[R] stiff delay differential equations

[Suzen, Mehmet]

Hello List,

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

__
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] list to matrix?

[R. Michael Weylandt]

On Wed, Dec 5, 2012 at 12:02 AM, arun  wrote:
> Hi,
>
>
>  p <- lapply(1:1e6, function(i)c(i, log2(i)))
>
>  system.time(z1 <- t(sapply(p,function(x)x)))
> #   user  system elapsed
>  # 2.568   0.048   2.619
>  system.time(z1 <- do.call(rbind,p))
> #   user  system elapsed
>  # 4.000   0.052   4.060
> A.K.

Thanks for that Arun -- I'll have to look into why rbind is so slow.

Some interesting notes:

1) On my machine

t(sapply(p, function(x) x))

is about 2x faster than

t(sapply(p, identity))

2) Similarly,

do.call(rbind, p)

is about 4x slower than

do.call(cbind, p)

which all goes to show, profiling is all-important (as Bill reminds me
often) and often counter-intuitive.

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] How do I get internal nodes of dendograms produced by R?

[Milan Bouchet-Valat]

Le mardi 04 décembre 2012 à 16:17 -0500, sagnik ray choudhury a écrit :
> I am using R for hierarchical clustering of a number of documents.
> 
> I have a distance matrix on which I have applied hclust method. When I plot
> the result of hclust method, I can see the dendogram plotted. What I need
> now is the dendogram stored as a tree in a data structure. My goal is to
> automatically label all internal nodes. For that, I need to know, which
> leaf nodes make a first level cluster, and which first level clusters make
> a second level cluster and so on. Is there a way in R to get this
> information?
You probably want to convert you 'hclust' object to a 'dendrogram'
object using as.dendrogram(), and work with that tree. See ?dendrogram,
in particular the "Details" section.


My two cents

__
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] Global variable in the C code used to create R extension

[Vineeth Mohan]

Hi ,

I am writing a function in C and i wish to call it from R.
Thanks to this tutorial , this is easily possible -
http://www.nceas.ucsb.edu/scicomp/usecases/CreateRPackageWithC

Now , i have a server written in R which listens on a port for requests and
process the requests.

Now what i need is to load a large amount of data in startup of the server.
And later this data should be available to any part of the R.

So basically i want to maintain a global pointer to the data , i have
loaded and access them globally from the C code within the R extension
which is executed from any wehre in the code.

How can i achieve this.

Thanks
  Vineeth

[[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] Problem in summary of var results

[Akhil dua]

Hello

I am running var on dataset

data1 is the name of my dataset
cn.chf  us.chf
2005-07-01 -1.18656633 -1.18656633
2005-07-04 -0.48835920 -0.48835920
2005-07-05 -0.01534272 -0.01534272
2005-07-06  0.08825279  0.08825279
2005-07-07  0.34223563  0.34223563
2005-07-08 -0.05776229 -0.05776229

commands which I am usings are

1) lag.var <- VARselect(data1, 10, type="both")$selection["AIC(n)"]
2) reg1 <- VAR(data1, p=lag.var)

but when I am saying summary(reg1)
and I am getting same error for irfs as well.

Its's giving me and error

Error in merge(lhs, rhs, all = FALSE) :
  error in evaluating the argument 'y' in selecting a method for function
'merge': Error: argument "rhs" is missing, with no default


Please help me how to solve this :-(

Regards
-- 
Akhil Dua
Consultant
National Institute of Public Finance and Policy

[[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] request

[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 active"NeosServer 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 active"NeosServer 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

[[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] Réponse automatique

[j . boutet]

Bonjour,



Je serais en congés jusqu'au Jeudi 6 Décembre.

Pour des raisons d'urgence, vous pourrez me contacter par téléphone au 06 46 34 
81 03.



Cordialement,



--



Jérôme Boutet

Conservatoire d'espaces naturels de Picardie

1, place Ginkgo - village Oasis

80 044 AMIENS cedex 

tél : 03 22 89 84 24 



> From: r-help-requ...@r-project.org
> Subject: R-help Digest, Vol 118, Issue 5
> Date: Wed, 05 Dec 2012 12:00:08 +0100

__
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] Trim

[Vasilchenko Aleksander]

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] how to assign factor level into each value

[Tammy Ma]

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] "0-30""51-75"   "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  CN111.67"101-150"
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.

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

[Jorge I Velez]

Hi Tammy,

Check ?cut and its examples.

HTH,
Jorge.-


On Wed, Dec 5, 2012 at 11:26 PM, Tammy Ma <> 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] "0-30""51-75"   "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  CN111.67"101-150"
> 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.
>

[[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] alternative to leaps command

[eliza botto]

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. 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
 
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.

