 (in your function fn).
Then

fna - function(alpha,beta){ beta*alpha/(1 - alpha) }

solves it. But this is not a standard R function.

Second, Yes:

and the standard R function is uniroot(). But you can only apply
it usefully if you first study the behaviour of your function fn(),
in rather careful detail.

a-10*(-50:50)
plot(a,fn(a),pch=+)

Clearly something extreme happens just to the left of a=0. So:

a - 0.025*(-100:0)
plot(a,fn(a),pch=+)

and so for this set of values of 'a' the previous behaviour
cannot be seen. So:

a - 0.01*(-100:100)+0.001
plot(a,fn(a),pch=+)

so the function goes very negative somewhere around a = -0.7.
But

fn(500)
[1] 0.996102

so it is positive for a=500. Now find (inspired by the latest
plot):

a[which(fn(a)  (-100))]
[1] -0.699

and now you can use uniroot:

uniroot(fn,c(-0.699,500))
\$root
[1] 0.001771128
\$f.root
[1] 2.379763e-05
\$iter
[1] 16
\$estim.prec
[1] 6.103516e-05

and, if that doesn't look precise enough:

uniroot(fn,c(-0.699,500),tol=1e-10)
\$root
[1] 0.001754386
\$f.root
[1] 1.354602e-14
\$iter
[1] 18
\$estim.prec
[1] 5e-11

Now compare with the function fna() that solves it directly:

fna(0.0025,0.7)
[1] 0.001754386

(so in fact it was worth increasing the precision for uniroot).

But the lesson to be drawn from all this is that for functions
like fn(), which have singularities (here at a = -0.7), the
blind application of root-finding functions may not work, since
they are not set up to explore the function is the kind of way
illustrated above. While there are procedures in the numerical
analysis world to handle this kind of thing, they tend to be
written for particular classes of function, and again you will
have to do a bit of exploration to find out which function to use.

And (while someone more knowledgeable may well disagree with me)
I suspect that these are not standard R funnctions.

Ted.

E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 25-Apr-07   Time: 09:31:29
-- XFMail --

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```

### [R] Permutations with samr

```I have a question regarding the samr package.

For a 2 class unpaired problem, with sample 1 of size N1 and sample 2
of size N2,  samr computes at most  (N1+N2)! permutations of the two
samples (if the user-supplied parameter nperms allows it). However,
there are only (N1+N2)/(N1!*N2!) DISTINCT permutations of the two
samples, so it seems to me that only these distinct permutations should
be taken into consideration. Of course, working with the (N1+N2)!
permutations and working with the (N1+N2)/(N1!*N2!) distinct
permutations will lead to the same correct result, but for a number of
permutations smaller than (N1+N2)/(N1!*N2!), the result will generally
be different.

I would be very  grateful if someone could provide me an explanation
for the samr choice of the permutations.

Isabelle

Dr. Isabelle Rivals - Maître de Conférences
Équipe de Statistique Appliquée - ESPCI
10 rue Vauquelin - 75231 PARIS Cedex 05
Tél : (00 33 1) 40 79 45 45
Fax : (00 33 1) 40 79 44 20

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```

### [R] omit y=zero line in histogram

```
Dear all,

hist ( ) plots a horizontal line at y=0 when the respective bin is empty. I
can deal with this by modifying the hist object before plotting it
(x\$density[x\$density == 0] - NA), but I'm sure I've seen a more elegant
way. Perhaps this was in truehist (MASS). I have looked but can't find it.
Does anyone know?

Best wishes

Paul
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### [R] RE : for loops

```You can see with this simple example.

matrix.t.test-function(mx){

p-dim(mx)[2]  #number of column in the matrix
n-dim(mx)[1]  #number of row

n.tests- p*(p-1)/2 #Number of tests to be done

tested.var -rep(,n.tests) #Keep rang of tested
column
r.t.stat-rep(0, n.tests)#contain t.stat
r.p.val -rep(0, n.tests)#contain p.values

ctst-1 #current test
for (i in 1:(p-1)){
for (j in (i+1):p){
r.t.stat[ctst]-t.test(mx[,i],mx[,j])\$statistic
r.p.val [ctst] -t.test(mx[,i],mx[,j])\$p.value
tested.var [ctst]-paste(i,-,j)
ctst-ctst+1
}
}

result-data.frame(tested.var,r.t.stat,r.p.val)
return(result)
}

matrix.t.test(matrix(rnorm(50),nr=10,nc=5))

--- [EMAIL PROTECTED]
[EMAIL PROTECTED] a écrit :

Hello everybody
I'm very new at using R so probably this is a very
stupid question.
I have a matrix of p columns and I have to
calculate for each of them the two sample
t-statistic and p-value and to save the results
into two different vectors.
I have divided my matrix into two submatrices:
submatrix A containing the first n1 rows (p
columns) and submatrix B containing the remaining
n2 (total rows=n1+n2).
How can I do this with for loop construction?
Friendly regards
Silvia

--
senza canone Telecom

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reproducible code.

Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.

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```

### [R] correlation table

```hello,
is it possible to create a correlation table between factors?

___

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```

### [R] unit testing frameworks for R

```Greetings!

After a quick look at current programming tools, especially with regards
to unit-testing frameworks, I've started looking at both butler and
RUnit.   I would be grateful to receieve real world development
experience and opinions with either/both.Please send to me directly
(yes, this IS my work email), I will summarize (named or anonymous, as
contributers desire) to the list.

(work email used, as this is applicable to my work rather than usual
hobbies for a change, and some people with good reason prefer truth in
requesting).

Best regards / Mit freundlichen Grüssen,
Anthony (Tony) Rossini
Novartis Pharma AG
MODELING  SIMULATION
Group Head a.i., EU Statistical Modeling
CHBS, WSJ-027.1.012
Novartis Pharma AG
Lichtstrasse 35
CH-4056 Basel
Switzerland
Phone: +41 61 324 4186
Fax: +41 61 324 3039
Cell: +41 79 367 4557
Email : [EMAIL PROTECTED]

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```

### [R] attributable risk

```Hi everybody,

Does anyone know a function to compute the attributable risk of a factor
in a logistic regression or a proportional hazard cox model (both with
confounding variables)?
I need also obtain the confidence interval.

Isaac Subirana.

La informaciÃ³ continguda en aquest missatge i en qualsevol fitxer
Si no Ã©s la persona a la qual anava dirigida aquesta informaciÃ³, si us
plau, notifiqui immediatament l'enviament erroni al remitent i esborri
el missatge. Qualsevol cÃ²pia, divulgaciÃ³, distribuciÃ³ o utilitzaciÃ³ no
prohibida en virtut de la legislaciÃ³ vigent.

La informaciÃ³n contenida en este mensaje y en cualquier fichero
destinatario. Si usted no es la persona a la cual iba dirigida esta
informaciÃ³n, por favor, notifique inmediatamente el envÃ­o errÃ³neo al
remitente y borre el mensaje. Cualquier copia, divulgaciÃ³n,
distribuciÃ³n o utilizaciÃ³n no autorizada de este correo electrÃ³nico y
de sus adjuntos estÃ¡ prohibida en virtud de la legislaciÃ³n vigente.

The information included in this e-mail and any attached files are
confidential and private. If you are not the intended recipient,
please notify the error to the sender and delete this message
immediately. Dissemination, forwarding or copying of this e-mail and
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### [R] regular expressions with grep() and negative indexing

```Dear R-helpers,

Does anyone know how to use regular expressions to return vector elements
that don't contain a word? For instance, if I have a vector
x - c(seal.0,seal.1-exclude)
I'd like to get back the elements which do not contain the word exclude,
using something like (I know this doesn't work) but:
grep([^(exclude)],x)

I can use
x[-grep(exclude,x)]
for this case but then if I use this expression in a recursive function, it
will not work for instances in which the vector contains no elements with
that word. For instance, if I have
x2 - c(dolphin.0,dolphin.1)
then
x2[-grep(exclude,x2)]
will give me 'character(0)'

I know I can accomplish this in several steps, for instance:
myfunc - function(x) {
iexclude - grep(exclude,x)
if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
# do stuff with x2 ...?
}

But this is embedded in a much larger function and I am trying to minimize
intermediate variable assignment (perhaps a futile effort). But if anyone
knows of an easy solution, I'd appreciate a tip.

Thanks very much!

Stephen

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### [R] Plotting minimum spanning tree in graph/RBGL

```I wonder if anyone could tell me how I can plot a Minimum Spanning Tree
using the functions provided in the graph and RBGL packages?

I am able to build the MST using the following set of commands:

library(graph)

library(RBGL)

y - dist(x)

g1 - new(distGraph,y)

g2 - mstree.kruskal(g1)

However, there doesn't seem to be a function that allows the command
plot(g2) to draw it.

Many thanks,

Andrew Wilson

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### [R] Help with saptial analysis (cluster)

```Hi R-users

I'm a beginner with R and statistics, so I need some help to start my data
analysis. I've been reading some docs and tutorials on R and cluster analysis.
I've got a large dataset (102000 points) with values of longitude, latitude and
temperature and want to see if I can find groups (clusters).

Following some tutorials I can look for principal components but get an error
with calculation of distances:

matriz.distancias-dist(comp.obs)
Error in vector(double, length) : specified vector size is too big (translated
from spanish)

So, my questions are: is the dataset too big? could you point me to any docs
explaining how to study spatially distributed data (lon,lat,data)?

___
Francisco Pastor
Meteorology department
Fundación CEAM
[EMAIL PROTECTED]
http://www.gva.es/ceamet
http://www.gva.es/ceam
Paterna, Valencia, Spain
___

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### Re: [R] Problem installing Rmpi with lam on SGI SLES9

```On 24/04/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Tue, 24 Apr 2007, Hendrik Fuß wrote:

Hi,

I've been trying here to install Rmpi on an SGI IA-64 machine with 64
processors, running SuSE Linux Enterprise Server 9, R 2.4.0 and
lam-mpi 7.1.3. While I've read of similar problems on this list, I
think I've got an entirely new set of error messages to contribute
(see below). I'm not sure what the actual error is and what the @gprel
relocation message is about. Any help greatly appreciated.

I don't know for sure, but on many 64-bit OSes you cannot link code from
static libraries into dynamic shared libraries, and that seems to be the
case with ia64 Linux.  Almost certainly you need to re-compile LAM with
-fPIC flags.

Yes, thanks a million, this solved the problem.

While Rmpi now works, there is another issue connected with this one:
I'm trying to use the papply package. However, when I do:

library(papply)
papply(list(1:10, 1:15, 1:20), sum)
1 slaves are spawned successfully. 0 failed.
master (rank 0, comm 1) of size 2 is running on: behemoth
slave1 (rank 1, comm 1) of size 2 is running on: behemoth
[1] Running serial version of papply\n

Papply only spawns one slave and then decides to run the serial
version instead. I'm not sure how to tell it to use all the 64
processors available.

Hendrik

--
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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```

### Re: [R] regular expressions with grep() and negative indexing

```Stephen Tucker wrote:
Dear R-helpers,

Does anyone know how to use regular expressions to return vector elements
that don't contain a word? For instance, if I have a vector
x - c(seal.0,seal.1-exclude)
I'd like to get back the elements which do not contain the word exclude,
using something like (I know this doesn't work) but:
grep([^(exclude)],x)

I can use
x[-grep(exclude,x)]
for this case but then if I use this expression in a recursive function, it
will not work for instances in which the vector contains no elements with
that word. For instance, if I have
x2 - c(dolphin.0,dolphin.1)
then
x2[-grep(exclude,x2)]
will give me 'character(0)'

I know I can accomplish this in several steps, for instance:
myfunc - function(x) {
iexclude - grep(exclude,x)
if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
# do stuff with x2 ...?
}

But this is embedded in a much larger function and I am trying to minimize
intermediate variable assignment (perhaps a futile effort). But if anyone
knows of an easy solution, I'd appreciate a tip.

It has come up a couple of times before, and yes, it is a bit of a pain.

Probably the quickest way out is

negIndex - function(i)

if(length(i))

-i

else

TRUE

--
O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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### Re: [R] regarding 3d Bar Plot

```
On 4/24/2007 9:38 AM, [EMAIL PROTECTED] wrote:

[EMAIL PROTECTED] wrote:

I have data in a two dimensional table. each row of the data adds
upto 100 ( hence they are percentages ).  it can be interpreted as
like this A - I are the matches and  P - X are the players. Thus
Player P scored 20% of the runs during this season in Match C, 60% in
Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
regards.

snip

Many years ago I picked up from the snews mailing list a
suite of functions for plotting 2D barplots (barplots with 2D
bases) written by a chap named Colin Goodall, from (at that
time) the University of Bristol and/or from Penn State.

I never actually did anything with this suite until
recently.  Seeing no replies to the enquiry about 3D
histograms,  I thought I'd try to get Goodal's code running
in R to see if it might solve guarav's problem.

The trouble is, all the guts of the procedure, *including*
the plotting are done from within Fortran.  The actual
plotting seems to be done through a call to a subroutine
``segmtz'' which is a piece of Splus software that does not
exist in R.

Is there an equivalent subroutine in R that could be called?
I dug around a bit but couldn't figure out what was going
on.  The function segments() simply calls
.Internal(segments(

I looked around a bit for corresponding C or Fortran code but
obviously didn't know how to look properly.

I think that the Fortran code could be translated into raw R
and the call to segmtz changed to a call to segments() ---
but this would seem to be a lot of work.

Can anyone suggest a reasonably simple way of replacing the
call to segmtz in the Fortran?

