date:20110126

Try this:

split(myDF, myDF$myField)

On Wed, Jan 26, 2011 at 8:18 AM, Aditya Bhagwat bhagwatadi...@gmail.comwrote:

 Dear all,

 I have a data frame 'myDf', in which one of the fields 'myField' can have
 several possible values. To extract the observations for which it has value
 A, I can do:

 subset(myDf, myField=A)

 However, when I try to do this within a loop, it doesn't work, it returns
 everything, and not a subset

 for (currField in c(A, B, C)){
   subset(myDf, myField=currField)
 }

 How should I modify the call of subset in the loop to make it work?

 Thanks for your help!

 Adi

 --
 Aditya Bhagwat

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Re: [R] Train error:: subscript out of bonds

2011-01-26 Thread Neeti


What I have understood in CARET train() method is that train() itself does
the model selection and tune the parameter. (please correct me if I am
wrong). That was my first motivation to select this package and method for
fitting the model. And use the parameter to e1071 svm() method and compare
the result.

fit1-train(train1,as.factor(trainset[,ncol(trainset)]),svmpoly,trControl
= trainControl((method = cv),10,verboseIter = F),tuneLength=3)

-- 
View this message in context: 
http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3237800.html
Sent from the R help mailing list archive at Nabble.com.

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[R] a problem with is.na

2011-01-26 Thread René Holst

Hello,

I have observed the following odd behavior of is.na( ) and hope someone
can give me an explanation
Example:
X1=rep(1:2,5)[-1]
X2=rep(1:5,rep(2,5))[-1]
y= runif(9)
y[3]=NA
xtabs(y~x1+x2)

Now

xtabs(is.na(y)~x1+x2) says that cell 2,2 is NA

   x2
x1  1 2
  1 0 0
  2 0 1
  3 0 0
  4 0 0
  5 0 0

Whereas  

xtabs(!is.na(y)~x1+x2) says that all but cell 1,1 and 2,2 are not NA
   x2
x1  1 2
  1 0 1
  2 1 0
  3 1 1
  4 1 1
  5 1 1 

An explanation will be much appriciated

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[R] [R-pkgs] New package versions for distr- and robast- families

2011-01-26 Thread Dr. Peter Ruckdeschel




New versions 2.3 of our distr-family of packages are now
available on CRAN.

(i.e.; startupmsg, SweaveListingUtils,
   distr, distrEx, distrDoc, distrEllipse,
   distrMod, distrSim, distrTEst, distrTeach)

Most importantly, we have included:

+ a quasi-MC trick by Nataliya Horbenko to better produce
  image distributions under complicated, not necessarily
  monotone transformations
+ enhanced function qqplot
+ (enhanced) support for GEV distribution
+ new functional kMad
+ as well as several bug fixes

For more details see the corresponding NEWS files
(e.g. news(package = distr)
or using function NEWS from package startupmsg
i.e. NEWS(distr)).

We are looking forward to getting your RFEs, bug reports 
or simple feedback,

Best,

Peter, Matthias, Nataliya




New versions 0.8 of our RobASt-family of packages are now
available on CRAN.

(i.e.; RandVar, RobAStBase, ROptEst,
   RobLox, ROptEstOld, ROptEstTS, RobRex,
   RobLoxBioC)

Most importantly, we have included:

+ a quasi-MC trick by Nataliya Horbenko to better produce
  random variables under complicated not necessarily
  monotone transformations

+ enhanced functions
   infoPlot, (plots relative information used for coordinates
  of a parameter estimator)
   ddPlot, (distance-distance plot)
   cniperPointPlot, (cniper concept for seemingly harmless
contamination behavior)
   qqplot (now gets outlier corrected versions)

+ new risks: asAnscombe, asL1, asL4
  for asymptotic L1 L4 risk, and optimal bias robust estimator,
   to given efficiency loss in ideal model

+ new helper methods makeIC
  to apply to functions or list of functions
  for easily producing (suboptimal) ICs

+ new function getReq for two ICs IC1 and IC2
  to compute a radius interval where IC1 is better
  than IC2 acc. to G-Risk

+ new function getMaxIneff() to compute,
  for any IC of class 'IC', the maximal inefficiency
  for radius r varying in [0,Inf)

+ as well as several bug fixes

For more details see the corresponding NEWS files
(e.g. news(package = RobAStBase)
or using function NEWS from package startupmsg
i.e. NEWS(RobAStBase)).

We are looking forward to getting your RFEs, bug reports 
or simple feedback,


Best,

Peter, Matthias, Nataliya




-- 
Dr. Peter Ruckdeschel, Abteilung Finanzmathematik, F3.17
Fraunhofer ITWM, Fraunhofer Platz 1, 67663 Kaiserslautern
Telefon:  +49 631/31600-4699
Fax:  +49 631/31600-5699
E-Mail :  peter.ruckdesc...@itwm.fraunhofer.de

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Re: [R] hwo to speed up aggregate

2011-01-26 Thread Dennis Murphy

If you have a large data frame, one option is package data.table. Try the
following:

library(data.table)
dt - data.table(df)
dt[, list(qsum = sum(quantity)), by ='client, date, name']
 client   date  name qsum
[1,]  1 2010-01-01   one   45
[2,]  1 2011-01-01   one 4500
[3,]  2 2010-01-01   two   40
[4,]  2 2011-01-01   two 4000
[5,]  3 2010-01-01 three   20
[6,]  3 2011-01-01 three 2000

BTW, the leading comma after the opening bracket is not a typo :)

For R versions = 2.11.x, aggregate() has a formula interface that saves a
fair bit of typing:

aggregate(quantity ~ client + date + name, data = df, FUN = sum)
  client   date  name quantity
1  1 2010-01-01   one   45
2  1 2011-01-01   one 4500
3  3 2010-01-01 three   20
4  3 2011-01-01 three 2000
5  2 2010-01-01   two   40
6  2 2011-01-01   two 4000

A third option is package plyr and function ddply():

library(plyr)
ddply(df, .(client, date, name), summarise, qsum = sum(quantity))
# same output as data.table

Hopefully one or more of these will improve your processing time.

Dennis

On Wed, Jan 26, 2011 at 2:39 AM, analys...@hotmail.com 
analys...@hotmail.com wrote:

 I have

  df
   quantity branch client   date  name
 110  1  1 2010-01-01   one
 220  2  1 2010-01-01   one
 330  3  2 2010-01-01   two
 415  4  1 2010-01-01   one
 510  5  2 2010-01-01   two
 620  6  3 2010-01-01 three
 7  1000  1  1 2011-01-01   one
 8  2000  2  1 2011-01-01   one
 9  3000  3  2 2011-01-01   two
 10 1500  4  1 2011-01-01   one
 11 1000  5  2 2011-01-01   two
 12 2000  6  3 2011-01-01 three

 I want to aggregate away the branch. I followed a suggestion by Gabor
 (thanks) and did

 
 aggregate(list(quantity=df$quantity),list(client=df$client,date=df$date),sum)
  client   date quantity
 1  1 2010-01-01   45
 2  2 2010-01-01   40
 3  3 2010-01-01   20
 4  1 2011-01-01 4500
 5  2 2011-01-01 4000
 6  3 2011-01-01 2000

 I want df$name also in the output and did what looked obvious:

 
 aggregate(list(quantity=df$quantity),list(client=df$client,date=df$date,name=df$name),sum)
  client   date  name quantity
 1  1 2010-01-01   one   45
 2  1 2011-01-01   one 4500
 3  3 2010-01-01 three   20
 4  3 2011-01-01 three 2000
 5  2 2010-01-01   two   40
 6  2 2011-01-01   two 4000

 It seems to work, but slows down tremendously for a dataframe with
 around a 1000 rows.

 Could anyone explain what is going on and suggest a way out?

 Thanks.

 __
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 http://www.R-project.org/posting-guide.html
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Re: [R] a problem with is.na

There isn't combination of c(1, 1), so is NA:

tapply(y, list(X1, X2), sum)

On Wed, Jan 26, 2011 at 9:04 AM, René Holst r...@constat.dk wrote:

 Hello,

 I have observed the following odd behavior of is.na( ) and hope someone
 can give me an explanation
 Example:
 X1=rep(1:2,5)[-1]
 X2=rep(1:5,rep(2,5))[-1]
 y= runif(9)
 y[3]=NA
 xtabs(y~x1+x2)

 Now

 xtabs(is.na(y)~x1+x2) says that cell 2,2 is NA

   x2
 x1  1 2
  1 0 0
  2 0 1
  3 0 0
  4 0 0
  5 0 0

 Whereas

 xtabs(!is.na(y)~x1+x2) says that all but cell 1,1 and 2,2 are not NA
   x2
 x1  1 2
  1 0 1
  2 1 0
  3 1 1
  4 1 1
  5 1 1

 An explanation will be much appriciated

 __
 R-help@r-project.org 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.




-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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[R] mvoutlier

2011-01-26 Thread Claudia Paladini


   Dear R-users,
   I used x.out=sign1(data,makeplot=TRUE) from the package mvoutlier to detect
   multivariate outliers.
   I would like to label the points in the resulting plot with the row names of
   my data set. But none of my attempts does lead to a result. Can anybody help
   me, please?
   Best regards
   Claudia


   Neu: WEB.DE De-Mail - Einfach wie E-Mail, sicher wie ein Brief!
   Jetzt De-Mail-Adresse reservieren: [1]https://produkte.web.de/go/demail02

References

   1. https://produkte.web.de/go/demail02
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Re: [R] hwo to speed up aggregate

Try this:

 unique(transform(df, quantity = ave(quantity, client, date, name, FUN =
sum), branch = NULL))

On Wed, Jan 26, 2011 at 8:39 AM, analys...@hotmail.com 
analys...@hotmail.com wrote:

 I have

  df
   quantity branch client   date  name
 110  1  1 2010-01-01   one
 220  2  1 2010-01-01   one
 330  3  2 2010-01-01   two
 415  4  1 2010-01-01   one
 510  5  2 2010-01-01   two
 620  6  3 2010-01-01 three
 7  1000  1  1 2011-01-01   one
 8  2000  2  1 2011-01-01   one
 9  3000  3  2 2011-01-01   two
 10 1500  4  1 2011-01-01   one
 11 1000  5  2 2011-01-01   two
 12 2000  6  3 2011-01-01 three

 I want to aggregate away the branch. I followed a suggestion by Gabor
 (thanks) and did

 
 aggregate(list(quantity=df$quantity),list(client=df$client,date=df$date),sum)
  client   date quantity
 1  1 2010-01-01   45
 2  2 2010-01-01   40
 3  3 2010-01-01   20
 4  1 2011-01-01 4500
 5  2 2011-01-01 4000
 6  3 2011-01-01 2000

 I want df$name also in the output and did what looked obvious:

 
 aggregate(list(quantity=df$quantity),list(client=df$client,date=df$date,name=df$name),sum)
  client   date  name quantity
 1  1 2010-01-01   one   45
 2  1 2011-01-01   one 4500
 3  3 2010-01-01 three   20
 4  3 2011-01-01 three 2000
 5  2 2010-01-01   two   40
 6  2 2011-01-01   two 4000

 It seems to work, but slows down tremendously for a dataframe with
 around a 1000 rows.

 Could anyone explain what is going on and suggest a way out?

 Thanks.

 __
 R-help@r-project.org 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.




-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] a problem with is.na

2011-01-26 Thread Martyn Byng

Hi,

This isn't an issue with is.na, you get the same if you use

aa = c(1,1,0,1,1,1,1,1,1)
bb = abs(aa - 1)
xtabs(aa~x1+x2)
xtabs(bb~x1+x2)

it is because you do not have any data in (1,1), i.e. there is no case where x1 
= 1 and x2 = 1 so xtabs is putting a zero in that cell

Hope this helps

Martyn

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of René Holst
Sent: 26 January 2011 11:05
To: r-help@r-project.org
Subject: [R] a problem with is.na

Hello,

I have observed the following odd behavior of is.na( ) and hope someone
can give me an explanation
Example:
X1=rep(1:2,5)[-1]
X2=rep(1:5,rep(2,5))[-1]
y= runif(9)
y[3]=NA
xtabs(y~x1+x2)

Now

xtabs(is.na(y)~x1+x2) says that cell 2,2 is NA

   x2
x1  1 2
  1 0 0
  2 0 1
  3 0 0
  4 0 0
  5 0 0

Whereas  

xtabs(!is.na(y)~x1+x2) says that all but cell 1,1 and 2,2 are not NA
   x2
x1  1 2
  1 0 1
  2 1 0
  3 1 1
  4 1 1
  5 1 1 

An explanation will be much appriciated

__
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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.


This e-mail has been scanned for all viruses by Star.\ _...{{dropped:12}}

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Re: [R] Sweave: \Sexpr{} inside ?

2011-01-26 Thread Duncan Murdoch


On 11-01-25 8:22 PM, zerfetzen wrote:

 Hi,
 Is it possible in Sweave to put \Sexpr{} inside?  This is a bad
 example, but here goes:

 results=hide
 Age- 5
 @

 
 x- \Sexpr{Age}
 @

 I'm trying to get it to display x- 5, rather than x- Age.  It's 
probably

 so obvious I'm going to feel sorry for having to ask, just the same, I'm
 stumped.  Any ideas?  Thanks.

No, you can't do that.  There are a couple of ways to do what you want. 
 Probably the easiest is to do what Sweave would do:


results=hide=
Age- 5
@

\begin{Schunk}
\begin{Sinput}
 x- \Sexpr{Age}
\end{Sinput}
\end{Schunk}

Duncan Murdoch

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Re: [R] MAtrix addressing



On Jan 26, 2011, at 2:47 AM, Alaios wrote:


The reason is the following image
http://img545.imageshack.us/i/maptoregion.jpg/
In the picture above you will find the indexes for each cell.

Also you will see that I place that matrix inside a x,y region that  
spans from -1 to 1. I am trying to write one function that will get  
as argument a (x,y) value x e[-1,1] y e[-1,1] and will return the  
indexes of that cell tha x,y value correspond to.




I really do not have a clue how I should try to approach that to  
solve it. So based on some version I had for 1-d vector I tried to  
extend it for 2-d. I used findInterval as a core to get results.
Unfortunately my code fails to produce accurate results as my  
approach 'assumes' (this is something inhereted by the find Interval  
function) that the numbering starts bottom left and goes high top  
right.

You will find my code below


If one wants to take an ordinary r matrix and reorder it in the manner  
you describe:


mtx2 - mtx[ nrow(mtx):1, ]

Whether that is an efficient way to get at the sokution your you  
programming task I cannot say. It sounds as though it has gotten too  
convoluted. I was not able to comprehend the overall goal from your  
problem description.







sr.map - function(sr){
# This function converts the s(x,y) matrix into a function x that  
spans #from -1 to 1 and y spans from -1 to 1.

# Input: sr a x,y matrix containing the shadowing values of a Region
breaksX - seq(from=-1, to = 1, length = nrow(sr) +1L )
breaksY - seq(from=-1, to = 1, length = ncol(sr) + 1L)
function(x,y){ # SPAGGETI CODE FOR EVER
indx - findInterval(x, breaksX,rightmost.closed=TRUE)
 indy - findInterval(y, breaksY,rightmost.closed=TRUE)
 c(indx,indy)
}
}



sr-matrix(data=seq(from=1,to=36),nrow=6,ncol=6,byrow=TRUE)
f.sr.map-sr.map((sr))
f.sr.map(-0.1,-0.1)
f.sr.map(0.1,0.1)



Best Regards
Alex





--- On Wed, 1/26/11, David Winsemius dwinsem...@comcast.net wrote:


From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] MAtrix addressing
To: Alaios ala...@yahoo.com
Cc: R-help@r-project.org
Date: Wednesday, January 26, 2011, 2:54 AM

On Jan 25, 2011, at 4:50 PM, Alaios wrote:


Hello
I would like to ask you if it is possible In R Cran to

change the default way of addressing a matrix.

for example
matrix(data=seq(from=1,to=4,nrow=2,ncol=2, by row

numbering) # not having R at this pc


will create something like the following
1 2
3 4

the way R address this matrix is from top left corner

moving to bottom right.

The cell numbers in that way are
1 2
3 4

IS it possible to change this default addresing number

to something that goes bottom left to top right? In this
simple case I want to have

3 4
1 2

Would that be possible?


Yes. it's possible but ... why?



I would like to thank y for your help
Regards
Alex

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and provide commented, minimal, self-contained,

reproducible code.

David Winsemius, MD
West Hartford, CT








David Winsemius, MD
West Hartford, CT

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[R] barchart panel.text add label value and percent

2011-01-26 Thread Christophe Bouffioux

Hello everybody,

i need some help to display text as label in my barchart
the label is the combination of x value + text
text= calculated percentage = per
it display properly the x value
but, wrongly repeats the text of the fisrt level LangueTXT factor on the
second

any solution?
Thanx very much
Christophe


 here is the code ##
library(lattice)
Langue - c(1, 1, 1, 2, 2, 2, 2)
n03interessantscore - c(1, 2, 3, 1, 2, 3, 4)
count - c(89, 148, 16, 88, 192, 28, 7)
sumcount - c(253, 253, 253, 315, 315, 315, 315)
per - c('35.2%','58.5%','6.3%','27.9%','61%','8.9%','2.2%')
LangueTXT - c('Nl','Nl','Nl','Fr','Fr','Fr','Fr')


databar - data.frame(Langue, n03interessantscore, count, sumcount, per,
LangueTXT)

barchart(n03interessantscore ~ count| LangueTXT, data=databar,
 layout=c(1,max(databar$Langue)), stack=TRUE, rectangles=TRUE,
horizontal=TRUE,
 ylab=Score,
 xlab=Count,
 col=grey,
 main=3.0z Trouvez-vous ce rapport intéressant?,
 border=NA,
 panel= function(y,x,...){panel.grid(h=0, v=-1, col=gray)
 Y - tapply(y, y, unique)
  panel.barchart(x,y,...)
  panel.text((x-0.1*x), Y, label= paste(round(x,0),'-',
databar$per), cex=0.9)}
 )

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Re: [R] 2 functions with same name - what to do to get the one I want



On Jan 26, 2011, at 5:38 AM, pdb wrote:



There seems to be 2 functions call ecdf...

http://lib.stat.cmu.edu/S/Harrell/help/Hmisc/html/ecdf.html


Apparently that used to be its name and that some time in the  
intervening 10 years the name was changed to Ecdf. The Statlib  
repository has rather ancient version of the Hmisc documentation.  
Notice that it gives Harrell's UVa address. You should not use that  
repository for documentation. Use the RSiteSearch:


http://search.r-project.org/cgi-bin/namazu.cgi?query=ecdfmax=100result=normalsort=scoreidxname=functions

--
David.



http://127.0.0.1:11885/library/stats/html/ecdf.html

How do I get the one ecdf {Hmisc} to run instead of the ecdf {stats}

A pointer in the right direction would be greatly appreciated.


