[R] How to open more windows to make more graphs at once!

2007-03-07 Thread Faramarzi Monireh

Dear R users,
I have a data frame (test) including five columns of upper (numeric), lower 
(numeric), observed (numeric), best_sim (numeric) and stname (factor with 80 
levels, each level with different length). Now I would like to write a short 
program to draw one graph as follow for each level of stname but I would like 
also to draw each time 12 graphs for the 12 levels of stname in the same 
graphic windows and save it as jpeg' file . This means at the end I will have 
7 (80 levels/12=7) graphic windows and 7 jpeg files each one with 12 graphs 
(the last one with 8 graphs) for the 12 levels of stname. I already wrote the 
following script to do it each time for 12 levels of stname but I have to 
change script each time for the another 12 levels [line 3 in the script for 
example: for( i in levels(test$stname)[12:24))] and I do not know how can I 
save the obtained graphs (seven graphic windows) as jpeg files (e.g. 
plot1.jpeg, plot2.jpeg and so on). As I have 45 dataset like this it would be 
gr!
 eat if somebody can help me to complete this script to do all together for a 
dataset using a script.
Thank you very much in advance for your cooperation,
Monireh


  
windows(9,9)
par(mfrow = c(3,4))
for( i in levels(test$stname)[1:12])
{ 
data- test[test$stname==i,]
xx - c(1:length(data$upper), length(data$upper):1)
yy - c(data$upper, rev(data$lower))
zz- data$observed
tt- data$Best_Sim
par(lab =c(10,15,2))
plot.jpeg- plot(xx,yy, type=n, xlim=c(min(xx), max(xx)), 
ylim=c(min(zz,yy,tt), max(yy,zz,tt)*1.4),
 main= i, xlab=Month (1990-2002),  ylab=Discharge(m3/s), font.axis=6)
polygon(xx, yy, col=green,  border = NA)
lines(zz, col=blue, lwd=1.5)
lines(tt,col=red, lwd=1.5) 
legend(length(zz)-60, max(yy,zz,tt)*1.45, c(Upper Limit, Lower Limit,  
Observed,Best etimation)
, lwd=c(10, 1,1.7,1.7), bty=n, col= c(green, white, blue,red))
 }

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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread Nicolas Mazziotta
Hello,

See
?Devices
?dev.print

e.g. 

X11() # opens a new screen device
plot(1:10) # plots points on the device
dev.print(jpeg, [args]) # prints from the X11 device to a jpeg file
dev.off() # close X11 device

or

jpeg([args]) # opens a jpeg device (file) for plotting
plot(1:10) # plots points on the device
dev.off() # closes connection to the jpeg device

Hope this helps.

Regards,


-- 
Nicolas Mazziotta

The contents of this e-mail, including any attachments, are ...{{dropped}}

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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread ONKELINX, Thierry
Creating more than one graphic windows is, as far as I know, not
possible in R. But it's no problem to run a script which create multiply
jpeg's. See ?jpeg on how to create jpeg's.

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:r-help-
 [EMAIL PROTECTED] Namens Faramarzi Monireh
 Verzonden: woensdag 7 maart 2007 9:39
 Aan: r-help@stat.math.ethz.ch
 Onderwerp: [R] How to open more windows to make more graphs at once!
 
 
 Dear R users,
 I have a data frame (test) including five columns of upper (numeric),
 lower (numeric), observed (numeric), best_sim (numeric) and stname
(factor
 with 80 levels, each level with different length). Now I would like to
 write a short program to draw one graph as follow for each level of
stname
 but I would like also to draw each time 12 graphs for the 12 levels of
 stname in the same graphic windows and save it as jpeg' file . This
means
 at the end I will have 7 (80 levels/12=7) graphic windows and 7 jpeg
files
 each one with 12 graphs (the last one with 8 graphs) for the 12 levels
of
 stname. I already wrote the following script to do it each time for 12
 levels of stname but I have to change script each time for the another
12
 levels [line 3 in the script for example: for( i in
 levels(test$stname)[12:24))] and I do not know how can I save the
obtained
 graphs (seven graphic windows) as jpeg files (e.g. plot1.jpeg,
plot2.jpeg
 and so on). As I have 45 dataset like this it would be gr!
  eat if somebody can help me to complete this script to do all
together
 for a dataset using a script.
 Thank you very much in advance for your cooperation,
 Monireh
 
 
 
 windows(9,9)
 par(mfrow = c(3,4))
 for( i in levels(test$stname)[1:12])
 {
 data- test[test$stname==i,]
 xx - c(1:length(data$upper), length(data$upper):1)
 yy - c(data$upper, rev(data$lower))
 zz- data$observed
 tt- data$Best_Sim
 par(lab =c(10,15,2))
 plot.jpeg- plot(xx,yy, type=n, xlim=c(min(xx), max(xx)),
 ylim=c(min(zz,yy,tt), max(yy,zz,tt)*1.4),
  main= i, xlab=Month (1990-2002),  ylab=Discharge(m3/s),
font.axis=6)
 polygon(xx, yy, col=green,  border = NA)
 lines(zz, col=blue, lwd=1.5)
 lines(tt,col=red, lwd=1.5)
 legend(length(zz)-60, max(yy,zz,tt)*1.45, c(Upper Limit, Lower
Limit,
  Observed,Best etimation)
 , lwd=c(10, 1,1.7,1.7), bty=n, col= c(green, white,
blue,red))
  }
 
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 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
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[R] where can I find Durbin-Waston test tables for Confidence Level 2.5% or 0.5%?

2007-03-07 Thread Michael
Hi all,

I am doing a two-sided DW test:

H0: rho = 0
H1: rho =/= 0

My understanding is that most test statistics tables are one-sided. It's the
way they created the table.

So from online, by doing Googling, I found a bunch of DW tables for
Confidence Level 5%.

Those tables can answer my two-sided question at 5x2 = 10% confidence level.

But what if I want two-sided test at 1% and 5% confidence level?

Then I need 0.5% and 2.5% tables on those one-sided table.

My sample size is 278, and the number of parameters is 2, these adds to the
hardship of finding a good table...

Could anybody give me some pointers of two-sided DW tables or 1-sided DW
table with 0.5% and 2.5% confidence levels?

Thanks a lot!

Moreover, I appreciate any pointers about electronic tables that I can use
in programs, I want to implement DW test myself, but let the program
searching a table automatically...

Thanks a lot!

[[alternative HTML version deleted]]

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[R] heatmap

2007-03-07 Thread XinMeng
Hello sir;
I use the function heatmap.2 to draw a heatmap of microarray data,which 
consists of logratios.

Q1
But the lengend shows the Z score and the corresponding color.But I want the 
legend to show the logratios and the corresponding color.How can I do it?

Q1
How can I control that cluster only applied to genes(rows) or samples(columns) 
via heatmap.2 function?Default is cluster applied to both rows and columns.

Thanks a lot!

My best!

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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread Gavin Simpson
On Wed, 2007-03-07 at 09:39 +0100, Faramarzi Monireh wrote:
 Dear R users,
 I have a data frame (test) including five columns of upper (numeric),
 lower (numeric), observed (numeric), best_sim (numeric) and stname
 (factor with 80 levels, each level with different length). Now I would
 like to write a short program to draw one graph as follow for each
 level of stname but I would like also to draw each time 12 graphs for
 the 12 levels of stname in the same graphic windows and save it as
 jpeg' file . This means at the end I will have 7 (80 levels/12=7)
 graphic windows and 7 jpeg files each one with 12 graphs (the last one
 with 8 graphs) for the 12 levels of stname. I already wrote the
 following script to do it each time for 12 levels of stname but I have
 to change script each time for the another 12 levels [line 3 in the
 script for example: for( i in levels(test$stname)[12:24))] and I do
 not know how can I save the obtained graphs (seven graphic windows) as
 jpeg files (e.g. plot1.jpeg, plot2.jpeg and so on). As I have 45
 dataset like this it would be gr!
  eat if somebody can help me to complete this script to do all
 together for a dataset using a script.
 Thank you very much in advance for your cooperation,
 Monireh
 

Hi Monireh,

I don't have your data set so I have generated some random data to
illustrate one approach to this.

## generate some data 
set.seed(1234)
dat - data.frame(upper = rnorm(100), lower = rnorm(100), 
  observed = rnorm(100), best_sim = rnorm(100), 
  stname = factor(gl(5, 20), labels = letters[1:5]))

## because this is going to be called 45 times, I've wrapped it in a
## function, foo()
## Note the filename arg. It contains %03d which means that R will
## insert a number and produce many jpegs, varying by this number
## e.g. myplot1.jpeg, myplot2.jpeg - see ?jpeg.
## the ... allow passing of arguments to jpeg
foo - function(x, filename = Rplot%03d.jpeg, ...) {
   ## start the jpeg device
   jpeg(filename = filename, ...)
   ## store the parameter defaults and set a 2 by 2 plot regions
   opar - par(mfrow = c(2,2))
   ## this insures that the device is closed and defaults restored on
   ## function exit
   on.exit({dev.off(); par(opar)})
   ## set up a loop to go over the levels of your factor
   for(i in levels(x$stname)) {
  ## do the plotting - here you need to add the plot commands
  ## you really want to use - these are just examples.
  plot(lower ~ upper, data = x, subset = stname == i)
  ## this just adds a lowess line, I use with() to make it easier
  ## to read.
  with(x, lines(lowess(upper[stname == i], lower[stname == i]), 
   col = red))
   }
   invisible()
}

## to use the function on the demo data
## uses default filename
foo(dat)

## or passing arguments to jpeg()
foo(dat, width = 600, height = 600, pointsize = 10)

## or using your own file name
foo(dat, filename = dataset1_%03d.jpeg, width = 600, height = 600,
pointsize = 10)

See ?jpeg to see why this works - the filename with %03d allows R to
produce several jpegs.

   
 windows(9,9)
 par(mfrow = c(3,4))
 for( i in levels(test$stname)[1:12])
 { 
 data- test[test$stname==i,]
 xx - c(1:length(data$upper), length(data$upper):1)
 yy - c(data$upper, rev(data$lower))
 zz- data$observed
 tt- data$Best_Sim
 par(lab =c(10,15,2))

In the line below, where you set the x- and y-limits, it would be
simpler and more readable to use range() instead of c(min(x), max(x) -
so your plot call could be:

plot.jpeg- plot(xx,yy, type=n, xlim= range(xx),  
 ylim=range(zz,yy,tt)*1.4), main= i, 
 xlab=Month (1990-2002),  
 ylab=Discharge(m3/s), 
 font.axis=6)

Also, you can format the y-label more nicely with:

ylab = expression(paste(Discharge (, m^-3 * s^{-1}, )))

HTH

G

-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Gavin Simpson [t] +44 (0)20 7679 0522
ECRC  [f] +44 (0)20 7679 0565
UCL Department of Geography
Pearson Building  [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street
London, UK[w] http://www.ucl.ac.uk/~ucfagls/
WC1E 6BT  [w] http://www.freshwaters.org.uk/
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] good procedure to estimate ARMA(p, q)?

2007-03-07 Thread Michael
Hi all,

I have some residuals from regression, and i suspect they have correlations
in them...

I am willing to cast the correlation into a ARMA(p, q) framework,

what's the best way to identify the most suitable p, and q, and fit ARMA(p,
q) model and then correct for the correlations in regression?

I know there are functions in R, I have used them before, but I just want to
see if I can do the whole procedure myself, just to improve my understanding
...

Please give me some pointers! Thanks a lot

[[alternative HTML version deleted]]

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


Re: [R] Sweave issue: quotes in verbatim-like output print incorrectly

2007-03-07 Thread Gavin Simpson
On Wed, 2007-03-07 at 15:33 +1000, Peter Dunn wrote:
 Hi all
 
 I love Sweave; use it all the time.
 
 But I recently received a new computer, and ever since I
 have had a problem I've never seen before.
 
 For example, I place the following in my Snw file:

Try this in the preamble of your Snw file:

\usepackage[utf8x]{inputenc}

(assuming you have the inputenc package installed and available). I'm
assuming you are now using a machine using UTF-8 for character
encodings. I used to get that output on my linux box (FC4 - 6) before I
added the above \usepackage statement.

HTH

G
 
 =
 sms - 
 read.table(http://www.sci.usq.edu.au/staff/dunn/Datasets/applications/popular/smsspeed.dat;,
 header=TRUE)
 attach(sms)
 
 sms.lm - lm( Time ~ Age*Phone, subset=(Age30) )
 summary(sms.lm)
 @
 
 Standard stuff.   The output appears in the corresponding LaTeX
 file as it should, in a verbatim-like environment as it should. 
 
 But since I have had this new machine, this line of output:
 
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
 
 appears in my resulting pdf document as
 
 Signif. codes: 0 ^a˘A¨Y***^a˘A´Z0.001 ^a˘A¨Y**^a˘A´Z0.01 ^a˘A¨Y*^a
 ˘A´Z0.05 ^a˘A¨Y.^a˘A´Z0.1 ^a˘A¨Y^a˘A´Z1
 
 In short, every quote is replaced by garbage.  This makes my
 output looks incredibly bad.  (This is true for all cases; the above
 is the output from my example.)
 
 I also imagine (hope!) there is a very simple fix.  Can anyone help me?
 
 Documents which used to produce the correct output document
 now do this, so it must be something to do with my machine 
 set up, or R set up, rather than the documents themselves, I guess.
 
 Any help appreciated.  I have no idea where to look for the solution
 (the FAQ. manuals and mailing archives were no help that I could see;
 happy to be corrected).
 
 P.
 
 
  version
_
 platform   i486-pc-linux-gnu
 arch   i486
 os linux-gnu
 system i486, linux-gnu
 status Patched
 major  2
 minor  4.0
 year   2006
 month  11
 day25
 svn rev39997
 language   R
 version.string R version 2.4.0 Patched (2006-11-25 r39997)
 
  sessionInfo()
 R version 2.4.0 Patched (2006-11-25 r39997)
 i486-pc-linux-gnu
 
 locale:
 LC_CTYPE=en_AU.UTF-8;LC_NUMERIC=C;LC_TIME=en_AU.UTF-8;LC_COLLATE=en_AU.UTF-8;LC_MONETARY=en_AU.UTF-8;LC_MESSAGES=en_AU.UTF-8;LC_PAPER=en_AU.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_AU.UTF-8;LC_IDENTIFICATION=C
 
 attached base packages:
 [1] stats graphics  grDevices utils datasets  methods
 [7] base
 
-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Gavin Simpson [t] +44 (0)20 7679 0522
ECRC  [f] +44 (0)20 7679 0565
UCL Department of Geography
Pearson Building  [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street
London, UK[w] http://www.ucl.ac.uk/~ucfagls/
WC1E 6BT  [w] http://www.freshwaters.org.uk/
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] info about R diffusion

2007-03-07 Thread diego pettena
Buongiorno a tutti
mi chiamo Diego Pettena e sono uno studente universitario iscritto alla
facoltà di economia. Ho da poco iniziato ad utilizzare il sitema R per
l'analisi statistica, tuttavia con la presente non intendo chiederVi
nulla riguardo al linguaggio o all'utilizzo del programma. Per questo mi
scuso sinceramente fin da subito se questa non è la sessione adatta a
cui inviare la mia richiesta.
Sto svolgendo un piccolo elaborato, sempre per scopi universitari,
concernente R e mi interesserebbe avere alcuni dati/statistiche
riguardanti la diffusione di R nel mondo (università, privati,
professionisti) magari, sempre se possibile, paragonati a quelli dei
principali sistemi concorrenti. Non ho bisogno di informazioni
estremamente dettagliate, mi basta avere una visione generale del grado
di diffusione del programma.
Se tali info non fossero in Vs possesso vi chiedo di perdonare
l'intrusione e vi porgo, comunque, i miei più sentiti ringraziamenti.
attendo vostre, cordiali saluti
Diego Pettena
 
 
Goodmorning all
I'm Diego Pettena and I'm a student of economics in the Verona's
University. I recently began to use R for statistical analysis; with
this mail I don't want to ask you informations about the language or the
use of the programme and I sincerely apologize if this is not the right
session where doing my request!
I'm doing a shortly article about R - for my studies - and I ask you if
are available some statistics concern the level of worldwide diffusion
of the programme (universities, private, professional..), and, ever if
is it possible, compared with the major competitors. 
If theese statistics are available I ask you to send me them by mail or
to say me where I can find them. Otherwise I pray you to apologize my
interference and I thank you.
Wiht my best regards, thanks a lot
Diego Pettena

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Re: [R] where can I find Durbin-Waston test tables for Confidence Level 2.5% or 0.5%?

2007-03-07 Thread Achim Zeileis
Michael:

 I am doing a two-sided DW test:

 H0: rho = 0
 H1: rho =/= 0

 My understanding is that most test statistics tables are one-sided. It's the
 way they created the table.

...because rho  0 is the alternative of interest in most applications.

 So from online, by doing Googling, I found a bunch of DW tables for
 Confidence Level 5%.

Using tables for the DW test is difficult because it's distribution 
depends on the particular set of regressors used. The tables of DW just 
give upper and lower bounds.

Back when the DW test was suggested, tables was the only way to make 
application of the test feasible. Today, you would either use the exact 
combination of chi-square distributions or an asymptotic approximation 
(both implemented in dwtest() from lmtest) or a bootstrap approximation 
(implemented in durbin.watson() from car). For 278 observations, the 
normal approximation should be sufficient.

hth,
Z

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Re: [R] Package RODBC

2007-03-07 Thread Wolfgang Raffelsberger
Dear Alberto,

please note that special characters (eg a space character) in the Excel 
sheet names mess up the simple way of querying provided by sqlFetch.

If you have a regular case of all sheets like Sheet1:

plan1 - sqlFetch(channel,Sheet1)   # should work


But if you have Sheet 1 ( similar..)  you have to use the command 
sqlQuery(), which means that you have to write a proper SQL query as 2nd 
argument that follows proper SQL syntax (starting with SELECT, 
etc...). If I wanted to combine this with sheet-names already read in 
variables/vectors I concatenate this into a single stringsimilar to your 
2nd code variant ... Of course you could also use grep() to search the 
position of a given sheet-name (the order of the sheets may be different 
that within Excel).

What you get with

plan1[,1]

depends on what you're reading.  In case that the 1st column is read as 
string, this is read by default as factor with n levels.  You can simply 
convert it using as.character() ...

Hope this helps,
Wolfgang

Mendiburu, Felipe (CIP) a écrit :
 Dear Alberto,

 channel - odbcConnectExcel(test.xls)
 name1 - tables[1, TABLE_NAME] # the name1 is Sheet1$
 it must be: 
 name1 - Sheet1
 plan1 - sqlFetch(channel, name1) is ok
 or
 plan1 - sqlFetch(channel, Sheet1)

 Regards,

 Felipe

 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] Behalf Of Alberto Monteiro
 Sent: Tuesday, March 06, 2007 9:37 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Package RODBC


 I have some questions about the RODBC package.

   library(RODBC)  # required for those who want to repeat these lines

 1st, I noticed that the following sequence does not work:

   channel - odbcConnextExcel(test.xls)
   tables - sqlTables(channel) 
   name1 - tables[1, TABLE_NAME]  # this should be the name
   plan1 - sqlFetch(channel, name1)  # bang!
   odbcClose(channel)

 However, I can circumvent this with:

   channel - odbcConnextExcel(test.xls)
   tables - sqlTables(channel) 
   name1 - tables[1, TABLE_NAME]  # this should be the name
   plan1 - sqlQuery(channel, sprintf(select * from [%s], name1))  # ok
   odbcClose(channel)

 2nd, it seems that only pure strings (which are not links to
 strings) and numerical values are correctly fetched or selected.
 Is this a bug?

 3rd, when do something like plan1[,1] a weird message about Levels
 appear. What is that?

 Alberto Monteiro


   


-- 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
. . . . .

Wolfgang Raffelsberger, PhD
Laboratoire de BioInformatique et Génomique Intégrative
IGBMC
1 rue Laurent Fries,  67404 Illkirch  Strasbourg,  France
Tel (+33) 388 65 3300 Fax (+33) 388 65 3276
[EMAIL PROTECTED]

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Re: [R] where can I find Durbin-Waston test tables for Confidence Level 2.5% or 0.5%?

2007-03-07 Thread Chuck Cleland
Michael wrote:
 Hi all,
 
 I am doing a two-sided DW test:
 
 H0: rho = 0
 H1: rho =/= 0
 
 My understanding is that most test statistics tables are one-sided. It's the
 way they created the table.
 
 So from online, by doing Googling, I found a bunch of DW tables for
 Confidence Level 5%.
 
 Those tables can answer my two-sided question at 5x2 = 10% confidence level.
 
 But what if I want two-sided test at 1% and 5% confidence level?
 
 Then I need 0.5% and 2.5% tables on those one-sided table.
 
 My sample size is 278, and the number of parameters is 2, these adds to the
 hardship of finding a good table...
 
