date:20091013

Try this;

DF - as.data.frame(matrix(rnorm(60), 10))
boxplot(DF, col = rep(seq(ncol(DF)) + 1, each = 2))

On Tue, Oct 13, 2009 at 2:11 AM, kennyPA tao...@yahoo.com wrote:

 Can anybody help me on how to boxplot multiple groups with different color?
 Say, I have 3 groups of data, each group with 2 boxes, and I'd like to have
 the following layout in the boxplot:

 red, red, green, green, blue, blue

 thanks in advance.
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Re: [R] Installing R on Ubuntu ( 8.10 ) ?

2009-10-13 Thread Dirk Eddelbuettel


On 13 October 2009 at 07:46, Robert Wilkins wrote:
| installing on Ubuntu, how to do it and have people found it to be glitchy?
| 
| which is easier , binary install or from source ?
| 
| With the source install, are you less likely to have a dependencies issue ?
| 
| ( Ubuntu does the GCC install seamlessly, but has no mention of R )

Really?  R has been part of every Ubuntu release. 

So install it from part of the distribution, or use the (usually more
current) CRAN repository for Ubuntu at

  http://cran.r-project.org/bin/linux/ubuntu/

where a detailed README tells you how to go about this.

Dirk

-- 
Three out of two people have difficulties with fractions.

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Re: [R] Counting

Try this:

table(Reduce(`==`, DF))

On Tue, Oct 13, 2009 at 9:20 AM, Ashta sewa...@gmail.com wrote:
 *Hi all,
 *

 *Assume that I have the following data set  with tow variables and I want
 count the number of observation with identical values
 *

 **

 *x1 x2*

 * 1   1 *

 * 1   0 *

 * 0   1*

 * 0   1*

 * 0   0*

 * 1   1*

 * 0   1
 *


 I want the  following output
 **

 *
 *

 *n1=3  # number of identical observation between x1 and x2 variables*

 *n2=4  # number of different observation*


 How do I do it in R?


 Thanks a lot




 **

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Re: [R] Counting

2009-10-13 Thread Julien Grassot

Or try this :

x1=c(1,1,0,0,0,1,0)
x2=c(1,0,1,1,0,1,1)

DF=cbind(x1,x2)

ifelse(DF[,1]==DF[,2],1,0)

your n1 is :
sum(ifelse(DF[,1]==DF[,2],1,0))

your n2 is :
dim(DF)[1]-sum(ifelse(DF[,1]==DF[,2],1,0))

Maybe not an elegant way, but hope this help.


Julien GRASSOT
Data Analysis Department
Flamel Technologies 
FRANCE 




-Message d'origine-
De : r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] De la 
part de Henrique Dallazuanna
Envoyé : mardi 13 octobre 2009 14:41
À : Ashta
Cc : R help
Objet : Re: [R] Counting

Try this:

table(Reduce(`==`, DF))

On Tue, Oct 13, 2009 at 9:20 AM, Ashta sewa...@gmail.com wrote:
 *Hi all,
 *

 *Assume that I have the following data set  with tow variables and I want
 count the number of observation with identical values
 *

 **

 *x1 x2*

 * 1   1 *

 * 1   0 *

 * 0   1*

 * 0   1*

 * 0   0*

 * 1   1*

 * 0   1
 *


 I want the  following output
 **

 *
 *

 *n1=3  # number of identical observation between x1 and x2 variables*

 *n2=4  # number of different observation*


 How do I do it in R?


 Thanks a lot




 **

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




-- 
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25° 25' 40 S 49° 16' 22 O

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Re: [R] SPSS long variable names

2009-10-13 Thread John Kane

library(Hmisc) spss.get may do it but it's been some time since I used it.

--- On Sat, 10/10/09, Orvalho Augusto orvaq...@gmail.com wrote:

 From: Orvalho Augusto orvaq...@gmail.com
 Subject: [R] SPSS long variable names
 To: r-help@r-project.org
 Received: Saturday, October 10, 2009, 12:14 PM
 Hello guys I am new to this list and
 for R too.

 I am wondering if there is a patch for the SPSS reading
 code on the
 foreign package, in order to be able to read long variable
 names.
 Right now read.spss() just trunc the names to 8
 characters.

 Or if someone could help me on other way:
 I have to process everyday a lot of SPSS Syntax Files and
 Dat files
 that come from one system that can only export data on
 through that
 way.

 I use PSPP to generate the spss data file (sav) that I read
 with R.
 From R I can export to MySQL, DBF and STATA to satisfy
 the needs of
 different guys here.

 The problem is the limit of 8 characters long on variable
 names.

 Can someone help on that?

 Caveman

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[R] How to draw more geographical boundaries

2009-10-13 Thread Angela Parenti

Dear R-users,

I am trying to plot in the same map both NUTS 2 (i.e. regional) and NUTS 3 
(i.e. provincial) boundaries for Italy
but I couldn't find how to plot more than one level boundary in the same map.//
//
I have shapefile both for regions and provinces in Italy and I tried to use the 
maptool and maps packages without
success.

Thank you in advance for any help.

Best regards,
Angela Parenti/
/


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Re: [R] lapply / mapply and assignments

2009-10-13 Thread Magnus Torfason


Thank you so much, relist and SIMPLIFY both work.

See more comments below ...

On 10/12/2009 5:35 PM, Charles C. Berry wrote:

On Mon, 12 Oct 2009, Magnus Torfason wrote:

I want to achieve the following:


 l - list( list(a=1,b=2), list(a=3,b=4))
 l[[]][a] - 5:6



See
?relist

something like:

relist( unlist( mapply( [-, l, a, 5:6) ), l )


Yes, this works exactly as needed. I had tried this, but I failed to 
notice that you could supply the skeleton argument separately. Thanks!


I had also tried

mapply( [-, l, a, c(5,6), SIMPLIFY=FALSE)

but failed to put SIMPLIFY in all-caps, and got an error message that I 
did not understand. It seems I was close on both methods, but not close 
enough.


Best,
Magnus

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Re: [R] Introduction to mark-recapture analysis in R?

2009-10-13 Thread Ben Bolker




  oops, library(sos) after install.packages(sos) and before
findFn(mark-recapture) ...
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Re: [R] Introduction to mark-recapture analysis in R?

2009-10-13 Thread Ben Bolker




Anne Link wrote:
 
 Normal021falsefalse
 false
 MicrosoftInternetExplorer4  
   
 
 Dear R-helpers,
 
  I was wondering whether there are any good books and/or website 
 links that introduce mark-recapture analysis in R. In particular, I am 
 interested in exploratory data analysis of resighting data and how to 
 create capture histories from dataframes in R.
 
 Thank you very much for your reply in advance!
 
  Cheers, 
 
 Anne
 -- 
 

Don't know about books, web sites, but the excellent new sos package
lists three packages with mark-recapture capabilities: Rcapture,
fishmethods,
and mra (install.packages(sos); findFn(mark-recapture)

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] SPSS long variable names

Thanks for the answer.

Hmisc uses read.spss from the foreign package. And so it does not
solve my trouble.

I need to read the long names on the SPSS dataset.

Caveman


On Tue, Oct 13, 2009 at 3:01 PM, John Kane jrkrid...@yahoo.ca wrote:
 library(Hmisc) spss.get may do it but it's been some time since I used it.

 --- On Sat, 10/10/09, Orvalho Augusto orvaq...@gmail.com wrote:

 From: Orvalho Augusto orvaq...@gmail.com
 Subject: [R] SPSS long variable names
 To: r-help@r-project.org
 Received: Saturday, October 10, 2009, 12:14 PM
 Hello guys I am new to this list and
 for R too.

 I am wondering if there is a patch for the SPSS reading
 code on the
 foreign package, in order to be able to read long variable
 names.
 Right now read.spss() just trunc the names to 8
 characters.

 Or if someone could help me on other way:
 I have to process everyday a lot of SPSS Syntax Files and
 Dat files
 that come from one system that can only export data on
 through that
 way.

 I use PSPP to generate the spss data file (sav) that I read
 with R.
 From R I can export to MySQL, DBF and STATA to satisfy
 the needs of
 different guys here.

 The problem is the limit of 8 characters long on variable
 names.

 Can someone help on that?

 Caveman

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



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[R] Identification of variables contributing to differences between factor in adonis

2009-10-13 Thread Paul Dennis


Dear all

I have used permutational multivariate analysis of variance (adonis in package 
vegan) based on Bray-Curtis distances  to assess the signifance of carbon, 
nitrogen and more complex nutrient amendments on soil microbial community 
structure (microbial fatty acids).

I have identified signifant effects of nutrients and would like to know how to 
identify the fatty acids (microbial markers) that are associated with these 
differences.  So far I have used Pierre Legrendre's Dufrene-Legendre Indicator 
Species Analysis duleg in package 'labdsv' to do this (Calculates the 
indicator value (fidelity and relative abundance) of species in 
clusters or types).  However, I am not familiar with the implementation of this 
method and am concerned about the relevance of the output as  the 'indicator 
species' are not probably not based on Bray-Curtis distances.

Is there another way to identify fatty acids contributing to differences 
between my nutrient amendment groups?  For example, the output of the 
permutational multivariate analysis of variance (adonis) has the following 
attributes:

 
aov.tab  6data.frame list   
call2-none-   call   
coefficients74   -none-   numeric
coef.sites   28   -none-  numeric
f.perms 999  -none-  numeric
model.matrix 28   -none-   numeric
terms3 termscall

Can I use the coefficients to select fatty acids that contribute to the 
differences between treatment? 

