date:20121205

Hi,
May be this helps:
dat1-read.table(text=
1 2006-11  NaN
2 2006-12  NaN
3 2006-10 0.1577647
4 2006-11  NaN
5 2006-12  NaN
6 2007-01  NaN
7 2007-02  NaN
8 2007-03 0.2956429
9 2007-01  NaN
10 2007-02  NaN
,sep=,header=FALSE,stringsAsFactors=FALSE)
res-dat1[seq(which(!is.na(dat1$V3))[1],which(!is.na(dat1$V3))[2],by=1),]
 res
#  V1  V2    V3
#3  3 2006-10 0.1577647
#4  4 2006-11   NaN
#5  5 2006-12   NaN
#6  6 2007-01   NaN
#7  7 2007-02   NaN
#8  8 2007-03 0.2956429
A.K.



- Original Message -
From: Vasilchenko Aleksander vasilchenko@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Wednesday, December 5, 2012 5:46 AM
Subject: [R] Trim

Hello,
I have a dataframe

1 2006-11       NaN
2 2006-12       NaN
3 2006-10 0.1577647
4 2006-11       NaN
5 2006-12       NaN
6 2007-01       NaN
7 2007-02       NaN
8 2007-03 0.2956429
9 2007-01       NaN
10 2007-02       NaN
I need to trim first and last NaN rows
Result -
1 2006-10 0.1577647
2 2006-11       NaN
3 2006-12       NaN
4 2007-01       NaN
5 2007-02       NaN
6 2007-03 0.2956429
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.


__
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] Makehelpneeded:No rule to make target `w'. Stop

2012-12-05 Thread Revathi Ramanadham

Hello,

I am Trying to Build a project, Makefile written so that outer make file calls 
the inner Makefiles to Build,
I am invoking the Makefile as  gmake depend  at the perent directory of project 
(Makefile here digs in and executes inner Makefiles to build the project)

.DEFAULT:
@for subdir in `ls */Makefile`; \
  do \
echo + Making $@ in `dirname $$subdir` ...; \
(cd `dirname $$subdir`; $(MAKE) $@); \
  done


the automatic variable $@ is set to depend, till some inner directories make 
process is going well and good ,after  some point after I could see the error 
Message as  *** No rule to make target `w'.  Stop.,
with google help I found reasons like

1.infinite recursion
 2.The value of $@ is set to w, as target now is w that won't be found in the 
Makefile so causing such an error.


Ø  I have gmake output I could see make entering in to directories only once 
and leaving from directories only once, so error is not because of recursion.

Ø  I could see @ := wp  when I issued gmake -p depend.

Could you suggest me the approach to solve this issue

Regards,
Revathi R




DISCLAIMER\ ==\ This e-mail ...{{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.

Re: [R] Trim



Hi,
I guess with only one rev() should also work:
 idx-cumsum(!is.na(dat$V2))*rev(cumsum(!is.na(dat$V2)))!=0
 dat[idx,]
#   V1    V2
#3 2006-10 0.1577647
#4 2006-11   NaN
#5 2006-12   NaN
#6 2007-01   NaN
#7 2007-02   NaN
#8 2007-03 0.2956429
A.K.


- Original Message -
From: Rui Barradas ruipbarra...@sapo.pt
To: Vasilchenko Aleksander vasilchenko@gmail.com
Cc: r-help@r-project.org
Sent: Wednesday, December 5, 2012 7:59 AM
Subject: Re: [R] Trim

Hello,

Try the following.

dat -
structure(list(V1 = structure(c(2L, 3L, 1L, 2L, 3L, 4L, 5L, 6L,
4L, 5L), .Label = c(2006-10, 2006-11, 2006-12, 2007-01,
2007-02, 2007-03), class = factor), V2 = c(NaN, NaN, 0.1577647,
NaN, NaN, NaN, NaN, 0.2956429, NaN, NaN)), .Names = c(V1, V2
), class = data.frame, row.names = c(1, 2, 3, 4, 5,
6, 7, 8, 9, 10))

idx - cumsum(!is.nan(dat$V2)) * rev(cumsum(rev(!is.nan(dat$V2 != 0
dat[idx, ]


Hope this helps,

Rui Barradas
Em 05-12-2012 10:46, Vasilchenko Aleksander escreveu:
 Hello,
 I have a dataframe

 1 2006-11       NaN
 2 2006-12       NaN
 3 2006-10 0.1577647
 4 2006-11       NaN
 5 2006-12       NaN
 6 2007-01       NaN
 7 2007-02       NaN
 8 2007-03 0.2956429
 9 2007-01       NaN
 10 2007-02       NaN
 I need to trim first and last NaN rows
 Result -
 1 2006-10 0.1577647
 2 2006-11       NaN
 3 2006-12       NaN
 4 2007-01       NaN
 5 2007-02       NaN
 6 2007-03 0.2956429
 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.

__
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-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] Trim

2012-12-05 Thread Berend Hasselman


On 05-12-2012, at 11:46, Vasilchenko Aleksander wrote:

 Hello,
 I have a dataframe
 
 1 2006-11   NaN
 2 2006-12   NaN
 3 2006-10 0.1577647
 4 2006-11   NaN
 5 2006-12   NaN
 6 2007-01   NaN
 7 2007-02   NaN
 8 2007-03 0.2956429
 9 2007-01   NaN
 10 2007-02   NaN
 I need to trim first and last NaN rows
 Result -
 1 2006-10 0.1577647
 2 2006-11   NaN
 3 2006-12   NaN
 4 2007-01   NaN
 5 2007-02   NaN
 6 2007-03 0.2956429

 dat -
+ structure(list(V1 = structure(c(2L, 3L, 1L, 2L, 3L, 4L, 5L, 6L,
+ 4L, 5L), .Label = c(2006-10, 2006-11, 2006-12, 2007-01,
+ 2007-02, 2007-03), class = factor), V2 = c(NaN, NaN, 0.1577647,
+ NaN, NaN, NaN, NaN, 0.2956429, NaN, NaN)), .Names = c(V1, V2
+ ), class = data.frame, row.names = c(1, 2, 3, 4, 5,
+ 6, 7, 8, 9, 10))

library(zoo)
na.trim(dat)

Berend

__
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] [R-pkgs] glm2 version 1.1

2012-12-05 Thread Ian Marschner

glm2 (1.1) is now available on CRAN.

glm2 fits generalized linear models using the same model specification
as glm, but with a modified fitting method that is more stable for
models that may fail to converge using glm (see: R Journal 3/2, 2011,
pp.12-15).

The previous version of glm2 had to be archived because it called a
Fortran routine no longer available in R = 2.15.1. This has been
addressed in the current version of glm2.

---
Ian Marschner
Professor, Dept. of Statistics
Macquarie University, Sydney,
Australia

___
R-packages mailing list
r-packa...@r-project.org
https://stat.ethz.ch/mailman/listinfo/r-packages

__
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] [R-pkgs] AQ-R 0.2 // realtime messaging.

2012-12-05 Thread Ulrich Staudinger

Hi there,

I am glad to announce AQ-R 0.2 has been successfully built and is 
available via install.packages(aqr, 
repos=http://R-Forge.R-project.org;).


The most important new feature is real-time messaging from within R.

AQ-R 0.2 enables you to send and receive byte[] messages within R 
through a STOMP compliant messaging server, such as the ActiveQuant 
Master Server. There are various STOMP protocol compliant servers [1].

You can build arbitrarily complex messaging infrastructures, where 
messages flow between R instances or other-language messaging components.

The key functions are:
aqSubscribe(aChannel) - subscribe to messages in a channel
aqWaitForData() - a blocking call that waits for data to arrive
aqPoll() - fetches all messages from the internal buffer (not the STOMP 
server)
aqSend(channel, text) - send a message to a channel

This extension does buffer incoming data between consecutive aqPoll() 
calls.

The core loop for using this should look like:

while(aqWaitForData()){
 data = aqPoll()
}


Version 0.2 is beta. I have thoroughly tested it on Windows and on 
Linux. I am looking for more testers and feedback.

Demo video at [2].

buzzI hope it opens up an easy venue to build distributed calculators 
without the hassle of going through MPI or anything similar, in a truely 
cross-platform approach, making R a viable citizen in  highly 
heterogenous message processing environments. /buzz



Cheers,
Ulrich




References:
[1] http://stomp.github.com/implementations.html#STOMP_1_0_Servers
[2] http://www.youtube.com/watch?v=h1gLgJOEWW0


-- 
Ulrich Staudinger, ActiveQuant GmbH

P: +41 79 702 05 95
E:ustaudin...@activequant.com

http://www.activequant.com
Connect online:https://www.xing.com/profile/Ulrich_Staudinger


[[alternative HTML version deleted]]

___
R-packages mailing list
r-packa...@r-project.org
https://stat.ethz.ch/mailman/listinfo/r-packages

__
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] alternative to leaps command

2012-12-05 Thread Bert Gunter

One correction, one comment, one suggestion:

On Wed, Dec 5, 2012 at 4:44 AM, eliza botto eliza_bo...@hotmail.com wrote:

Dear UseRs,

I wanted to know that i have been using leaps for the proper models
selection for my work. I read so many articles from internet which
categorically outlined that leaps command for model selection should
never be used as its not efficient.

No. That is not the problem. The problem is that it efficiently produces
nonsense, as your next line indicates.

Moreover, as a matter of fact, i personally noticed that the accuracy of
leaps package is very much questionable.

Is there an alternative package, which is more efficient

You are asking about model selection/building, which is a central theme in
statistics. There is no simple answer to your question.

Moreover, this is not the place for such discussions. I strongly recommend
that you consult a local statistician, as you are clearly out of your
depth. Failing that, post to a statistics list like stats.stackexchange.com
.

-- Bert

your help will be deeply acknowledged...

warm regards

eliza
[[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.

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

[[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] MBA with FP Growth

2012-12-05 Thread Libardo López Guzmán

Somebody knows about a MBA (Market Basket Analysis) implementation using
FP-Growth and Arules in R?. I'm no programmer, instead I am a user of this
application and given the computational cost of the APRIORI algorithm, this
does not work well with big Datasets like what you find in the real world.

TIA,

-- 
LIBARDO LÓPEZ GUZMÁN

QR 
vCARDhttp://chart.apis.google.com/chart?cht=qrchs=350x350chld=Mchoe=UTF-8chl=BEGIN%3AVCARD%0AN%3ALIBARDO+LOPEZ+GUZMAN%0ATEL%3A3176397899%0AEMAIL%3Alilopezg%40gmail.com%0AADR%3ABOGOTA+COLOMBIA%0ANOTE%3ABI%2C+DM%2C+Marketing+Especialist%0AEND%3AVCARD

[[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.

Re: [R] Help for a function

2012-12-05 Thread Rui Barradas


Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the 
loop it never exits.

And

res1[j] -(a*h)
res2 -sum( res1[j])

is equivalent to

res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:

On 12/05/2012 01:01 AM, anoumou wrote:

Hello all,
I need a help.
I am modeling a disease and a create a R function like that:
...
But i do not get the results,i try by all means but i d'ont 
understant the

problem.


Hi anoumou,
Your function provides almost no indication of what two of its five 
arguments are supposed to be. If, with a certain degree of optimistic 
inference, we suppose x to be a data frame organized as shown below 
the function. t, i and CONTAGIEUX must be the appropriately 
named columns of the data frame. This leaves two columns, Symptomes 
and Incubation. Say we flip a coin to decide which of these to 
assign to r, and with all of our degrees of freedom gone, we assume 
that the other is h. We are forced to the conclusion that a is a 
nuisance argument, added to throw us off the scent.


Peering within the function, we notice that the date format is wrong, 
braces are unmatched and that our data frame is alphabetically ordered 
by name of country to no purpose whatsoever. The best I can do here is 
to make the function potentially able to do something if you can work 
out what to do with it.


Lambda-function (x,date1,r,h,a) {
 ndate1 - as.Date(date1, %Y-%m-%d)
 t1 - as.numeric(ndate1)
 x[order(x$i),]
 xt -x[,t]
 xi -x[,i]
 CONTAGIEUX -x[,CONTAGIEUX]
 while ( t1  min(xt) ){
  for (i in 1:length(xi) ){
   for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
res1[j] -(a*h)
res2 -sum( res1[j])
   }
  }
 lambda[i] - r*res2
 }
 x-data.frame(x,lambda)
 x
}

Jim

__
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-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] Help for a function

2012-12-05 Thread teko maurice

Thanks you all

 
Maurice TEKO

 
Biostatisticien,Doctorant.

Université de Liège
Département des Sciences et Gestion de l'environnement
185 avenue de Longwy
6700 Arlon (Belgique)
Mail :teko_maur...@yahoo.fr
:anoumou.tekoahate...@ulg.ac.be
 http://www.ulg.ac.be
  




 De : Rui Barradas ruipbarra...@sapo.pt
À : Jim Lemon j...@bitwrit.com.au 

Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function

Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the loop it 
never exits.
And

    res1[j] -(a*h)
    res2 -sum( res1[j])

is equivalent to

    res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:
 On 12/05/2012 01:01 AM, anoumou wrote:
 Hello all,
 I need a help.
 I am modeling a disease and a create a R function like that:
 ...
 But i do not get the results,i try by all means but i d'ont understant the
 problem.
 
 Hi anoumou,
 Your function provides almost no indication of what two of its five arguments 
 are supposed to be. If, with a certain degree of optimistic inference, we 
 suppose x to be a data frame organized as shown below the function. t, 
 i and CONTAGIEUX must be the appropriately named columns of the data 
 frame. This leaves two columns, Symptomes and Incubation. Say we flip a 
 coin to decide which of these to assign to r, and with all of our degrees 
 of freedom gone, we assume that the other is h. We are forced to the 
 conclusion that a is a nuisance argument, added to throw us off the scent.
 
 Peering within the function, we notice that the date format is wrong, braces 
 are unmatched and that our data frame is alphabetically ordered by name of 
 country to no purpose whatsoever. The best I can do here is to make the 
 function potentially able to do something if you can work out what to do with 
 it.
 
 Lambda-function (x,date1,r,h,a) {
  ndate1 - as.Date(date1, %Y-%m-%d)
  t1 - as.numeric(ndate1)
  x[order(x$i),]
  xt -x[,t]
  xi -x[,i]
  CONTAGIEUX -x[,CONTAGIEUX]
  while ( t1  min(xt) ){
   for (i in 1:length(xi) ){
    for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
     res1[j] -(a*h)
     res2 -sum( res1[j])
    }
   }
  lambda[i] - r*res2
  }
  x-data.frame(x,lambda)
  x
 }
 
 Jim
 
 __
 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.
[[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.

Re: [R] Help for a function

2012-12-05 Thread teko maurice



Rui ,
I discover all this when running and rerunning my function one and thousand 
times and per line.
Now it's perfect, and work so well.
Thanks.