 Could anybody give me some pointers of two-sided DW tables or 1-sided DW
 table with 0.5% and 2.5% confidence levels?
 
 Thanks a lot!
 
 Moreover, I appreciate any pointers about electronic tables that I can use
 in programs, I want to implement DW test myself, but let the program
 searching a table automatically...

  Are you aware of the implementations of this test in the car, lmtest,
and fMultivar packages?

RSiteSearch(Durbin-Watson, restrict=function) finds those functions.

 Thanks a lot!
 
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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread Ted Harding
On 07-Mar-07 ONKELINX, Thierry wrote:
 Creating more than one graphic windows is, as far as I know, not
 possible in R.

It is if you are running R on Linux (in which the X Windowing
System -- or X -- is the standard graphical system). As it
says in ?X11:

  This can only be done on machines that run X.

Then each call to X11() opens a new graphics window, so you
can have as many as you like.

And, using the display parameter in X11(), if you have monitors
for more than one machine on your desk you can throw each window
to a monitor of your choice, giving you multiple graphics windows
on multiple screens, all from the same run of R.

But, as to whether/to what extent X or equivalent is available for
MS Windows, that is another question on which I have no expertise.

Ted.

 But it's no problem to run a script which create
 multiply
 jpeg's. See ?jpeg on how to create jpeg's.
 
 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:r-help-
 [EMAIL PROTECTED] Namens Faramarzi Monireh
 Verzonden: woensdag 7 maart 2007 9:39
 Aan: r-help@stat.math.ethz.ch
 Onderwerp: [R] How to open more windows to make more graphs at once!
 
 
 Dear R users,
 I have a data frame (test) including five columns of upper (numeric),
 lower (numeric), observed (numeric), best_sim (numeric) and stname
 (factor
 with 80 levels, each level with different length). Now I would like to
 write a short program to draw one graph as follow for each level of
 stname
 but I would like also to draw each time 12 graphs for the 12 levels of
 stname in the same graphic windows and save it as jpeg' file . This
 means
 at the end I will have 7 (80 levels/12=7) graphic windows and 7 jpeg
 files
 each one with 12 graphs (the last one with 8 graphs) for the 12 levels
 of
 stname. I already wrote the following script to do it each time for 12
 levels of stname but I have to change script each time for the another
 12
 levels [line 3 in the script for example: for( i in
 levels(test$stname)[12:24))] and I do not know how can I save the
 obtained
 graphs (seven graphic windows) as jpeg files (e.g. plot1.jpeg,
 plot2.jpeg
 and so on). As I have 45 dataset like this it would be gr!
  eat if somebody can help me to complete this script to do all
 together
 for a dataset using a script.
 Thank you very much in advance for your cooperation,
 Monireh
 
 
 
 windows(9,9)
 par(mfrow = c(3,4))
 for( i in levels(test$stname)[1:12])
 {
 data- test[test$stname==i,]
 xx - c(1:length(data$upper), length(data$upper):1)
 yy - c(data$upper, rev(data$lower))
 zz- data$observed
 tt- data$Best_Sim
 par(lab =c(10,15,2))
 plot.jpeg- plot(xx,yy, type=n, xlim=c(min(xx), max(xx)),
 ylim=c(min(zz,yy,tt), max(yy,zz,tt)*1.4),
  main= i, xlab=Month (1990-2002),  ylab=Discharge(m3/s),
 font.axis=6)
 polygon(xx, yy, col=green,  border = NA)
 lines(zz, col=blue, lwd=1.5)
 lines(tt,col=red, lwd=1.5)
 legend(length(zz)-60, max(yy,zz,tt)*1.45, c(Upper Limit, Lower
 Limit,
  Observed,Best etimation)
 , lwd=c(10, 1,1.7,1.7), bty=n, col= c(green, white,
 blue,red))
  }
 
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 guide.html
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E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 07-Mar-07   Time: 09:55:12
-- XFMail --

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Re: [R] R and SAS proc format

2007-03-07 Thread Ulrike Grömping
The down side to R's factor solution:

The numerical values of factors are always 1 to number of levels. Thus, it

can be tough and requires great care to work with studies that have both

numerical values different from this and value labels. This situation is

currently not well-supported by R.



Regards, Ulrike



P.S.: I fully agree with Frank regarding the annoyance one sometimes

encounters with formats in SAS!

 You can add an attribute to a variable.  In the sas.get function in the 
 Hmisc package for example, when importing SAS variables that have PROC 
 FORMAT value labels, an attribute 'sas.codes' keeps the original codes; 
 these can be retrieved using sas.codes(variable name).  This could be 
 done outside the SAS import context also. 
 
 Frank 
 -- 
 Frank E Harrell Jr   Professor and Chair           School of Medicine 
                      Department of Biostatistics   Vanderbilt University

Frank,

are these attributes preserved when merging or subsetting a data frame?
Are they used in R packages other than Hmisc and Design (e.g. in a simple table 
request)?

If this is the case, my wishlist items 8658 and 8659 
(http://bugs.r-project.org/cgi-bin/R/wishlist?id=8658;user=guest, 
http://bugs.r-project.org/cgi-bin/R/wishlist?id=8659;user=guest) can be closed. 
Otherwise, I maintain the opinion that there are workarounds but that R is not 
satisfactorily able to handle this type of data.

Regards, Ulrike

--- End of Original Message ---
 

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Re: [R] Generic distributions

2007-03-07 Thread Alberto Monteiro
Matthias Kohl wrote:
 
 in distr you can do:
 
 library(distr)
 N - Norm(mean = 1, sd = 2)
 p(N)(0.5)
 r(N)(100)
 
 !!! not: p(N, 0.5) or r(N, 100) !!!
 A detailed description of package distr is given in package distrDoc.
 
 library(distrDoc)
 vignette(distr)
 
Thanks!!! This is almost perfect. It even has (some) arithmetics!!!

z1 - Norm(mean = 1, sd= 0.6)
z2 - Norm(mean = 2, sd= 0.8)
z1+z2

Distribution Object of Class: Norm
mean :  3 
sd :  1 
Warning message:
arithmetics on distributions are understood as operations on r.v.'s
see 'distrARITH()'; for switching off this warning see '?distroptions' in: 
print(object) 
 
Alberto Monteiro

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Re: [R] Sweave issue: quotes in verbatim-like output print incorrectly

2007-03-07 Thread Patrick Drechsler
Gavin Simpson [EMAIL PROTECTED] writes:

 On Wed, 2007-03-07 at 15:33 +1000, Peter Dunn wrote:
 But I recently received a new computer, and ever since I
 have had a problem I've never seen before.
 
 For example, I place the following in my Snw file:

 Try this in the preamble of your Snw file:

 \usepackage[utf8x]{inputenc}

\usepackage[utf8]{inputenc}

should suffic for this. Also, utf8 is newer than utf8x. So if you only
need the ISO 8859 subset of Unicode, utf8 should serve your needs.

HTH

Patrick
-- 
I never used a logarithm in my life, and could not undertake to
extract the square root of four without misgivings.
(Georg Bernhard Shaw)

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Re: [R] Identifying points in a plot that have duplicate values

2007-03-07 Thread Antony Unwin
Use the iplots package.  The interactive querying works effectively  
for largish datasets (over 50,000 cases) where sunflower plots may not.

Antony Unwin

Professor of Computer-Oriented Statistics and Data Analysis,
University of Augsburg,
86135 Augsburg, Germany
Tel: + 49 821 5982218
http://stats.math.uni-augsburg.de/



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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread Alberto Monteiro
Ted Harding wrote:

 Creating more than one graphic windows is, as far as I know, not
 possible in R.
  
 But, as to whether/to what extent X or equivalent is available for
 MS Windows, that is another question on which I have no expertise.
 
X11() seems to work for Windows XP.

Alberto Monteiro

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[R] anova applied to a lme object

2007-03-07 Thread Berta
Hi R-users,

when carrying out a multiple regression, say lm(y~x1+x2), we can use an 
anova of the regression with summary.aov(lm(y~x1+x2)), and afterwards 
evaluate the relative contribution of each variable using the global Sum of 
Sq of the regression and the Sum of Sq of the simple regression y~x1.

Now I would like to incorporate a random effect in the model, as some data 
correspond to the same region and others not:  mylme- lme(y~x1+x2, random= 
~1|as.factor(region)). I would like to know, if possible, which is the 
contribution of each variable to the global variability. Using anova(mylme) 
produce an anova table (without the Sum of Sq column), but I am not sure how 
can I derive the contribution of each variable from it, or even whether it 
is nonsense to try, nor can I derive a measure of how much variability is 
left unexplained.

Sorry for the type of question, but I did not find a simple solution and 
some researchers I work with love to have relative contributions to global 
variability.

Thanks a lot in advance,

Berta





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[R] Multi-line plots with matrices in R

2007-03-07 Thread Joseph Wakeling
Hello all,

I'm a new user of R, experienced with Octave/MATLAB and therefore
struggling a bit with the new syntax.

One of the easy things in Octave or MATLAB is to plot multiple lines or
 sets of points by using a matrix where either the columns or the rows
contain the y-values to be plotted.  Both packages automatically give
each line/points their own unique colour, character etc.

I'm wondering how I get the same functionality in R.  For example, if X
is a vector of x-values and Y is a matrix whose rows contain the
y-values, I can do,

apply(Y,1,lines,x=X)

... but of course everything is all in black, with the same type of line
or points.  I'd like each line to have its own unique colour and/or style.

Another thing I'd like clarification on is the ability to update an
existing plot.  For example if I do,

plot.window(xlim=c(0,100),ylim=c(0,1))

and then after plotting data decide I want ylim=c(0,0.5), how do I
update the graphic?  A new plot.window() command does nothing.

Many thanks,

-- Joe

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Re: [R] Multi-line plots with matrices in R

2007-03-07 Thread Petr Pikal
Hi

see matplot, matlines.

or use forbidden for cycle.

for (i in 1:n) lines(x,y[,i], col=i)

or if you want to use more colours use built in rainbow, topo.colors 
or generate your own set.

Regards
Petr


On 7 Mar 2007 at 12:30, Joseph Wakeling wrote:

Date sent:  Wed, 07 Mar 2007 12:30:46 +
From:   Joseph Wakeling [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] Multi-line plots with matrices in R

 Hello all,
 
 I'm a new user of R, experienced with Octave/MATLAB and therefore
 struggling a bit with the new syntax.
 
 One of the easy things in Octave or MATLAB is to plot multiple lines
 or
  sets of points by using a matrix where either the columns or the rows
 contain the y-values to be plotted.  Both packages automatically give
 each line/points their own unique colour, character etc.
 
 I'm wondering how I get the same functionality in R.  For example, if
 X is a vector of x-values and Y is a matrix whose rows contain the
 y-values, I can do,
 
 apply(Y,1,lines,x=X)
 
 ... but of course everything is all in black, with the same type of
 line or points.  I'd like each line to have its own unique colour
 and/or style.
 
 Another thing I'd like clarification on is the ability to update an
 existing plot.  For example if I do,
 
 plot.window(xlim=c(0,100),ylim=c(0,1))
 
 and then after plotting data decide I want ylim=c(0,0.5), how do I
 update the graphic?  A new plot.window() command does nothing.
 
 Many thanks,
 
 -- Joe
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html and provide commented,
 minimal, self-contained, reproducible code.

Petr Pikal
[EMAIL PROTECTED]

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Re: [R] Multi-line plots with matrices in R

2007-03-07 Thread Gavin Simpson
On Wed, 2007-03-07 at 12:30 +, Joseph Wakeling wrote:
 Hello all,
 
 I'm a new user of R, experienced with Octave/MATLAB and therefore
 struggling a bit with the new syntax.
 
 One of the easy things in Octave or MATLAB is to plot multiple lines or
  sets of points by using a matrix where either the columns or the rows
 contain the y-values to be plotted.  Both packages automatically give
 each line/points their own unique colour, character etc.
 
 I'm wondering how I get the same functionality in R.  For example, if X
 is a vector of x-values and Y is a matrix whose rows contain the
 y-values, I can do,
 
 apply(Y,1,lines,x=X)

You want maplot here. See ?matplot  but here is an example:

## generate some data to use, a matrix of Y values
## and a vector of x indices.
mat - matrix(runif(100), ncol = 5)
vec - seq(1, 100, length = 20)

## plot it using matplot
matplot(vec, mat, type = l) # type = l to get lines

There is also matlines() and matpoints() for adding lines and points to
existing plots.

 
 ... but of course everything is all in black, with the same type of line
 or points.  I'd like each line to have its own unique colour and/or style.
 
 Another thing I'd like clarification on is the ability to update an
 existing plot.  For example if I do,
 
 plot.window(xlim=c(0,100),ylim=c(0,1))

Standard graphics in R are not modifiable after being plotted. You need
to re-plot. When plotting data, I rarely need plot.window. This is what
I would do:

x - 1:100 * runif(100)
y - seq(0,1, length = 100) * runif(100)

plot(x, y, xlim = c(0, 100), ylim = c(0, 1))

# now change the limits
plot(x, y, xlim = c(0, 100), ylim = c(0, 0.5))

 
 and then after plotting data decide I want ylim=c(0,0.5), how do I
 update the graphic?  A new plot.window() command does nothing.

But it does:

opar - par(mfrow = c(1,2))
plot(x, y, xlim = c(0, 100), ylim = c(0, 0.5))
plot(x, y, xlim = c(0, 100), ylim = c(0, 1))
plot.window(xlim = c(0, 100), ylim = c(0, 0.5))
points(x, y, col = red)
par(opar)

The points on the left plot correspond exactly to the points in red on
the right plot. The axis limits have changed, but because the axes have
already been labelled, these are not updated. We can illustrate this by
adding axes to the top and right of that plot

opar - par(mfrow = c(1,2), mar = c(5,4,4,4) + 0.1)
plot(x, y, xlim = c(0, 100), ylim = c(0, 0.5))
plot(x, y, xlim = c(0, 100), ylim = c(0, 1))
plot.window(xlim = c(0, 100), ylim = c(0, 0.5))
points(x, y, col = red)
axis(3)
axis(4)
par(opar)

Note the changed axis range in the right-hand margin. The problem is
that you can't use plot.window to achieve what you want, not that
plot.window doesn't do anything.

 
 Many thanks,
 
 -- Joe

HTH

G

-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
 UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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[R] R-2.4.1 AIX build patch.

2007-03-07 Thread Ei-ji Nakama
Dear AIX useR's.

Will you test it in various environment?
There is not xlf in the environment where I can approach.

cf.
  32bit, 64bit, --enable-R-shlib, --enable-BLAS-shlib ...various cases!

http://prs.ism.ac.jp/%7enakama/AIX/
-- 
EI-JI Nakama  [EMAIL PROTECTED]
\u4e2d\u9593\u6804\u6cbb  [EMAIL PROTECTED]

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Re: [R] Package RODBC

2007-03-07 Thread Mendiburu, Felipe \(CIP\)
Dear Alberto,

It is better to assign a name to an area of data and not to use 
the name of the sheet, because this can have graphs and other data. 
If you this interested can see:
http://tarwi.lamolina.edu.pe/~fmendiburu/Rsolutions.htm 

I hope that this also helps,
Felipe.

-Original Message-
From: Wolfgang Raffelsberger [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 07, 2007 4:50 AM
To: Mendiburu, Felipe (CIP)
Cc: Alberto Monteiro; r-help@stat.math.ethz.ch
Subject: Re: [R] Package RODBC


Dear Alberto,

please note that special characters (eg a space character) in the Excel 
sheet names mess up the simple way of querying provided by sqlFetch.

If you have a regular case of all sheets like Sheet1:

plan1 - sqlFetch(channel,Sheet1)   # should work


But if you have Sheet 1 ( similar..)  you have to use the command 
sqlQuery(), which means that you have to write a proper SQL query as 2nd 
argument that follows proper SQL syntax (starting with SELECT, 
etc...). If I wanted to combine this with sheet-names already read in 
variables/vectors I concatenate this into a single stringsimilar to your 
2nd code variant ... Of course you could also use grep() to search the 
position of a given sheet-name (the order of the sheets may be different 
that within Excel).

What you get with

plan1[,1]

depends on what you're reading.  In case that the 1st column is read as 
string, this is read by default as factor with n levels.  You can simply 
convert it using as.character() ...

Hope this helps,
Wolfgang

Mendiburu, Felipe (CIP) a écrit :
 Dear Alberto,

 channel - odbcConnectExcel(test.xls)
 name1 - tables[1, TABLE_NAME] # the name1 is Sheet1$
 it must be: 
 name1 - Sheet1
 plan1 - sqlFetch(channel, name1) is ok
 or
 plan1 - sqlFetch(channel, Sheet1)

 Regards,

 Felipe

 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] Behalf Of Alberto Monteiro
 Sent: Tuesday, March 06, 2007 9:37 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Package RODBC


 I have some questions about the RODBC package.

   library(RODBC)  # required for those who want to repeat these lines

 1st, I noticed that the following sequence does not work:

   channel - odbcConnextExcel(test.xls)
   tables - sqlTables(channel) 
   name1 - tables[1, TABLE_NAME]  # this should be the name
   plan1 - sqlFetch(channel, name1)  # bang!
   odbcClose(channel)

 However, I can circumvent this with:

   channel - odbcConnextExcel(test.xls)
   tables - sqlTables(channel) 
   name1 - tables[1, TABLE_NAME]  # this should be the name
   plan1 - sqlQuery(channel, sprintf(select * from [%s], name1))  # ok
   odbcClose(channel)

 2nd, it seems that only pure strings (which are not links to
 strings) and numerical values are correctly fetched or selected.
 Is this a bug?

 3rd, when do something like plan1[,1] a weird message about Levels
 appear. What is that?

 Alberto Monteiro


   


-- 

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
. . . . .

Wolfgang Raffelsberger, PhD
Laboratoire de BioInformatique et Génomique Intégrative
IGBMC
1 rue Laurent Fries,  67404 Illkirch  Strasbourg,  France
Tel (+33) 388 65 3300 Fax (+33) 388 65 3276
[EMAIL PROTECTED]

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[R] Appropriate error distribution

2007-03-07 Thread Cristina Gomes
Hi,
My name is Cristina. I'm interested in studying which continuos predictor 
variables (such as grooming received, rank, etc.) affect grooming given, as 
well a continuos variable. I'm having problems finding an appropriate family 
distribution to fit the GLMM I'm doing. The response variable, grooming 
given, has many zeros which does not allow me to use a gamma distribution. I 
tried with a poisson but since I had to convert the data to integers I fear 
loosing a lot of information (and anyway my data is originally continuos and 
not counts). I found in the help archive that someone with a similar problem 
(continuos data and many zeros) was adviced to use Tweedie models. I don't 
know anything about this and wouldn't kow if this is appropriate in my case 
and possible to apply to generalized linear mixed models.
I'm wondering if anybody could provide me with any good insights on what 
distribution I could us and if, in the case of tweedie being a good option, 
if it can be used with GLMM's..
Thanks a lot,
Cristina.

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Re: [R] Memory Limits in Ubuntu Linux

2007-03-07 Thread Bos, Roger
David,

I wouldn't give up on windows so fast.  Many people have gotten the 3Gb
switch to work. One used to have to modify the header of the Rgui.exe
program to use the switch, but now the binary comes ready for that, so
its really quite easy.  I would like to hear more about why its not
working for you.

As for Linux, I use FC5 for which there is a 64-bit binary.  But there
are also 64-bit binaries for other distros.  The 32-bit and 64-bit
binaries are in different directories, so you should have no trouble
telling them apart.  

I have heard good things about Ubuntu--mainly that its very easy to
use--but FC5 has been pretty easy to learn too and I use the KDE desktop
which gives me Kate as a text editor.  You can open a terminal window in
Kate to run R and set up a key like F10 to send the code from the editor
to R.  Its not quite as good as my Windows setup with Tinn-R, but almost
as good.

Thanks,

Roger


-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] 
Sent: Tuesday, March 06, 2007 5:37 PM
To: Bos, Roger
Subject: RE: [R] Memory Limits in Ubuntu Linux

Thanks for your prompt reply!

The windows 3GB switch is quite problematic - it was not useable on my
machine, and there are comments about these problems around the net.
Thus, on to Linux. My machine has 4Gig, and some megabytes are grabbed
by my Asus motherboard, leaving some 3.56 Gig. 

So if I understand your suggestion, try the 64-bit version of Ubuntu
(based on Debian but I had better luck with the video part of the
install) and then use the corresponding image from CRAN. My fear is that
the CRAN Ubuntu version might be 32-bit - any idea how to find out
before I embark on another install? Which Linux do you have - you
described some significant success with getting large jobs to run.

And yes, I've worked hard to save memory by tweaking the code.


Thanks again.