Thanks

Paul  
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Re: [R] RPostgreSQL and needed .dlls

2009-10-13 Thread Uwe Ligges




Josuah Rechtsteiner wrote:

Dear List,

I am trying to connect from R 2.9.2 on Win XP SP3 to a remotely 
installed PostgreSQL DB (8.3.7 on Ubuntu Server 9.04). Everything seems 
to be properly installed, as I can connect to the DB from within Excel 
and RKWard (running on another machine).
But regarding R on the Win XP, I cannot load RPostgreSQL. libpq.dll is 
missing. What can I do now, without having to install PostgreSQL on this 
Windows, where I do not need it as I have my remote DB?


Well, you need libpq.dll for RPostgreSQL in any case since the package 
links dynamically against that library. Or connect with another method 
to your database.


Uwe Ligges





Thanks in advance

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Re: [R] Use R -- term and logo copyright?


Perhaps:

Yooze R!

On second thought, probably not quite right for King's College.

Then?

dig R!

--  
David


On Oct 9, 2009, at 3:46 PM, Marianne Promberger wrote:


Dear list,

I would like to start some R workshops at King's College London, and
to do so, I would like to use the Use R! logo at
http://www.agrocampus-ouest.fr/math/useR-2009//useR%21%202008_fichiers/useR-middle.png

Since it seems to be difficult to get a shell account at KCL, I also  
went

ahead and registered use-r.org.uk and am starting to put together a
website at kcl.use-r.org.uk.

I really like the Use R! slogan, which seems to be used by the R
user conferences and Springer (the latter without the exclamation
mark).

Even more, I really *really* like Use R! logo.  I think it is very
elegant indeed! Kudos to whoever designed it.

However, I'm completely in the dark about copyright issues of the logo
and the slogan.

Can I use (a) the logo and/or (b) the slogan for the KCL R workshops?
I think it is quite clear from my website that this is neither about
the Springer book series nor about an R user conference.

I enquired with the agrocampus-ouest.fr website about the logo but was
pointed to r-project.org and the R development core team. I thought in
that case it might be best to ask here to have the answer publicly
available -- sorry if I overlooked the information online somewhere.

Marianne

--
Marianne Promberger PhD, King's College London
http://promberger.info
R version 2.9.2 (2009-08-24)
Ubuntu 9.04

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] Function to find prime numbers

2009-10-13 Thread Thomas Lumley


On Tue, 13 Oct 2009, AJ83 wrote:



I need to create a function to find all the prime numbers in an array. Can
anyone point me in the right direction?


It depends a bit on how big the numbers are.  If the array is large but the numbers are not very large the 
fastest approach is probably to create a vector of small primes and use

   my_array %in% smallprimes.

For example, the first 1000 primes are at 
http://primes.utm.edu/lists/small/1000.txt
and the first 1 are on the same site.


  -thomas

Thomas Lumley   Assoc. Professor, Biostatistics
tlum...@u.washington.eduUniversity of Washington, Seattle

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[R] update.formula drop interaction terms

2009-10-13 Thread Eleni Rapsomaniki

Dear R users,

How do I drop multiplication terms from a formula using update?
e.g.
forml=as.formula(Surv(time, status) ~ x1+x2+A*x3+A*x4+B*x5+strata(sex))

#I would like to drop all instances of variable A (the main effect and its 
interactions). The following:
updated.forml=update(forml, ~ . -A)

#gives me this:
#Surv(time, status) ~ x1 + x2 + x3 + x4 + B + x5 + strata(sex) + A:x3 + A:x4 + 
B:x5

#but I want this:
#updated.forml=as.formula(Surv(time, status) ~ x1+x2+x3+x4+B*x5+strata(sex))

Any ideas?
Thanks in advance

Eleni Rapsomaniki

Research Associate
Strangeways Research Laboratory
Department of Public Health and Primary Care
University of Cambridge
 

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Re: [R] splitting dataframe, assign to new dataframe, add new rows to new dataframe

2009-10-13 Thread hadley wickham

On Tue, Oct 13, 2009 at 6:57 AM, Ista Zahn istaz...@gmail.com wrote:
 I'm sure there's a really cool way to do this with plyr, although I
 don't know if my particular plyr version is much better. Anyway here
 it is:

 cmbine - read.csv(textConnection('names, mass, classes
 apple,0.50,1
 tiger,100.00,2
 pencil,0.01,3
 chicken,1.00,2
 banana,0.15,1
 pear,0.30,1'))

 library(plyr)

 dfl - list()

 for(i in 1:max(cmbine$classes)) {
  dfl[[i]] - ddply(cmbine, .(classes), function(x) {x[i,]})
 }

Here's another approach:

cmbine - read.csv(textConnection('names, mass, classes
apple,0.50,1
tiger,100.00,2
pencil,0.01,3
chicken,1.00,2
banana,0.15,1
pear,0.30,1'))

cmbine - ddply(cmbine, classes, transform, i = seq_along(names))
dlply(cmbine, i)

Hadley

-- 
http://had.co.nz/

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[R] Creating a list of empty lists

2009-10-13 Thread Magnus Torfason

Well here is one more brain-teaser related to assigning stuff into a 
list of list. What if I need to create a new list of empty lists? I have 
actually got a solution to this problem:


l = list(list())
for ( i in sequence(length-1) )
{
l = list(unlist(l,recursive=FALSE), list())
}

But it is not very neat to do this in a loop. Are there any cuter ways 
to do this?


Best,
Magnus

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Re: [R] svy / weighted regression

2009-10-13 Thread Thomas Lumley

I think there is a much simpler explanation.

The survey design object has eight observations, two per country. With a sample size of two per
country it is hardly surprising that country-specific estimates are not very precise. The actual data has
hundreds of thousands of observations per country, so it will have more precise estimates.

Grouping the data doesn't make a difference for model-based glm estimation, where it is simply a
computational convenience. It *does* make a difference for design-based estimation, because it
changes the design.

-thomas

On Tue, 13 Oct 2009, Laust wrote:

Dear David,

Thanks again for your input! I realize that I did a bad job of
explaining this in my first email, but the setup is that in Finland
persons who die are sampled with a different probability (1) from
those who live (.5). This was done by the Finnish data protection
authorities to protect individuals against identification. In the rest
of the countries everyone is sampled with a probability of 1. The data
that I am supplying to R is summarized data for each country
stratified by case status. Another way of organizing the data would
be:

# creating data
listc - c(Denmark,Finland,Norway,Sweden)
listw - c(1,2,1,1)
listd - c(1000,1000,1000,2000)
listt - c(755000,505000,905000,191)
list.cwdt - c(listc, listw, listd, listt)
country2 - data.frame(country=listc,weight=listw,deaths=listd,time=listt)

I hope that it is clearer now that for no value of the independent
variable 'country' is the rate going to be zero. I think this was also
not the case in my original example, but this was obscured by my poor
communication- R-skills. But if data is organized this way then
sampling weight of 2 for Finland should only be applied to the
time-variable that contains person years at risk and *not* to the
number of deaths, which would complicate matters further. I would know
how to get this to work in R or in any other statistical package.
Perhaps it is - as Peter Dalgaard suggested - the estimation of the
dispersion parameter by the survey package that is causing trouble,
not the data example eo ipso. Or perhaps I am just using survey in a
wrong way.

Best
Laust

Post doc. Laust Mortensen, PhD
Epidemiology Unit
University of Southern Denmark

On Mon, Oct 12, 2009 at 3:32 PM, David Winsemius dwinsem...@comcast.net wrote:

I think you are missing the point. You have 4 zero death counts associated
with much higher person years of exposure followed by 4 death counts in the
thousands associated with lower degrees of exposures. It seems unlikely that
these are real data as there are not cohorts that would exhibit such lower
death-rates. So it appears that in setting up your test case, you have
created an impossibly unrealistic test problem.

--
David

On Oct 12, 2009, at 9:12 AM, Laust wrote:

Dear Peter,

Thanks for the input. The zero rates in some strata occurs because
sampling depended on case status: In Finland only 50% of the non-cases
were sampled, while all others were sampled with 100% probability.

Best
Laust

On Sat, Oct 10, 2009 at 11:02 AM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:

Sorry, forgot to reply all...

Laust wrote:

Dear list,

I am trying to set up a propensity-weighted regression using the
survey package. Most of my population is sampled with a sampling
probability of one (that is, I have the full population). However, for
a subset of the data I have only a 50% sample of the full population.
In previous work on the data, I analyzed these data using SAS and
STATA. In those packages I used a propensity weight of 1/[sampling
probability] in various generalized linear regression-procedures, but
I am having trouble setting this up. I bet the solution is simple, but
I’m a R newbie. Code to illustrate my problem below.

Hi Laust,

You probably need the package author to explain fully, but as far as I
can see, the crux is that a dispersion parameter is being used, based on
Pearson residuals, even in the Poisson case (i.e. you effectively get
the same result as with quasipoisson()).

I don't know what the rationale is for this, but it is clear that with
your data, an estimated dispersion parameter is going to be large. E.g.
the data has both 0 cases in 75 person-years and 1000 cases in 5000
person-years for Denmark, and in your model they are supposed to have
the same Poisson rate.

summary.svyglm starts off with

est.disp - TRUE

and AFAICS there is no way it can get set to FALSE. Knowing Thomas,
there is probably a perfectly good reason not to just set the dispersion
to 1, but I don't get it either...

Thanks
Laust

# loading survey
library(survey)

# creating data
listc -

c(Denmark,Finland,Norway,Sweden,Denmark,Finland,Norway,Sweden)
listw - c(1,2,1,1,1,1,1,1)
listd - c(0,0,0,0,1000,1000,1000,2000)
listt - c(75,50,90,190,5000,5000,5000,1)
list.cwdt - c(listc, listw, listd, listt)
country -

Re: [R] Use R -- term and logo copyright?