I don't know how to do what you want, but I'd suggest working in R code
rather than Fortran.  I did write a hist3d function for the djmrgl
package (based on hist), mostly to show off the graphics, but haven't
found it useful enough to port to rgl.  Here's a quick port, not good
enough to use, but maybe it will give you a starting point.

Duncan Murdoch

hist3d -
function (x, y, xbreaks, ybreaks, freq= NULL, probability = !freq,
include.lowest= TRUE,
right= TRUE,
xlim = range(xbreaks), ylim = range(ybreaks), zlim = NULL,
xlab = xname, ylab = yname, zlab,
plot = TRUE, top = TRUE, nclass = NULL, ...)
{
if (!is.numeric(x) | !is.numeric(y))
stop(`x' and `y' must be numeric)
xname - deparse(substitute(x))
yname - deparse(substitute(y))
n - length(x - x[!is.na(x)])
use.xbr - !missing(xbreaks)
if(use.xbr) {
if(!missing(nclass))
warning(`nclass' not used when `xbreaks' specified)
}
else if(!is.null(nclass)  length(nclass) == 1)
xbreaks - nclass
use.xbr - use.xbr  (nB - length(xbreaks))  1
if(use.xbr)
xbreaks - sort(xbreaks)
else {  # construct vector of breaks
rx - range(x)
nnb -
if(missing(xbreaks)) 1 + log2(n)
else {  # breaks = `nclass'
if (is.na(xbreaks) | xbreaks  2)
stop(invalid number of xbreaks)
xbreaks
}
xbreaks - pretty (rx, n = nnb, min.n=1, eps.corr = 2)
}
nxB - length(xbreaks)
if(nxB = 1) ##-- Impossible !
stop(paste(hist3d: error, xbreaks=,format(xbreaks)))

storage.mode(x) - double
storage.mode(xbreaks) - double
use.ybr - !missing(ybreaks)
if(use.ybr) {
if(!missing(nclass))
warning(`nclass' not used when `ybreaks' specified)
}
else if(!is.null(nclass)  length(nclass) == 1)
ybreaks - nclass
use.ybr - use.ybr  (nB - length(ybreaks))  1
if(use.ybr)
ybreaks - sort(ybreaks)
else {  # construct vector of breaks
ry - range(y)
nnb -
if(missing(ybreaks)) 1 + log2(n)
else {  # breaks = `nclass'
if (is.na(ybreaks) | ybreaks  2)
stop(invalid number of ybreaks)
ybreaks
}
ybreaks - pretty (ry, n = nnb, min.n=1, eps.corr = 2)
}
nyB - length(ybreaks)
if(nyB = 1) ##-- Impossible !
stop(paste(hist3d: error, ybreaks=,format(ybreaks)))

storage.mode(y) - double
storage.mode(ybreaks) - double
counts - table(cut(x,xbreaks),cut(y,ybreaks))
if (sum(counts)  n)
stop(some data not counted; maybe breaks do not span range of data)
xh - diff(xbreaks)
if ```

### Re: [R] regular expressions with grep() and negative indexing

```Find the ones that match and then remove them from the full set with 'setdiff'.

x - c(seal.0,seal.1-exclude)
x.match - grep(exclude, x)  # find matches
x.match
[1] 2
setdiff(seq_along(x), x.match)  # exclude the matches
[1] 1

On 4/25/07, Peter Dalgaard [EMAIL PROTECTED] wrote:
Stephen Tucker wrote:
Dear R-helpers,

Does anyone know how to use regular expressions to return vector elements
that don't contain a word? For instance, if I have a vector
x - c(seal.0,seal.1-exclude)
I'd like to get back the elements which do not contain the word exclude,
using something like (I know this doesn't work) but:
grep([^(exclude)],x)

I can use
x[-grep(exclude,x)]
for this case but then if I use this expression in a recursive function, it
will not work for instances in which the vector contains no elements with
that word. For instance, if I have
x2 - c(dolphin.0,dolphin.1)
then
x2[-grep(exclude,x2)]
will give me 'character(0)'

I know I can accomplish this in several steps, for instance:
myfunc - function(x) {
iexclude - grep(exclude,x)
if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
# do stuff with x2 ...?
}

But this is embedded in a much larger function and I am trying to minimize
intermediate variable assignment (perhaps a futile effort). But if anyone
knows of an easy solution, I'd appreciate a tip.

It has come up a couple of times before, and yes, it is a bit of a pain.

Probably the quickest way out is

negIndex - function(i)

if(length(i))

-i

else

TRUE

--
O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

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```

### Re: [R] regarding 3d Bar Plot

```
Hi Duncan

I am restating the problem and thanks you for sending me such a good
function histogram in 3d. Thanks for that but i think my problem has been
misinterpreted. I just wanted a simple 3d bar Plot. Although I have not
written anything for R but i will surely like to contribute to R and if i
can assist someone in writing then it would be a great help to me.

Problem was :-)

I have data in a two dimensional table. each row of the data adds upto 100

( hence they are percentages ).
it can be interpreted as like this A - I are the matches and  P - X are
the players. Thus Player P scored 20% of the runs during this season in
Match C, 60% in Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
Matches and Z axis as the percentage(0 - 100%)

A   B   C   D   E   F   G   H   I
P   0   0   20  60  0   0   20  0   0
Q   0   16.8674726.907631   11.646586   0
12.449799   0.8032129   0   31.325301
R   0   59.649123   10.526316   12.280702   0   0
1.7543860   15.789474
S   3.57909807  20.281556   33.404915   7.31329 0.584586
5.9651631.1930327   0   27.678358
T   0   0   0   0   0   0   0   0   0
U   0   9.09090927.272727   18.181818   0
36.363636   0   0   9.090909
V   0   33.33   33.33   0   0   33.33
0   0   0
W   0   2.1881841.09409236.105033   0
44.420131   5.2516411   0   10.940919
X   0.05994234  51.550409   16.304315   6.9976680
17.383277   0.5994234   0.4741439   6.630821

-gaurav

Duncan Murdoch [EMAIL PROTECTED]
25-04-07 04:42 PM

To
[EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
Subject
Re: [R] regarding 3d Bar Plot

On 4/24/2007 9:38 AM, [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:

I have data in a two dimensional table. each row of the data adds
upto 100 ( hence they are percentages ).  it can be interpreted as
like this A - I are the matches and  P - X are the players. Thus
Player P scored 20% of the runs during this season in Match C, 60% in
Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
regards.

snip

Many years ago I picked up from the snews mailing list a
suite of functions for plotting 2D barplots (barplots
with 2D
bases) written by a chap named Colin Goodall, from (at
that
time) the University of Bristol and/or from Penn State.

I never actually did anything with this suite until
recently.  Seeing no replies to the enquiry about 3D
histograms,  I thought I'd try to get Goodal's code
running
in R to see if it might solve guarav's problem.

The trouble is, all the guts of the procedure,
*including*
the plotting are done from within Fortran.  The actual
plotting seems to be done through a call to a subroutine
``segmtz'' which is a piece of Splus software that does
not
exist in R.

Is there an equivalent subroutine in R that could be
called?
I dug around a bit but couldn't figure out what was going
on.  The function segments() simply calls
.Internal(segments(

I looked around a bit for corresponding C or Fortran code
but
obviously didn't know how to look properly.

I think that the Fortran code could be translated into
raw R
and the call to segmtz changed to a call to segments()
---
but this would seem to be a lot of work.

Can anyone suggest a reasonably simple way of replacing
the
call to segmtz in the Fortran?

I don't know how to do what you want, but I'd suggest working in R code
rather than Fortran.  I did write a hist3d function for the djmrgl
package (based on hist), mostly to show off the graphics, but haven't
found it useful enough to port to rgl.  Here's a quick port, not good
enough to use, but maybe it will give you a starting point.

Duncan Murdoch

hist3d -
function (x, y, xbreaks, ybreaks, freq= NULL, probability = !freq,
include.lowest= TRUE,
right= TRUE,
xlim = range(xbreaks), ylim = range(ybreaks), zlim = NULL,
xlab = xname, ylab = yname, zlab,
plot = TRUE, top ```

### Re: [R] correlation table

```Possible: yes - just calcuate correlation of as.numeric(your.factors)
Meaningful: no
It will depend on the coding for your factors, which may be absolutely
arbitrally...

Petr

elyakhlifi mustapha napsal(a):
hello,
is it possible to create a correlation table between factors?

___

[[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.

--
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [R] regarding 3d Bar Plot --- correction.

```
Hi Rolf,

If it is possible then please share the code as i am not able to locate
any pointers. Thanks in advance :-) cheers and chiao

regards
-gaurav

[EMAIL PROTECTED]
24-04-07 07:27 PM

To
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
cc

Subject
Re: [R] regarding 3d Bar Plot --- correction.

I mis-spoke.  It seems I had two collections of functions in the same
directory.  One by Colin Goodall, and one by David Scott (I have no
record of where he is/was located).  It is the *latter* collection
that does all its work from within Fortran.

I'll have another look at what Colin Goodall actually wrote to see if
it could be useful to guarav.

cheers,

Rolf Turner
[EMAIL PROTECTED]

DISCLAIMER AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}

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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [R] Coercing data types for use in model.frame

```You could try to follow the code in the dyn package.  It intercepts
model.frame calls involving time series objects so that it can align
lagged objects:

e.g.
z - ts(seq(10)^2)
library(dyn)
dyn\$lm(z ~ lag(z, -1))

It transforms the last line above to:

dyn(lm(dyn(z ~ lag(z, -1)))

and the inner dyn then produces a formula with class c(dyn, model.frame)
so that model.frame.dyn can intercept the call while the outer dyn adds
dyn to the class of the result so that anova.dyn, predict.dyn, etc. can be
used to intercept the result.

Thus for any lm-like function you just preface it with dyn\$ as shown and
you get automatically alignment of time series or in your case you would
interception of the mChoice variables.

On 4/24/07, Frank E Harrell Jr [EMAIL PROTECTED] wrote:
In the Hmisc package there is a new data class 'mChoice' for multiple
choice variables.  There are format and as.numeric methods (the latter
creates a matrix of dummy variables).  mChoice variables are not allowed
by model.frame.  Is there a way to specify a conversion function that
model.frame will use automatically?  I would use as.factor here.
model.frame does not seem to use as.data.frame.foo for individual variables.

Thanks
Frank
--
Frank E Harrell Jr   Professor and Chair   School of Medicine
Department of Biostatistics   Vanderbilt University

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```

### Re: [R] RE : for loops

```I know that a looping solution was requested, but this is exactly what
apply() should be used for...

Petr

justin bem napsal(a):
You can see with this simple example.

matrix.t.test-function(mx){

p-dim(mx)[2]  #number of column in the matrix
n-dim(mx)[1]  #number of row

n.tests- p*(p-1)/2 #Number of tests to be done

tested.var -rep(,n.tests) #Keep rang of tested
column
r.t.stat-rep(0, n.tests)#contain t.stat
r.p.val -rep(0, n.tests)#contain p.values

ctst-1 #current test
for (i in 1:(p-1)){
for (j in (i+1):p){
r.t.stat[ctst]-t.test(mx[,i],mx[,j])\$statistic
r.p.val [ctst] -t.test(mx[,i],mx[,j])\$p.value
tested.var [ctst]-paste(i,-,j)
ctst-ctst+1
}
}

result-data.frame(tested.var,r.t.stat,r.p.val)
return(result)
}

matrix.t.test(matrix(rnorm(50),nr=10,nc=5))

--- [EMAIL PROTECTED]
[EMAIL PROTECTED] a écrit :

Hello everybody
I'm very new at using R so probably this is a very
stupid question.
I have a matrix of p columns and I have to
calculate for each of them the two sample
t-statistic and p-value and to save the results
into two different vectors.
I have divided my matrix into two submatrices:
submatrix A containing the first n1 rows (p
columns) and submatrix B containing the remaining
n2 (total rows=n1+n2).
How can I do this with for loop construction?
Friendly regards
Silvia

--
senza canone Telecom

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and provide commented, minimal, self-contained,
reproducible code.

Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.

__
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and provide commented, minimal, self-contained, reproducible code.

--
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [R] A very simple question

```If by objects you mean the packages you installed and assuming you
are on Windows then there are two batch files movedir.bat and copydir.bat
in the batchfiles distribution that will either move them from an older version
(they won't be available to the older version any more but its much faster and
they won't be taking up space twice) or copy them (they will now be in both
versions so you can use them from either but its slower and they will be taking

you should be sure to read.

On 4/25/07, David Kaplan [EMAIL PROTECTED] wrote:
Hi all,

I just loaded R 2.50.  I want this version to bring the objects from the
previous version.  How do I do that?

Thanks

David

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```

### Re: [R] Help with saptial analysis (cluster)

```Dear Fransico,

The distance matrix would be 102000 x 102000. So it would contain 1040400
values. If you need one bit for each value, this would requier 9,7 GB. So the
distance matrix won't fit in the RAM of your computer.

Cheers,

Thierry

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology
and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be

Do not put your faith in what statistics say until you have carefully
considered what they do not say.  ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney

-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Namens Francisco Pastor
Verzonden: woensdag 25 april 2007 12:34
Aan: r-help@stat.math.ethz.ch
Onderwerp: [R] Help with saptial analysis (cluster)

Hi R-users

I'm a beginner with R and statistics, so I need some help to
start my data analysis. I've been reading some docs and
tutorials on R and cluster analysis.
I've got a large dataset (102000 points) with values of
longitude, latitude and temperature and want to see if I can
find groups (clusters).

Following some tutorials I can look for principal components
but get an error with calculation of distances:

matriz.distancias-dist(comp.obs)
Error in vector(double, length) : specified vector size is
too big (translated from spanish)

So, my questions are: is the dataset too big? could you point
me to any docs explaining how to study spatially distributed
data (lon,lat,data)?