À : Rui Barradas ruipbarra...@sapo.pt; Jim Lemon j...@bitwrit.com.au
Cc : r-help@r-project.org r-help@r-project.org 
Envoyé le : Mercredi 5 décembre 2012 16h45
Objet : Re: [R] Help for a function


Thanks you all
 
Maurice TEKO

 
Biostatisticien,Doctorant.

Université de Liège
Département des Sciences et Gestion de l'environnement
185 avenue de Longwy
6700 Arlon (Belgique)

:anoumou.tekoahate...@ulg.ac.be
 http://www.ulg.ac.be
  




 De : Rui Barradas ruipbarra...@sapo.pt
À : Jim Lemon j...@bitwrit.com.au 

Envoyé le : Mercredi 5 décembre 2012 16h26
Objet : Re: [R] Help for a function

Hello,

Also, t1 and min(xt) do not vary inside the loop so if it enters the loop it 
never
 exits.
And

    res1[j] -(a*h)
    res2 -sum( res1[j])

is equivalent to

    res2 - a*h

so the inner-most loop is not needed at all.

Hope this helps,

Rui Barradas
Em 05-12-2012 04:20, Jim Lemon escreveu:
 On 12/05/2012 01:01 AM, anoumou wrote:
 Hello all,
 I need a help.
 I am modeling a disease and a create a R function like that:
 ...
 But i do not get the results,i try by all means but i d'ont understant the
 problem.
 
 Hi anoumou,
 Your function provides almost no indication of what two of its five arguments 
 are supposed to be. If, with a certain degree of optimistic inference, we 
 suppose x to be a data frame organized as shown below the function. t, 
 i and CONTAGIEUX must be the appropriately named columns of the data 
 frame. This leaves two columns,
 Symptomes and Incubation. Say we flip a coin to decide which of these to 
assign to r, and with all of our degrees of freedom gone, we assume that the 
other is h. We are forced to the conclusion that a is a nuisance argument, 
added to throw us off the scent.
 
 Peering within the function, we notice that the date format is wrong, braces 
 are unmatched and that our data frame is alphabetically ordered by name of 
 country to no purpose whatsoever. The best I can do here is to make the 
 function potentially able to do something if you can work out what to do with 
 it.
 
 Lambda-function (x,date1,r,h,a) {
  ndate1 - as.Date(date1, %Y-%m-%d)
  t1 - as.numeric(ndate1)
  x[order(x$i),]
  xt -x[,t]
  xi -x[,i]
  CONTAGIEUX -x[,CONTAGIEUX]
  while ( t1  min(xt) ){
   for (i in 1:length(xi)
 ){
    for (j in 1:CONTAGIEUX[length(CONTAGIEUX)]){
     res1[j] -(a*h)
     res2 -sum( res1[j])
    }
   }
  lambda[i] - r*res2
  }
  x-data.frame(x,lambda)
  x
 }
 
 Jim
 
 __
 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.
[[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] Reminder: useR! 2013 call for tutorials

2012-12-05 Thread Virgilio Gómez-Rubio

Dear all,

This is a reminder that the deadline for submitting a tutorial proposal
for the useR! 2013 conference is December 15, 2012. See details below.

Best wishes,

Virgilio

---


We are pleased to announce that the R user conference

   useR! 2013

is scheduled for July 10-12, 2013, and will take place at the
University of Castilla-La Mancha, Albacete, Spain.

As for the predecessor conferences, the program will consist of two
parts: invited lectures and user-contributed sessions (abstract
submission will be available soon). Prior to the
conference, there will be tutorials on R (proposals for tutorials should
be sent before December 15, 2012, see below).

CONFIRMED INVITED SPEAKERS

María Jesús Bayarri, José Manuel Benítez-Sánchez, Havard Rue, Steve
Scott and Hadley Wickham

USER-CONTRIBUTED SESSIONS

The conference will feature both talks and posters illustrating the use
of R in practice. Contributions are welcome that introduce recent
developments in the R Project (including CRAN packages), demonstrate
applications of R in areas of current interest, or otherwise engage and
inspire participants in their use of R.

PRE-CONFERENCE TUTORIALS

Before the official program, half-day tutorials will be offered on
Tuesday, July 9.

We invite R users to submit proposals for three hour tutorials on
special topics regarding R. The proposals should give a brief
description of the tutorial, including goals, detailed outline,
justification of why the tutorial is important, background knowledge
required and potential attendees. The proposals should be sent before
December 15, 2012 to useR-2013_at_R-project.org.

A web page offering more information on the `useR!' conference is
available at

   http://www.R-project.org/useR-2013

We hope to see you in Albacete!

The organizing committee:

Esteban Alfaro, José Luis Alfaro, M. Teresa Alonso, Emilio L. Cano, 
Gonzalo García-Donato, Matías Gámez, Noelia García, Virgilio
Gómez-Rubio, María José Haro, Eva J. Moreno and Francisco Parreño. 

__
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] Incidence Matrix of Experimental Design Using R language Program

2012-12-05 Thread Zaheer Abbas

Hi
I am working on educational assignment to produce an *Incidence matrix *from
a *BIB design *using R language software.
 I found a web page *http://wiki.math.yorku.ca/index.php/R:_Incidence_matrix
* about the problem. But it produces Data matrix instead of
Incidence matrix. can you please help me out using R software.

thanks and regards

zaheer

*following are the sample codes of my assignment  with resultant Incidence
Matrix as mentioned in above link, but this is, according to my knowledge,
Data matrix of any experimental design. *

 b=4   #Number of Blocks

 t=4   #Number of Column

 z=c(1,2,3)#Shift

m=NULL

 y=c(0)

 w=c(y,cumsum(z) %%t)

print(The Initial Block is=)

print(w)



 p=seq(from=0, to=t-1, by=1)

 l=NULL



   for(i in 1:b)

 {

 for(j in 1:t)

{

l=c(l,rep((w[i]+p[j]+t)%% t))

 }

  }



x= matrix(c(l),nrow=b,ncol=t,byrow = TRUE)

print(The Design Layout is )

print.matrix - function(x){



write.table(format(x, justify=right),



 row.names=F,col.names=F,quote=F

})



print(x)

*BIB design Output*

 [,1] [,2] [,3] [,4]

[1,]0123

[2,]1230

[3,]3012

[4,]2301

vec - c(as.matrix(x)) # converting matrix into Vector for incidence
correct matrix

a=t(x)%*%x

print.matrix - function(a){



write.table(format(a, justify=right),



row.names=F, col.names=F, quote=F)

}



print(XX )

print(a)

A=contrasts( as.factor(x), contrasts =FALSE)[ as.factor(x),]

print.matrix - function(A){



write.table(format(A, justify=right),



row.names=F, col.names=F, quote=F)

}

print(The Incidence Matrix is=)

   print(A)


*OUTPUT*

1 0 0 0

0 1 0 0

0 0 0 1

0 0 1 0

0 1 0 0

0 0 1 0

1 0 0 0

0 0 0 1

0 0 1 0

0 0 0 1

0 1 0 0

1 0 0 0

0 0 0 1

1 0 0 0

0 0 1 0

0 1 0 0


*I need this *

* 1   1  1  1*

* 1   1  1  1*

* 1   1  1  1*

* 1   1  1  1*

[[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] How can I unsubscrie to this R forum¿?

2012-12-05 Thread Sébastien Morant

Hi,
even though the help and the community was a greta plus and a very welcome
experience, I wanted to know I can I exit it?
thanks for you support.
BR

[[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] Using multcomp::glht() with Anova object

Hello everyone,

I've conducted a Type III repeated-measures ANOVA using Anova() from the
car package, based on the suggestions at
http://blog.gribblelab.org/2009/03/09/repeated-measures-anova-using-r/(option
3) and
http://languagescience.umd.edu/wiki/EEG#ERP_ANOVA_in_R. My ANOVA has two
factors: Condition (3 levels) and Region (6 levels) and their interaction.
Below is code to run the Anova and get the model object (the data are at
https://docs.google.com/open?id=0B6-m45Jvl3ZmOVFTYVpZYV9sUUk).


data - as.matrix( read.table( file=EEGpriming.txt, header=T, sep=\t) )

Condition - c( rep(Cond1, 6), rep(Cond2, 6), rep(Cond3, 6) )
Region - factor( rep( c(la, ma, ra, lp, mp, rp), 3 ) )

library(car)

model.Anova - Anova( lm(eeg ~ 1), idata=data.frame(Condition, Region),
idesign=~Condition*Region ) )

I'm now trying to do post-hoc comparisons between all levels of Condition
using glht() from the multcomp package. I do this using the way that has
worked for me with lmer() objects in the past: set up a contrast matrix and
then feed it to glht():

designmatrix - matrix(0, nrow=3, ncol=18)
rownames(designmatrix) - c(Cond1_minus_Cond2, Cond3_minus_Cond1,
Cond3_minus_Cond2)
designmatrix[1, 1:6] - 1
designmatrix[1, 7:12] - -1
designmatrix[2, 1:6] - -1
designmatrix[2, 13:18] - 1
designmatrix[3, 13:18] - 1
designmatrix[3, 7:12] - -1

library(multcomp)

glht( model.Anova, linfct=designmatrix )

However, when I give glht() an Anova object, I get the following error:

 glht( model.anova, linfct=designmatrix )
Error in UseMethod(vcov) :
  no applicable method for 'vcov' applied to an object of class Anova.mlm
Error in modelparm.default(model, ...) :
  no âvcovâ method for âmodelâ found!

Does anyone know how if it's possible to use glht with an Anova object? (I
guess I can get these same comparisons by re-doing the anova using aov() or
lme(), since I don't need the Huynh-Feldt corrections in the post-hoc
tests...but I am just curious if i can do the post-hocs directly on my
Anova object.)

Best,
Steve Politzer-Ahles

-- 
Stephen Politzer-Ahles
University of Kansas
Linguistics Department
http://people.ku.edu/~sjpa/

[[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.

Re: [R] How can I unsubscrie to this R forum¿?

2012-12-05 Thread Jeff Newmiller

Follow the link in the footer of any message from the list to the web page that 
lets you modify your list subscription.
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Sébastien Morant sebmor...@gmail.com wrote:

Hi,
even though the help and the community was a greta plus and a very
welcome
experience, I wanted to know I can I exit it?
thanks for you support.
BR

   [[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-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] reformatting some data



On Dec 4, 2012, at 1:44 PM, arun wrote:


Hi,
You can also do this:
dat1-structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L),  
X3.Hydroxybutyrate =

 c(4e-04,
 5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04,
 3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04,
 5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate,
 X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L,
 365L, 371L, 377L), class = data.frame)

datM-melt(dat1,id.var=group)


An explicit call to library(reshape2) might be in order before a melt  
call.




xtabs(value~variable+group,data=datM)
#   group
#variable 2  3  4
 # X3.Hydroxybutyrate0.0007 0.0005 0.0019
 # X3.Hydroxyisovalerate 0.0004 0.0003 0.0012
 # ADP   0.0007 0.0006 0.0023


This has implicitly summed the entires from items by group values.  
That might or might not have been what the OP wanted. xtabs does not  
allow a function to be specified. (It's output is expected to be a  
contingency table of counts so it only makes sense to sum values.) The  
aggregate function allows either choice within the base package:


I see that arun used `mean` as his function in an earlier post using  
reshape2::dcast


 aggregate(dat1[-1] , dat1['group'], FUN=mean)
  group X3.Hydroxybutyrate X3.Hydroxyisovalerate  ADP
1 2   0.000700 4e-04 0.000700
2 3   0.000500 3e-04 0.000600
3 4   0.000475 3e-04 0.000575

This is the aggregate method for summation:

 aggregate(dat1[-1] , dat1['group'], FUN=sum)
  group X3.Hydroxybutyrate X3.HydroxyisovalerateADP
1 2 0.00070.0004 0.0007
2 3 0.00050.0003 0.0006
3 4 0.00190.0012 0.0023


A.K.



- Original Message -
From: Charles Determan Jr deter...@umn.edu
To: r-help@r-project.org
Cc:
Sent: Tuesday, December 4, 2012 4:17 PM
Subject: [R] reformatting some data

Hello,

I am trying to reformat some data so that it is organized by group  
in the

columns.  The data currently looks like this:

   group X3.Hydroxybutyrate X3.Hydroxyisovalerate   ADP
347 4  4e-04 3e-04   
5e-04
353 3  5e-04 3e-04   
6e-04
359 4  4e-04 3e-04   
6e-04
365 4  6e-04 3e-04   
5e-04
371 4  5e-04 3e-04   
7e-04
377 2  7e-04 4e-04   
7e-04


I would like to reformat it so it is like this:

2  3   4
var1
var2
var3


I realize that there unequal numbers in each group but I would like to
none-the-less if possible.
Here is a subset of the data:

structure(list(group = c(4L, 3L, 4L, 4L, 4L, 2L), X3.Hydroxybutyrate =
c(4e-04,
5e-04, 4e-04, 6e-04, 5e-04, 7e-04), X3.Hydroxyisovalerate = c(3e-04,
3e-04, 3e-04, 3e-04, 3e-04, 4e-04), ADP = c(5e-04, 6e-04, 6e-04,
5e-04, 7e-04, 7e-04)), .Names = c(group, X3.Hydroxybutyrate,
X3.Hydroxyisovalerate, ADP), row.names = c(347L, 353L, 359L,
365L, 371L, 377L), class = data.frame)




David Winsemius, MD
Alameda, CA, USA

__
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] Using multcomp::glht() with Anova object

2012-12-05 Thread John Fox

Dear Steve,

Usually the best place to look for information about functions in the car 
package, along with the help files for the package, is the book with which the 
package is associated. In this case, there's an on-line appendix to the book on 
multivariate linear models which describes how to use the linearHypothesis() 
function in the car package with repeated measures: sec. 3.2 of 
http://socserv.socsci.mcmaster.ca/jfox/Books/Companion/appendix/Appendix-Multivariate-Linear-Models.pdf.
 I think that this should help you do what you want.