Tried to instal Hmisc but got this message, so I assume I have it


utils:::menuInstallPkgs()

Warning: package 'Hmisc' is in use and will not be installed



ran the demo from Hmisc with no luck...


set.seed(1)
ch - rnorm(1000, 200, 40)
ecdf(ch, xlab=Serum Cholesterol)

Error in ecdf(ch, xlab = Serum Cholesterol) :
 unused argument(s) (xlab = Serum Cholesterol)


ran the sample code from stats and it worked...


x - rnorm(12)
Fn - ecdf(x)
Fn # a *function*

Empirical CDF
Call: ecdf(x)
x[1:12] = -1.9123, -1.6626, -1.2468,  ..., 1.1119,  1.135

Fn(x)  # returns the percentiles for x

[1] 1. 0.9167 0. 0.6667 0.5833 0.1667
0.7500 0.0833 0.2500 0.8333 0.4167 0.5000

tt - seq(-2,2, by = 0.1)
12 * Fn(tt) # Fn is a 'simple' function {with values k/12}
[1]  0  1  1  1  2  2  2  2  3  3  3  3  4  4  4  5  5  5  6  6  6   
7  7  8

8  8  8  8  8  9 10 10 12 12 12 12 12 12 12 12 12




--
View this message in context: 
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Re: [R] mvoutlier

2011-01-26 Thread stephen sefick

I would look into ggplot2.  I use this quite frequently to do what you  
are talking about, and also for most of my plotting.  Hadley has done  
a wonderful job with this package.

kindest regards,

Stephen

On Jan 26, 2011, at 3:48 AM, Claudia Paladini wrote:



  Dear R-users,
  I used x.out=sign1(data,makeplot=TRUE) from the package mvoutlier  
to detect

  multivariate outliers.
  I would like to label the points in the resulting plot with the  
row names of
  my data set. But none of my attempts does lead to a result. Can  
anybody help

  me, please?
  Best regards
  Claudia


  Neu: WEB.DE De-Mail - Einfach wie E-Mail, sicher wie ein Brief!
  Jetzt De-Mail-Adresse reservieren: [1]https://produkte.web.de/go/demail02

References

  1. https://produkte.web.de/go/demail02
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[R] Projecting data onto a NMS plot

2011-01-26 Thread cystis


Hello all and thanks for the help.

I am analyzing a dataset using MetaMDS and I would like to project some
extra samples onto the plot such that  the extra samples do not play a role
in defining the axes. I have been thinking of different ways of doing this
and I was wondering if anyone had a suggestion for an easy way to do this 
(possibly a way to weight samples in the analysis?)

Thanks
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[R] Problem with setMethod in R package

2011-01-26 Thread Pascal A. Niklaus


Dear all,

My apologies for re-posting this question, but I have not found any 
solution to my problem so far. I also think that my post may have been 
overseen due to the posting time and high traffic on this list.


I experience a problem in implementing a S4 class in a package and would 
appreciate your help in this matter.


The class is implemented using Rcpp, which works well. I then extend the 
class with R functions, adding them as methods with a call to 
setMethod(...). When I do this outside the package (after loading the 
package with library(...)), everything works smoothly. However, when I 
try to incorporate the call to setMethod(...) in the package code 
itself, the only way I get it to work is to include the call in 
.onLoad(...) or .onAttach(...). This works, but when loading the library 
I get the messages creating new generic function for plot in 
.GlobalEnv, as well as no definition for class Rcpp_rothC...


Again, the code works, but the messages let me presume that I add the 
methods in the wrong way or miss-specify name spaces, class names, or 
something else.


Thank you for your advice.

Pascal

(The error message and the relevant parts of the package files are 
pasted below, together with my related questions as comments.)




showMethods(plot)

Function plot:
 not a generic function  # I am surprised about this --
   # why isn't plot a generic function?


library(RrothC2)

Loading required package: Rcpp
Loading required package: pascal
Creating a new generic function for plot in .GlobalEnv
in method for ‘plot’ with signature ‘x=Rcpp_rothC,y=missing’: no 
definition for class Rcpp_rothC

   # class seems not to be known
   # at this stage.
   # but where should I put setMethod()?

r1 - new(rothC$rothC);
class(r1)# class is known by now

[1] Rcpp_rothC
attr(,package)
[1] .GlobalEnv

plot(r1) # plot is dispatched correctly



showMethods(plot)

Function: plot (package graphics) # now plot from graphics shows up
x=ANY, y=ANY  # why was it missing before?
x=Rcpp_rothC, y=missing


# FILE rcpp_rothC_module.h #

class rothC { ... }

# FILE rcpp_rothC_module.cpp #

using namespace Rcpp ;

RCPP_MODULE(rothC) {
  class_rothC(rothC)
.constructor()
   ...
}

# FILE rcpp_rothC.R #

Rcpp_rothC.plot - function(x,y,...) { ... }

## FILE zzz.R ##

.NAMESPACE - environment()

rothC - new( Module )

.onLoad - function(libname, pkgname) {
  unlockBinding( rothC , .NAMESPACE )
  assign( rothC,  Module( rothC ), .NAMESPACE )

setMethod(f=plot,signature(x=Rcpp_rothC,y=missing),definition=Rcpp_rothC.plot,where=.GlobalEnv);
  lockBinding( rothC, .NAMESPACE )
}

.onAttach - function(libname, pkgname) {}

.onUnload - function(libname, pkgname) { gc(); }

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Re: [R] Counting number of rows with two criteria in dataframe

2011-01-26 Thread Hadley Wickham

On Wed, Jan 26, 2011 at 5:27 AM, Dennis Murphy djmu...@gmail.com wrote:
 Hi:

 Here are two more candidates, using the plyr and data.table packages:

 library(plyr)
 ddply(X, .(x, y), function(d) length(unique(d$z)))
  x y V1
 1 1 1  2
 2 1 2  2
 3 2 3  2
 4 2 4  2
 5 3 5  2
 6 3 6  2

 The function counts the number of unique z values in each sub-data frame
 with the same x and y values. The argument d in the anonymous function is a
 data frame object.

Another approach is to use the much faster count function:

count(unique(X))

Hadley

-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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Re: [R] Failing to install {rggobi} on win-7 R 2.12.0

Hello Prof Brian Ripley, Yihui and Tom,

Thank you for your suggestions.  It seemed to have made some differences in
the error massages - but rggobi still fails to load.

Steps taken:
1) I removed the old GTK (through the uninstall interface)
2) I ran  library(RGtk2) which downloaded the new GTK-runtime
version 2.22.0-2010-10-21 (instead of the one I got from ggobi, which
was 2.12.9-2).
3) I downloaded both
ftp://ftp.zlatkovic.com/libxml/libxml2-2.7.7.win32.zip and
ftp://ftp.zlatkovic.com/libxml/iconv-1.9.2.win32.zip
Unzipped them, and moved their dll's (from their bin directory), into -
C:\Program Files (x86)\GTK2-Runtime\bin
4) I then tried starting rggobi:  library(rggobi)  and got the following
error massages:

Error 1:

the program can't start because
libgdk-win32-2.0-0.dll
is missing from your computer.
Try reinstalling the program to fix this problem.

It then tried to reinstall GTK, and after I refused to, it sent the second
Error massage:

the program can't start because
libfreetype-6.dll
is missing from your computer.
Try reinstalling the program to fix this problem.



Any suggestions what else I should try?

Many thanks for helping,
Tal




Contact
Details:---
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--




On Wed, Jan 26, 2011 at 9:17 AM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote:

 On Tue, 25 Jan 2011, Tom La Bone wrote:

  I recall that my problem on Windows was related to having a number of
 stray
 versions of GTK+ installed. I went back and deleted all versions and
 reinstalled the latest GTK+ and that seemed to fix things. However, when I
 went to do any work of substance ggobi locked up and became unresponsive.
 Never did get it working right on Windows. Had much more luck with R/ggobi
 on Ubuntu 10.10.


 I've just been setting rggobi up for our classroom.  It seems that on
 Windows we now need to use Rgui in SDI mode to run rggobi without lookups.
  (That was not the case last year, so it might be due to the change in GTK+
 version or it might be due to the change from XP to x64 Windows 7 on those
 machines.)

 The rggobi binary on CRAN extras is statically linked against everything
 except GTK+, but the www.ggobi.org ggobi DLL needs both GTK+ DLLs and
 libxml2.dll (which needs iconv.dll and zlib1.dll). Late last year there was
 a problem in that GTK+ and libxml2.dll needed different zlib1.dll's, but
 AFAICS this is now resolved by using
 ftp://ftp.zlatkovic.com/libxml/libxml2-2.7.7.win32.zip and
 ftp://ftp.zlatkovic.com/libxml/iconv-1.9.2.win32.zip.  (Unpack those and
 drop the DLLs into somewhere on your path, e.g. the GTK+ bin directory.)

 We've had a lot of trouble over zlib1.dll: those prepared from zLib 1.2.3
 and 1.2.5 are incompatible.  The whole point of the '1' in the name is to
 change the name in that case!  I suspect very few of those benefitting from
 Windows binary packages have any idea how much work goes into circumventing
 such issues.


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


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[[alternative HTML version deleted]]

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[R] Fwd: MAtrix addressing

Begin forwarded message:

From: David Winsemius dwinsem...@comcast.net
Date: January 26, 2011 8:32:30 AM EST
To: Alaios ala...@yahoo.com
Subject: Re: [R] MAtrix addressing

On Jan 26, 2011, at 7:58 AM, Alaios wrote:

Unfortunately right now is convoluted... by I was trying to find  
some solution.

Bring again this picture in front of u
http://img545.imageshack.us/i/maptoregion.jpg/

Consider f a function that gets as input the coords
so
f(-1,-1) should return the value of the bottom left point
f(1,1) should return the value of the top right point.

A.For me an area is approximated by a matrix so each cell of the  
matrix corresponds to a fixed value in a small sub-area.
So When my function gets coords like f(0,0.3) should find the  
corresponding sub-area.

B. This is want to do
I also have a data structure called matrix that in every cell has  
the appropriate values of that area.

A+B = Combine these two and have a function that returns the  
appropriate value of a subarea given its coords. Attention my area  
spans from -1 to 1 in y plane and from -1 to 1 in the x plane.

 mtx - matrix(seq(1:36), nrow=6, byrow=TRUE,  
dimnames=list(x=seq(-1, 1, length=7)[-7], y=seq(-1, 1, length=7)[-7]) )

 mtx
   y
x-1 -0.667 -0.333  0  
0.333 0.667
 -1  1  2  3   
4 5 6
 -0.667  7  8  9  
101112
 -0.333 13 14 15  
161718
 0  19 20 21  
222324
 0.333  25 26 27  
282930
 0.667  31 32 33  
343536

 fnfind - function(x,y) mtx[ findInterval(x,  
c(as.numeric(rownames(mtx)), 1)),
+  findInterval(y,  
c(as.numeric(colnames(mtx)), 1))]

 fnfind(.5,.5)
[1] 29
 fnfind(-.5,-.5)
[1] 8

This could obviously be made more compact, but the current form allows  
simple modification of the length and endpoints of x and y.

Was it clearer this way? (Why is always so hard to me to explain  
even simple tasks?)

Regards
Alex

--- On Wed, 1/26/11, David Winsemius dwinsem...@comcast.net wrote:

From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] MAtrix addressing
To: Alaios ala...@yahoo.com
Cc: R-help@r-project.org
Date: Wednesday, January 26, 2011, 12:49 PM

On Jan 26, 2011, at 2:47 AM, Alaios wrote:

The reason is the following image
http://img545.imageshack.us/i/maptoregion.jpg/
In the picture above you will find the indexes for

each cell.

Also you will see that I place that matrix inside a

x,y region that spans from -1 to 1. I am trying to write one
function that will get as argument a (x,y) value x e[-1,1] y
e[-1,1] and will return the indexes of that cell tha x,y
value correspond to.

I really do not have a clue how I should try to

approach that to solve it. So based on some version I had
for 1-d vector I tried to extend it for 2-d. I used
findInterval as a core to get results.

Unfortunately my code fails to produce accurate

results as my approach 'assumes' (this is something
inhereted by the find Interval function) that the numbering
starts bottom left and goes high top right.

You will find my code below

If one wants to take an ordinary r matrix and reorder it in
the manner you describe:

mtx2 - mtx[ nrow(mtx):1, ]

Whether that is an efficient way to get at the sokution
your you programming task I cannot say. It sounds as though
it has gotten too convoluted. I was not able to comprehend
the overall goal from your problem description.

sr.map - function(sr){
# This function converts the s(x,y) matrix into a

function x that spans #from -1 to 1 and y spans from -1 to
1.

# Input: sr a x,y matrix containing the shadowing

values of a Region

breaksX - seq(from=-1, to

= 1, length = nrow(sr) +1L )

breaksY - seq(from=-1, to

= 1, length = ncol(sr) + 1L)

function(x,y){ # SPAGGETI CODE

FOR EVER

indx -

findInterval(x, breaksX,rightmost.closed=TRUE)

indy -

findInterval(y, breaksY,rightmost.closed=TRUE)

c(indx,indy)
}
}

sr-matrix(data=seq(from=1,to=36),nrow=6,ncol=6,byrow=TRUE)

f.sr.map-sr.map((sr))
f.sr.map(-0.1,-0.1)
f.sr.map(0.1,0.1)

Best Regards
Alex

--- On Wed, 1/26/11, David Winsemius dwinsem...@comcast.net

wrote:

From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] MAtrix addressing
To: Alaios ala...@yahoo.com
Cc: R-help@r-project.org
Date: Wednesday, January 26, 2011, 2:54 AM

On Jan 25, 2011, at 4:50 PM, Alaios wrote:

Hello
I would like to ask you if it is possible In R

Cran to

change the default way of addressing a matrix.

for example

Re: [R] Problem with setMethod in R package

2011-01-26 Thread Martin Morgan

On 01/26/2011 05:08 AM, Pascal A. Niklaus wrote:
 Dear all,
 
 My apologies for re-posting this question, but I have not found any
 solution to my problem so far. I also think that my post may have been
 overseen due to the posting time and high traffic on this list.
 
 I experience a problem in implementing a S4 class in a package and would
 appreciate your help in this matter.
 
 The class is implemented using Rcpp, which works well. I then extend the
 class with R functions, adding them as methods with a call to
 setMethod(...). When I do this outside the package (after loading the
 package with library(...)), everything works smoothly. However, when I
 try to incorporate the call to setMethod(...) in the package code
 itself, the only way I get it to work is to include the call in
 .onLoad(...) or .onAttach(...). This works, but when loading the library
 I get the messages creating new generic function for plot in
 .GlobalEnv, as well as no definition for class Rcpp_rothC...
 
 Again, the code works, but the messages let me presume that I add the
 methods in the wrong way or miss-specify name spaces, class names, or
 something else.
 
 Thank you for your advice.
 
 Pascal
 
 (The error message and the relevant parts of the package files are
 pasted below, together with my related questions as comments.)
 
 
 showMethods(plot)
 Function plot:
  not a generic function  # I am surprised about this --
# why isn't plot a generic function?
 
 library(RrothC2)
 Loading required package: Rcpp
 Loading required package: pascal
 Creating a new generic function for plot in .GlobalEnv
 in method for ‘plot’ with signature ‘x=Rcpp_rothC,y=missing’: no
 definition for class Rcpp_rothC
# class seems not to be known
# at this stage.
# but where should I put setMethod()?
 r1 - new(rothC$rothC);
 class(r1)# class is known by now
 [1] Rcpp_rothC
 attr(,package)
 [1] .GlobalEnv
 plot(r1) # plot is dispatched correctly
 
 showMethods(plot)
 Function: plot (package graphics) # now plot from graphics shows up
 x=ANY, y=ANY  # why was it missing before?
 x=Rcpp_rothC, y=missing
 
 
 # FILE rcpp_rothC_module.h #
 
 class rothC { ... }
 
 # FILE rcpp_rothC_module.cpp #
 
 using namespace Rcpp ;
 
 RCPP_MODULE(rothC) {
   class_rothC(rothC)
 .constructor()
...
 }
 
 # FILE rcpp_rothC.R #
 
 Rcpp_rothC.plot - function(x,y,...) { ... }
 
 ## FILE zzz.R ##
 
 .NAMESPACE - environment()
 
 rothC - new( Module )
 
 .onLoad - function(libname, pkgname) {
   unlockBinding( rothC , .NAMESPACE )
   assign( rothC,  Module( rothC ), .NAMESPACE )
 
 setMethod(f=plot,signature(x=Rcpp_rothC,y=missing),definition=Rcpp_rothC.plot,where=.GlobalEnv);
 
   lockBinding( rothC, .NAMESPACE )
 }
 
 .onAttach - function(libname, pkgname) {}
 
 .onUnload - function(libname, pkgname) { gc(); }

Hi Pascal --

Normally one would arrange code (e.g., using the Collate: field of the
DESCRIPTION file to define S4 classes, generics, then methods, equivalent to

setClass(Foo)
setGeneric(plot)  ## promote plot to S4 generic
setMethod(plot, c(x=Foo, y=missing), Rcpp_rothC.plot)

One would not normally specify a 'where' argument to setMethod; by
default the method is created in the 'top' environment at the time the
package is installed, which is the package name space. It is not usually
necessary to include setMethod and friends in .onLoad.