__
_
Francisco Pastor
Meteorology department
Fundación CEAM
[EMAIL PROTECTED]
http://www.gva.es/ceamet
http://www.gva.es/ceam
Paterna, Valencia, Spain
__
_

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R-help@stat.math.ethz.ch mailing list
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```

### Re: [R] prelim.norm() function not working

```
Thank you very much, that was indeed the problem. (And now that I read
more carefully the help page, it did in fact say that the input was a
data matrix and not a data frame.)

Brant
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Prof Brian Ripley
Sent: Wednesday, April 25, 2007 12:12 AM
To: Brant Inman
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] prelim.norm() function not working

Looks like you have a data frame where you need a matrix.  (The same
issue occurs in most of Joe Schafer's packages, e.g. mix.)

Try as.matrix(usnews).

On Tue, 24 Apr 2007, Brant Inman wrote:

R-experts:
I am trying to reproduce some of Paul Allison's results in his little
green book on missing data (Sage 2002).  The dataset for which I am
having problems, usnews, can be found at:
http://www.ats.ucla.edu/stat/books/md/default.htm.  I am working on a
Windows machine with R 2.5 installed, all packages up-to-date.
The problem has to do with the prelim.norm() function of the package
norm.   Specifically, I need to use this pre-processing function to
later use the EM algorithm and DA procedures in the norm package.  I
am getting an error with the following code.
--
pre - prelim.norm(usnews)

Error in as.double.default(list(csat = c(972L, 961L, NA, 881L, NA, NA,
:
(list) object cannot be coerced to 'double'

-
I have read the previous postings and I am wondering if the problem
with prelim.norm is the size of the usnews dataset or the amount of
missing data.

dim(usnews)
[1] 13027

Does anyone have any ideas?  If not, are there alternatives to norm
for implementing the MLE and EM methods of dealing with missing data?

Thanks,

Brant Inman
Mayo Clinic

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--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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```

### [R] program avail. for simulating spatial patterns?

```Hi all,

I am wondering if there is a function available in R for simulating
spatial distribution of objects (plants in this case) in order to
simulate sampling of a population .  Specifically, I would like to be
able to change the spatial correlation of individuals.  I don't want
to reinvent the wheel if it already exists.

Thanks,

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```

### [R] assigning two conditions to grep()

```Hi,
i have a problem in assigning 2 conditions to grep()  ,
my data look like this:
DA 24 N7 Rad= 3.4 20 Sac= 0.93 Acc= 4.76
DA 24 N7 Rad= 3.4 14 Sac= 0.65 Acc= 3.33
DA 24 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
DA 24 N7 Rad= 3.4 11 Sac= 0.51 Acc= 2.62
DG 23 N7 Rad= 3.4  8 Sac= 0.37 Acc= 1.91
DG 23 N7 Rad= 3.4  5 Sac= 0.23 Acc= 1.19
DG 23 N7 Rad= 3.4  0 Sac= 0.00 Acc= 0.00
DG 23 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
DG 23 O6 Rad= 3.3  1 Sac= 0.04 Acc=  0.22
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
(it's a data.frame)

at first i wanted all the line begining with A 24:
data[grep(^24, data\$V2)]
this works
and than i wanted to exctract all the lines with G23 N7,
neither the column 23 and the column N7 are unique
so i tried this
data[grep(^23*N7, data),]
but doesn't work
not either
x[(grep(^N7, as.character(x\$V3))) (grep(^23, x\$V2)),]
he just returns everything.

thank u for any help,
josephine

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```

### Re: [R] regarding 3d Bar Plot

```On 4/25/2007 7:56 AM, [EMAIL PROTECTED] wrote:
Hi Duncan

I am restating the problem and thanks you for sending me such a good
function histogram in 3d. Thanks for that but i think my problem has been
misinterpreted. I just wanted a simple 3d bar Plot. Although I have not
written anything for R but i will surely like to contribute to R and if i
can assist someone in writing then it would be a great help to me.

Problem was :-)

I have data in a two dimensional table. each row of the data adds upto 100

( hence they are percentages ).
it can be interpreted as like this A - I are the matches and  P - X are
the players. Thus Player P scored 20% of the runs during this season in
Match C, 60% in Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
Matches and Z axis as the percentage(0 - 100%)

The plot.histogram function I sent does most of what you want.  The
hist3d function calculates the matrix of counts that it plots, and
plot.histogram plots the resulting bar chart.

Duncan Murdoch

A   B   C   D   E   F   G   H   I
P   0   0   20  60  0   0   20  0   0
Q   0   16.8674726.907631   11.646586   0
12.449799   0.8032129   0   31.325301
R   0   59.649123   10.526316   12.280702   0   0
1.7543860   15.789474
S   3.57909807  20.281556   33.404915   7.31329 0.584586
5.9651631.1930327   0   27.678358
T   0   0   0   0   0   0   0   0   0
U   0   9.09090927.272727   18.181818   0
36.363636   0   0   9.090909
V   0   33.33   33.33   0   0   33.33
0   0   0
W   0   2.1881841.09409236.105033   0
44.420131   5.2516411   0   10.940919
X   0.05994234  51.550409   16.304315   6.9976680
17.383277   0.5994234   0.4741439   6.630821

-gaurav

Duncan Murdoch [EMAIL PROTECTED]
25-04-07 04:42 PM

To
[EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
Subject
Re: [R] regarding 3d Bar Plot

On 4/24/2007 9:38 AM, [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] wrote:

I have data in a two dimensional table. each row of the data adds
upto 100 ( hence they are percentages ).  it can be interpreted as
like this A - I are the matches and  P - X are the players. Thus
Player P scored 20% of the runs during this season in Match C, 60% in
Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
regards.
snip

Many years ago I picked up from the snews mailing list a
suite of functions for plotting 2D barplots (barplots
with 2D
bases) written by a chap named Colin Goodall, from (at
that
time) the University of Bristol and/or from Penn State.

I never actually did anything with this suite until
recently.  Seeing no replies to the enquiry about 3D
histograms,  I thought I'd try to get Goodal's code
running
in R to see if it might solve guarav's problem.

The trouble is, all the guts of the procedure,
*including*
the plotting are done from within Fortran.  The actual
plotting seems to be done through a call to a subroutine
``segmtz'' which is a piece of Splus software that does
not
exist in R.

Is there an equivalent subroutine in R that could be
called?
I dug around a bit but couldn't figure out what was going
on.  The function segments() simply calls
.Internal(segments(

I looked around a bit for corresponding C or Fortran code
but
obviously didn't know how to look properly.

I think that the Fortran code could be translated into
raw R
and the call to segmtz changed to a call to segments()
---
but this would seem to be a lot of work.

Can anyone suggest a reasonably simple way of replacing
the
call to segmtz in the Fortran?

I don't know how to do what you want, but I'd suggest working in R code
rather than Fortran.  I did write a hist3d function for the djmrgl
package (based on hist), mostly to show off the graphics, but haven't
found it useful enough to port to rgl.  Here's a quick port, not good
enough to use, but maybe it will give you a ```

### Re: [R] program avail. for simulating spatial patterns?

```Have a look at packages in the spatial taskview (spatstat, splancs).

Cheers,

Thierry

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be

Do not put your faith in what statistics say until you have carefully
considered what they do not say.  ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney

-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED]
Verzonden: woensdag 25 april 2007 15:03
Aan: r-help@stat.math.ethz.ch
Onderwerp: [R] program avail. for simulating spatial patterns?

Hi all,

I am wondering if there is a function available in R for
simulating spatial distribution of objects (plants in this
case) in order to simulate sampling of a population .
Specifically, I would like to be able to change the spatial
correlation of individuals.  I don't want to reinvent the

Thanks,

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```

### Re: [R] regarding 3d Bar Plot

``` I am restating the problem and thanks you for sending me such a good
function histogram in 3d. Thanks for that but i think my problem has been
misinterpreted. I just wanted a simple 3d bar Plot. Although I have not
written anything for R but i will surely like to contribute to R and if i
can assist someone in writing then it would be a great help to me.

Problem was :-)

I have data in a two dimensional table. each row of the data adds upto 100

( hence they are percentages ).
it can be interpreted as like this A - I are the matches and  P - X are
the players. Thus Player P scored 20% of the runs during this season in
Match C, 60% in Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
Matches and Z axis as the percentage(0 - 100%)

I suggest that you don't use a 3d bar chart.  3d bar charts are
generally hard to interpret for two reasons: large bars will obscure
small bars behind them, and it is very difficult to judge the true
length of the bars. I suggest you try creating a series of 2d bar
charts instead - you are far more likely to be able to interpret them
easily.

For this data, you might also want to look into fluctuation diagrams.

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```

### [R] R CMD CHECK and require() / library() methods

```Hi,

I have a piece of code that decides at runtime whether to load a data
package (and which package to load).
This is then done with a call to:

library(x)

(where x is a character variable containing the package name).

This causes R CMD check to throw out a warning:
'library' or 'required' calls not declared from:
x

Does anyone have any suggestions as to a fix or workaround for this?

Crispin

This email is confidential and intended solely for the use o...{{dropped}}

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```

### [R] How to identify and exclude the outliers with R?

```Hello, everyone,

I want to ask a simple question.
If I have a set  of data,and I want to identify how many outliers there are
in the data.Which packages and functions can I use?

Thanks.

Shao chunxuan.

[[alternative HTML version deleted]]

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```

### [R] creating random numbers

``` l want to create a  column of 1 and 2 randomly what command should l use
eg
treatment  strata

1  1
2  0
1  1
2  1
2  0
2  1
2  0
1  0
these should be created randomly

secondly if l have something like
for (i in 1:n ) df =c(1,rexp(1,.001),rexp(1,.002)) can l add an if statement
in  side the c maybe to compare the 2 exponential  numbers to create  another
variable

-

[[alternative HTML version deleted]]

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```

### Re: [R] regarding 3d Bar Plot

```On 4/25/2007 9:26 AM, hadley wickham wrote:
I am restating the problem and thanks you for sending me such a good
function histogram in 3d. Thanks for that but i think my problem has been
misinterpreted. I just wanted a simple 3d bar Plot. Although I have not
written anything for R but i will surely like to contribute to R and if i
can assist someone in writing then it would be a great help to me.

Problem was :-)

I have data in a two dimensional table. each row of the data adds upto 100

( hence they are percentages ).
it can be interpreted as like this A - I are the matches and  P - X are
the players. Thus Player P scored 20% of the runs during this season in
Match C, 60% in Match D and remaining 20% in Match G.

I want to plot 3-d bar plot, where X axis have players, Y axis have
Matches and Z axis as the percentage(0 - 100%)

I suggest that you don't use a 3d bar chart.  3d bar charts are
generally hard to interpret for two reasons: large bars will obscure
small bars behind them, and it is very difficult to judge the true
length of the bars. I suggest you try creating a series of 2d bar
charts instead - you are far more likely to be able to interpret them
easily.

I agree with this.

Duncan Murdoch

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```

### Re: [R] NA and NaN randomForest

```Hi Clayton,

If you use the formula interface, then it should do what you want:

R library(randomForest)
randomForest 4.5-18
Type rfNews() to see new features/changes/bug fixes.
R iris1 - iris[-(1:5),]
R iris2 - iris[1:5,]
R iris2[1, 3] - NA
R iris2[3, 1] - NA
R iris.rf - randomForest(Species ~ ., iris1)
R predict(iris.rf, iris2[-5])
[1] NA   setosa NA   setosa setosa
Levels: setosa versicolor virginica

The problem, of course, is that the formula interface is not suitable
for data with large number of variables.  I'll look into doing the same
thing in the default method.

Andy

From: [EMAIL PROTECTED]

Dear R-help,

This is about randomForest's handling of NA and NaNs in test set data.
Currently, if the test set data contains an NA or NaN then
predict.randomForest will skip that row in the output.

I would like to change that behavior to outputting an NA.

Can this be done with flags to randomForest?
If not can some sort of wrapper be built to put the NAs back in?

thanks,

Clayton
_

CONFIDENTIALITY NOTICE\ \ The information contained in this
...{{dropped}}

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### Re: [R] creating random numbers

```sample(1:2, 10, replace=TRUE)

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of raymond chiruka
Sent: Wednesday, April 25, 2007 9:45 AM
To: r-help@stat.math.ethz.ch
Subject: [R] creating random numbers

l want to create a  column of 1 and 2 randomly what
command should l use
eg
treatment  strata

1  1
2  0
1  1
2  1
2  0
2  1
2  0
1  0
these should be created randomly

secondly if l have something like
for (i in 1:n ) df =c(1,rexp(1,.001),rexp(1,.002)) can l
add an if statement in  side the c maybe to compare the 2
exponential  numbers to create  another  variable

-

[[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.

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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [R] Random Number Generator of Park and Miller

```On Tue, 24 Apr 2007, gracezhang wrote:

Hi,

I failed to search for R package providing random number generator of Park
and Miller.
Anyone know any R package supporting this kind of function?

rng.lcg-function(x,p1=16807,p2=0,N=2147483647){(x*p1+p2)%%N}

Dave
--
Dr. David Forrest
[EMAIL PROTECTED](804)684-7900w
[EMAIL PROTECTED] (804)642-0662h
http://maplepark.com/~drf5n/

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```

### [R] print format - fixed number of digits

```Hi,

is it possible to format real (double) numbers in a data frame to an
exact format?
I need something like format(..., digits=5) but this is a suggestion
only and I need exactly 5 digits. Example where there are more than 5
digits printed follows.

