date:20160331

Re: [R] [DKIM] Batch Installer for R [SEC=UNCLASSIFIED]

2016-03-31 Thread Li Jin

Hi Tobias,
Here is something I acquired from this mailing list some years ago. It works 
well for me:

#---run in previous version (e.g. R 3.1.0)
packages <- installed.packages()[,"Package"]
save(packages, file="Rpackages_R3.1.0")

#---run in new version
load("Rpackages_R3.1.0")

for (p in setdiff(packages, installed.packages()[,"Package"]))
install.packages(p)

Hope this helps.

Jin

-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Tobias Knuth
Sent: Friday, 1 April 2016 12:05 AM
To: r-help@r-project.org
Subject: [DKIM] [R] Batch Installer for R

Hi everyone,

in Python, you can run pip install -r filename to install all packages listed 
in the file. Is there something similar to R? If not, isn't it quite easy to 
write?

For me, it would be much easier to work on projects with other people if I 
could just install all dependencies with one line in a generalised manner.

Did anybody try something like that before me?

Best,
Tobias

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Re: [R] Filtering based on the occurrence

2016-03-31 Thread Farnoosh Sheikhi via R-help

Hi Jim, 
Thank you tons for your help. The code worked perfectly :) Best,Farnoosh

 

On Wednesday, March 30, 2016 1:13 AM, Jim Lemon  
wrote:
 

 Hi Farnoosh,
Despite my deep suspicion that this answer will solve a useless
problem, try this:

last_subject<-0
keep_deps<-c("B","D","F")
keep_rows<-NULL
for(rowindex in 1:dim(df)[1]) {
 if(df[rowindex,"Subject"] != last_subject) {
  last_subject<-df[rowindex,"Subject"]
  start_keeping<-0
 }
 if(df[rowindex,"deps"] %in% keep_deps) start_keeping<-1
 if(start_keeping) keep_rows<-c(keep_rows,rowindex)
}
final<-matrix(unlist(lapply(df[keep_rows,],as.character)),ncol=3)

I find it terribly hard to ignore puzzles.

Jim


On Wed, Mar 30, 2016 at 10:52 AM, Farnoosh Sheikhi via R-help
 wrote:
> Hello,
> I have a data set similar to below and I wanted to keep the observations 
> after the first occurrence of these department: "B", "D", "F".For example for 
> ID=2, the observation with deps=B and anything after will be kept in the 
> data. For ID=3, observations with deps=D and anything after will be included.
> Subject<- c("2", "2", "2", "3", "3", "3", "4", "4", "5", "5", "5", 
> "5")dates<-seq(as.Date('2011-01-01'),as.Date('2011-01-12'),by = 1) 
> deps<-c("A", "B", "C", "C", "D", "A", "F", "G", "A", "F", "A", "D")df <- 
> data.frame(Subject, dates, deps)df
> The final data should look like this:final<-c("2 2011-01-02    B","2 
> 2011-01-03    C","3 2011-01-05    D","3 2011-01-06    A","4 2011-01-07    
> F","4 2011-01-08    G","5 2011-01-10    F","5 2011-01-11    A","5 2011-01-12  
>   D") Thank you tons for your help.
> Farnoosh
>
>
>        [[alternative HTML version deleted]]
>
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Re: [R] Compute the Gini coefficient

2016-03-31 Thread Marine Regis

Hello,

Thank you very much for your help. 

How can I draw a Lorenz curve with several replications ?

Here is an example with 4 replications:

hosts=c(23,31,19,10,7,7,3,
39,40,8,3,6,2,2,
47,17,8,10,6,11,1,
30,30,10,0,15,15,0)
parasites=rep(seq(from=0,to=6,by=1),4)
replications=c(rep(1,7),rep(2,7),rep(3,7),rep(4,7))
test <- cbind(parasites,hosts,replications)

Should I calculate the average frequency of hosts (replication mean values) and 
next calculate the cumulative percentage of hosts from the average frequency ? 

Thank you very much for your time.
Have a nice day.
Marine 

De : Achim Zeileis 
Envoyé : mercredi 30 mars 2016 12:05
À : Erich Neuwirth
Cc : Marine Regis; r-help@r-project.org
Objet : Re: [R] Compute the Gini coefficient

On Wed, 30 Mar 2016, Erich Neuwirth wrote:

>
>> On 30 Mar 2016, at 02:53, Marine Regis  wrote:
>>
>> Hello,
>>
>> I would like to build a Lorenz curve and calculate a Gini coefficient in 
>> order to find how much parasites does the top 20% most infected hosts 
>> support.
>>
>> Here is my data set:
>>
>> Number of parasites per host:
>> parasites = c(0,1,2,3,4,5,6,7,8,9,10)
>>
>> Number of hosts associated with each number of parasites given above:
>> hosts = c(18,20,28,19,16,10,3,1,0,0,0)
>>
>> To represent the Lorenz curve:
>> I manually calculated the cumulative percentage of parasites and hosts:
>>
>> cumul_parasites <- cumsum(parasites)/max(cumsum(parasites))
>> cumul_hosts <- cumsum(hosts)/max(cumsum(hosts))
>> plot(cumul_hosts, cumul_parasites, type= "l?)
>
>
> Your values in hosts are frequencies. So you need to calculate
>
> cumul_hosts = cumsum(hosts)/sum(hosts)
> cumul_parasites = cumsum(hosts*parasites)/sum(parasites)

That's what I thought as well but Marine explicitly said that the 'host'
are _not_ weights. Hence I was confused what this would actually mean.