On 6 Mar 2007 at 16:51, Bos, Roger wrote:

 David,
 
 First of all, under Windows you can get about 3GB available to R by 
 using the /3Gb switch in your boot.ini file, assuming you have 4Gb of 
 memory installed on your windows machine.  Using that method, I have 
 seen the memory using of my R process get as big as 2.7Gb in task 
 manager.  What's important, of course, is contiguous space, as you 
 mentioned.  There, you may want to check your code closely and make 
 sure that its memory usage is as efficient as possible and you are 
 storing the minimal amount you need for each run.  If you don't need 
 an object for a while consider writing it to disk and reading it back
in later.
 
 Second, AFAIK to get any benefit from more memory is Linux you have to

 go to the 64bit version.  I am a Linux newbie too, so I choose to use 
 one of the pre-compiled binaries available on CRAN.  In other words, 
 you shouldn't have to compile anything yourself.  How much memory do 
 you have on your Linux box?  I have 16Gb and I know I have ran stuff 
 that wouldn't run on my 4Gb windows box.
 
 HTH,
 
 Roger
 
 
 
 
 
  
 
 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of 
 [EMAIL PROTECTED]
 Sent: Tuesday, March 06, 2007 3:44 PM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Memory Limits in Ubuntu Linux
 
 I am an R user trying to get around the 2Gig memory limit in Windows, 
 so here I am days later with a working Ubuntu, and R under Ubuntu. But

 - the memory problems seem worse than ever. R code that worked under 
 windows fails, unable to allocate memory.
 
 Searching around the web, it appears that the problem may be the 
 ability to find contguous memory for my big vectors, but a fresh boot 
 of Ubuntu does not help either.
 
 Which way to go?
 
 1) Try to install 64-bit version for bigger address space. Would this 
 help? Is this workable for my Athlon 64 Dual-core? (the live cd seems 
 to work but I never got it to boot after a disk install, but then the 
 386 version was no better until I learned more about Grub...I could 
 try again if this might solve the
 problem)
 
 2) Recompile R to get bigger memory capability? (I'll have to 
 cross-post to some R forums too) This will be a challenge for a Linux 
 newbie...like me.
 
 3) Any other suggestions? My goal is to create a bigger neural network

 than fits in my Windows R version.
 --
 David Katz
  www.davidkatzconsulting.com
541 482-1137
 
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Re: [R] Autogenerate tags in tag=value pairs for

2007-03-07 Thread Gabor Grothendieck
Try na.locf from the zoo package and then use merge with specified suffixes:

library(zoo)
f - function(x) {
   rownames(x) - NULL
   merge(x, na.locf(x[-1], na.rm = FALSE), by = 0, suffixes = c(, .by))[-1]
}
do.call(rbind, by(x, x$id, f))


On 3/7/07, Jon Olav Vik [EMAIL PROTECTED] wrote:
 Dear list,

 Is there a way to programmatically specify tag names for the ... (ellipsis)
 part of the argument list to a function? In other words, a way to do this:

 x - data.frame(A=1:5)

 if the name A was not hardcoded but given by a variable, and without
 resorting to:

 x - data.frame(1:5)
 names(x) - A


 A longer example describing my actual problem follows. Thanks in advance for
 any help.

 Best regards,
 Jon Olav


 I want to use function transformBy() in package doBy. The key is that the ...
 Further arguments of the form tag=value require tag to be specified,
 otherwise the output does not include the results of my groupwise 
 calculations.

 Quoting the documentation:
  transformBy(doBy)
  Function to make groupwise transformations of data
  by applying the transform function to subsets of data.
 
  Usage
  transformBy(formula, data, ...)
 
  Arguments
  formula A formula with only a right hand side, see examples below
  data A data frame
  ... Further arguments of the form tag=value

 ### example ###

 # a function to replace NAs with the last non-NA value from above
 filldown - function(x) {
notna - !is.na(x) # elements with values
ix - cumsum(notna) # index to previous element (but zeros where we need 
 NA)
ix[ix==0] - NA # use [NA] as index to produce NA in output
return(x[notna][ix]) # for each: return previous value if found, else NA
 }
 # illustration of how it works
 tmp - c(NA,NA,1,NA,3,NA,NA)
 cbind(tmp,filldown(tmp))

 # I now want to apply filldown() to subsets of a data frame
 # and I want it to work on several columns

 # generate a data frame for illustration,
 # with a few non-NA values scattered round
 set.seed(5) # repeatable example
 x - data.frame(id = rep(1:4,each=6), v1=NA, v2=NA)
 ix - which(runif(nrow(x))0.75)
 x[ix,2] - rpois(length(ix),5)
 ix - which(runif(nrow(x))0.75)
 x[ix,3] - rpois(length(ix),5)
 x

 library(doBy)
 # the hard way -- works as required,
 # but I would like not having to hardcode column names v1 etc.
 transformBy(~id,data=x,v1.fd = filldown(v1),v2.fd = filldown(v2))

 # does not work because
 # output includes only columns explicitly mentioned in the ... argument
 transformBy(~id,data=x,function(y) lapply(y,filldown))

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Re: [R] R and SAS proc format

2007-03-07 Thread Frank E Harrell Jr
Ulrike Grömping wrote:
 
 
The down side to R's factor solution: 
 
The numerical values of factors are always 1 to number of levels. Thus, it
 
can be tough and requires great care to work with studies that have both
 
numerical values different from this and value labels. This situation is
 
currently not well-supported by R.
 

 
Regards, Ulrike
 

 
P.S.: I fully agree with Frank regarding the annoyance one sometimes
 
encounters with formats in SAS! 
 
   You can add an attribute to a variable.  In the sas.get function in the
   Hmisc package for example, when importing SAS variables that have PROC
   FORMAT value labels, an attribute 'sas.codes' keeps the original codes;
   these can be retrieved using sas.codes(variable name).  This could be
   done outside the SAS import context also.
  
   Frank
   --
   Frank E Harrell Jr   Professor and Chair   School of Medicine
 Department of Biostatistics   Vanderbilt 
 University
 
 Frank,
 
 are these attributes preserved when merging or subsetting a data frame?
 Are they used in R packages other than Hmisc and Design (e.g. in a 
 simple table request)?

no; would need to add functions like those that are used by the Hmisc 
label or impute functions.  And they are not used outside Hmisc/Design. 
  In fact I have little need for them as I always find the final labels 
as the key to analysis.

 
 If this is the case, my wishlist items 8658 and 8659 
 (http://bugs.r-project.org/cgi-bin/R/wishlist?id=8658;user=guest, 
 http://bugs.r-project.org/cgi-bin/R/wishlist?id=8659;user=guest) can be 
 closed.
 Otherwise, I maintain the opinion that there are workarounds but that R 
 is not satisfactorily able to handle this type of data.

R gives the framework for doing this elegantly but the user has an 
overhead of implementing new methods for such attributes.

Cheers
Frank

 
 Regards, Ulrike
 
 
 *--- End of Original Message ---*


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

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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread James W. MacDonald
Alberto Monteiro wrote:
 Ted Harding wrote:
 
Creating more than one graphic windows is, as far as I know, not
possible in R.

 
But, as to whether/to what extent X or equivalent is available for
MS Windows, that is another question on which I have no expertise.

 
 X11() seems to work for Windows XP.

Although I believe the preferred method is windows().

Best,

Jim


 
 Alberto Monteiro
 
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-- 
James W. MacDonald, M.S.
Biostatistician
Affymetrix and cDNA Microarray Core
University of Michigan Cancer Center
1500 E. Medical Center Drive
7410 CCGC
Ann Arbor MI 48109
734-647-5623


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Re: [R] R and SAS proc format

2007-03-07 Thread Peter Dalgaard
Jason Barnhart wrote:
 - Original Message - 
 From: John Kane [EMAIL PROTECTED]
 To: lamack lamack [EMAIL PROTECTED]; R-help@stat.math.ethz.ch
 Sent: Tuesday, March 06, 2007 2:13 PM
 Subject: Re: [R] R and SAS proc format


   
 --- lamack lamack [EMAIL PROTECTED] wrote:

 
 Dear all, Is there an R equivalent to SAS's proc
 format?
   
 What does the SAS PROC FORMAT do?
 

 It formats or reformats data in the SAS system.
   

Slightly more precisely: It creates user-defined formats, which are
subsequently associated with variables and used for reading, printing,
tabulating, and analyzing data. It is akin to R's factor()
constructions, but not quite. For one thing, SAS's formats are separate
entities - same format can be used for many variables, whereas R's
factors have the formatting coded as a part of the object. For related
reasons, a variable in SAS can have more distinct values than there are
value labesl for, etc. 
 It looks this:

 proc format; value kanefmt 1='A' 2='B' 3='C' 4='X' 5='Throw me 
 out';
 data temp; do i=1 to 10; kanevar=put(i,kanefmt.); output; end;
 proc print; run;

 And produces this:

 Obs i  kanevar
   1 1A
   2 2B
   3 3C
   4 4X
   5 5Throw me out
   6 6   6
   7 7   7
   8 8   8
   9 9   9
  1010  10


 But it is more robust than what is shown here.



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


-- 
   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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[R] Calculating confidence limits on acf graphs

2007-03-07 Thread Gladwin, Philip [CIB-FI]
Hello,
I was wondering if anybody could help me with this?

I have plotted an acf function for a time series and am very happy with it.
Now I am interested in calculating for myself the two values for the confidence
intervals that are plotted on the graph of the acf.

The confidence intervals do not appear to be returned from the acf function (is 
this true?).

So far I haven't managed to calculate them myself.  Can anybody help?

Phil,

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[R] Power calculation for detecting linear trend

2007-03-07 Thread Meesters, Erik
Dear people,
I've a problem in doing a power calculation. In Fryer and Nicholson
(1993), ICES J. mar. Sci. 50: 161-168 page 164 an example is given with
the following characteristics
T=5, points in time
R=5, replicates
Var.within=0.1
q=10, a 10% increase per year
The degrees of freedom for the test are calculated as Vl=T*R-2=23 and
the non-centrality parameter Dl=4.54.
Using this they get a power of 0.53, but the result that I'm getting is
0.05472242.

I've tried this several ways in R, but I'm not able to come up with the
same number. Am I doing something wrong in the calculation of the power?
Here's my code:

T-5
R-5
sigmasq-0.1
q-10
Vl-(T*R)-2
Dl-(R*(T-1)*T*(T+1)/(12*sigmasq))*(log(1+(q/100)))^2 #Dl result is
still similar

power.1-1-pf(qf(.95,(T*R-2),1,ncp=0),(T*R-2),1,ncp=Dl)

Thank you for any suggestions/help.

I'm using R2.4.1, on windowsXP.

Erik Meesters   




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[R] I need some help

2007-03-07 Thread Luis Garavito
Hi,

I have a little problem with the installation of a new packages. The
installation of R software is correct, but my server required authentication
for use it, and for load a new package directly from R it is not possible.
Is there a code or process for server authentication (put my login and
password) in R for download directly the packages?

The best regards,
-- 
Luis Antonio Garavito
Facultad de Ingeniería
Universidad Nacional de Colombia
Te. 3165000   Ext.  13525

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[R] Fwd: Package-RODBC-MSACCESS

2007-03-07 Thread j.joshua thomas
I have used RODBC to get the database i can view the tables in RGUI-2.4.1
how can i query the records in R
i tried with sqlQuery

need some help

JJ

-- 
Lecturer J. Joshua Thomas
KDU College Penang Campus
Research Student,
University Sains Malaysia


-- 
Lecturer J. Joshua Thomas
KDU College Penang Campus
Research Student,
University Sains Malaysia

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[R] hopach

2007-03-07 Thread Vallejo, Roger
Dear R users,

I thought this might be useful for users of the R package HOPACH
(Windows version). Using R 2.4.1 in Windows XP, I found that the
functions: makeoutput, boot2fuzzy and hopach2tree is not recognized by
HOPACH versions 1.4.3 and 1.8.0. However, it works perfect with HOPACH
version 1.6.0 downloaded from K.S. Pollard's site at UCDavis. Thanks.

Roger

 

 

Roger L. Vallejo, Ph.D.

Computational Biologist  Geneticist

U.S. Department of Agriculture, ARS  

National Center for Cool  Cold Water Aquaculture

11861 Leetown Road

Kearneysville, WV 25430

Voice:(304) 724-8340 Ext. 2141

Email:   [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] 

 


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[R] rattle- MSACCESS database problem

2007-03-07 Thread j.joshua thomas
library(RGtk2)
library(rattle)
rattle()

click the ODBC option it as the DSN i am a bit confused with this i already
put my *.mdb file in C:drive
i try put the DSN name as Microsoft Access driver, in the appropriate text
box but i couldnt locate the table

i tried the other way round open- locate the *.mdb in C:drive couldnt
locate

i tried RODBC aswell, but i want to use rattle to Data mine my database

need someone's help



-- 
Lecturer J. Joshua Thomas
KDU College Penang Campus
Research Student,
University Sains Malaysia

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Re: [R] Calculating confidence limits on acf graphs

2007-03-07 Thread Petr Klasterecky
Hmm, this does not seem to be over-documented :-)
But try
?plot.acf
and
getAnywhere(plot.acf)

Then you can find in the code how the values are actually calculated.
Petr

Gladwin, Philip [CIB-FI] napsal(a):
 Hello,
 I was wondering if anybody could help me with this?
 
 I have plotted an acf function for a time series and am very happy with it.
 Now I am interested in calculating for myself the two values for the 
 confidence
 intervals that are plotted on the graph of the acf.
 
 The confidence intervals do not appear to be returned from the acf function 
 (is this true?).
 
 So far I haven't managed to calculate them myself.  Can anybody help?
 
 Phil,
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
Petr Klasterecky
Dept. of Probability and Statistics
Charles University in Prague
Czech Republic

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Re: [R] R and SAS proc format

2007-03-07 Thread Carlos J. Gil Bellosta
On 3/7/07, Peter Dalgaard [EMAIL PROTECTED] wrote:
 Jason Barnhart wrote:
  - Original Message -
  From: John Kane [EMAIL PROTECTED]
  To: lamack lamack [EMAIL PROTECTED]; R-help@stat.math.ethz.ch
  Sent: Tuesday, March 06, 2007 2:13 PM
  Subject: Re: [R] R and SAS proc format
 
 
 
  --- lamack lamack [EMAIL PROTECTED] wrote:
 
 
  Dear all, Is there an R equivalent to SAS's proc
  format?
 
  What does the SAS PROC FORMAT do?
 
 
  It formats or reformats data in the SAS system.
 

 Slightly more precisely: It creates user-defined formats, which are
 subsequently associated with variables and used for reading, printing,
 tabulating, and analyzing data. It is akin to R's factor()
 constructions, but not quite. For one thing, SAS's formats are separate
 entities - same format can be used for many variables, whereas R's
 factors have the formatting coded as a part of the object. For related
 reasons, a variable in SAS can have more distinct values than there are
 value labesl for, etc.
  It looks this:
 
  proc format; value kanefmt 1='A' 2='B' 3='C' 4='X' 5='Throw me
  out';
  data temp; do i=1 to 10; kanevar=put(i,kanefmt.); output; end;
  proc print; run;
 
  And produces this:
 
  Obs i  kanevar
1 1A
2 2B
3 3C
4 4X
5 5Throw me out
6 6   6
7 7   7
8 8   8
9 9   9
   1010  10
 
 
  But it is more robust than what is shown here.
 
 
 
 
  __
  R-help@stat.math.ethz.ch mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide
  http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 
 
 
  __
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  PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
  and provide commented, minimal, self-contained, reproducible code.
 


 --
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

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Also, SAS formats are used as a (somewhat cumbersome) replacement for
dictionary data structures. Starting from SAS 9.1 (I believe), hash
tables can be used within data steps for the same purpose (albeit
still cumbersome).

In this regard, not only formats but also lists could be a replacement
for them. They can be used as a way to get key-value mappings.

These key-value mappings (I mean, these kind of data structures) are
very handy tools. I have used both factors and lists for some kind of
ad hoc replacement for these data structures. Hasn't anybody
considered the posibility of having these data structures implemented
in R in a much python-like or java-like touch and feel?

Regards,

Carlos J. Gil Bellosta
http://www.datanalytics.com

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[R] compiling latest version of R

2007-03-07 Thread Jenny Barnes
Dear R-help community,

I have had trouble in the past installing the latest version of R: we got the 
errors shown below (the computer specifications and version of R are below 
that). Does anybody have tips for compiling the latest version of R so that I 
can avoid these errors?

configure
make
...
...
...

f90: CODE: 0 WORDS, DATA: 0 WORDS
gcc -G -L/usr/local/lib -o stats.so init.o kmeans.o  ansari.o bandwidths.o
chisq
sim.o d2x2xk.o fexact.o kendall.o ks.o  line.o smooth.o  prho.o swilk.o 
ksmooth
.o loessc.o isoreg.o Srunmed.o Trunmed.o  dblcen.o distance.o
hclust-utils.o  nl
s.o  HoltWinters.o PPsum.o arima.o burg.o filter.o  mAR.o pacf.o starma.o
port.o
 family.o sbart.o bsplvd.o bvalue.o bvalus.o loessf.o ppr.o qsbart.o 
sgram.o si
nerp.o sslvrg.o stxwx.o  hclust.o kmns.o  eureka.o stl.o portsrc.o
-L../../../..
/lib -lRblas  -lg2c -lm -lgcc_s
mkdir ../../../../library/stats/libs
building package 'datasets'
mkdir ../../../library/datasets
mkdir ../../../library/datasets/R
mkdir ../../../library/datasets/data
Error in dyn.load(x, as.logical(local), as.logical(now)) :
unable to load shared library
'/tmp/R-2.4.0/library/stats/libs/stats.so'
:
  ld.so.1: R: fatal: relocation error: file
/tmp/R-2.4.0/library/stats/libs/stat
s.so: symbol __i_abs: referenced symbol not found
Execution halted
*** Error code 1


These are my specifications:

platform   sparc-sun-solaris2.10 
arch   sparc 
os solaris2.10   
system sparc, solaris2.10
status   
major  2 
minor  3.1   
year   2006  
month  06
day01
svn rev38247 
language   R 
version.string Version 2.3.1 (2006-06-01)


Many thanks for your time in reading this problem, I look forward to hearing 
your suggestions and advice,

Jenny


~~
Jennifer Barnes
PhD student: long range drought prediction 
Climate Extremes Group
Department of Space and Climate Physics
University College London
Holmbury St Mary 
Dorking, Surrey, RH5 6NT
Tel: 01483 204149
Mob: 07916 139187
Web: http://climate.mssl.ucl.ac.uk

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Re: [R] Fwd: Package-RODBC-MSACCESS

2007-03-07 Thread Roland Rau
Hi,

On 3/7/07, j.joshua thomas [EMAIL PROTECTED] wrote:

 I have used RODBC to get the database i can view the tables in RGUI-2.4.1
 how can i query the records in R
 i tried with sqlQuery

 need some help

 A bit of code showing what you did would help others to track down what
may have gone wrong.
Nevertheless, this is code I used under R 2.3.1 on WinXP:
library(RODBC)
myaccessDB - c:/data/db1.mdb
connectedMSAcc - odbcConnectAccess(myaccessDB)
sqlTables(connectedMSAcc) # all the tables in the Access DB
onetable - sqlFetch(connectedMSAcc, lexp) # fetching one table called
'lexp'
myQuery - SELECT Country, y2001 FROM lexp WHERE y2001  82;
myresults - sqlQuery(connectedMSAcc, myQuery)
myresults


I hope this helps?

Best,
Roland

[[alternative HTML version deleted]]

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[R] transform R function

2007-03-07 Thread lamack lamack
Dear all, Why the transform function does not accept two statistics 
functions?

a = data.frame(matrix(rnorm(20),ncol=2))

transform(a,M.1=mean(X1),M.2=mean(X2)) # does not works

#while:

transform(a,M.1=mean(X1),M2=log(abs(X2))) #works

Best regards

JL

_
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e agora com rede social http://spaces.live.com/

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[R] No years() function?

2007-03-07 Thread Sérgio Nunes
Hi,

I'm trying to aggregate date values using the aggregate function. For example:

aggregate(data,by=list(weekdays(LM),months(LM)),FUN=length)

I would also like to aggregate by year but there seems to be no
years() function.
Should there be one? Is there any alternative choice?

Also, a hours() function would be great. Any tip on this?

Thanks in advance!
Sérgio Nunes

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Re: [R] Multi-line plots with matrices in R

2007-03-07 Thread Joseph Wakeling
Gavin Simpson wrote:
 You want maplot here. See ?matplot  but here is an example:

Great!  Thanks to you and Petr for pointing this out, it's exactly what
I wanted.  Petr's other suggestions look interesting and I'll explore
them at length later.

 Note the changed axis range in the right-hand margin. The problem is
 that you can't use plot.window to achieve what you want, not that
 plot.window doesn't do anything.

Ahhh, I see.  So, it does not affect what has already been plotted, but
affects how new material is inserted into the plot area.  Entering

plot.window(xlim=c(0,100),ylim=c(0,0.5))
axis(1)
axis(2)
plot.window(xlim=c(0,100),ylim=c(0,1))
axis(2)

... is instructive. :-)

So, _is_ there a command which will rearrange the existing plotted
items, including axes?  Or does R require that I have a good idea of the
space in which I want to plot from the start?