I can do it via
cbind(as.numeric(sprintf(%.5f,column_1)),
as.numeric(sprintf(%.5f,column_2)),...)
but it is really annoying and there must be an easier solution.

Petr

x - as.data.frame(matrix(rnorm(6),nrow=3)/100)

x
V1   V2
1  0.002640759 -0.002335782
2 -0.003960130  0.010373135
3 -0.007079349 -0.005792717

format(x,digits=5,scientific=FALSE)
V1 V2
1  0.0026408 -0.0023358
2 -0.0039601  0.0103731
3 -0.0070793 -0.0057927

--
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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### [R] Split column of concatenated data

```Hi

I have a column of concatenated information stored in an RG object in the limma
package and I need to split this information and then paste the first two
pieces of data in each case back into two columns of the RG object.

This is how I am currently doing this

gene.info.split-strsplit(RG\$genes\$Name,,,fixed=TRUE)

for (h in 1: length(gene.info.split)){
RG\$genes\$ID[h]-gene.info.split[h][[1]][1]
RG\$genes\$Name[h]-gene.info.split[h][[1]][2]
}

However, this is very slow and presumably 'messy'. The problem is that there
are an inconsistent number of comma separated entries in the original Name
column so I cannot do

gene.info.split-as.data.frame(strsplit(RG\$genes\$Name,,,fixed=TRUE))

because I get the error message

Error in data.frame(c(OligoCy3, SP Control poplar 48pin, A24, no length
information :
arguments imply differing number of rows: 4, 6, 5, 1

I also can't figure out how to usefully put the [list] data into a matrix (my
ignorance I am sure).

Ideally I would be able to put each comma separated item into a column and then
simply paste the first and second columns over the RG\$genes\$Name and
RG\$genes\$ID columns respectively (and do away with the for loop).

Some cases in the original RG\$genes\$Name has only one piece of information (ie
no commas) so I would need a way to fill any blanks with an NA value

If anyone can help me, it would be much appreciated

Nat Street

---
Nathaniel Street
University of Southampton
Plants and Environment Lab
School of Biological Sciences
Basset Crescent East
Southampton
SO16 7PX
tel: +44 (0) 2380 594268
fax: +44 (0) 2380 594269
[EMAIL PROTECTED]

http://www.populus.biol.soton.ac.uk/~nat
http://del.icio.us/n.r.street

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```

### [R] barchart producing incorrect number of barcharts when columns renamed

```Hi everybody,

I'm having problems with the barchart command in the lattice package.

I'm creating barcharts from matrices with with anything from 20 to 71
columns. When I leave the column names alone, that is they are set in
the read.table command (and inherited by subsequent commands) the
correct number barcharts is created by the print(barchart(...))
command. However, when I reset the column names by means of a scan
command, the number of barcharts drawn by the same command is
incorrect: it is always too few. The scan commands produce lists the
same length as the number of columns for which I want barcharts.

In total I've got four pages with barcharts. The numbers in the table
below indicate the number of barcharts per page. The numbers without
() are the number of barcharts that I expect (and get when I don't
reset the column titles). The numbers in () are the numbers of
barcharts I get when I reset the column titles.

Not ClippedClipped
Errs   20 (18) 71 (46)
Stops 32 (24) 46 (36)

The following is the code used to create the barcharts with sample
text output below it.

library('lattice')
rm(list=ls())

textFontSize=6;
pdf('../data/cocaineBarcharts.pdf', paper='a4')
fontsize=trellis.par.get(fontsize);
fontsize\$text=textFontSize;
trellis.par.set(fontsize, fontsize);

resultsDirs=c(../data/Group.results.noclipping, ../data/
Group.results.clipped);
#resultsDirs=c(../data/Group.results.clipped);

for (resultsDir in resultsDirs) {
cat(resultsDir, \n)
if (any(grep(clipped, resultsDir)))
{
clipping=(Clipped);
}
else
{
clipping=(NOT Clipped);
}

roi.errs.names=names(roi.errs);
# ctrl
roi.errs.ctrl-roi.errs[roi.errs[,Group]==ctrl, 4:length
(roi.errs)]
roi.errs.ctrl.subjects=roi.errs[roi.errs[,Group]==ctrl, 2]
# short
roi.errs.short-roi.errs[roi.errs[,Group]==short, 4:length
(roi.errs)]
roi.errs.short.subjects=roi.errs[roi.errs[,Group]==short, 2]
# long
roi.errs.long-roi.errs[roi.errs[,Group]==long, 4:length
(roi.errs)]
roi.errs.long.subjects=roi.errs[roi.errs[,Group]==long, 2]

sep=);
roi.stops.names=names(roi.stops);
# ctrl
roi.stops.ctrl-roi.stops[roi.stops[,Group]==ctrl, 4:length
(roi.stops)]
roi.stops.ctrl.subjects=roi.stops[roi.stops[,Group]==ctrl, 2]
# short
roi.stops.short-roi.stops[roi.stops[,Group]==short, 4:length
(roi.stops)]
roi.stops.short.subjects=roi.stops[roi.stops[,Group]==short, 2]
# long
roi.stops.long-roi.stops[roi.stops[,Group]==long, 4:length
(roi.stops)]
roi.stops.long.subjects=roi.stops[roi.stops[,Group]==long, 2]

#matrixToPlot=as.matrix(roi.errs.ctrl[1:5,])
#yylim=c(floor(min(matrixToPlot)), ceiling(max(matrixToPlot)))
#barplot(matrixToPlot, col=c(2:6), beside=T, ylim=yylim,
names.arg=colnames(roi.errs.ctrl),
#border=c(2:6), legend.text=roi.errs\$Subject[1:5])

roi.errs.ctrl.matrix=as.matrix(roi.errs.ctrl)
roi.errs.short.matrix=as.matrix(roi.errs.short)
roi.errs.long.matrix=as.matrix(roi.errs.long)

roi.stops.ctrl.matrix=as.matrix(roi.stops.ctrl)
roi.stops.short.matrix=as.matrix(roi.stops.short)
roi.stops.long.matrix=as.matrix(roi.stops.long)

#
### errors
#

#  pdf(paste(resultsDir, 'errorsByGroup.pdf', sep=/), paper='a4')
#  fontsize=trellis.par.get(fontsize);
#  fontsize\$text=textFontSize;
#  trellis.par.set(fontsize, fontsize);

roi.errs.ctrl.means=colMeans(roi.errs.ctrl.matrix)
roi.errs.short.means=colMeans(roi.errs.short.matrix)
roi.errs.long.means=colMeans(roi.errs.long.matrix)
yylim=c(floor(min(roi.errs[, 4:length(roi.errs)])), ceiling(max
(roi.errs[, 4:length(roi.errs)])))

errs.Means=rbind(roi.errs.ctrl.means, roi.errs.short.means,
roi.errs.long.means)
rownames(errs.Means)=c('control', 'short', 'long')
cat(errs.Means dimensions before col name change , dim
(errs.Means), \n);
colnames(errs.Means) = scan(paste(resultsDir,
clusterLocations.errs.csv, sep=/), sep=,, what=character)
cat(errs.Means dimensions after col name change , dim
(errs.Means), \n);

print(barchart(errs.Means, groups=rownames(errs.Means), xlab='Mean
Intensity',
main=paste(Mean Cluster Intensity for Errors,
clipping),
ylab='Group', col=rainbow(3), border=rainbow(3)))
#  dev.off()

```

### Re: [R] simulate values

```Hello,

Yes I tried arima.sim and everything works fine. Thanks for the help!

Alin Soare

2007/4/25, Leeds, Mark (IED) [EMAIL PROTECTED]:

he's just being a wise guy bcause I'm sure you meant to have an
epsilon_t on the end. And he
Surely knows that also. Did you
Check out ?arima.sim.

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Uwe Ligges
Sent: Wednesday, April 25, 2007 3:15 AM
To: Soare Marcian-Alin
Cc: R-help@stat.math.ethz.ch
Subject: Re: [R] simulate values

Soare Marcian-Alin wrote:
Hello,

I want to simulate 100 values of the ARMA Process with this function:

x[i] = 0.5 * x[i-1] + 0.2 * x[i-2] + x[i] + 0.9 * x[i-1] + 0.2 *
x[i-2] +
0.3 * x[i-3]

There is no kind of noise in your model, hence no need to do any
simulation so far ...

Uwe Ligges

which possibilities do I have?

Alin Soare

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### [R] levelplot and unequal cell sizes

```I am using levelplot() from lattice with grids that have unequal cell
sizes.  This means that the boundary between two cells is not always
half-way between nodes, as levelplot() assumes.  The result is that some
cell sizes are rendered incorrectly, which can be painfully obvious if
using relatively large cells.  Is there any work-around?  I am using the
conditioning capability of lattice and therefore image() would not be a
good way to go.

Thanks,
Scott Waichler
Pacific Northwest National Laboratory
scott.waichler _at_ pnl.gov

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```

### Re: [R] assigning two conditions to grep()

```janek0 wrote:

Dnia 25-04-2007, śro o godzinie 15:04 +0200, Abi Ghanem josephine
napisał(a):

use data[grep(^24|N7, data\$V2)]
and see ?regexp

but actually my problem is that the column containing 23 is in data\$V2
and the one containing N7 is in data\$V3,
so the line doen't work i just have all the line containing G23 N7  and
G23 O6
and i want to separate the two.

josephine

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### Re: [R] R CMD CHECK and require() / library() methods

```On Wed, 25 Apr 2007, Crispin Miller wrote:

Hi,

I have a piece of code that decides at runtime whether to load a data
package (and which package to load).
This is then done with a call to:

library(x)

(where x is a character variable containing the package name).

This causes R CMD check to throw out a warning:
'library' or 'required' calls not declared from:
x

Which version of R is this?  All I can find say 'require'.

Does anyone have any suggestions as to a fix or workaround for this?

That call should be

library(x, character.only=TRUE)

and that will in R 2.5.0 stop the warning AFAIK.

--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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```

### Re: [R] How to solve difficult equations?

```At 03:15 AM 4/25/2007, francogrex wrote:

This below is not solvable with uniroot to find a:
fn=function(a){
b=(0.7/a)-a
(1/(a+b+1))-0.0025
}
uniroot(fn,c(-500,500))  gives
Error in uniroot(fn, c(-500, 500)) : f() values at end points not of
opposite sign

I read R-help posts and someone wrote a function:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html
but it is not very precise. Is there any 'standard function in R that can
solve this? thanks.

Actually, if you're solving fn(a)==0, then some trivial algebra leads
to a linear equation with a=0.001754.

Why are you trying to solve this numerically? Is it a single instance
of a larger, more general problem?

Robert A. LaBudde, PhD, PAS, Dpl. ACAFS  e-mail: [EMAIL PROTECTED]
Least Cost Formulations, Ltd.URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239Fax: 757-467-2947

Vere scire est per causas scire

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### Re: [R] creating random numbers

```?sample
sample(c(1,2),10,replace=TRUE)

P.

raymond chiruka napsal(a):
l want to create a  column of 1 and 2 randomly what command should l use
eg
treatment  strata

1  1
2  0
1  1
2  1
2  0
2  1
2  0
1  0
these should be created randomly

secondly if l have something like
for (i in 1:n ) df =c(1,rexp(1,.001),rexp(1,.002)) can l add an if
statement in  side the c maybe to compare the 2 exponential  numbers to
create  another  variable

-

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--
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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### Re: [R] R CMD CHECK and require() / library() methods

```Many thanks!

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Prof
Brian Ripley
Sent: 25 April 2007 16:03
To: Crispin Miller
Cc: R-help@stat.math.ethz.ch
Subject: Re: [R] R CMD CHECK and require() / library() methods

On Wed, 25 Apr 2007, Crispin Miller wrote:

Hi,

I have a piece of code that decides at runtime whether to
package (and which package to load).
This is then done with a call to:

library(x)

(where x is a character variable containing the package name).

This causes R CMD check to throw out a warning:
'library' or 'required' calls not declared from:
x

Which version of R is this?  All I can find say 'require'.

My mistake - it was a typo it says: 'require'

Does anyone have any suggestions as to a fix or workaround for this?

That call should be

library(x, character.only=TRUE)

and that will in R 2.5.0 stop the warning AFAIK.

It works - much appreciated...