Using the "ineq" package you can also do
plot(Lc(parasites, hosts))

> The Lorenz curves starts at (0,0), so to draw it, you need to extend these 
> vectors
>
> cumul_hosts = c(0,cumul_hosts)
> cumul_parasites = c(0,cumul_parasites)
>
> plot(cumul_hosts,cum9l_parasites,type=?l?)
>
>
> The Gini coefficient can be calculated as
> library(reldist)
> gini(parasites,hosts)
>
>
> If you want to check, you can ?recreate? the original data (number of 
> parasited for each host) with
>
> num_parasites = rep(parasites,hosts)
>
> and
> gini(num_parasites)
>
> will also give you the Gini coefficient you want.
>
>
>
>>
>
>>> From this Lorenz curve, how can I calculate the Gini coefficient with the 
>>> function "gini" in R (package reldist) given that the vector "hosts" is not 
>>> a vector of weights ?
>>
>> Thank you very much for your help.
>> Have a nice day
>> Marine
>>
>>
>>  [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/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] Ask if an object will respond to a function or method

2016-03-31 Thread David Winsemius


> On Mar 31, 2016, at 1:00 PM, Paul Johnson  wrote:
> 
> In the rockchalk package, I want to provide functions for regression
> objects that are "well behaved." If an object responds to the methods
> that lm or glm objects can handle, like coef(), nobs(), and summary(),
> I want to be able to handle the same thing.
> 
> It is more difficult than expected to ask a given fitted model object
> "do you respond to these functions: coef(), nobs(), summary()." How
> would you do it?
> 
> I tried this with the methods() function but learned that all methods
> that a class can perform are not listed.  I'll demonstrate with a
> regression "zz" that is created by the example in the plm package.
> The coef() function succeeds on the zz object, but coef is not listed
> in the list of methods that the function can carry out.
> 
>> library(plm)
>> example(plm)
> 
>> class(zz)
> [1] "plm""panelmodel"
>> methods(class = "plm")
> [1] ercomp  fixef   has.intercept   model.matrix
> [5] pFtest  plmtest plotpmodel.response
> [9] pooltestpredict residuals   summary
> [13] vcovBK  vcovDC  vcovG   vcovHC
> [17] vcovNW  vcovSCC
> see '?methods' for accessing help and source code
>> methods(class = "panelmodel")
> [1] deviance  df.residual   fittedhas.intercept index
> [6] nobs  pbgtest   pbsytest  pcdtest   pdim
> [11] pdwtest   phtestprint pwartest  pwfdtest
> [16] pwtestresiduals terms updatevcov
> see '?methods' for accessing help and source code
>> coef(zz)
>   log(pcap)  log(pc) log(emp)unemp
> -0.026149654  0.292006925  0.768159473 -0.005297741
> 
> I don't understand why coef(zz) succeeds but coef is not listed as a method.

There is a coef.default method. (Also an S$  coef,ANY-method )

> 
> Right now, I'm contemplating this:
> 
> zz1 < - try(coef(zz))
> if (inherits(zz1, "try-error")) stop("Your model has no coef method")
> 
> This seems like a bad workaround because I have to actually run the
> function in order to find out if the function exists. That might be
> time consuming for some summary() methods.
> 
> pj
> 
> -- 
> Paul E. Johnson
> Professor, Political ScienceDirector
> 1541 Lilac Lane, Room 504  Center for Research Methods
> University of Kansas University of Kansas
> http://pj.freefaculty.org  http://crmda.ku.edu
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/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
Alameda, CA, USA

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Re: [R] reduced set of alternatives in package mlogit

2016-03-31 Thread Bert Gunter

code? example data?  We can only guess based on your vague post.

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

Moreover, this sounds like a statistical question, not a question
about R programming, and so might be more appropriate for a
statistical list like stats.stackexchange.com  .

Cheers,
Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Thu, Mar 31, 2016 at 2:51 PM, Jose Marcos Ferraro
 wrote:
> I'm trying to estimate a multinomial logit model  but in some choices only 
> alternatives from a subset of all possible alternatives can be chosen.
> At the moment I get around it by creating "dummy" variables to mean the 
> alternative is not available and let it estimate this coefficient as highly 
> negative. Is there a better way to do it?
>
> [[alternative HTML version deleted]]
>
> __
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> and provide commented, minimal, self-contained, reproducible code.

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[R-es] R igraph

2016-03-31 Thread Javier Marcuzzi

Estimados

Copio un mensaje de error que da igraph, es más fácil verlo que explicarlo.

> fastgreedy.community(udatos)
Error in .Call("R_igraph_community_fastgreedy", graph, as.logical(merges),  : 
  At fast_community.c:553 : fast-greedy community finding works only on graphs 
without multiple edges, Invalid value
> cluster_louvain(udatos)
IGRAPH clustering multi level, groups: 7, mod: 0.26
+ groups:
  $`1`
   [1] "Ficha 1127"  "Ficha 1309"  "Ficha 1530" 
 "Ficha 603"   "mes 2"

Los datos están de esta forma: (udatos son los mismos pero -  udatos 
<-as.undirected(datos.network))
> head(get.data.frame(dIgraph.network))
fromto  Descripcion A   Descripcion B   Relacion
1 Ficha 1022mes 10  Ficha   Mes Ficha 
Mes
2 Ficha 1024mes 8   Ficha   Mes Ficha 
Mes
3 Ficha 1028mes 2   Ficha   Mes Ficha 
Mes
…

Esperando que sea entendible mi correo, ¿Dónde tendré el problema? ¿En la 
preparación de los datos?
Desde ya, gracias.

Javier Rubén Marcuzzi


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[R] reduced set of alternatives in package mlogit

2016-03-31 Thread Jose Marcos Ferraro

I'm trying to estimate a multinomial logit model  but in some choices only 
alternatives from a subset of all possible alternatives can be chosen.
At the moment I get around it by creating "dummy" variables to mean the 
alternative is not available and let it estimate this coefficient as highly 
negative. Is there a better way to do it?