I hope this helps,
 John


John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/

On Wed, 5 Dec 2012 11:09:44 -0600
 Stephen Politzer-Ahles politzerahl...@gmail.com wrote:
 Hello everyone,
 
 I've conducted a Type III repeated-measures ANOVA using Anova() from the
 car package, based on the suggestions at
 http://blog.gribblelab.org/2009/03/09/repeated-measures-anova-using-r/(option
 3) and
 http://languagescience.umd.edu/wiki/EEG#ERP_ANOVA_in_R. My ANOVA has two
 factors: Condition (3 levels) and Region (6 levels) and their interaction.
 Below is code to run the Anova and get the model object (the data are at
 https://docs.google.com/open?id=0B6-m45Jvl3ZmOVFTYVpZYV9sUUk).
 
 
 data - as.matrix( read.table( file=EEGpriming.txt, header=T, sep=\t) )
 
 Condition - c( rep(Cond1, 6), rep(Cond2, 6), rep(Cond3, 6) )
 Region - factor( rep( c(la, ma, ra, lp, mp, rp), 3 ) )
 
 library(car)
 
 model.Anova - Anova( lm(eeg ~ 1), idata=data.frame(Condition, Region),
 idesign=~Condition*Region ) )
 
 I'm now trying to do post-hoc comparisons between all levels of Condition
 using glht() from the multcomp package. I do this using the way that has
 worked for me with lmer() objects in the past: set up a contrast matrix and
 then feed it to glht():
 
 designmatrix - matrix(0, nrow=3, ncol=18)
 rownames(designmatrix) - c(Cond1_minus_Cond2, Cond3_minus_Cond1,
 Cond3_minus_Cond2)
 designmatrix[1, 1:6] - 1
 designmatrix[1, 7:12] - -1
 designmatrix[2, 1:6] - -1
 designmatrix[2, 13:18] - 1
 designmatrix[3, 13:18] - 1
 designmatrix[3, 7:12] - -1
 
 library(multcomp)
 
 glht( model.Anova, linfct=designmatrix )
 
 However, when I give glht() an Anova object, I get the following error:
 
  glht( model.anova, linfct=designmatrix )
 Error in UseMethod(vcov) :
   no applicable method for 'vcov' applied to an object of class Anova.mlm
 Error in modelparm.default(model, ...) :
   no âvcovâ method for âmodelâ found!
 
 Does anyone know how if it's possible to use glht with an Anova object? (I
 guess I can get these same comparisons by re-doing the anova using aov() or
 lme(), since I don't need the Huynh-Feldt corrections in the post-hoc
 tests...but I am just curious if i can do the post-hocs directly on my
 Anova object.)
 
 Best,
 Steve Politzer-Ahles
 
 -- 
 Stephen Politzer-Ahles
 University of Kansas
 Linguistics Department
 http://people.ku.edu/~sjpa/
 
   [[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.

Re: [R] Using multcomp::glht() with Anova object

Thank you John, I'll take a look at that!
Best,
Steve

On Wed, Dec 5, 2012 at 11:39 AM, John Fox j...@mcmaster.ca wrote:

 Dear Steve,

 Usually the best place to look for information about functions in the car 
 package, along with the help files for the package, is the book with which 
 the package is associated. In this case, there's an on-line appendix to the 
 book on multivariate linear models which describes how to use the 
 linearHypothesis() function in the car package with repeated measures: sec. 
 3.2 of 
 http://socserv.socsci.mcmaster.ca/jfox/Books/Companion/appendix/Appendix-Multivariate-Linear-Models.pdf.
  I think that this should help you do what you want.

 I hope this helps,
  John

 
 John Fox
 Sen. William McMaster Prof. of Social Statistics
 Department of Sociology
 McMaster University
 Hamilton, Ontario, Canada
 http://socserv.mcmaster.ca/jfox/

__
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] Import multiple data frames and combine them using cbind



On Dec 4, 2012, at 8:56 PM, Gyanendra Pokharel wrote:


Thanks Dennis,
Your code also produces same as Jim's but I am not looking that one,  
I need
to use cbind so that finally I will get the data frame of size  
200X320

(i,e, 200 X(16X20)).
Thanks


Here we are a day later and your question risks sinking beneath the  
waves because you have not realized that you are implicitly making  
impossible demands on your volunteer consultants. Please review your  
postings and responses and try to put yourself in the position of  
someone who doesn't have the capacity to read your mind or see your  
screen.


--
David.



On Tue, Dec 4, 2012 at 10:46 PM, Dennis Murphy djmu...@gmail.com  
wrote:



In addition to Jim's reply, had you used the plyr package, you could
have done this in one shot:

library(plyr)

DF - ldply(llply(temp, read.table), rbind)

The inner llply call is equivalent to your lapply() and the outer
ldply() is equivalent to Jim's do.call() code.

Dennis

On Tue, Dec 4, 2012 at 6:37 PM, Gyanendra Pokharel
gyanendra.pokha...@gmail.com wrote:

Hi group,

I imported  16 data frames using the function list.files

temp - list.files(path=...)
myfiles = lapply(temp, read.table,sep = )

Now I have 16 data set imported in R window.

I want to combine them by row and tried some thing like (Here I am
considering only 20 columns)

for(i in 1:16){
   data- cbind(myfiles[[i]][,1:20])

}


but it returns only first data set. I can combine them using

data - cbind(myfiles[[1]][,1:20],myfiles[[2]] 
[1:20],...)


But  I want in a loop so that I can make the efficient code.

Any kind of suggestion will be great for me.

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.




[[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.


David Winsemius, MD
Alameda, CA, USA

__
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] What is print print print ?



On Dec 5, 2012, at 5:03 AM, Vladimir eremeev wrote:


Hi all.

What is print print print?

I don't see output of the print command in for loop and have found  
this

link:
http://stackoverflow.com/questions/1816200/chisq-test-doesnt-print-results-when-in-a-loop

It describes a problem, similar to mine.
My problem. I want to execute print command in for loop.

If I copy for loop body with print() and paste it to console, I  
don't see

any output.



Code we, want code.



If I type the command, I do see the output.
If I paste the command by parts I also see the output.

If I construct the command by picking parts of the history with the up
arrow, I also don't see any output.


Code?



Search for print print print still didn't bring anything useful.


The questioner on SO was executing print chisq.test print and he was  
advised (and apparently  understood ) that he need to be executing  
print print print. Get it now?


If you have a specific problem to be addressed then you should post  
the code, but I think using Rhelp to explain cryptic answers on SO  
might be expecting a bit much.


--
David Winsemius, MD
Alameda, CA, USA

__
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] data manipulation between vector and matrix

Dear list,
I was curious how to subtract a vector from matrix?

Say, I have

mat - matrix(1:40, nrow=20, ncol=2)

x -c(1,2)

I want,

x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
of x against column elements in mat.

But x-mat won't do it.

Thanks,

Mike

[[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.

Re: [R] How can I unsubscrie to this R forum¿?



On Dec 5, 2012, at 8:33 AM, Sébastien Morant wrote:


Hi,
even though the help and the community was a greta plus and a very  
welcome

experience, I wanted to know I can I exit it?
thanks for you support.


The same place you signed up  ... and the link is at the bottom of  
every posting to Rhelp.



BR

[[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.


David Winsemius, MD
Alameda, CA, USA

__
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] Import multiple data frames and combine them using cbind

2012-12-05 Thread Gyanendra Pokharel

I am sorry David, I don't mean to give the answer of the impossible
questions. Just you can say impossible. If there is no way to do what I
explained, that's fine, we do have other alternatives what I wrote in the
beginning.

Thank you so much.



On Wed, Dec 5, 2012 at 1:12 PM, David Winsemius dwinsem...@comcast.netwrote:

 beneath

[[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.

Re: [R] data manipulation between vector and matrix

The only solution I found was
x-t(mu)

Is there a better way?
Mike

On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:

 Dear list,
 I was curious how to subtract a vector from matrix?

 Say, I have

 mat - matrix(1:40, nrow=20, ncol=2)

 x -c(1,2)

 I want,

 x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
 of x against column elements in mat.

 But x-mat won't do it.

 Thanks,

 Mike



[[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.

Re: [R] data manipulation between vector and matrix

2012-12-05 Thread Sarah Goslee

Hi,

On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
 Dear list,
 I was curious how to subtract a vector from matrix?

 Say, I have

 mat - matrix(1:40, nrow=20, ncol=2)

 x -c(1,2)

Thanks for the actual reproducible example.

 I want,

 x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
 of x against column elements in mat.

 But x-mat won't do it.

This will (note the modification to get x - mat):
 sweep(-mat, 2, x, +)
  [,1] [,2]
 [1,]0  -19
 [2,]   -1  -20
 [3,]   -2  -21
 [4,]   -3  -22
 [5,]   -4  -23
etc.

--
Sarah Goslee
http://www.functionaldiversity.org

__
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] Import multiple data frames and combine them using cbind


On Dec 5, 2012, at 10:36 AM, Gyanendra Pokharel wrote:

 I am sorry David, I don't mean to give the answer of the impossible  
 questions. Just you can say impossible. If there is no way to do  
 what I explained, that's fine, we do have other alternatives what I  
 wrote in the beginning.


I did not say it was impossible to do what you are thinking you want  
to do, only that it is impossible to read your mind. You need to to  
describe what you are thinking in greater detail.

 Thank you so much.



 On Wed, Dec 5, 2012 at 1:12 PM, David Winsemius dwinsem...@comcast.net 
  wrote:
 beneath


David Winsemius, MD
Alameda, CA, USA


[[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.

Re: [R] data manipulation between vector and matrix

Thanks, Sarah.  First time heard about sweep(), it worked just the way I
wanted.
Mike

On Wed, Dec 5, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.com wrote:

 Hi,

 On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
  Dear list,
  I was curious how to subtract a vector from matrix?
 
  Say, I have
 
  mat - matrix(1:40, nrow=20, ncol=2)
 
  x -c(1,2)

 Thanks for the actual reproducible example.

  I want,
 
  x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
  of x against column elements in mat.
 
  But x-mat won't do it.

 This will (note the modification to get x - mat):
  sweep(-mat, 2, x, +)
   [,1] [,2]
  [1,]0  -19
  [2,]   -1  -20
  [3,]   -2  -21
  [4,]   -3  -22
  [5,]   -4  -23
 etc.

 --
 Sarah Goslee
 http://www.functionaldiversity.org


[[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.

Re: [R] Import multiple data frames and combine them using cbind

2012-12-05 Thread Berend Hasselman


On 05-12-2012, at 03:37, Gyanendra Pokharel wrote:

 Hi group,
 
 I imported  16 data frames using the function list.files
 
 temp - list.files(path=...)
 myfiles = lapply(temp, read.table,sep = )
 
 Now I have 16 data set imported in R window.
 
 I want to combine them by row and tried some thing like (Here I am
 considering only 20 columns)
 
 for(i in 1:16){
data- cbind(myfiles[[i]][,1:20])
 
 }
 
 
 but it returns only first data set. I can combine them using
 
 data - cbind(myfiles[[1]][,1:20],myfiles[[2]][1:20],...)
 
 But  I want in a loop so that I can make the efficient code.

Slightly guessing what you want.
Toy example

set.seed(123)
mydf - lapply(1:4,FUN=function(dname) data.frame(x=round(runif(10),2), 
y=round(runif(10),2),z=round(runif(10),2)))
mydf

do.call(cbind,lapply(mydf, function(df) df[,1:2]))


Berend

__
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] Import multiple data frames and combine them using cbind

2012-12-05 Thread Gyanendra Pokharel

Thanks Berend, your Idea is great, that,s what I was looking.

Thanks again

On Wed, Dec 5, 2012 at 2:06 PM, Berend Hasselman b...@xs4all.nl wrote:

 do.call(cbind,lapply(mydf, function(df) df[,1:2]))

[[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.

Re: [R] fitting a gamma frailty model (coxph)

2012-12-05 Thread Terry Therneau


I looked at your data:
 table(x, cluster)
 1  2  3  4  5  6
  0  0 48  0 48 48  0
  1 48  0 48  0  0 48

Your covariate x is perfectly predicted by the cluster variable.
If you fit a fixed effects model:
  coxph(Surv(time, event) ~ factor(cluster) +x)

then the x variable is declared redundant.  When the variance of the random 
effect is
sufficiently large, the same happens in the gamma model when the variance is sufficiently 
large.  Your model approaches this limit, and the solution fails.  As mentioned in the 
manual page, the coxme function is now preferred.


Last, your particular error message is caused by an invalid value for sparse.  I'll add 
a check to the program.

You likely want sparse=10 to force non-sparse computation.

Terry Therneau



On 12/04/2012 05:00 AM, r-help-requ...@r-project.org wrote:

Dear all,

I have a data 
sethttp://yaap.it/paste/c11b9fdcfd68d02b#gIVtLrrme3MaiQd9hHy1zcTjRq7VsVQ8eAZ2fol1lUc=with
6 clusters, each containing 48 (possibly censored, in which case
event = 0) survival times. The x column contains a binary explanatory
variable. I try to describe that data with a gamma frailty model as follows:

library(survival)

mod- coxph(Surv(time, event) ~
x + frailty.gamma(cluster, eps=1e-10, method=em, sparse=0),
   outer.max=1000, iter.max=1,
   data=data)

Here is the error message:

Error in if (history[2, 3]  (history[1, 3] + 1)) theta-
mean(history[1:2,  :
   missing value where TRUE/FALSE needed

Does anyone have an idea on how to debug?

Yours sincerely,
Marco


__
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] loading using Ryacas

2012-12-05 Thread Alicia Ellis

I'm having trouble loading Ryacas.  I've downloaded and extracted Ryacas
0.2-9 (also tried Ryacas ) and yacas 1.0.63, have the latest version of R
and have tried the following (this works for installing other packages):

install.packages(Ryacas)
library(Ryacas)
install.packages(yacas)
library(yacas)

and get this:
Warning message:
package yacas is not available (for R version 2.15.2)

Got the same thing for R version 2.15.1.

Suggestions?  I'm fairly new to R and could use some detailed
instructions.  Working on Mac OSx 10.8.2

Thanks!
-- 
Alicia Ellis
Postdoc
Gund Institute for Ecological Economics
University of Vermont
617 Main Street
Burlington, VT  05405
(802) 656-1046
http://www.uvm.edu/~aellis5/
 http://entomology.ucdavis.edu/faculty/scott/aellis/

[[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] Changing data frame column headings


  I have a reshaped data frame with value column headings concatenated from
two column headings in the melted data frame. I want to change all 56
headings in a single command, but 'names' allows me to change only one at a
time. In Hadley's 2007 article on reshape in the Journal of Statistical
Software he mentions a 'rename' function, but I cannot find that.

  Is there a way to change all data frame column headings in a single
command?

Rich

__
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] Changing data frame column headings

2012-12-05 Thread R. Michael Weylandt

x - data.frame(a = 1:5, b = rnorm(5))

names(x) - LETTERS[2:1]

print(x)

seems to work.

Can you be more explicit about your problem?

Michael

On Wed, Dec 5, 2012 at 6:51 PM, Rich Shepard rshep...@appl-ecosys.com wrote:
   I have a reshaped data frame with value column headings concatenated from
 two column headings in the melted data frame. I want to change all 56
 headings in a single command, but 'names' allows me to change only one at a
 time. In Hadley's 2007 article on reshape in the Journal of Statistical
 Software he mentions a 'rename' function, but I cannot find that.

   Is there a way to change all data frame column headings in a single
 command?

 Rich

 __
 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-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] In factor analysis in the psych package, how can I work out which factors the columns in $scores relate to? How do I know what each of the scores is scoring?