2009-10-13 Thread hadley wickham

 Can I use (a) the logo and/or (b) the slogan for the KCL R workshops?
 I think it is quite clear from my website that this is neither about
 the Springer book series nor about an R user conference.

 No, sorry, the logo/name should be used exclusively for the Springer series
 and the R User Conferences. For the latter see
  http://www.R-project.org/conferences.html

 Hence, the logo of your workshop and preferably also its URL should be
 different. You may use R in the title though and have the usual R logo on
 the workshop page.

You can certainly claim copyright on the logo, but through what right
do you claim ownership of the term useR! ?

Hadley

-- 
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Re: [R] Creating a list of empty lists

2009-10-13 Thread Romain Francois


On 10/13/2009 03:48 PM, Magnus Torfason wrote:

l = list(list())
 for ( i in sequence(length-1) )
 {
 l = list(unlist(l,recursive=FALSE), list())
 }


About this :

 rep( list(list()), 3 )
[[1]]
list()

[[2]]
list()

[[3]]
list()

Romain

--
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Professional R Enthusiast
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Re: [R] Creating a list of empty lists

2009-10-13 Thread Magnus Torfason


Live and learn ...

Thank you!

On 10/13/2009 9:57 AM, Romain Francois wrote:

On 10/13/2009 03:48 PM, Magnus Torfason wrote:

l = list(list())
 for ( i in sequence(length-1) )
 {
 l = list(unlist(l,recursive=FALSE), list())
 }


About this :

  rep( list(list()), 3 )
[[1]]
list()

[[2]]
list()

[[3]]
list()

Romain



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Re: [R] Creating a list of empty lists

Try this:

replicate(3, list())

On Tue, Oct 13, 2009 at 10:48 AM, Magnus Torfason
zulutime@gmail.com wrote:
 Well here is one more brain-teaser related to assigning stuff into a list of
 list. What if I need to create a new list of empty lists? I have actually
 got a solution to this problem:

    l = list(list())
    for ( i in sequence(length-1) )
    {
        l = list(unlist(l,recursive=FALSE), list())
    }

 But it is not very neat to do this in a loop. Are there any cuter ways to do
 this?

 Best,
 Magnus

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[R] How to calculate average correlation coefficient of a correlation matrix ?

2009-10-13 Thread Amit Kumar

Hi! All,
I have large correlation matrix Cor. I wish to calculate average
correlation coefficient for this matrix.
Is there any function in R to do this?
Thanks in advance.

Amit

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[R] multivariate multiple nonlinear regression

2009-10-13 Thread Yingyun Liu

Hello, 
I have several dataframes of identical structure. Each dataframe has one
dependent variable and several independent variables. Two different
functions are used to describe the relationships among all the
dataframes. Some parameters that are to be obtained from regression are
shared by all dataframes. Is there a way to do the regression
simultaneously with all dataframes in R? Can nls() do this?
Thanks

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Re: [R] Nelder-Mead with output of simplex vertices

2009-10-13 Thread Ben Bolker



bartjoosen wrote:
 
 Hi,
 
 Is it possible to share the code on this list?
 I'm also interested (and maybe others to)
 Or are you planning to make a package?
 
 Best regards
 
 Bart
 
 
 
 Ted.Harding-2 wrote:
 
 On 12-Oct-09 13:33:17, Ted Harding wrote:
 On 12-Oct-09 13:24:01, Terry Therneau wrote:
 -- begin included 
 Greetings!
 I want to follow the evolution of a Nelder-Mead function
 minimisation (a function of 2 variables). Hence each simplex
 will have 3 vertices.
 
 Therefore I would like to have a function which can output
 the coordinates of the 3 vertices after each new simplex
 is generated. However, there seems to be no way (which I can
 detect) of extracting this information from optim() (the 'trace'
 argument to 'control' does not seem to have provision for this,
 according to '?optim', and I have tried it out without success).
 
 --- end include -
 
  Why not put a cat() statement into fn, the function that you supply
 which optim is calling?  That will give the vertices that it tries one
 by one.
 
 Terry T.
 
 That's neat and simple! It hadn't occurred to me. Thanks!
 Ted.
 
 And, 10 seconds after posting, I realised why it hadn't -- there
 would be no visible association between the vertex and the simplex
 (in this instance the triangle) that it belongs to.
 
 In other words, which two other points in the preceding sequence,
 along with the current one, make up the triangle being tested?
 
 Given the complexity of the Nelder-Mead process, it would be very
 tricky indeed to try to track this through the sequence of vertices
 which cat() would output.
 
 As it happens, Ben Bolker kindly sent me code which he wrote (see
 his earlier mail) which does do this nicely, since it has an output
 option within the Nelder-Mead routine itself -- at which point,
 the routine itself knows what the simplex is.
 
 Ted.
 
 
 E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
 Fax-to-email: +44 (0)870 094 0861
 Date: 12-Oct-09   Time: 14:45:00
 -- XFMail --
 
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The reason I haven't shared is that the code is a translation from
_Numerical Recipes in C_.
Therefore I'm uncertain about its redistribution status.  If it were a
straight transcription
rather than a translation, it would be un-redistributable (I feel bound to
honor Press et al's
redistribution policy, even though it's really annoying: see
http://mingus.as.arizona.edu/~bjw/software/boycottnr.html).  Because it's
a translation of their C implementation of a public-domain
algorithm, it's less clear to me whether this is allowed or not (any
intellectual property
lawyers lurking on the list should feel free to chime in here!); my
compromise is that I'm
willing to send the code on request, but won't post it to the list.  

If people really want an R-only implementation of Nelder-Mead, it would
presumably
take someone not very long to translate from an unencumbered source (the
source code
in R, or the source code in GSL, or the description of the original
algorithm) ...

  cheers
Ben Bolker

-- 
View this message in context: 
http://www.nabble.com/Nelder-Mead-with-output-of-simplex-vertices-tp25838572p25874021.html
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Re: [R] SPSS long variable names

2009-10-13 Thread joris meys

Sorry to be so blunt, but I cannot believe PSPP can't save a dataset
as a .csv file for example. That should be the prefered format to
transport a dataset to any other statistical package, including R. csv
files are universal.

Cheers
Joris

On Tue, Oct 13, 2009 at 3:30 PM, Orvalho Augusto orvaq...@gmail.com wrote:
 Thanks for the answer.

 Hmisc uses read.spss from the foreign package. And so it does not
 solve my trouble.

 I need to read the long names on the SPSS dataset.

 Caveman


 On Tue, Oct 13, 2009 at 3:01 PM, John Kane jrkrid...@yahoo.ca wrote:
 library(Hmisc) spss.get may do it but it's been some time since I used it.

 --- On Sat, 10/10/09, Orvalho Augusto orvaq...@gmail.com wrote:

 From: Orvalho Augusto orvaq...@gmail.com
 Subject: [R] SPSS long variable names
 To: r-help@r-project.org
 Received: Saturday, October 10, 2009, 12:14 PM
 Hello guys I am new to this list and
 for R too.

 I am wondering if there is a patch for the SPSS reading
 code on the
 foreign package, in order to be able to read long variable
 names.
 Right now read.spss() just trunc the names to 8
 characters.

 Or if someone could help me on other way:
 I have to process everyday a lot of SPSS Syntax Files and
 Dat files
 that come from one system that can only export data on
 through that
 way.

 I use PSPP to generate the spss data file (sav) that I read
 with R.
 From R I can export to MySQL, DBF and STATA to satisfy
 the needs of
 different guys here.

 The problem is the limit of 8 characters long on variable
 names.

 Can someone help on that?

 Caveman

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Re: [R] How to calculate average correlation coefficient of a correlation matrix ?

2009-10-13 Thread Chuck Cleland

On 10/13/2009 10:13 AM, Amit Kumar wrote:
 Hi! All,
 I have large correlation matrix Cor. I wish to calculate average
 correlation coefficient for this matrix.
 Is there any function in R to do this?
 Thanks in advance.

cormat - cor(iris[,1:4])

corlowtri - cormat[lower.tri(cormat)]

corlowtri
[1] -0.1175698  0.8717538  0.8179411 -0.4284401 -0.3661259  0.9628654

mean(corlowtri)
[1] 0.2900708

mean(abs(corlowtri))
[1] 0.594116

avgcor - function(x){mean(abs(x[lower.tri(x)]))}

avgcor(cor(iris[,1:4]))
[1] 0.594116

 Amit
 
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-- 
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Re: [R] SPSS long variable names

Thanks guys for the greaty ideia!!

I am fool because I did not realize that before.

Caveman


On Tue, Oct 13, 2009 at 4:41 PM, joris meys jorism...@gmail.com wrote:
 Just for clarity : the csv format will solve your problem, as the
 restrictions on the variable names will only depend on the program you
 use to load them. I never experienced problems with variable names
 using csv to switch datasets between SPlus, R, SAS and SPSS.

 Cheers
 Joris

 On Tue, Oct 13, 2009 at 4:35 PM, joris meys jorism...@gmail.com wrote:
 Sorry to be so blunt, but I cannot believe PSPP can't save a dataset
 as a .csv file for example. That should be the prefered format to
 transport a dataset to any other statistical package, including R. csv
 files are universal.

 Cheers
 Joris

 On Tue, Oct 13, 2009 at 3:30 PM, Orvalho Augusto orvaq...@gmail.com wrote:
 Thanks for the answer.

 Hmisc uses read.spss from the foreign package. And so it does not
 solve my trouble.

 I need to read the long names on the SPSS dataset.

 Caveman


 On Tue, Oct 13, 2009 at 3:01 PM, John Kane jrkrid...@yahoo.ca wrote:
 library(Hmisc) spss.get may do it but it's been some time since I used it.