[[alternative HTML version deleted]]

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[R] Installing specific version of R

2016-03-31 Thread Sandhya Muloth Vijayachandran

Hi,

When  I am trying to  install specific version of R ,  using  apt-get -y 
--force-yes install r-base=3.2.1-4precise0 r-recommended=3.2.1-4precise0 
r-base-dev=3.2.1-4precise0,  I do see  correct version  being set up in in the 
logs , but the installed version shows me 3.2.4.
Is there anything else to be done to install specific version?

$version.string
[1] "R version 3.2.4 Revised (2016-03-16 r70336)"

Setting up r-recommended (3.2.1-4precise0) ...
Setting up r-base (3.2.1-4precise0) ...
Setting up liblzma-dev (5.1.1alpha+20110809-3) ...
Setting up r-doc-html (3.2.4-revised-1precise0) ...
Setting up x11-xserver-utils (7.6+3) ...
Setting up xbitmaps (1.1.1-1) ...
Setting up xterm (271-1ubuntu2.1) ...
update-alternatives: using /usr/bin/xterm to provide 
/usr/bin/x-terminal-emulator (x-terminal-emulator) in auto mode.
update-alternatives: using /usr/bin/uxterm to provide 
/usr/bin/x-terminal-emulator (x-terminal-emulator) in auto mode.
update-alternatives: using /usr/bin/lxterm to provide 
/usr/bin/x-terminal-emulator (x-terminal-emulator) in auto mode.
Setting up r-base-html (3.2.4-revised-1precise0) ...
Setting up libstdc++6-4.6-dev (4.6.3-1ubuntu5) ...
Setting up g++-4.6 (4.6.3-1ubuntu5) ...
Setting up g++ (4:4.6.3-1ubuntu5) ...
update-alternatives: using /usr/bin/g++ to provide /usr/bin/c++ (c++) in auto 
mode.
Setting up build-essential (11.5ubuntu2.1) ...
Setting up libwww-perl (6.03-1) ...
Setting up libxml-parser-perl (2.41-1build1) ...
Setting up intltool (0.50.2-2) ...
Setting up dh-translations (116) ...
Setting up cdbs (0.4.100ubuntu2) ...
Setting up r-base-dev (3.2.1-4precise0) ...
Setting up liblwp-protocol-https-perl (6.02-1) ...

Thanks!
Sandhya


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Re: [R] Ask if an object will respond to a function or method

2016-03-31 Thread Martin Morgan




On 03/31/2016 04:00 PM, Paul Johnson wrote:

In the rockchalk package, I want to provide functions for regression
objects that are "well behaved." If an object responds to the methods
that lm or glm objects can handle, like coef(), nobs(), and summary(),
I want to be able to handle the same thing.

It is more difficult than expected to ask a given fitted model object
"do you respond to these functions: coef(), nobs(), summary()." How
would you do it?

I tried this with the methods() function but learned that all methods
that a class can perform are not listed.  I'll demonstrate with a
regression "zz" that is created by the example in the plm package.
The coef() function succeeds on the zz object, but coef is not listed
in the list of methods that the function can carry out.


library(plm)
example(plm)



class(zz)

[1] "plm""panelmodel"

methods(class = "plm")

  [1] ercomp  fixef   has.intercept   model.matrix
  [5] pFtest  plmtest plotpmodel.response
  [9] pooltestpredict residuals   summary
[13] vcovBK  vcovDC  vcovG   vcovHC
[17] vcovNW  vcovSCC
see '?methods' for accessing help and source code

methods(class = "panelmodel")

  [1] deviance  df.residual   fittedhas.intercept index
  [6] nobs  pbgtest   pbsytest  pcdtest   pdim
[11] pdwtest   phtestprint pwartest  pwfdtest
[16] pwtestresiduals terms updatevcov
see '?methods' for accessing help and source code

coef(zz)

log(pcap)  log(pc) log(emp)unemp
-0.026149654  0.292006925  0.768159473 -0.005297741

I don't understand why coef(zz) succeeds but coef is not listed as a method.


coef(zz) finds stats:::coef.default, which happens to do the right thing 
for zz but also 'works' (returns without an error) for things that don't 
have coefficients, e.g., coef(data.frame()).


stats:::coef.default is

> stats:::coef.default
function (object, ...)
object$coefficients

Maybe fail on use, rather than trying to guess up-front that the object 
is fully appropriate?


Martin Morgan



Right now, I'm contemplating this:

zz1 < - try(coef(zz))
if (inherits(zz1, "try-error")) stop("Your model has no coef method")

This seems like a bad workaround because I have to actually run the
function in order to find out if the function exists. That might be
time consuming for some summary() methods.

pj




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Re: [R] converting a class dataframe (chars) to transaction class

2016-03-31 Thread Giorgio Garziano

A more specific reproducible example.

set.seed(1023)
library(arules)

# starting from a dataframe whose fields are characters (see stringsAsFactors = 
FALSE), as asked
products <- c("P1", "P2", "P3", "P4", "P5", "P6", "P7", "P8", "P9", "P10")
mydf <- data.frame(user = sample(LETTERS[1:20], 100, replace=T),
   prod = sample(products, 100, replace=T),
   stringsAsFactors=FALSE)
str(mydf)

# convert to factors
mydf$user <- factor(mydf$user)
mydf$prod <- factor(mydf$prod)

# splitting by user
prodlist <- split(x=mydf[,"prod"], f=mydf$user)
prodlist

# remove duplicates
prodlist <- lapply(prodlist, unique)
prod.trans <- as(prodlist, "transactions")
itemFrequency(prod.trans)

# generating rules
rules <- apriori(prod.trans,parameter=list(support=.1, confidence=.5))
inspect(rules)

--

Best,

GG





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[R] Ask if an object will respond to a function or method

2016-03-31 Thread Paul Johnson

In the rockchalk package, I want to provide functions for regression
objects that are "well behaved." If an object responds to the methods
that lm or glm objects can handle, like coef(), nobs(), and summary(),
I want to be able to handle the same thing.