Oh, and a quick cosmetic query---I notice that the axes when created are
spaced apart somewhat so the axis lines do not meet at the plot origin.
 Is there a way to alter this so that the outline of the box, and the
extreme values of the axis, match up?

Thanks again,

-- Joe

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Re: [R] Power calculation for detecting linear trend

2007-03-07 Thread Peter Dalgaard
Meesters, Erik wrote:
 Dear people,
 I've a problem in doing a power calculation. In Fryer and Nicholson
 (1993), ICES J. mar. Sci. 50: 161-168 page 164 an example is given with
 the following characteristics
 T=5, points in time
 R=5, replicates
 Var.within=0.1
 q=10, a 10% increase per year
 The degrees of freedom for the test are calculated as Vl=T*R-2=23 and
 the non-centrality parameter Dl=4.54.
 Using this they get a power of 0.53, but the result that I'm getting is
 0.05472242.

 I've tried this several ways in R, but I'm not able to come up with the
 same number. Am I doing something wrong in the calculation of the power?
 Here's my code:

 T-5
 R-5
 sigmasq-0.1
 q-10
 Vl-(T*R)-2
 Dl-(R*(T-1)*T*(T+1)/(12*sigmasq))*(log(1+(q/100)))^2 #Dl result is
 still similar

 power.1-1-pf(qf(.95,(T*R-2),1,ncp=0),(T*R-2),1,ncp=Dl)

 Thank you for any suggestions/help.
   
I think your DF are upside-down:

 power.1-1-pf(qf(.95,1,(T*R-2),ncp=0),1,(T*R-2),ncp=Dl)
 power.1
[1] 0.532651



-- 
   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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[R] Plotting a broken line?

2007-03-07 Thread Aldi Kraja
Hi,

Is there a smart way in the R graphs to create a line that is broken in 
intervals based on the indicator given below.
following is a small test graph

Location,indicator,otherinfo
1.2,1,2.2
2.5,1,2.5
3.7,1,2.3
20.1,2,4.3

22.5,2,5.2
25.0,2,3.4
27.3,2,2.2

35.1,3,3.4
37.0,3,7.2
38.0,3,6.1
40.1,3,5.4
52.9,3,3.3

Right now in the plot the line is continuous, but I would like to have 
it broken based on the indicator. If the line of the plot reaches the 
last observation of indicator=1 then the line needs to stop; the next 
line will start at location 22.5 and continue up top 27.3; the next line 
goes from 35.1 up to 52.9.

  x-read.table(file='c:\\aldi\\testgraph.csv',sep=',',header=T)
  x
   Location indicator otherinfo
1   1.2 1   2.2
2   2.5 1   2.5
3   3.7 1   2.3
4  20.1 2   4.3
5  22.5 2   5.2
6  25.0 2   3.4
7  27.3 2   2.2
8  35.1 3   3.4
9  37.0 3   7.2
10 38.0 3   6.1
11 40.1 3   5.4
12 52.9 3   3.3

  
plot(x$Location,x$indicator,type='l',xlim=c(0,max(x$Location)),ylim=c(0,max(x$indicator,x$otherinfo)))
  points(x$Location,x$otherinfo)

TIA,
Aldi

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Re: [R] Multi-line plots with matrices in R

2007-03-07 Thread Gavin Simpson
On Wed, 2007-03-07 at 15:11 +, Joseph Wakeling wrote:
 Gavin Simpson wrote:
  You want maplot here. See ?matplot  but here is an example:
 
 Great!  Thanks to you and Petr for pointing this out, it's exactly what
 I wanted.  Petr's other suggestions look interesting and I'll explore
 them at length later.
 
  Note the changed axis range in the right-hand margin. The problem is
  that you can't use plot.window to achieve what you want, not that
  plot.window doesn't do anything.
 
 Ahhh, I see.  So, it does not affect what has already been plotted, but
 affects how new material is inserted into the plot area.  Entering
 
 plot.window(xlim=c(0,100),ylim=c(0,0.5))
 axis(1)
 axis(2)
 plot.window(xlim=c(0,100),ylim=c(0,1))
 axis(2)
 
 ... is instructive. :-)
 
 So, _is_ there a command which will rearrange the existing plotted
 items, including axes?  Or does R require that I have a good idea of the
 space in which I want to plot from the start?

Not with the standard R graphics - think of the graphics window as a
piece of paper and if you draw anything on it you have done so in
permanent ink. If something needs changing you need a new sheet of paper
and have to redraw the lot. Most people I know write their code in some
text editor and send (or copy paste) it into R. It is an easy matter to
edit one or two bits of your code to tweak the display and re-plot...

I think you can modify lattice graphics objects and just plot (print
really) them again - but again you are really redrawing the whole plot
from scratch. IIRC grid might be able to do some of what you are looking
for.

 
 Oh, and a quick cosmetic query---I notice that the axes when created are
 spaced apart somewhat so the axis lines do not meet at the plot origin.
  Is there a way to alter this so that the outline of the box, and the
 extreme values of the axis, match up?
 

Look at ?par and xaxs and yaxs. E.g.

plot(1:10, xaxs = i, yaxs = i)

G

 Thanks again,
 
 -- Joe
 
-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
 UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%

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Re: [R] How to open more windows to make more graphs at once!

2007-03-07 Thread José Rafael Ferrer Paris
Dear Monireh,
try using lattice:


library(lattice)
set.seed(1234)
dat - data.frame(months=rep(1:10,80),upper = rnorm(800)+1, 
  lower = rnorm(800)-1, 
  observed = rnorm(800), best.sim = rnorm(800), 
  stname = factor(gl(80, 10)))

jpeg(filename = Rplot%03d.jpeg)
xyplot(best.sim+observed+lower+upper~months|stname,dat,
   layout=c(3,4),type=b,auto.key=T)
dev.off()

It should produce almost exactly what you want. Lattice is a very
powerful tool for creating multiple graphics. You can customize the
individual plots within the lattice using panel and prepanel functions,
take a look at the documentation of the library and the documentation of
xyplot and panel.xyplot. Lattice is a little bit more complex than
normal plots in R, so you would have to spend more time in learning
how to use its functionality, but it is worth trying.

have a lot of fun

JR


El mié, 07-03-2007 a las 09:39 +0100, Faramarzi Monireh escribió:
 Dear R users,
 I have a data frame (test) including five columns of upper (numeric), lower 
 (numeric), observed (numeric), best_sim (numeric) and stname (factor with 80 
 levels, each level with different length). Now I would like to write a short 
 program to draw one graph as follow for each level of stname but I would like 
 also to draw each time 12 graphs for the 12 levels of stname in the same 
 graphic windows and save it as jpeg' file . This means at the end I will 
 have 7 (80 levels/12=7) graphic windows and 7 jpeg files each one with 12 
 graphs (the last one with 8 graphs) for the 12 levels of stname. I already 
 wrote the following script to do it each time for 12 levels of stname but I 
 have to change script each time for the another 12 levels [line 3 in the 
 script for example: for( i in levels(test$stname)[12:24))] and I do not know 
 how can I save the obtained graphs (seven graphic windows) as jpeg files 
 (e.g. plot1.jpeg, plot2.jpeg and so on). As I have 45 dataset like this it 
 would be gr!
  eat if somebody can help me to complete this script to do all together for a 
 dataset using a script.
 Thank you very much in advance for your cooperation,
 Monireh
 
 
   
 windows(9,9)
 par(mfrow = c(3,4))
 for( i in levels(test$stname)[1:12])
 { 
 data- test[test$stname==i,]
 xx - c(1:length(data$upper), length(data$upper):1)
 yy - c(data$upper, rev(data$lower))
 zz- data$observed
 tt- data$Best_Sim
 par(lab =c(10,15,2))
 plot.jpeg- plot(xx,yy, type=n, xlim=c(min(xx), max(xx)), 
 ylim=c(min(zz,yy,tt), max(yy,zz,tt)*1.4),
  main= i, xlab=Month (1990-2002),  ylab=Discharge(m3/s), font.axis=6)
 polygon(xx, yy, col=green,  border = NA)
 lines(zz, col=blue, lwd=1.5)
 lines(tt,col=red, lwd=1.5) 
 legend(length(zz)-60, max(yy,zz,tt)*1.45, c(Upper Limit, Lower Limit,  
 Observed,Best etimation)
 , lwd=c(10, 1,1.7,1.7), bty=n, col= c(green, white, blue,red))
  }
 
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[R] Two-way Unbalanced multiple sample ANOVA

2007-03-07 Thread Gregory Hughes
Hello all,

I was wondering if anyone could help me formulate a Two-way ANOVA for 
unbalanced multiple sample data?

We have a new study method aimed to help students to study for tests 
using computers. (I am a computer scientists, hence my 
soon-to-be-apparent lack of statistical knowledge).

To test this study method we devised a user study where 30 participant 
attended 2 lectures, lecture1 and lecture2. Two test were created, test1 
and test2.

test1 corresponds to the material in lecture1 and test2 corresponds to 
the material in lecture2.

The 30 participants were split into two groups, group1 and group2.

group1 used our new study method to review for lecture1 and their 
existing study method to review the material from lecture2
group2 used our new study method to review for lecture2 and their 
existing study method to review the material from lecture1

Each group then took the two test.

This is a repeated measure experiment because we have 2 exam scores for 
each participant, one using our new method to study and one not using 
our new method to study.

The data is unbalanced because participants did not take the same test 
twice.

 From what I understand balanced data would look like
IDTEST SYSTEM SCORE
1   11 80
1   10 70
1   21 90
1   20 95
2   11 70
2   10 75
2   21 80
2   20 75

But instead our data look like this:
IDTEST SYSTEM SCORE
1   11 80
1   20 95
2   10 75
2   21 80

So participant 2 never took test1 using our system.

Anyway, I want to look to see if our new study method had an impact one 
test results. Also, I want to see if the test number had an impact on 
the exam results.

Here is some sample data:


 dataSet - data.frame(
particID=factor(c(1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8)),
whichExam=factor(c(1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2)),
studyMethod=factor(c(1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1)),
score=c(90,80,75,70,70,58,73,68,69,87,68,79,80,80,99,95))


 From what I have read this should be how to compute and ANOVA on this data:


  summary(aov(score~whichExam*studyMethod+Error(particID),data=dataSet))

Error: particID
  Df  Sum Sq Mean Sq F value Pr(F)
whichExam:studyMethod  1  333.06  333.06  1.8211 0.2259
Residuals  6 1097.38  182.90  

Error: Within
Df  Sum Sq Mean Sq F value  Pr(F) 
whichExam1   3.062   3.062  0.1072 0.75445 
studyMethod  1 203.062 203.062  7.1094 0.03721 *
Residuals6 171.375  28.562 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1



Is this correct way do do an ANOVA test for this data?
 From what I can tell this means that the study method did have a 
statistically significant impact on the scores, is that correct? This 
also shows that it did not matter which test the subject took, meaning 
that the two test were equally difficult.


What exactly do the titles Error ... mean?
What are Residuals?

Can anyone recommend a good book on R which covers this information, all 
I can find are books on SPSS?

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Re: [R] anova applied to a lme object

2007-03-07 Thread José Rafael Ferrer Paris
The variances of the random effects and the residual variances are given
by the summary function. Maybe VarCorr or varcomp gives you the answer
you are looking for:

library(nlme)
library(ape)
?VarCorr
?ape

JR
El mié, 07-03-2007 a las 13:09 +0100, Berta escribió:
 Hi R-users,
 
 when carrying out a multiple regression, say lm(y~x1+x2), we can use an 
 anova of the regression with summary.aov(lm(y~x1+x2)), and afterwards 
 evaluate the relative contribution of each variable using the global Sum of 
 Sq of the regression and the Sum of Sq of the simple regression y~x1.
 
 Now I would like to incorporate a random effect in the model, as some data 
 correspond to the same region and others not:  mylme- lme(y~x1+x2, random= 
 ~1|as.factor(region)). I would like to know, if possible, which is the 
 contribution of each variable to the global variability. Using anova(mylme) 
 produce an anova table (without the Sum of Sq column), but I am not sure how 
 can I derive the contribution of each variable from it, or even whether it 
 is nonsense to try, nor can I derive a measure of how much variability is 
 left unexplained.
 
 Sorry for the type of question, but I did not find a simple solution and 
 some researchers I work with love to have relative contributions to global 
 variability.
 
 Thanks a lot in advance,
 
 Berta
 
 
 
 
 
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-- 
Dipl.-Biol. JR Ferrer Paris
~~~
Laboratorio de Biología de Organismos --- Centro de Ecología
Instituto Venezolano de Investigaciones Científicas (IVIC) 
Apdo. 21827, Caracas 1020-A 
República Bolivariana de Venezuela

Tel: (+58-212) 504-1452
Fax: (+58-212) 504-1088

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Re: [R] compiling latest version of R

2007-03-07 Thread Uwe Ligges


Jenny Barnes wrote:
 Dear R-help community,
 
 I have had trouble in the past installing the latest version of R: we got the 
 errors shown below (the computer specifications and version of R are below 
 that). Does anybody have tips for compiling the latest version of R so that I 
 can avoid these errors?

1. 2.4.1 is recent, not 2.3.1 as shown at the bottom.
2. You have some R-2.4.0 in your path or set in your R_LIBS, remove that 
one first.

Uwe Ligges


 
 configure
 make
 ...
 ...
 ...
 
 f90: CODE: 0 WORDS, DATA: 0 WORDS
 gcc -G -L/usr/local/lib -o stats.so init.o kmeans.o  ansari.o bandwidths.o
 chisq
 sim.o d2x2xk.o fexact.o kendall.o ks.o  line.o smooth.o  prho.o swilk.o 
 ksmooth
 .o loessc.o isoreg.o Srunmed.o Trunmed.o  dblcen.o distance.o
 hclust-utils.o  nl
 s.o  HoltWinters.o PPsum.o arima.o burg.o filter.o  mAR.o pacf.o starma.o
 port.o
  family.o sbart.o bsplvd.o bvalue.o bvalus.o loessf.o ppr.o qsbart.o 
 sgram.o si
 nerp.o sslvrg.o stxwx.o  hclust.o kmns.o  eureka.o stl.o portsrc.o
 -L../../../..
 /lib -lRblas  -lg2c -lm -lgcc_s
 mkdir ../../../../library/stats/libs
 building package 'datasets'
 mkdir ../../../library/datasets
 mkdir ../../../library/datasets/R
 mkdir ../../../library/datasets/data
 Error in dyn.load(x, as.logical(local), as.logical(now)) :
 unable to load shared library
 '/tmp/R-2.4.0/library/stats/libs/stats.so'
 :
   ld.so.1: R: fatal: relocation error: file
 /tmp/R-2.4.0/library/stats/libs/stat
 s.so: symbol __i_abs: referenced symbol not found
 Execution halted
 *** Error code 1
 
 
 These are my specifications:
 
 platform   sparc-sun-solaris2.10 
 arch   sparc 
 os solaris2.10   
 system sparc, solaris2.10
 status   
 major  2 
 minor  3.1   
 year   2006  
 month  06
 day01
 svn rev38247 
 language   R 
 version.string Version 2.3.1 (2006-06-01)
 
 
 Many thanks for your time in reading this problem, I look forward to hearing 
 your suggestions and advice,
 
 Jenny
 
 
 ~~
 Jennifer Barnes
 PhD student: long range drought prediction 
 Climate Extremes Group
 Department of Space and Climate Physics
 University College London
 Holmbury St Mary 
 Dorking, Surrey, RH5 6NT
 Tel: 01483 204149
 Mob: 07916 139187
 Web: http://climate.mssl.ucl.ac.uk
 
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Re: [R] anova applied to a lme object

2007-03-07 Thread Berta
Thanks José Rafael, I will try with library(ape) (at the moment I cannot 
load it).

VarCorr gives  the variance estimates for the random effect and the error 
terms. However, what I am looking for is a measure of the explained 
proportion of variance, such as it is R2 in regression models, and more 
precisely, I am looking for a measure of the explained proprotion of 
variance of each of the variables considered (continuous variables and other 
with random slope). For example, Snijders and Bosker (2003) pg 102 dedicate 
a chapter in their book to  how much does the multilevel model explain 
(chapter 7) and derive formulaes for R_1 and R_2  (variance in the first and 
second level respectively). Things seem to get complicated when a slope 
random effect is included in the model, as in my case.  It seems that 
package HLM provides the necessary estimates.

I will have a look at library(ape), thanks for the suggestion.

The book I mention is: Snijders, TAB and Bosker RJ (2003). Multilevel 
Analysis. An introduction to basic and advanced multilevel modeling. SAGE, 
London.

Berta




- Original Message - 
From: José Rafael Ferrer Paris [EMAIL PROTECTED]
To: Berta [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Wednesday, March 07, 2007 5:16 PM
Subject: Re: [R] anova applied to a lme object


 The variances of the random effects and the residual variances are given
 by the summary function. Maybe VarCorr or varcomp gives you the answer
 you are looking for:

 library(nlme)
 library(ape)
 ?VarCorr
 ?ape

 JR
 El mié, 07-03-2007 a las 13:09 +0100, Berta escribió:
 Hi R-users,

 when carrying out a multiple regression, say lm(y~x1+x2), we can use an
 anova of the regression with summary.aov(lm(y~x1+x2)), and afterwards
 evaluate the relative contribution of each variable using the global Sum 
 of
 Sq of the regression and the Sum of Sq of the simple regression y~x1.

 Now I would like to incorporate a random effect in the model, as some 
 data
 correspond to the same region and others not:  mylme- lme(y~x1+x2, 
 random=
 ~1|as.factor(region)). I would like to know, if possible, which is the
 contribution of each variable to the global variability. Using 
 anova(mylme)
 produce an anova table (without the Sum of Sq column), but I am not sure 
 how
 can I derive the contribution of each variable from it, or even whether 
 it
 is nonsense to try, nor can I derive a measure of how much variability is
 left unexplained.

 Sorry for the type of question, but I did not find a simple solution and
 some researchers I work with love to have relative contributions to 
 global
 variability.

 Thanks a lot in advance,

 Berta



 

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 -- 
 Dipl.-Biol. JR Ferrer Paris
 ~~~
 Laboratorio de Biología de Organismos --- Centro de Ecología
 Instituto Venezolano de Investigaciones Científicas (IVIC)
 Apdo. 21827, Caracas 1020-A
 República Bolivariana de Venezuela

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 Fax: (+58-212) 504-1088

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Re: [R] Memory Limits in Ubuntu Linux

2007-03-07 Thread davidkat
Thanks for the tips, Roger.

fyi: When I added /3GB to the boot.ini, the resulting desktop was incomplete 
and locked - no chance to try starting R. Searching the web lead me to 
believe that this was possibly a dead-end, so I abandoned this effort. Any 
hints on getting this to work, anyone? 



On 7 Mar 2007 at 8:26, Bos, Roger wrote:

 David,
 
 I wouldn't give up on windows so fast.  Many people have gotten the 3Gb
 switch to work. One used to have to modify the header of the Rgui.exe
 program to use the switch, but now the binary comes ready for that, so
 its really quite easy.  I would like to hear more about why its not
 working for you.
 
 As for Linux, I use FC5 for which there is a 64-bit binary.  But there
 are also 64-bit binaries for other distros.  The 32-bit and 64-bit
 binaries are in different directories, so you should have no trouble
 telling them apart.  
 
 I have heard good things about Ubuntu--mainly that its very easy to
 use--but FC5 has been pretty easy to learn too and I use the KDE desktop
 which gives me Kate as a text editor.  You can open a terminal window in
 Kate to run R and set up a key like F10 to send the code from the editor
 to R.  Its not quite as good as my Windows setup with Tinn-R, but almost
 as good.
 
 Thanks,
 
 Roger
 
 
 -Original Message-
 From: [EMAIL PROTECTED]
 [mailto:[EMAIL PROTECTED] 
 Sent: Tuesday, March 06, 2007 5:37 PM
 To: Bos, Roger
 Subject: RE: [R] Memory Limits in Ubuntu Linux
 
 Thanks for your prompt reply!
 
 The windows 3GB switch is quite problematic - it was not useable on my
 machine, and there are comments about these problems around the net.
 Thus, on to Linux. My machine has 4Gig, and some megabytes are grabbed
 by my Asus motherboard, leaving some 3.56 Gig. 
 
 So if I understand your suggestion, try the 64-bit version of Ubuntu
 (based on Debian but I had better luck with the video part of the
 install) and then use the corresponding image from CRAN. My fear is that
 the CRAN Ubuntu version might be 32-bit - any idea how to find out
 before I embark on another install? Which Linux do you have - you
 described some significant success with getting large jobs to run.
 
 And yes, I've worked hard to save memory by tweaking the code.
 
 
 Thanks again.
 