 --- On Sat, 10/10/09, Orvalho Augusto orvaq...@gmail.com wrote:

 From: Orvalho Augusto orvaq...@gmail.com
 Subject: [R] SPSS long variable names
 To: r-help@r-project.org
 Received: Saturday, October 10, 2009, 12:14 PM
 Hello guys I am new to this list and
 for R too.

 I am wondering if there is a patch for the SPSS reading
 code on the
 foreign package, in order to be able to read long variable
 names.
 Right now read.spss() just trunc the names to 8
 characters.

 Or if someone could help me on other way:
 I have to process everyday a lot of SPSS Syntax Files and
 Dat files
 that come from one system that can only export data on
 through that
 way.

 I use PSPP to generate the spss data file (sav) that I read
 with R.
 From R I can export to MySQL, DBF and STATA to satisfy
 the needs of
 different guys here.

 The problem is the limit of 8 characters long on variable
 names.

 Can someone help on that?

 Caveman

 __
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 mailing list
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 http://www.R-project.org/posting-guide.html
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Re: [R] Nelder-Mead with output of simplex vertices

2009-10-13 Thread bartjoosen


Hi,

Is it possible to share the code on this list?
I'm also interested (and maybe others to)
Or are you planning to make a package?

Best regards

Bart



Ted.Harding-2 wrote:
 
 On 12-Oct-09 13:33:17, Ted Harding wrote:
 On 12-Oct-09 13:24:01, Terry Therneau wrote:
 -- begin included 
 Greetings!
 I want to follow the evolution of a Nelder-Mead function
 minimisation (a function of 2 variables). Hence each simplex
 will have 3 vertices.
 
 Therefore I would like to have a function which can output
 the coordinates of the 3 vertices after each new simplex
 is generated. However, there seems to be no way (which I can
 detect) of extracting this information from optim() (the 'trace'
 argument to 'control' does not seem to have provision for this,
 according to '?optim', and I have tried it out without success).
 
 --- end include -
 
  Why not put a cat() statement into fn, the function that you supply
 which optim is calling?  That will give the vertices that it tries one
 by one.
 
 Terry T.
 
 That's neat and simple! It hadn't occurred to me. Thanks!
 Ted.
 
 And, 10 seconds after posting, I realised why it hadn't -- there
 would be no visible association between the vertex and the simplex
 (in this instance the triangle) that it belongs to.
 
 In other words, which two other points in the preceding sequence,
 along with the current one, make up the triangle being tested?
 
 Given the complexity of the Nelder-Mead process, it would be very
 tricky indeed to try to track this through the sequence of vertices
 which cat() would output.
 
 As it happens, Ben Bolker kindly sent me code which he wrote (see
 his earlier mail) which does do this nicely, since it has an output
 option within the Nelder-Mead routine itself -- at which point,
 the routine itself knows what the simplex is.
 
 Ted.
 
 
 E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk
 Fax-to-email: +44 (0)870 094 0861
 Date: 12-Oct-09   Time: 14:45:00
 -- XFMail --
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
 

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Re: [R] SPSS long variable names

2009-10-13 Thread Robert Baer

- Original Message - 
From: Robert Baer rb...@atsu.edu

To: Orvalho Augusto orvaq...@gmail.com
Sent: Tuesday, October 13, 2009 9:52 AM
Subject: Re: [R] SPSS long variable names

I am wondering if there is a patch for the SPSS reading
code on the
foreign package, in order to be able to read long variable
names.
Right now read.spss() just trunc the names to 8
characters.
This sequence seems to access the long filenames for me if I know what you 
are asking for:

library('foreign')
a-read.spss('fil.sav')
lnames - attr(a,variable.labels,exact=FALSE)

Rob

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Re: [R] svy / weighted regression



On Oct 13, 2009, at 6:07 AM, Laust wrote:


Dear David,

Thanks again for your input! I realize that I did a bad job of
explaining this in my first email, but the setup is that in Finland
persons who die are sampled with a different probability (1) from
those who live (.5). This was done by the Finnish data protection
authorities to protect individuals against identification. In the rest
of the countries everyone is sampled with a probability of 1. The data
that I am supplying to R is summarized data for each country
stratified by case status. Another way of organizing the data would
be:

# creating data
listc - c(Denmark,Finland,Norway,Sweden)
listw - c(1,2,1,1)
listd - c(1000,1000,1000,2000)
listt - c(755000,505000,905000,191)
list.cwdt - c(listc, listw, listd, listt)
country2 -  
data.frame(country=listc,weight=listw,deaths=listd,time=listt)


I hope that it is clearer now that for no value of the independent
variable 'country' is the rate going to be zero.


It is clearer now, and I think you were correct in believing that  
should not have been the problem, so please accept my apologies. The  
denominators and numerators should have been properly summed prior to  
estimation.



I think this was also
not the case in my original example, but this was obscured by my poor
communication-  R-skills. But if data is organized this way then
sampling weight of 2 for Finland should only be applied to the
time-variable that contains person years at risk and *not* to the
number of deaths, which would complicate matters further. I would know
how to get this to work in R or in any other statistical package.




Perhaps it is - as Peter Dalgaard suggested - the estimation of the
dispersion parameter by the survey package that is causing trouble,
not the data example eo ipso. Or perhaps I am just using survey in a
wrong way.


I think it is likely that we are now both using it incorrectly, but my  
efforts are also creating nonsense. From the help page I thought that  
the formula in svydesign might be need to be the country variable ...  
wrong. Or that the weights might need to be the inverse of what you  
had used ...wrong. Or that you ought to use quasipoisson for the  
family  wrong again.


Lumley is preparing a book to accompany the package but that is still  
several months away from release. He and Norm Breslow also published a  
paper very recently in the American Journal of Epidemiology on the  
using of survey sampling for analysis of case-cohort designs (of which  
your problem seems to be an exceedingly simple example, albeit only in  
one of the four strata.) I don't have access to the original paper at  
the moment, but perhaps you are in an academic setting where such  
access would be routine.


Or probably even more efficient would be to shoot a letter to Thomas  
Lumley.


--
David



Best
Laust


Post doc. Laust Mortensen, PhD
Epidemiology Unit
University of Southern Denmark

On Mon, Oct 12, 2009 at 3:32 PM, David Winsemius dwinsem...@comcast.net 
 wrote:
I think you are missing the point. You have 4 zero death counts  
associated
with much higher person years of exposure followed by 4 death  
counts in the
thousands associated with lower degrees of exposures. It seems  
unlikely that
these are real data as there are not cohorts that would exhibit  
such lower
death-rates. So it appears that in setting up your test case, you  
have

created an impossibly unrealistic test problem.

--
David


On Oct 12, 2009, at 9:12 AM, Laust wrote:


Dear Peter,

Thanks for the input. The zero rates in some strata occurs because
sampling depended on case status: In Finland only 50% of the non- 
cases

were sampled, while all others were sampled with 100% probability.

Best
Laust

On Sat, Oct 10, 2009 at 11:02 AM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:


Sorry, forgot to reply all...

Laust wrote:


Dear list,

I am trying to set up a propensity-weighted regression using the
survey package. Most of my population is sampled with a sampling
probability of one (that is, I have the full population).  
However, for
a subset of the data I have only a 50% sample of the full  
population.

In previous work on the data, I analyzed these data using SAS and
STATA. In those packages I used a propensity weight of 1/[sampling
probability] in various generalized linear regression- 
procedures, but
I am having trouble setting this up. I bet the solution is  
simple, but

I’m a R newbie. Code to illustrate my problem below.


Hi Laust,

You probably need the package author to explain fully, but as far  
as I
can see, the crux is that a dispersion parameter is being used,  
based on
Pearson residuals, even in the Poisson case (i.e. you effectively  
get

the same result as with quasipoisson()).

I don't know what the rationale is for this, but it is clear that  
with
your data, an estimated dispersion parameter is going to be  
large. E.g.
the data has both 0 cases in 75 person-years and 1000 cases  
in

Re: [R] How to choose a proper smoothing spline in GAM of mgcv package?

2009-10-13 Thread Simon Wood


 I have 5 datasets. I would like to choose a basis spline with same knots in
 GAM function in order to obtain same basis function for 5 datasets.
 Moreover, the basis spline is used to for an interaction of two covarites.
The `knots' argument to `gam' allows you to fix the knot locations used with a 
basis, and thereby obtain the same basis for each analysis. 


 I used cr in one covariate, but it can only smooth w.r.t 1 covariate. Can
 anyone give me some suggestion about how to choose a proper smoothing
 spline (bs='?') and knots for two covariates?
You can use the tp basis. Again use `knots' to supply the same set of knots 
for each dataset. For the tp basis I would pool you samples and take a 
largish (up to 1000) random sample of covariate pairs to use as the `knots'. 
The tp basis does not use the knot locations directly as knots, but rather 
as the starting point point for finding an optimal eigen-basis for the 
smoother (the only exception is if you supply exactly the same number of 
knots as the basis dimension).  

Alternatively use a tensor product of cr smooths for bivariate smoothing: 
see ?te. Again, supplying the same `knots' for all analyses fixes the basis 
used.


Finally, with some loss of computational efficiency, you can just fit all the 
data at once. Simply combine all the data frames, adding a column containing 
a five level factor variable indicating which original data set the data 
relate to (call it set) then something like:

gam(y~s(x,z,by=set)+set)

will produce one smooth for each level of set. They will all use the same 
basis. You can force them to all have the same smoothing parameter as well 
with something like:

gam((y~s(x,z,by=set,id=1)+set)

The same thing works for `te' terms.

best,
Simon



 Thanks a lot.