It is more difficult than expected to ask a given fitted model object
"do you respond to these functions: coef(), nobs(), summary()." How
would you do it?

I tried this with the methods() function but learned that all methods
that a class can perform are not listed.  I'll demonstrate with a
regression "zz" that is created by the example in the plm package.
The coef() function succeeds on the zz object, but coef is not listed
in the list of methods that the function can carry out.

> library(plm)
> example(plm)

> class(zz)
[1] "plm""panelmodel"
> methods(class = "plm")
 [1] ercomp  fixef   has.intercept   model.matrix
 [5] pFtest  plmtest plotpmodel.response
 [9] pooltestpredict residuals   summary
[13] vcovBK  vcovDC  vcovG   vcovHC
[17] vcovNW  vcovSCC
see '?methods' for accessing help and source code
> methods(class = "panelmodel")
 [1] deviance  df.residual   fittedhas.intercept index
 [6] nobs  pbgtest   pbsytest  pcdtest   pdim
[11] pdwtest   phtestprint pwartest  pwfdtest
[16] pwtestresiduals terms updatevcov
see '?methods' for accessing help and source code
> coef(zz)
   log(pcap)  log(pc) log(emp)unemp
-0.026149654  0.292006925  0.768159473 -0.005297741

I don't understand why coef(zz) succeeds but coef is not listed as a method.

Right now, I'm contemplating this:

zz1 < - try(coef(zz))
if (inherits(zz1, "try-error")) stop("Your model has no coef method")

This seems like a bad workaround because I have to actually run the
function in order to find out if the function exists. That might be
time consuming for some summary() methods.

pj

-- 
Paul E. Johnson
Professor, Political ScienceDirector
1541 Lilac Lane, Room 504  Center for Research Methods
University of Kansas University of Kansas
http://pj.freefaculty.org  http://crmda.ku.edu

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Re: [R] Batch Installer for R

2016-03-31 Thread Federman, Douglas

Does the package packrat do what you want?

--
Better name for the general practitioner might be multispecialist. 
~Martin H. Fischer (1879-1962)


-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Tobias Knuth
Sent: Thursday, March 31, 2016 9:05 AM
To: r-help@r-project.org
Subject: [R] Batch Installer for R

Hi everyone,

in Python, you can run pip install -r filename to install all packages
listed in the file. Is there something similar to R? If not, isn't it quite
easy to write?

For me, it would be much easier to work on projects with other people if I
could just install all dependencies with one line in a generalised manner.

Did anybody try something like that before me?

Best,
Tobias

[[alternative HTML version deleted]]

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[R] Batch Installer for R

2016-03-31 Thread Tobias Knuth

Hi everyone,

in Python, you can run pip install -r filename to install all packages
listed in the file. Is there something similar to R? If not, isn't it quite
easy to write?

For me, it would be much easier to work on projects with other people if I
could just install all dependencies with one line in a generalised manner.

Did anybody try something like that before me?

Best,
Tobias

[[alternative HTML version deleted]]

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Re: [R-es] usar R a traves de la web

2016-03-31 Thread eric


Genial que se pueda usar un enlace de dropbox, muchas gracias,

saludos,

eric.




On 03/31/2016 06:32 AM, Jesús Para Fernández wrote:

Si tus datos no son muy confidenciales, puedes subirlos a dropbox o
similar y usar el enlace que dicha plataforma te proporciona.

Un saludo
Jesús

From: j.para.fernan...@hotmail.com
To: ericconchamu...@gmail.com; r-help-es@r-project.org
Date: Thu, 31 Mar 2016 11:30:45 +0200
Subject: Re: [R-es] usar R a traves de la web

Lo unicio que funciona es subir los datos a la web y luego usar el codigo:

datos<-read.table("http://mate.dm.uba.ar/~meszre/datos_regresion/ejemploint.txt",header=T)




Subject: Re: [R-es] usar R a traves de la web
To: j.para.fernan...@hotmail.com; r-help-es@r-project.org
From: ericconchamu...@gmail.com
Date: Sun, 6 Mar 2016 12:28:51 -0300

Hola Jesus, muchas gracias, esta opcion se ve bastante bien, aunque no
encuentro un modo de subir mis archivos de datos, pero debe poderse, no ?

Muchas gracias,

Eric.




On 03/05/2016 09:03 AM, Jes�s Para Fern�ndez wrote:
> Buscas algo asi?
>
>http://www.r-fiddle.org/#/
>
>
>
>  > From: ericconchamu...@gmail.com
>  > To: r-help-es@r-project.org
>  > Date: Sat, 5 Mar 2016 09:01:42 -0300
>  > Subject: [R-es] usar R a traves de la web
>  >
>  > Estimada comunidad, para mi trabajo uso latex y R normalmente, ahora
>  > debo viajar sin mi portatil, pero tengo la opcion de llevar un peque�o
>  > tablet (con android) ... para suplir latex he estado usando
>  > www.overleaf.com y trabaja excelente, practicamente todos los paquetes
>  > que uso estan disponibles ahi ... pero no he encontrado algo similar
> para R.
>  >
>  > Saben ustedes si existe algun proyecto que permita usar R en la web ? ya
>  > he mirado muchas opciones
>  >
> 
(https://nsaunders.wordpress.com/2009/11/30/a-brief-survey-of-r-web-interfaces/)
>
>  > como las de Rstudio (tengo android, no puedo instalar R en el tablet),
>  > tampoco tengo la opcion de instalar un RApache por ejemplo, pero ninguna
>  > me sirve ... necesito una pagina en que simplemente entre, pegue el
>  > codigo y me devuelva las salidas sin tener que instalar nada localmente.
>  >
>  > Muchas gracias,
>  >
>  > Eric.
>  >
>  >
>  >
>  > --
>  > Forest Engineer
>  > Master in Environmental and Natural Resource Economics
>  > Ph.D. student in Sciences of Natural Resources at La Frontera University
>  > Member in AguaDeTemu2030, citizen movement for Temuco with green city
>  > standards for living
>  >
>  > Nota: Las tildes se han omitido para asegurar compatibilidad con algunos
>  > lectores de correo.
>  >
>  > ___
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>  > R-help-es@r-project.org
>  >https://stat.ethz.ch/mailman/listinfo/r-help-es

--
Forest Engineer
Master in Environmental and Natural Resource Economics
Ph.D. student in Sciences of Natural Resources at La Frontera University
Member in AguaDeTemu2030, citizen movement for Temuco with green city
standards for living

Nota: Las tildes se han omitido para asegurar compatibilidad con algunos
lectores de correo.