Crispin

This email is confidential and intended solely for the use o...{{dropped}}

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```

### Re: [R] regular expressions with grep() and negative indexing

```Peter Dalgaard wrote:
Stephen Tucker wrote:

Dear R-helpers,

Does anyone know how to use regular expressions to return vector elements
that don't contain a word? For instance, if I have a vector
x - c(seal.0,seal.1-exclude)
I'd like to get back the elements which do not contain the word exclude,
using something like (I know this doesn't work) but:
grep([^(exclude)],x)

I can use
x[-grep(exclude,x)]
for this case but then if I use this expression in a recursive function, it
will not work for instances in which the vector contains no elements with
that word. For instance, if I have
x2 - c(dolphin.0,dolphin.1)
then
x2[-grep(exclude,x2)]
will give me 'character(0)'

I know I can accomplish this in several steps, for instance:
myfunc - function(x) {
iexclude - grep(exclude,x)
if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
# do stuff with x2 ...?
}

But this is embedded in a much larger function and I am trying to minimize
intermediate variable assignment (perhaps a futile effort). But if anyone
knows of an easy solution, I'd appreciate a tip.

It has come up a couple of times before, and yes, it is a bit of a pain.

Probably the quickest way out is

negIndex - function(i)

if(length(i))

-i

else

TRUE

... which of course needs braces if typed on the command line

negIndex - function(i)
{
if(length(i))
-i
else
TRUE
}

And I should probably also have said that it works like this:

x2 - c(dolphin.0,dolphin.1)
x2[-grep(exclude,x2)]
character(0)
x2[negIndex(grep(exclude,x2))]
[1] dolphin.0 dolphin.1

--
O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - ([EMAIL PROTECTED])  FAX: (+45) 35327907

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### Re: [R] assigning two conditions to grep()

```On 25/04/07, Abi Ghanem josephine [EMAIL PROTECTED] wrote:

Hi,
i have a problem in assigning 2 conditions to grep()  ,
my data look like this:
DA 24 N7 Rad= 3.4 20 Sac= 0.93 Acc= 4.76
DA 24 N7 Rad= 3.4 14 Sac= 0.65 Acc= 3.33
DA 24 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
DA 24 N7 Rad= 3.4 11 Sac= 0.51 Acc= 2.62
DG 23 N7 Rad= 3.4  8 Sac= 0.37 Acc= 1.91
DG 23 N7 Rad= 3.4  5 Sac= 0.23 Acc= 1.19
DG 23 N7 Rad= 3.4  0 Sac= 0.00 Acc= 0.00
DG 23 N7 Rad= 3.4  3 Sac= 0.14 Acc= 0.71
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
DG 23 O6 Rad= 3.3  1 Sac= 0.04 Acc=  0.22
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
DG 23 O6 Rad= 3.3  0 Sac= 0.00 Acc=  0.00
(it's a data.frame)

at first i wanted all the line begining with A 24:
data[grep(^24, data\$V2)]
this works
and than i wanted to exctract all the lines with G23 N7,
neither the column 23 and the column N7 are unique
so i tried this
data[grep(^23*N7, data),]
but doesn't work

data[ intersect( grep(^24, data\$V2), grep(N7,data\$V3) ) , ]
?

C.

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```

### [R] Getting Confused

```Hello,

I'm getting confused with my experience of R installing.

I had R installed on January without any trouble. (I just had to install
gcc4.1.1)

Now I'd like to install a packages which requires tcl/tk. So basically I
need to reconfigure and re install R right after having installed
tcl/tk.

So I installed tcl/tk I run the process to install R but I receive this
error :

checking for dummy main to link with Fortran libraries...

none

checking for Fortran name-mangling scheme... configure:

error: cannot compile a simple Fortran program See `config.log' for more
details.

I checked in the config.log and the fact is that there's no fortran
compiler installed. But don't gcc already have a fortran compiler in
it?

If somebody could help I would be thankful and especially if somebody
has a clue why it worked without any error before and now yes.

Thanks a lot

Julien Steiner

[[alternative HTML version deleted]]

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```

### [R] Box Ljung Statistics

```
Hi All R Experts,

I met with below mentioned statistics in paper Stock Index Volatility
Forecasting with High Frequency Data
by Eugenie Hol, Siem Jan Koopman
http://ideas.repec.org/p/dgr/uvatin/20020068.html

I would like to ask that what is Box-Ljung portmantacau statistic based
on N squared autocorrelation ?
Is it same as Box-Ljung Statistics of stats package ?
Further, please tell me how to compute it ?

I have a return series of an Index.
in the paper for S  P 100:-)

Sayonara With Smile  With Warm Regards :-)

G a u r a v   Y a d a v
Assistant Manager,
Economic Research  Surveillance Department,
Clearing Corporation Of India Limited.

Address: 5th, 6th, 7th Floor, Trade Wing 'C',  Kamala City, S.B. Marg,
Mumbai - 400 013
Telephone(Office): - +91 022 6663 9398 ,  Mobile(Personal) (0)9821286118
Email(Office) :- [EMAIL PROTECTED] ,  Email(Personal) :-
[EMAIL PROTECTED]

DISCLAIMER AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}

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```

### [R] Unsubscription Confirmation

```Thank you for subscribing. You have now unsubscribed and no more messages will
be sent.

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```

### Re: [R] NA and NaN randomForest

```Hi Andy,

It worked for classification, but not regression. For example:

iris1 - iris[-(1:5),]
iris2 - iris[(1:5),]
iris2[1,3] - NA
iris2[3,1] - NA
iris_sum - sum (iris\$Sepal.Length + iris\$Sepa.Width + iris\$Petal.Length
+ iris\$Petal.Width)
iris_sum1 -  iris_sum[-(1:5)]
iris_sum2 -  iris_sum[(1:5)]
iris_sum.rf - randomForest (iris_sum1 ~ ., iris1[,c(1:4)])
predict (iris_sum.rf, iris2[-5])
predict (iris_sum.rf, iris2[-5])
[1]  9.556591  9.589573 10.104155

# Just to be clear I was hoping for behavior like the linear model has:

iris_sum.lm - lm (iris_sum1 ~ ., iris1[,c(1:4)])
predict (iris_sum.lm, iris2[-5])
12345
NA  9.5   NA  9.4 10.2

In the event that this is not available in the regression part of
randomForest, is a work around possible?

thanks,

Clayton

Liaw, Andy [EMAIL PROTECTED]
04/25/2007 09:59 AM

To
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
cc

Subject
RE: [R] NA and NaN randomForest

Hi Clayton,

If you use the formula interface, then it should do what you want:

R library(randomForest)
randomForest 4.5-18
Type rfNews() to see new features/changes/bug fixes.
R iris1 - iris[-(1:5),]
R iris2 - iris[1:5,]
R iris2[1, 3] - NA
R iris2[3, 1] - NA
R iris.rf - randomForest(Species ~ ., iris1)
R predict(iris.rf, iris2[-5])
[1] NA   setosa NA   setosa setosa
Levels: setosa versicolor virginica

The problem, of course, is that the formula interface is not suitable
for data with large number of variables.  I'll look into doing the same
thing in the default method.

Andy

From: [EMAIL PROTECTED]

Dear R-help,

This is about randomForest's handling of NA and NaNs in test set data.
Currently, if the test set data contains an NA or NaN then
predict.randomForest will skip that row in the output.

I would like to change that behavior to outputting an NA.

Can this be done with flags to randomForest?
If not can some sort of wrapper be built to put the NAs back in?

thanks,

Clayton
_

CONFIDENTIALITY NOTICE\ \ The information contained in this
...{{dropped}}

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### Re: [R] regular expressions with grep() and negative indexing

```I use regexpr() instead of grep() in cases like this, e.g.:

x2[regexpr(exclude,x2)==-1]

(regexpr returns a vector of the same length as character vector given
it, so there's no problem with it returning a zero length vector)

-- Tony Plate

Peter Dalgaard wrote:
Stephen Tucker wrote:
Dear R-helpers,

Does anyone know how to use regular expressions to return vector elements
that don't contain a word? For instance, if I have a vector
x - c(seal.0,seal.1-exclude)
I'd like to get back the elements which do not contain the word exclude,
using something like (I know this doesn't work) but:
grep([^(exclude)],x)

I can use
x[-grep(exclude,x)]
for this case but then if I use this expression in a recursive function, it
will not work for instances in which the vector contains no elements with
that word. For instance, if I have
x2 - c(dolphin.0,dolphin.1)
then
x2[-grep(exclude,x2)]
will give me 'character(0)'

I know I can accomplish this in several steps, for instance:
myfunc - function(x) {
iexclude - grep(exclude,x)
if(length(iexclude)  0) x2 - x[-iexclude] else x2 - x
# do stuff with x2 ...?
}

But this is embedded in a much larger function and I am trying to minimize
intermediate variable assignment (perhaps a futile effort). But if anyone
knows of an easy solution, I'd appreciate a tip.

It has come up a couple of times before, and yes, it is a bit of a pain.

Probably the quickest way out is

negIndex - function(i)

if(length(i))

-i

else

TRUE

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### [R] R News, volume 7, issue 1 is now available

```
Hi

The October 2006 issue of R News is now available on CRAN under the

Torsten
(on behalf of the R News Editorial Board)

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```

### Re: [R] How to identify and exclude the outliers with R?

```It depends on the nature of your data set. There is a package simply called
'outliers', which has the Grubbs/Dixon/Cochran tests. There is also the
Bonferroni outlier test in 'car' package. I'm sure there are more in the
hundreds of packages on CRAN.

HTH

Horace

Shao [EMAIL PROTECTED] 4/25/2007 6:27:37 AM
Hello, everyone,

I want to ask a simple question.
If I have a set  of data,and I want to identify how many outliers there are
in the data.Which packages and functions can I use?

Thanks.

Shao chunxuan.

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### Re: [R] Getting Confused [Broadcast]

```If you are serious in getting useful help, please do try to follow
suggestions in the Posting Guide.  You have not told us anything about
your OS, the versions of R you tried to install, and exactly what you
typed to build/install them.

Many Linux distro by default do not install the Fortran part of GCC, so
don't be surprised if that's the case for you (if you are trying to do
this on some version of Linux).

Andy

From: Steiner, Julien

Hello,

I'm getting confused with my experience of R installing.

I had R installed on January without any trouble. (I just had
to install
gcc4.1.1)

Now I'd like to install a packages which requires tcl/tk. So
basically I
need to reconfigure and re install R right after having installed
tcl/tk.

So I installed tcl/tk I run the process to install R but I
error :

checking for dummy main to link with Fortran libraries...

none

checking for Fortran name-mangling scheme... configure:

error: cannot compile a simple Fortran program See
`config.log' for more
details.

I checked in the config.log and the fact is that there's no fortran
compiler installed. But don't gcc already have a fortran compiler in
it?

If somebody could help I would be thankful and especially if somebody
has a clue why it worked without any error before and now yes.

Thanks a lot

Julien Steiner

[[alternative HTML version deleted]]

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### Re: [R] R News, volume 7, issue 1 is now available

```Torsten Hothorn [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]

The October 2006 issue of R News is now available on CRAN under the

R News
http://cran.r-project.org/doc/Rnews/

April 2007 Issue:
http://cran.r-project.org/doc/Rnews/Rnews_2007-1.pdf

efg

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### Re: [R] Problem installing Rmpi with lam on SGI SLES9

```Hendrik,

Are you starting the lam daemons before starting R?

% lamboot

You might need to specify a 'hosts' argument to lamboot. The default
way Rmpi calls lamboot is with no arguments, and this might simply
create a single lam daemon.

I don't usually use papply, but glancing at it's code suggests that it
does require(Rmpi) and then decides based on the result of
mpi.comm.size what to do. So to debug, load Rmpi and try
mpi.comm.size. As a work-around, I think it should be possible to

library(Rmpi)
mpi.spawn.Rslaves(nslaves=64) # maybe a little bold, initially!

before making your first papply call.

Hope that helps

Martin

Hendrik Fuß [EMAIL PROTECTED] writes:

On 24/04/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Tue, 24 Apr 2007, Hendrik Fuß wrote:

Hi,

I've been trying here to install Rmpi on an SGI IA-64 machine with 64
processors, running SuSE Linux Enterprise Server 9, R 2.4.0 and
lam-mpi 7.1.3. While I've read of similar problems on this list, I
think I've got an entirely new set of error messages to contribute
(see below). I'm not sure what the actual error is and what the @gprel
relocation message is about. Any help greatly appreciated.

I don't know for sure, but on many 64-bit OSes you cannot link code from
static libraries into dynamic shared libraries, and that seems to be the
case with ia64 Linux.  Almost certainly you need to re-compile LAM with
-fPIC flags.

Yes, thanks a million, this solved the problem.

While Rmpi now works, there is another issue connected with this one:
I'm trying to use the papply package. However, when I do:

library(papply)
papply(list(1:10, 1:15, 1:20), sum)
1 slaves are spawned successfully. 0 failed.
master (rank 0, comm 1) of size 2 is running on: behemoth
slave1 (rank 1, comm 1) of size 2 is running on: behemoth
[1] Running serial version of papply\n

Papply only spawns one slave and then decides to run the serial
version instead. I'm not sure how to tell it to use all the 64
processors available.

Hendrik

--
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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http://bioconductor.org

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### Re: [R] levelplot and unequal cell sizes

```On 4/25/07, Waichler, Scott R [EMAIL PROTECTED] wrote:
I am using levelplot() from lattice with grids that have unequal cell
sizes.  This means that the boundary between two cells is not always
half-way between nodes, as levelplot() assumes.  The result is that some
cell sizes are rendered incorrectly, which can be painfully obvious if
using relatively large cells.  Is there any work-around?  I am using the
conditioning capability of lattice and therefore image() would not be a
good way to go.

You might be able to use the tile plot in ggplot, which allows you to
specify the size of each tile (it assumes they're all the same size by
your data, I could provide a worked example.

Regards,

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### Re: [R] R News, volume 7, issue 1 is now available

```Earl F. Glynn wrote:
Torsten Hothorn [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]

The October 2006 issue of R News is now available on CRAN under the

R News
http://cran.r-project.org/doc/Rnews/

April 2007 Issue:
http://cran.r-project.org/doc/Rnews/Rnews_2007-1.pdf

efg

... And maybe the TOC:

Editorial

Viewing Binary Files with the hexView Package
FlexMix: An R Package for Finite Mixture Modelling
Using R to Perform the AMMI Analysis on Agriculture Variety Trials
Inferences for Ratios of Normal Means
Working with Unknown Values
A New Package for Fitting Random Effect Models
Augmenting R with Unix Tools
POT: Modelling Peaks Over a Threshold
Backtests

Review of John Verzani’s Book Using R for Introductory Statistics
DSC 2007
New Journal: Annals of Applied Statistics
Forthcoming Events: useR! 2007
Changes in R 2.5.0
Changes on CRAN
R Foundation News
R News Referees 2006

--
Mango Solutions
data analysis that delivers

Tel:  +44(0) 1249 467 467
Fax:  +44(0) 1249 467 468
Mob:  +44(0) 7813 526 123

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### Re: [R] Coercing data types for use in model.frame

```Prof Brian Ripley wrote:
Moved to R-devel

What is the 'data class'?  In particular what is its underlying type?
And where in model.frame[.default] are you trying to use it (in the
formula, data, in ..., etc).

This is an example of where some reproducible code and the error
messages would be very helpful indeed.

Brian

Brian,

Sorry - this was one of those too late in the day errors.  The problem
was in a function called just before model.frame.  model.frame seems to
work fine with an object of class c('mChoice', 'labelled').  It keeps
mChoice variables as mChoice.  After model.frame is finished I'll change
such variables to factors or matrices.

Frank

On Tue, 24 Apr 2007, Frank E Harrell Jr wrote:

In the Hmisc package there is a new data class 'mChoice' for multiple
choice variables.  There are format and as.numeric methods (the latter
creates a matrix of dummy variables).  mChoice variables are not allowed
by model.frame.  Is there a way to specify a conversion function that
model.frame will use automatically?  I would use as.factor here.
model.frame does not seem to use as.data.frame.foo for individual
variables.

Thanks
Frank

--
Frank E Harrell Jr   Professor and Chair   School of Medicine
Department of Biostatistics   Vanderbilt University

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```

### Re: [R] Box Ljung Statistics

```Gaurav,

I met with below mentioned statistics in paper Stock Index Volatility
Forecasting with High Frequency Data
by Eugenie Hol, Siem Jan Koopman
http://ideas.repec.org/p/dgr/uvatin/20020068.html

I would like to ask that what is Box-Ljung portmantacau statistic based
on N squared autocorrelation ?
Is it same as Box-Ljung Statistics of stats package ?