I do not use Rcpp, but one would normally have to write R code to wrap
the C++ object, e.g., an S4 class with an 'externalptr' slot to contain
the address of the object, plus methods to interact with the object.
This part of the problem sounds like a question for the Rcpp help list.

Martin
 
 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide
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-- 
Computational Biology
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109

Location: M1-B861
Telephone: 206 667-2793

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Re: [R] Help Derivate for Nonlinear Growth Models

2011-01-26 Thread Ben Bolker

acocac acocac at gmail.com writes:

 
 
 Hi!! Im doing my graduated work in Onion Curves Growth with Nonlinear Models,
 I'm amateur in R so i have  doubt how i put or program next models,
 
 http://r.789695.n4.nabble.com/file/n3236748/96629508.png 
 
 Also, i cant derivate for Gauss Model, and Richard Model dont have funtion,
 If someone could help me, i was so grate,
 

  You need more help than we can give you here ... You need to 
use the ?nls function. I think if you are going to be doing a serious
project fitting nonlinear growth models, you should probably learn or
refresh enough calculus so that you can compute the derivatives yourself
(R can do some small bit of analytic differentiation -- see ?D -- but
it doesn't simplify at all, so the answers are quite often uglier than
if you did the computation by hand). 
  You don't absolutely need the derivative to use nls(), although it
helps a lot.

  I would also suggest going to Google scholar and
searching for 'nls growth curve Bates to find some papers that
have used this approach.

  If you need to post again, please read the posting guide and
show us how far you have managed to get on your own.

  good luck,
   Ben Bolker

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Re: [R] Train error:: subscript out of bonds

2011-01-26 Thread Max Kuhn

Sort of. It lets you define a grid of candidate values to test and to
define the rule to choose the best. For some models, it is each to
come up with default values that work well (e.g. RBF SVM's, PLS, KNN)
while others are more data dependent. In the latter case, the defaults
may not work well.

MAx

On Wed, Jan 26, 2011 at 5:45 AM, Neeti nikkiha...@gmail.com wrote:

 What I have understood in CARET train() method is that train() itself does
 the model selection and tune the parameter. (please correct me if I am
 wrong). That was my first motivation to select this package and method for
 fitting the model. And use the parameter to e1071 svm() method and compare
 the result.

 fit1-train(train1,as.factor(trainset[,ncol(trainset)]),svmpoly,trControl
 = trainControl((method = cv),10,verboseIter = F),tuneLength=3)

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3237800.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org 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.




-- 

Max

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Re: [R] Failing to install {rggobi} on win-7 R 2.12.0

2011-01-26 Thread Prof Brian Ripley


Your GTK+ installation is not being found: check your PATH.

On Wed, 26 Jan 2011, Tal Galili wrote:


Hello Prof Brian Ripley, Yihui and Tom,

Thank you for your suggestions.  It seemed to have made some differences in
the error massages - but rggobi still fails to load.

Steps taken:
1) I removed the old GTK (through the uninstall interface)
2) I ran  library(RGtk2) which downloaded the new GTK-runtime
version 2.22.0-2010-10-21 (instead of the one I got from ggobi, which
was 2.12.9-2).
3) I downloaded both
ftp://ftp.zlatkovic.com/libxml/libxml2-2.7.7.win32.zip and 
ftp://ftp.zlatkovic.com/libxml/iconv-1.9.2.win32.zip
Unzipped them, and moved their dll's (from their bin directory), into - 
C:\Program Files (x86)\GTK2-Runtime\bin
4) I then tried starting rggobi:  library(rggobi)  and got the following
error massages:

Error 1:
  the program can't start because 
libgdk-win32-2.0-0.dll
is missing from your computer.
Try reinstalling the program to fix this problem.

It then tried to reinstall GTK, and after I refused to, it sent the second
Error massage:
  the program can't start because 
libfreetype-6.dll
is missing from your computer.
Try reinstalling the program to fix this problem.



Any suggestions what else I should try?

Many thanks for helping,
Tal




Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
---




On Wed, Jan 26, 2011 at 9:17 AM, Prof Brian Ripley rip...@stats.ox.ac.uk
wrote:
  On Tue, 25 Jan 2011, Tom La Bone wrote:

I recall that my problem on Windows was related to
having a number of stray
versions of GTK+ installed. I went back and deleted
all versions and
reinstalled the latest GTK+ and that seemed to fix
things. However, when I
went to do any work of substance ggobi locked up and
became unresponsive.
Never did get it working right on Windows. Had much
more luck with R/ggobi
on Ubuntu 10.10.


I've just been setting rggobi up for our classroom.  It seems that on
Windows we now need to use Rgui in SDI mode to run rggobi without
lookups.  (That was not the case last year, so it might be due to the
change in GTK+ version or it might be due to the change from XP to x64
Windows 7 on those machines.)

The rggobi binary on CRAN extras is statically linked against
everything except GTK+, but the www.ggobi.org ggobi DLL needs both
GTK+ DLLs and libxml2.dll (which needs iconv.dll and zlib1.dll). Late
last year there was a problem in that GTK+ and libxml2.dll needed
different zlib1.dll's, but AFAICS this is now resolved by using
ftp://ftp.zlatkovic.com/libxml/libxml2-2.7.7.win32.zip and
ftp://ftp.zlatkovic.com/libxml/iconv-1.9.2.win32.zip.  (Unpack those
and drop the DLLs into somewhere on your path, e.g. the GTK+ bin
directory.)

We've had a lot of trouble over zlib1.dll: those prepared from zLib
1.2.3 and 1.2.5 are incompatible.  The whole point of the '1' in the
name is to change the name in that case!  I suspect very few of those
benefitting from Windows binary packages have any idea how much work
goes into circumventing such issues.


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


__
R-help@r-project.org 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.






--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595__
R-help@r-project.org 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.

Re: [R] How to call subset in a for loop?

2011-01-26 Thread David Reiner

you were caught by the '=' versus '==' error ;-)
and Henrique's elegant one-liner avoids the problem altogether.

-- David

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Henrique Dallazuanna
Sent: Wednesday, January 26, 2011 6:00 AM
To: Aditya Bhagwat
Cc: r-help@r-project.org
Subject: Re: [R] How to call subset in a for loop?

Try this:

split(myDF, myDF$myField)

On Wed, Jan 26, 2011 at 8:18 AM, Aditya Bhagwat bhagwatadi...@gmail.comwrote:

 Dear all,

 I have a data frame 'myDf', in which one of the fields 'myField' can have
 several possible values. To extract the observations for which it has value
 A, I can do:

 subset(myDf, myField=A)

 However, when I try to do this within a loop, it doesn't work, it returns
 everything, and not a subset

 for (currField in c(A, B, C)){
   subset(myDf, myField=currField)
 }

 How should I modify the call of subset in the loop to make it work?

 Thanks for your help!

 Adi

 --
 Aditya Bhagwat

[[alternative HTML version deleted]]

 __
 R-help@r-project.org 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.

--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

[[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] coxme and random factors

2011-01-26 Thread Terry Therneau

 --- begin inclusion ---
Response variable:  survival (death)
Factor 1:   treatment (4 levels)
Factor 2:   sex (male / female)
Random effects 1:   person nested within day (2 people did the  
experiment over 2   days)
Random effects 2:   box nested within treatment (animals were kept in  
boxes in groups of 6, and there were multiple boxes per treatment)

I've read the introductions to coxme by Terry Therneau, and something  
like the following is what I think I should use:

model1-coxme(Surv(death,censor)~treatment*sex+(1|day/person)+(1| 
treatment/box))

--- End inclusion ---

That looks right to me.  Your questions:
1: How to test: As you guessed, fit the model without one of the random
effects and compare the integrated likelihood for the two fits.  The
usual is it a chisq or sum of chisquares question from random effects
models applies -- the simple chisq test will be conservative.

2: (treatment |box) term.  For a factor variable such as treatment a
term (1 | treatment/box) specifies a (random) coefficient for each
treatment by box combination.  The term (treatment|box) is asking for
exactly the same thing, but coxme currently does not support asking for
it in that way.

3. I do not have an extension of cox.zph to the mixed effects model, in
either theory or code.  Residuals methods for coxme would be an
important addition and is on my to-do list. (But as my wife would point
out, so is a bathroom remodel and she isn't holding her breath.)

__
R-help@r-project.org 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] lattice draw.key(): position of key in panels

2011-01-26 Thread Boris.Vasiliev

Good morning,

This problem was already addressed in a previous post:

https://stat.ethz.ch/pipermail/r-help/2009-February/187244.html

In the call to draw.key() use
'vp=viewport(x=unit(0.1,npc),y=unit(0.1,npc))'.
Prior to calling viewport() make sure grid package is loaded.

Apologies for cluttering the mail list.
Regards,
Boris Vasiliev.


 -Original Message-
 From: Vasiliev B@CEFCOM HQ@Ottawa-Hull 
 Sent: Tuesday, 25, January, 2011 16:22 PM
 To: 'r-help@r-project.org'
 Subject: lattice draw.key(): position of key in panels
 
 Good afternoon,
 
 I am working on a plot that requires custom legends to be 
 placed in some panels of the plot; other panels do not 
 contain legends.  The problem that I run into is positioning 
 of the legend in individual panels. In particular, the 'x' 
 and 'y' elements of the key-list are ignored by draw.key() 
 when it is called from inside a panel function.  As a result, 
 the legend is placed in the middle of the panel.  The example 
 below illustrates this problem.
 
 df - data.frame(x=c(1,1),y=c(1,2),type=c(A,B))
 
 panel.xyplot.x - function(...) {
   # draw data
   panel.xyplot(...)
 
   # create key-list
   pnlid - panel.number()
   lbl - ifelse(pnlid==1,AA,BB)
   pts - Rows(trellis.par.get(superpose.symbol),pnlid)
   key - list(points=pts,text=list(lbl),x=0.1,y=0.9,corner=c(0,1))
 
   # draw key
   draw.key(key,draw=TRUE)
 }
 
 oltc - xyplot(y~x|type,data=df,panel=panel.xyplot.x)
 print(oltc)
 
 I tried using 'vp=current.viewport()' in the call to 
 draw.key() but it did not help.  Can anybody suggest the 
 proper way to specify position of the key-list so that it is 
 respected by draw.key() when called within a panel function?
 
 Sincerely,
 Boris Vasiliev.

__
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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] Failing to install {rggobi} on win-7 R 2.12.0

I checked it using:

Sys.getenv(PATH)

And the output includes the PATH to the GTK2 installation (it's the last
item in the following list):

C:\\Windows\\system32;C:\\Windows;C:\\Windows\\System32\\Wbem;C:\\Windows\\System32\\WindowsPowerShell\\v1.0\\;C:\\Program
Files (x86)\\Common Files\\Ulead Systems\\MPEG;C:\\Program
Files\\TortoiseGit\\bin;C:\\Program Files
(x86)\\QuickTime\\QTSystem\\;C:\\Program Files (x86)\\ggobi;C:\\Program
Files (x86)\\GTK2-Runtime\\bin


What else might I try?


(Thanks)


Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Wed, Jan 26, 2011 at 4:23 PM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote:

 Your GTK+ installation is not being found: check your PATH.


 On Wed, 26 Jan 2011, Tal Galili wrote:

  Hello Prof Brian Ripley, Yihui and Tom,

 Thank you for your suggestions.  It seemed to have made some differences
 in
 the error massages - but rggobi still fails to load.

 Steps taken:
 1) I removed the old GTK (through the uninstall interface)
 2) I ran  library(RGtk2) which downloaded the new GTK-runtime
 version 2.22.0-2010-10-21 (instead of the one I got from ggobi, which
 was 2.12.9-2).
 3) I downloaded both
 ftp://ftp.zlatkovic.com/libxml/libxml2-2.7.7.win32.zip and
 ftp://ftp.zlatkovic.com/libxml/iconv-1.9.2.win32.zip
 Unzipped them, and moved their dll's (from their bin directory), into -
 C:\Program Files (x86)\GTK2-Runtime\bin
 4) I then tried starting rggobi:  library(rggobi)  and got the following
 error massages:

 Error 1:
  the program can't start because
 libgdk-win32-2.0-0.dll
 is missing from your computer.
 Try reinstalling the program to fix this problem.

 It then tried to reinstall GTK, and after I refused to, it sent the second
 Error massage:
  the program can't start because
 libfreetype-6.dll
 is missing from your computer.
 Try reinstalling the program to fix this problem.



 Any suggestions what else I should try?

 Many thanks for helping,
 Tal




 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 ---
 ---




 On Wed, Jan 26, 2011 at 9:17 AM, Prof Brian Ripley rip...@stats.ox.ac.uk
 
 wrote:
  On Tue, 25 Jan 2011, Tom La Bone wrote:

I recall that my problem on Windows was related to
having a number of stray
versions of GTK+ installed. I went back and deleted
all versions and
reinstalled the latest GTK+ and that seemed to fix
things. However, when I
went to do any work of substance ggobi locked up and
became unresponsive.
Never did get it working right on Windows. Had much
more luck with R/ggobi
on Ubuntu 10.10.


 I've just been setting rggobi up for our classroom.  It seems that on
 Windows we now need to use Rgui in SDI mode to run rggobi without
 lookups.  (That was not the case last year, so it might be due to the
 change in GTK+ version or it might be due to the change from XP to x64
 Windows 7 on those machines.)

 The rggobi binary on CRAN extras is statically linked against
 everything except GTK+, but the www.ggobi.org ggobi DLL needs both
 GTK+ DLLs and libxml2.dll (which needs iconv.dll and zlib1.dll). Late
 last year there was a problem in that GTK+ and libxml2.dll needed
 different zlib1.dll's, but AFAICS this is now resolved by using
 ftp://ftp.zlatkovic.com/libxml/libxml2-2.7.7.win32.zip and
 ftp://ftp.zlatkovic.com/libxml/iconv-1.9.2.win32.zip.  (Unpack those
 and drop the DLLs into somewhere on your path, e.g. the GTK+ bin
 directory.)

 We've had a lot of trouble over zlib1.dll: those prepared from zLib
 1.2.3 and 1.2.5 are incompatible.  The whole point of the '1' in the
 name is to change the name in that case!  I suspect very few of those
 benefitting from Windows binary packages have any idea how much work
 goes into circumventing such issues.


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


 __
 R-help@r-project.org 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,

Re: [R] Train error:: subscript out of bonds

2011-01-26 Thread Neeti


Thank you so much for your reply. In my case it is giving error in some seed
value for example if I set seed value to 357 this gives an error. Does train
have some specific seed range?
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3238197.html
Sent from the R help mailing list archive at Nabble.com.

__
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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] barplot with varaible-width bars

2011-01-26 Thread Bill Pikounis

Hi Larry,
If I understand correctly, your barplot() call dispatches to the
method function barplot.default()  to do the work. Looking at the
definition of that function and your specific call, it seems that
around line 51 of barplot.default(),  the value of the width argument
is truncated:

width - rep(width, length.out = NR)

where NR - nrow(height)  is defined a bit earlier around line 44. So
in the execution width takes on the value

[1] 417 153

which seems to explain the same width pieces across pairs.

Just quick and dirty I copied the function barplot.default in the
workspace to an editor, renamed it as mod.barplot.default() and then
commented out line 51 and added the line there: width - width which
seems at though it could actually be left out as long as beside=TRUE
is kept in the call. Then I created mod.barplot.default() as a working
function, and this call

mod.barplot.default(yy[,2*1:5], las=1, width=yy[,(2*1:5)-1],
space=c(.1,.5) ,beside=TRUE)

looks like it might provide what you wanted.

Hope that helps.

Bill


-
Bill Pikounis
http://billpikounis.net/



On Tue, Jan 25, 2011 at 10:47, Gould, A. Lawrence larry_go...@merck.com wrote:
 I would like to produce a bar plot with varying-width bars.  Here is an 
 example to illustrate:

 ww - c(417,153,0.0216,0.0065,556,256,0.0162,0.0117,
 +  726,379,0.0358,0.0501,786,502,0.0496,0.0837,
 +  892,591,0.0785,0.0795)
 yy-t(t(array(ww,c(2,10

 barplot(yy[,2*1:5],las=1,space=c(.1,.5),beside=T)

 produces a barplot of 5 pairs of bars that are of equal width

 barplot(yy[,2*1:5],las=1,width=c(yy[,(2*1:5)-1]),space=c(.1,.5),beside=T)

 makes the bars in each pair of unequal width, but the two widths do not vary 
 from pair to pair

 I would like the width of each bar to be proportional to its corresponding 
 value in the width statement of this last call of barplot, like what I think 
 could be done with the mulbar function of SPlus.  Can I do this with barplot 
 itself, or is this something for which lattice or ggplot 2 is needed?  And, 
 if so, what would typical code look like?

 Thanks for your help.

 Larry Gould


 Notice:  This e-mail message, together with any attachme...{{dropped:14}}

 __
 R-help@r-project.org 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.


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R-help@r-project.org 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] plot with 2 y axes

2011-01-26 Thread Greg Snow

Why do you need the line to overlay the bars?  Which bars are touched by the 
line is just a quirk of scaling and could easily change with the scales.  All 
the overlay does is to make it harder to read, why not jut have 2 panels 
aligned on the x-axis but with the line plot above the bar plot?

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Mario Beolco
 Sent: Tuesday, January 25, 2011 5:31 PM
 To: r-help@r-project.org
 Subject: [R] plot with 2 y axes
 
 Dear R users,
 
 apologies for the total beginner's question. I would like to create a
 barchart for some temperature values with the y axis on the right hand
 side of the plot. On this plot would like to overlay some time series
 data  (in the form of a line) for some other variable called Index.
 The y axis for this latter variable should be on the left hand side of
 the plot.
 