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Forest Engineer
Master in Environmental and Natural Resource Economics
Ph.D. student in Sciences of Natural Resources at La Frontera University
Member in AguaDeTemu2030, citizen movement for Temuco with green city 
standards for living


Nota: Las tildes se han omitido para asegurar compatibilidad con algunos 
lectores de correo.


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Re: [R] Convergence issues when using ns splines (pkg: spline) in Cox model (coxph) even when changing coxph.control

2016-03-31 Thread Therneau, Terry M., Ph.D.

Thanks to David for pointing this out.  The "time dependent covariates" vignette in the 
survival package has a section on time dependent coefficients that talks directly about 
this issue.  In short, the following model is simply wrong:

 coxph(Surv(time, status) ~ trt + prior + karno + I(karno * log(time)), 
data=veteran)

People try this often as a way to create the time dependent covariate  Karnofsky * log(t), 
which is often put forwards as a way to deal with non-proportional hazards.  To do this 
correctly you have to use the tt() functionality in coxph to move the computation out of 
the model statement:

  coxph(Surv(time, status) ~ trt + prior + karno + tt(karno), data=veteran,
tt = function(x, time, ...) x*log(time))


BTW the following SAS code is also wrong:
 proc phreg data=veteran;
 model time * status(0) = trt + prior + karno* time;

SAS does the right thing, however, if you move the computation off the model 
line.
  model time * status(0) = trt + karno + zzz;
  zzz = karno * time;

The quote "SAS does it but R fails" comes at me moderately often in this context.  The 
reason is that SAS won't LET you put a phrase like "log(time)" into the model statement, 
so people end up doing the right thing, but by accident.


Terry T.


On 03/30/2016 05:28 PM, Göran Broström wrote:



On 2016-03-30 23:06, David Winsemius wrote:



On Mar 29, 2016, at 1:47 PM, Jennifer Wu, Miss
 wrote:

Hi,

I am currently using R v3.2.3 and on Windows 10 OS 64Bit.

I am having convergence issues when I use coxph with a interaction
term (glarg*bca_py) and interaction term with the restricted cubic
spline (glarg*bca_time_ns). I use survival and spline package to
create the Cox model and cubic splines respectively. Without the
interaction term and/or spline, I have no convergence problem. I
read some forums about changing the iterations and I have but it
did not work. I was just wondering if I am using the inter.max and
outer.max appropriately. I read the survival manual, other R-help
and stackoverflow pages and it suggested changing the iterations
but it doesn't specify what is the max I can go. I ran something
similar in SAS and did not run into a convergence problem.

This is my code:

bca_time_ns <- ns(ins_ca$bca_py, knots=3,
Boundary.knots=range(2,5,10)) test <- ins_ca$glarg*ins_ca$bca_py
test1 <- ins_ca$glarg*bca_time_ns


In your `coxph` call the variable 'bca_py' is the survival time and


Right David: I didn't notice that the 'missing main effect' in fact was part of 
the
survival object! And as you say: Time to rethink the whole model.

Göran


yet here you are constructing not just one but two interactions (one
of which is a vector but the other one a matrix) between 'glarg' and
your survival times. Is this some sort of effort to identify a
violation of proportionality over the course of a study?

Broström sagely points out that these interactions are not in the
data-object and subsequent efforts to refer to them may be confounded
by the multiple environments from which data would be coming into the
model. Better to have everything come in from the data-object.

The fact that SAS did not have a problem with this rather
self-referential or circular model may be a poor reflection on SAS
rather than on the survival package. Unlike Therneau or Broström who
asked for data, I suggest the problem lies with the model
construction and you should be reading what Therneau has written
about identification of non-proportionality and identification of
time dependence of effects. See Chapter 6 of his "Modeling Survival
Data".





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Re: [R] ts or xts with high-frequency data within a year

2016-03-31 Thread Bert Gunter

TYPO on TYPO!

It should be X[,3]