 
 On 6 Mar 2007 at 16:51, Bos, Roger wrote:
 
  David,
  
  First of all, under Windows you can get about 3GB available to R by 
  using the /3Gb switch in your boot.ini file, assuming you have 4Gb of 
  memory installed on your windows machine.  Using that method, I have 
  seen the memory using of my R process get as big as 2.7Gb in task 
  manager.  What's important, of course, is contiguous space, as you 
  mentioned.  There, you may want to check your code closely and make 
  sure that its memory usage is as efficient as possible and you are 
  storing the minimal amount you need for each run.  If you don't need 
  an object for a while consider writing it to disk and reading it back
 in later.
  
  Second, AFAIK to get any benefit from more memory is Linux you have to
 
  go to the 64bit version.  I am a Linux newbie too, so I choose to use 
  one of the pre-compiled binaries available on CRAN.  In other words, 
  you shouldn't have to compile anything yourself.  How much memory do 
  you have on your Linux box?  I have 16Gb and I know I have ran stuff 
  that wouldn't run on my 4Gb windows box.
  
  HTH,
  
  Roger
  
  
  
  
  
   
  
  -Original Message-
  From: [EMAIL PROTECTED] 
  [mailto:[EMAIL PROTECTED] On Behalf Of 
  [EMAIL PROTECTED]
  Sent: Tuesday, March 06, 2007 3:44 PM
  To: r-help@stat.math.ethz.ch
  Subject: [R] Memory Limits in Ubuntu Linux
  
  I am an R user trying to get around the 2Gig memory limit in Windows, 
  so here I am days later with a working Ubuntu, and R under Ubuntu. But
 
  - the memory problems seem worse than ever. R code that worked under 
  windows fails, unable to allocate memory.
  
  Searching around the web, it appears that the problem may be the 
  ability to find contguous memory for my big vectors, but a fresh boot 
  of Ubuntu does not help either.
  
  Which way to go?
  
  1) Try to install 64-bit version for bigger address space. Would this 
  help? Is this workable for my Athlon 64 Dual-core? (the live cd seems 
  to work but I never got it to boot after a disk install, but then the 
  386 version was no better until I learned more about Grub...I could 
  try again if this might solve the
  problem)
  
  2) Recompile R to get bigger memory capability? (I'll have to 
  cross-post to some R forums too) This will be a challenge for a Linux 
  newbie...like me.
  
  3) Any other suggestions? My goal is to create a bigger neural network
 
  than fits in my Windows R version.
  --
  David Katz
   www.davidkatzconsulting.com
 541 482-1137
  
  [[alternative HTML version deleted]]
  
  __
  R-help@stat.math.ethz.ch mailing list
  

Re: [R] Plotting a broken line?

2007-03-07 Thread Greg Snow
If you insert an NA (or row of NA's) into the data at each place you
want a break (after indicator increases), then the regular plot with
type='l' will break the line for you.

Is this what you want?

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
 
 

 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of Aldi Kraja
 Sent: Wednesday, March 07, 2007 9:21 AM
 To: r-help@stat.math.ethz.ch
 Subject: [R] Plotting a broken line?
 
 Hi,
 
 Is there a smart way in the R graphs to create a line that is 
 broken in intervals based on the indicator given below.
 following is a small test graph
 
 Location,indicator,otherinfo
 1.2,1,2.2
 2.5,1,2.5
 3.7,1,2.3
 20.1,2,4.3
 
 22.5,2,5.2
 25.0,2,3.4
 27.3,2,2.2
 
 35.1,3,3.4
 37.0,3,7.2
 38.0,3,6.1
 40.1,3,5.4
 52.9,3,3.3
 
 Right now in the plot the line is continuous, but I would 
 like to have it broken based on the indicator. If the line of 
 the plot reaches the last observation of indicator=1 then the 
 line needs to stop; the next line will start at location 22.5 
 and continue up top 27.3; the next line goes from 35.1 up to 52.9.
 
   x-read.table(file='c:\\aldi\\testgraph.csv',sep=',',header=T)
   x
Location indicator otherinfo
 1   1.2 1   2.2
 2   2.5 1   2.5
 3   3.7 1   2.3
 4  20.1 2   4.3
 5  22.5 2   5.2
 6  25.0 2   3.4
 7  27.3 2   2.2
 8  35.1 3   3.4
 9  37.0 3   7.2
 10 38.0 3   6.1
 11 40.1 3   5.4
 12 52.9 3   3.3
 
  
 plot(x$Location,x$indicator,type='l',xlim=c(0,max(x$Location))
 ,ylim=c(0,max(x$indicator,x$otherinfo)))
   points(x$Location,x$otherinfo)
 
 TIA,
 Aldi
 
 --
 
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 https://stat.ethz.ch/mailman/listinfo/r-help
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Re: [R] Plotting a broken line?

2007-03-07 Thread Petr Pikal
Hi

you shall probably cooperate with segments, so you need to extract 
start and end points for your lines e.g.

 x-c(1:6, 10:15,20:25)
 y-rep(c(1,2,3), each=6)
 plot(x,y, type=l)
 plot(x,y)
 

segments(sapply(split(x,y), min),1:3, sapply(split(x,y),max),1:3)

Regards
Petr



On 7 Mar 2007 at 10:21, Aldi Kraja wrote:

Date sent:  Wed, 07 Mar 2007 10:21:06 -0600
From:   Aldi Kraja [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] Plotting a broken line?
Send reply to:  [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]

 Hi,
 
 Is there a smart way in the R graphs to create a line that is broken
 in intervals based on the indicator given below. following is a small
 test graph
 
 Location,indicator,otherinfo
 1.2,1,2.2
 2.5,1,2.5
 3.7,1,2.3
 20.1,2,4.3
 
 22.5,2,5.2
 25.0,2,3.4
 27.3,2,2.2
 
 35.1,3,3.4
 37.0,3,7.2
 38.0,3,6.1
 40.1,3,5.4
 52.9,3,3.3
 
 Right now in the plot the line is continuous, but I would like to have
 it broken based on the indicator. If the line of the plot reaches the
 last observation of indicator=1 then the line needs to stop; the next
 line will start at location 22.5 and continue up top 27.3; the next
 line goes from 35.1 up to 52.9.
 
   x-read.table(file='c:\\aldi\\testgraph.csv',sep=',',header=T)  x
Location indicator otherinfo
 1   1.2 1   2.2
 2   2.5 1   2.5
 3   3.7 1   2.3
 4  20.1 2   4.3
 5  22.5 2   5.2
 6  25.0 2   3.4
 7  27.3 2   2.2
 8  35.1 3   3.4
 9  37.0 3   7.2
 10 38.0 3   6.1
 11 40.1 3   5.4
 12 52.9 3   3.3
 
   
 plot(x$Location,x$indicator,type='l',xlim=c(0,max(x$Location)),ylim=c(
 0,max(x$indicator,x$otherinfo)))
   points(x$Location,x$otherinfo)
 
 TIA,
 Aldi
 
 --
 
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 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html and provide commented,
 minimal, self-contained, reproducible code.

Petr Pikal
[EMAIL PROTECTED]

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[R] error installing packages

2007-03-07 Thread Bricklemyer, Ross S

I was finally able to get R to 'configure', 'make', and 'install' on Mandriva 
2007.  Itried to install gnomeGUI and I received an error.  See below.  At what 
step do I make R a shared library?  Where did I go wrong?

Ross

==
downloaded 74Kb

* Installing *Frontend* package 'gnomeGUI' ...
Using R Installation in R_HOME=/usr/local/lib64/R
R was not built as a shared library
Need a shared R library
ERROR: configuration failed for package 'gnomeGUI'
* Removing '/usr/local/lib64/R/library/gnomeGUI'

The downloaded packages are in
/root/tmp/RtmpkHUeyA/downloaded_packages
Warning message:
installation of package 'gnomeGUI' had non-zero exit status in: 
install.packages(c(gnomeGUI))
=

[[alternative HTML version deleted]]

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[R] No fit statistics for some models using sem

2007-03-07 Thread Ista Zahn
Hi,

New to both R and SEM, so this may be a very simple question. I am  
trying to run a very simple path analysis using the sem package.  
There are 2 exogenous (FARSCH, LOCUS10) and 2 endogenous (T_ATTENT,  
RMTEST) observed variables in the model.  The idea is that T_ATTENT  
mediates the effect of FARSCH and LOCUS10 on RMTEST. The RAM  
specification I used is

FARSCH - T_ATTENT, y1x1, NA
LOCUS10 - T_ATTENT, y1x2, NA
FARSCH - RMTEST10, y2x1, NA
LOCUS10 - RMTEST10, y2x2, NA
T_ATTENT - RMTEST10, y2y1, NA
FARSCH - FARSCH, x1x1, NA
LOCUS10 - LOCUS10, x2x2, NA
T_ATTENT - T_ATTENT, y1y1, NA
RMTEST10 - RMTEST10, y2y2, NA
LOCUS10 - FARSCH, x2x1, NA

This model runs, but using the summary function does not return the  
usual model fit statistics, only the following:

Model Chisquare =  0   Df =  0 Pr(Chisq) = NA
  Chisquare (null model) =  8526.8   Df =  6
  Goodness-of-fit index =  1
  BIC =  0

If I omit the last line from the RAM specification(i.e., delete  
LOCUS10 - FARSCH, x2x1, NA), I DO get all the usual statistics:

  Model Chisquare =  1303.7   Df =  1 Pr(Chisq) = 0
  Chisquare (null model) =  8526.8   Df =  6
  Goodness-of-fit index =  0.95864
  Adjusted goodness-of-fit index =  0.58639
  RMSEA index =  0.30029   90% CI: (NA, NA)
  Bentler-Bonnett NFI =  0.84711
  Tucker-Lewis NNFI =  0.082726
  Bentler CFI =  0.84712
  BIC =  1294.1

My understanding is the you should always put in the correlation  
between exogenous predictors, but when I do this I don't get fit  
statistics. Can anyone help me understand what is happening here?

Thank you,

Ista

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[R] Fitting Data to tCopula

2007-03-07 Thread aat

Hello,

Has anyone successfully fit empirical data to a tCopula using the fitCopula
function?  If so, are there ways to pick intelligent starting values to
avoid the errors such as a minor matrix not being positive definite and the
initial value of 'vmmin' not  being finite?

I've been able to fit a normal Copula to my data, but am having diffuculty
with the tCopula.  Any suggestions are greatly appreciated.  

Thank you.

Adam
  
-- 
View this message in context: 
http://www.nabble.com/Fitting-Data-to-tCopula-tf3363801.html#a9358456
Sent from the R help mailing list archive at Nabble.com.

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[R] Failure to run mcsamp() in package arm

2007-03-07 Thread Michael Kubovy
Dear r-helpers,

I can run the examples on the mcsamp help page. For example:

  M1 - lmer (y1 ~ x + (1|group))
  (M1.sim - mcsamp (M1))
fit using lmer,
3 chains, each with 1000 iterations (first 500 discarded)
n.sims = 1500 iterations saved
   mean  sd 2.5%  25%  50%  75% 97.5% Rhat n.eff
beta.(Intercept)   0.1 0.7 -1.2 -0.3  0.1  0.5   1.4  1.0  1500
beta.x 2.5 0.4  1.7  2.2  2.5  2.7   3.2  1.0  1500
sigma.y3.8 0.3  3.3  3.6  3.7  3.9   4.3  1.061
sigma.grop.(In)1.5 0.8  0.0  1.0  1.4  1.9   3.3  1.412
eta.group.(Intercept)[1]   0.0 1.0 -2.1 -0.5  0.0  0.6   2.0  1.0  1500
eta.group.(Intercept)[2]   1.0 1.1 -0.9  0.2  0.9  1.7   3.4  1.059
eta.group.(Intercept)[3]  -1.3 1.2 -4.0 -2.0 -1.3 -0.4   0.5  1.066
eta.group.(Intercept)[4]   1.3 1.1 -0.6  0.4  1.1  2.0   3.7  1.143
eta.group.(Intercept)[5]  -0.7 1.0 -3.0 -1.4 -0.6  0.0   1.2  1.0   120
eta.group.(Intercept)[6]   1.5 1.2 -0.3  0.6  1.4  2.2   4.0  1.049
eta.group.(Intercept)[7]   0.3 1.0 -1.7 -0.3  0.1  0.8   2.5  1.0   440
eta.group.(Intercept)[8]  -1.6 1.2 -4.0 -2.4 -1.5 -0.6   0.3  1.141
eta.group.(Intercept)[9]   0.4 1.0 -1.6 -0.2  0.2  0.9   2.7  1.0   180
eta.group.(Intercept)[10] -1.0 1.1 -3.3 -1.6 -0.9 -0.2   0.8  1.086

For each parameter, n.eff is a crude measure of effective sample size,
and Rhat is the potential scale reduction factor (at convergence,  
Rhat=1).

But when I try to do this with my own data I get an error:

  display(e7.lmer2)
lmer(formula = baLO ~ I(baRatio - 0.985) + delta + (1 + I(baRatio -   
0.985) + delta | subject), data = e7)
coef.est coef.se
(Intercept)-0.19 0.06
I(baRatio - 0.985) -4.95 0.74
delta   0.41 0.06
Error terms:
Groups   Name   Std.Dev. Corr
subject  (Intercept)0.13
   I(baRatio - 0.985) 2.57  0.45
   delta  0.22 -0.12 -0.94
Residual0.39
number of obs: 494, groups: subject, 13
deviance = 551.4

  e7.sim - mcsamp(e7.lmer2)
Error in as.bugs.array(sims, program = lmer, n.iter = n.iter,  
n.burnin = n.burnin,  :
error in parameter sigma. in parameters.to.save

I would appreciate a pointer to what the problem might be.
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
 McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/

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Re: [R] transform R function

2007-03-07 Thread Peter Dalgaard
lamack lamack wrote:
 Dear all, Why the transform function does not accept two statistics 
 functions?

 a = data.frame(matrix(rnorm(20),ncol=2))

 transform(a,M.1=mean(X1),M.2=mean(X2)) # does not works

 #while:

 transform(a,M.1=mean(X1),M2=log(abs(X2))) #works

   
It's a variation of this effect:

data.frame(airquality, list(x=1))#works
data.frame(airquality, list(x=1, y=2)) #works not

Not quite sure what the logic of that is

Of course transform() isn't really intended to handle anything but 
transformed vectors of the same length as the original.

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Re: [R] No fit statistics for some models using sem

2007-03-07 Thread David Barron
It's not the correlation as such that is the problem; it's because you
only have 10 degrees of freedom available with four observed
variables, and you are estimating 10 parameters, which is why you get
a chi square of zero.  When you remove any one free parameter (such as
the correlation), the model becomes identified.

On 07/03/07, Ista Zahn [EMAIL PROTECTED] wrote:
 Hi,

 New to both R and SEM, so this may be a very simple question. I am
 trying to run a very simple path analysis using the sem package.
 There are 2 exogenous (FARSCH, LOCUS10) and 2 endogenous (T_ATTENT,
 RMTEST) observed variables in the model.  The idea is that T_ATTENT
 mediates the effect of FARSCH and LOCUS10 on RMTEST. The RAM
 specification I used is

 FARSCH - T_ATTENT, y1x1, NA
 LOCUS10 - T_ATTENT, y1x2, NA
 FARSCH - RMTEST10, y2x1, NA
 LOCUS10 - RMTEST10, y2x2, NA
 T_ATTENT - RMTEST10, y2y1, NA
 FARSCH - FARSCH, x1x1, NA
 LOCUS10 - LOCUS10, x2x2, NA
 T_ATTENT - T_ATTENT, y1y1, NA
 RMTEST10 - RMTEST10, y2y2, NA
 LOCUS10 - FARSCH, x2x1, NA

 This model runs, but using the summary function does not return the
 usual model fit statistics, only the following:

 Model Chisquare =  0   Df =  0 Pr(Chisq) = NA
   Chisquare (null model) =  8526.8   Df =  6
   Goodness-of-fit index =  1
   BIC =  0

 If I omit the last line from the RAM specification(i.e., delete
 LOCUS10 - FARSCH, x2x1, NA), I DO get all the usual statistics:

   Model Chisquare =  1303.7   Df =  1 Pr(Chisq) = 0
   Chisquare (null model) =  8526.8   Df =  6
   Goodness-of-fit index =  0.95864
   Adjusted goodness-of-fit index =  0.58639
   RMSEA index =  0.30029   90% CI: (NA, NA)
   Bentler-Bonnett NFI =  0.84711
   Tucker-Lewis NNFI =  0.082726
   Bentler CFI =  0.84712
   BIC =  1294.1

 My understanding is the you should always put in the correlation
 between exogenous predictors, but when I do this I don't get fit
 statistics. Can anyone help me understand what is happening here?

 Thank you,

 Ista

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.



-- 
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP

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Re: [R] Plotting a broken line?

2007-03-07 Thread Aldi Kraja
Hi Greg,

Thank you for your response and a previous posting about Macros in R.

Thank you also to Ken Knoblouch (Ken had the same idea as Greg's and 
Peter Pikal (who proposed the use of segments function).
There is only a technical specific that when applying max function to 
find the limit of y one has to use it with

max(x$indicator, na.rm =TRUE))

It worked!!!
  x
   Location indicator otherinfo
1   1.2 1   2.2
2   2.5 1   2.5
3   3.7 1   2.3
4   3.7NANA
5  20.1 2   4.3
6  22.5 2   5.2
7  25.0 2   3.4
8  27.3 2   2.2
9  27.3NANA
10 35.1 3   3.4
11 37.0 3   7.2
12 38.0 3   6.1
13 40.1 3   5.4
14 52.9 3   3.3

Aldi

Greg Snow wrote:

If you insert an NA (or row of NA's) into the data at each place you
want a break (after indicator increases), then the regular plot with
type='l' will break the line for you.

Is this what you want?

  


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Re: [R] No fit statistics for some models using sem

2007-03-07 Thread John Fox
Dear David and Ista,

I haven't looked at this model carefully, but the fact that the df are
0 suggests that the model is just-identified and therefore necessarily
perfectly reproduces the covariances among the observed variables.
Removing a parameter would over-identify the model, making possible the
computation of the missing fit statistics.

Regards,
 John

On Wed, 7 Mar 2007 18:31:09 +
 David Barron [EMAIL PROTECTED] wrote:
 It's not the correlation as such that is the problem; it's because
 you
 only have 10 degrees of freedom available with four observed
 variables, and you are estimating 10 parameters, which is why you get
 a chi square of zero.  When you remove any one free parameter (such
 as
 the correlation), the model becomes identified.
 
 On 07/03/07, Ista Zahn [EMAIL PROTECTED] wrote:
  Hi,
 
  New to both R and SEM, so this may be a very simple question. I am
  trying to run a very simple path analysis using the sem package.
  There are 2 exogenous (FARSCH, LOCUS10) and 2 endogenous (T_ATTENT,
  RMTEST) observed variables in the model.  The idea is that T_ATTENT
  mediates the effect of FARSCH and LOCUS10 on RMTEST. The RAM
  specification I used is
 
  FARSCH - T_ATTENT, y1x1, NA
  LOCUS10 - T_ATTENT, y1x2, NA
  FARSCH - RMTEST10, y2x1, NA
  LOCUS10 - RMTEST10, y2x2, NA
  T_ATTENT - RMTEST10, y2y1, NA
  FARSCH - FARSCH, x1x1, NA
  LOCUS10 - LOCUS10, x2x2, NA
  T_ATTENT - T_ATTENT, y1y1, NA
  RMTEST10 - RMTEST10, y2y2, NA
  LOCUS10 - FARSCH, x2x1, NA
 
  This model runs, but using the summary function does not return the
  usual model fit statistics, only the following:
 
  Model Chisquare =  0   Df =  0 Pr(Chisq) = NA
Chisquare (null model) =  8526.8   Df =  6
Goodness-of-fit index =  1
BIC =  0
 
  If I omit the last line from the RAM specification(i.e., delete
  LOCUS10 - FARSCH, x2x1, NA), I DO get all the usual statistics:
 
Model Chisquare =  1303.7   Df =  1 Pr(Chisq) = 0
Chisquare (null model) =  8526.8   Df =  6
Goodness-of-fit index =  0.95864
Adjusted goodness-of-fit index =  0.58639
RMSEA index =  0.30029   90% CI: (NA, NA)
Bentler-Bonnett NFI =  0.84711
Tucker-Lewis NNFI =  0.082726
Bentler CFI =  0.84712
BIC =  1294.1
 
  My understanding is the you should always put in the correlation
  between exogenous predictors, but when I do this I don't get fit
  statistics. Can anyone help me understand what is happening here?
 
  Thank you,
 
  Ista
 
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  PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
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 -- 
 =
 David Barron
 Said Business School
 University of Oxford
 Park End Street
 Oxford OX1 1HP
 
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 R-help@stat.math.ethz.ch mailing list
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 PLEASE do read the posting guide
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John Fox
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

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[R] sqlSave help!