Yes, it seems the same. But note that the paper computes the
statistic for the raw data and also for the squared data.

Further, please tell me how to compute it ?

If you mean R, use the Box.test() function. If you mean theory,
see any good book on time series like Brockwell and Davis'
Introduction to Time Series and Forecasting.

I have a return series of an Index.
in the paper for S  P 100:-)

I can't help you here since I don't have the 5-minute data used in
the paper.

Rogerio

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### [R] Scripting graph generation

```Hi,

I'm looking to automate the generation of some graphs in R.  I can't seem to
figure out how to script R, and change the output device of hist() or plot()
to create a .gif or .png file.  This seems like something that is probably
really simple, and I've just overlooked the call do do it.  Can anyone point
me in the right direction, or maybe send a sample script?

thanks,
--Mike Huber

[[alternative HTML version deleted]]

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```

### [R] GLS terminology question not related to R

```This is a terminology question not related to R. The literature often
says that OLS is inefficient relative to GLS if the residuals in
the system are correlated ( and the RHS sides of each are not identical
). Does this mean that OLS overestimates residual and coefficient
variances , underestimates them or just gets them wrong and the
direction is not known ? Thanks.

This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}

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```

### [R] [R-pkgs] new package adegenet

```The new package *adegenet* (linked to the ade4 package for multivariate
analysis) has been released on CRAN.

Its main focus is on molecular marker data handling for multivariate
analysis.

Adegenet offers data import/export functions (from GENETIX, Genepop,
Fstat and to the packages genetics and hierfstat) as well as several
methods as well as to usual population genetics methods (HWE tests,
G-statistic tests, ...).

Suggestions, questions and contributions are welcome !

Regards,

Thibaut Jombart.
--
##
Thibaut JOMBART
CNRS UMR 5558 - Laboratoire de Biométrie et Biologie Evolutive
Universite Lyon 1
43 bd du 11 novembre 1918
69622 Villeurbanne Cedex
Tél. : 04.72.43.29.35
Fax : 04.72.43.13.88
[EMAIL PROTECTED]
http://biomserv.univ-lyon1.fr/sitelabo/pageperso.php?id_personne=178

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### Re: [R] GLS terminology question not related to R

```On Wed, 25 Apr 2007, Leeds, Mark (IED) wrote:

This is a terminology question not related to R. The literature often
says that OLS is inefficient relative to GLS if the residuals in
the system are correlated ( and the RHS sides of each are not identical
). Does this mean that OLS overestimates residual and coefficient
variances , underestimates them or just gets them wrong and the
direction is not known ? Thanks.

It does not mean either.

It means that the true variance of the OLS estimates is greater than the
true variance of the GLS estimates.

A separate issue is whether the estimated variance of an OLS estimator is
greater or less than the true variance of the OLS estimator.  This can go
either way.

-thomas

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```

### Re: [R] levelplot and unequal cell sizes

```On 4/25/07, Waichler, Scott R [EMAIL PROTECTED] wrote:
I am using levelplot() from lattice with grids that have unequal cell
sizes.  This means that the boundary between two cells is not always
half-way between nodes, as levelplot() assumes.

levelplot() is not supposed to make any such assumptions. Can you

-Deepayan

The result is that some
cell sizes are rendered incorrectly, which can be painfully obvious if
using relatively large cells.  Is there any work-around?  I am using the
conditioning capability of lattice and therefore image() would not be a
good way to go.

Thanks,
Scott Waichler
Pacific Northwest National Laboratory
scott.waichler _at_ pnl.gov

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```

### [R] help

```Hi all,

I have 2 questions:

1)How do I calculate the mean on an imported txt file? I've imported the
file below and that's what it looks like imported. How do I then calcuate
the mean, median, or mode on the column LeafArea using the desktop R
package?

Any help would be greatly appreciated!!

Thanks,

Nat

LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness
1 13 0.12   0.12   0.21  0.00
2 13 0.17   0.17   0.36  0.00
3 13 0.21   0.05   0.47  0.16
4 13 0.11   0.14   0.23  0.21
5 23 0.03   0.27   0.16  0.60
6 23 0.08   0.20   0.15  0.75
7 23 0.22   0.05   0.24  1.09
8 23 0.20   0.10   0.26  1.30
9 23 0.18   0.10   0.33  1.33
1023 0.14   0.07   0.19  1.36
1123 0.16   0.13   0.22  1.38
1223 0.18   0.06   0.25  1.39
1323 0.05   0.00   0.07  1.40
1423 0.11   0.01   0.21  1.41
1523 0.22   0.04   0.31  1.41
1623 0.09   0.10   0.13  1.44
1723 0.09   0.10   0.13  1.44
1823 0.13   0.08   0.19  1.46
1923 0.15   0.13   0.22  1.47
2023 0.15   0.03   0.22  1.47
2123 0.21   0.01   0.31  1.48
2213 0.21   0.14   0.32  1.50
2323 0.10   0.00   0.15  1.50
2413 0.26   0.60   0.40  1.53
2523 0.12   0.18   0.20  1.54
2623 0.20   0.15   0.31  1.55
2713 0.19   0.16   0.31  1.60
2813 0.13   0.00   0.21  1.62
2913 0.13   0.01   0.21  1.62
3013 0.37   0.27   0.60  1.62
3123 0.11   0.09   0.18  1.64
3223 0.14   0.00   0.23  1.64
3323 0.15   0.08   0.21  1.64
3423 0.20   0.10   0.33  1.65
3523 0.15   0.01  -0.25  1.67
3623 0.17   0.06   0.29  1.67
3723 0.13   0.08   0.22  1.69
3813 0.16   0.31   0.27  1.70
3913 0.21   0.01   0.40  1.70
4013 0.14   0.07   0.29  1.71
4123 0.14   0.00   0.24  1.71
4223 0.21   0.14   0.35  1.71
4323 0.11   0.09   0.19  1.73
4423 0.15   0.01   0.26  1.73
4523 0.19   0.11   0.33  1.74
4610 0.28   0.27   0.50  1.79
4713 0.10   0.01   0.18  1.80
4823 0.05   0.00   0.09  1.80
4913 0.12   0.11   0.22  1.83
5013 0.20   0.05   0.37  1.85
5123 0.14   0.14   0.26  1.86
5213 0.15   0.07   0.28  1.87
5313 0.15   0.01   0.28  1.87
5423 0.12   0.08   0.23  1.92
5523 0.15   0.00   0.29  1.93
5613 0.17   0.00   0.34  2.00
5713 0.21   0.02   0.42  2.00
5823 0.13   0.08   0.26  2.00
5913 0.16   0.06   0.32  2.05
6013 0.14   0.14   0.29  2.07
6123 0.12   0.08   0.25  2.08
6213 0.17   0.06   0.36  2.12
6313 0.13   0.08   0.28  2.13
6413 0.20   0.10   0.43  2.15
6513 0.26   0.08   0.56  2.15
6613 0.20   0.10   0.44  2.20
6713 0.19   0.11   0.42  2.21
6813 0.08   0.00   0.18  2.25
6913 0.12   0.00   0.27  2.25
7013 0.12   0.08   0.27  ```

### Re: [R] Scripting graph generation

```On Wed, 2007-04-25 at 13:21 -0400, Mike Huber wrote:
Hi,

I'm looking to automate the generation of some graphs in R.  I can't seem to
figure out how to script R, and change the output device of hist() or plot()
to create a .gif or .png file.  This seems like something that is probably
really simple, and I've just overlooked the call do do it.  Can anyone point
me in the right direction, or maybe send a sample script?

To plot multiple PNG files

m - matrix(runif(100*10), nrow=10)
png(filename = 'plot%03d.png')
for (i in 1:nrow(m)) {
plot(m[i,])
}
dev.off()

To plot multiple graphs as individual pages of a PDF file

m - matrix(runif(100*10), nrow=10)
pdf(file = 'plot.pdf')
for (i in 1:nrow(m)) {
plot(m[i,])
}
dev.off()

HTH,

---
Rajarshi Guha [EMAIL PROTECTED]
GPG Fingerprint: 0CCA 8EE2 2EEB 25E2 AB04 06F7 1BB9 E634 9B87 56EE
---
In matrimony, to hesitate is sometimes to be saved.
-- Butler

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```

### [R] Самый доступный и недорогой метод похудения

```   «ÇÎËÎÒÀß ÑÅÐÜÃÀ» - ÈÇÁÀÂËÅÍÈÅ ÎÒ ËÈØÍÈÕ ÊÈËÎÃÐÀÌÌÎÂ (4 ÊÃ Â ÌÅÑ.)!

ÌÈËËÈÎÍÛ ËÞÄÅÉ   ÈÇÁÀÂÈËÈÑÜ ÎÒ ËÈØÍÈÕ ÊÈËÎÃÐÀÌÌÎÂ ÁËÀÃÎÄÀÐß ÇÎËÎÒÎÉ_ÑÅÐÜÃÅ
ÁÅÇ ÔÈÇÈ×ÅÑÊÈÕ ÓÏÐÀÆÍÅÍÈÉ, ÈÑÒßÇÀÞÙÈÕ ÄÈÅÒ, ÒÀÁËÅÒÎÊ!!

ÑÒÎÈÌÎÑÒÜ «ÇÎËÎÒÎÉ ÑÅÐÜÃÈ» ÂÑÅÃÎ 3000 ÐÓÁ.!!

Ïðèíöèï ìåòîäà Çîëîòàÿ Ñåðüãà ñîçäàí îñíîâûâàÿñü íà ñåêðåòàõ òðàäèöèîííîé
Êèòàéñêîé ìåäèöèíû. «Çîëîòàÿ Ñåðüãà» âîçäåéñòâóåò íà áèîëîãè÷åñêè àêòèâíûå
òî÷êè óøíîé ðàêîâèíû, óñòàíàâëèâàåòñÿ íà äëèòåëüíîå âðåìÿ.  Â èòîãå ïîäàâëÿåòñÿ
àïïåòèò. Ìîçã ïîëó÷àåò ñèãíàë, ÷òî ÷åëîâåê ñûò.

ÏÐÅÈÌÓÙÅÑÒÂÀ ÄÀÍÍÎÃÎ ÌÅÒÎÄÀ:
Ñàìûé äîñòóïíûé è íåäîðîãîé ìåòîä ïîõóäåíèÿ. Âû èçáàâèòåñü îò ïðîáëåìíûõ
çîí. Ñíèæåíèå âåñà îò 3 äî 10 êã çà îäèí ìåñÿö. Ïîäàâëÿåòñÿ ÷óâñòâî ãîëîäà
è àïïåòèò. Ïîõóäåòü ëåãêî  Âû ïîñåùàåòå ñïåöèàëèñòà âñåãî 1 ðàç. Âàì íå
ïðèäåòñÿ çàíèìàòüñÿ ñïîðòîì è ïðèíèìàòü òàáëåòêè. Ïîñëå îêîí÷àíèÿ êóðñà
âåñ íå íàáèðàåòñÿ.
Äåéñòâåííîñòü äàííîãî ñïîñîáà çàâèñèò îò ñòðîåíèÿ æèðîâîé òêàíè, íàñëåäñòâåííîé
ñêëîííîñòè, îñîáåííîñòåé îðãàíèçìà, è â êàæäîì ñëó÷àå îïðåäåëÿåòñÿ ïåðñîíàëüíî.
Óçíàéòå ïîäðîáíîñòè ïî òåëåôîíàì!!!

Íàø íîìåð òåëåôîíà (495) 739`34_39 â ëþáîå âðåìÿ
ìåòðî Òóøèíñêàÿ, óë. Ìèòèíñêàÿ, ä.43, ì-í Ìèòèíî
Ñò.Ì. «Ëþáëèíî» (100 ì îò ìåòðî), óë. Íîâîðîññèéñêàÿ, ä.28

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```

### [R] dmnorm not meant for 1024-dimensional data?

```Hello,

I have some data generated by a simple mixture of Gaussians (more like
K-means) and (as a test) am using dmnorm to calculate the probability
of each data point coming from each Gaussian.  However, I get only
zero probabilities.

This code works in low dimensions (tried 2 and 3 already).  I have run
into many implementations that do not work in high dimension, but I
thought that I was safe with dmnorm because it has an option to
compute the log of the probability.

So, is dmnorm not intended to be used with data of such high dimensionality?

Thank you,

dan elliott

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```

### Re: [R] negative number to positive number

```Quoting H. Paul Benton [EMAIL PROTECTED]:

Hello all,

I know this is a pretty easy question but I can't find it in S poetry or
R help.