 An example of what I would like to obtain:
 
 https://sites.google.com/site/graphtests1/
 
 I have tried to do this using ggplot2 and this where I have got (for
 data see at the bottom of the e-mail):
 
 none-theme_blank()
 p-ggplot(tmp3,aes(x=year,y=Temperature))
 p1-p+geom_bar(stat=identity,fill=#9ACD32,colour=#00)
 p1 + geom_line(data=tmp3, aes(x=year, y=Index),
 colour=black,size=1)+opts(legend.position=none,panel.grid.major=non
 e,panel.grid.minor=none)+opts(panel.border=none)+theme_bw(base_size=20)
 
 This code does not do what I want because the Temperature y axis
 should be on the left hand side and the the y axis for the other
 variable called Index is not even there (should in theory be on the
 left hand side). I also get the following warning message when I run
 that code I get Warning message:Stacking not well defined when ymin
 != 0. (Should I worry about this?).
 
 I do not know whether ggplot2 can is the best package for creating the
 type of plot that I want. I would, however, be very grateful for any
 suggestions on to improve the above code or on how I could use other
 packages to create the plot I want.
 
 thanks!
 
 Mario
 
 
 
 year,Temperature,Index
 1966,2.9,1
 1967,4.5,1.24
 1968,1.9,1.46
 1969,1,1.37
 1970,2.9,1.87
 1971,4.3,2.66
 1972,3.9,3.07
 1973,4.3,3.91
 1974,4.9,4.16
 1975,4.4,4.32
 1976,4.5,2.52
 1977,2,2.44
 1978,2.8,2.18
 1979,-0.4,1.18
 1980,2.3,1.93
 1981,3,2.13
 1982,0.3,1.92
 1983,1.7,2.24
 1984,3.3,2.01
 1985,0.8,1.89
 1986,-1.1,0.66
 1987,0.8,1.01
 1988,4.9,1.5
 1989,5.2,2.11
 1990,4.9,2.02
 1991,1.5,0.7
 1992,3.7,0.75
 1993,3.6,1.28
 1994,3.2,1.37
 1995,4.8,2.01
 1996,2.3,1.54
 1997,2.5,2
 1998,5.2,2.07
 1999,5.3,2.11
 2000,4.9,2.42
 2001,3.2,2.29
 2002,3.6,2.15
 2003,3.9,2.21
 2004,4.8,2.14
 2005,4.3,2.33
 2006,3.7,1.89
 2007,5.8,2.03
 2008,4.9,2.58
 
 __
 R-help@r-project.org 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.

[R] Greek letters in CairoPDF

2011-01-26 Thread Eduardo de Oliveira Horta

Hello there,

Straight to the point: it seems that CairoPDF from package Cairo
cannot handle greek letters from expression(). For example,

 eta = seq(from=-pi, to=pi, length=100)
 f = sin(eta)^2
 pdf(file = temp_pdf.pdf)
 plot(eta, f, type=l, main=expression(f(eta)==sin(eta)^2), 
 xlab=expression(eta), ylab=expression(f(eta)))
 dev.off()

gives the expected result, but

 require(Cairo)
 CairoPDF(file = temp_CairoPDF.pdf)
 plot(eta, f, type=l, main=expression(f(eta)==sin(eta)^2), 
 xlab=expression(eta), ylab=expression(f(eta)))
 dev.off()

leaves a blank where it should display the etas. Any ideas here?
(session info below)

Thanks in advance and best regards,

Eduardo



 sessionInfo()
R version 2.11.1 (2010-05-31)
i386-pc-mingw32

locale:
[1] LC_COLLATE=Portuguese_Brazil.1252  LC_CTYPE=Portuguese_Brazil.1252
[3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C
[5] LC_TIME=Portuguese_Brazil.1252

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

other attached packages:
[1] Cairo_1.4-5

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

Re: [R] write.table -- maintain decimal places

2011-01-26 Thread Jim Moon

I am using:
R version 2.11.1 (2010-05-31)
It is good to know that it works in 2.12.1

Jim

-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca] 
Sent: Tuesday, January 25, 2011 5:57 PM
To: Jim Moon
Cc: r-help@r-project.org
Subject: Re: [R] write.table -- maintain decimal places

On 2011-01-25 17:22, Jim Moon wrote:
 Thank you for the response, Peter.

 The approach:
 write.table(format(df, 
 drop0trailing=FALSE),file='df.txt',quote=F,sep='\t',row.names=F)
 surprisingly still results in some loss of trailing 0's.

What version of R?
I'm using R version 2.12.1 Patched (2010-12-27 r53883)
and it works for me.

Peter Ehlers

 df:
 EFFECT2  PVALUE
 1 0.0230.88080
 2 -0.260  0.08641
 3 -0.114  0.45200

 df.txt:
 EFFECT2PVALUE
 0.023  8.808e-01
 -0.26  8.641e-02
 -0.114 4.520e-01

 -Original Message-
 From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
 Sent: Tuesday, January 25, 2011 5:09 PM
 To: Jim Moon
 Cc: r-help@r-project.org
 Subject: Re: [R] write.table -- maintain decimal places

 On 2011-01-25 16:16, Jim Moon wrote:
 Hello, All,

 How can I maintain the decimal places when using write.table()?

 Jim

 e.g.

 df:
 EFFECT2  PVALUE
 1 0.0230.88080
 2 -0.260  0.08641
 3 -0.114  0.45200

 write.table(df,file='df.txt',quote=F,sep='\t',row.names=F)

write.table(format(df, drop0trailing=FALSE), )

 Peter Ehlers

 df.txt:
 EFFECT2PVALUE
 0.023  0.8808
 -0.26  0.08641
 -0.114 0.452

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Re: [R] write.table -- maintain decimal places

2011-01-26 Thread Jim Moon

Great.  Thank you, Peter!

-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca] 
Sent: Tuesday, January 25, 2011 7:26 PM
To: Jim Moon
Cc: r-help@r-project.org
Subject: Re: [R] write.table -- maintain decimal places

On 2011-01-25 17:22, Jim Moon wrote:
 Thank you for the response, Peter.

 The approach:
 write.table(format(df, 
 drop0trailing=FALSE),file='df.txt',quote=F,sep='\t',row.names=F)
 surprisingly still results in some loss of trailing 0's.

Here are a couple more (essentially identical) ways:

# 1.
  dfm - within(df, {
EFFECT2 - sprintf(%6.3f, EFFECT2)
PVALUE  - sprintf(%7.5f, PVALUE)
})

# 2.
  dfm - within(df, {
   EFFECT2 - formatC(EFFECT2, format=f, digits=3)
   PVALUE  - formatC(PVALUE,  format=f, digits=5)
   })

write.table(dfm, file='dfm.txt', quote=FALSE, sep='\t', row.names=FALSE)

Peter Ehlers

 df:
 EFFECT2  PVALUE
 1 0.0230.88080
 2 -0.260  0.08641
 3 -0.114  0.45200

 df.txt:
 EFFECT2PVALUE
 0.023  8.808e-01
 -0.26  8.641e-02
 -0.114 4.520e-01

 -Original Message-
 From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
 Sent: Tuesday, January 25, 2011 5:09 PM
 To: Jim Moon
 Cc: r-help@r-project.org
 Subject: Re: [R] write.table -- maintain decimal places

 On 2011-01-25 16:16, Jim Moon wrote:
 Hello, All,

 How can I maintain the decimal places when using write.table()?

 Jim

 e.g.

 df:
 EFFECT2  PVALUE
 1 0.0230.88080
 2 -0.260  0.08641
 3 -0.114  0.45200

 write.table(df,file='df.txt',quote=F,sep='\t',row.names=F)

write.table(format(df, drop0trailing=FALSE), )

 Peter Ehlers

 df.txt:
 EFFECT2PVALUE
 0.023  0.8808
 -0.26  0.08641
 -0.114 0.452

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Re: [R] Counting number of rows with two criteria in dataframe

2011-01-26 Thread Ryan Utz

Hadley and Dennis:

THANK YOU THANK YOU! This is exactly what I was looking for.

Ryan


On Wed, Jan 26, 2011 at 5:27 AM, Dennis Murphy djmu...@gmail.com wrote:
  Hi:
 
  Here are two more candidates, using the plyr and data.table packages:
 
  library(plyr)
  ddply(X, .(x, y), function(d) length(unique(d$z)))
   x y V1
  1 1 1  2
  2 1 2  2
  3 2 3  2
  4 2 4  2
  5 3 5  2
  6 3 6  2
 
  The function counts the number of unique z values in each sub-data frame
  with the same x and y values. The argument d in the anonymous function is
 a
  data frame object.

 Another approach is to use the much faster count function:

 count(unique(X))

 Hadley

 --
 Assistant Professor / Dobelman Family Junior Chair
 Department of Statistics / Rice University
 http://had.co.nz/




-- 
Ryan Utz
Postdoctoral research scholar
University of California, Santa Barbara
(724) 272 7769

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[R] Making up a graph and its equation which better fits two groups of data

2011-01-26 Thread Jimmy Martina


Dear R-folks:
 
I have a group of data ('x' and 'y' axis), but I'd like to know how to draw a 
graph which would fits my data in a better way, I also need its equation. Could 
you give me any R-rutine ideas?.
 
I thank you in advance for your kind support.   
  
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Re: [R] plot with 2 y axes

2011-01-26 Thread kirtau


I hope this is what you are looking for. you will have to add your own colors
and such.

year = c(1966:2008)
tempur =
c(2.9,4.5,1.9,1,2.9,4.3,3.9,4.3,4.9,4.4,4.5,2,2.8,-.4,2.3,3,.3,1.7,3.3,.8,-1.1,.8,4.9,5.2,4.9,1.5,3.7,3.6,3.2,4.8,2.3,2.5,5.2,5.3,4.9,3.2,3.6,3.9,4.8,4.3,3.7,5.8,4.9)
indx =
c(1,1.24,1.46,1.37,1.87,2.66,3.07,3.91,4.16,4.32,2.52,2.44,2.18,1.18,1.93,2.13,1.92,2.24,2.01,1.89,.66,1.01,1.5,2.11,2.02,.7,.75,1.28,1.37,2.01,1.54,2,2.07,2.11,2.42,2.29,2.15,2.21,2.14,2.33,1.89,2.03,2.58)
data1 = data.frame(year,tempur,indx)

# create new graphing window
windows()
# create margins for variable names
par(mar = c(5,5,5,5))
# create bar plot
barplot(data1$tempur, names = data1$year, xlab = year, ylab = Temp)
# Allows for plotting on same charts (kinda like overlay)
par(new=T)
# plot line points
plot(data1$year,data1$indx, xaxt = n, yaxt = n, xlab = , ylab = )
# add lines
lines(data1$year,data1$indx)
# adds axis for second plot to right hand side
axis(side = 4)
# adds second y axis variable name to right hand side
mtext(Index,side = 4, line = 3)
# quits plotting on the current plotting window
par(new = F)
-- 
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[R] build interval

2011-01-26 Thread Rustamali Manesiya

Hello,

I have some question on chron

I currently doing this

t1 - chron(,11:30:00)
t2 - chron(,11:45:00)
tt - seq(t1,t2,by=times(00:00:01))

tt has 901 values (15 minutes * 60 secs) and then

x1 - rnorm(1:901)
x2 - rnorm(1:901)
x3 - rnorm(1:901)


df -  data.frame(tt, x1, x2, x3)

I would like to write a function such that I can divide df vector into any
interval,
For e.g 10 secs, 5 secs, 15 secs etc..

How can I achieve this. Is there a way to subtract seconds on a chron object
For e.g.

#This should subtract 20 seconds, but doesn't work, looks like it is
subtracting days
newtime - df$tt - 20

Please help.

Rusty

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[R] removing outlier function / dataset update

2011-01-26 Thread kirtau


Hi,

I have a few lines of code that will remove outliers for a regression test
based on the studentized residuals being above or below 3, -3. I have to do
this multiple times and have attempted to create a function to lessen the
amount of copying, pasting and replacing. 

I run into trouble with the function and receiving the error Error in
`$-.data.frame`(`*tmp*`, varpredicted, value = c(0.114285714285714,  : 
  replacement has 20 rows, data has 19


any help would be appreciated. a list of code is listed below. 

Thank you for your time!

x = c(1:20)
y = c(1,3,4,2,5,6,18,8,10,8,11,13,14,14,15,85,17,19,19,20)
data1 = data.frame(x,y)

# remove outliers for regression by studentized residuals being greater than
3
data1$predicted = predict(lm(data1$y~data1$x))
data1$stdres = rstudent(lm(data1$y~data1$x));
i=length(which(data1$stdres3|data1$stdres -3))
while(i = 1){
remove-which(data1$stdres3|data1$stdres -3)
print(data1[remove,])
data1 = data1[-remove,]
data1$predicted = predict(lm(data1$y~data1$x))
data1$stdres = rstudent(lm(data1$y~data1$x))
i = with(data1,length(which(stdres3|stdres -3)))
 }

# attemp to create a function to perfom same idea as above
rm.outliers = function(dataset,var1, var2) {

  dataset$varpredicted = predict(lm(var1~var2))
  dataset$varstdres = rstudent(lm(var1~var2))
  i = length(which(dataset$varstdres  3 | dataset$varstdres  -3))
  while(i = 1){
 removed = which(dataset$varstdres  3 | dataset$varstdres  -3)
 print(dataset[removed,])
 dataset = dataset[-removed,]
 dataset$varpredicted = predict(lm(var1~var2))
   dataset$varstdres = rstudent(lm(var1~var2))
 i = with(dataset,length(varstdres  3 | varstdres  -3))
   }
}
-- 
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[R] How to calculate p-value for Kolmogorov Smirnov test statistics?

2011-01-26 Thread saray


Although I saw this issue being discussed many times before, I still
did not find the answer to:
why does R can not calculate p-values for data with ties (i.e. -
sample with two or more values the same)?

Can anyone elaborate some details about how does R calculate the p-
values for the Kolmogorov Smirnov test statistics?

I can understand the theoretical problem that continuous distributions
do not generate ties, but again - why isn't it possible to calculate
accurate p-values with ties? how does it work? what is the procedure
to calculate p-values for the Kolmogorov Smirnov test statistics?

Thank you advance,
Saray
-- 
View this message in context: 
http://r.789695.n4.nabble.com/How-to-calculate-p-value-for-Kolmogorov-Smirnov-test-statistics-tp3238293p3238293.html
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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] Combing forest plots

2011-01-26 Thread Ross, Stephanie

Hi All,

I am trying to combine two forest plots on the same page using the forestplot 
function in the rmeta package. Once I use the par() function to combine my 
plots on the same page, I find that my two plots are overlaying each other. 
Does anyone have any suggestions on how to  fix this?

Thanks!



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

[R] Extracting the terms from an rpart object

Hello all,

I wish to extract the terms from an rpart object.
Specifically, I would like to be able to know what is the response variable
(so I could do some manipulation on it).
But in general, such a method for rpart will also need to handle a . case
(see fit2)

Here are two simple examples:

fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
fit1$call
fit2 - rpart(Kyphosis ~ ., data=kyphosis)
fit2$call


Is there anything prettier then using string manipulation?


Thanks.





Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--

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

[R] Inconsistencies in the rpart.object help file?

Hello all,

I'm was going through the help for
?rpart.object
And noticed some inconsistencies, Some might be a mistake in the help file
and some might be my misunderstanding.

The help in the section:
value - frame (first paragraph), states that:

  yval, the fitted value of the response at each node, *and splits, a two
 column matrix of left and right split labels for each node. *


But from looking at the object, for example

fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
fit1$frame

  var  n wt dev yval complexity ncompete nsurrogateyval2.1
 yval2.2yval2.3yval2.4yval2.5

1   Start 81 81  171 0.176470592  1  1.000
64.000 17.000  0.7901235  0.2098765

2   Start 62 62   61 0.019607842  2  1.000
56.000  6.000  0.9032258  0.0967742

4  leaf 29 29   01 0.01000  0  1.000
29.000  0.000  1.000  0.000

I can't see any splits column.

I'm also not sure I understand what the yval2 columns signify (even that I
read what it says in the help).

p.s: I hope I sent it to the correct e-mail.

Best,
Tal





Contact
Details:---
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Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--

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Re: [R] How to calculate p-value for Kolmogorov Smirnov test statistics?

The answers to these questions can be found either by looking at the
code or by reviewing what has been said about this in prior postings
to R-help.

--
David.

On Jan 26, 2011, at 10:40 AM, saray wrote:

Although I saw this issue being discussed many times before, I still
did not find the answer to:
why does R can not calculate p-values for data with ties (i.e. -
sample with two or more values the same)?

Can anyone elaborate some details about how does R calculate the p-
values for the Kolmogorov Smirnov test statistics?

I can understand the theoretical problem that continuous distributions
do not generate ties, but again - why isn't it possible to calculate
accurate p-values with ties? how does it work? what is the procedure
to calculate p-values for the Kolmogorov Smirnov test statistics?

Thank you advance,
Saray
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-calculate-p-value-for-Kolmogorov-Smirnov-test-statistics-tp3238293p3238293.html
Sent from the R help mailing list archive at Nabble.com.

__
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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.

David Winsemius, MD
West Hartford, CT

Re: [R] Extracting the terms from an rpart object

2011-01-26 Thread William Dunlap

Take a look at the output of
   terms(fit2)
In particular
   tm - terms(fit2)
   attr(tm, response)
is 1 if there is a response and
   variables - as.list(attr(tm, variables))[-1]
   variables[[1]]
gives the response expression if there is one.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Tal Galili
 Sent: Wednesday, January 26, 2011 9:33 AM
 To: r-help@r-project.org
 Subject: [R] Extracting the terms from an rpart object
 
 Hello all,
 
 I wish to extract the terms from an rpart object.
 Specifically, I would like to be able to know what is the 
 response variable
 (so I could do some manipulation on it).
 But in general, such a method for rpart will also need to 
 handle a . case
 (see fit2)
 
 Here are two simple examples:
 
 fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
 fit1$call
 fit2 - rpart(Kyphosis ~ ., data=kyphosis)
 fit2$call
 
 
 Is there anything prettier then using string manipulation?
 