 Lee

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Re: [R] Function to find prime numbers

2009-10-13 Thread Barry Rowlingson

On Tue, Oct 13, 2009 at 2:41 PM, Thomas Lumley tlum...@u.washington.edu wrote:
 On Tue, 13 Oct 2009, AJ83 wrote:


 I need to create a function to find all the prime numbers in an array. Can
 anyone point me in the right direction?

 This almost sounds like a homework problem to me... So here's a
solution that you can happily present to a tutor - if you can explain
how it works, then you deserve full marks!

primer=function(v){
  
return(regexpr(^1$|^(11+?)\\1+$,unlist(lapply(v,function(z){paste(rep(1,z),sep='',collapse='')})),perl=TRUE)
== -1)
}

Test:

  (1:30)[primer(1:30)]
 [1]  2  3  5  7 11 13 17 19 23 29

I'm not sure how big a number this works for

R golf anyone?

Barry

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Re: [R] lapply() reccursively



On Oct 13, 2009, at 7:33 AM, Kaveh Vakili wrote:



Hi all,

I was wondering whether it is possible to use the lapply() function
to alter the value of the input, something in the spirit of :

a1-runif(100)
a2-function(i){
a1[i]-a1[i-1]*a1[i];a1[i]
}
a3-lapply(2:100,a2)


Neither a1 nor 2:100 are lists, so it would seem that sapply would be  
more appropriate.




Something akin to a for() loop, but using the lapply() infrastructure.
I haven't been able to get rapply() to do this.


You did not specify what the correct answer should look like, but I  
get no error after changing the l to an s and the output is a  
vector rather than a list. I got no error with the lapply version so  
it remains unclear  what problem you are experiencing.


 a1-runif(100)
 a2-function(i){
+ a1[i]-a1[i-1]*a1[i];a1[i]
+ }
 a3-sapply(2:100,a2)
 a3
 [1] 2.990506e-01 2.957213e-02 3.343994e-02 7.234998e-01 2.036053e-01  
1.850228e-01 2.355974e-01 3.295134e-01
 [9] 3.206837e-01 1.073884e-02 1.121334e-02 1.368814e-01 1.381827e-01  
3.426581e-01 3.683766e-01 2.096506e-01

snipped



The reason is that the real a2 function is a difficult function  
that only needs to be evaluated if the value of a1[i-1] meets some  
criteria.


Then maybe you should only apply it when those criteria are met?



Thanks in advance,

__

--

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Heritage Laboratories
West Hartford, CT

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Re: [R] SPSS long variable names

No!

That is variable labels.

Caveman


On Tue, Oct 13, 2009 at 4:52 PM, Robert  Baer rb...@atsu.edu wrote:
 I am wondering if there is a patch for the SPSS reading
 code on the
 foreign package, in order to be able to read long variable
 names.
 Right now read.spss() just trunc the names to 8
 characters.

 This sequence seems to access the long filenames for me if I know what you
 are asking for:

 library('foreign')
 a-read.spss('fil.sav')
 lnames - attr(a,variable.labels,exact=FALSE)

 Rob




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Re: [R] Counting

2009-10-13 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Ashta
 Sent: Tuesday, October 13, 2009 5:20 AM
 To: R help
 Subject: [R] Counting

 *Hi all,
 *

 *Assume that I have the following data set  with tow 
 variables and I want
 count the number of observation with identical values
 *

 **

 *x1 x2*

 * 1   1 *

 * 1   0 *

 * 0   1*

 * 0   1*

 * 0   0*

 * 1   1*

 * 0   1
 *

 I want the  following output
 **

 *
 *

 *n1=3  # number of identical observation between x1 and x2 variables*

 *n2=4  # number of different observation*

sum() converts TRUE to 1 and FALSE to 0 so the following works
   n1 - sum(x1 == x2)
   n2 - sum(x1 != x2)

You can also use table() to get both numbers in one vector.  In the
following I make table's input a factor (a) to make sure that both the
== and != counts are in the table even if one count is zero and (b)
to put them in the order you asked for, TRUE then FALSE:
  n12 - table(factor(x1==x2, levels=c(TRUE,FALSE)))
  n1 - n12[1]
  n2 - n12[2]

If there may be missing values in the data then you have to decide how
to handle them.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 How do I do it in R?

 Thanks a lot

 **

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[R] Free Introductory R Course Taught Over the Web

2009-10-13 Thread Paul H Geissler

Free Introductory R Course Taught Over the Web

The course is designed for natural resource managers and is open to all 
who are interested without charge. Audio of the presentations is available 
either using your computer speakers and optional microphone or headset or 
by calling a phone bridge long distance. Live video of the presenter's 
computer screen is available over the web. You can also share your 
computer's screen with other participants when asking a question or making 
a point. An audio and video recording of the presentations and discussion 
will be available on our FTP site after the presentations.  There are no 
specific prerequisites but some knowledge of statistics would be helpful. 
A basic knowledge of computers and the internet will be assumed. 

Please forward this notice to those who may be interested.

The course will start Monday, November 9.  There will be presentations on 
Mondays and Wednesdays, and a lab on Tuesdays for two hours.  The times 
will be: Hawaii 9:00-11:00, Alaska 10:00-12:00, Pacific 11:00-1:00, 
Mountain 12:00-2:00, Central 1:00-3:00, Eastern 2:00-4:00, UTC 7:00-9:00. 
The course will continue until we finish the outline: 2-3 weeks if you 
only continue through the GUI interface (menu) portion, perhaps 6-8 weeks 
including more advanced statistical analyses. 

Links:
You can register at 
http://www.fort.usgs.gov/brdscience/courseRegister.aspx
The course website: http://www.fort.usgs.gov/brdscience/learnR.htm
Last year's course website: 
http://www.fort.usgs.gov/brdscience/learnR08.htm

The course is presented by the US Geological Survey, Status and Trends 
Program (Paul Geissler, paul_geiss...@usgs.gov) and the National Park 
Service, Inventory and Monitoring Program (Tom Philippi, 
tom_phili...@nps.gov).  Please contact us for more information.  Comments 
and suggestions will be very welcome. 

Cheers,
Paul
---
Paul H. Geissler, Ph.D.
USGS Status  Trends of Biological Resources Program
Coordinator, National Park Monitoring Project
Assistant Program Coordinator
USGS Fort Collins Science Center
2150 Centre Ave., Building C
Fort Collins, CO 80526-8118
970-226-9482, FAX 970-226-9452
paul_geiss...@usgs.gov (please do NOT send e-mail to Paul E. Geissler 
pgeiss...@usgs.gov)

It is easy to lie with statistics.
It is hard to tell the truth without statistics.
Andrejs Dunkels, quoted by Maindonald  Braun
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Re: [R] lapply() reccursively

2009-10-13 Thread hadley wickham

 Neither a1 nor 2:100 are lists, so it would seem that sapply would be more
 appropriate.

The difference between lapply and sapply is the output, not the input.

Hadley

-- 
http://had.co.nz/

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Re: [R] SPSS long variable names

Sadlly exportation to csv or another format is not implemented yet:
http://www.gnu.org/software/pspp/manual/html_node/Not-Implemented.html

Caveman


On Tue, Oct 13, 2009 at 4:41 PM, joris meys jorism...@gmail.com wrote:
 Just for clarity : the csv format will solve your problem, as the
 restrictions on the variable names will only depend on the program you
 use to load them. I never experienced problems with variable names
 using csv to switch datasets between SPlus, R, SAS and SPSS.

 Cheers
 Joris

 On Tue, Oct 13, 2009 at 4:35 PM, joris meys jorism...@gmail.com wrote:
 Sorry to be so blunt, but I cannot believe PSPP can't save a dataset
 as a .csv file for example. That should be the prefered format to
 transport a dataset to any other statistical package, including R. csv
 files are universal.

 Cheers
 Joris

 On Tue, Oct 13, 2009 at 3:30 PM, Orvalho Augusto orvaq...@gmail.com wrote:
 Thanks for the answer.

 Hmisc uses read.spss from the foreign package. And so it does not
 solve my trouble.

 I need to read the long names on the SPSS dataset.

 Caveman


 On Tue, Oct 13, 2009 at 3:01 PM, John Kane jrkrid...@yahoo.ca wrote:
 library(Hmisc) spss.get may do it but it's been some time since I used it.

 --- On Sat, 10/10/09, Orvalho Augusto orvaq...@gmail.com wrote:

 From: Orvalho Augusto orvaq...@gmail.com
 Subject: [R] SPSS long variable names
 To: r-help@r-project.org
 Received: Saturday, October 10, 2009, 12:14 PM
 Hello guys I am new to this list and
 for R too.

 I am wondering if there is a patch for the SPSS reading
 code on the
 foreign package, in order to be able to read long variable
 names.
 Right now read.spss() just trunc the names to 8
 characters.

 Or if someone could help me on other way:
 I have to process everyday a lot of SPSS Syntax Files and
 Dat files
 that come from one system that can only export data on
 through that
 way.

 I use PSPP to generate the spss data file (sav) that I read
 with R.
 From R I can export to MySQL, DBF and STATA to satisfy
 the needs of
 different guys here.

 The problem is the limit of 8 characters long on variable
 names.

 Can someone help on that?

 Caveman

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



      __
 The new Internet Explorer® 8 - Faster, safer, easier.  Optimized for 
 Yahoo!  Get it Now for Free! at 
 http://downloads.yahoo.com/ca/internetexplorer/


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Re: [R] Introduction to mark-recapture analysis in R?

On Oct 13, 2009, at 8:00 AM, Anne-Katrin Link wrote:

I was wondering whether there are any good books and/or
website

links that introduce mark-recapture analysis in R. In particular, I am
interested in exploratory data analysis of resighting data and how to
create capture histories from dataframes in R.