2012-12-05 Thread Brent Caldwell

Hi
I have used fa() to perform a factor analysis of a psychological battery which 
is thought to have 11 factors.  I can identify which factors the loadings 
relate to easily enough because I can see which items are loading onto each of 
the columns in the $loading output.  However, how can I identify which items or 
loadings are being used to create each of the columns in the $scores output?  I 
have used generalised linear models which have shown that some of the scores 
are significant predictors of treatment outcome, but I can't work out which of 
the 11 factors they are scoring?

when I export to csv the $loadings I get the following columns in this order:
 MR4, MR3, MR2, MR6, MR7, MR5, MR1, MR8, MR10, MR9, MR11 

When I export to csv the $scores I get the following columns in this order:
V1, V2, V3, V4, V5, V6, V7, V8, V9, V10,V11 

Are the scores in the column V1 derived from the loadings in column MR1 ?  
Or are the scores in the first column of $scores (ie V1) derived from the 
loadings in the first column of $loadings (ie MR4)?

Thank you very much for your guidance, and your patience with my confusion
Best wishes
Brent


Brent Caldwell, MBChB, DPH, MPH.
Research Fellow
Department of Medicine
University of Otago, Wellington
New Zealand
U  brent.caldw...@otago.ac.nz
b  04 918 6041
   021 87 22 64
  23a Mein Street, PO Box 7343, Newtown, WELLINGTON 6021, NEW ZEALAND
brent.ower.caldwell
  Zonnic study

__
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] loading using Ryacas



On Dec 5, 2012, at 10:21 AM, Alicia Ellis wrote:

I'm having trouble loading Ryacas.  I've downloaded and extracted  
Ryacas
0.2-9 (also tried Ryacas ) and yacas 1.0.63, have the latest version  
of R
and have tried the following (this works for installing other  
packages):


install.packages(Ryacas)
library(Ryacas)
install.packages(yacas)


Yacas is an external program and on the Mac (and Linux) needs to be  
installed outside of R.




library(yacas)

and get this:
Warning message:
package yacas is not available (for R version 2.15.2)

Got the same thing for R version 2.15.1.

Suggestions?  I'm fairly new to R and could use some detailed
instructions.  Working on Mac OSx 10.8.2

Thanks!
--
Alicia Ellis
Postdoc
Gund Institute for Ecological Economics
University of Vermont
617 Main Street
Burlington, VT  05405
(802) 656-1046
http://www.uvm.edu/~aellis5/
http://entomology.ucdavis.edu/faculty/scott/aellis/

[[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.


David Winsemius, MD
Alameda, CA, USA

__
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] Changing data frame column headings

Hi,
I am not sure why ?rename() is not working.
 a - list(a = 1, b = 2, c = 3)
     rename(a, c(b = a, c = b, a=c)) 
A.K.




- Original Message -
From: Rich Shepard rshep...@appl-ecosys.com
To: r-help@r-project.org
Cc: 
Sent: Wednesday, December 5, 2012 1:51 PM
Subject: [R] Changing data frame column headings

  I have a reshaped data frame with value column headings concatenated from
two column headings in the melted data frame. I want to change all 56
headings in a single command, but 'names' allows me to change only one at a
time. In Hadley's 2007 article on reshape in the Journal of Statistical
Software he mentions a 'rename' function, but I cannot find that.

  Is there a way to change all data frame column headings in a single
command?

Rich

__
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-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] duplicated() with long vectors

Hello,

duplicated() does not seem to work for a long vector. For example, if
you download the data from
https://docs.google.com/open?id=0B6-m45Jvl3ZmNmpaSlJWMXo5bmc (a vector
with about 12,000 numbers) and then run the following code which does
duplicated() over the whole vector but just shows the last 30
elements:

data.frame( tail(verylong, 30), tail(duplicated(verylong), 30) )

you'll see that at the end of the very long vector everything is
listed as a duplicate of the preceding element (even though it
shouldn't be). On the other hand, if you run the following code which
just takes out the last 30 elements of the vector and does duplicated
on them:

data.frame( tail(verylong, 30), duplicated(tail(verylong, 30)) )

you get the correct results (FALSE shows up wherever the value in the
first column changes). Does anyone know why this happens, and if
there's a fix? I notice the documentation for duplicated() says: Long
vectors are supported for the default method of duplicated, but may
only be usable if nmax is supplied.  But I've tried running this with
a high value of nmax given, and it still gives me the same problem.

So far the only way I've figured out to get this duplicated()-like
vector is to use a for loop going through one item at a time, but that
takes about a minute to run.

Best,
Steve Politzer-Ahles

--
Stephen Politzer-Ahles
University of Kansas
Linguistics Department
http://people.ku.edu/~sjpa/

__
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] duplicated() with long vectors

2012-12-05 Thread Sarah Goslee

Hi,

duplicated() doesn't just look at consecutive values, but anywhere in
the object. Since your 12320-element vector has only 48 separate
values, and all of them occur before the last 30 elements, so
duplicated() returns TRUE.

You might be looking for something involving rle(). What are you
trying to accomplish?

Sarah

On Wed, Dec 5, 2012 at 3:53 PM, Stephen Politzer-Ahles
politzerahl...@gmail.com wrote:
 Hello,

 duplicated() does not seem to work for a long vector. For example, if
 you download the data from
 https://docs.google.com/open?id=0B6-m45Jvl3ZmNmpaSlJWMXo5bmc (a vector
 with about 12,000 numbers) and then run the following code which does
 duplicated() over the whole vector but just shows the last 30
 elements:

 data.frame( tail(verylong, 30), tail(duplicated(verylong), 30) )

 you'll see that at the end of the very long vector everything is
 listed as a duplicate of the preceding element (even though it
 shouldn't be). On the other hand, if you run the following code which
 just takes out the last 30 elements of the vector and does duplicated
 on them:

 data.frame( tail(verylong, 30), duplicated(tail(verylong, 30)) )

 you get the correct results (FALSE shows up wherever the value in the
 first column changes). Does anyone know why this happens, and if
 there's a fix? I notice the documentation for duplicated() says: Long
 vectors are supported for the default method of duplicated, but may
 only be usable if nmax is supplied.  But I've tried running this with
 a high value of nmax given, and it still gives me the same problem.

 So far the only way I've figured out to get this duplicated()-like
 vector is to use a for loop going through one item at a time, but that
 takes about a minute to run.

 Best,
 Steve Politzer-Ahles



--
Sarah Goslee
http://www.functionaldiversity.org

__
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] data manipulation between vector and matrix

thanks, I knew about apply, but did not you you can put plus signs with
quotes.  That's a cool tricky,
Mike

On Wed, Dec 5, 2012 at 4:05 PM, arun smartpink...@yahoo.com wrote:

 HI,
 In addition to ?sweep(), you can use

 apply(-mat,1,`+`,x)

 #or
 library(plyr)
 aaply(-mat,1,+,x)


 A.K.






 - Original Message -
 From: C W tmrs...@gmail.com
 To: Sarah Goslee sarah.gos...@gmail.com
 Cc: r-help r-help@r-project.org
 Sent: Wednesday, December 5, 2012 1:51 PM
 Subject: Re: [R] data manipulation between vector and matrix

 Thanks, Sarah.  First time heard about sweep(), it worked just the way I
 wanted.
 Mike

 On Wed, Dec 5, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.com
 wrote:

  Hi,
 
  On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
   Dear list,
   I was curious how to subtract a vector from matrix?
  
   Say, I have
  
   mat - matrix(1:40, nrow=20, ncol=2)
  
   x -c(1,2)
 
  Thanks for the actual reproducible example.
 
   I want,
  
   x-mat[1,] and x-mat[2,], and so on... Basically, subtract column
 elements
   of x against column elements in mat.
  
   But x-mat won't do it.
 
  This will (note the modification to get x - mat):
   sweep(-mat, 2, x, +)
[,1] [,2]
   [1,]0  -19
   [2,]   -1  -20
   [3,]   -2  -21
   [4,]   -3  -22
   [5,]   -4  -23
  etc.
 
  --
  Sarah Goslee
  http://www.functionaldiversity.org
 

 [[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.



[[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.

Re: [R] how to assign factor level into each value

2012-12-05 Thread jim holtman

?cut

 x - read.table(text =  Country   Price
+ 1  CN 44.25
+ 2  CN 21.07
+ 3  CN 92.70
+ 4  CN 47.41
+ 5  CN111.67
+ 6  CN 50.57, as.is = TRUE, header = TRUE)
 x$levels - cut(x$Price, breaks = c(0,30,50,75,100,150,200,300,400,500,Inf))
 x
  Country  Pricelevels
1  CN  44.25   (30,50]
2  CN  21.07(0,30]
3  CN  92.70  (75,100]
4  CN  47.41   (30,50]
5  CN 111.67 (100,150]
6  CN  50.57   (50,75]




On Wed, Dec 5, 2012 at 7:26 AM, Tammy Ma metal_lical...@live.com wrote:

 HI, All

 I met the following problem. I dont know how to handle it.

   Country   Price
 1  CN 44.25
 2  CN 21.07
 3  CN 92.70
 4  CN 47.41
 5  CN111.67
 6  CN 50.57


 I want to create the 3rd colume with different factor levels:
  [1] 0-3051-75   31-50   76-100  101-150 151-200 201-300 
 500+
  [9] 301-400 401-500

 then the final result which I want is:

   Country   Pricelevels
 1  CN 44.25  31-50
 2  CN 21.07  0-30
 3  CN 92.70 76-100
 4  CN 47.41   31-50
 5  CN111.67101-150
 6  CN 50.57 51-75


 How can I do this?

 Thanks.

 Tammy



 [[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.



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

__
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] Changing data frame column headings


On Wed, 5 Dec 2012, R. Michael Weylandt wrote:


Can you be more explicit about your problem?


Michael,

  Data frame contains water chemistry data; site, date, parameter, value.
The column names after dcast() are, for example, alk_quant, ph_quant,
tds_quant. I wanted to remove the '_quant' from each column header.

  Using names() with a string vector of the new names did not work, so I
specified each one and names() made the changes.

Rich

__
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] Changing data frame column headings


On Wed, 5 Dec 2012, arun wrote:


I am not sure why ?rename() is not working.
 a - list(a = 1, b = 2, c = 3)
     rename(a, c(b = a, c = b, a=c))


  I have the reshape2 library loaded and ?rename did not find the help page.
Are the parentheses required in the command?

Thanks,

Rich

__
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] In factor analysis in the psych package, how can I work out which factors the columns in $scores relate to? How do I know what each of the scores is scoring?

2012-12-05 Thread David L Carlson

Without seeing what options you have specified in your call to fa(), it is
not possible to answer the question. There are detailed discussions in ?fa
and ?factor.scores in the psych package, but for the final word you should
probably contact the package maintainer: 

Package: psych
Version: 1.2.8
Date: 2012-08-08
Title: Procedures for Psychological, Psychometric, and Personality
Research
Author: William Revelle reve...@northwestern.edu
Maintainer: William Revelle reve...@northwestern.edu

--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Brent Caldwell
 Sent: Wednesday, December 05, 2012 2:53 PM
 To: r-help@R-project.org
 Subject: [R] In factor analysis in the psych package, how can I work
 out which factors the columns in $scores relate to? How do I know what
 each of the scores is scoring?
 
 Hi
 I have used fa() to perform a factor analysis of a psychological
 battery which is thought to have 11 factors.  I can identify which
 factors the loadings relate to easily enough because I can see which
 items are loading onto each of the columns in the $loading output.
 However, how can I identify which items or loadings are being used to
 create each of the columns in the $scores output?  I have used
 generalised linear models which have shown that some of the scores are
 significant predictors of treatment outcome, but I can't work out which
 of the 11 factors they are scoring?
 
 when I export to csv the $loadings I get the following columns in this
 order:
  MR4, MR3, MR2, MR6, MR7, MR5, MR1, MR8, MR10, MR9, MR11
 
 When I export to csv the $scores I get the following columns in this
 order:
 V1, V2, V3, V4, V5, V6, V7, V8, V9, V10,V11
 
 Are the scores in the column V1 derived from the loadings in column MR1
 ?
 Or are the scores in the first column of $scores (ie V1) derived from
 the loadings in the first column of $loadings (ie MR4)?
 
 Thank you very much for your guidance, and your patience with my
 confusion
 Best wishes
 Brent
 
 
 Brent Caldwell, MBChB, DPH, MPH.
 Research Fellow
 Department of Medicine
 University of Otago, Wellington
 New Zealand
 U  brent.caldw...@otago.ac.nz
 b  04 918 6041
021 87 22 64
   23a Mein Street, PO Box 7343, Newtown, WELLINGTON 6021, NEW ZEALAND
 brent.ower.caldwell
   Zonnic study
 
 __
 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-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] data manipulation between vector and matrix

HI,
In addition to ?sweep(), you can use

apply(-mat,1,`+`,x) 

#or
library(plyr)
aaply(-mat,1,+,x) 


A.K.






- Original Message -
From: C W tmrs...@gmail.com
To: Sarah Goslee sarah.gos...@gmail.com
Cc: r-help r-help@r-project.org
Sent: Wednesday, December 5, 2012 1:51 PM
Subject: Re: [R] data manipulation between vector and matrix

Thanks, Sarah.  First time heard about sweep(), it worked just the way I
wanted.
Mike

On Wed, Dec 5, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.com wrote:

 Hi,

 On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
  Dear list,
  I was curious how to subtract a vector from matrix?
 
  Say, I have
 
  mat - matrix(1:40, nrow=20, ncol=2)
 
  x -c(1,2)

 Thanks for the actual reproducible example.

  I want,
 
  x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
  of x against column elements in mat.
 
  But x-mat won't do it.

 This will (note the modification to get x - mat):
  sweep(-mat, 2, x, +)
       [,1] [,2]
  [1,]    0  -19
  [2,]   -1  -20
  [3,]   -2  -21
  [4,]   -3  -22
  [5,]   -4  -23
 etc.

 --
 Sarah Goslee
 http://www.functionaldiversity.org


    [[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-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] Changing data frame column headings

2012-12-05 Thread R. Michael Weylandt

On Wed, Dec 5, 2012 at 9:23 PM, Rich Shepard rshep...@appl-ecosys.com wrote:
 On Wed, 5 Dec 2012, R. Michael Weylandt wrote:

 Can you be more explicit about your problem?


 Michael,

   Data frame contains water chemistry data; site, date, parameter, value.
 The column names after dcast() are, for example, alk_quant, ph_quant,
 tds_quant. I wanted to remove the '_quant' from each column header.

   Using names() with a string vector of the new names did not work, so I
 specified each one and names() made the changes.

 Rich

I'm afraid I'm still having trouble visualizing -- perhaps you could
work up a reproducible example?