 
 Thanks.
 
 
 
 
 
 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il 
 (Hebrew) |
 www.r-statistics.com (English)
 --
 
 
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 __
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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-help@r-project.org mailing list
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Re: [R] Extracting the terms from an rpart object

Try this:

all.vars(terms(fit1))
all.vars(terms(fit2))


On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili tal.gal...@gmail.com wrote:

 Hello all,

 I wish to extract the terms from an rpart object.
 Specifically, I would like to be able to know what is the response variable
 (so I could do some manipulation on it).
 But in general, such a method for rpart will also need to handle a . case
 (see fit2)

 Here are two simple examples:

 fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
 fit1$call
 fit2 - rpart(Kyphosis ~ ., data=kyphosis)
 fit2$call


 Is there anything prettier then using string manipulation?


 Thanks.





 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 --

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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] removing outlier function / dataset update

2011-01-26 Thread Ista Zahn

Hi,
x and y are being picked up from your global environment, not from the
x and y in dataset. Here is a version that seems to work:

rm.outliers = function(dataset,var1, var2) {

dataset$varpredicted = predict(lm(as.formula(paste(var1, var2,
sep= ~ )), data=dataset))
dataset$varstdres = rstudent(lm(as.formula(paste(var1, var2, sep=
~ )), data=dataset))
i = length(which(dataset$varstdres  3 | dataset$varstdres  -3))
while(i = 1){
removed = which(dataset$varstdres  3 | dataset$varstdres  -3)
print(dataset[removed,])
dataset = dataset[-removed,]
dataset$varpredicted = predict(lm(as.formula(paste(var1, var2,
sep= ~ )), data=dataset))
dataset$varstdres = rstudent(lm(as.formula(paste(var1, var2,
sep= ~ )), data=dataset))
i = with(dataset,length(varstdres  3 | varstdres  -3))
}
}


Best,
Ista

On Wed, Jan 26, 2011 at 11:36 AM, kirtau kir...@live.com wrote:

 Hi,

 I have a few lines of code that will remove outliers for a regression test
 based on the studentized residuals being above or below 3, -3. I have to do
 this multiple times and have attempted to create a function to lessen the
 amount of copying, pasting and replacing.

 I run into trouble with the function and receiving the error Error in
 `$-.data.frame`(`*tmp*`, varpredicted, value = c(0.114285714285714,  :
  replacement has 20 rows, data has 19
 

 any help would be appreciated. a list of code is listed below.

 Thank you for your time!

 x = c(1:20)
 y = c(1,3,4,2,5,6,18,8,10,8,11,13,14,14,15,85,17,19,19,20)
 data1 = data.frame(x,y)

 # remove outliers for regression by studentized residuals being greater than
 3
 data1$predicted = predict(lm(data1$y~data1$x))
 data1$stdres = rstudent(lm(data1$y~data1$x));
 i=length(which(data1$stdres3|data1$stdres -3))
 while(i = 1){
        remove-which(data1$stdres3|data1$stdres -3)
        print(data1[remove,])
        data1 = data1[-remove,]
        data1$predicted = predict(lm(data1$y~data1$x))
        data1$stdres = rstudent(lm(data1$y~data1$x))
        i = with(data1,length(which(stdres3|stdres -3)))
  }

 # attemp to create a function to perfom same idea as above
 rm.outliers = function(dataset,var1, var2) {

  dataset$varpredicted = predict(lm(var1~var2))
  dataset$varstdres = rstudent(lm(var1~var2))
  i = length(which(dataset$varstdres  3 | dataset$varstdres  -3))
  while(i = 1){
         removed = which(dataset$varstdres  3 | dataset$varstdres  -3)
         print(dataset[removed,])
         dataset = dataset[-removed,]
         dataset$varpredicted = predict(lm(var1~var2))
   dataset$varstdres = rstudent(lm(var1~var2))
         i = with(dataset,length(varstdres  3 | varstdres  -3))
   }
 }
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/removing-outlier-function-dataset-update-tp3238394p3238394.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org 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.




-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] Extracting the terms from an rpart object

Thanks Henrique, exactly what I was looking for.


Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Wed, Jan 26, 2011 at 7:40 PM, Henrique Dallazuanna www...@gmail.comwrote:

 Try this:

 all.vars(terms(fit1))
 all.vars(terms(fit2))


 On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili tal.gal...@gmail.com wrote:

 Hello all,

 I wish to extract the terms from an rpart object.
 Specifically, I would like to be able to know what is the response
 variable
 (so I could do some manipulation on it).
 But in general, such a method for rpart will also need to handle a .
 case
 (see fit2)

 Here are two simple examples:

 fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
 fit1$call
 fit2 - rpart(Kyphosis ~ ., data=kyphosis)
 fit2$call


 Is there anything prettier then using string manipulation?


 Thanks.





 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 --

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 __
 R-help@r-project.org 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.




 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


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

Re: [R] Train error:: subscript out of bonds

2011-01-26 Thread Max Kuhn

No. Any valid seed should work. In this case, train() should on;y be
using it to determine which training set samples are in the CV or
bootstrap data sets.

Max

On Wed, Jan 26, 2011 at 9:56 AM, Neeti nikkiha...@gmail.com wrote:

 Thank you so much for your reply. In my case it is giving error in some seed
 value for example if I set seed value to 357 this gives an error. Does train
 have some specific seed range?
 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3238197.html
 Sent from the R help mailing list archive at Nabble.com.

 __
 R-help@r-project.org 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.




-- 

Max

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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Extracting the terms from an rpart object

Another (similar) question,

If I now want to know the name of the data argument used, is there an easy
way for me to access it?

I'm trying to use something like:
eval(parse(text = all.vars(terms(fit1))[1]))

Which (of course) wouldn't work, since the response variable is only
available in the data used by rpart (specifically the kyphosis dataset)

Thanks upfront.

Tal




Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Wed, Jan 26, 2011 at 7:40 PM, Henrique Dallazuanna www...@gmail.comwrote:

 Try this:

 all.vars(terms(fit1))
 all.vars(terms(fit2))


 On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili tal.gal...@gmail.com wrote:

 Hello all,

 I wish to extract the terms from an rpart object.
 Specifically, I would like to be able to know what is the response
 variable
 (so I could do some manipulation on it).
 But in general, such a method for rpart will also need to handle a .
 case
 (see fit2)

 Here are two simple examples:

 fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
 fit1$call
 fit2 - rpart(Kyphosis ~ ., data=kyphosis)
 fit2$call


 Is there anything prettier then using string manipulation?


 Thanks.





 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 --

[[alternative HTML version deleted]]

 __
 R-help@r-project.org 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.




 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


[[alternative HTML version deleted]]

__
R-help@r-project.org 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] adding error bars

2011-01-26 Thread ogbos okike

Dear all,
I am trying to add error bars on a boxplot but have encountered an error as
indicated below. Is there a package I need to install or a library I have to
load before this goes please.
Thanks for any idea.
Ogbos

x-replicate(20,rnorm(50))
 boxplot(x,notch=TRUE,main=Notched boxplot with error bars)
 error.bars(x,add=TRUE)
Error: could not find function error.bars

[[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] Extracting the terms from an rpart object

2011-01-26 Thread William Dunlap

Note that all.vars(terms(fit)) only looks at the
formula in the terms object and throws away all
the analysis done by rpart's call to terms(formula,data).
Here is a contrived example of that approach failing:

   ageThreshold - 50
   fit3 - rpart(Kyphosis==present ~ (AgeageThreshold) + log(Number, 
base=2) + Start, data=kyphosis)
   all.vars(terms(fit3))
  [1] Kyphosis Age  ageThreshold Number   Start

Looking at the attributes of the terms object
tells you what I think you want:

   attr(terms(fit3), response) # 1=there is a response variable, 0=no 
response
  [1] 1
   as.list(attr(terms(fit3), variables))[-1]
  [[1]]
  Kyphosis == present

  [[2]]
  Age  ageThreshold

  [[3]]
  log(Number, base = 2)

  [[4]]
  Start

rpart doesn't allow interaction terms (x:y), but if it did
you would want to look at the factors attribute to see
which items in the variables lists are in each term of
the expanded formula.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Tal Galili
 Sent: Wednesday, January 26, 2011 10:07 AM
 To: Henrique Dallazuanna
 Cc: r-help@r-project.org
 Subject: Re: [R] Extracting the terms from an rpart object
 
 Thanks Henrique, exactly what I was looking for.
 
 
 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il 
 (Hebrew) |
 www.r-statistics.com (English)
 --
 
 
 
 
 
 On Wed, Jan 26, 2011 at 7:40 PM, Henrique Dallazuanna 
 www...@gmail.comwrote:
 
  Try this:
 
  all.vars(terms(fit1))
  all.vars(terms(fit2))
 
 
  On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili 
 tal.gal...@gmail.com wrote:
 
  Hello all,
 
  I wish to extract the terms from an rpart object.
  Specifically, I would like to be able to know what is the response
  variable
  (so I could do some manipulation on it).
  But in general, such a method for rpart will also need to 
 handle a .
  case
  (see fit2)
 
  Here are two simple examples:
 
  fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
  fit1$call
  fit2 - rpart(Kyphosis ~ ., data=kyphosis)
  fit2$call
 
 
  Is there anything prettier then using string manipulation?
 
 
  Thanks.
 
 
 
 
 
  Contact
  Details:---
  Contact me: tal.gal...@gmail.com |  972-52-7275845
  Read me: www.talgalili.com (Hebrew) | 
 www.biostatistics.co.il (Hebrew) |
  www.r-statistics.com (English)
 
  
 --
 
 
 [[alternative HTML version deleted]]
 
  __
  R-help@r-project.org 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.
 
 
 
 
  --
  Henrique Dallazuanna
  Curitiba-Paraná-Brasil
  25° 25' 40 S 49° 16' 22 O
 
 
   [[alternative HTML version deleted]]
 
 

__
R-help@r-project.org 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] adding error bars

2011-01-26 Thread Joshua Wiley

Hi,

You will find package sos has some handy functions for searching for
functions/packages:

###
install.packages(sos)
require(sos)
findFn(error.bars)
###

Perhaps the psych package has the error.bars() function you are thinking of?

As a side note, I am not sure it makes much sense to add error bars to
a boxplot---it already includes 25th/75th percentiles (at least
approximately, some use hinges, etc. but same general idea), and
typically minimum and maximum scores.  What would the error bars add
to that picture that would aid a viewer in understanding the
distribution of your data or the difference between different sets of
data?  In other words, is the information conveyed over and above the
boxplots large enough to warrant the additional clutter, possible
confusion, and necessary key/description in your figure caption so
viewers know what they are looking at?  Just something to think about.

Cheers,

Josh

On Wed, Jan 26, 2011 at 10:04 AM, ogbos okike ogbos.ok...@gmail.com wrote:

 Dear all,
 I am trying to add error bars on a boxplot but have encountered an error as
 indicated below. Is there a package I need to install or a library I have to
 load before this goes please.
 Thanks for any idea.
 Ogbos

 x-replicate(20,rnorm(50))
  boxplot(x,notch=TRUE,main=Notched boxplot with error bars)
  error.bars(x,add=TRUE)
 Error: could not find function error.bars

        [[alternative HTML version deleted]]

 __
 R-help@r-project.org 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.



--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

__
R-help@r-project.org 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] how to get loglik parameter from splm package?

2011-01-26 Thread Millo Giovanni

;^)

you're right: we'll add it soon. Keep an eye on R-forge.

All the best,
Giovanni

-- original message 

Message: 89
Date: Tue, 25 Jan 2011 17:14:00 -0800 (PST)
From: zhaowei zao_...@msn.com
To: r-help@r-project.org
Subject: Re: [R] how to get loglik parameter from splm package?
Message-ID: 1296004440124-3237303.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii


thank  Millo  for your very valuable reply

The political situation in Italy may be better than in  china.
And wish logik function could be supported in FE models too.
If then ,Splm will be more welcome.

thanks for your all works for useR.



Millo Giovanni wrote:
 
 Dear useR,
 
 although I admit that getting the log likelihood is important, you
must
 concede that obtaining the parameter estimates is not bad either.
 Regarding craze, well there are crazier things in the world than
this,
 just look at the political situation in Italy.
 
 Anyway, the loglik has always been there, although it wasn't exported
 (hence the NULL value). In the most recent versions of 'splm' we have
 made it available, at least for RE models, through the standard way: a
 logLik() method. Usage:
 
 logLik(yourmodel)
 
 You can download it from R-forge, as usual.
 
 Best,
 Giovanni
 
  original message --
 
 Message: 42
 Date: Mon, 24 Jan 2011 06:59:39 -0800 (PST)
 From: zhaowei zao_...@msn.com
 To: r-help@r-project.org
 Subject: [R] how to get loglik parameter from splm package?
 Message-ID: 1295881179014-3234185.p...@n4.nabble.com
 Content-Type: text/plain; charset=us-ascii
 
 
 splm package is a r  implemention of spatial panel data models.
 and the loglik paremeter is most important infomation for splm
methods.
 but  i found the loglik always been null ,it's craze to get right
 estimation 
 in splm with null  loglik.
 Any one knows the splm package and can get the right loglik ? please
 help
 me.
 
 thanks
 -- 
 View this message in context:

http://r.789695.n4.nabble.com/how-to-get-loglik-parameter-from-splm-pack
 age-tp3234185p3234185.html
 Sent from the R help mailing list archive at Nabble.com.
 
 --- end original message -
 
 Giovanni Millo
 Research Dept.,
 Assicurazioni Generali SpA
 Via Machiavelli 4, 
 34132 Trieste (Italy)
 tel. +39 040 671184 
 fax  +39 040 671160 
 
  
 Ai sensi del D.Lgs. 196/2003 si precisa che le
informazi...{{dropped:13}}
 
 __
 R-help@r-project.org 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.
 
 

-- 
View this message in context:
http://r.789695.n4.nabble.com/how-to-get-loglik-parameter-from-splm-pack
age-tp3234192p3237303.html
Sent from the R help mailing list archive at Nabble.com.
-- end original message -

Giovanni Millo
Research Dept.,
Assicurazioni Generali SpA
Via Machiavelli 4, 
34132 Trieste (Italy)
tel. +39 040 671184 
fax  +39 040 671160 

__
R-help@r-project.org 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.

Re: [R] adding error bars


Harrell's wiki/website has material on so-called dynamite plots

http://biostat.mc.vanderbilt.edu/twiki/bin/view/Main/DynamitePlots


Ben Bolker has a page on them as well:

http://emdbolker.wikidot.com/blog:dynamite

--  
David.



On Jan 26, 2011, at 1:26 PM, Joshua Wiley wrote:


Hi,

You will find package sos has some handy functions for searching for
functions/packages:

###
install.packages(sos)
require(sos)
findFn(error.bars)
###

Perhaps the psych package has the error.bars() function you are  
thinking of?


As a side note, I am not sure it makes much sense to add error bars to
a boxplot---it already includes 25th/75th percentiles (at least
approximately, some use hinges, etc. but same general idea), and
typically minimum and maximum scores.  What would the error bars add
to that picture that would aid a viewer in understanding the
distribution of your data or the difference between different sets of
data?  In other words, is the information conveyed over and above the
boxplots large enough to warrant the additional clutter, possible
confusion, and necessary key/description in your figure caption so
viewers know what they are looking at?  Just something to think about.

Cheers,

Josh

On Wed, Jan 26, 2011 at 10:04 AM, ogbos okike  
ogbos.ok...@gmail.com wrote:


Dear all,
I am trying to add error bars on a boxplot but have encountered an  
error as
indicated below. Is there a package I need to install or a library  
I have to

load before this goes please.
Thanks for any idea.
Ogbos

x-replicate(20,rnorm(50))
 boxplot(x,notch=TRUE,main=Notched boxplot with error bars)
 error.bars(x,add=TRUE)
Error: could not find function error.bars

   [[alternative HTML version deleted]]

__
R-help@r-project.org 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.




--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

__
R-help@r-project.org 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 Winsemius, MD
West Hartford, CT

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

[R] using popbio, reduce number of digits in image2 plot

2011-01-26 Thread Tali


Hello All.

I am using the image2 plot function in the popbio package to create 6
elasticity analyses. I am trying to reduce the number of significant digits
that displays -- from 3 digits to 2. I tried rounding my original matrices,
but one is comprised primarily of zeros, and the cut / breaks options
returns an error message. (see code and error message below)

Here are the elasticity values that I'm working with. Matrix 2 2005-2006
is the one that is creating the problem. I tried fiddling with cut, breaks,
digit, rounding...to no avail.

Further, is there an easy way to turn the top side labels 90 degrees? 

Thank you Chris Stubbens for this wonderful code!