Thank you very much for your reply in advance!

Have you looked at the citations you get in the package that come back
from the obvious search strategy?

http://search.r-project.org/cgi-bin/namazu.cgi?query=%22capture-recapture%22max=100result=normalsort=scoreidxname=functionsidxname=Rhelp08idxname=views

And I thought I remembered a posting from the author of:

Ecological Models and Data in R (Hardcover) by Benjamin M. Bolker

saying that such methods were discussed and exemplified. But I
cannot find the link to confirm my memory on this point. (And now that
I check to see if more knowledgeable persons might have already
answered this, I see that Bolker himself says he doesn't know about
books, so my memory may have been manufactured.

http://finzi.psych.upenn.edu/R/library/Rcapture/doc/RcaptureJSS.pdf

And I am not a biologist so have no experience on what would be the
best book.

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] time grid for survfit Survival function outputs

2009-10-13 Thread Muhtar Osman

Dear All,

Maybe it is a silly question. But I wasn't able to find it from manual
or R site search.
I was wondering what is the corresponding time axis for survival
function outputs in survfit. I think it is survfit(...)$time,
but not 100% sure.
If it is, is it possible we could make survival function outputs on
the pre-specified time grid with fixed increment and fixed length.

Thank you so much.

Regards,

MJO

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Re: [R] SPSS long variable names

2009-10-13 Thread Frank E Harrell Jr


Orvalho Augusto wrote:

Sadlly exportation to csv or another format is not implemented yet:
http://www.gnu.org/software/pspp/manual/html_node/Not-Implemented.html

Caveman


That would not solve the problem anyway because you would not get labels 
and other variable attributes.


Frank




On Tue, Oct 13, 2009 at 4:41 PM, joris meys jorism...@gmail.com wrote:

Just for clarity : the csv format will solve your problem, as the
restrictions on the variable names will only depend on the program you
use to load them. I never experienced problems with variable names
using csv to switch datasets between SPlus, R, SAS and SPSS.

Cheers
Joris

On Tue, Oct 13, 2009 at 4:35 PM, joris meys jorism...@gmail.com wrote:

Sorry to be so blunt, but I cannot believe PSPP can't save a dataset
as a .csv file for example. That should be the prefered format to
transport a dataset to any other statistical package, including R. csv
files are universal.

Cheers
Joris

On Tue, Oct 13, 2009 at 3:30 PM, Orvalho Augusto orvaq...@gmail.com wrote:

Thanks for the answer.

Hmisc uses read.spss from the foreign package. And so it does not
solve my trouble.

I need to read the long names on the SPSS dataset.

Caveman


On Tue, Oct 13, 2009 at 3:01 PM, John Kane jrkrid...@yahoo.ca wrote:

library(Hmisc) spss.get may do it but it's been some time since I used it.

--- On Sat, 10/10/09, Orvalho Augusto orvaq...@gmail.com wrote:


From: Orvalho Augusto orvaq...@gmail.com
Subject: [R] SPSS long variable names
To: r-help@r-project.org
Received: Saturday, October 10, 2009, 12:14 PM
Hello guys I am new to this list and
for R too.

I am wondering if there is a patch for the SPSS reading
code on the
foreign package, in order to be able to read long variable
names.
Right now read.spss() just trunc the names to 8
characters.

Or if someone could help me on other way:
I have to process everyday a lot of SPSS Syntax Files and
Dat files
that come from one system that can only export data on
through that
way.

I use PSPP to generate the spss data file (sav) that I read
with R.
From R I can export to MySQL, DBF and STATA to satisfy
the needs of
different guys here.

The problem is the limit of 8 characters long on variable
names.

Can someone help on that?

Caveman

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mailing list
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reproducible code.





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

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[R] how to have tkchooseDirectory resize in windows?

2009-10-13 Thread David Gattrell

R-2.8.0  / tcltk8.5
In windows, Rgui.exe has a directory browser that can be resized, but when I
call
tkchooseDirectory(), it is a fixed size.  In linux, when I call
tkchooseDirectory() it
can be resized.
How do I get a windows version that I can resize?

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Re: [R] how to output profile plots for groups using lattice package

2009-10-13 Thread George Kalema

A million thanks Peter, subsetting fixed it.
George

On Mon, Oct 12, 2009 at 11:51 PM, Peter Ehlers ehl...@ucalgary.ca wrote:

 Hi George,

 Your problem is not with xyplot, but with the NA occurrences in
 your data. Try adding

  subset = {!is.na(MSE)},

 to your xyplot call, or (better), subset the data before
 calling xyplot.

  -Peter Ehlers


 George Kalema wrote:

 Hi Peter (and anyone else willing to help me out),
 Many thanks for your help. Having used your code plus a few other
 modifications, I only get the points plotted but without the two lines. I
 just cannot figure out what the problem is.

 My code is as follows:

 library(lattice)
 datos2 - subset(datos, samplesize != 10  parm != Theta0)
 unq - sort(unique(datos2$samplesize))
 datos2$fsamplesize - factor(datos2$samplesize, labels = paste(Sample
 size =, unq))
 datos2$parm - factor(datos2$parm, levels = c(Intercept, time,
 trt, time*trt))
 tp1.sim - xyplot(MSE ~ ntimes | fsamplesize + parm, group = group, data
 = datos2,
type = b, lty = 1:2, pch = 1:2,
scales = list(x = list(at = c(2, 4, 8, 16)), alternating = 1),
as.table = TRUE, key = list(text = list(c(GNA, PNA)), points =
 list(pch = 1:2))
 )
 plot(tp1.sim)

 I have attached my real dataset (called datos) as well.

 Kind appreciations to your efforts.

 George


 On Wed, Oct 7, 2009 at 9:20 AM, Peter Ehlers ehl...@ucalgary.ca wrote:

  see below

 George Kalema wrote:

  Dear R users,
 I am trying to have an xyplot of a data set which has the following
 variables:
 case (n=10,20,30)
 parameter (parm=a,b)
 group (grp=g1,g2)
 y (y values)
 x (x=2,4,8)

 My plot should be parameter by case such that I have 2 rows (each row=
 each
 parameter) and 3 columns (each column=each case). My R-code is as
 follows
 but I am not able to get what I want to:

 tp1.sim - xyplot(y~ x | case + parm , group=group, data = data, lty =
 1:4
 ,
 pch = 1:4)
 print(tp1.sim)

 How can I have two lines (for g1 and g2) in each plot (each box)?

  include the type=b argument

  How do I label the x-axis with only values 2, 4, 8?
 include the scales= argument or make x a factor

  How do I label each column with the corresponding case number?
 make 'case' a factor

 The following should do what you want:

 xyplot(y ~ x | factor(case) + parm, group=group, data=data,
   type='b', lty=1:2, pch=1:2,
   scales=list(x=list(at=c(2,4,8)))
 )

 I don't understand why you want 4 line types/point chars.

  -Peter Ehlers


  My hypothetical data set is as follows:

 parm x case y group
 a 2 10 0.03 g1
 b 2 10 0.02 g1
 a 4 10 0.03 g1
 b 4 10 0.02 g1
 a 8 10 0.03 g1
 b 8 10 0.02 g1
 a 2 20 0.03 g1
 b 2 20 0.02 g1
 a 4 20 0.03 g1
 b 4 20 0.02 g1
 a 8 20 0.03 g1
 b 8 20 0.02 g1
 a 2 30 0.03 g1
 b 2 30 0.02 g1
 a 4 30 0.03 g1
 b 4 30 0.02 g1
 a 8 30 0.03 g1
 b 8 30 0.02 g1
 a 2 10 0.13 g2
 b 2 10 0.12 g2
 a 4 10 0.13 g2
 b 4 10 0.12 g2
 a 8 10 0.13 g2
 b 8 10 0.12 g2
 a 2 20 0.13 g2
 b 2 20 0.12 g2
 a 4 20 0.13 g2
 b 4 20 0.12 g2
 a 8 20 0.13 g2
 b 8 20 0.12 g2
 a 2 30 0.13 g2
 b 2 30 0.12 g2
 a 4 30 0.13 g2
 b 4 30 0.12 g2
 a 8 30 0.13 g2
 b 8 30 0.12 g2

 Many thanks in advance for your response.

 George

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-- 

George Williams KALEMA,

Schapenstraat 37/282,
3000 Leuven,
Belgium.

Cell: +32 495 33 13 02


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[R] histogram

2009-10-13 Thread Dmitry Gospodaryov


Dear R developers,How I can build a histogram from matrix:

0 0.5 1

0.25 34 43 65
1 23 35 54
4 22 29 42
10 21 22 29
20 15 17 20

(first string is represented names of columns,
first column is represented names of rows)
where names of columns should be x-axis labels; respectively
to this, I want to have three groups of bars (5 bars in each group)?
Y values should be represented by values given in the core of
matrix. Names of the rows should be in a legend, and should
represent the each of 5 bars (in group) name.
I would also try to build filled contour, however, i can't
ask the program to consider column and rownames like
true values, not only like labels. So, column names should
be the y-values, while row names should be the x-values.
Values placed in the core of matrix should be z-values.

With regard, Dmitry.

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[R] response surface designs

2009-10-13 Thread Dmitry Gospodaryov


How I can obtain graphics for response surface design,
basing on data:
x: 0.25, 1, 4, 10, 20.
y: 0, 0.5, 1
z1 (for y = 0): 45, 35, 25, 15, 10.
z2 (for y = 0.5): 50, 45, 36, 21, 17.
z3 (for y = 1): 37, 34, 22, 17, 11?
z-values should be scaled in colour (e. g. from red to blue).
I consider to use packages rsm and graphics
(with grDevices). I suggest to use Box-Behnken design
in rsm and filled.contour in graphics. However,
I do not know how I can combain these two opportunities.
I am not familiar in response surface analysis, and able
to be wrong in the selection of the model. Thank you
for any reply. With regard, Dmitry.