How can I make a negative number positive. Such as
-5 to be +5
I tried +(-5), but that didn't work.

So no, I don't mean taking a -5^2 just to get a positive number.
This is in a function so it's not just -5 it's x. :)

Thanks,

Paul

how about just multiplying it by -1??? :-)

-5*(-1)

Jose

--
Dr. Jose I. de las Heras  Email: [EMAIL PROTECTED]
The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131 6513374
Institute for Cell  Molecular BiologyFax:   +44 (0)131 6507360
University of Edinburgh
Edinburgh EH9 3JR
UK

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```

### [R] Help on 'find.BIB' function

```Hello everyone,

I am trying to use the 'find.BIB' function to construct a balanced incomplete
block design.  When I ran the example given in the help file
(find.BIB(10,30,4)), I obtained the following error message:

Error in optBlock(~., withinData = factor(1:trt), blocksize = rep(k, b)) :

I investigated what the optBlock function is doing but the manual / help
files did not give me any information regarding the above error.

I hope somebody can help me regarding this matter.

Best regards,

Jason Parcon

-

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```

### Re: [R] barchart producing incorrect number of barcharts when columns renamed

```On 4/25/07, Colm G. Connolly [EMAIL PROTECTED] wrote:
Hi everybody,

I'm having problems with the barchart command in the lattice package.

[... ... ...]

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You seem to have missed this footer that appears in every r-help
message. Your code is not reproducible, and not minimal by a long,
long, shot.

-Deepayan

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```

### Re: [R] Help on 'find.BIB' function

```Jason Parcon wrote:
Hello everyone,

I am trying to use the 'find.BIB' function to construct a balanced
incomplete block design.  When I ran the example given in the help file
(find.BIB(10,30,4)), I obtained the following error message:

Error in optBlock(~., withinData = factor(1:trt), blocksize = rep(k, b)) :

I investigated what the optBlock function is doing but the manual / help
files did not give me any information regarding the above error.

I hope somebody can help me regarding this matter.

The following seems to work for me:

library(crossdes)

set.seed(671969)

find.BIB(10,30,4)
[,1] [,2] [,3] [,4]
[1,]457   10
[2,]123   10
[3,]156   10
[4,]289   10
[5,]3567
[6,]349   10
[7,]1589
[8,]1679
[9,]1247
[10,]268   10
[11,]2357
[12,]1679
[13,]267   10
[14,]1239
[15,]2568
[16,]2459
[17,]3468
[18,]158   10
[19,]2478
[20,]369   10
[21,]1246
[22,]378   10
[23,]2359
[24,]145   10
[25,]4689
[26,]479   10
[27,]1378
[28,]3456
[29,]5789
[30,]1348

I get the same error you report if I don't do the set.seed() step.

sessionInfo()
R version 2.4.1 Patched (2007-03-31 r41127)
i386-pc-mingw32

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252

attached base packages:
[1] stats graphics  grDevices utils datasets  methods
[7] base

other attached packages:
crossdes  MASSgtools AlgDesign
1.0-7  7.2-33   2.3.1   1.0-7

Best regards,

Jason Parcon

-

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--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894

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```

### [R] help

```Hi,

Would anyone know how to calculate the modal value of LeafArea?

Thank-you very much!!

Nat

__

Hi all,

I have 2 questions:

1)How do I calculate the mean on an imported txt file? I've imported the
file below and that's what it looks like imported. How do I then calcuate
the mean, median, or mode on the column LeafArea using the desktop R
package?

Any help would be greatly appreciated!!

Thanks,

Nat

LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness
1 13 0.12   0.12   0.21  0.00
2 13 0.17   0.17   0.36  0.00
3 13 0.21   0.05   0.47  0.16
4 13 0.11   0.14   0.23  0.21
5 23 0.03   0.27   0.16  0.60
6 23 0.08   0.20   0.15  0.75
7 23 0.22   0.05   0.24  1.09
8 23 0.20   0.10   0.26  1.30
9 23 0.18   0.10   0.33  1.33
1023 0.14   0.07   0.19  1.36
1123 0.16   0.13   0.22  1.38
1223 0.18   0.06   0.25  1.39
1323 0.05   0.00   0.07  1.40
1423 0.11   0.01   0.21  1.41
1523 0.22   0.04   0.31  1.41
1623 0.09   0.10   0.13  1.44
1723 0.09   0.10   0.13  1.44
1823 0.13   0.08   0.19  1.46
1923 0.15   0.13   0.22  1.47
2023 0.15   0.03   0.22  1.47
2123 0.21   0.01   0.31  1.48
2213 0.21   0.14   0.32  1.50
2323 0.10   0.00   0.15  1.50
2413 0.26   0.60   0.40  1.53
2523 0.12   0.18   0.20  1.54
2623 0.20   0.15   0.31  1.55
2713 0.19   0.16   0.31  1.60
2813 0.13   0.00   0.21  1.62
2913 0.13   0.01   0.21  1.62
3013 0.37   0.27   0.60  1.62
3123 0.11   0.09   0.18  1.64
3223 0.14   0.00   0.23  1.64
3323 0.15   0.08   0.21  1.64
3423 0.20   0.10   0.33  1.65
3523 0.15   0.01  -0.25  1.67
3623 0.17   0.06   0.29  1.67
3723 0.13   0.08   0.22  1.69
3813 0.16   0.31   0.27  1.70
3913 0.21   0.01   0.40  1.70
4013 0.14   0.07   0.29  1.71
4123 0.14   0.00   0.24  1.71
4223 0.21   0.14   0.35  1.71
4323 0.11   0.09   0.19  1.73
4423 0.15   0.01   0.26  1.73
4523 0.19   0.11   0.33  1.74
4610 0.28   0.27   0.50  1.79
4713 0.10   0.01   0.18  1.80
4823 0.05   0.00   0.09  1.80
4913 0.12   0.11   0.22  1.83
5013 0.20   0.05   0.37  1.85
5123 0.14   0.14   0.26  1.86
5213 0.15   0.07   0.28  1.87
5313 0.15   0.01   0.28  1.87
5423 0.12   0.08   0.23  1.92
5523 0.15   0.00   0.29  1.93
5613 0.17   0.00   0.34  2.00
5713 0.21   0.02   0.42  2.00
5823 0.13   0.08   0.26  2.00
5913 0.16   0.06   0.32  2.05
6013 0.14   0.14   0.29  2.07
6123 0.12   0.08   0.25  2.08
6213 0.17   0.06   0.36  2.12
6313 0.13   0.08   0.28  2.13
6413 0.20   0.10   0.43  2.15
6513 0.26   0.08   0.56  2.15
6613 0.20   0.10   0.44  2.20
6713 0.19   0.11   0.42  2.21
6813 0.08   0.00   0.18  2.25
691  ```

### Re: [R] negative number to positive number

```?abs

Clint BowmanINTERNET:   [EMAIL PROTECTED]
Air Dispersion Modeler  INTERNET:   [EMAIL PROTECTED]
Air Quality Program VOICE:  (360) 407-6815
Department of Ecology   FAX:(360) 407-7534

USPS:   PO Box 47600, Olympia, WA 98504-7600
Parcels:300 Desmond Drive, Lacey, WA 98503-1274

On Wed, 25 Apr 2007 [EMAIL PROTECTED] wrote:

Quoting H. Paul Benton [EMAIL PROTECTED]:

Hello all,

I know this is a pretty easy question but I can't find it in S poetry or
R help.

How can I make a negative number positive. Such as
-5 to be +5
I tried +(-5), but that didn't work.

So no, I don't mean taking a -5^2 just to get a positive number.
This is in a function so it's not just -5 it's x. :)

Thanks,

Paul

how about just multiplying it by -1??? :-)

-5*(-1)

Jose

--
Dr. Jose I. de las Heras  Email: [EMAIL PROTECTED]
The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131 6513374
Institute for Cell  Molecular BiologyFax:   +44 (0)131 6507360
University of Edinburgh
Edinburgh EH9 3JR
UK

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### Re: [R] Problem installing Rmpi with lam on SGI SLES9

```On 25/04/07, Martin Morgan [EMAIL PROTECTED] wrote:
Hendrik Fuß [EMAIL PROTECTED] writes:
I'm trying to use the papply package. However, when I do:

library(papply)
papply(list(1:10, 1:15, 1:20), sum)
1 slaves are spawned successfully. 0 failed.
master (rank 0, comm 1) of size 2 is running on: behemoth
slave1 (rank 1, comm 1) of size 2 is running on: behemoth
[1] Running serial version of papply\n

Papply only spawns one slave and then decides to run the serial
version instead. I'm not sure how to tell it to use all the 64
processors available.

Hendrik,

Are you starting the lam daemons before starting R?

% lamboot

You might need to specify a 'hosts' argument to lamboot. The default
way Rmpi calls lamboot is with no arguments, and this might simply
create a single lam daemon.

Thanks, that was a pointer in the right direction.

The solution is to edit the file /etc/lam/lam-bhost.def and specify
the number of cpus (see man bhost)

cheers
Hendrik

--
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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```

### Re: [R] GLS terminology question not related to R

```Thomas Lumley wrote:
On Wed, 25 Apr 2007, Leeds, Mark (IED) wrote:

This is a terminology question not related to R. The literature often
says that OLS is inefficient relative to GLS if the residuals in
the system are correlated ( and the RHS sides of each are not identical
). Does this mean that OLS overestimates residual and coefficient
variances , underestimates them or just gets them wrong and the
direction is not known ? Thanks.

It does not mean either.

It means that the true variance of the OLS estimates is greater than the
true variance of the GLS estimates.

Yes, and to complicate things further that is not necessarily true if
many parameters go into determining the variances and covariances
necessary for GLS. (Cue recent discussion comparing T^2 and F tests).
A separate issue is whether the estimated variance of an OLS estimator is
greater or less than the true variance of the OLS estimator.  This can go
either way.

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```

### Re: [R] Fwd: dmnorm not meant for 1024-dimensional data?

```On Wed, Apr 25, 2007 at 01:25:09PM -0500, Daniel Elliott wrote:
Hello.  Thank you for your mnormt package.  Below is an email I sent to
R-help regarding the function dmnorm.

Hello.

You have not provided information on what specifically you
have tried as for parameters and evaluation points, so I have
chosen a particularly simple case, namely the 1024-dimensional
density with independent standard marginals, using this code

d- 1024
x - 1
const- 0.5*log(2*pi)
log.pdf -0
for (i in 1:d)   log.pdf - log.pdf - x^2/2 -const
#
S - diag(d)
mu- rep(0,d)
X - rep(x,d)
log.pdf2 - dmnorm(X,mu,S, log=TRUE)
cat(log.pdf, log.pdf2, abs(log.pdf-log.pdf2),\n)

-1453 -1453 2.137e-11
which seems quite decent to me. Obviously, you must not take exp()
of this log-density, as otherwise a 0 is indeed produced, but
the reason is not in the package, rather in the possibility of
representing that number using standard floating-point
numerical devices.

best regards,

Thank you.

- dan elliott

-- Forwarded message --
From: Daniel Elliott [EMAIL PROTECTED]
Date: Apr 25, 2007 1:21 PM
Subject: dmnorm not meant for 1024-dimensional data?
To: r-help@stat.math.ethz.ch

Hello,

I have some data generated by a simple mixture of Gaussians (more like
K-means) and (as a test) am using dmnorm to calculate the probability
of each data point coming from each Gaussian.  However, I get only
zero probabilities.

This code works in low dimensions (tried 2 and 3 already).  I have run
into many implementations that do not work in high dimension, but I
thought that I was safe with dmnorm because it has an option to
compute the log of the probability.

So, is dmnorm not intended to be used with data of such high dimensionality?

Thank you,

dan elliott

--
Dipart.Scienze Statistiche, Università di Padova, Italia
tel. +39 049 8274147,  http://azzalini.stat.unipd.it/

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```

### [R] applying rbind to list elements

```Hi,

I have a list of n data.frames (or matrices) which I would like to
convert to a single data.frame using rbind:

x - rbind( l[[1]], l[[2]], l[[3]], l[[4]], ..., l[[n]] )

Is there a simple way to do this?

thanks
Hendrik

--
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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```

### Re: [R] applying rbind to list elements

```On 4/25/2007 4:09 PM, Hendrik Fuß wrote:
Hi,

I have a list of n data.frames (or matrices) which I would like to
convert to a single data.frame using rbind:

x - rbind( l[[1]], l[[2]], l[[3]], l[[4]], ..., l[[n]] )

Is there a simple way to do this?

do.call(rbind, l).