2007-03-07 Thread leesan

Hi Everyone,
I'm so confused.  I've been trying to save data to a table but I keep
getting an error that says the table
does not exist and at other times saying that it does. So here are some
statements:

 sqlQuery(channel, select top 1 * from
 TestDB.[SILICON\\holouis1].clep_tier_shift)
  State NB Change_Number
1IL 2005-02-08 7

It exists and I can get data from it, but if I try to use fetch or columns:

 sqlFetch(channel, TestDB.[SILICON\\holouis1].clep_tier_shift)
Error in odbcTableExists(channel, sqtable) : 
'TestDB.[SILICON\holouis1].clep_tier_shift': table not found on
channel

 sqlColumns(channel, TestDB.[SILICON\\holouis1].clep_tier_shift)
Error in sqlColumns(channel, TestDB.[SILICON\\holouis1].clep_tier_shift) : 
'TestDB.[SILICON\holouis1].clep_tier_shift': table not found on
channel

Now if I try to save, it says error: table exists already even though I set
append = T...

 sqlQuery(channel, select top 1 * from
 TestDB.[SILICON\\holouis1].clep_tier_shift) - nuts
 sqlSave(channel, dat= nuts,
 tablename=TestDB.[SILICON\\holouis1].clep_tier_shift,safer= T)
Error in sqlSave(channel, dat = nuts, tablename =
TestDB.[SILICON\\holouis1].clep_tier_shift,  : 
[RODBC] ERROR: Could not SQLExecDirect
S0001 2714 [Microsoft][ODBC SQL Server Driver][SQL Server]There is already
an object named 'clep_tier_shift' in the database.

any help would be much appreciated, thanks!

-- 
View this message in context: 
http://www.nabble.com/sqlSave-help%21-tf3364399.html#a9360420
Sent from the R help mailing list archive at Nabble.com.

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[R] Query about using setdiff

2007-03-07 Thread lalitha viswanath
Hi
I have two dataframes
names(DF1) = c(id, val1, val2);

names(DF2) = c(id2);

Ids in DF2 are a complete subset of those in DF1

How can I extract entries from DF1 where id NOT IN
DF2.

I tried setdiff(DF1, DF2); setdiff(DF1$id, DF2$id),
etc.
Although the latter eliminates the ids as required, I
dont know how to extract val1 and val2 for the
resultant set.


Thanks
Lalitha 


 

8:00? 8:25? 8:40? Find a flick in no time

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[R] Download packages problem.

2007-03-07 Thread Luis Garavito
Hi,

I have a little problem with the installation of a new packages. The
installation of R software is correct, but my server required authentication
for use it, and for load a new package directly from R it is not possible.
Is there a code or process for server authentication (put my login and
password) in R for download directly the packages?

The best regards,
-- 
Luis Antonio Garavito

[[alternative HTML version deleted]]

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Re: [R] Memory Limits in Ubuntu Linux

2007-03-07 Thread Bos, Roger
David,

Here is what my boot.ini file looks like:

[boot loader]
timeout=5
default=multi(0)disk(0)rdisk(0)partition(1)\WINDOWS
[operating systems]
multi(0)disk(0)rdisk(0)partition(1)\WINDOWS=Microsoft Windows XP
Professional /noexecute=optin /fastdetect /3gb

The easiest way to edit the boot.ini file is My
Computer/Properties/Advanced/Startup  Recovery/Edit; add the /3gb and
reboot.  I know that a messed up boot.ini file can be a real pain.  I
posted what mine looks like so you can compare yours, but I wouldn't
suggest making any changes to your boot.ini except at the very end of
the last line.

HTH,

Roger


 

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] 
Sent: Wednesday, March 07, 2007 12:27 PM
To: Bos, Roger
Cc: r-help@stat.math.ethz.ch
Subject: RE: [R] Memory Limits in Ubuntu Linux

Thanks for the tips, Roger.

fyi: When I added /3GB to the boot.ini, the resulting desktop was
incomplete and locked - no chance to try starting R. Searching the web
lead me to believe that this was possibly a dead-end, so I abandoned
this effort. Any hints on getting this to work, anyone? 



On 7 Mar 2007 at 8:26, Bos, Roger wrote:

 David,
 
 I wouldn't give up on windows so fast.  Many people have gotten the 
 3Gb switch to work. One used to have to modify the header of the 
 Rgui.exe program to use the switch, but now the binary comes ready for

 that, so its really quite easy.  I would like to hear more about why 
 its not working for you.
 
 As for Linux, I use FC5 for which there is a 64-bit binary.  But there

 are also 64-bit binaries for other distros.  The 32-bit and 64-bit 
 binaries are in different directories, so you should have no trouble 
 telling them apart.
 
 I have heard good things about Ubuntu--mainly that its very easy to 
 use--but FC5 has been pretty easy to learn too and I use the KDE 
 desktop which gives me Kate as a text editor.  You can open a terminal

 window in Kate to run R and set up a key like F10 to send the code 
 from the editor to R.  Its not quite as good as my Windows setup with 
 Tinn-R, but almost as good.
 
 Thanks,
 
 Roger
 
 
 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED]
 Sent: Tuesday, March 06, 2007 5:37 PM
 To: Bos, Roger
 Subject: RE: [R] Memory Limits in Ubuntu Linux
 
 Thanks for your prompt reply!
 
 The windows 3GB switch is quite problematic - it was not useable on my

 machine, and there are comments about these problems around the net.
 Thus, on to Linux. My machine has 4Gig, and some megabytes are grabbed

 by my Asus motherboard, leaving some 3.56 Gig.
 
 So if I understand your suggestion, try the 64-bit version of Ubuntu 
 (based on Debian but I had better luck with the video part of the
 install) and then use the corresponding image from CRAN. My fear is 
 that the CRAN Ubuntu version might be 32-bit - any idea how to find 
 out before I embark on another install? Which Linux do you have - you 
 described some significant success with getting large jobs to run.
 
 And yes, I've worked hard to save memory by tweaking the code.
 
 
 Thanks again.
 
 
 On 6 Mar 2007 at 16:51, Bos, Roger wrote:
 
  David,
  
  First of all, under Windows you can get about 3GB available to R by 
  using the /3Gb switch in your boot.ini file, assuming you have 4Gb 
  of memory installed on your windows machine.  Using that method, I 
  have seen the memory using of my R process get as big as 2.7Gb in 
  task manager.  What's important, of course, is contiguous space, as 
  you mentioned.  There, you may want to check your code closely and 
  make sure that its memory usage is as efficient as possible and you 
  are storing the minimal amount you need for each run.  If you don't 
  need an object for a while consider writing it to disk and reading 
  it back
 in later.
  
  Second, AFAIK to get any benefit from more memory is Linux you have 
  to
 
  go to the 64bit version.  I am a Linux newbie too, so I choose to 
  use one of the pre-compiled binaries available on CRAN.  In other 
  words, you shouldn't have to compile anything yourself.  How much 
  memory do you have on your Linux box?  I have 16Gb and I know I have

  ran stuff that wouldn't run on my 4Gb windows box.
  
  HTH,
  
  Roger
  
  
  
  
  
   
  
  -Original Message-
  From: [EMAIL PROTECTED] 
  [mailto:[EMAIL PROTECTED] On Behalf Of 
  [EMAIL PROTECTED]
  Sent: Tuesday, March 06, 2007 3:44 PM
  To: r-help@stat.math.ethz.ch
  Subject: [R] Memory Limits in Ubuntu Linux
  
  I am an R user trying to get around the 2Gig memory limit in 
  Windows, so here I am days later with a working Ubuntu, and R under 
  Ubuntu. But
 
  - the memory problems seem worse than ever. R code that worked under

  windows fails, unable to allocate memory.
  
  Searching around the web, it appears that the problem may be the 
  ability to find contguous memory for my big vectors, but a fresh 
  boot of Ubuntu does not help either.
  
  Which way to go?
  
  1) 

Re: [R] Query about using setdiff

2007-03-07 Thread Dimitris Rizopoulos
try something along these lines (untested):

DF1[DF1$id %in% DF2$id2, c(val1, val2)]
DF1[!DF1$id %in% DF2$id2, c(val1, val2)]


I hope it helps.

Best,
Dimitris


Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
  http://www.student.kuleuven.be/~m0390867/dimitris.htm


Quoting lalitha viswanath [EMAIL PROTECTED]:

 Hi
 I have two dataframes
 names(DF1) = c(id, val1, val2);

 names(DF2) = c(id2);

 Ids in DF2 are a complete subset of those in DF1

 How can I extract entries from DF1 where id NOT IN
 DF2.

 I tried setdiff(DF1, DF2); setdiff(DF1$id, DF2$id),
 etc.
 Although the latter eliminates the ids as required, I
 dont know how to extract val1 and val2 for the
 resultant set.


 Thanks
 Lalitha



 
 8:00? 8:25? 8:40? Find a flick in no time

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Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

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[R] ODP: Plotting a broken line?

2007-03-07 Thread Bojanowski, M.J. \(Michal\)
Hi Aldi,

Yet another way to do what you want. 'd' is your data frame. You cannot easily 
modify the attributes of the lines however...

# just to set up coordinates
plot( d$Location, d$otherinfo, type=n) 
# plot the lines
invisible(by(d, d$indicator, function(dd) lines(dd$Location, dd$otherinfo)))




*
*** Please note that my e-mail address has changed to [EMAIL PROTECTED]
*** Please update you address books accordingly. Thank you!
*

Michal Bojanowski
ICS / Department of Sociology
Utrecht University
Heidelberglaan 2; 3584 CS Utrecht
Room 1428
[EMAIL PROTECTED]
http://www.fss.uu.nl/soc/bojanowski/



-Wiadomo¶æ oryginalna-
Od: [EMAIL PROTECTED] w imieniu Aldi Kraja
Wys³ano: ¦r 2007-03-07 17:21
Do: r-help@stat.math.ethz.ch
Temat: [R] Plotting a broken line?
 
Hi,

Is there a smart way in the R graphs to create a line that is broken in 
intervals based on the indicator given below.
following is a small test graph

Location,indicator,otherinfo
1.2,1,2.2
2.5,1,2.5
3.7,1,2.3
20.1,2,4.3

22.5,2,5.2
25.0,2,3.4
27.3,2,2.2

35.1,3,3.4
37.0,3,7.2
38.0,3,6.1
40.1,3,5.4
52.9,3,3.3

Right now in the plot the line is continuous, but I would like to have 
it broken based on the indicator. If the line of the plot reaches the 
last observation of indicator=1 then the line needs to stop; the next 
line will start at location 22.5 and continue up top 27.3; the next line 
goes from 35.1 up to 52.9.

  x-read.table(file='c:\\aldi\\testgraph.csv',sep=',',header=T)
  x
   Location indicator otherinfo
1   1.2 1   2.2
2   2.5 1   2.5
3   3.7 1   2.3
4  20.1 2   4.3
5  22.5 2   5.2
6  25.0 2   3.4
7  27.3 2   2.2
8  35.1 3   3.4
9  37.0 3   7.2
10 38.0 3   6.1
11 40.1 3   5.4
12 52.9 3   3.3

  
plot(x$Location,x$indicator,type='l',xlim=c(0,max(x$Location)),ylim=c(0,max(x$indicator,x$otherinfo)))
  points(x$Location,x$otherinfo)

TIA,
Aldi

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[R] C to R

2007-03-07 Thread Heloise Mattos
I`m doing some functions on C that gives me the x and y coordinates.
I`d like to now how I can get these coordinates (both are a vector of
number) on R to that I can make a graphic.
I`ve already made a package with my functions, so I just wanna how
about how to get the coordinates.

Thanks, Heloise.

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Re: [R] Download packages problem.

2007-03-07 Thread Seth Falcon
Luis Garavito [EMAIL PROTECTED] writes:

 Hi,

 I have a little problem with the installation of a new packages. The
 installation of R software is correct, but my server required authentication
 for use it, and for load a new package directly from R it is not possible.
 Is there a code or process for server authentication (put my login and
 password) in R for download directly the packages?

Yes.





Without telling us what operating system, what version of R, and some
specifics of what you are trying and the error messages you are
seeing, it is rather difficult to imagine that you will get any useful
help.

Please have a look at the posting guide and send an update.

+ seth

-- 
Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center
http://bioconductor.org

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[R] ATLAS for Pentium D and Pentium Duo Core?

2007-03-07 Thread Jonathan Zhang
Dear all,

  I wanted to use ATLAS to increase the computational speed of my
program. I have installed R in a straightfoward way with .exe file
(not building from source).

  However, after getting the P4 ATLAS Rblas.dll file from the CRAN
directory bin/windows/contrib/ATLAS, and replacing the default
Rblas.dll file in my R home directory under bin/, the speed actually
decreased by 3% compared with the default Rblas file!!

  So what am I doing wrong? Is the above the correct procedure for
using ATLAS? Or, is my machine not a Pentium 4 and hence the ATLAS
file that I got from CRAN is not appropriate?

  If the issue lies in the processor, then please kindly let me know
where to get the correct ATLAS file for the following computers:

 IThe first one is Pentium D 3.2 ghz, the second is a MacBook 1.83ghz
dual-boot and I am running R in windows.

  Thank you!

Jonathan Zhang

Marketing Division
Columbia Business School

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Re: [R] Download packages problem.

2007-03-07 Thread Petr Klasterecky
Luis,

posting a message several times usually won't help you to get answers...
You did not tell us what kind of error you obtain, what packages did you 
try to install and how, what operating system are you using etc. To be 
honest, I have no idea what kind of server can be involved - R does not 
need any server to run on and the CRAN ftp severs work with anonymous 
logins. Please read the posting guide (link given below in the footer) 
and follow it next time.
Petr

Luis Garavito napsal(a):
 Hi,
 
 I have a little problem with the installation of a new packages. The
 installation of R software is correct, but my server required authentication
 for use it, and for load a new package directly from R it is not possible.
 Is there a code or process for server authentication (put my login and
 password) in R for download directly the packages?
 
 The best regards,

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

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[R] Read data with different column lengths

2007-03-07 Thread Hofert Marius
Dear r-help users,

I have the following simple problem: Reading data from a file. The  
file is a .txt file exported (save as...) from Excel (see below for  
an example). The Excel file consists of two header rows (first row  
consists of ticker symbols of stocks, the second row consists of  
column explanations (Date,Px Last), followed by several rows of  
data. Now forget about the first two rows, I can deal with that (read  
separately, then extract the actual ticker symbols ADS,  
ALV, ...). For reading the rest, I tried several things, for example:
data=read.table(infile,quote=,fill=T,dec=,,skip=2,colClasses=rep(c 
(character,numeric),ntickers))
or
data=matrix(scan(file=infile,what=rep(c 
(character,numeric),ntickers),dec=,,skip=2),ncol=2*ntickers,byrow= 
T)
where infile specifies the path to the input file and ntickers is  
the number of ticker-columns in the data set, so in the example  
below, ntickers=2.

Both ways of reading the data work perfectly fine if all columns have  
the same length (i.e. the same number of filled rows), so if the data  
is given in a (filled) rectangular form. Now, as you can imagine,  
there are days when one stock is traded but not the other... so,  
there might be columns that do not have the same number of filled  
rows (see below, for the stock with ticker symbol ADS, only 3  
trading days are shown, so this column is shorter than the data  
column for the stock ALV). Now, if I export such a structure to  
a .txt file, then all (by default) blank fields will be replaced by  
\t, i.e. tabs. Both reading procedures as give above have problems  
as they either display that the number of rows/columns do not fit  
together or as they read the table, but some cells are shifted to the  
left (for the example below, the entry 07/02/05134,7 appears in  
the empty field of the stock ADS which is of course not what we want).
So the simple question is: How do I read such a structure?
Can there be a simple solution? The problem is simply that empty  
cells are replace by \t which are then ignored for reading. So how  
do we distinguish between the empty cells that are given between the  
columns and the empty cells that actually fill a column to have the  
same length as other columns. Of course I could manually put in a  
certain character (e.g. a *) to fill in the gaps, but the data set  
is simply too large. If it helps, these blank fields only appear in  
the end of each column, not in the middle.

As I work on a Mac (OS X 10.4), it was not possible (at least to me)  
to read the data directly from the Excel file vial the library RODBC  
or read.xls.

Note, that the same problem arises, when I export the Excel file as  
a .csv, then all blank fields are separated by ; instead of \t  
and the reading procedure can also not decide if the field  
corresponds to an empty separating column or actually to a column  
with given entries, but which is simply not as long as another column  
in the file.

Hope, you can help. I would really appreciate it.

Best regards.

Marius

Excel example (I hope it's displayed correctly, the entry in the last  
row should be aligned with the last column):

ADS GY Equity   ALV GY Equity   
DatePx Last DatePx Last
07/02/0441,395  07/01/31130,234
07/02/0542,134  07/02/01133,353
07/02/0641,875  07/02/04133,824
07/02/05134,734

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[R] how to avoid to overwrite object

2007-03-07 Thread Aimin Yan
Dear R list,
I have a question in R, it could be very simple, but I don't know how to do it?

for example:
I assign 6 to x in beginning of of my R script code
  x-6
..
After many line code, I forget using x variable before, I use x 
again, and do assignment like this
  x-45
  x
[1] 45

then value 6 of previous x is replaced by 45.

I am wondering if there are some way R can give me warning like
x is used before, overwrite it or not? when I use x again?

Thanks,

Aimin

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Re: [R] Failure to run mcsamp() in package arm

2007-03-07 Thread Michael Kubovy
More problems. If I run
sim(fm1 - lmer(Reaction ~ Days + (Days|Subject), sleepstudy))
from the lmer() help page.

I get the error
Error in mvrnorm(n.sims, bhat[j, ], V.beta) :
'Sigma' is not positive definite

On Mar 7, 2007, at 1:30 PM, Michael Kubovy wrote:

 Dear r-helpers,

 I can run the examples on the mcsamp help page. For example:
 
 M1 - lmer (y1 ~ x + (1|group))
 (M1.sim - mcsamp (M1))
 fit using lmer,
 3 chains, each with 1000 iterations (first 500 discarded)
 n.sims = 1500 iterations saved
mean  sd 2.5%  25%  50%  75% 97.5% Rhat  
 n.eff
 beta.(Intercept)   0.1 0.7 -1.2 -0.3  0.1  0.5   1.4  1.0   
 1500
 beta.x 2.5 0.4  1.7  2.2  2.5  2.7   3.2  1.0   
 1500
 sigma.y3.8 0.3  3.3  3.6  3.7  3.9   4.3   
 1.061
 sigma.grop.(In)1.5 0.8  0.0  1.0  1.4  1.9   3.3   
 1.412
 eta.group.(Intercept)[1]   0.0 1.0 -2.1 -0.5  0.0  0.6   2.0  1.0   
 1500
 eta.group.(Intercept)[2]   1.0 1.1 -0.9  0.2  0.9  1.7   3.4   
 1.059
 eta.group.(Intercept)[3]  -1.3 1.2 -4.0 -2.0 -1.3 -0.4   0.5   
 1.066
 eta.group.(Intercept)[4]   1.3 1.1 -0.6  0.4  1.1  2.0   3.7   
 1.143
 eta.group.(Intercept)[5]  -0.7 1.0 -3.0 -1.4 -0.6  0.0   1.2  1.0
 120
 eta.group.(Intercept)[6]   1.5 1.2 -0.3  0.6  1.4  2.2   4.0   
 1.049
 eta.group.(Intercept)[7]   0.3 1.0 -1.7 -0.3  0.1  0.8   2.5  1.0
 440
 eta.group.(Intercept)[8]  -1.6 1.2 -4.0 -2.4 -1.5 -0.6   0.3   
 1.141
 eta.group.(Intercept)[9]   0.4 1.0 -1.6 -0.2  0.2  0.9   2.7  1.0
 180
 eta.group.(Intercept)[10] -1.0 1.1 -3.3 -1.6 -0.9 -0.2   0.8   
 1.086

 For each parameter, n.eff is a crude measure of effective sample size,
 and Rhat is the potential scale reduction factor (at convergence,
 Rhat=1).
 
 But when I try to do this with my own data I get an error:
 
 display(e7.lmer2)
 lmer(formula = baLO ~ I(baRatio - 0.985) + delta + (1 + I(baRatio -
 0.985) + delta | subject), data = e7)
 coef.est coef.se
 (Intercept)-0.19 0.06
 I(baRatio - 0.985) -4.95 0.74
 delta   0.41 0.06
 Error terms:
 Groups   Name   Std.Dev. Corr
 subject  (Intercept)0.13
I(baRatio - 0.985) 2.57  0.45
delta  0.22 -0.12 -0.94
 Residual0.39
 number of obs: 494, groups: subject, 13
 deviance = 551.4

 e7.sim - mcsamp(e7.lmer2)
 Error in as.bugs.array(sims, program = lmer, n.iter = n.iter,
 n.burnin = n.burnin,  :
   error in parameter sigma. in parameters.to.save
 
 I would appreciate a pointer to what the problem might be.
 _
 Professor Michael Kubovy
 University of Virginia
 Department of Psychology
 USPS: P.O.Box 400400Charlottesville, VA 22904-4400
 Parcels:Room 102Gilmer Hall
  McCormick RoadCharlottesville, VA 22903
 Office:B011+1-434-982-4729
 Lab:B019+1-434-982-4751
 Fax:+1-434-982-4766
 WWW:http://www.people.virginia.edu/~mk9y/

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting- 
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] No years() function?