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

Michael

__
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] data manipulation between vector and matrix

Thanks for the benchmark.  I actually wanted to go with the winner, except
the x-t(mat) output is very different than the others.
Mike

On Wed, Dec 5, 2012 at 4:40 PM, arun smartpink...@yahoo.com wrote:

 Hi,

 By comparing the different methods:
 set.seed(5)
  mat1-matrix(sample(1:1e6,1e6,replace=TRUE),ncol=1)
  set.seed(25)
  x-sample(1:1e6,1,replace=TRUE)
  system.time(z1-sweep(-mat1,2,x,+))
 #   user  system elapsed
  # 0.076   0.000   0.069
  system.time(z2-apply(-mat1,1,`+`,x))
  #  user  system elapsed
  # 0.036   0.000   0.031
  system.time(z3-aaply(-mat1,1,`+`,x))
 #   user  system elapsed
 #  1.880   0.000   1.704
  system.time(z4- x-t(mat1))  #winner
 #   user  system elapsed
  # 0.004   0.000   0.007
  system.time(z5- t(x-t(mat1)))
 #   user  system elapsed
 #  0.008   0.000   0.009


 A.K.





 
 From: C W tmrs...@gmail.com
 To: arun smartpink...@yahoo.com
 Cc: R help r-help@r-project.org; Sarah Goslee sarah.gos...@gmail.com
 Sent: Wednesday, December 5, 2012 4:11 PM
 Subject: Re: [R] data manipulation between vector and matrix


 thanks, I knew about apply, but did not you you can put plus signs with
 quotes.  That's a cool tricky,
 Mike


 On Wed, Dec 5, 2012 at 4:05 PM, arun smartpink...@yahoo.com wrote:

 HI,
 In addition to ?sweep(), you can use
 
 apply(-mat,1,`+`,x)
 
 #or
 library(plyr)
 aaply(-mat,1,+,x)
 
 
 A.K.
 
 
 
 
 
 
 
 - Original Message -
 From: C W tmrs...@gmail.com
 To: Sarah Goslee sarah.gos...@gmail.com
 Cc: r-help r-help@r-project.org
 Sent: Wednesday, December 5, 2012 1:51 PM
 Subject: Re: [R] data manipulation between vector and matrix
 
 Thanks, Sarah.  First time heard about sweep(), it worked just the way I
 wanted.
 Mike
 
 On Wed, Dec 5, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.com
 wrote:
 
  Hi,
 
  On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
   Dear list,
   I was curious how to subtract a vector from matrix?
  
   Say, I have
  
   mat - matrix(1:40, nrow=20, ncol=2)
  
   x -c(1,2)
 
  Thanks for the actual reproducible example.
 
   I want,
  
   x-mat[1,] and x-mat[2,], and so on... Basically, subtract column
 elements
   of x against column elements in mat.
  
   But x-mat won't do it.
 
  This will (note the modification to get x - mat):
   sweep(-mat, 2, x, +)
[,1] [,2]
   [1,]0  -19
   [2,]   -1  -20
   [3,]   -2  -21
   [4,]   -3  -22
   [5,]   -4  -23
  etc.
 
  --
  Sarah Goslee
  http://www.functionaldiversity.org
 
 
 [[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.
 
 


[[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.

Re: [R] Changing data frame column headings

2012-12-05 Thread David L Carlson

If you check, the help files, you will find rename in reshape and plyr, but
not reshape2. But you have never shown us the command you used with names()
and what didn't work:

 a - data.frame(alk_quant=rnorm(5, 5), ph_quant= rnorm(5, 5), 
+ tds_quant=rnorm(5, 5))
 a
  alk_quant ph_quant tds_quant
1  6.569293 5.494560  5.521039
2  4.854873 5.612902  5.235817
3  4.636218 5.116499  5.973769
4  5.430009 6.273394  3.511017
5  5.714755 6.876349  4.035907
 names(a) - gsub(_quant, , names(a))
 a
   alk   ph  tds
1 6.569293 5.494560 5.521039
2 4.854873 5.612902 5.235817
3 4.636218 5.116499 5.973769
4 5.430009 6.273394 3.511017
5 5.714755 6.876349 4.035907

--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352




 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Rich Shepard
 Sent: Wednesday, December 05, 2012 3:25 PM
 To: R help
 Subject: Re: [R] Changing data frame column headings
 
 On Wed, 5 Dec 2012, arun wrote:
 
  I am not sure why ?rename() is not working.
   a - list(a = 1, b = 2, c = 3)
       rename(a, c(b = a, c = b, a=c))
 
I have the reshape2 library loaded and ?rename did not find the help
 page.
 Are the parentheses required in the command?
 
 Thanks,
 
 Rich
 
 __
 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-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] Changing data frame column headings


On Wed, 5 Dec 2012, arun wrote:


You can get ?rename() by either loading library(reshape) or library(plyr).


A.K.

  I wondered about that, but assumed that reshape functions would also be
found in reshape2. I now know that's not the case. :-)

Much appreciated,

Rich

__
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] stiff delay differential equations

2012-12-05 Thread Ben Bolker

Suzen, Mehmet msuzen at gmail.com writes:

 
 Hello List,
 
 Can you recommend me if odeSolve can handle stiff delay differential equations
 with discontinuities? Or any other package?
 Best,
 -m
 
 


  I don't know, but I think you're probably best off trying it out for
yourself and seeing how it works -- the start-up cost shouldn't
be *that* high. Looking at

http://finzi.psych.upenn.edu/R/library/deSolve/html/dede.html

is useful.

 Also,

library(sos)
findFn({delay differential equation})

 might be helpful.

__
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] duplicated() with long vectors

Hi Sarah,

Thanks a lot for your explanation. I was mistakenly under the
impression that duplicated() only looked at immediately preceding
element, not all preceding elements.

What I was trying to do was get a vector saying, for each item,
whether that item is the same as the preceding item. Now that I think
of it, I could do this easily by copying the vector, shifting it over
one (by removing the first element and adding something to the end),
and then just compare the elements of the two vectors directly.

Best,
Steve

On Wed, Dec 5, 2012 at 3:08 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
 Hi,

 duplicated() doesn't just look at consecutive values, but anywhere in
 the object. Since your 12320-element vector has only 48 separate
 values, and all of them occur before the last 30 elements, so
 duplicated() returns TRUE.

 You might be looking for something involving rle(). What are you
 trying to accomplish?

 Sarah

 On Wed, Dec 5, 2012 at 3:53 PM, Stephen Politzer-Ahles
 politzerahl...@gmail.com wrote:
 Hello,

 duplicated() does not seem to work for a long vector. For example, if
 you download the data from
 https://docs.google.com/open?id=0B6-m45Jvl3ZmNmpaSlJWMXo5bmc (a vector
 with about 12,000 numbers) and then run the following code which does
 duplicated() over the whole vector but just shows the last 30
 elements:

 data.frame( tail(verylong, 30), tail(duplicated(verylong), 30) )

 you'll see that at the end of the very long vector everything is
 listed as a duplicate of the preceding element (even though it
 shouldn't be). On the other hand, if you run the following code which
 just takes out the last 30 elements of the vector and does duplicated
 on them:

 data.frame( tail(verylong, 30), duplicated(tail(verylong, 30)) )

 you get the correct results (FALSE shows up wherever the value in the
 first column changes). Does anyone know why this happens, and if
 there's a fix? I notice the documentation for duplicated() says: Long
 vectors are supported for the default method of duplicated, but may
 only be usable if nmax is supplied.  But I've tried running this with
 a high value of nmax given, and it still gives me the same problem.

 So far the only way I've figured out to get this duplicated()-like
 vector is to use a for loop going through one item at a time, but that
 takes about a minute to run.

 Best,
 Steve Politzer-Ahles



 --
 Sarah Goslee
 http://www.functionaldiversity.org



-- 
Stephen Politzer-Ahles
University of Kansas
Linguistics Department
http://people.ku.edu/~sjpa/

__
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] duplicated() with long vectors

2012-12-05 Thread Sarah Goslee

 What I was trying to do was get a vector saying, for each item,
 whether that item is the same as the preceding item. Now that I think
 of it, I could do this easily by copying the vector, shifting it over
 one (by removing the first element and adding something to the end),
 and then just compare the elements of the two vectors directly.

Right. Did you look at rle() yet?

Though for your particular simple case,

 system.time(verylong[1:(n-1)] == verylong[2:n])
   user  system elapsed
  0.001   0.000   0.002

is nearly instantaneous.



On Wed, Dec 5, 2012 at 5:04 PM, Stephen Politzer-Ahles
politzerahl...@gmail.com wrote:
 Hi Sarah,

 Thanks a lot for your explanation. I was mistakenly under the
 impression that duplicated() only looked at immediately preceding
 element, not all preceding elements.

 What I was trying to do was get a vector saying, for each item,
 whether that item is the same as the preceding item. Now that I think
 of it, I could do this easily by copying the vector, shifting it over
 one (by removing the first element and adding something to the end),
 and then just compare the elements of the two vectors directly.

 Best,
 Steve

 On Wed, Dec 5, 2012 at 3:08 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
 Hi,

 duplicated() doesn't just look at consecutive values, but anywhere in
 the object. Since your 12320-element vector has only 48 separate
 values, and all of them occur before the last 30 elements, so
 duplicated() returns TRUE.

 You might be looking for something involving rle(). What are you
 trying to accomplish?

 Sarah

 On Wed, Dec 5, 2012 at 3:53 PM, Stephen Politzer-Ahles
 politzerahl...@gmail.com wrote:
 Hello,

 duplicated() does not seem to work for a long vector. For example, if
 you download the data from
 https://docs.google.com/open?id=0B6-m45Jvl3ZmNmpaSlJWMXo5bmc (a vector
 with about 12,000 numbers) and then run the following code which does
 duplicated() over the whole vector but just shows the last 30
 elements:

 data.frame( tail(verylong, 30), tail(duplicated(verylong), 30) )

 you'll see that at the end of the very long vector everything is
 listed as a duplicate of the preceding element (even though it
 shouldn't be). On the other hand, if you run the following code which
 just takes out the last 30 elements of the vector and does duplicated
 on them:

 data.frame( tail(verylong, 30), duplicated(tail(verylong, 30)) )

 you get the correct results (FALSE shows up wherever the value in the
 first column changes). Does anyone know why this happens, and if
 there's a fix? I notice the documentation for duplicated() says: Long
 vectors are supported for the default method of duplicated, but may
 only be usable if nmax is supplied.  But I've tried running this with
 a high value of nmax given, and it still gives me the same problem.

 So far the only way I've figured out to get this duplicated()-like
 vector is to use a for loop going through one item at a time, but that
 takes about a minute to run.

 Best,
 Steve Politzer-Ahles




--
Sarah Goslee
http://www.functionaldiversity.org

__
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] Changing data frame column headings


On Wed, 5 Dec 2012, David L Carlson wrote:


names(a) - gsub(_quant, , names(a))
a


David,

  I did not pick that up from the names() help page. Thanks for the insight.

Rich

__
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] error reading xlsm file with read.xls


On Dec 4, 2012, at 10:05 AM, Sebastian Kruk wrote:

 Dear all,
 
 I cannot reading a .xlsm file using read.xls.
 

That doesn't surprise me. It's a macro format. Why should R be reading Excel 
macros?


 I executed:
 
 read.xls(resultados.xlsm,
 colNames = TRUE,
 sheet = 1,
 type = data.frame,
 from = 1,
 rowNames = NA,
 colClasses = character,
 checkNames = TRUE,
 dateTime = numeric,
 naStrings = NA,
 stringsAsFactors = F)
 Error:
 
 Call(ReadXls, file, colNames, sheet, type, from, rowNames, :
 Incorrect number of arguments (11), expecting 10 for 'ReadXls'
 If I just write
 
 read.xls(resultados.xlsm)
 It give me the same error.

-- 

David Winsemius, MD
Alameda, CA, USA

__
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] duplicated() with long vectors

2012-12-05 Thread Prof Brian Ripley


On 05/12/2012 21:08, Sarah Goslee wrote:

Hi,

duplicated() doesn't just look at consecutive values, but anywhere in
the object. Since your 12320-element vector has only 48 separate
values, and all of them occur before the last 30 elements, so
duplicated() returns TRUE.

You might be looking for something involving rle(). What are you
trying to accomplish?


And BTW, 'long vector' is a technical term in R: not 12,000, but more 
than 2 billion elements.  You will hear it a lot more in the run-up to 
the next 'minor' release of R (currently R-devel, maybe 2.16.0-to-be, 
which is the only version from which that quote comes that I am aware of).


The posting guide asked for 'at a minimum' information: if you are using 
an unreleased development version of R you really must tell us (and 
should not be reporting to the R-help list).




Sarah

On Wed, Dec 5, 2012 at 3:53 PM, Stephen Politzer-Ahles
politzerahl...@gmail.com wrote:

Hello,

duplicated() does not seem to work for a long vector. For example, if
you download the data from
https://docs.google.com/open?id=0B6-m45Jvl3ZmNmpaSlJWMXo5bmc (a vector
with about 12,000 numbers) and then run the following code which does
duplicated() over the whole vector but just shows the last 30
elements:

data.frame( tail(verylong, 30), tail(duplicated(verylong), 30) )

you'll see that at the end of the very long vector everything is
listed as a duplicate of the preceding element (even though it
shouldn't be). On the other hand, if you run the following code which
just takes out the last 30 elements of the vector and does duplicated
on them:

data.frame( tail(verylong, 30), duplicated(tail(verylong, 30)) )

you get the correct results (FALSE shows up wherever the value in the
first column changes). Does anyone know why this happens, and if
there's a fix? I notice the documentation for duplicated() says: Long
vectors are supported for the default method of duplicated, but may
only be usable if nmax is supplied.  But I've tried running this with
a high value of nmax given, and it still gives me the same problem.

So far the only way I've figured out to get this duplicated()-like
vector is to use a for loop going through one item at a time, but that
takes about a minute to run.

Best,
Steve Politzer-Ahles



--
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 reset R settings to original state other than remove .Rdata and .Rhistory

2012-12-05 Thread Greg Snow

Read the ?Startup page, it documents the files that are read on startup and
how to skip them, it may suggest another way to start a fresh session
without deleting files to see if that works for you, it also gives other
files or sources that you may investigate to find the differences in the 2
machines.


On Tue, Dec 4, 2012 at 9:53 AM, Tian3507 dbb_t...@hotmail.com wrote:


 Dear all,
 Do you know how to return all the R settings to original state? Other than
 .Rdata and .Rhistory

 Some weid thing happened to my machine.  I was trying to get shaded
 confidence band ploted using survplot from rms liberary.

 It worked on one machine, but not on the other. I tried unstall R and
 reinstall R and remove. Rdata and .Rhisotory. Nothing helped so far.

 Thanks for your input.
 Best regards,
 Hong

 Codes are
 library(rms)
 data generation
 n - 1000
 set.seed(731)
 age - 50 + 12*rnorm(n)
 label(age) - Age
 sex - factor(sample(c('male','female'), n, TRUE))
 cens - 15*runif(n)
 h - .02*exp(.04*(age-50)+.8*(sex=='female'))
 dt - -log(runif(n))/h
 label(dt) - 'Follow-up Time'
 e - ifelse(dt = cens,1,0)
 dt - pmin(dt, cens)
 units(dt) - 'Year'
 dd - datadist(age, sex)
 options(datadist='dd')
 S - Surv(dt,e)
 tte-data.frame(dt,e,age,sex)
 male group KM curve with bands#
 f - survfit(Surv(dt,e)~1, data=tte,subset=sex=='male')
 survplot(f,conf=bands)
 female group KM curve with bands#
 g - survfit(Surv(dt,e)~1, data=tte,subset=sex!='male')
 survplot(g,conf=bands, col=3, add=T)
 [[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.