Duncan Murdoch

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```

### Re: [R] applying rbind to list elements

```I knew there would be a simple solution.

thanks everybody.

On 25/04/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 4/25/2007 4:09 PM, Hendrik Fuß wrote:
Hi,

I have a list of n data.frames (or matrices) which I would like to
convert to a single data.frame using rbind:

x - rbind( l[[1]], l[[2]], l[[3]], l[[4]], ..., l[[n]] )

Is there a simple way to do this?

do.call(rbind, l).

Duncan Murdoch

--
Hendrik Fuß
PhD student
Systems Biology Research Group

University of Ulster, School of Biomedical Sciences
Cromore Road, Coleraine, BT52 1SA, Northern Ireland

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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [R] levelplot and unequal cell sizes

```Hadley and Deepayan,

Thank you for responding.  Here is a simple example of what I'm talking
about.  It is a grid that is 5 cells wide by 2 cells tall.  The width of
the cells in the x-direction is variable; the cells at either end have
width = 4 units, and the three cells in the middle have width = 2 units.
My objective is to have the color contour boundaries fall on the cell
boundaries instead of equidistant between cell nodes.  In the plot, I
want the cyan/blue and orange/gray boundaries to be located at the red
cell boundary lines.  Also, the colored regions should extend to the
ends of the domain (x = 0, 14).

library(lattice)

x.node - rep(c(2, 5, 7, 9, 12), 2)
y.node - c(rep(0.5, 5), rep(1.5, 5))
z - rep(1:5, 2)
contour.levels - seq(0.5, 5.5, by=1)
x.cell.boundary - c(0, 4, 6, 8, 10, 14)
contour.colors - c(cyan, blue, green, orange, gray)

print(
levelplot(z ~ x.node * y.node,
panel = function(z,...) {
panel.levelplot(z,...)
panel.abline(v = x.cell.boundary, col=red)
},
xlim = range(x.cell.boundary),
at=contour.levels,
colorkey = list(space=top, width=1, height=0.9,
at=1:5,
col=contour.colors,
labels=list(labels=z, at=z)
),
col.regions=contour.colors,
region = T,
contour = F
)
)

Thanks,
Scott Waichler
Pacific Northwest National Laboratory
scott.waichler _at_ pnl.gov
http://hydrology.pnl.gov

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```

### [R] dist label names

```Hello,

I am trying to do a multi-dimensional scaling of the World Bank's quality
of governance indicators for the Balkan region. I am having trouble
labelling my plot. Could some kind person help me out. How do I set the
attribute Label by a variable (say, Code)? At present I get this:

qog.dist-dist(Balkans.data, method = euclidean, diag = FALSE, upper =
FALSE)
labels(qog.dist)
[1] 1  2  3  4  5  6  7  8  9  10

I know this must be a really basic questions, but none of the 5-6 books on
my shelf nor a web search have proved much help. In case you hadn't
guessed, I am pretty new to R.

Lindsay

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```

### Re: [R] applying rbind to list elements

```do.call(rbind, l)

or, in the case of matrices, using the abind package:

abind(l, along=1)

library(abind)
l - list(matrix(1:6, ncol=2), matrix(11:14, ncol=2))
abind(l, along=1)
[,1] [,2]
[1,]14
[2,]25
[3,]36
[4,]   11   13
[5,]   12   14

Hendrik Fuß wrote:
Hi,

I have a list of n data.frames (or matrices) which I would like to
convert to a single data.frame using rbind:

x - rbind( l[[1]], l[[2]], l[[3]], l[[4]], ..., l[[n]] )

Is there a simple way to do this?

thanks
Hendrik

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and provide commented, minimal, self-contained, reproducible code.

```

### [R] Sum of specific column

```I have a data set that I have imported (not sure if that makes a
difference) and I would like to calculate the sum of only specific
columns.  I have tried
colSums(dataset, by=list(dataset\$col5), dims=1) and I get an error of
unused arguments
I have also tried
aggregate(dataset, by=list(dataset\$col5), sum) and I get the error that
sum is not meaningful for factors.

I want to only calculate the sum for specific columns because some of
the columns have words in them and I have not been able to find anything
else that would help or why these errors are occuring.
Jacquie

[[alternative HTML version deleted]]

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```

### [R] aggregate similar to SPSS

```Hi,

Does anyone know if: with R can you take a set of numbers and aggregate
them like you can in SPSS? For example, could you calculate the percentage
of people who smoke based on a dataset like the following:

smoke = 1
non-smoke = 2

variable
1
1
1
2
2
1
1
1
2
2
2
2
2
2

When aggregated, SPSS can tell you what percentage of persons are smokers
based on the frequency of 1's and 2's. Can R statistical package do a
similar thing?

Thanks,

Nat

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and provide commented, minimal, self-contained, reproducible code.

```

### Re: [R] levelplot and unequal cell sizes

```On 4/25/07, Waichler, Scott R [EMAIL PROTECTED] wrote:

Thank you for responding.  Here is a simple example of what I'm talking
about.  It is a grid that is 5 cells wide by 2 cells tall.  The width of
the cells in the x-direction is variable; the cells at either end have
width = 4 units, and the three cells in the middle have width = 2 units.
My objective is to have the color contour boundaries fall on the cell
boundaries instead of equidistant between cell nodes.  In the plot, I
want the cyan/blue and orange/gray boundaries to be located at the red
cell boundary lines.  Also, the colored regions should extend to the
ends of the domain (x = 0, 14).

library(lattice)

x.node - rep(c(2, 5, 7, 9, 12), 2)
y.node - c(rep(0.5, 5), rep(1.5, 5))
z - rep(1:5, 2)
contour.levels - seq(0.5, 5.5, by=1)
x.cell.boundary - c(0, 4, 6, 8, 10, 14)
contour.colors - c(cyan, blue, green, orange, gray)

print(
levelplot(z ~ x.node * y.node,
panel = function(z,...) {
panel.levelplot(z,...)
panel.abline(v = x.cell.boundary, col=red)
},
xlim = range(x.cell.boundary),
at=contour.levels,
colorkey = list(space=top, width=1, height=0.9,
at=1:5,
col=contour.colors,
labels=list(labels=z, at=z)
),
col.regions=contour.colors,
region = T,
contour = F
)
)

You are right, panel.levelplot is indeed assuming that the boundaries
are between consecutive midpoints. There is no built in way around
that; there simply isn't enough information available to the panel
function.

The cleanest solution, in principle, is to write your own panel
function that ends up calling panel.polygon or grid.polygon.
panel.levelplot is a good starting point (the only tricky part is
getting the colors right, almost everything else you can get rid of).
Maybe Hadley will have a simpler solution.

Here's a possible implementation using a panel function:

my.panel.levelplot -
function (x, y, z, subscripts, at = pretty(z),
col.regions = regions\$col, ...,
w, h)
{
regions - trellis.par.get(regions)
numcol - length(at) - 1
numcol.r - length(col.regions)
col.regions - if (numcol.r = numcol)
rep(col.regions, length = numcol)
else col.regions[floor(1+(1:numcol-1) * (numcol.r-1)/(numcol-1))]
zcol - findInterval(z, at, rightmost.closed = TRUE)
x - as.numeric(x[subscripts])
y - as.numeric(y[subscripts])
z - as.numeric(z[subscripts])
w - as.numeric(w[subscripts])
h - as.numeric(h[subscripts])
zcol - as.numeric(zcol[subscripts])
print(data.frame(z, x.node, y.node, w.node, h.node, col.regions[zcol]))
panel.rect(x = x, y = y, width = w, height = h,
col = col.regions[zcol], ...)
}

x.node - rep(c(2, 5, 7, 9, 12), 2)
y.node - c(rep(0.5, 5), rep(1.5, 5))
z - rep(1:5, 2)
contour.levels - seq(0.5, 5.5, by=1)
x.cell.boundary - c(0, 4, 6, 8, 10, 14)
contour.colors - c(cyan, blue, green, orange, gray)

w.node - rep(diff(x.cell.boundary), 2)
h.node - rep(1, 10)

levelplot(z ~ x.node * y.node, h = h.node, w = w.node,
panel = function(...) {
my.panel.levelplot(...)
panel.abline(v = x.cell.boundary, col=red)
},
xlim = range(x.cell.boundary),
at=contour.levels,
colorkey =
list(space=top, width=1, height=0.9,
at=contour.levels,
col=contour.colors,
labels=list(labels=z, at=z)),
col.regions=contour.colors)

-Deepayan

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```

### [R] Biostatistician Opportunities at Vanderbilt

```The Department of Biostatistics at Vanderbilt University's School of
Medicine has openings for biostatisticians at all levels.  Details and
application procedures may be found at
http://biostat.mc.vanderbilt.edu/JobOpenings .  For M.S. and B.S.
biostatisticians we are especially interested in statisticians
proficient in R, S-Plus, or Stata.  We have faculty positions available
at the Assistant, Associate, and Professor levels.

Frank Harrell
Chairman, Dept. of Biostatistics

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```

### Re: [R] aggregate similar to SPSS

```?table

On Wednesday 25 April 2007 14:32, Natalie O'Toole wrote:
Hi,

Does anyone know if: with R can you take a set of numbers and aggregate
them like you can in SPSS? For example, could you calculate the percentage
of people who smoke based on a dataset like the following:

smoke = 1
non-smoke = 2

variable
1
1
1
2
2
1
1
1
2
2
2
2
2
2

When aggregated, SPSS can tell you what percentage of persons are smokers
based on the frequency of 1's and 2's. Can R statistical package do a
similar thing?

Thanks,

Nat

__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
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self-contained, reproducible code.

--
Dylan Beaudette
Soil Resource Laboratory
http://casoilresource.lawr.ucdavis.edu/
University of California at Davis
530.754.7341

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```

### Re: [R] aggregate similar to SPSS

```Hi Nat,

can I suggest, without offending, that you purchase and read Peter
Dalgaard's Introductory Statistics with R or Michael Crawley's
Statistics: An Introduction using R or Venables and Ripley's Modern
Applied Statistics with S or Maindonald and Braun's Data Analysis
and Graphics Using R: An Example-based Approach,

or

http://cran.r-project.org/doc/manuals/R-intro.pdf

or one of the numerous contributed documents at

http://cran.r-project.org/other-docs.html

?

I hope that this helps,

Andrew.

On Wed, Apr 25, 2007 at 03:32:11PM -0600, Natalie O'Toole wrote:
Hi,

Does anyone know if: with R can you take a set of numbers and aggregate
them like you can in SPSS? For example, could you calculate the percentage
of people who smoke based on a dataset like the following:

smoke = 1
non-smoke = 2

variable
1
1
1
2
2
1
1
1
2
2
2
2
2
2

When aggregated, SPSS can tell you what percentage of persons are smokers
based on the frequency of 1's and 2's. Can R statistical package do a
similar thing?

Thanks,

Nat

__
R-help@stat.math.ethz.ch mailing list
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and provide commented, minimal, self-contained, reproducible code.

--
Andrew Robinson
Department of Mathematics and StatisticsTel: +61-3-8344-9763
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/

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```

### [R] ANOVA results in R conflicting with results in other software packages

```Hi,

I'm wrestling with an analysis of a dataset, which I previously
analyzed in SYSTAT, but am now converting to R and was doing a
re-analysis. I noticed, however, that the same model yields different
results (different sums of squares) from the two programs. I first
thought this might be because the two programs use different
calculations to get the sums of squares, but the problem persisted
even after I specified type III sums of squares. Can anyone help me
by clarifying why there is this discrepancy?

The data table is:

A   yes  35  21
A   yes  30  13
A   no   73 -6
A   yes 22   3
C   yes  19 -1
A   no  53  1
C   no   48 -27
A   yes  32  26
A   yes 14   1
A   no   83 42
A   yes  19 -3
A   no  66   -7
C   no  69  -14
A   yes  30 30
C   no   69 -22
A   yes  10  6
C   no  65  -15
A   yes  11 4
A   yes  15 15
A   no  77  30
C   yes 11   11
A   no  48   -4
C   yes 29  -4
A   yes 0   0
C   no  69   -2
A   yes 10   -40
C   yes  8  -6
C   no   91 -2
C   no  65  13
A   yes 12   0
C   yes 16   -26
C   yes 38  -12
A   no  43  20
C   no  81   -7
A   yes  9  9
C   no  100 25
A   yes 18   12
C   yes 27   -6
A   yes 11   -3

The dialogue in R is as follows:
library(car)

attach(nn)
ls(2)

Call:
lm(formula = maladapt ~ host * increase * size2)

Coefficients:
(Intercept)hostC
increase size2yes
59.54144 17.13828
0.34487-44.41381
hostC:increase   hostC:size2yes
increase:size2yes  hostC:increase:size2yes
0.30449-12.50558
0.03766 -0.90697

Anova(fm, type=III)
Anova Table (Type III tests)

Sum Sq Df  F valuePr(F)
(Intercept) 18348.5  1 152.9683 1.595e-13 ***
host  920.9  1   7.6774  0.009366 **
increase  278.4  1   2.3210  0.137773
size27447.0  1  62.0841 6.806e-09 ***
host:increase 105.1  1   0.8758  0.356584
host:size2266.9  1   2.2252  0.145880
increase:size2  2.0  1   0.0171  0.896902
host:increase:size2   332.3  1   2.7703  0.106108
Residuals3718.4 31
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Contrast this with the results from SYSTAT

SourceSum-of-SquaresdfMean-SquareF-ratioP
HOST\$808.9491808.9496.7440.014
SIZE2\$17525.418117525.418146.1060.000
INCREASE540.5791540.5794.5070.042
SIZE2\$*HOST\$266.9151266.9152.2250.146
SIZE2\$*INCREASE279.3891279.3892.3290.137
HOST\$*INCREASE35.869135.8690.2990.588
SIZE2\$*HOST\$*INCREASE332.2931332.2932.7700.106
Error3718.44131119.950

I've been trying to find anything in the documentation for anova()
that would give a default that is different from what is in SYSTAT,
but part of the problem is that SYSTAT is somewhat opaque as to its
calculations, so it is hard to contrast the two. I would really
really welcome feedback as to what may cause this discrepancy.

Thanks very much for your help,

Dan Bolnick
Section of Integrative Biology
University of Texas at Austin
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