Re: [R] Trim

[Rui Barradas]

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] What is "print print print" ?

[Vladimir eremeev]

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.

-- 
Best regards,
 Vladimir  mailto:wl2...@gmail.com

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

[R] catching errors in RNetCDF

[Jannis]

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.

Re: [R] Trim

[arun]

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 
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

[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

[arun]

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 
To: Vasilchenko Aleksander 
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

[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

[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.

[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].

I 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. 



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

[Bert Gunter]

One correction, one comment, one suggestion:

On Wed, Dec 5, 2012 at 4:44 AM, eliza botto  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

[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 
vCARD

[[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

[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

[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 
À : Jim Lemon  

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

[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 ; Jim Lemon 
Cc : "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 
À : Jim Lemon  

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

[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

[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("X’X ")

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¿?

[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

[Stephen Politzer-Ahles]

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 ‘vcov’ method for ‘model’ 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¿?

[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:Basics: ##.#.   ##.#.  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"  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

[David Winsemius]

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 
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

[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 
.
 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  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 ‘vcov’ method for ‘model’ 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

[Stephen Politzer-Ahles]

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

On Wed, Dec 5, 2012 at 11:39 AM, John Fox  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 
> .
>  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"

[David Winsemius]

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   
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
 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" ?

[David Winsemius]

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

[C W]

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¿?

[David Winsemius]

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"

[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 wrote:

> 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

[C W]

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  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

[Sarah Goslee]

Hi,

On Wed, Dec 5, 2012 at 1:30 PM, C W  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"

[David Winsemius]

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  > 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

[C W]

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  wrote:

> Hi,
>
> On Wed, Dec 5, 2012 at 1:30 PM, C W  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"

[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"

[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  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)

[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 
setwith
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

[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/
 

[[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

[Rich Shepard]

  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

[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  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?

[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

[David Winsemius]

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/


[[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

[arun]

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 
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

[Stephen Politzer-Ahles]

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

[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
 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

[C W]

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  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 
> To: Sarah Goslee 
> Cc: r-help 
> 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 
> wrote:
>
> > Hi,
> >
> > On Wed, Dec 5, 2012 at 1:30 PM, C W  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

[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  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] "0-30""51-75"   "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  CN111.67"101-150"
> 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

[Rich Shepard]

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

[Rich Shepard]

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?

[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 
Maintainer: William Revelle 

--
David L Carlson
Associate Professor of Anthropology
Texas A&M 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

[arun]

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 
To: Sarah Goslee 
Cc: r-help 
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  wrote:

> Hi,
>
> On Wed, Dec 5, 2012 at 1:30 PM, C W  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

[R. Michael Weylandt]

On Wed, Dec 5, 2012 at 9:23 PM, Rich Shepard  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

[C W]

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  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 
> To: arun 
> Cc: R help ; Sarah Goslee 
> 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  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 
> >To: Sarah Goslee 
> >Cc: r-help 
> >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 
> wrote:
> >
> >> Hi,
> >>
> >> On Wed, Dec 5, 2012 at 1:30 PM, C W  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

[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 A&M 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

[Rich Shepard]

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

[Ben Bolker]

Suzen, Mehmet  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

[Stephen Politzer-Ahles]

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  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
>  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

[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
 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  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
>>  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

[Rich Shepard]

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

[David Winsemius]

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

[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
 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

[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  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

[Stephen Politzer-Ahles]

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  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
>>  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

[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" ?

[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

[David Winsemius]

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?

[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 
Maintainer: William Revelle 

--
David L Carlson
Associate Professor of Anthropology
Texas A&M 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?

[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 
= S􀀀1. 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 
> Maintainer: William Revelle 
> 
> --
> David L Carlson
> Associate Professor of Anthropology
> Texas A&M 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

Re: [R] catching errors in RNetCDF

[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

[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

[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

[David Winsemius]

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?

[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?

[Steve Lianoglou]

Hi,

On Thu, Dec 6, 2012 at 12:01 AM, Worik R  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

[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  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 
> To: "r-help@r-project.org help" 
> 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?

[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  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

[arun]

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




- Original Message -
From: Rich Shepard 
To: R help 
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

[arun]

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 
To: arun  
Cc: R help ; Sarah Goslee  
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  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 
>To: Sarah Goslee 
>Cc: r-help 
>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  wrote:
>
>> Hi,
>>
>> On Wed, Dec 5, 2012 at 1:30 PM, C W  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

[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 wrote:

> 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 active"NeosServer 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 active"NeosServer 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

[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

[arun]

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 
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

[arun]

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 
To: arun  
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  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 
>To: arun 
>Cc: R help ; Sarah Goslee 
>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  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 
>>To: arun 
>>Cc: R help ; Sarah Goslee 
>>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  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 
>>>To: Sarah Goslee 
>>>Cc: r-help 
>>>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  wrote:
>>>
 Hi,

 On Wed, Dec 5, 2012 at 1:30 PM, C W  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

[arun]

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 
To: "r-help@r-project.org help" 
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.