Cheers,
Bert
Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, Mar 31, 2016 at 9:47 AM, Bert Gunter  wrote:
> Inline.
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, Mar 30, 2016 at 5:37 PM, Ryan Utz  wrote:
>> Bill, Josh, and Bert,
>>
>> Thanks for your responses. I still can't quite get this when I use actual
>> dates. Here's an example of what is going wrong:
>>
>> X=as.data.frame(1:6000)
>> X[2]=seq.POSIXt(ISOdate(2015,11,1),by='hour',length.out=6000)
>> X[3]=sample(100,size=6000,replace=T)
>>
>> Y=xts(X[,3],order.by=X[,2])
>> decompose(Y)
>>
>> Z=ts(X[,2],frequency=24*365)
>> plot(decompose(Z))
>
> ## TYPO!  It should be X[,2].
>
> ***WARNING***
> I am not a time series expert. The remarks below are therefore subject
> to confirmation or correction by those who are. Caveat emptor!
>
> Also, you seem to be somewhat confused about how the seasonal
> decomposition in decompose() works.
>
> My understanding from what you said is that your unit of time is days
> and that you have samples every 15 minutes, but that you wish to
> estimate hourly effects. If this is correct, you first need to first
> combine (e.g. via by() ) your 4 observations into a single summary
> (e.g. average or median, maybe) so that you have an hourly time
> series. Run this through ts() with frequency = 24: your period is
> daily = 24 hours and you have 24 observations/period.
>
> (Note that you could use the original data to get 96 15 minute
> estimates instead of hourly estimates).
>
> When you run this through decompose(), it will give you 24 "seasonal"
> effects/period, the hourly effects. The "trend" component will try to
> remove longer term effects and trends over time.
>
> Note, however, this is probably **not** what you want. I suspect you
> want both hourly effects and daily or monthly effects to reflect
> annual seasonalities as well as daily. decompose() cannot do this,
> apparently, so you should look for an alternative solution. Better
> yet, consult with a local time series expert to help you figure out
> what you might want and how to get it in R.
>
> Cheers,
> Bert
>
>
>>
>> When I specify an actual date/time (rather than just a number as Bill
>> posited), it does not like anything short of a year. This seems like I'm
>> overlooking something obvious, but I can't get this for the life of me...
>>
>> Thanks for your time,
>> r
>>
>>
>> On Wed, Mar 30, 2016 at 1:03 PM, William Dunlap  wrote:
>>>
>>> You said you specified frequency=96 when you constructed the time
>>> series, but when I do that the decomposition looks reasonable:
>>>
>>> > time <- seq(0,9,by=1/96) # 15-minute intervals, assuming time unit is
>>> > day
>>> > measurement <- sqrt(time) + 1/(1.2+sin(time*2*pi)) +
>>> > rnorm(length(time),0,.3)
>>> > plot(decompose(ts(measurement, frequency=96)))
>>>
>>> How is your code different from the above?
>>>
>>>
>>>
>>> Bill Dunlap
>>> TIBCO Software
>>> wdunlap tibco.com
>>>
>>> On Wed, Mar 30, 2016 at 8:03 AM, Ryan Utz  wrote:

 Hello,

 I have a time series that represents data sampled every 15-minutes. The
 data currently run from November through February, 8623 total readings.
 There are definitely daily periodic trends and non-stationary long-term
 trends. I would love to decompose this using basic time series analysis.

 However, every time I attempt decomposition, I get the

 Error in decompose( ) : time series has no or less than 2 periods

 Is it only possible to do basic time-series analysis if you have a year
 or
 more worth of data? That seems absurd to me, since there is definite
 periodicity and the data are a time series. I have tried every manner of
 specifying frequency= with no luck (96 does not work). All manner of
 searching for help has turned up fruitless.

 Can I only do this after I wait another year or two?

 Thanks,
 Ryan

 --

 Ryan Utz, Ph.D.
 Assistant professor of water resources
 *chatham**UNIVERSITY*
 Home/Cell: (724) 272-7769

 [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>
>>
>>
>> --
>>
>> Ryan Utz, Ph.D.
>> Assistant professor of water resources
>> chathamUNIVERSITY
>> Home/Cell: (724) 272-7769
>>

Re: [R] ts or xts with high-frequency data within a year

2016-03-31 Thread Bert Gunter

Inline.

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Wed, Mar 30, 2016 at 5:37 PM, Ryan Utz  wrote:
> Bill, Josh, and Bert,
>
> Thanks for your responses. I still can't quite get this when I use actual
> dates. Here's an example of what is going wrong:
>
> X=as.data.frame(1:6000)
> X[2]=seq.POSIXt(ISOdate(2015,11,1),by='hour',length.out=6000)
> X[3]=sample(100,size=6000,replace=T)
>
> Y=xts(X[,3],order.by=X[,2])
> decompose(Y)
>
> Z=ts(X[,2],frequency=24*365)
> plot(decompose(Z))

## TYPO!  It should be X[,2].

***WARNING***
I am not a time series expert. The remarks below are therefore subject
to confirmation or correction by those who are. Caveat emptor!

Also, you seem to be somewhat confused about how the seasonal
decomposition in decompose() works.

My understanding from what you said is that your unit of time is days
and that you have samples every 15 minutes, but that you wish to
estimate hourly effects. If this is correct, you first need to first
combine (e.g. via by() ) your 4 observations into a single summary
(e.g. average or median, maybe) so that you have an hourly time
series. Run this through ts() with frequency = 24: your period is
daily = 24 hours and you have 24 observations/period.

(Note that you could use the original data to get 96 15 minute
estimates instead of hourly estimates).

When you run this through decompose(), it will give you 24 "seasonal"
effects/period, the hourly effects. The "trend" component will try to
remove longer term effects and trends over time.

Note, however, this is probably **not** what you want. I suspect you
want both hourly effects and daily or monthly effects to reflect
annual seasonalities as well as daily. decompose() cannot do this,
apparently, so you should look for an alternative solution. Better
yet, consult with a local time series expert to help you figure out
what you might want and how to get it in R.