TV, PhD Candidate
Center for Marine Biodiversity and Conservation
Scripps Institution of Oceanography

elast:

, , 2004-2005

1   2   3   4
1 0.017558594 0.003972943 0.002701352 0.006443492
2 0.011770698 0.106806350 0.020465566 0.022990744
3 0.001347089 0.047278966 0.258967762 0.061338978
4 0.0 0.003975099 0.086798115 0.347584256

, , 2005-2006

  1234
1 0 0.00e+00 0.00e+00 0.00e+00
2 0 6.872613e-17 2.641284e-17 3.763301e-17
3 0 0.00e+00 0.00e+00 0.00e+00
4 0 0.00e+00 0.00e+00 1.00e+00

, , 2006.1-2007

1   23   4
1 0.006634579 0.002471288 0.0001750284 0.001891018
2 0.004537334 0.040648910 0.0076802263 0.014082075
3 0.0 0.023828347 0.1431152824 0.015278433
4 0.0 0.0 0.0312515256 0.708405954

, , 2007-2008

1   2   3  4
1 0.022147389 0.007198642 0.001271820 0.0264
2 0.015296663 0.065937859 0.00541 0.01685541
3 0.004286438 0.030397868 0.157415674 0.04280021
4 0.0 0.0 0.070768252 0.54906670

, , 2008-2009

   1  2   3   4
1 0.02359622 0.00377114 0.001521911 0.007169308
2 0.01246236 0.05323351 0.009987896 0.014909244
3 0. 0.03358836 0.060984960 0.016624312
4 0. 0. 0.038702865 0.723447911

, , 2009-2010

1   23   4
1 0.005384555 0.002185377 0.0006226765 0.004539465
2 0.007347518 0.032320771 0.0065127002 0.010417766
3 0.0 0.022092608 0.0947136415 0.059574250
4 0.0 0.0 0.0745314819 0.679757189


layout(matrix(1:6,2,3,byrow=TRUE)); 
for (i in 1:6) {
image2(round(elast[,,i],2),
col = gray(seq(1,.4,-.1)))
}

Error in cut.default(log10(x), breaks) : 'breaks' are not unique


-- 
View this message in context: 
http://r.789695.n4.nabble.com/using-popbio-reduce-number-of-digits-in-image2-plot-tp3238622p3238622.html
Sent from the R help mailing list archive at Nabble.com.

__
R-help@r-project.org 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] Extracting the terms from an rpart object

Try this:

as.character(as.list(fit1$call)$data)

On Wed, Jan 26, 2011 at 4:12 PM, Tal Galili tal.gal...@gmail.com wrote:

 Another (similar) question,

 If I now want to know the name of the data argument used, is there an
 easy way for me to access it?

 I'm trying to use something like:
 eval(parse(text = all.vars(terms(fit1))[1]))

 Which (of course) wouldn't work, since the response variable is only
 available in the data used by rpart (specifically the kyphosis dataset)

 Thanks upfront.

 Tal




 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 --




 On Wed, Jan 26, 2011 at 7:40 PM, Henrique Dallazuanna www...@gmail.comwrote:

 Try this:

 all.vars(terms(fit1))
 all.vars(terms(fit2))


 On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili tal.gal...@gmail.com wrote:

 Hello all,

 I wish to extract the terms from an rpart object.
 Specifically, I would like to be able to know what is the response
 variable
 (so I could do some manipulation on it).
 But in general, such a method for rpart will also need to handle a .
 case
 (see fit2)

 Here are two simple examples:

 fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
 fit1$call
 fit2 - rpart(Kyphosis ~ ., data=kyphosis)
 fit2$call


 Is there anything prettier then using string manipulation?


 Thanks.





 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 --

[[alternative HTML version deleted]]

 __
 R-help@r-project.org 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.




 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O





-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

[[alternative HTML version deleted]]

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

[R] hmm.discnp hidden markov model

2011-01-26 Thread Qasim Javed

Hi all,

I am using a discrete Hidden Markov Model with discrete observations in
order to detect a sequence of integers. I am using the hmm.discnp package.

I am using the following code:

signature - c(-89, -98, -90, -84, -77, -75, -64, -60, -58, -55, -56, -57,
-57, -63, -77, -81, -82, -91, -85, -89, -93)

quant - length(-110:-6)

# Initialize and train the hmm with the observed sequence mentioned above.
# yval lists the possible values for observations and K is the number of
hidden states.
my_hmm - hmm(y=signature, yval=c(-110:-6), K=5)

print(my_hmm)


The above shows that the HMM was trained using signature and the values
seem to be intuitive.

My question is more a fundamental  one in regards to understanding HMMs. I
know I should use more examples of the above sequences to train the HMM in
order to make it more robust. Assuming, that the HMM is trained good enough,
I can use the viterbi algorithm to find the most probable sequence of hidden
states. However, what I really want to find out is whether a particular
observed sequence is modeled by my HMM (created above). There seems to be a
viterbi() function in hmm.discnp and also mps() but both of them give them
most probable hidden state sequence, whereas, I want the probability of a
particular observed sequence, that is, the likelihood for an arbitrary
observed sequence. This is typically solved using the solution to
Evaluation Problem in HMMs, but I do not see a function in hmm.discnp for
calculating this.

Am I missing something?

Thanks for the help.

[[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] Extracting the terms from an rpart object

Exactly what I needed Henrique,
Thank you.


Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Wed, Jan 26, 2011 at 8:26 PM, Henrique Dallazuanna www...@gmail.comwrote:

 Try this:

 as.character(as.list(fit1$call)$data)


 On Wed, Jan 26, 2011 at 4:12 PM, Tal Galili tal.gal...@gmail.com wrote:

 Another (similar) question,

 If I now want to know the name of the data argument used, is there an
 easy way for me to access it?

 I'm trying to use something like:
 eval(parse(text = all.vars(terms(fit1))[1]))

 Which (of course) wouldn't work, since the response variable is only
 available in the data used by rpart (specifically the kyphosis dataset)

 Thanks upfront.

 Tal




 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)

 --




 On Wed, Jan 26, 2011 at 7:40 PM, Henrique Dallazuanna 
 www...@gmail.comwrote:

 Try this:

 all.vars(terms(fit1))
 all.vars(terms(fit2))


 On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili tal.gal...@gmail.comwrote:

 Hello all,

 I wish to extract the terms from an rpart object.
 Specifically, I would like to be able to know what is the response
 variable
 (so I could do some manipulation on it).
 But in general, such a method for rpart will also need to handle a .
 case
 (see fit2)

 Here are two simple examples:

 fit1 - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
 fit1$call
 fit2 - rpart(Kyphosis ~ ., data=kyphosis)
 fit2$call


 Is there anything prettier then using string manipulation?


 Thanks.





 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew)
 |
 www.r-statistics.com (English)

 --

[[alternative HTML version deleted]]

 __
 R-help@r-project.org 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.




 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O





 --
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O


[[alternative HTML version deleted]]

__
R-help@r-project.org 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] Extracting the terms from an rpart object

2011-01-26 Thread William Dunlap

I would leave off the as.character().  With it you get
things like:
   f - rpart(Kyphosis ~ ., kyphosis[-3])
   as.list(f$call)$data
  kyphosis[-3]
   as.character(as.list(f$call)$data)
  [1] [kyphosis -3
Expressions like quote(kyphosis[-3]) are much
easier to analyze as expressions that as vectors
of character strings.  E.g., you can use all.vars
on an expression to see what variables are mentioned
in it.
 
When it is time to print the expression you may
need to use deparse() to turn it into a readable
vector of strings.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 




From: Tal Galili [mailto:tal.gal...@gmail.com] 
Sent: Wednesday, January 26, 2011 11:35 AM
To: Henrique Dallazuanna
Cc: r-help@r-project.org; William Dunlap
Subject: Re: [R] Extracting the terms from an rpart object


Exactly what I needed Henrique, 
Thank you.


Contact 
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) 
| www.r-statistics.com (English)

--





On Wed, Jan 26, 2011 at 8:26 PM, Henrique Dallazuanna 
www...@gmail.com wrote:


Try this:

as.character(as.list(fit1$call)$data) 


On Wed, Jan 26, 2011 at 4:12 PM, Tal Galili 
tal.gal...@gmail.com wrote:


Another (similar) question, 

If I now want to know the name of the data argument 
used, is there an easy way for me to access it?

I'm trying to use something like:
eval(parse(text = all.vars(terms(fit1))[1]))

Which (of course) wouldn't work, since the response 
variable is only available in the data used by rpart (specifically the 
kyphosis dataset)

Thanks upfront.


Tal




Contact 
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | 
www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English)

--





On Wed, Jan 26, 2011 at 7:40 PM, Henrique Dallazuanna 
www...@gmail.com wrote:


Try this:

all.vars(terms(fit1))
all.vars(terms(fit2))



On Wed, Jan 26, 2011 at 3:33 PM, Tal Galili 
tal.gal...@gmail.com wrote:


Hello all,

I wish to extract the terms from an 
rpart object.
Specifically, I would like to be able 
to know what is the response variable
(so I could do some manipulation on it).
But in general, such a method for rpart 
will also need to handle a . case
(see fit2)

Here are two simple examples:

fit1 - rpart(Kyphosis ~ Age + Number + 
Start, data=kyphosis)
fit1$call
fit2 - rpart(Kyphosis ~ ., 
data=kyphosis)
fit2$call


Is there anything prettier then using 
string manipulation?


Thanks.





Contact

[R] how could I choose subvector of a vector?

2011-01-26 Thread Akram Khaleghei Ghosheh balagh

Hello ;
How could I choose subvector of a vctor. for example if v=c(1,2,5,0,1), how
could I chose the (1,2) or (1,2,5).
thanks;

[[alternative HTML version deleted]]

__
R-help@r-project.org 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] Creating a new variable from existing ones

2011-01-26 Thread Melanie Zoelck

Hi,

I am relatively new to R and have a question regarding code. I have a data set 
which has data organised by location (site names, which are factors). I now 
want to add a new variable Region (this will be non numerical, as it will be 
names). Each region will contain locations. So for example:

Region: WestCoast
Locations in Region WestCoast: Tralee, Carradale and so on...

So each Region will contain different Locations, so that location can be nested 
within Region if needed.

I have found info on how to create different Age categories for example but my 
input is not numeric, but it is names or rather factors with a number of levels 
(which are place names).

I would be grateful for any assistance you may be able to provide.

Thank you!

Melanie Zoelck
86 Gleann Na Ri 
Renmore
Galway
Ireland

Tel.: +353 91 746086
Movil/Mobile: +353 857246196
E-Mail: mzoe...@mail.infocanarias.com
[[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] how could I choose subvector of a vector?



On Jan 26, 2011, at 2:42 PM, Akram Khaleghei Ghosheh balagh wrote:


Hello ;
How could I choose subvector of a vctor. for example if  
v=c(1,2,5,0,1), how

could I chose the (1,2) or (1,2,5).


?[

v[ c(1,2,5) ]


thanks;

[[alternative HTML version deleted]]

__
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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.


David Winsemius, MD
West Hartford, CT

__
R-help@r-project.org 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] Creating a new variable from existing ones



On Jan 26, 2011, at 2:06 PM, Melanie Zoelck wrote:


Hi,

I am relatively new to R and have a question regarding code. I have  
a data set which has data organised by location (site names, which  
are factors). I now want to add a new variable Region (this will be  
non numerical, as it will be names). Each region will contain  
locations. So for example:


Region: WestCoast
Locations in Region WestCoast: Tralee, Carradale and so on...


Presumably you have some sort of table with location and region  
and another table with perhaps persons and location. You would  
merge these two tables with ... wait for it  merge:


?merge


In the future it is considered more helpful to offer a tiny data  
example constructed with R code to allow more complete illustrations  
to be offered. Please read the Posting Guide.




So each Region will contain different Locations, so that location  
can be nested within Region if needed.


I have found info on how to create different Age categories for  
example but my input is not numeric, but it is names or rather  
factors with a number of levels (which are place names).


I would be grateful for any assistance you may be able to provide.

Thank you!

Melanie Zoelck
86 Gleann Na Ri
Renmore
Galway
Ireland



David Winsemius, MD
West Hartford, CT

__
R-help@r-project.org 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.

Re: [R] how could I choose subvector of a vector?



On Jan 26, 2011, at 2:48 PM, David Winsemius wrote:



On Jan 26, 2011, at 2:42 PM, Akram Khaleghei Ghosheh balagh wrote:


Hello ;
How could I choose subvector of a vctor. for example if  
v=c(1,2,5,0,1), how

could I chose the (1,2) or (1,2,5).


?[

v[ c(1,2,5) ]


I didn't notice at first that you wanted the values 1,2,5 ...(thought  
you wanted the 1st 2nd and 5th (obviously not correct with only a 4  
element vector)... so your values would be accessed with


v[1:3]





thanks;

[[alternative HTML version deleted]]


David Winsemius, MD
West Hartford, CT

__
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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] applying a set of rules to each row

2011-01-26 Thread KATSCHKE, ADRIAN CIV DFAS

All,

I would like to apply a set of rules to each row of the sample data set
below. The rule sets are the guidelines for determining an individual's
date for retirement eligibility. The rules are found in this document,
http://www.opm.gov/feddata/RetirementPaperFinal_v4.pdf. I am only
interested in the top two categories for retirement eligibility, the
CSRS and FERS plans. 

The data set has four variables Date of Birth (DOB), service computation
date (srvCompDT), retirement plan (retirePlan), and the age at which the
employee entered federal service (ageFedStart). The service computation
date is used to compute the date eligible for retirement. The retirement
plan indicates what system the employee is enrolled under.

The data does contain a few other retirement plans, for now I want to
just ignore those plans. I have labeled plans as 1-CSRS and 2-FERS, and
3-Other. My first attempt at applying the rules was through a complex
nesting of ifelse statements, this was not very successful and quite
difficult to follow. I then wrote a function and tried using apply
unsuccessfully. The function is shown below.

I would like to put a short script or function together that would allow
for an efficient application of the rules to each of the employees. I am
trying to avoid a loop, because my data set is quite large, and I may
need to update my data set regularly and re-run the analysis and reports
that will come from this work.

Any advice or guidance on building the function or code to apply the
rules would be quite helpful.

retireHelp -
structure(list(DOB = structure(c(-6642, -5134, -3444, -5598, 
-4356, 5737, -4894, -1951, -2950, 2467, 6945, 4908, -7930, -7236, 
-7727, -77, 4158, -7892, -6028, -7132, -5959, 2309, -2494, -3513, 
-383, -216, -3369, -5861, 3674, -10265, -8986, -5023, -4862, 
1526, -1022, 2175, -11790, -278, -7275, -5084, -1842, 430, -2220, 
-7444, 440, 4285, -7812, 3335, -7271, -6825, -1098, -1670, -10219, 
-7131, 5963, 704, -7662, 4219, -2813, 5147, -7334, -8223, -5922, 
-7497, -9276, -1291, -11640, -5631, 518, -7268, -2105, -5901, 
-690, -8146, -7059, 133, 1176, -6091, -2895, -6020, -4724, -3616, 
-5059, -8253, -2604, -12400, -4776, -3671, -9326, -7000, -5574, 
-3248, 4255, -1358, -6255, 8, -7115, -1701, -5227, 9, -517, -8674, 
-2554, -4069, -2077, -9872, -6534, 2970, -8307, -3020, -1343, 
-8897, -2304, -7424, 2078, -8274, -5559, -, -9262, -8473, 
-4088, -2429, -8006, -1091, 5015, 2765, 4036, 3101, -3743, 5103, 
-10018, -12095, -7646, -5966, -6208, -5784, -1325, -4288, -1665, 
-1409, 4685, -7881, -3413, 2738, -2201, 1217, -5113, 206, -1292, 
-1725, 10, -2978, -1895, -830, -105, -2395, -3496, -8244, -9956, 
-6494, -4678, -4077, 575, 2013, -3411, 3824, -4356, 4523, -5836, 
-6350, -5337, -41, -2001, -6632, -970, -6790, -2828, -4061, 476, 
5854, -9648, -4227, 850, 2619, -7747, -2672, 4069, -12618, -6898, 
-4178, -1772, -1643, -2064, -157, 4551, -8688, -6087, -2040, 
-7239, -783), format = m/d/y, origin = structure(c(1, 1, 1970
), .Names = c(month, day, year)), class = c(dates, times
)), srvCompDT = structure(c(743, 12429, 3585, 4364, 13227, 13578, 
13591, 8585, 9587, 13913, 14753, 13247, 2246, 1439, 8845, 7018, 
12625, -552, 5688, 7080, 13255, 13549, 12709, 13969, 13997, 9532, 
13689, 1226, 13549, 4093, 13423, 13801, 3181, 14809, 13353, 9457, 
7745, 8986, 4759, 4486, 6449, 11172, 8669, 3344, 13745, 12275, 
5081, 13605, 8006, 3048, 6330, 13521, 5254, 1733, 14095, 8516, 
4848, 13521, 5970, 14697, 8291, 139, 11435, 3567, 8961, 5775, 
3602, 1409, 11577, 12163, 12258, 13156, 9472, 7963, 1362, 10332, 
9557, 3997, 7509, 4691, 3133, 5877, 6782, 11449, 13283, 8040, 
11565, 3425, 7860, 1790, 10778, 13199, 12625, 5889, 3317, 9831, 
1068, 8040, 7123, 9104, 12836, 7928, 12764, 8922, 5324, -1004, 
1806, 10263, 5635, 10310, 5625, 8861, 14613, 3896, 10316, 5725, 
12751, 6113, 2997, 112, 5707, 4987, -1018, 8055, 13885, 13073, 
14585, 14865, 14935, 14390, 9735, 7654, 4557, 661, 1638, 1112, 
14011, 3086, 7032, 13942, 13325, 6735, 13900, 12673, 10148, 14193, 
14767, 8447, 6114, 10688, 13544, 7106, 8587, 14753, 7886, 12280, 
11946, 13662, 3332, 2108, 13977, 6203, 8369, 13857, 8369, 11486, 
8306, 12466, 12639, 7270, 4325, 13843, 14026, 14039, 6147, 7676, 
5781, 7038, 9187, 14640, 6174, 11491, 13913, 13787, 13465, 8854, 
13152, 1826, 1412, 4317, 5794, 5548, 8951, 12947, 12639, 5345, 
5961, 4637, 6465, 13717), format = m/d/y, origin = structure(c(1, 
1, 1970), .Names = c(month, day, year)), class = c(dates, 
times)), retirePlan = c(1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
1, 3, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 
2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 
2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 3, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 
3, 2, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 
1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 
2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
1, 2, 2,

[R] print() required sometimes in interactive use of console

2011-01-26 Thread Giles Crane


Surprising behavior:
Most R users are aware that print()
must be used inside functions to gain
output on the console.