-- 
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com

[[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.

Re: [R] duplicated() with long vectors

Sorry, that's my mistake, I should not have said 'long vector'; mine
is just a normal vector. I'm not actually using a development version.

Best,
Steve

On Wed, Dec 5, 2012 at 4:22 PM, Prof Brian Ripley rip...@stats.ox.ac.uk wrote:


 And BTW, 'long vector' is a technical term in R: not 12,000, but more than 2
 billion elements.  You will hear it a lot more in the run-up to the next
 'minor' release of R (currently R-devel, maybe 2.16.0-to-be, which is the
 only version from which that quote comes that I am aware of).

 The posting guide asked for 'at a minimum' information: if you are using an
 unreleased development version of R you really must tell us (and should not
 be reporting to the R-help list).



 Sarah

 On Wed, Dec 5, 2012 at 3:53 PM, Stephen Politzer-Ahles
 politzerahl...@gmail.com wrote:

 Hello,

 duplicated() does not seem to work for a long vector. For example, if
 you download the data from
 https://docs.google.com/open?id=0B6-m45Jvl3ZmNmpaSlJWMXo5bmc (a vector
 with about 12,000 numbers) and then run the following code which does
 duplicated() over the whole vector but just shows the last 30
 elements:

 data.frame( tail(verylong, 30), tail(duplicated(verylong), 30) )

 you'll see that at the end of the very long vector everything is
 listed as a duplicate of the preceding element (even though it
 shouldn't be). On the other hand, if you run the following code which
 just takes out the last 30 elements of the vector and does duplicated
 on them:

 data.frame( tail(verylong, 30), duplicated(tail(verylong, 30)) )

 you get the correct results (FALSE shows up wherever the value in the
 first column changes). Does anyone know why this happens, and if
 there's a fix? I notice the documentation for duplicated() says: Long
 vectors are supported for the default method of duplicated, but may
 only be usable if nmax is supplied.  But I've tried running this with
 a high value of nmax given, and it still gives me the same problem.

 So far the only way I've figured out to get this duplicated()-like
 vector is to use a for loop going through one item at a time, but that
 takes about a minute to run.

 Best,
 Steve Politzer-Ahles



 --
 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



-- 
Stephen Politzer-Ahles
University of Kansas
Linguistics Department
http://people.ku.edu/~sjpa/

__
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] stiff delay differential equations

2012-12-05 Thread Thomas Petzoldt



On 12/5/2012 9:15 AM, Suzen, Mehmet wrote:

Hello List,

Can you recommend me if odeSolve can handle stiff delay differential
equations with discontinuities? Or any other package? Best, -m


Package deSolve (the successor of odesolve) can candle:

- stiff differential equations: yes (solvers: lsoda, lsode, vode, vode,
radau)

- delay differential equations: yes (function dede, that calls lsoda,
lsode, vode, radau, ...)

- discontinuities: yes (events and root finding)

A combination is also possible, but please keep in mind that this can
become difficult by definition, so that it is good modelling practise
to start with a simple system first - and to introduce all the
complications only when necessary.

I am not aware that any other package contains all the requested
features, but if please let me know if there is something new and we'll
add it to the DifferentialEquations task view.

Thomas Petzoldt

__
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] What is print print print ?

2012-12-05 Thread Jim Lemon


On 12/06/2012 12:03 AM, Vladimir eremeev wrote:

Hi all.

What is print print print?

I don't see output of the print command in for loop and have found this
link:
http://stackoverflow.com/questions/1816200/chisq-test-doesnt-print-results-when-in-a-loop

It describes a problem, similar to mine.
My problem. I want to execute print command in for loop.

If I copy for loop body with print() and paste it to console, I don't see
any output.

If I type the command, I do see the output.
If I paste the command by parts I also see the output.

If I construct the command by picking parts of the history with the up
arrow, I also don't see any output.

Search for print print print still didn't bring anything useful.


Hi Vladimir,
Your problem is twofold. First, the print command simply tries to 
display whatever is passed as the first argument to it. Invoking:


print()

prints nothing, as there is nothing passed to display. Second, the print 
command is often used in R scripts because most functions will display 
their return values when invoked directly in the R console, but not when 
invoked within a for loop. The problem that you mentioned was due to 
the chisq.test function displaying its return value on its own, but not 
within the loop. The print print print comment was intended to mean:


print(rownames(cl.vs.Onerall)[i])
print(chisq.test(observed, p=expected.fr))
print(--)

in the loop so that the successive return values would be displayed. Try 
it for yourself with this simple example:


x-1:5
length(x)
for(i in 1:3) length(x)
for(i in 1:3) print(length(x))

Jim

__
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] error reading xlsm file with read.xls


On Dec 5, 2012, at 2:19 PM, David Winsemius wrote:

 
 On Dec 4, 2012, at 10:05 AM, Sebastian Kruk wrote:
 
 Dear all,
 
 I cannot reading a .xlsm file using read.xls.
 
 
 That doesn't surprise me. It's a macro format. Why should R be reading Excel 
 macros?
 
 
 I executed:
 
 read.xls(resultados.xlsm,
 colNames = TRUE,
 sheet = 1,
 type = data.frame,
 from = 1,
 rowNames = NA,
 colClasses = character,
 checkNames = TRUE,
 dateTime = numeric,
 naStrings = NA,
 stringsAsFactors = F)

The other point I would make is that you should be looking at the help page for 
read.xls and making sure that on the off-chance that the authors did intend for 
you to have access to Excel macros for some readsin in R tha thtey were using 
the same aruments. The error message says you are having problems with you 
argument list and looking at the help page makes me think you are using 
read.tabe arguments for read.xls. Not a wise idea.This is the list of formals:

read.xlsx(file, sheetIndex, sheetName=NULL, rowIndex=NULL,
  colIndex=NULL, as.data.frame=TRUE, header=TRUE, colClasses=NA,
  keepFormulas=FALSE, encoding=unknown, ...)

And it says the 'dots' are sent to data.frame rather than to read.table.

 Error:
 
 Call(ReadXls, file, colNames, sheet, type, from, rowNames, :
 Incorrect number of arguments (11), expecting 10 for 'ReadXls'
 If I just write
 
 read.xls(resultados.xlsm)
 It give me the same error.
 
 -- 
 
 David Winsemius, MD
 Alameda, CA, USA
 
 __
 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
Alameda, CA, USA

__
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] In factor analysis in the psych package, how can I work out which factors the columns in $scores relate to? How do I know what each of the scores is scoring?

2012-12-05 Thread Brent Caldwell

Dear Prof Carlson
Thank you very much.  As far as I can tell, factor.scores requires an object 
from class grm, itm, rason, or tpm; which I think are part of the irt package 
with which I am unfamiliar.
I had used the psych package to do the following factor analysis:
fa.11factors.rawdata - fa(WISDMrawdataframe,nfactors=11)
# WISDMrawdataframe has 37 columns and 2408 rows, and contain 1051 rows that 
are NA (having NA in this dataframe allows the raw data to have the same length 
as my vector of ID numbers so I can work out which participants the data belong 
to)
fa.11factors.rawdata$loadings #on the computer screen this gives me the 
loadings with column namesMR4MR3MR2MR6MR7MR5MR1
MR8MR10   MR9MR11
fa.11factors.rawdata$scores # on the computer screen this gives me the scores 
with column names  [,1]   [,2][,3]   [,4]
[,5]   [,6]   [,7]   [,8] [,9]  [,10]   [,11]
Am I safe in assuming that the scores in the first column relate to the 
loadings in the column called MR4?  
If this is a safe assumption, then I can work out which factor the scores 
relate to, by looking at which items load onto MR4MR3MR2MR6MR7  
  MR5MR1MR8MR10   MR9MR11
Thank you, I've taken your advice and cc'd this to Prof William Revelle
Thank you
Best wishes
Brent

-Original Message-
From: David L Carlson [mailto:dcarl...@tamu.edu] 
Sent: Thursday, 6 December 2012 10:33 a.m.
To: Brent Caldwell; 'r-help@R-project.org'
Subject: RE: [R] In factor analysis in the psych package, how can I work out 
which factors the columns in $scores relate to? How do I know what each of the 
scores is scoring?

Without seeing what options you have specified in your call to fa(), it is not 
possible to answer the question. There are detailed discussions in ?fa and 
?factor.scores in the psych package, but for the final word you should probably 
contact the package maintainer: 

Package: psych
Version: 1.2.8
Date: 2012-08-08
Title: Procedures for Psychological, Psychometric, and Personality
Research
Author: William Revelle reve...@northwestern.edu
Maintainer: William Revelle reve...@northwestern.edu

--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- 
 project.org] On Behalf Of Brent Caldwell
 Sent: Wednesday, December 05, 2012 2:53 PM
 To: r-help@R-project.org
 Subject: [R] In factor analysis in the psych package, how can I work 
 out which factors the columns in $scores relate to? How do I know what 
 each of the scores is scoring?
 
 Hi
 I have used fa() to perform a factor analysis of a psychological 
 battery which is thought to have 11 factors.  I can identify which 
 factors the loadings relate to easily enough because I can see which 
 items are loading onto each of the columns in the $loading output.
 However, how can I identify which items or loadings are being used to 
 create each of the columns in the $scores output?  I have used 
 generalised linear models which have shown that some of the scores are 
 significant predictors of treatment outcome, but I can't work out 
 which of the 11 factors they are scoring?
 
 when I export to csv the $loadings I get the following columns in this
 order:
  MR4, MR3, MR2, MR6, MR7, MR5, MR1, MR8, MR10, MR9, MR11
 
 When I export to csv the $scores I get the following columns in this
 order:
 V1, V2, V3, V4, V5, V6, V7, V8, V9, V10,V11
 
 Are the scores in the column V1 derived from the loadings in column 
 MR1 ?
 Or are the scores in the first column of $scores (ie V1) derived from 
 the loadings in the first column of $loadings (ie MR4)?
 
 Thank you very much for your guidance, and your patience with my 
 confusion Best wishes Brent
 
 
 Brent Caldwell, MBChB, DPH, MPH.
 Research Fellow
 Department of Medicine
 University of Otago, Wellington
 New Zealand
 U  brent.caldw...@otago.ac.nz
 b  04 918 6041
021 87 22 64
   23a Mein Street, PO Box 7343, Newtown, WELLINGTON 6021, NEW ZEALAND
 brent.ower.caldwell
   Zonnic study
 
 __
 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-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] In factor analysis in the psych package, how can I work out which factors the columns in $scores relate to? How do I know what each of the scores is scoring?

2012-12-05 Thread David L Carlson

I think you are safe in assuming that the first factor is used to compute the 
first factor score column. The factor.scores() procedure is used by fa() to 
compute the factor scores so it provides details regarding how they are 
computed beyond the information in the fa() help page. As I understand the help 
file, the default rotation is oblimin so 

In the oblique case, the factor loadings are referred to as Pattern 
coefficients and are related to the Structure coefficients by S = P and thus P 
= S1. When estimating factor scores, fa and factanal differ in that fa finds 
the factors from the Structure matrix and factanal seems to find them from the 
Pattern matrix. Thus, although in the orthogonal case, fa and factanal agree 
perfectly in their factor score estimates, they do not agree in the case of 
oblique factors. Setting oblique.scores = FALSE will produce factor score 
estimate that match those of factanal.

So you might want to look at $Structure.

The $scores part of the output is a matrix without column names (they are just 
numbered). The V1, V2, . . . is added by the write.csv() function and is not 
part of the actual output from fa(). 

---
David

 -Original Message-
 From: Brent Caldwell [mailto:brent.caldw...@otago.ac.nz]
 Sent: Wednesday, December 05, 2012 7:08 PM
 To: dcarl...@tamu.edu; 'r-help@R-project.org'
 Cc: William R Revelle (reve...@northwestern.edu)
 Subject: RE: [R] In factor analysis in the psych package, how can I
 work out which factors the columns in $scores relate to? How do I know
 what each of the scores is scoring?
 
 Dear Prof Carlson
 Thank you very much.  As far as I can tell, factor.scores requires an
 object from class grm, itm, rason, or tpm; which I think are part of
 the irt package with which I am unfamiliar.
 I had used the psych package to do the following factor analysis:
 fa.11factors.rawdata - fa(WISDMrawdataframe,nfactors=11)
 # WISDMrawdataframe has 37 columns and 2408 rows, and contain 1051 rows
 that are NA (having NA in this dataframe allows the raw data to have
 the same length as my vector of ID numbers so I can work out which
 participants the data belong to)
 fa.11factors.rawdata$loadings #on the computer screen this gives me the
 loadings with column namesMR4MR3MR2MR6MR7MR5
 MR1MR8MR10   MR9MR11
 fa.11factors.rawdata$scores # on the computer screen this gives me the
 scores with column names  [,1]   [,2][,3]
 [,4][,5]   [,6]   [,7]   [,8] [,9]
 [,10]   [,11]
 Am I safe in assuming that the scores in the first column relate to the
 loadings in the column called MR4?
 If this is a safe assumption, then I can work out which factor the
 scores relate to, by looking at which items load onto MR4MR3MR2
 MR6MR7MR5MR1MR8MR10   MR9MR11
 Thank you, I've taken your advice and cc'd this to Prof William Revelle
 Thank you
 Best wishes
 Brent
 
 -Original Message-
 From: David L Carlson [mailto:dcarl...@tamu.edu]
 Sent: Thursday, 6 December 2012 10:33 a.m.
 To: Brent Caldwell; 'r-help@R-project.org'
 Subject: RE: [R] In factor analysis in the psych package, how can I
 work out which factors the columns in $scores relate to? How do I know
 what each of the scores is scoring?
 
 Without seeing what options you have specified in your call to fa(), it
 is not possible to answer the question. There are detailed discussions
 in ?fa and ?factor.scores in the psych package, but for the final word
 you should probably contact the package maintainer:
 
 Package: psych
 Version: 1.2.8
 Date: 2012-08-08
 Title: Procedures for Psychological, Psychometric, and Personality
 Research
 Author: William Revelle reve...@northwestern.edu
 Maintainer: William Revelle reve...@northwestern.edu
 
 --
 David L Carlson
 Associate Professor of Anthropology
 Texas AM University
 College Station, TX 77843-4352
 
  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
  project.org] On Behalf Of Brent Caldwell
  Sent: Wednesday, December 05, 2012 2:53 PM
  To: r-help@R-project.org
  Subject: [R] In factor analysis in the psych package, how can I work
  out which factors the columns in $scores relate to? How do I know
 what
  each of the scores is scoring?
 