Cheers,
Bert

>
> When I specify an actual date/time (rather than just a number as Bill
> posited), it does not like anything short of a year. This seems like I'm
> overlooking something obvious, but I can't get this for the life of me...
>
> Thanks for your time,
> r
>
>
> On Wed, Mar 30, 2016 at 1:03 PM, William Dunlap  wrote:
>>
>> You said you specified frequency=96 when you constructed the time
>> series, but when I do that the decomposition looks reasonable:
>>
>> > time <- seq(0,9,by=1/96) # 15-minute intervals, assuming time unit is
>> > day
>> > measurement <- sqrt(time) + 1/(1.2+sin(time*2*pi)) +
>> > rnorm(length(time),0,.3)
>> > plot(decompose(ts(measurement, frequency=96)))
>>
>> How is your code different from the above?
>>
>>
>>
>> Bill Dunlap
>> TIBCO Software
>> wdunlap tibco.com
>>
>> On Wed, Mar 30, 2016 at 8:03 AM, Ryan Utz  wrote:
>>>
>>> Hello,
>>>
>>> I have a time series that represents data sampled every 15-minutes. The
>>> data currently run from November through February, 8623 total readings.
>>> There are definitely daily periodic trends and non-stationary long-term
>>> trends. I would love to decompose this using basic time series analysis.
>>>
>>> However, every time I attempt decomposition, I get the
>>>
>>> Error in decompose( ) : time series has no or less than 2 periods
>>>
>>> Is it only possible to do basic time-series analysis if you have a year
>>> or
>>> more worth of data? That seems absurd to me, since there is definite
>>> periodicity and the data are a time series. I have tried every manner of
>>> specifying frequency= with no luck (96 does not work). All manner of
>>> searching for help has turned up fruitless.
>>>
>>> Can I only do this after I wait another year or two?
>>>
>>> Thanks,
>>> Ryan
>>>
>>> --
>>>
>>> Ryan Utz, Ph.D.
>>> Assistant professor of water resources
>>> *chatham**UNIVERSITY*
>>> Home/Cell: (724) 272-7769
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
>
>
> --
>
> Ryan Utz, Ph.D.
> Assistant professor of water resources
> chathamUNIVERSITY
> Home/Cell: (724) 272-7769
>

__
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Re: [R-es] Bootstrap de días seguidos

2016-03-31 Thread Carlos Ortega

Hola,

Yo lo que he entendido es que quieres generar los 20 días siguientes a una
fecha dada.
Una forma de hacerlo es así:


> as.Date("2000-01-01") + 0:10
 [1] "2000-01-01" "2000-01-02" "2000-01-03" "2000-01-04" "2000-01-05"
"2000-01-06"
 [7] "2000-01-07" "2000-01-08" "2000-01-09" "2000-01-10" "2000-01-11"

Más detalles aquí:

http://stackoverflow.com/questions/7536306/creating-a-unique-sequence-of-dates

Saludos,
Carlos Ortega
www.qualityexcellence.es


El 31 de marzo de 2016, 10:46, Jesús Para Fernández <
j.para.fernan...@hotmail.com> escribió:

> Buenas a todos,
>
> Lo primero agradecer todas las respuesta sque tuve en el tema de Bootstrap
> dataframe, que por estar de baja no he podido agradecer.
>
> De aquel tema salió una sugerencia que me parece muy interesante y que a
> dia de hoy no soy capaz de hacer de una manera optima.
>
> Lo que quiero hacer es coger un dia al azar de todo el periodo, y a
> partuir de ese dia, coger por ejemplo los 20 dias siguientes.
>
> Recuerdo que para cogerlos al azar hacia lo siguiente:
>
> set.seed(121)
> final<-0
> nuevo<-0
> for(i in 1:10){
> nuevo<-sample(datos$pedidos,replace=T)
> final[i]<-sum(nuevo[1:20])
> }
>
> donde aqui estoy cogiendo los 20 dias al azar.
>
> ¿Como haria para coger estos 20 dias seguidos??
>
> Gracias
> Jesús
>
>
>
>
> [[alternative HTML version deleted]]
>
>
> ___
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> R-help-es@r-project.org
> https://stat.ethz.ch/mailman/listinfo/r-help-es
>



-- 
Saludos,
Carlos Ortega
www.qualityexcellence.es

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Re: [R-es] Bootstrap de días seguidos

2016-03-31 Thread Luisfo Chiroque via R-help-es

Buenas,

para concatenar (rbind) los resultados de ‘muestras’ a un data.frame podrías 
hacer lo siguiente:
df.muestras <- data.frame(do.call(rbind, muestras))

Un saludo,
Luisfo


> El 31 mar 2016, a las 18:00, Jesús Para Fernández 
>  escribió:
> 
> Es justo lo que buscaba. Muchas gracias. 
> 
> Una unica cosa, como puedo pasar a un data.frame el resultado muestras, del 
> lapply???
> 
> 
> 
> 
> 
>> To: r-help-es@r-project.org
>> From: canadasre...@gmail.com
>> Date: Thu, 31 Mar 2016 15:18:55 +0200
>> Subject: Re: [R-es] Bootstrap de días seguidos
>> 
>> Hola.
>> 
>> No sé si buscas algo parecido a esto
>> 
>> datos <- data.frame(v1 = rnorm(1000, 2, 5), v2 = rnorm(1000) )
>> 
>> # numero de puntos aleatorios
>> n.puntos <- 20
>> puntos <- replicate(n.puntos, sample(nrow(datos), 1, replace = T) )
>> 
>> puntos
>>  [1] 348  52 520 675 574 303 264 678 749  29 310 691 460 114 892 903 
>> 335 984 207 964
>> 
>> # muestras de 21 filas
>>  k <- 20
>> muestras <- lapply(puntos, function(x) datos[x:(x+k),])
>> 
>> # muestras es una lista con k data.frames, el primero serán los datos de 
>> la fila 348 hasta la368
>> muestras[[1]]
>> v1  v2
>> 348 -1.8855298  1.67022010
>> 349  8.3539108 -0.75856401
>> 350  3.1723330 -0.15722935
>> 351  2.5871373  1.30962887
>> 352  4.0801806 -0.22205638
>> 353  8.7792425  1.92769400
>> 354  1.8023941  0.60780632
>> 355 -4.4542464 -0.30940621
>> 356  1.4032584 -1.22315174
>> 357 -1.1669957 -0.36789523
>> 358  0.8834993 -0.51625882
>> 359 -4.4173234  0.35013974
>> 360 -6.2964411  0.64394556
>> 361  0.4808418  1.41868648
>> 362  0.6912628 -0.29357748
>> 363 -4.1933794  0.90492395
>> 364 -9.3685116  0.08371681
>> 365  1.3305264 -0.18474498
>> 366  2.9247997  1.24475278
>> 367  8.8120307  0.48149808
>> 368  8.0995250  1.30719019
>> 
>> El 31/03/16 a las 10:46, Jesús Para Fernández escribió:
>>> Buenas a todos,
>>> 
>>> Lo primero agradecer todas las respuesta sque tuve en el tema de Bootstrap 
>>> dataframe, que por estar de baja no he podido agradecer.
>>> 
>>> De aquel tema sali� una sugerencia que me parece muy interesante y que a 
>>> dia de hoy no soy capaz de hacer de una manera optima.
>>> 
>>> Lo que quiero hacer es coger un dia al azar de todo el periodo, y a partuir 
>>> de ese dia, coger por ejemplo los 20 dias siguientes.
>>> 
>>> Recuerdo que para cogerlos al azar hacia lo siguiente:
>>> 
>>> set.seed(121)
>>> final<-0
>>> nuevo<-0
>>> for(i in 1:10){
>>> nuevo<-sample(datos$pedidos,replace=T)
>>> final[i]<-sum(nuevo[1:20])
>>> }
>>> 
>>> donde aqui estoy cogiendo los 20 dias al azar.
>>> 
>>> �Como haria para coger estos 20 dias seguidos??
>>> 
>>> Gracias
>>> Jes�s
>>> 
>>> 
>>> 
>>> 
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Re: [R-es] Bootstrap de días seguidos