2007-03-07 Thread Ben Bolker
Sérgio Nunes snunes at gmail.com writes:

 
 Hi,
 
 I'm trying to aggregate date values using the aggregate function. For example:
 
 aggregate(data,by=list(weekdays(LM),months(LM)),FUN=length)
 
 I would also like to aggregate by year but there seems to be no
 years() function.
 Should there be one? Is there any alternative choice?
 
 Also, a hours() function would be great. Any tip on this?
 
 Thanks in advance!
 Sérgio Nunes
 

  Well, working by analogy with the existing
functions, this might work (not messing with
setting up an S3 default though):

 apropos(weekdays)
[1] weekdaysweekdays.Date   weekdays.POSIXt

 weekdays.Date
function (x, abbreviate = FALSE)
format(x, ifelse(abbreviate, %a, %A))
environment: namespace:base


d1 = Sys.time()

years - function(x,abbreviate=FALSE) {
  as.numeric(format(x, ifelse(abbreviate, %y, %Y)))
}

hours - function(x) {
   as.numeric(format(x,%H))
}

years(d1); hours(d1)

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Re: [R] C to R

2007-03-07 Thread Andrew Robinson
Hi Heloise,

there's a manual online to help you with this - see here

http://cran.r-project.org/doc/manuals/R-exts.html

Cheers

Andrew

On Wed, Mar 07, 2007 at 05:29:10PM -0300, Heloise Mattos wrote:
 I`m doing some functions on C that gives me the x and y coordinates.
 I`d like to now how I can get these coordinates (both are a vector of
 number) on R to that I can make a graphic.
 I`ve already made a package with my functions, so I just wanna how
 about how to get the coordinates.
 
 Thanks, Heloise.
 
 __
 R-help@stat.math.ethz.ch mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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-- 
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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Re: [R] No years() function?

2007-03-07 Thread José Rafael Ferrer Paris
From the help of weekdays:

Note:

 Other components such as the day of the month or the year are very
 easy to compute: just use 'as.POSIXlt' and extract the relevant
 component.

Yet another option:

help(package=chron)

JR

El mié, 07-03-2007 a las 15:35 +, Sérgio Nunes escribió:
 Hi,
 
 I'm trying to aggregate date values using the aggregate function. For example:
 
 aggregate(data,by=list(weekdays(LM),months(LM)),FUN=length)
 
 I would also like to aggregate by year but there seems to be no
 years() function.
 Should there be one? Is there any alternative choice?
 
 Also, a hours() function would be great. Any tip on this?
 