  Hi
  I have used fa() to perform a factor analysis of a psychological
  battery which is thought to have 11 factors.  I can identify which
  factors the loadings relate to easily enough because I can see which
  items are loading onto each of the columns in the $loading output.
  However, how can I identify which items or loadings are being used to
  create each of the columns in the $scores output?  I have used
  generalised linear models which have shown that some of the scores
 are
  significant predictors of treatment outcome, but I can't work out
  which of the 11 factors they are scoring?
 
  when I

Re: [R] catching errors in RNetCDF

2012-12-05 Thread Pascal Oettli


Hi,

When I try your example, I simply get an error:

 library(RNetCDF)
 con - create.nc('test.nc')
 test - try(var.get.nc(con, 'dummy'))
Error : NetCDF: Variable not found

Regards,
Pascal


Le 05/12/2012 22:08, Jannis a écrit :

Dear R community,


I quite frequently run into errors while using the RNetCDF package which
do not seem to be recognised as normal R errors and, hence, do not stop
the execution of the code making it hard to debug the code.

Consider, for example:

library(RNetCDF)
con - create.nc('test.nc')
test - try(var.get.nc(con, 'dummy'))


In this case, some sort of error message is printed to the screen, but R
does not recognise this as an error. Is there any way to solve this?

I contacted the author of the package but it seems that there will be no
solution from that side.


Any Ideas?


Cheers
Jannis

__
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-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] Assignment of values with different indexes

2012-12-05 Thread Brian Feeny


I would like to take the values of observations and map them to a new index.  I 
am not sure how to accomplish this.  The result would look like so:

x[1,2,3,4,5,6,7,8,9,10]
becomes
y[2,4,6,8,10,12,14,16,18,20]

The newindex would not necessarily be this sequence, but a sequence I have 
stored in a vector, so it could be all kinds of values.  here is what happens:

 x - rnorm(10)
 myindex - seq(from = 1,to = 20, by = 2)
 y - numeric()
 y[myindex] - x
 y
 [1] -0.03745988  NA -0.09078822  NA  0.92484413  NA  
0.32057426  NA
 [9]  0.01536279  NA  0.02200198  NA  0.37535438  NA  
1.46606535  NA
[17]  1.44855796  NA -0.05048738

So yes, it maps the values to my new indexes, but I have NA's.  The result I 
want would look like this instead:


 [1] -0.03745988  
 [3] -0.09078822   
 [5] 0.92484413   
 [7] 0.32057426 
 [9]  0.01536279   
 [11] 0.02200198   
 [13] 0.37535438   
 [15] 1.46606535 
 [17]  1.44855796  
 [19] -0.05048738


and remove the NA's.  I tried this with na.omit() on x, but it looks like so:

 x - rnorm(10)
 myindex - seq(from = 1,to = 20, by = 2)
 y - numeric()
 y[myindex] - na.omit(x)
 y
 [1]  0.87399523  NA -0.39908184  NA  0.14583051  NA  
0.01850755  NA
 [9] -0.47413632  NA  0.88410517  NA -1.64939190  NA  
0.57650807  NA
[17]  0.44016971  NA -0.56313802

Brian

__
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] Assignment of values with different indexes

2012-12-05 Thread Pascal Oettli


Hello,

I am sorry, but I sincerely don't understand what you are trying to do.

Regards,
Pascal

Le 06/12/2012 11:47, Brian Feeny a écrit :


I would like to take the values of observations and map them to a new index.  I 
am not sure how to accomplish this.  The result would look like so:

x[1,2,3,4,5,6,7,8,9,10]
becomes
y[2,4,6,8,10,12,14,16,18,20]

The newindex would not necessarily be this sequence, but a sequence I have 
stored in a vector, so it could be all kinds of values.  here is what happens:


x - rnorm(10)
myindex - seq(from = 1,to = 20, by = 2)
y - numeric()
y[myindex] - x
y

  [1] -0.03745988  NA -0.09078822  NA  0.92484413  NA  
0.32057426  NA
  [9]  0.01536279  NA  0.02200198  NA  0.37535438  NA  
1.46606535  NA
[17]  1.44855796  NA -0.05048738

So yes, it maps the values to my new indexes, but I have NA's.  The result I 
want would look like this instead:


  [1] -0.03745988
  [3] -0.09078822
  [5] 0.92484413
  [7] 0.32057426
  [9]  0.01536279
  [11] 0.02200198
  [13] 0.37535438
  [15] 1.46606535
  [17]  1.44855796
  [19] -0.05048738


and remove the NA's.  I tried this with na.omit() on x, but it looks like so:


x - rnorm(10)
myindex - seq(from = 1,to = 20, by = 2)
y - numeric()
y[myindex] - na.omit(x)
y

  [1]  0.87399523  NA -0.39908184  NA  0.14583051  NA  
0.01850755  NA
  [9] -0.47413632  NA  0.88410517  NA -1.64939190  NA  
0.57650807  NA
[17]  0.44016971  NA -0.56313802

Brian

__
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-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] Assignment of values with different indexes



On Dec 5, 2012, at 6:47 PM, Brian Feeny wrote:



I would like to take the values of observations and map them to a  
new index.  I am not sure how to accomplish this.  The result would  
look like so:


x[1,2,3,4,5,6,7,8,9,10]
becomes
y[2,4,6,8,10,12,14,16,18,20]


This suggests to me that you are having difficulty transfering your  
Matlab knowldege to R.





The newindex would not necessarily be this sequence, but a  
sequence I have stored in a vector, so it could be all kinds of  
values.  here is what happens:



x - rnorm(10)
myindex - seq(from = 1,to = 20, by = 2)
y - numeric()
y[myindex] - x
y
[1] -0.03745988  NA -0.09078822  NA   
0.92484413  NA  0.32057426  NA
[9]  0.01536279  NA  0.02200198  NA   
0.37535438  NA  1.46606535  NA

[17]  1.44855796  NA -0.05048738

So yes, it maps the values to my new indexes, but I have NA's.  The  
result I want would look like this instead:


The only structure in R that allows NULL values is a list. Matrices  
need to have either value or NA.





[1] -0.03745988
[3] -0.09078822
[5] 0.92484413
[7] 0.32057426
[9]  0.01536279
[11] 0.02200198
[13] 0.37535438
[15] 1.46606535
[17]  1.44855796
[19] -0.05048738


and remove the NA's.  I tried this with na.omit() on x, but it looks  
like so:



x - rnorm(10)
myindex - seq(from = 1,to = 20, by = 2)
y - numeric()
y[myindex] - na.omit(x)
y
[1]  0.87399523  NA -0.39908184  NA   
0.14583051  NA  0.01850755  NA
[9] -0.47413632  NA  0.88410517  NA  
-1.64939190  NA  0.57650807  NA

[17]  0.44016971  NA -0.56313802


However it will not display the way you were inteniding:

 y - list()
 y[seq(1,20, by=2)] - 1:10
 y
[[1]]
[1] 1

[[2]]
NULL

[[3]]
[1] 2

[[4]]
NULL

[[5]]
[1] 3

snipped

--
David.


Brian

__
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
Alameda, CA, USA

__
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] Line numbers with errors and warnings?

2012-12-05 Thread Worik R

If you `source(test.R, keep.source=FALSE)`, you will see that the
 line number is not reported.


Not always.

I have code that uses sapply to call another function and all I get back is
the line of the sapply.

Useful but in the 21st century I do think I could get more aid from the
runtime and compiler to help me.

I am giving up.

In the short term I am giving yup on expecting line numbers (running with
anything other than  options(error=NULL) is a pain for all sorts of
reasons) and in the medium term I am giving up on R.

It is sad.  R is very powerful and I can say a lot with very little.  But I
have been using it intensely for 3 years, written (and debugged) thousands
of lines of code.  It is exactly what I need *but* the debugging facilities
are too primitive and the time gained in design and implementation is lost,
with interest, in debugging.

IMO R is very suitable for simple tasks that would be complex in other
languages.  But once it gets complex in R, it is too hard.

I am not saying R is very bad -  far from it.  But for ambitious projects
it cannot compete without proper debugging facilities.

It may be that I am not good enough at programming - fair enough. But I am
what I am, and I have been beating my head against this for far too long.

Having large amounts of code in R that I use every day I am not running
away at high speed.  But I am moving away.

If R gets modern debugging (lets say up to where gdb was in the early
1990s) I'll be back with a big smile on my face.

But I cannot dictate the priorities of the R maintainers, and I will not
try.

peace
Worik Stanton
Dunedin
New Zealand

[[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.

Re: [R] Line numbers with errors and warnings?

2012-12-05 Thread Steve Lianoglou

Hi,

On Thu, Dec 6, 2012 at 12:01 AM, Worik R wor...@gmail.com wrote:


 If you `source(test.R, keep.source=FALSE)`, you will see that the
 line number is not reported.


 Not always.

 I have code that uses sapply to call another function and all I get back is
 the line of the sapply.

The function that is being called inside the sapply that throws the
error -- is it in a different package?

If you reinstall *that* package w/ `options(keep.source.pkg=TRUE)` (or
R_KEEP_PKG_SOURCE=yes in your environment if installing from cmd line
(see ?options)), does that help?

If not -- could you provide, in a similar fashion as I did w/ the
BadPackage on github from an earlier message in this thread, an
example that recapitulates this no-line-number-on-error problem and
point out where/how it happens so we can also trigger it and see? (I
have a function that does xxx is hard for anybody else to help you
with).

Also, emacs/ess also has tracebug:

http://code.google.com/p/ess-tracebug/

which may be useful.

anyway ... if you are leaving R for greener pastures, do us a favor
and send us an email w/ an update if you find Nirvana in another
language ;-)

-steve

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

__
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] Assignment of values with different indexes

2012-12-05 Thread Brian Feeny


No, because it does not assign the indexes of myindex.

If its not possible, which I am assuming its not, thats OK.  I thought that if 
I had say 10 observations, sequentially ordered (or any order, it doesn't 
matter), and I wanted to assign them specific indexes, and not have NA's, that 
it was possible.
I am OK with knowing that I can assign them the specific indexes, and that 
there will be empty spots, which are marked NA.  Most functions I would need to 
use can handle NA's by telling the function to ignore. I appreciate all the 
help that has been given.

Brian

On Dec 5, 2012, at 11:49 PM, arun smartpink...@yahoo.com wrote:

 
 
 Hi,
 
 Would it be okay to use:
  y-na.omit(y[myindex]-x)
 y
 # [1] -1.36025132 -0.57529211  1.18132359  0.41038489  1.83108252 -0.03563686
  #[7]  1.25267314  1.08311857  1.56973422 -0.30752939
 
 A.K.
 
 
 - Original Message -
 From: Brian Feeny bfe...@mac.com
 To: r-help@r-project.org help r-help@r-project.org
 Cc: 
 Sent: Wednesday, December 5, 2012 9:47 PM
 Subject: [R] Assignment of values with different indexes
 
 
 I would like to take the values of observations and map them to a new index.  
 I am not sure how to accomplish this.  The result would look like so:
 
 x[1,2,3,4,5,6,7,8,9,10]
 becomes
 y[2,4,6,8,10,12,14,16,18,20]
 
 The newindex would not necessarily be this sequence, but a sequence I have 
 stored in a vector, so it could be all kinds of values.  here is what happens:
 
 x - rnorm(10)
 myindex - seq(from = 1,to = 20, by = 2)
 y - numeric()
 y[myindex] - x
 y
 [1] -0.03745988  NA -0.09078822  NA  0.92484413  NA  
 0.32057426  NA
 [9]  0.01536279  NA  0.02200198  NA  0.37535438  NA  
 1.46606535  NA
 [17]  1.44855796  NA -0.05048738
 
 So yes, it maps the values to my new indexes, but I have NA's.  The result I 
 want would look like this instead:
 
 
 [1] -0.03745988  
 [3] -0.09078822  
 [5] 0.92484413  
 [7] 0.32057426
 [9]  0.01536279  
 [11] 0.02200198  
 [13] 0.37535438  
 [15] 1.46606535
 [17]  1.44855796  
 [19] -0.05048738
 
 
 and remove the NA's.  I tried this with na.omit() on x, but it looks like so:
 
 x - rnorm(10)
 myindex - seq(from = 1,to = 20, by = 2)
 y - numeric()
 y[myindex] - na.omit(x)
 y
 [1]  0.87399523  NA -0.39908184  NA  0.14583051  NA  
 0.01850755  NA
 [9] -0.47413632  NA  0.88410517  NA -1.64939190  NA  
 0.57650807  NA
 [17]  0.44016971  NA -0.56313802
 
 Brian
 
 __
 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-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] Line numbers with errors and warnings?

2012-12-05 Thread Worik R

Bert

R is compiled.  Compile and go cycle is quick, but it is compiled.  Not to
a machine executable but to an object the R runtinme can interperet.

Steve

It is all a bit hard.  I can make progress the old fashioned way with debug
messages and conditional execution and dozens of other techniques.  From
the 1960s!

Do not get me wrong.  i accept the situation.  I see why it is like it is.
I am being too ambitious with my R code.

As for Nirvana.  I'll not announce it here, that would be rude!

Frankly I think I'll be using Perl for data processing and C++ for heavy
lifting and the data structures I need.

I might take a quick look at Julia: http://julialang.org/  but life is
short, new languages hard.

I will not abandon R, but I have learnt some respect - I'll use R
appropriately in the future.

W


On Thu, Dec 6, 2012 at 6:33 PM, Steve Lianoglou 
mailinglist.honey...@gmail.com wrote:

 Hi,

 On Thu, Dec 6, 2012 at 12:01 AM, Worik R wor...@gmail.com wrote:
 
 
  If you `source(test.R, keep.source=FALSE)`, you will see that the
  line number is not reported.
 
 
  Not always.
 
  I have code that uses sapply to call another function and all I get back
 is
  the line of the sapply.

 The function that is being called inside the sapply that throws the
 error -- is it in a different package?

 If you reinstall *that* package w/ `options(keep.source.pkg=TRUE)` (or
 R_KEEP_PKG_SOURCE=yes in your environment if installing from cmd line
 (see ?options)), does that help?

 If not -- could you provide, in a similar fashion as I did w/ the
 BadPackage on github from an earlier message in this thread, an
 example that recapitulates this no-line-number-on-error problem and
 point out where/how it happens so we can also trigger it and see? (I
 have a function that does xxx is hard for anybody else to help you
 with).

 Also, emacs/ess also has tracebug:

 http://code.google.com/p/ess-tracebug/

 which may be useful.

 anyway ... if you are leaving R for greener pastures, do us a favor
 and send us an email w/ an update if you find Nirvana in another
 language ;-)

 -steve

 --
 Steve Lianoglou
 Graduate Student: Computational Systems Biology
  | Memorial Sloan-Kettering Cancer Center
  | Weill Medical College of Cornell University
 Contact Info: http://cbio.mskcc.org/~lianos/contact


[[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.