Apparently, print() is sometimes required
when interactively using the console.
For example, the followingmay be
entered into the R console with different results.

sample(1:8,8)  #prints a permutation of 1 to 8

for(i in 1:5)   #does not
sample(1:8,8)



Cordially,
Giles Crane

__
R-help@r-project.org 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] applying a set of rules to each row

2011-01-26 Thread Bert Gunter

If I understand you correctly, you want ?ifelse, which works on the
full logical vectors of rules applied to the variables, not
ifelse, which works on only a single logical.

-- Bert Gunter

On Wed, Jan 26, 2011 at 12:18 PM, KATSCHKE, ADRIAN CIV DFAS
adrian.katsc...@dfas.mil wrote:
 All,

 I would like to apply a set of rules to each row of the sample data set
 below. The rule sets are the guidelines for determining an individual's
 date for retirement eligibility. The rules are found in this document,
 http://www.opm.gov/feddata/RetirementPaperFinal_v4.pdf. I am only
 interested in the top two categories for retirement eligibility, the
 CSRS and FERS plans.

 The data set has four variables Date of Birth (DOB), service computation
 date (srvCompDT), retirement plan (retirePlan), and the age at which the
 employee entered federal service (ageFedStart). The service computation
 date is used to compute the date eligible for retirement. The retirement
 plan indicates what system the employee is enrolled under.

 The data does contain a few other retirement plans, for now I want to
 just ignore those plans. I have labeled plans as 1-CSRS and 2-FERS, and
 3-Other. My first attempt at applying the rules was through a complex
 nesting of ifelse statements, this was not very successful and quite
 difficult to follow. I then wrote a function and tried using apply
 unsuccessfully. The function is shown below.

 I would like to put a short script or function together that would allow
 for an efficient application of the rules to each of the employees. I am
 trying to avoid a loop, because my data set is quite large, and I may
 need to update my data set regularly and re-run the analysis and reports
 that will come from this work.

 Any advice or guidance on building the function or code to apply the
 rules would be quite helpful.

 retireHelp -
 structure(list(DOB = structure(c(-6642, -5134, -3444, -5598,
 -4356, 5737, -4894, -1951, -2950, 2467, 6945, 4908, -7930, -7236,
 -7727, -77, 4158, -7892, -6028, -7132, -5959, 2309, -2494, -3513,
 -383, -216, -3369, -5861, 3674, -10265, -8986, -5023, -4862,
 1526, -1022, 2175, -11790, -278, -7275, -5084, -1842, 430, -2220,
 -7444, 440, 4285, -7812, 3335, -7271, -6825, -1098, -1670, -10219,
 -7131, 5963, 704, -7662, 4219, -2813, 5147, -7334, -8223, -5922,
 -7497, -9276, -1291, -11640, -5631, 518, -7268, -2105, -5901,
 -690, -8146, -7059, 133, 1176, -6091, -2895, -6020, -4724, -3616,
 -5059, -8253, -2604, -12400, -4776, -3671, -9326, -7000, -5574,
 -3248, 4255, -1358, -6255, 8, -7115, -1701, -5227, 9, -517, -8674,
 -2554, -4069, -2077, -9872, -6534, 2970, -8307, -3020, -1343,
 -8897, -2304, -7424, 2078, -8274, -5559, -, -9262, -8473,
 -4088, -2429, -8006, -1091, 5015, 2765, 4036, 3101, -3743, 5103,
 -10018, -12095, -7646, -5966, -6208, -5784, -1325, -4288, -1665,
 -1409, 4685, -7881, -3413, 2738, -2201, 1217, -5113, 206, -1292,
 -1725, 10, -2978, -1895, -830, -105, -2395, -3496, -8244, -9956,
 -6494, -4678, -4077, 575, 2013, -3411, 3824, -4356, 4523, -5836,
 -6350, -5337, -41, -2001, -6632, -970, -6790, -2828, -4061, 476,
 5854, -9648, -4227, 850, 2619, -7747, -2672, 4069, -12618, -6898,
 -4178, -1772, -1643, -2064, -157, 4551, -8688, -6087, -2040,
 -7239, -783), format = m/d/y, origin = structure(c(1, 1, 1970
 ), .Names = c(month, day, year)), class = c(dates, times
 )), srvCompDT = structure(c(743, 12429, 3585, 4364, 13227, 13578,
 13591, 8585, 9587, 13913, 14753, 13247, 2246, 1439, 8845, 7018,
 12625, -552, 5688, 7080, 13255, 13549, 12709, 13969, 13997, 9532,
 13689, 1226, 13549, 4093, 13423, 13801, 3181, 14809, 13353, 9457,
 7745, 8986, 4759, 4486, 6449, 11172, 8669, 3344, 13745, 12275,
 5081, 13605, 8006, 3048, 6330, 13521, 5254, 1733, 14095, 8516,
 4848, 13521, 5970, 14697, 8291, 139, 11435, 3567, 8961, 5775,
 3602, 1409, 11577, 12163, 12258, 13156, 9472, 7963, 1362, 10332,
 9557, 3997, 7509, 4691, 3133, 5877, 6782, 11449, 13283, 8040,
 11565, 3425, 7860, 1790, 10778, 13199, 12625, 5889, 3317, 9831,
 1068, 8040, 7123, 9104, 12836, 7928, 12764, 8922, 5324, -1004,
 1806, 10263, 5635, 10310, 5625, 8861, 14613, 3896, 10316, 5725,
 12751, 6113, 2997, 112, 5707, 4987, -1018, 8055, 13885, 13073,
 14585, 14865, 14935, 14390, 9735, 7654, 4557, 661, 1638, 1112,
 14011, 3086, 7032, 13942, 13325, 6735, 13900, 12673, 10148, 14193,
 14767, 8447, 6114, 10688, 13544, 7106, 8587, 14753, 7886, 12280,
 11946, 13662, 3332, 2108, 13977, 6203, 8369, 13857, 8369, 11486,
 8306, 12466, 12639, 7270, 4325, 13843, 14026, 14039, 6147, 7676,
 5781, 7038, 9187, 14640, 6174, 11491, 13913, 13787, 13465, 8854,
 13152, 1826, 1412, 4317, 5794, 5548, 8951, 12947, 12639, 5345,
 5961, 4637, 6465, 13717), format = m/d/y, origin = structure(c(1,
 1, 1970), .Names = c(month, day, year)), class = c(dates,
 times)), retirePlan = c(1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
 1, 3, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1,
 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1,
 2, 2, 2, 2, 2,

Re: [R] print() required sometimes in interactive use of console

2011-01-26 Thread Duncan Murdoch


On 26/01/2011 3:27 PM, Giles Crane wrote:

Surprising behavior:
Most R users are aware that print()
must be used inside functions to gain
output on the console.

Apparently, print() is sometimes required
when interactively using the console.
For example, the followingmay be
entered into the R console with different results.

sample(1:8,8)  #prints a permutation of 1 to 8

for(i in 1:5)   #does not
  sample(1:8,8)


What you say is right, but I think your mental model of what is going on 
is wrong.  The rule for printing is simple:  the result of an expression 
entered at the console is automatically printed unless it is marked 
invisible.


The result of for() {} is marked invisible, so it doesn't print.

The result of x - 1 is marked invisible, so it doesn't print.

Expressions in functions are not entered at the console, so they don't 
print.


You can explicitly mark something as invisible, and it won't print.  For 
example,


invisible(sample(1:8, 8))

won't print.  Though it doesn't use the R function invisible(), that's 
essentially what happens in a for loop:  it returns NULL, but marked 
invisible, so it doesn't print.  (I seem to recall that in older 
versions it would return the last value of the loop, but that was 
changed some time ago, or I'm thinking of some other language.)


There's a function called withVisible() that returns a value and its 
visibility; you can experiment with that to see what's going on.  For 
example,


 withVisible(x - 1)
$value
[1] 1

$visible
[1] FALSE

Duncan Murdoch

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Re: [R] applying a set of rules to each row

2011-01-26 Thread KATSCHKE, ADRIAN CIV DFAS

Yes. That is exactly what I would like to have running. Here is the first 
attempt I made at using a nested ?ifelse statement for one of the retirement 
plans. The variables are all there but with different names. ageYOSstart is 
ageFedStart, SCDCivLeave is srvCompDT. I haven't gotten this working. I am not 
sure that it is the correct way to do what I would like.

## Regular retirement eligibility date for FERS employees
retData.All$regRetireDT2[retData.All$retireSystem == FERS] - 
with(retData.All[retData.All$retireSystem == FERS,],
 ifelse(DOB  01/01/53, ## Born before 1953 
minimum retirement age of 55
ifelse(ageYOSstart  26, 
dates(DOB+(365.25*55)),
   ifelse((ageYOSstart = 26  
ageYOSstart  31), dates(SCDCivLeave*(365.25*30)),
  ifelse((ageYOSstart = 31 
 ageYOSstart  41), dates(DOB+(365.25*60)),
 
ifelse((ageYOSstart = 41  ageYOSstart  43),

dates(SCDCivLeave+(365.25*20)),

ifelse((ageYOSstart = 43  ageYOSstart  58),
   
dates(DOB+(365.25*62)),
   
ifelse(ageYOSstart = 58,

  dates(SCDCivLeave+(365.25*5)), NA)),
 ifelse((DOB  12/31/69  DOB  
01/01/53), ## Born between 1953 and 1969 MRA of 56
 ifelse(ageYOSstart  27, 
dates(DOB+(365.25*56)),
ifelse((ageYOSstart = 
27  ageYOSstart  31),

dates(SCDCivLeave+(365.25*30)),

ifelse((ageYOSstart = 31  ageYOSstart  41),
   
dates(DOB+(365.25*60)),
   
ifelse((ageYOSstart = 41  ageYOSstart  43),
  
dates(SCDCivLeave+(365.25*20)),
  
ifelse((ageYOSstart = 43  ageYOSstart  58),

 dates(DOB+(365.25*62)),

 ifelse(ageYOSstart = 58,

dates(SCDCivLeave+(365.25*5)),

NA))),
 ifelse(DOB = 01/01/69, ## Born 
after 1969 Min Retire Age of 57
ifelse(ageYOSstart  28, 
dates(DOB+(365.25*57)),
   ifelse((ageYOSstart = 
28  ageYOSstart  31),
  
dates(SCDCivLeave+(365.25*30)),
  
ifelse((ageYOSstart = 31  ageYOSstart  41),
 
dates(DOB+(365.25*20)),
 
ifelse((ageYOSstart = 41  ageYOSstart  43),

dates(SCDCivLeave+(365.25*20)),

ifelse((ageYOSstart = 43  ageYOSstart  57),

   dates(DOB+(365.25*62)),

   ifelse(ageYOSstart = 58,

  dates(SCDCivLeave+(365.25*5)),

  NA))), NA))

Adrian 


If I understand you correctly, you want ?ifelse, which works on the
full logical vectors of rules applied to the variables, not
ifelse, which works on only a single logical.

-- Bert Gunter

On Wed, Jan 26, 2011 at 12:18 PM, KATSCHKE, ADRIAN CIV DFAS
adrian.katsc...@dfas.mil wrote:
 All,

 I would like to apply a set of rules to each row of the sample data set
 below. The rule sets are the guidelines for determining an individual's
 date for retirement eligibility. The rules are found in this document,

Re: [R] applying a set of rules to each row

I remember something about the degree of nesting of ifelse calls being  
limited to 7 deep (???)  that makes me worry about this approach. You  
may want to look at the arules package or the data.table package or  
the sqldf package for approaches that are specifically constructed  
with this sort of processing in mind.


--
David.

On Jan 26, 2011, at 3:42 PM, KATSCHKE, ADRIAN CIV DFAS wrote:

Yes. That is exactly what I would like to have running. Here is the  
first attempt I made at using a nested ?ifelse statement for one of  
the retirement plans. The variables are all there but with different  
names. ageYOSstart is ageFedStart, SCDCivLeave is srvCompDT. I  
haven't gotten this working. I am not sure that it is the correct  
way to do what I would like.


## Regular retirement eligibility date for FERS employees
retData.All$regRetireDT2[retData.All$retireSystem == FERS] -  
with(retData.All[retData.All$retireSystem == FERS,],
ifelse(DOB  01/01/53, ## Born  
before 1953 minimum retirement age of 55
   ifelse(ageYOSstart  26,  
dates(DOB+(365.25*55)),
  ifelse((ageYOSstart =  
26  ageYOSstart  31), dates(SCDCivLeave*(365.25*30)),
  
ifelse((ageYOSstart = 31  ageYOSstart  41), dates(DOB+(365.25*60)),
 
ifelse((ageYOSstart = 41  ageYOSstart  43),

dates(SCDCivLeave+(365.25*20)),

ifelse((ageYOSstart = 43  ageYOSstart  58),
  dates 
(DOB+(365.25*62)),
  ifelse 
(ageYOSstart = 58,
 dates 
(SCDCivLeave+(365.25*5)), NA)),
ifelse((DOB  12/31/69   
DOB  01/01/53), ## Born between 1953 and 1969 MRA of 56
ifelse(ageYOSstart   
27, dates(DOB+(365.25*56)),

ifelse((ageYOSstart = 27  ageYOSstart  31),

dates(SCDCivLeave+(365.25*30)),

ifelse((ageYOSstart = 31  ageYOSstart  41),
  dates 
(DOB+(365.25*60)),
  ifelse 
((ageYOSstart = 41  ageYOSstart  43),
 dates 
(SCDCivLeave+(365.25*20)),
 ifelse 
((ageYOSstart = 43  ageYOSstart  58),
dates 
(DOB+(365.25*62)),
ifelse 
(ageYOSstart = 58,
   dates 
(SCDCivLeave+(365.25*5)),
   NA 
))),
ifelse(DOB = 01/01/69, ##  
Born after 1969 Min Retire Age of 57
   ifelse(ageYOSstart   
28, dates(DOB+(365.25*57)),
   
ifelse((ageYOSstart = 28  ageYOSstart  31),
  
dates(SCDCivLeave+(365.25*30)),
  
ifelse((ageYOSstart = 31  ageYOSstart  41),
 
dates(DOB+(365.25*20)),
 
ifelse((ageYOSstart = 41  ageYOSstart  43),
   dates 
(SCDCivLeave+(365.25*20)),
   ifelse 
((ageYOSstart = 43  ageYOSstart  57),
  dates 
(DOB+(365.25*62)),
  ifelse 
(ageYOSstart = 58,
 dates 
(SCDCivLeave+(365.25*5)),
 NA 
))), NA))


Adrian



If I understand you correctly, you want ?ifelse, which works on the
full logical vectors of rules applied to the variables, not
ifelse,

Re: [R] NA replacing

2011-01-26 Thread andrija djurovic

Thanks Dennis this is what I was looking for.



On Wed, Jan 26, 2011 at 4:14 AM, Dennis Murphy djmu...@gmail.com wrote:

 Hi:

 Here's one approach:

 f - function(df) {
  rs - with(na.exclude(df), tapply(y, strata, sum)/tapply(x, strata,
 sum))
  u - transform(subset(df, is.na(y)), y = x * rs[strata])
  transform(df, y = replace(y, u$id, u$y))
   }

 f(df)

 The function works as follows:

 (1) With the rows of the data frame where y is not missing,
  find the sum(y)/sum(x) ratio in each stratum. rs is a vector whose
 length is
  the number of strata. (Hopefully, all of your x-sums are nonzero...)
 If you have
  missing x values in your real data, you need to think about how you
 want to
  handle them.
 (2) In a sub-data frame u containing the missing y's, replace them with the
 value of
  x times the value of rs corresponding to its stratum.
 (3) Replace the missing y's in df with the y's from u, matching on id
 numbers. (This
  is a by-product of subset(), BTW.)

 HTH,
 Dennis

 On Tue, Jan 25, 2011 at 9:40 AM, andrija djurovic djandr...@gmail.comwrote:

 Hello R user,

 I have following data frame:

 df=data.frame(id=c(1:10),strata=rep(c(1,2),c(5,5)),y=c(
 10,12,10,NA,15,70,NA,NA,55,100),x=c(3,4,5,7,4,10,12,8,3,15))

 and I would like to replace NA's with:

 instead of first NA
  tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[1]*
 *7 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[1]
 where 7 is the value of x (id=4) in strata 1 where y=NA

 instead of second NA
 tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[2]*
 *12 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[2]
 where 12 is the value of x (id=7) in strata 2 where y=NA

 instead of third NA tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[2]*
 *
 8 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[2]
 where 8 is the value of x(id=8) in strata 2 where y=NA.

 So, I would like to replace NA inside the stratas on above explained way.

 Does anyone know how to do this?

 thanks in advance

 Andrija

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[[alternative HTML version deleted]]

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[R] Can not invoke maxent() of library(dismo) in Mac OSX

2011-01-26 Thread Mao Jianfeng

Dear R-helpers,

I can not invoke maxent() in Mac OSX. Could you give me any directions
on that? Thank you in advance.