 Thanks in advance!
 Sérgio Nunes
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
-- 
Dipl.-Biol. JR Ferrer Paris
~~~
Laboratorio de Biología de Organismos --- Centro de Ecología
Instituto Venezolano de Investigaciones Científicas (IVIC) 
Apdo. 21827, Caracas 1020-A 
República Bolivariana de Venezuela

Tel: (+58-212) 504-1452
Fax: (+58-212) 504-1088

email: [EMAIL PROTECTED]
clave-gpg: 2C260A95

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[R] hwo can i get a vector that...

2007-03-07 Thread bunny , lautloscrew.com
dear all,

how can i get a vector that shows the number of the column of matrix  
that contains the maximum of the row ??
can´t believe in need a loop for this...

i have a  100 x 3 matrix and want to get a 100 x 1 vector with values  
1,2,3 .

there must be a simple solution. i just cannot find it. i think am  
searching on the wrong end.

thx for help in advance.

m.

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Re: [R] hwo can i get a vector that...

2007-03-07 Thread bunny , lautloscrew.com



 dear all,

 how can i get a vector that shows the number of the column of  
 matrix that contains the maximum of the row ??
 can´t believe in need a loop for this...

 i have a  100 x 3 matrix and want to get a 100 x 1 vector with  
 values 1,2,3 .

 there must be a simple solution. i just cannot find it. i think am  
 searching on the wrong end.

 thx for help in advance.

 m.


EDIT: ok,  i know the following by now :)

apply(for18[,-1], 1, max, na.rm=T)

but this doesn´t get me the number of the column - which is what i  
need... 
[[alternative HTML version deleted]]

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


Re: [R] hwo can i get a vector that...

2007-03-07 Thread Petr Klasterecky
?apply
?which.max

  m - matrix(rnorm(12),nrow=4)
  m
  apply(m,1,which.max)

Petr

bunny , lautloscrew.com napsal(a):
 dear all,
 
 how can i get a vector that shows the number of the column of matrix  
 that contains the maximum of the row ??
 can´t believe in need a loop for this...
 
 i have a  100 x 3 matrix and want to get a 100 x 1 vector with values  
 1,2,3 .
 
 there must be a simple solution. i just cannot find it. i think am  
 searching on the wrong end.
 
 thx for help in advance.
 
 m.
 
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 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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Re: [R] hwo can i get a vector that...

2007-03-07 Thread Bert Gunter
apply(yourMatrix,1,which.max) 


Bert Gunter
Nonclinical Statistics
7-7374

-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of bunny ,
lautloscrew.com
Sent: Wednesday, March 07, 2007 2:12 PM
To: r-help@stat.math.ethz.ch
Subject: [R] hwo can i get a vector that...

dear all,

how can i get a vector that shows the number of the column of matrix  
that contains the maximum of the row ??
can´t believe in need a loop for this...

i have a  100 x 3 matrix and want to get a 100 x 1 vector with values  
1,2,3 .

there must be a simple solution. i just cannot find it. i think am  
searching on the wrong end.

thx for help in advance.

m.

__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


Re: [R] hwo can i get a vector that...

2007-03-07 Thread Scionforbai
 can´t believe in need a loop for this...
No, you don't ;)

 want to get a 100 x 1 vector
Has each row unique values? If yes:

mat - matrix(rnorm(300),nr=100)
vet - apply(mat,1,function(x) {return(which(x==max(x)))})



scionforbai

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Re: [R] Failure to run mcsamp() in package arm

2007-03-07 Thread Michael Kubovy
Andrew Robinson has gently chided me for not including more  
information. So here goes:
R version 2.4.1 (2006-12-18)
powerpc-apple-darwin8.8.0

locale:
C

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

other attached packages:
 foreign car arm   R2WinBUGSlme4   
Matrix lattice
0.8-18 1.2-11.0-13 2.0-4 0.9975-13  
0.9975-11   0.14-16
MASS JGR  iplots  JavaGD   rJava
7.2-321.4-15 1.0-5 0.3-50.4-14

On Mar 7, 2007, at 4:30 PM, Michael Kubovy wrote:

 More problems. If I run
 sim(fm1 - lmer(Reaction ~ Days + (Days|Subject), sleepstudy))
 from the lmer() help page.

 I get the error
 Error in mvrnorm(n.sims, bhat[j, ], V.beta) :
   'Sigma' is not positive definite

 On Mar 7, 2007, at 1:30 PM, Michael Kubovy wrote:

 Dear r-helpers,

 I can run the examples on the mcsamp help page. For example:
 
 M1 - lmer (y1 ~ x + (1|group))
 (M1.sim - mcsamp (M1))
 fit using lmer,
 3 chains, each with 1000 iterations (first 500 discarded)
 n.sims = 1500 iterations saved
mean  sd 2.5%  25%  50%  75% 97.5% Rhat
 n.eff
 beta.(Intercept)   0.1 0.7 -1.2 -0.3  0.1  0.5   1.4  1.0
 1500
 beta.x 2.5 0.4  1.7  2.2  2.5  2.7   3.2  1.0
 1500
 sigma.y3.8 0.3  3.3  3.6  3.7  3.9   4.3
 1.061
 sigma.grop.(In)1.5 0.8  0.0  1.0  1.4  1.9   3.3
 1.412
 eta.group.(Intercept)[1]   0.0 1.0 -2.1 -0.5  0.0  0.6   2.0  1.0
 1500
 eta.group.(Intercept)[2]   1.0 1.1 -0.9  0.2  0.9  1.7   3.4
 1.059
 eta.group.(Intercept)[3]  -1.3 1.2 -4.0 -2.0 -1.3 -0.4   0.5
 1.066
 eta.group.(Intercept)[4]   1.3 1.1 -0.6  0.4  1.1  2.0   3.7
 1.143
 eta.group.(Intercept)[5]  -0.7 1.0 -3.0 -1.4 -0.6  0.0   1.2  1.0
 120
 eta.group.(Intercept)[6]   1.5 1.2 -0.3  0.6  1.4  2.2   4.0
 1.049
 eta.group.(Intercept)[7]   0.3 1.0 -1.7 -0.3  0.1  0.8   2.5  1.0
 440
 eta.group.(Intercept)[8]  -1.6 1.2 -4.0 -2.4 -1.5 -0.6   0.3
 1.141
 eta.group.(Intercept)[9]   0.4 1.0 -1.6 -0.2  0.2  0.9   2.7  1.0
 180
 eta.group.(Intercept)[10] -1.0 1.1 -3.3 -1.6 -0.9 -0.2   0.8
 1.086

 For each parameter, n.eff is a crude measure of effective sample  
 size,
 and Rhat is the potential scale reduction factor (at convergence,
 Rhat=1).
 
 But when I try to do this with my own data I get an error:
 
 display(e7.lmer2)
 lmer(formula = baLO ~ I(baRatio - 0.985) + delta + (1 + I(baRatio -
 0.985) + delta | subject), data = e7)
 coef.est coef.se
 (Intercept)-0.19 0.06
 I(baRatio - 0.985) -4.95 0.74
 delta   0.41 0.06
 Error terms:
 Groups   Name   Std.Dev. Corr
 subject  (Intercept)0.13
I(baRatio - 0.985) 2.57  0.45
delta  0.22 -0.12 -0.94
 Residual0.39
 number of obs: 494, groups: subject, 13
 deviance = 551.4

 e7.sim - mcsamp(e7.lmer2)
 Error in as.bugs.array(sims, program = lmer, n.iter = n.iter,
 n.burnin = n.burnin,  :
  error in parameter sigma. in parameters.to.save
 
 I would appreciate a pointer to what the problem might be.
 _
 Professor Michael Kubovy
 University of Virginia
 Department of Psychology
 USPS: P.O.Box 400400Charlottesville, VA 22904-4400
 Parcels:Room 102Gilmer Hall
  McCormick RoadCharlottesville, VA 22903
 Office:B011+1-434-982-4729
 Lab:B019+1-434-982-4751
 Fax:+1-434-982-4766
 WWW:http://www.people.virginia.edu/~mk9y/

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting- 
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] hwo can i get a vector that...

2007-03-07 Thread Peter Alspach

Check out which.max 

Peter Alspach


 -Original Message-
 From: [EMAIL PROTECTED] 
 [mailto:[EMAIL PROTECTED] On Behalf Of bunny 
 , lautloscrew.com
 Sent: Thursday, 8 March 2007 11:20 a.m.
 To: R-help@stat.math.ethz.ch
 Subject: Re: [R] hwo can i get a vector that...
 
 
 
 
  dear all,
 
  how can i get a vector that shows the number of the column 
 of matrix 
  that contains the maximum of the row ??
  can´t believe in need a loop for this...
 
  i have a  100 x 3 matrix and want to get a 100 x 1 vector 
 with values 
  1,2,3 .
 
  there must be a simple solution. i just cannot find it. i think am 
  searching on the wrong end.
 
  thx for help in advance.
 
  m.
 
 
 EDIT: ok,  i know the following by now :)
 
 apply(for18[,-1], 1, max, na.rm=T)
 
 but this doesn´t get me the number of the column - which is 
 what i need... 
   [[alternative HTML version deleted]]
 
 
 

__

The contents of this e-mail are privileged and/or confidenti...{{dropped}}

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Re: [R] good procedure to estimate ARMA(p, q)?

2007-03-07 Thread Gad Abraham
Michael wrote:
 Hi all,
 
 I have some residuals from regression, and i suspect they have correlations
 in them...
 
 I am willing to cast the correlation into a ARMA(p, q) framework,
 
 what's the best way to identify the most suitable p, and q, and fit ARMA(p,
 q) model and then correct for the correlations in regression?
 
 I know there are functions in R, I have used them before, but I just want to
 see if I can do the whole procedure myself, just to improve my understanding
 ...
 
 Please give me some pointers! Thanks a lot

I'm assuming the data is a time series, otherwise ARIMA models might not 
be applicable here.

I think identifying the order of ARIMA models is something of an art, 
because most real world models aren't as clean and simple as textbook 
examples. When you have several similar models, each with its own 
strengths and weaknesses, which one is best?

In short, you want to make sure your series is stationary, look at its 
ACF and PACF, then try different values of p and q based on that, and 
finally look at the residuals (autocorrelation, distribution, etc).

This is basically the Box-Jenkins methodology. The most accessible 
descriptions I've seen are in Forecasting: Methods and Applications by 
Makridakis, Wheelwright and Hyndman (chapter 7), and Forecasting with 
Univariate Box-Jenkins Models by Pankratz.

Cheers,
Gad

-- 
Gad Abraham
Department of Mathematics and Statistics
The University of Melbourne
Parkville 3010, Victoria, Australia
email: [EMAIL PROTECTED]
web: http://www.ms.unimelb.edu.au/~gabraham

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Re: [R] how to avoid to overwrite object

2007-03-07 Thread Gabor Grothendieck
Decompose your code into small understandable functions.

On 3/7/07, Aimin Yan [EMAIL PROTECTED] wrote:
 Dear R list,
 I have a question in R, it could be very simple, but I don't know how to do 
 it?

 for example:
 I assign 6 to x in beginning of of my R script code
   x-6
 ..
 After many line code, I forget using x variable before, I use x
 again, and do assignment like this
   x-45
   x
 [1] 45

 then value 6 of previous x is replaced by 45.

 I am wondering if there are some way R can give me warning like
 x is used before, overwrite it or not? when I use x again?

 Thanks,

 Aimin

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 R-help@stat.math.ethz.ch mailing list
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Re: [R] hwo can i get a vector that...

2007-03-07 Thread Hong Ooi

___


max.col sounds like what you're after.


-- 
Hong Ooi
Senior Research Analyst, IAG Limited
388 George St, Sydney NSW 2000
+61 (2) 9292 1566
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of bunny , 
lautloscrew.com
Sent: Thursday, 8 March 2007 9:12 AM
To: r-help@stat.math.ethz.ch
Subject: [R] hwo can i get a vector that...

dear all,

how can i get a vector that shows the number of the column of matrix  
that contains the maximum of the row ??
can´t believe in need a loop for this...

i have a  100 x 3 matrix and want to get a 100 x 1 vector with values  
1,2,3 .

there must be a simple solution. i just cannot find it. i think am  
searching on the wrong end.

thx for help in advance.

m.

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[R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work

2007-03-07 Thread ALAN SMITH
Hello R users,

Problem...I do not understand how to use aggregate,by, or the
appropriate apply to perform a function on data with more than one
factor on unbalanced data...
I have a data frame in the long format that does not contain balanced
data. The ID is a unique identifier corresponding to the experimental
unit that will later be examined by ANOVA, T-tests etc. Y is the data
generated from the experiment.  The factors represent the differences
between each sample or run measured.

str(mydata)  ### sample of table at bottom of email ###
'data.frame':   129982 obs. of  6 variables:
 $ ID: num  7 7 7 7 7 7 8 8 8 8 ...
 $ time : Factor w/ 2 levels 120hr,24hr: 1 1 1 1 2 2 2 1 1 1 ...
 $ treatment: Factor w/ 2 levels control,trt: 1 1 1 2 2 1 1 2 1 1 ...
 $ expREP   : Factor w/ 3 levels expREP1,expREP2,..: 1 1 1 3 1 1 1 1 2 2 ...
 $ techREP  : Factor w/ 3 levels techREP1,techREP2,..: 3 2 1 1 1 3
1 3 3 2 ...
 $ Y : num  14.4 14.1 14.2 13.8 14.1 ...

Could someone please help with doing something like the following
1. I would like to find the median for each unique combination of
factors using the data in the   long format (like finding the median
of a single column of data).
2. Create a new column where the median is repeated for the number of
rows of the unique factor combination
3. I would like to learn the most efficient way to do this because I
want to avoid recreating the table from scratch with many commands
like the series below. I will have to perform this operation on many
different data sets some, with many more factors then this example.

### help me learn to use an apply or other command that will do the
following #
m0-mydata$cpdID[mydata$time==24hr  mydata$treatment==control 
mydata$expREP==expREP1  mydata$techREP==techREP1]
m1-mydata$Y[mydata$time==24hr  mydata$treatment==control 
mydata$expREP==expREP1  mydata$techREP==techREP1]
m2-median(m1)
m3-cbind(ID=m0,time=rep(24hr,length(m1)),
treatment=rep(control,length(m1)), expREP=rep(expREP1,length(m1)),
techREP=rep(techREP1,length(m1)),Y=m1,Y50=rep(m2,length(m1)))
# I would like to avoid writing the above hundreds of times ##

I am able to reshape into wide format and then find the column
medians. However restacking the data and regenerating the factors
becomes very very messy on data sets with 150 columns.  I am able to
preform this analysis is SAS easily using BY, but I would like to know
how to do it in R.


I have tried these commands in a number of different variations with
no luck and similar error messages
 test1-aggregate(mydata[,-1],
list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
,median, na.rm=T)
Error in median.default(X[[1]], ...) : need numeric data ### Y in numeric

test1-by(mydata[,-1],
list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
,median, na.rm=T)
Error in median.default(data[x, ], ...) : need numeric data

Thanks
Alan
winXP R 2.4.1


#Example data frame##
mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8,
8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23,
23, 23, 23, 23, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33,
33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33,
33, 33, 33, 33, 33, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42,
42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42,
42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42,
42, 42, 42, 42, 42, 42, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47,
47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47), time = structure(as.integer(c(1,
1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2,
1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2,
1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2,
2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 2,
2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2,
2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2,
2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1,
1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2,
2, 2, 1, 2, 2, 1, 2, 2, 1, 2)), .Label = c(120hr, 24hr), class = factor),
treatment = structure(as.integer(c(1, 1, 1, 2, 2, 1, 1, 2,
1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1,
2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2,
1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1,
1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2,
2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2,
1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 2,
2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 2, 1,
1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 

Re: [R] Read data with different column lengths

2007-03-07 Thread jim holtman
I would have thought that a csv file written out by Excel would have looked
like this for your data:

ADS GY Equity, ,ALV GY Equity,
Date,   Px Last,Date  , Px Last
07/02/04,   41,395  ,07/01/31  , 130,234
07/02/05,   42,134  ,07/02/01  , 133,353
07/02/06,   41,875  ,07/02/04  , 133,824
 , , 07/02/05, 134,734

Here we have commas separating the fields and this will read in the data:

 x - read.csv(/tempxx.csv, skip=2, header=FALSE, as.is=TRUE)
 str(x)
'data.frame':   4 obs. of  4 variables:
 $ V1: chr  07/02/04 07/02/05 07/02/06  
 $ V2: chr 41,395  42,134  41,875
   
 $ V3: chr  07/01/31   07/02/01   07/02/0407/02/05
 $ V4: chr   130,234  133,353  133,824  134,734
 x
 V1  V2 V3   V4
1  07/02/0441,395   07/01/31130,234
2  07/02/0542,134   07/02/01133,353
3  07/02/0641,875   07/02/04133,824
4 07/02/05  134,734


You will have to delete the ',' from the numeric fields, The 'dec=,' on
your command would have read '133,353' in as '133.353' which I don't think
is what you want.  You can delete the commas and convert to numeric with the
following:

 x$V4 - as.numeric(gsub(',', '', x$V4))
 x
 V1  V2 V3 V4
1  07/02/0441,395   07/01/31   130234
2  07/02/0542,134   07/02/01   133353
3  07/02/0641,875   07/02/04   133824
4 07/02/05 134734


So take a look at the options that you have with writing out the Excel data;
you should be able to get either tab or comma delimited.  If not, you can
try reading in each line as a character string and then using 'substr' to
split it apart assuming that you have constant width columns.


On 3/7/07, Hofert Marius [EMAIL PROTECTED] wrote:

 Dear r-help users,

 I have the following simple problem: Reading data from a file. The
 file is a .txt file exported (save as...) from Excel (see below for
 an example). The Excel file consists of two header rows (first row
 consists of ticker symbols of stocks, the second row consists of
 column explanations (Date,Px Last), followed by several rows of
 data. Now forget about the first two rows, I can deal with that (read
 separately, then extract the actual ticker symbols ADS,
 ALV, ...). For reading the rest, I tried several things, for example:
 data=read.table(infile,quote=,fill=T,dec=,,skip=2,colClasses=rep(c
 (character,numeric),ntickers))
 or
 data=matrix(scan(file=infile,what=rep(c
 (character,numeric),ntickers),dec=,,skip=2),ncol=2*ntickers,byrow=
 T)
 where infile specifies the path to the input file and ntickers is
 the number of ticker-columns in the data set, so in the example
 below, ntickers=2.

 Both ways of reading the data work perfectly fine if all columns have
 the same length (i.e. the same number of filled rows), so if the data
 is given in a (filled) rectangular form. Now, as you can imagine,
 there are days when one stock is traded but not the other... so,
 there might be columns that do not have the same number of filled
 rows (see below, for the stock with ticker symbol ADS, only 3
 trading days are shown, so this column is shorter than the data
 column for the stock ALV). Now, if I export such a structure to
 a .txt file, then all (by default) blank fields will be replaced by
 \t, i.e. tabs. Both reading procedures as give above have problems
 as they either display that the number of rows/columns do not fit
 together or as they read the table, but some cells are shifted to the
 left (for the example below, the entry 07/02/05134,7 appears in
 the empty field of the stock ADS which is of course not what we want).
 So the simple question is: How do I read such a structure?
 Can there be a simple solution? The problem is simply that empty
 cells are replace by \t which are then ignored for reading. So how
 do we distinguish between the empty cells that are given between the
 columns and the empty cells that actually fill a column to have the
 same length as other columns. Of course I could manually put in a
 certain character (e.g. a *) to fill in the gaps, but the data set
 is simply too large. If it helps, these blank fields only appear in
 the end of each column, not in the middle.

 As I work on a Mac (OS X 10.4), it was not possible (at least to me)
 to read the data directly from the Excel file vial the library RODBC
 or read.xls.

 Note, that the same problem arises, when I export the Excel file as
 a .csv, then all blank fields are separated by ; instead of \t
 and the reading procedure can also not decide if the field
 corresponds to an empty separating column or actually to a column
 with given entries, but which 

Re: [R] how to apply functions to unbalanced data in long format by factors......cant get by or aggregate to work

2007-03-07 Thread jim holtman
Here is one way of doing it:

 # create the rows for each unique combination
 x.split - split(seq(nrow(mydata)), list(mydata$time, mydata$treatment,
+ mydata$expREP, mydata$techREP), drop=TRUE)
 # now go through the list of indices and add the median
 mydata$Y50 - 0   # add the dummy median column
 for (i in x.split){
+ mydata$Y50[i] - median(mydata$Y[i])  # median for each group
+ }
 head(mydata,20)
   cpdID  time treatment  expREP  techREPY  Y50
1  7 120hr   control expREP1 techREP3 14.42331 15.74599
2  7 120hr   control expREP1 techREP2 14.05282 15.10810
3  7 120hr   control expREP1 techREP1 14.22814 14.63248
4  7 120hr   trt expREP3 techREP1 13.84921 15.08641
5  7  24hr   trt expREP1 techREP1 14.07648 15.17235
6  7  24hr   control expREP1 techREP3 14.21394 14.63314
7  8  24hr   control expREP1 techREP1 14.33992 14.81328
8  8 120hr   trt expREP1 techREP3 14.36256 15.34493
9  8 120hr   control expREP2 techREP3 14.01419 15.14270
10 8 120hr   control expREP2 techREP2 14.09803 15.10079
11 8 120hr   trt expREP1 techREP2 14.31522 15.39152
12 8 120hr   trt expREP1 techREP1 14.45288 14.65430
13 8  24hr   control expREP2 techREP1 13.95919 14.71188
14 8  24hr   trt expREP2 techREP3 14.40640 14.36332
15 8  24hr   trt expREP1 techREP2 14.04038 14.42856
16 8  24hr   control expREP3 techREP3 14.22859 15.08463
17 8  24hr   trt expREP1 techREP3 14.12598 14.53840
18 8  24hr   trt expREP3 techREP1 13.59257 14.69984
19 8  24hr   trt expREP3 techREP2 13.58308 14.85730
2010 120hr   control expREP3 techREP1 13.02808 14.07448




On 3/7/07, ALAN SMITH [EMAIL PROTECTED] wrote:

 Hello R users,

 Problem...I do not understand how to use aggregate,by, or the
 appropriate apply to perform a function on data with more than one
 factor on unbalanced data...
 I have a data frame in the long format that does not contain balanced
 data. The ID is a unique identifier corresponding to the experimental
 unit that will later be examined by ANOVA, T-tests etc. Y is the data
 generated from the experiment.  The factors represent the differences
 between each sample or run measured.

 str(mydata)  ### sample of table at bottom of email ###
 'data.frame':   129982 obs. of  6 variables:
 $ ID: num  7 7 7 7 7 7 8 8 8 8 ...
 $ time : Factor w/ 2 levels 120hr,24hr: 1 1 1 1 2 2 2 1 1 1 ...
 $ treatment: Factor w/ 2 levels control,trt: 1 1 1 2 2 1 1 2 1 1 ...
 $ expREP   : Factor w/ 3 levels expREP1,expREP2,..: 1 1 1 3 1 1 1 1 2
 2 ...
 $ techREP  : Factor w/ 3 levels techREP1,techREP2,..: 3 2 1 1 1 3
 1 3 3 2 ...
 $ Y : num  14.4 14.1 14.2 13.8 14.1 ...

 Could someone please help with doing something like the following
 1. I would like to find the median for each unique combination of
 factors using the data in the   long format (like finding the median
 of a single column of data).
 2. Create a new column where the median is repeated for the number of
 rows of the unique factor combination
 3. I would like to learn the most efficient way to do this because I
 want to avoid recreating the table from scratch with many commands
 like the series below. I will have to perform this operation on many
 different data sets some, with many more factors then this example.

 ### help me learn to use an apply or other command that will do the
 following #
 m0-mydata$cpdID[mydata$time==24hr  mydata$treatment==control 
 mydata$expREP==expREP1  mydata$techREP==techREP1]
 m1-mydata$Y[mydata$time==24hr  mydata$treatment==control 
 mydata$expREP==expREP1  mydata$techREP==techREP1]
 m2-median(m1)
 m3-cbind(ID=m0,time=rep(24hr,length(m1)),
 treatment=rep(control,length(m1)), expREP=rep(expREP1,length(m1)),
 techREP=rep(techREP1,length(m1)),Y=m1,Y50=rep(m2,length(m1)))
 # I would like to avoid writing the above hundreds of times ##

 I am able to reshape into wide format and then find the column
 medians. However restacking the data and regenerating the factors
 becomes very very messy on data sets with 150 columns.  I am able to
 preform this analysis is SAS easily using BY, but I would like to know
 how to do it in R.


 I have tried these commands in a number of different variations with
 no luck and similar error messages
 test1-aggregate(mydata[,-1],
 list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
 ,median, na.rm=T)
 Error in median.default(X[[1]], ...) : need numeric data ### Y in
 numeric

 test1-by(mydata[,-1],
 list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
 ,median, na.rm=T)
 Error in median.default(data[x, ], ...) : need numeric data

 Thanks
 Alan
 winXP R 2.4.1


 #Example data frame##
 mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8,
 8, 8, 8, 8,
 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
 10, 10, 10, 10, 10, 10, 19, 19, 19, 

[R] Some problems with X11

2007-03-07 Thread Rafael Rosolem

Hi,

I am really new with R, so I don't know anything about it. I have 
written a script (attached) which tries to do really basic stuff (such 
as computing basic statistics and basic plots). When I try to plot a 
histogram and pairs, for example, I get the following message:


 source(project.R)
Loading required package: sp

-
Analysis of geostatistical data
For an Introduction to geoR go to http://www.est.ufpr.br/geoR
geoR version 1.6-13 (built on 2006/12/26) is now loaded
-

Error in title(main = main, sub = sub, xlab = xlab, ylab = ylab, ...) :
X11 font at size 8 could not be loaded


I have seen some threads about this problem, though none of them really 
states what should be done in order to solve that problem (At least it 
was not clear for me!).


This is the R version I have:
 version
   _
platform   i486-pc-linux-gnu
arch   i486
os linux-gnu
system i486, linux-gnu
status
major  2
minor  4.1
year   2006
month  12
day18
svn rev40228
language   R
version.string R version 2.4.1 (2006-12-18)

I am also running R on a Ubuntu Linux Edgy distro. The interesting thing 
is that I have this problem on my desktop only. I also have Ubuntu Edgy 
installed on my laptop and it is working just fine!


Thank you

R


#  #
# MATH574 - Geostatistics  #
# Prof. Dr. Donald Myers   #
#  #
# Term-project: Anaylizing climatological data from Sao Paulo State,   #
# Brazil:  #
#- Annual precipitation (mm); and  #
#- Annual mean temperature (oC).   #
#  #
# * elevation data is also provided#
#  #
# Author: Rafael Rosolem ([EMAIL PROTECTED])  #
# Department: Hydrology and Water Resources#
# The University of Arizona#
#  #
# Date: 02.14.2007 #
# Last update: 02.19.2007  #
#  #


# Opening libraries   

library(geoR) # geoR package
library(gstat) # gstat package
library(scatterplot3d) # scatterplot3d package

# Setting the path and file name   

path - /home/rafael/math574/term-project/
input - dataset_rafael.dat

# Opening data file   

data - read.table(paste(path,input,sep=),header = TRUE)

# Converting data object to geodata class

data_geoR - as.geodata(data,coords.col = c(1,2),data.col = c(3,4,5))

# Basics statistics

data_summary - summary(data_geoR) # Data summary
data_stdev - sd(data_geoR$data) # Standard deviation
data_cor - cor(data_geoR$data) # Data correlation

# Histograms

X11(display = ,width = 9,height = 4,pointsize = 12,bg = transparent)
#png(file=paste(path,data_hist.png,sep=),height=576,width=576,
#res=600,bg=transparent)
   par(mfrow = c(1,3)) # 1x3 graph
   temp_hist -  hist(data_geoR$data[,temp],breaks = c(17:26),
  col = red,border = black,
  main = NULL,
 xlim = c(min(data_geoR$data[,temp])-1,
 max(data_geoR$data[,temp])+1),ylim = NULL,
 xlab = Temperature (oC),ylab = Freq) # Temperature
   prec_hist -  hist(data_geoR$data[,prec],breaks = c(250*0:14),
  col = darkblue,border = black,
 main = NULL,
 xlim = c(min(data_geoR$data[,prec])-500,
 max(data_geoR$data[,prec])+500),ylim = NULL,
 xlab = Precipitation (mm),ylab = Freq) # Precipitation
   elev_hist - 

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__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


[R] sink with R-code

2007-03-07 Thread Cooley, Scott K
I have the same question that Eusebio had:

Is there a function similar to sink that redirect also R code to a
file that
is:

sink(R001)
x - c(2,-6,-4,8,5,4,1,3,4,-9,0,1)
A - matrix(x, ncol=3)
A
A.prima - t(A)
A.prima
dim(A)
dim(A.prima)
sink()

create a file R001 with contents:

--
 [,1] [,2] [,3]
[1,]254
[2,]   -64   -9
[3,]   -410
[4,]831
 [,1] [,2] [,3] [,4]
[1,]2   -6   -48
[2,]5413
[3,]4   -901
[1] 4 3
[1] 3 4



and what I want is a file with:


 x - c(2,-6,-4,8,5,4,1,3,4,-9,0,1)
 A - matrix(x, ncol=3)
 A
 [,1] [,2] [,3]
[1,]254
[2,]   -64   -9
[3,]   -410
[4,]831
 A.prima - t(A)
 A.prima
 [,1] [,2] [,3] [,4]
[1,]2   -6   -48
[2,]5413
[3,]4   -901
--
Any hint will be appreciated

Eusebio


Scott K. Cooley
Statistical Sciences, K6-08
Battelle--Pacific Northwest Division
Pacific Northwest National Laboratory
P.O. Box 999
Richland, WA  99352
Phone: 509-375-3604
FAX: 509-375-2604
Email: [EMAIL PROTECTED] 



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Re: [R] Sweave issue: quotes in verbatim-like output print incorrectly

2007-03-07 Thread Peter Dunn
Thanks to Gavin Simpson and Patrick Drechsler:
The solution was simple.  (No idea how I would have
discovered these answers without this R mailing list!)

Many thanks.

P.

Gavin Simpson [EMAIL PROTECTED] writes:

 On Wed, 2007-03-07 at 15:33 +1000, Peter Dunn wrote:
 But I recently received a new computer, and ever since I
 have had a problem I've never seen before.
 
 For example, I place the following in my Snw file:

 Try this in the preamble of your Snw file:

 \usepackage[utf8x]{inputenc}

\usepackage[utf8]{inputenc}

should suffic for this. Also, utf8 is newer than utf8x. So if you only
need the ISO 8859 subset of Unicode, utf8 should serve your needs.

HTH

Patrick
-- 
I never used a logarithm in my life, and could not undertake to
extract the square root of four without misgivings.
(Georg Bernhard Shaw)

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Re: [R] how to apply functions to unbalanced data in long format byfactors......cant get by or aggregate to work

2007-03-07 Thread Mendiburu, Felipe \(CIP\)
Dear Alan,
 
I think that podria to be of utility the function tapply.stat () of the package 
agricolae.
see ?tapply.stat
Regards,
Felipe.
 
for example:
library(agricolae)
attach(mydata)
set1-tapply.stat(mydata[,2:5],Y,median)
set2-tapply.stat(time,Y,function(x) median(x))
set3-tapply.stat(mydata[,c(2,3)],Y,function(x) median(x))
set2
   timeY
1 120hr 14.94159
2  24hr 14.81914
set3
   time treatmentY
1 120hr   control 15.31974
2 120hr   trt 14.82851
3  24hr   control 15.03627
4  24hr   trt 14.70249




From: [EMAIL PROTECTED] on behalf of ALAN SMITH
Sent: Wed 3/7/2007 6:25 PM
To: r-help@stat.math.ethz.ch
Subject: [R] how to apply functions to unbalanced data in long format 
byfactors..cant get by or aggregate to work



Hello R users,

Problem...I do not understand how to use aggregate,by, or the
appropriate apply to perform a function on data with more than one
factor on unbalanced data...
I have a data frame in the long format that does not contain balanced
data. The ID is a unique identifier corresponding to the experimental
unit that will later be examined by ANOVA, T-tests etc. Y is the data
generated from the experiment.  The factors represent the differences
between each sample or run measured.

str(mydata)  ### sample of table at bottom of email ###
'data.frame':   129982 obs. of  6 variables:
 $ ID: num  7 7 7 7 7 7 8 8 8 8 ...
 $ time : Factor w/ 2 levels 120hr,24hr: 1 1 1 1 2 2 2 1 1 1 ...
 $ treatment: Factor w/ 2 levels control,trt: 1 1 1 2 2 1 1 2 1 1 ...
 $ expREP   : Factor w/ 3 levels expREP1,expREP2,..: 1 1 1 3 1 1 1 1 2 2 ...
 $ techREP  : Factor w/ 3 levels techREP1,techREP2,..: 3 2 1 1 1 3
1 3 3 2 ...
 $ Y : num  14.4 14.1 14.2 13.8 14.1 ...

Could someone please help with doing something like the following
1. I would like to find the median for each unique combination of
factors using the data in the   long format (like finding the median
of a single column of data).
2. Create a new column where the median is repeated for the number of
rows of the unique factor combination
3. I would like to learn the most efficient way to do this because I
want to avoid recreating the table from scratch with many commands
like the series below. I will have to perform this operation on many
different data sets some, with many more factors then this example.

### help me learn to use an apply or other command that will do the
following #
m0-mydata$cpdID[mydata$time==24hr  mydata$treatment==control 
mydata$expREP==expREP1  mydata$techREP==techREP1]
m1-mydata$Y[mydata$time==24hr  mydata$treatment==control 
mydata$expREP==expREP1  mydata$techREP==techREP1]
m2-median(m1)
m3-cbind(ID=m0,time=rep(24hr,length(m1)),
treatment=rep(control,length(m1)), expREP=rep(expREP1,length(m1)),
techREP=rep(techREP1,length(m1)),Y=m1,Y50=rep(m2,length(m1)))
# I would like to avoid writing the above hundreds of times ##

I am able to reshape into wide format and then find the column
medians. However restacking the data and regenerating the factors
becomes very very messy on data sets with 150 columns.  I am able to
preform this analysis is SAS easily using BY, but I would like to know
how to do it in R.


I have tried these commands in a number of different variations with
no luck and similar error messages
 test1-aggregate(mydata[,-1],
list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
,median, na.rm=T)
Error in median.default(X[[1]], ...) : need numeric data ### Y in numeric

test1-by(mydata[,-1],
list(mydata$time,mydata$treatment,mydata$expREP,mydata$techREP)
,median, na.rm=T)
Error in median.default(data[x, ], ...) : need numeric data

Thanks
Alan
winXP R 2.4.1


#Example data frame##
mydata-as.data.frame(structure(list(cpdID = c(7, 7, 7, 7, 7, 7, 8, 8,
8, 8, 8, 8,
8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23,
23, 23, 23, 23, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33,
33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33,
33, 33, 33, 33, 33, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42,
42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42,
42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42, 42,
42, 42, 42, 42, 42, 42, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47,
47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47), time = structure(as.integer(c(1,
1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2,
1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2, 1, 2, 2,
1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 2, 2,
2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 2,
2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2,
2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 

[R] reading a text file with a stray carriage return

2007-03-07 Thread Walter R. Paczkowski

   Hi,
   I'm  hoping someone has a suggestion for handling a simple problem.  A
   client  gave  me a comma separated value file (call it x.csv) that has
   an  id  and name and address for about 25,000 people (25,000 records).
   I used read.table to read it, but then discovered that there are stray
   carriage returns on several records.  This plays havoc with read.table
   since it starts a new input line when it sees the carriage return.  In
   short, the read is all wrong.
   I thought I could write a simple function to parse a line and write it
   back  out,  character by character.  If a carriage return is found, it
   would  simply  be  ignored on the writing back out part.  But how do I
   identify a carriage return?  What is the code or symbol?  Is there any
   easier  way  to  rid the file of carriage returns in the middle of the
   input lines?
   Any help is appreciated.
   Walt Paczkowski

   _
   Walter R. Paczkowski, Ph.D.
   Data Analytics Corp.
   44 Hamilton Lane
   Plainsboro, NJ  08536
   (V) 609-936-8999
   (F) 609-936-3733
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Re: [R] Passing command line parameters to a script

2007-03-07 Thread Oleg Sklyar
#!/bin/sh
echo 'a=${1}; b=${2}; source(myRcodeUsing_a_and_b.R); ' | R 
--vanilla --quiet

and you can run this from shell like:

./callR valueOfa valueOfb

Best, Oleg

akintayo holder wrote:
 Hi,
  Does any one know if it is possible to create an R script that can use
 command line parameters. I can execute an R script from the command line,
 but I  cannot figure out how to pass parameters to the script. The only
 resources I have found seem somewhat involved or incomplete.
 
 Any help is appreciated.
 Akintayo
 
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-- 
Dr Oleg Sklyar | EBI-EMBL, Cambridge CB10 1SD, UK | +44-1223-494466

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Re: [R] reading a text file with a stray carriage return

2007-03-07 Thread jim holtman
How do you define a carriage return in the middle of a line if a carriage
return is also used to delimit a line?  One of the things you can do is to
use 'count.fields' to determine the number of fields in each line.  For
those lines that are not the right length, you could combine them together
with a 'paste' command when you write them out.

On 3/7/07, Walter R. Paczkowski [EMAIL PROTECTED] wrote:


   Hi,
   I'm  hoping someone has a suggestion for handling a simple problem.  A
   client  gave  me a comma separated value file (call it x.csv) that has
   an  id  and name and address for about 25,000 people (25,000 records).
   I used read.table to read it, but then discovered that there are stray
   carriage returns on several records.  This plays havoc with read.table
   since it starts a new input line when it sees the carriage return.  In
   short, the read is all wrong.
   I thought I could write a simple function to parse a line and write it
   back  out,  character by character.  If a carriage return is found, it
   would  simply  be  ignored on the writing back out part.  But how do I
   identify a carriage return?  What is the code or symbol?  Is there any
   easier  way  to  rid the file of carriage returns in the middle of the
   input lines?
   Any help is appreciated.
   Walt Paczkowski

   _
   Walter R. Paczkowski, Ph.D.
   Data Analytics Corp.
   44 Hamilton Lane
   Plainsboro, NJ  08536
   (V) 609-936-8999
   (F) 609-936-3733
 __
 R-help@stat.math.ethz.ch mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
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 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.




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

What is the problem you are trying to solve?

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[R] Calling Optim() from C

2007-03-07 Thread Mervyn G Marasinghe
Hello:

I am sure this question was dealt with several years ago.  Is the function
vmmin()  available  from Rmath Standalone? If not is it possible  to call
optim()  or  nlm()  from Rmath in C.  Thank you.

Mervyn

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