Re: [R] Changing data frame column headings

Hi Rich,
You can get ?rename() by either loading library(reshape) or library(plyr).
A.K.

- Original Message -
From: Rich Shepard rshep...@appl-ecosys.com
To: R help r-help@r-project.org
Cc: 
Sent: Wednesday, December 5, 2012 4:24 PM
Subject: Re: [R] Changing data frame column headings

On Wed, 5 Dec 2012, arun wrote:

 I am not sure why ?rename() is not working.
  a - list(a = 1, b = 2, c = 3)
      rename(a, c(b = a, c = b, a=c))

  I have the reshape2 library loaded and ?rename did not find the help page.
Are the parentheses required in the command?

Thanks,

Rich

__
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-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] data manipulation between vector and matrix

Hi,

By comparing the different methods:
set.seed(5)
 mat1-matrix(sample(1:1e6,1e6,replace=TRUE),ncol=1)
 set.seed(25)
 x-sample(1:1e6,1,replace=TRUE)
 system.time(z1-sweep(-mat1,2,x,+))
#   user  system elapsed 
 # 0.076   0.000   0.069 
 system.time(z2-apply(-mat1,1,`+`,x))
 #  user  system elapsed 
 # 0.036   0.000   0.031 
 system.time(z3-aaply(-mat1,1,`+`,x))
#   user  system elapsed 
#  1.880   0.000   1.704 
 system.time(z4- x-t(mat1))  #winner
#   user  system elapsed 
 # 0.004   0.000   0.007 
 system.time(z5- t(x-t(mat1)))
#   user  system elapsed 
#  0.008   0.000   0.009 


A.K.






From: C W tmrs...@gmail.com
To: arun smartpink...@yahoo.com 
Cc: R help r-help@r-project.org; Sarah Goslee sarah.gos...@gmail.com 
Sent: Wednesday, December 5, 2012 4:11 PM
Subject: Re: [R] data manipulation between vector and matrix


thanks, I knew about apply, but did not you you can put plus signs with quotes. 
 That's a cool tricky,
Mike


On Wed, Dec 5, 2012 at 4:05 PM, arun smartpink...@yahoo.com wrote:

HI,
In addition to ?sweep(), you can use

apply(-mat,1,`+`,x)

#or
library(plyr)
aaply(-mat,1,+,x)


A.K.







- Original Message -
From: C W tmrs...@gmail.com
To: Sarah Goslee sarah.gos...@gmail.com
Cc: r-help r-help@r-project.org
Sent: Wednesday, December 5, 2012 1:51 PM
Subject: Re: [R] data manipulation between vector and matrix

Thanks, Sarah.  First time heard about sweep(), it worked just the way I
wanted.
Mike

On Wed, Dec 5, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.com wrote:

 Hi,

 On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
  Dear list,
  I was curious how to subtract a vector from matrix?
 
  Say, I have
 
  mat - matrix(1:40, nrow=20, ncol=2)
 
  x -c(1,2)

 Thanks for the actual reproducible example.

  I want,
 
  x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
  of x against column elements in mat.
 
  But x-mat won't do it.

 This will (note the modification to get x - mat):
  sweep(-mat, 2, x, +)
       [,1] [,2]
  [1,]    0  -19
  [2,]   -1  -20
  [3,]   -2  -21
  [4,]   -3  -22
  [5,]   -4  -23
 etc.

 --
 Sarah Goslee
 http://www.functionaldiversity.org


    [[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-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] request

2012-12-05 Thread Rubel Das

Dear Dr. Bernhard
Thank you very much for quick reply. One good news is that i can connect
with Neos server. I appreciate your suggestion to change 'protocols' to
'proxy'. after assigning proxy, i can connect with my desired website.
Thankyou very much

With best regards
Rubel


On Thu, Dec 6, 2012 at 3:32 AM, Bernhard Pfaff bernh...@pfaffikus.dewrote:

 Hello Rubel,

 many thanks for your email. Diagnosing your problem is a bit hard
 without further information, hence I can only provide you with some
 pointers for further checking/testing:

 1) You specify implictly a port for the proxy. Is this the correct one?
 2) Specify, 'proxy' as a list element for your proxy. See
 listCurlOptions() on what is available and can be included in your list
 object for 'curlopts'.
 3) Can you establish a connection to some other site by means of the
 provided functions in RCurl and your chosen setting?

 Best,
 Bernhard


 Am Mittwoch, den 05.12.2012, 19:15 +0900 schrieb Rubel Das:
  Dear Dr. bernhard
  cc. r-help
  Thank you very much for deverlopping rneos package. I read the
  document of rneos.
  however, due to my inability, i could not figure-out  how to connect
  with neos server from R environment.
 
 
  let me explain the steps, i took. my laptop is using wireless of my
  laboratory. to connect the internet, i need proxy address and specific
  port (that i have mentioned in protocols in CreateNeosComm). However,
  the proxy server does not have any password though to access wifi i
  need password.
  After inserting the protocols as shown in Input (below), i have go the
  output as shown in below.
 
 
 
 
 
  input
  library (rneos)
  NC-CreateNeosComm(curlopts=list(httpheader=c('content-type'=text/xml,
 'User-Agent'=R), protocols=proxy.noc.titech.ac.jp:3128,
 port=3332),curlhandle=getCurlHandle())
  #to check the neos server is activeNeosServer is alive is returned
  Nping(convert=T, nc=NC)
 
 
  output
   library (rneos)
  
  NC-CreateNeosComm(curlopts=list(httpheader=c('content-type'=text/xml,
 'User-Agent'=R), protocols=proxy.noc.titech.ac.jp:3128,
 port=3332),curlhandle=getCurlHandle())
  Error in match(x, table, nomatch = 0L) :
object 'proxy.noc.titech.ac.jp' not found
   #to check the neos server is activeNeosServer is alive is returned
   Nping(convert=T, nc=NC)
  Error in .postForm(curl, .opts, .params, style) :
Stale CURL handle being passed to libcurl
  Error in curlSetOpt(writefunction = NULL, curl = curl) :
Stale CURL handle being passed to libcurl
 
 
 
 
  Since this procedure is not working, would you suggest me how can i
  fix the problem?
  i appreciate your help.
 
 
 
 
  --
  With best Regards
  Rubel Das
 
 
  Hanaoka Research Group/Transport Planning
  Dept. of International Development Engineering
  Tokyo Institute of Technology
  2-12-1-I4-12, O-okayama, Meguro-ku, Tokyo, 152-8550, Japan
 
 
  email: rubeldas2...@gmail.com
  rubel...@tp.ide.titech.ac.jp
 
 





-- 
With best Regards
Rubel Das

Hanaoka Research Group/Transport Planning
Dept. of International Development Engineering
Tokyo Institute of Technology
2-12-1-I4-12, O-okayama, Meguro-ku, Tokyo, 152-8550, Japan


email: rubeldas2...@gmail.com
rubel...@tp.ide.titech.ac.jp

[[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] bootstrap based confidence band

2012-12-05 Thread Rebecca

I'm trying to find a bootstrap based confidence band for a linear model.
I have created a data set with X and Y
X=runif(n,-1.25,1.25)
e=rnorm(n,0,1)
Y=exp(3*X)+5*sin((30*X)/(2*pi))+2*e
fit=lm(Y~X)
summary(fit)
 
I define a bootstrap function named PairedBootstrap which is not listed here. 
Than I try many ways to find the confidence band. One way is to predict Y using 
the model I get above for 1000 times.
Data.Orig=data.frame(cbind(X,Y))
B=1000
Boot.Result=matrix(nrow=B,ncol=n)
for ( b in 1:B){
 Data.Orig.Boot=PairedBootstrap(Data.Orig)
  fit.Boot=predict(fit,newdata=Data.Orig.Boot,type=response)
 Boot.Result[b,]=fit.Boot
 }
 
And try to find 95% confidence interval for 1000 copies of y corresponding to 
each x.
ConfidenceSet.Pointwise=function(Boot.Result,alpha){
  n=ncol(Boot.Result)
  B=nrow(Boot.Result)
  SetBounds=matrix(ncol=2,nrow=n)
  for(j in 1:n){
    Result.Sort=sort(Boot.Result[,j])
  SetBounds[j,1]=Result.Sort[floor(0.5*B*alpha)]
   SetBounds[j,2]=Result.Sort[ceiling(B*(1-0.5*alpha))]
  }
  return(SetBounds)
  }

And then  try to line them up. But the result is not what I want.
alpha=0.05
 Boot.Pointwise=ConfidenceSet.Pointwise(Boot.Result,alpha)
 lines(Data.Orig[,1],Boot.Pointwise[,1], col=2, lwd=3)
 
Could someone tell where I'm wrong, or is there another better way to do it?
Thank you so much!

[[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.

Re: [R] Changing data frame column headings

Hi,
This could be done using ?gsub()
set.seed(5)
dat1-data.frame(alk_quant=sample(1:15,6,replace=TRUE),ph_quant=sample(5:9,6,replace=TRUE),tds_quant=sample(10:20,6,replace=TRUE))
 names(dat1)-gsub((.*)\\_.*,\\1,names(dat1))
 head(dat1,2)
#  alk ph tds
#1   4  7  13
#2  11  9  16
A.K.




- Original Message -
From: Rich Shepard rshep...@appl-ecosys.com
To: r-help@r-project.org
Cc: 
Sent: Wednesday, December 5, 2012 4:23 PM
Subject: Re: [R] Changing data frame column headings

On Wed, 5 Dec 2012, R. Michael Weylandt wrote:

 Can you be more explicit about your problem?

Michael,

  Data frame contains water chemistry data; site, date, parameter, value.
The column names after dcast() are, for example, alk_quant, ph_quant,
tds_quant. I wanted to remove the '_quant' from each column header.

  Using names() with a string vector of the new names did not work, so I
specified each one and names() made the changes.

Rich

__
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-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] data manipulation between vector and matrix

HI,
The option z5 takes care of it.
z5-t(x-t(mat)) #still faster than ?sweep()
 dim(z5)
[1] 20  2
 identical(sweep(-mat,2,x,+),z5)
#[1] TRUE


A.K.






From: C W tmrs...@gmail.com
To: arun smartpink...@yahoo.com 
Sent: Wednesday, December 5, 2012 5:09 PM
Subject: Re: [R] data manipulation between vector and matrix


Hi Arun,
Sorry, I might be a little unclear with my words.

The dimensions are different. This is what I got:
 x-t(mat)
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]    0   -1   -2   -3   -4   -5   -6   -7   -8    -9   -10   -11   -12   -13
[2,]  -19  -20  -21  -22  -23  -24  -25  -26  -27   -28   -29   -30   -31   -32
     [,15] [,16] [,17] [,18] [,19] [,20]
[1,]   -14   -15   -16   -17   -18   -19
[2,]   -33   -34   -35   -36   -37   -38
 sweep(-mat, 2, x, +)
      [,1] [,2]
 [1,]    0  -19
 [2,]   -1  -20
 [3,]   -2  -21
 [4,]   -3  -22
 [5,]   -4  -23
 [6,]   -5  -24
 [7,]   -6  -25
 [8,]   -7  -26
 [9,]   -8  -27
[10,]   -9  -28
[11,]  -10  -29
[12,]  -11  -30
[13,]  -12  -31
[14,]  -13  -32
[15,]  -14  -33
[16,]  -15  -34
[17,]  -16  -35
[18,]  -17  -36
[19,]  -18  -37
[20,]  -19  -38
 dim(x-t(mat))
[1]  2 20
 dim(sweep(-mat, 2, x, +))
[1] 20  2

On Wed, Dec 5, 2012 at 4:55 PM, arun smartpink...@yahoo.com wrote:

HI Mike,
I didn't understand except the x-t(mat) output is very different than the 
others.  Are you saying that it needs to be transposed?  BTW, that was z5.

A.K.








From: C W tmrs...@gmail.com
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; Sarah Goslee sarah.gos...@gmail.com
Sent: Wednesday, December 5, 2012 4:47 PM

Subject: Re: [R] data manipulation between vector and matrix


Thanks for the benchmark.  I actually wanted to go with the winner, except the 
x-t(mat) output is very different than the others.
Mike


On Wed, Dec 5, 2012 at 4:40 PM, arun smartpink...@yahoo.com wrote:

Hi,

By comparing the different methods:
set.seed(5)
 mat1-matrix(sample(1:1e6,1e6,replace=TRUE),ncol=1)
 set.seed(25)
 x-sample(1:1e6,1,replace=TRUE)
 system.time(z1-sweep(-mat1,2,x,+))
#   user  system elapsed
 # 0.076   0.000   0.069
 system.time(z2-apply(-mat1,1,`+`,x))
 #  user  system elapsed
 # 0.036   0.000   0.031
 system.time(z3-aaply(-mat1,1,`+`,x))
#   user  system elapsed
#  1.880   0.000   1.704
 system.time(z4- x-t(mat1))  #winner
#   user  system elapsed
 # 0.004   0.000   0.007
 system.time(z5- t(x-t(mat1)))
#   user  system elapsed
#  0.008   0.000   0.009


A.K.







From: C W tmrs...@gmail.com
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; Sarah Goslee sarah.gos...@gmail.com
Sent: Wednesday, December 5, 2012 4:11 PM

Subject: Re: [R] data manipulation between vector and matrix


thanks, I knew about apply, but did not you you can put plus signs with 
quotes.  That's a cool tricky,
Mike


On Wed, Dec 5, 2012 at 4:05 PM, arun smartpink...@yahoo.com wrote:

HI,
In addition to ?sweep(), you can use

apply(-mat,1,`+`,x)

#or
library(plyr)
aaply(-mat,1,+,x)


A.K.







- Original Message -
From: C W tmrs...@gmail.com
To: Sarah Goslee sarah.gos...@gmail.com
Cc: r-help r-help@r-project.org
Sent: Wednesday, December 5, 2012 1:51 PM
Subject: Re: [R] data manipulation between vector and matrix

Thanks, Sarah.  First time heard about sweep(), it worked just the way I
wanted.
Mike

On Wed, Dec 5, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.com wrote:

 Hi,

 On Wed, Dec 5, 2012 at 1:30 PM, C W tmrs...@gmail.com wrote:
  Dear list,
  I was curious how to subtract a vector from matrix?
 
  Say, I have
 
  mat - matrix(1:40, nrow=20, ncol=2)
 
  x -c(1,2)

 Thanks for the actual reproducible example.

  I want,
 
  x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements
  of x against column elements in mat.
 
  But x-mat won't do it.

 This will (note the modification to get x - mat):
  sweep(-mat, 2, x, +)
       [,1] [,2]
  [1,]    0  -19
  [2,]   -1  -20
  [3,]   -2  -21
  [4,]   -3  -22
  [5,]   -4  -23
 etc.

 --
 Sarah Goslee
 http://www.functionaldiversity.org


    [[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-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] Assignment of values with different indexes