[R] gamcheck doubts

[Tania Mendo Aguilar]

Dear All,

I am fitting scallop count data to negative binomial GAMs. I have two 
significant parameters that explain 43%of the deviance. The adjusted r square 
is 0.25. The gam.check function gives me the figure attached. In the graph of 
linear predictor vs. residuals there seems to be more negative residual values 
than positive. Is that telling me that the fit is underestimating the response? 
Can I accept this model?


Tania Mendo
PhD Candidate | Institute for Marine and Antarctic Studies (IMAS)
University of Tasmania

__
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] nlme starting values are not the correct length

[PIKAL Petr]

Hi

see inline

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Lara Reichmann
> Sent: Wednesday, December 05, 2012 1:53 AM
> To: r-h...@lists.r-project.org
> Subject: [R] nlme starting values are not the correct length
> 
> Dear R community,
> 
> I am trying to fit an nlme model where I want to estimate the fixed
> effects of two treatments on the parameters on the following equation
> Photo~(a*(1-exp(-c*PARi/a)))-b I was able to fit a simple model without
> covariates following the method described in Mixed-Effects Methods and
> Classes for S and S-PLUS, version 3.0, but when I add the covariates, I
> get the error " starting values for the fixed component are not the
> correct length"
> 
> My data has the following structure "Subject" "Species" "Fert" "Photo"
> "PARi" , where several "Photo" measurements where taken on the same
> subject by changing "PARi", 4 Species levels and 2 Fert levels, there
> are 31 Subjects (one missing value), and 323 observations
> 
> DATA extract
> 
> Subject   Species FertPhoto   PARi
> 1 bb  1   22.5389 1499.3307
> 1 bb  1   21.881  1248.913
> 1 bb  1   21.2862 999.3387
> 1 bb  1   20.5836 799.9308
> 1 bb  1   19.3758 601.1412
> 1 bb  1   15.5915 399.815
> 1 bb  1   8.7978  200.1087
> 1 bb  1   4.4347  99.686
> 1 bb  1   2.0387  49.7842
> 1 bb  1   -1.4854 0.0576
> 2 sw  0   6.782   1500.5337
> 2 sw  0   7.1432  1249.2749
> 2 sw  0   7.3319  1000.9891
> 2 sw  0   7.5848  799.1752
> 2 sw  0   7.1882  599.5544
> 2 sw  0   6.809   399.988
> 2 sw  0   5.3877  198.7574
> 2 sw  0   3.5104  100.7115
> 2 sw  0   0.8856  50.7015
> 2 sw  0   -1.121  0.0569
> 3 jg  1   16.0827 2000.4941
> 3 jg  1   16.0236 1501.1957
> 3 jg  1   16.3818 1248.9551
> 3 jg  1   16.7815 1499.6414
> 3 jg  1   17.175  2000.6851
> 3 jg  1   16.6529 1000.2707
> 3 jg  1   15.7987 799.676
> 3 jg  1   15.5437 598.9409
> 3 jg  1   11.7683 400.7715
> 3 jg  1   4.89200.7468
> 3 jg  1   4.1294  100.9664
> 3 jg  1   1.6008  50.9254
> 3 jg  1   -0.89   0.5347
> 4 sw  1   25.2889 2000.1454
> 4 sw  1   24.7284 1499.6191
> 4 sw  1   24.3637 1249.7523
> 4 sw  1   23.3523 1000.0944
> 4 sw  1   21.6057 800.2209
> 4 sw  1   18.8926 599.7022
> 4 sw  1   14.6598 398.9366
> 4 sw  1   7.7182  201.5697
> 4 sw  1   3.4775  100.5139
> 4 sw  1   1.169   49.7045
> 4 sw  1   -1.3558 1.6914
> 5 jg  0   6.1626  2000.9351
> 5 jg  0   7.5573  1499.6581
> 5 jg  0   7.7129  1249.5073
> 5 jg  0   7.442   1000.7276
> 5 jg  0   7.5135  799.1286
> 5 jg  0   7.1559  599.5568
> 5 jg  0   6.8161  400.3576
> 5 jg  0   4.0097  199.7442
> 5 jg  0   2.7202  101.1253
> 5 jg  0   1.0746  51.1787
> 5 jg  0   -0.5913 0.975
> 
> 
> 
> This works so far:
> 
> lightresponse<-groupedData(Photo~PARi|Subject,data=lightr,outer = ~
> Species * Fert,labels = list(x = "PAR", y = "CO2 uptake rate"),units =
> list(x = "(photon s-1)", y = "(umol/m?2 s)")) Photo.resp<-
> function(PARi,A,B,C)A*(1-exp(-C*PARi/A))-B
> Photo.resp<-deriv ((~A *(1-exp(-C*PARi/A))-
> B),c("A","B","C"),function(PARi,A,B,C){})
> 
> >
> >lightresp.fit1<-
> nlme(model=Photo~Photo.resp(PARi,A,B,C),fixed=A+B+C~1,d
> >ata=lightresponse,start=c(30,-5,0.1))#fitting  nlme without any
> >covariates
> >lightresp.fit1
> 
> OUTPUT
> > lightresp.fit1
> Nonlinear mixed-effects model fit by maximum likelihood
>   Model: Photo ~ Photo.resp(PARi, A, B, C)
>   Data: lightresponse
>   Log-likelihood: -494.5926
>   Fixed: A + B + C ~ 1
>   A   B   C
> 24.89334793  1.77983637  0.06499634
> 
> Random effects:
>  Formula: list(A ~ 1, B ~ 1, C ~ 1)
>  Level: Subject2
>  Structure: General positive-definite, Log-Cholesky parametrization
>  StdDev  Corr
> A10.67382785 A B
> B 0.52572012 1.000
> C 0.01433605 0.371 0.384
> Residual  0.71900020
> 
> Number of Observations: 323
> Number of Groups: 31
> 
> ##Now, I want to test the effect of Species and Fert, I don't fully
> understand how to modify the start vector, as I tried several options
> and no one seems to be correct. Do the number of levels in each factor
> matter? In that case 4 Species and 2 Fert levels, I would need 6
> initial parameters x 3? This didn't work either
> 
> >lightresp.fit2<-nlme(model=Photo~Photo.resp(PARi,A,B,C),fixed=A+B+C ~
> >Species*Fert,random=A+B+C