Here is my info:

# (1) the error
  me - maxent(predictors, occtrain, factors='biome')
me - maxent(predictors, occtrain, factors='biome')
Error in .jcall(mxe, S, fit, c(autorun, -e, afn, -o, dirout,  :
  java.lang.NoClassDefFoundError: Could not initialize class
density.DirectorySelect

# (2) the variables for maxent: predictors, and occtrain
 summary(predictors)
summary(predictors)
Cells:  35712
NAs  :  0 0 0 0 0 0 0 0 0

1  2  3  4 56 7 8  9
Min.-23.00.00.00.0  97.0 -207  64.0 -66.0  1.000
1st Qu. 171.0  715.5  317.0   30.0 306.0   29 132.0 213.0  1.000
Median  230.0 1271.0  513.0   87.0 322.0  131 183.0 249.0  4.000
Mean209.3 1364.0  570.3  136.3 314.4  103 210.4 225.9  5.135
3rd Qu. 257.0 1879.0  834.0  216.0 336.0  188 275.0 262.0  8.000
Max.289.0 7091.0 2458.0 1496.0 422.0  242 449.0 322.0 14.000
summary based on a sample of 5000 cells, which is 14.0008960573477 %
of all cells
 summary(occtrain)
summary(occtrain)
  lon  lat
 Min.   :-85.93   Min.   :-23.450
 1st Qu.:-79.47   1st Qu.: -2.750
 Median :-75.58   Median :  5.300
 Mean   :-72.22   Mean   :  2.823
 3rd Qu.:-65.40   3rd Qu.:  9.150
 Max.   :-46.73   Max.   : 13.950

# (3) my OS and R version
 version
version
   _
platform   x86_64-apple-darwin9.8.0
arch   x86_64
os darwin9.8.0
system x86_64, darwin9.8.0
status
major  2
minor  12.1
year   2010
month  12
day16
svn rev53855
language   R
version.string R version 2.12.1 (2010-12-16)
 sessionInfo()
sessionInfo()
R version 2.12.1 (2010-12-16)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

locale:
[1] en_US/en_US/en_US/C/en_US/en_US

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

other attached packages:
[1] evaluate_0.3  dismo_0.5-14  rJava_0.8-8   raster_1.7-29 sp_0.9-76

loaded via a namespace (and not attached):
[1] grid_2.12.1 lattice_0.19-17 plyr_1.4stringr_0.4
[5] tools_2.12.1

-- 
Jian-Feng, Mao

the Institute of Botany,
Chinese Academy of Botany,

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Re: [R] applying a set of rules to each row

2011-01-26 Thread Bert Gunter

... or perhaps just break things up with assignments and do it in stages.

-- Bert

On Wed, Jan 26, 2011 at 12:52 PM, David Winsemius
dwinsem...@comcast.net wrote:
 I remember something about the degree of nesting of ifelse calls being
 limited to 7 deep (???)  that makes me worry about this approach. You may
 want to look at the arules package or the data.table package or the sqldf
 package for approaches that are specifically constructed with this sort of
 processing in mind.

 --
 David.

 On Jan 26, 2011, at 3:42 PM, KATSCHKE, ADRIAN CIV DFAS wrote:

 Yes. That is exactly what I would like to have running. Here is the first
 attempt I made at using a nested ?ifelse statement for one of the retirement
 plans. The variables are all there but with different names. ageYOSstart is
 ageFedStart, SCDCivLeave is srvCompDT. I haven't gotten this working. I am
 not sure that it is the correct way to do what I would like.

 ## Regular retirement eligibility date for FERS employees
 retData.All$regRetireDT2[retData.All$retireSystem == FERS] -
 with(retData.All[retData.All$retireSystem == FERS,],
                                ifelse(DOB  01/01/53, ## Born before
 1953 minimum retirement age of 55
                                       ifelse(ageYOSstart  26,
 dates(DOB+(365.25*55)),
                                              ifelse((ageYOSstart = 26 
 ageYOSstart  31), dates(SCDCivLeave*(365.25*30)),
                                                     ifelse((ageYOSstart =
 31  ageYOSstart  41), dates(DOB+(365.25*60)),

  ifelse((ageYOSstart = 41  ageYOSstart  43),

 dates(SCDCivLeave+(365.25*20)),

 ifelse((ageYOSstart = 43  ageYOSstart  58),

  dates(DOB+(365.25*62)),

  ifelse(ageYOSstart = 58,

       dates(SCDCivLeave+(365.25*5)), NA)),
                                        ifelse((DOB  12/31/69  DOB 
 01/01/53), ## Born between 1953 and 1969 MRA of 56
                                                ifelse(ageYOSstart  27,
 dates(DOB+(365.25*56)),
                                                       ifelse((ageYOSstart
 = 27  ageYOSstart  31),

 dates(SCDCivLeave+(365.25*30)),

 ifelse((ageYOSstart = 31  ageYOSstart  41),

  dates(DOB+(365.25*60)),

  ifelse((ageYOSstart = 41  ageYOSstart  43),

   dates(SCDCivLeave+(365.25*20)),

   ifelse((ageYOSstart = 43  ageYOSstart  58),

          dates(DOB+(365.25*62)),

          ifelse(ageYOSstart = 58,

                 dates(SCDCivLeave+(365.25*5)),

                 NA))),
                                        ifelse(DOB = 01/01/69, ## Born
 after 1969 Min Retire Age of 57
                                               ifelse(ageYOSstart  28,
 dates(DOB+(365.25*57)),
                                                      ifelse((ageYOSstart
 = 28  ageYOSstart  31),

 dates(SCDCivLeave+(365.25*30)),

 ifelse((ageYOSstart = 31  ageYOSstart  41),

  dates(DOB+(365.25*20)),

  ifelse((ageYOSstart = 41  ageYOSstart  43),

 dates(SCDCivLeave+(365.25*20)),

 ifelse((ageYOSstart = 43  ageYOSstart  57),

        dates(DOB+(365.25*62)),

        ifelse(ageYOSstart = 58,

               dates(SCDCivLeave+(365.25*5)),

               NA))), NA))

 Adrian


 If I understand you correctly, you want ?ifelse, which works on the
 full logical vectors of rules applied to the variables, not
 ifelse, which works on only a single logical.

 -- Bert Gunter

 On Wed, Jan 26, 2011 at 12:18 PM, KATSCHKE, ADRIAN CIV DFAS
 adrian.katsc...@dfas.mil wrote:

 All,

 I would like to apply a set of rules to each row of the sample data set
 below. The rule sets are the guidelines for determining an individual's
 date for retirement eligibility. The rules are found in this document,
 http://www.opm.gov/feddata/RetirementPaperFinal_v4.pdf. I am only
 interested in the top two categories for retirement eligibility, the
 CSRS and FERS plans.

 The data set has four variables Date of Birth (DOB), service computation
 date (srvCompDT), retirement plan (retirePlan), and the age at which the
 employee entered federal service (ageFedStart). The service computation
 date is used to compute the date eligible for retirement. The retirement
 plan indicates what system the employee is enrolled under.

 The data does contain a few other retirement plans, for now I want to
 just ignore those plans. I have labeled plans as 1-CSRS and 2-FERS, and
 3-Other. My first attempt at applying the rules was through a complex
 nesting of ifelse statements, this was not very successful and quite
 difficult to follow. I then wrote a function and tried using apply
 unsuccessfully. The function is shown below.

 I would like to put a short script or function together that would allow
 for an efficient application of the rules to each of the employees. I am
 trying to avoid a loop, because my data set is quite large, and I may
 need to update my data set regularly and re-run the analysis and reports
 that will come from this work.

 Any advice or guidance on building the function or code to apply the

Re: [R] Sweave: \Sexpr{} inside ?

2011-01-26 Thread zerfetzen


Thanks Duncan, that helps.  It successfully displays what I'm looking for,
but it is not executing it.  In a previous code chunk, it notes the time it
took to run something, and in the successive code chunk, it runs something
else where the previous time is now a parameter, but I'd like it to
numerically display that previous time in the new chunk, rather than a
variable name I create for it behind the scenes.  Is this possible?  Many
thanks.
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Sweave-Sexpr-inside-tp3237313p3238764.html
Sent from the R help mailing list archive at Nabble.com.

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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] baseline hazard function

2011-01-26 Thread Simon Kiss

Dear colleagues, I have the following dataset.  It is modelled on the data 
included in Box-Seteffenheiser and Jones Event History Modelling
Using the following code, I try to find the baseline hazard function 

haz_1-muhaz(bpa$time, bpa$censored, subset=(bpa$year==2010 | bpa$ban==1), 
min.time=1, max.time=3)

I think I'm doing everything right, but what I don't understand is how to 
derive a duration dependency coefficient rom the values contained in the muhaz 
object as per Box-Steffenheiser and Jones' recommendations in Ch. 5 of Event 
History Modelling.

I get the following summary(haz_1)
Number of Observations .. 50
Censored Observations ... 43
Method used . Local
Boundary Correction Type  Left and Right
Kernel type . Epanechnikov
Minimum Time  1
Maximum Time  3
Number of minimization points ... 51
Number of estimation points . 101
Pilot Bandwidth . 0.25
Smoothing Bandwidth . 1.27
Minimum IMSE  6716.9


Can anyone provide any advice?
Yours, Simon Kiss

'data.frame':   147 obs. of  7 variables:
 $ state   : Factor w/ 50 levels Alabama,Alaska,..: 1 1 1 2 2 2 3 3 3 4 ...
 $ partisan: Factor w/ 3 levels democrat,mixed,..: 1 1 1 2 2 2 3 3 3 1 ...
 $ ban : num  0 0 0 0 0 0 0 0 0 0 ...
 $ year: num  2008 2009 2010 2008 2009 ...
 $ news: num  1.67 1.67 0 2 0 ...
 $ time: num  1 2 3 1 2 3 1 2 3 1 ...
 $ censored: num  0 0 0 0 0 0 0 0 0 0 ...


*
Simon J. Kiss, PhD
Assistant Professor, Wilfrid Laurier University
73 George Street
Brantford, Ontario, Canada
N3T 2C9
Cell: +1 519 761 7606

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[R] [R-pkgs] RGtk2Extras package for dataframe editing and easy dialog creation

2011-01-26 Thread Taverner, Thomas

Dear useRs,

This is to announce the RGtk2Extras R package is available on CRAN in version 
0.5.1.

This package provides useful extras for R programmers who wish to create 
graphic user interfaces. It is based on GTK, using Michael Lawrence's RGtk2 
package and John Verzani's gWidgets, and some ideas from John Verzani's traitr 
package.

The first major feature of RGtk2Extras, the run.dialog function, is an 
interface for creating simple front-ends to R functions using a terse markup 
scheme. No GUI knowledge is required; if you can write an R function, you can 
create a dialog for it.

An example:

# A function with one argument, N

  Histogram = function(N) hist(rnorm(N))

# Dialog markup list for the function Histogram
# Create the main dialog label with the first label= (optional)
# Then specify N.integerItem=50, where 
#   (1) N is the function argument
#   (2) integerItem is the type of widget to use, see ?run.dialog
#   (3) the value 50 is the default
# Then with the second label= add a label for the N item (optional)

  Histogram.dialog = list(label = Histogram of N points,
N.integerItem = 50, label = Value of N)
  
# Run the dialog.
# The returned list has elements args for dialog arguments
# and retval for the return value.
# With auto.assign=TRUE, the return value is stored in Histogram_output.
# run.dialog also does error handling, interrupts and an optional 
# progress bar dialog for long running tasks, see ?run.dialog

  a = run.dialog(Histogram)

A more complex demo example, thanks to Graham Williams:

  demo(MakeAngle)


The second feature of RGtk2Extras is gtkDfEdit, an editable spreadsheet widget 
designed for editing data matrices and data frames. It's also based on RGtk2. 
This was released earlier as RGtk2DfEdit which was then folded into RGtk2Extras.

The gtkDfEdit spreadsheet is quite full featured and has been designed to be 
familiar to Excel users, while allowing most of the data frames and factor 
operations that are possible from the R command line.

Data frame columns are type-aware and there is a factor editor. Most functions 
can be accessed from right clicking on data columns or cells. There is undo, 
cross platform copy/paste, sorting, cell filling, and data loading 
capabilities. See ?gtkDfEdit.

The widget provided by gtkDfEdit() can be integrated into larger RGtk2 based 
user interfaces, or its standalone wrapper dfedit() can be used as a straight 
replacement for edit.data.frame().

There is also a basic API for manipulating and binding event handlers to the 
spreadsheet. See ?gtkDfEditDoTask and ?gtkDfEditSetActionHandler.

# Edit the iris data frame
# Note the blank row at the bottom to allow pasting into rows below.
x = dfedit(iris)

# Example user interactions
1. Right click the Species column or column header to see or change the data 
type. Click Edit Factors to change factor levels or ordering.

2. Right click a column header and then click Sort to sort columns

3. Right click column header and change data type by selecting 
Character/Integer etc. Factors can be changed to integer levels (To Factor 
Levels) or integers (To Factor Ordering)

4. Right click column header to insert or delete columns or change column name.

5. Select a submatrix of numbers within iris and press the = key to bring 
up the Command Editor, then type hist in the command field and OK to create 
a histogram of the submatrix; function(x) hist(x) works as well.

6. Select a submatrix within iris and press Ctrl-V or right-click and select 
Copy to copy into an external spreadsheet; Ctrl-C works to paste into the data 
frame. Ctrl-Shift-C copies submatrix and row and column names. (The equivalent 
Mac keys should also work.)

7. Select a range of cells within Species column and right-click to select 
Fill in Cycles to fill cyclic factors.

8. Right click left hand corner cell to open CSV file into editor or save as 
file.

9. Right click left hand corner cell and Default Columns to set columns to 
default Excel-style column headers.

10. Ctrl-Z to undo previous edits.
# End of examples

RGtk2Extras is still in a beta stage of development. It is hosted on 
r-forge.r-project.org/R/?group_id=924. Comments and suggestions appreciated; 
email thomas.taverner _AT_ pnl.gov

Thanks to the entire R community, particularly the R Development Core Team, 
Michael Lawrence and John Verzani, and also Graham Williams and Iago Conde for 
comments.

Tom

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Re: [R] Sweave: \Sexpr{} inside ?

2011-01-26 Thread Duncan Murdoch


On 11-01-26 3:43 PM, zerfetzen wrote:


Thanks Duncan, that helps.  It successfully displays what I'm looking for,
but it is not executing it.  In a previous code chunk, it notes the time it
took to run something, and in the successive code chunk, it runs something
else where the previous time is now a parameter, but I'd like it to
numerically display that previous time in the new chunk, rather than a
variable name I create for it behind the scenes.  Is this possible?  Many
thanks.


Of course it's possible, but it'll need ugly trickery.  Just do every 
calculation in a hidden chunk, then use code like I did to fake a 
display of it.


Duncan Murdoch

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Re: [R] Making up a graph and its equation which better fits two groups of data

2011-01-26 Thread Greg Snow

Your faith in our ability to read your mind is apparently much higher than our 
actual ability to do so.  What is the nature of your data? what question are 
you trying to answer? What type of equation do you want? What do you mean by 
better?  Better than what?

Maybe the esp package (pre-alpha) can give some additional insights:

 source('g:/R/esp/esp.R')
 esp()
[1] dogie perm syllable brunt wore majestic Messiah luxuriously meretricious


OK, that doesn't help me much, maybe someone else can get something out of 
that, but in the meantime I would suggest reading the posting guide and adding 
the info it suggests to make a question that we have a chance of giving a 
meaningful answer to.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Jimmy Martina
 Sent: Wednesday, January 26, 2011 9:05 AM
 To: R
 Subject: [R] Making up a graph and its equation which better fits two
 groups of data
 
 
 Dear R-folks:
 
 I have a group of data ('x' and 'y' axis), but I'd like to know how to
 draw a graph which would fits my data in a better way, I also need its
 equation. Could you give me any R-rutine ideas?.
 
 I thank you in advance for your kind support.
 
   [[alternative HTML version deleted]]
 
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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
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[R] Trouble variable scoping a function writing with get()

2011-01-26 Thread Pat Schmitz

I am having trouble with variable scoping inside/outside functions.  I
want to use get() to grab a named and quoted variable as an input to a
function, however the function can't find the variable when it is entered
into the function call, only when it is named in the main environment.

I obviously am not so clear on variable scoping, and ?get really doesn't
clear up my confusion.

I am sure that there are better, more appropriate ways to write a
function...

Please enlighten me,
Pat

# Example code

dat - expand.grid(a = factor(c(a, b)), b=1:10)

# Function that requires get()
ex - function(data, response){
 library(plyr)
 output - ddply(data,
.(a),
summarize,
 res = sum(get(response))
)
return(output)
}


out - ex(data = dat, response = b)
# Error in get(response) : object 'response' not found

# However if I name reponse outside of the function, it is found by the
function
response = b
out - ex(data = dat, response = b)
out
# out
#  a res
#1 a  55
#2 b  55


-- 
Patrick Schmitz
Graduate Student
Plant Biology
1206 West Gregory Drive
RM 1500

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Re: [R] Trouble variable scoping a function writing with get()

2011-01-26 Thread Bert Gunter

It's looking for an object named b in the frame of the function.
There is none. You need to specify

get(response, pos = parent.frame())

## or maybe even

get(response, pos= globalenv())

HOWEVER, this is **exactly** why you shouldn't do this! Instead of
passing in the name of the object, pass in the object itself (response
= b, not response = b) and forget about using get(). Then it
**would** work as writted without getting tangled up in scoping. The
whole point local (to the function) scope and call by value is to
avoid exactly the sort of mess that you've gotten yourself into. Stick
with R's programming paradigms in R rather than trying to impose
others and life will be simpler and sweeter.

-- Bert



On Wed, Jan 26, 2011 at 2:21 PM, Pat Schmitz pschm...@illinois.edu wrote:
 I am having trouble with variable scoping inside/outside functions.  I
 want to use get() to grab a named and quoted variable as an input to a
 function, however the function can't find the variable when it is entered
 into the function call, only when it is named in the main environment.

 I obviously am not so clear on variable scoping, and ?get really doesn't
 clear up my confusion.

 I am sure that there are better, more appropriate ways to write a
 function...

 Please enlighten me,
 Pat

 # Example code

 dat - expand.grid(a = factor(c(a, b)), b=1:10)

 # Function that requires get()
 ex - function(data, response){
  library(plyr)
  output - ddply(data,
 .(a),
 summarize,
  res = sum(get(response))
 )
 return(output)
 }


 out - ex(data = dat, response = b)
 # Error in get(response) : object 'response' not found

 # However if I name reponse outside of the function, it is found by the
 function
 response = b
 out - ex(data = dat, response = b)
 out
 # out
 #  a res
 #1 a  55
 #2 b  55


 --
 Patrick Schmitz
 Graduate Student
 Plant Biology
 1206 West Gregory Drive
 RM 1500

        [[alternative HTML version deleted]]

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 R-help@r-project.org 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.




-- 
Bert Gunter
Genentech Nonclinical Biostatistics

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[R] boxplot - code for labeling outliers - any suggestions for improvements?