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[R] can anybody give suggestion on how to use survreg with 3-parameter weibull

2009-10-13 Thread Josephine Sari

Hi,

 

I am a beginner in R. I would like to know if there is any trick in using
survreg with 3-parameter weibull?

I would like to do survival analysis of failure time but not using the
2-parameter weibull which is available as one of the options in the survreg
but instead I would like to see the result when it is modelled with
3-parameter weibull.

I don't find the syntax. May be I miss the manual/explanation.

 

Thank you in advanced.

 

Best regards,

Josephine Sari


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[R] decostand

2009-10-13 Thread Rosa Manrique

Hi:
I do not know why the function decostand is not found in my vegan library, I 
have downloaded the package recently, and it seems to work well..Do you have 
any suggestion?
Rosa.
[[alternative HTML version deleted]]

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Re: [R] Introduction to mark-recapture analysis in R?

2009-10-13 Thread Jeff Laake

Another package that is not on CRAN or other R depository is RMark.  It 
works with the software MARK which does the computation.  You can find 
RMark at  http://www.phidot.org/software/mark/rmark/

and MARK from http://www.phidot.org/software/mark/index.html
From the same site you'll find a very comprehensive electronic book 
(that describes how to use MARK 
(http://www.phidot.org/software/mark/docs/book/) and it has an appendix 
(http://www.phidot.org/software/mark/docs/book/pdf/app_3.pdf)that 
describes use of RMark.  You may also want to look at the RMark workshop 
notes 
(http://www.phidot.org/software/mark/rmark/RMarkWorkshopNotes.pdf). 
http://www.phidot.org/software/mark/rmark/RMarkWorkshopNotes.pdf


RMark does contain some routines that do CJS and JS models solely in R 
(see ?cjs and ?js) but they are experimental at this point.  RMark was 
not put on CRAN because with the above exceptions it does require MARK 
which is a separate piece of FORTRAN software that runs only in WINDOWS, 
although it possible to obtain a LINUX version with some work.  MARK is 
freely available software as an executable but is not open-source per 
se.  Source for RMark is available from the above sites.


With regard to creating capture histories in R, that is easily 
accomplished with the table function.  For example, if x is a dataframe 
of capture/recapture events with fields ID (unique identifier) and 
Occasion  is a  factor variable for the capture occasion then


chmat=with (x, table(ID,Occasion))

will create a count of captures by ID and Occasion.  If an animal can be 
caught more than once per occasion then add:


chmat[chmat0]=1

Then you can change to capture history strings with:

apply(chmat,1,paste,sep=)

Here is a simple example (with nonsense data)

x=data.frame(ID=floor(10*runif(100))+1,Occasion=floor(5*runif(100))+1)
chmat=with (x, table(ID,Occasion))
chmat[chmat0]=1
ch=apply(chmat,1,paste,collapse=)
table(ch)

If you wanted to include individual covariates you would not use the 
last table(ch) statement but would tie back to any individual data.


regards --jeff






Anne-Katrin Link wrote:
Normal021falsefalse
false
MicrosoftInternetExplorer4  
  

Dear R-helpers,

 I was wondering whether there are any good books and/or website 
links that introduce mark-recapture analysis in R. In particular, I am 
interested in exploratory data analysis of resighting data and how to 
create capture histories from dataframes in R.

Thank you very much for your reply in advance!

 Cheers, 

Anne
  



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Re: [R] Introduction to mark-recapture analysis in R?

2009-10-13 Thread Kingsford Jones

see

https://stat.ethz.ch/pipermail/r-sig-ecology/2008-May/000160.html


hth,
Kingsford

On Tue, Oct 13, 2009 at 6:00 AM, Anne-Katrin Link anne.l...@gmx.de wrote:
 Normal        0        21                        false        false
 false
 MicrosoftInternetExplorer4


 Dear R-helpers,

          I was wondering whether there are any good books and/or website
 links that introduce mark-recapture analysis in R. In particular, I am
 interested in exploratory data analysis of resighting data and how to
 create capture histories from dataframes in R.

 Thank you very much for your reply in advance!

          Cheers,

 Anne
 --
 Jetzt kostenlos herunterladen: Internet Explorer 8 und Mozilla Firefox 3.5 -
 sicherer, schneller und einfacher! http://portal.gmx.net/de/go/atbrowser

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Re: [R] lapply() reccursively

2009-10-13 Thread Charles C. Berry


On Tue, 13 Oct 2009, Kaveh Vakili wrote:



Hi all,

I was wondering whether it is possible to use the lapply() function
to alter the value of the input, something in the spirit of :

a1-runif(100)
a2-function(i){
a1[i]-a1[i-1]*a1[i];a1[i]
}
a3-lapply(2:100,a2)

Something akin to a for() loop, but using the lapply() infrastructure.
I haven't been able to get rapply() to do this.


Maybe you want to check out

?Reduce

For the example above, something like

a3 -  Reduce( *, a1, accumulate = TRUE )


HTH,

Chuck



The reason is that the real a2 function is a difficult function that only 
needs to be evaluated if the value of a1[i-1] meets some criteria.

Thanks in advance,

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



Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:cbe...@tajo.ucsd.edu   UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

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Re: [R] Selecting initial numerals

2009-10-13 Thread Dieter Menne




PDXRugger wrote:
 
 II just want to create a new object with the first two numerals of the
 data. Not sure why this isnt working, consider the following:
 
 EmpEst$naics=c(238321, 624410, 484121 ,238911, 81, 531110, 621399,
 541613,
 524210 ,236115 ,811121 ,236115 ,236115 ,621610 ,814110 ,812320)
 
 
 EmpEst$naics2-formatC(EmpEst$naics %% 1e2, width=2, flag=, mode
 =integer)
 #RESULT:Warning message:
 #In Ops.factor(EmpEst$naics, 100) : %% not meaningful for factors
 
 

It always good to make a complete example; the above code does not run. If I
do a guess how it could have looked like, there is no warning.

Dieter

# what is empest?
EmpEst = data.frame(x=1:16)
EmpEst$naics=c(238321, 624410, 484121 ,238911, 81, 531110, 621399,
541613,
524210 ,236115 ,811121 ,236115 ,236115 ,621610 ,814110 ,812320)

EmpEst$naics2-formatC(EmpEst$naics %% 1e2, width=2, flag=, mode
=integer)
# no warning

-- 
View this message in context: 
http://www.nabble.com/Selecting-initial-numerals-tp25876664p25876826.html
Sent from the R help mailing list archive at Nabble.com.

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[R] replacing period with a space

2009-10-13 Thread Dimitri Liakhovitski

Dear R-ers!

I have x as a variable in a data frame x.

x-data.frame(x=c(aa.bb,cc.dd.ee))
x$x-as.character(x$x)
x

I am sorry for such a simple question - but how can I replace all
periods in x$x with spaces?

sub('.', ' ', x$x) - removes all letters to the left of each period...

Thanks a lot for your advice!

-- 
Dimitri Liakhovitski
Ninah.com
dimitri.liakhovit...@ninah.com

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Re: [R] replacing period with a space

2009-10-13 Thread Sarah Goslee

As you've discovered, the . means something special in regular
expressions (and R's version of them). You need to escape it with \\:

 x-data.frame(x=c(aa.bb,cc.dd.ee))
 x$x-as.character(x$x)
 x
 x
1aa.bb
2 cc.dd.ee
 sub(\\.,  , x$x)
[1] aa bbcc dd.ee
 gsub(\\.,  , x$x)
[1] aa bbcc dd ee

And to change all, you need gsub() rather than sub().

Sarah

On Tue, Oct 13, 2009 at 1:26 PM, Dimitri Liakhovitski ld7...@gmail.com wrote:
 Dear R-ers!

 I have x as a variable in a data frame x.

 x-data.frame(x=c(aa.bb,cc.dd.ee))
 x$x-as.character(x$x)
 x

 I am sorry for such a simple question - but how can I replace all
 periods in x$x with spaces?

 sub('.', ' ', x$x) - removes all letters to the left of each period...

 Thanks a lot for your advice!




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

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Re: [R] replacing period with a space

You need escape the period:

gsub(\\.,  , x$x)

On Tue, Oct 13, 2009 at 2:26 PM, Dimitri Liakhovitski ld7...@gmail.com wrote:
 Dear R-ers!

 I have x as a variable in a data frame x.

 x-data.frame(x=c(aa.bb,cc.dd.ee))
 x$x-as.character(x$x)
 x

 I am sorry for such a simple question - but how can I replace all
 periods in x$x with spaces?

 sub('.', ' ', x$x) - removes all letters to the left of each period...

 Thanks a lot for your advice!

 --
 Dimitri Liakhovitski
 Ninah.com
 dimitri.liakhovit...@ninah.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.




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

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Re: [R] replacing period with a space

2009-10-13 Thread Dimitri Liakhovitski

Thanks a lot for your help, Henrique and Sarah!
Dimitri

On Tue, Oct 13, 2009 at 1:37 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
 As you've discovered, the . means something special in regular
 expressions (and R's version of them). You need to escape it with \\:

 x-data.frame(x=c(aa.bb,cc.dd.ee))
 x$x-as.character(x$x)
 x
         x
 1    aa.bb
 2 cc.dd.ee
 sub(\\.,  , x$x)
 [1] aa bb    cc dd.ee
 gsub(\\.,  , x$x)
 [1] aa bb    cc dd ee

 And to change all, you need gsub() rather than sub().