Re: [R] What is "print print print" ?

[Vladimir eremeev]

Yes, now I see.
It's ridiculous that I haven't noticed this elementary error, which I know
of for over 10 years. :)

Now my problem, it's another question.

Here is the code and data, as requested.

Sample data, I've copied a part of text file:

 Charge Strobe [1]
 Charge FC [2]
 Charge SG [3]

 Carcass FC  [1]
 Carcass Strobe [2]
 Carcass SG  [3]

 ChartStrobe  [1]
 ChartFC  [2]
 ChartSG  [3]

 Boy  Strobe [1]
 Boy   FC  [2]
 Boy   SG  [3]

 Wires   Strobe  [1]
 Wires   SG  [2]
 Wires   FC  [3]

Now I read them and make last column numeric:

votes<-read.table("comparison.txt", col.names=c("Object", "Method", "Vote"))
votes$Rank <- as.numeric(votes$Vote)

Now I want to see for each method how many it has first, second and last places:

for(m in levels(votes$Metod)) {
print(c(m, hist(votes[with(votes, Method == m), "Rank"], breaks=0:3, 
plot=FALSE)$counts))
}

Oups... And now I've found my error, thanks to Opera's spell checker.
I have missed the letter 'h' in Method.

-- 
Best regards,
 Vladimirmailto:wl2...@gmail.com

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

[R] Incorrect DST time changes in DateTimeClasses

[Andrew Digby]

Can anyone please shed any light on why R DateTimeClasses give weird times for 
when daylight saving time information changes, and which aren't consistent with 
the OS?

Example:

Expected result: in New Zealand DST stopped (NZDT -> NZST) at 03:00 
NZDT on 2010-04-04, as confirmed by the OS time zone info (OS X 10.8.2):

zdump -v /etc/localtime
/etc/localtime  Sat Apr  3 13:59:59 2010 UTC = Sun Apr  4 
02:59:59 2010 NZDT isdst=1
/etc/localtime  Sat Apr  3 14:00:00 2010 UTC = Sun Apr  4 
02:00:00 2010 NZST isdst=0

Result in R: R has DST changing at 02:26:08, instead of 03:00!

> a<-as.POSIXlt('2010-04-04 02:26:07',tz="NZ")
> a$isdst
[1] 1
> b<-as.POSIXlt('2010-04-04 02:26:08',tz="NZ")
> b$isdst
[1] 0


So does R get its DST information from somewhere else? Any suggestions would be 
much appreciated!

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.