 Sarah

 On Tue, Oct 13, 2009 at 1:26 PM, Dimitri Liakhovitski ld7...@gmail.com 
 wrote:
 Dear R-ers!

 I have x as a variable in a data frame x.

 x-data.frame(x=c(aa.bb,cc.dd.ee))
 x$x-as.character(x$x)
 x

 I am sorry for such a simple question - but how can I replace all
 periods in x$x with spaces?

 sub('.', ' ', x$x) - removes all letters to the left of each period...

 Thanks a lot for your advice!




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




-- 
Dimitri Liakhovitski
Ninah.com
dimitri.liakhovit...@ninah.com

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Re: [R] decostand

2009-10-13 Thread Jorge Ivan Velez

Hi Rosa,
It works for me on an R-fresh session:

 require(vegan)
Loading required package: vegan
This is vegan 1.15-4
 ?decostand
 sessionInfo()
R version 2.9.2 RC (2009-08-23 r49375)
i386-pc-mingw32

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

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

other attached packages:
[1] vegan_1.15-4

Perhaps if you told us your OS and  vegan package's version we might be of
more help.

Best,
Jorge


On Tue, Oct 13, 2009 at 11:06 AM, Rosa Manrique  wrote:

 Hi:
 I do not know why the function decostand is not found in my vegan library,
 I have downloaded the package recently, and it seems to work well..Do you
 have any suggestion?
 Rosa.
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 http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]

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Re: [R] lapply() reccursively

2009-10-13 Thread Kaveh Vakili


Thanks, Chuck's answer is the closest to what i want (gives the same result as 
cumprod()) ...but using this function seems actually slower than the loop (is 
it normal ?):

a1-runif(10)
cadd-function(x) Reduce(*, x, accumulate = TRUE)
looop-function(a1){
j-length(a1)
for(i in 2:j){
a1[i]-a1[i-1]*a1[i]
}
a1
}
 
 system.time(cadd(a1))
   user  system elapsed 
  1.344   0.004   1.353 
 system.time(cumprod(a1))
   user  system elapsed 
  0.004   0.000   0.002 
 system.time(looop(a1))
   user  system elapsed 
  0.772   0.000   0.775 
 


On Tue, 13 Oct 2009, Kaveh Vakili wrote:


 Hi all,

 I was wondering whether it is possible to use the lapply() function
 to alter the value of the input, something in the spirit of :

 a1-runif(100)
 a2-function(i){
 a1[i]-a1[i-1]*a1[i];a1[i]
 }
 a3-lapply(2:100,a2)

 Something akin to a for() loop, but using the lapply() infrastructure.
 I haven't been able to get rapply() to do this.

Maybe you want to check out

  ?Reduce

For the example above, something like

a3 -  Reduce( *, a1, accumulate = TRUE )


HTH,

Chuck


 The reason is that the real a2 function is a difficult function that only 
 needs to be evaluated if the value of a1[i-1] meets some criteria.

 Thanks in advance,

 __
 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.


Charles C. Berry(858) 534-2098
 Dept of Family/Preventive 
 Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901









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[R] How to specify an ARMA(1, [1,4]) model?

2009-10-13 Thread Len Vir

Hi,

I'm trying to model an ARMA(1,[1,4]),
i.e. I want only lags 1 and 4 of the Moving Average part.
It's the '[1,4]' part that is giving me a problem.

I've tried different arma's and arima's in different packages, namely:
packages tseries, fArma, FinTS, timeSeries, TSA, Zelig, ds1, forecast


For example, with package FinTS:

 ( ARIMA(y, order=c(1,0,c(1,4)))  )
Error in arima(x = x, order = order, seasonal = seasonal, xreg = xreg,  :
  'order' must be a non-negative numeric vector of length 3

Using ARIMA(1,0,1) with a seasonal argument for lag 4
does not get me any further.


With package Zelig I got:

 (  zelig(Diff(lppi,1) ~ one + lag.y(1) + lag.eps(1) + lag.eps(4) ,
model=arima  , data=Q)  )
Error in model.frame.default(mf$formula, data) :
  invalid type (list) for variable 'lag.eps(1)'

I get basically the same kind of answers with other packages
and with different configurations.

Thanks for any advice,

len

[[alternative HTML version deleted]]

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Re: [R] How to specify an ARMA(1, [1,4]) model?

2009-10-13 Thread Duncan Murdoch


On 10/13/2009 2:35 PM, Len Vir wrote:

Hi,

I'm trying to model an ARMA(1,[1,4]),
i.e. I want only lags 1 and 4 of the Moving Average part.
It's the '[1,4]' part that is giving me a problem.

I've tried different arma's and arima's in different packages, namely:
packages tseries, fArma, FinTS, timeSeries, TSA, Zelig, ds1, forecast


For example, with package FinTS:


( ARIMA(y, order=c(1,0,c(1,4)))  )

Error in arima(x = x, order = order, seasonal = seasonal, xreg = xreg,  :
  'order' must be a non-negative numeric vector of length 3

Using ARIMA(1,0,1) with a seasonal argument for lag 4
does not get me any further.


What's wrong with

arima(x, order=c(1,0,1), seasonal=list(order=c(0,0,1), period=4))

using stats::arima?

Duncan Murdoch




With package Zelig I got:


(  zelig(Diff(lppi,1) ~ one + lag.y(1) + lag.eps(1) + lag.eps(4) ,

model=arima  , data=Q)  )
Error in model.frame.default(mf$formula, data) :
  invalid type (list) for variable 'lag.eps(1)'

I get basically the same kind of answers with other packages
and with different configurations.

Thanks for any advice,

len

[[alternative HTML version deleted]]

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Re: [R] replacing period with a space

2009-10-13 Thread Jason Rupert

Here is one more that works:
gsub(., ,Start.Time, fixed = TRUE)

fixed = TRUE really helps in a lot of instances for removing specific 
characters without accidently angering the regular expression gods. 

Enjoy. 



--- On Tue, 10/13/09, Dimitri Liakhovitski ld7...@gmail.com wrote:

 From: Dimitri Liakhovitski ld7...@gmail.com
 Subject: [R] replacing period with a space
 To: R-Help List r-h...@stat.math.ethz.ch
 Date: Tuesday, October 13, 2009, 12:26 PM
 Dear R-ers!
 
 I have x as a variable in a data frame x.
 
 x-data.frame(x=c(aa.bb,cc.dd.ee))
 x$x-as.character(x$x)
 x
 
 I am sorry for such a simple question - but how can I
 replace all
 periods in x$x with spaces?
 
 sub('.', ' ', x$x) - removes all letters to the left of
 each period...
 
 Thanks a lot for your advice!
 
 -- 
 Dimitri Liakhovitski
 Ninah.com
 dimitri.liakhovit...@ninah.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-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] time grid for survfit Survival function outputs



On Oct 13, 2009, at 12:03 PM, Muhtar Osman wrote:


Dear All,

Maybe it is a silly question. But I wasn't able to find it from manual
or R site search.


After library(survival) , the description of survfit objects will be  
found with:


?survfit.object


I was wondering what is the corresponding time axis for survival
function outputs in survfit. I think it is survfit(...)$time,
but not 100% sure.


What makes you doubt this?


If it is, is it possible we could make survival function outputs on
the pre-specified time grid with fixed increment and fixed length.


With the appropriate supply of a dataset, it should be possible   
for various meanings of pre-specified and fixed increment and  
fixed length, all of which at the moment are ambiguous at the  
moment. That is why the Posting Guide strongly suggests both an R  
encoded example and a clear specification of the desired output.




Thank you so much.

Regards,

MJO


--

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] Greater than less than in ifelse

2009-10-13 Thread Richardson, Patrick

I'm trying to categorize a continuous variable (yes, I know that's horrible, 
but I'm trying to reproduce some exercises from a textbook) and don't really 
know an efficient way to do this.

I have a data frame that looks like:

   surv_time relapse sex log_WBC rx
1 35   0   11.45  0
2 34   0   11.47  0
3 32   0   12.20  0
4 32   0   12.53  0

And I'm trying to categorize log_WBC into:

(0-2.30) = low
(2.31-3.00)= medium
(3.00) = high

I've used an ifelse statement such as:

anderson$log_WBC - ifelse(anderson$log_WBC2.30,low,anderson$log_WBC)

Is there a way to use greater than less than syntax within the context of 
an ifelse statement? Or can someone point me to a function that will do this 
easier.

Many Thanks,

Patrick

This email message, including any attachments, is for th...{{dropped:6}}

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Re: [R] Greater than less than in ifelse

2009-10-13 Thread Jorge Ivan Velez

Dear Patrick,
Take a look at ?cut for some ideas.

HTH,
Jorge


On Tue, Oct 13, 2009 at 3:37 PM, Richardson, Patrick  wrote:

 I'm trying to categorize a continuous variable (yes, I know that's
 horrible, but I'm trying to reproduce some exercises from a textbook) and
 don't really know an efficient way to do this.

 I have a data frame that looks like:

   surv_time relapse sex log_WBC rx
 1 35   0   11.45  0
 2 34   0   11.47  0
 3 32   0   12.20  0
 4 32   0   12.53  0

 And I'm trying to categorize log_WBC into:

 (0-2.30) = low
 (2.31-3.00)= medium
 (3.00) = high

 I've used an ifelse statement such as:

 anderson$log_WBC - ifelse(anderson$log_WBC2.30,low,anderson$log_WBC)

 Is there a way to use greater than less than syntax within the context
 of an ifelse statement? Or can someone point me to a function that will do
 this easier.

 Many Thanks,

 Patrick

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Re: [R] Greater than less than in ifelse