2016-03-31 Thread Jose Luis Cañadas Reche

Hola.

No sé si buscas algo parecido a esto

datos <- data.frame(v1 = rnorm(1000, 2, 5), v2 = rnorm(1000) )

# numero de puntos aleatorios
n.puntos <- 20
puntos <- replicate(n.puntos, sample(nrow(datos), 1, replace = T) )

puntos
  [1] 348  52 520 675 574 303 264 678 749  29 310 691 460 114 892 903 
335 984 207 964

# muestras de 21 filas
  k <- 20
muestras <- lapply(puntos, function(x) datos[x:(x+k),])

# muestras es una lista con k data.frames, el primero serán los datos de 
la fila 348 hasta la368
muestras[[1]]
 v1  v2
348 -1.8855298  1.67022010
349  8.3539108 -0.75856401
350  3.1723330 -0.15722935
351  2.5871373  1.30962887
352  4.0801806 -0.22205638
353  8.7792425  1.92769400
354  1.8023941  0.60780632
355 -4.4542464 -0.30940621
356  1.4032584 -1.22315174
357 -1.1669957 -0.36789523
358  0.8834993 -0.51625882
359 -4.4173234  0.35013974
360 -6.2964411  0.64394556
361  0.4808418  1.41868648
362  0.6912628 -0.29357748
363 -4.1933794  0.90492395
364 -9.3685116  0.08371681
365  1.3305264 -0.18474498
366  2.9247997  1.24475278
367  8.8120307  0.48149808
368  8.0995250  1.30719019

El 31/03/16 a las 10:46, Jesús Para Fernández escribió:
> Buenas a todos,
>
> Lo primero agradecer todas las respuesta sque tuve en el tema de Bootstrap 
> dataframe, que por estar de baja no he podido agradecer.
>
> De aquel tema sali� una sugerencia que me parece muy interesante y que a dia 
> de hoy no soy capaz de hacer de una manera optima.
>
> Lo que quiero hacer es coger un dia al azar de todo el periodo, y a partuir 
> de ese dia, coger por ejemplo los 20 dias siguientes.
>
> Recuerdo que para cogerlos al azar hacia lo siguiente:
>
> set.seed(121)
> final<-0
> nuevo<-0
> for(i in 1:10){
> nuevo<-sample(datos$pedidos,replace=T)
> final[i]<-sum(nuevo[1:20])
> }
>
> donde aqui estoy cogiendo los 20 dias al azar.
>
> �Como haria para coger estos 20 dias seguidos??
>
> Gracias
> Jes�s
>
>
>
>   
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>
>
>
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Re: [R] ts or xts with high-frequency data within a year

2016-03-31 Thread Giorgio Garziano

Ryan,

>From "decompose()" source code, two conditions can trigger the error message:

   "time series has no or less than 2 periods"

based on the frequency value, specifically:

  1. f <= 1
  2. length(na.omit(x)) < 2 * f

It appears to me that your reproducible code has got a typo error, it should be:

  Z=ts(X[,3],frequency=24*365)

I mean X[,3] in place of X[,2].


Further, frequency=24*365, 2*frequency > 6000 = length(X[,3]), and that 
triggers the second condition.

>From R console:

> decompose
function (x, type = c("additive", "multiplicative"), filter = NULL)
{
type <- match.arg(type)
l <- length(x)
f <- frequency(x)
if (f <= 1 || length(na.omit(x)) < 2 * f)
stop("time series has no or less than 2 periods")


--

Best,

GG



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[R] function to carry out Bootstrap LRT with poLCA results

2016-03-31 Thread Alain D. via R-help

Dear List, 

I would like to determine the optimal number of latent classes (polytomous data)
using Bootstrap LRT. poLCA does not provide such a possibility and I am not
enough into programming to modify the code. Is there any other way to do this,
e.g. use a poLCA object with some other package?

Thank you for help.

Best wishes

Alain

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[R-es] Bootstrap de días seguidos

2016-03-31 Thread Jesús Para Fernández

Buenas a todos,

Lo primero agradecer todas las respuesta sque tuve en el tema de Bootstrap 
dataframe, que por estar de baja no he podido agradecer.

De aquel tema sali� una sugerencia que me parece muy interesante y que a dia de 
hoy no soy capaz de hacer de una manera optima. 

Lo que quiero hacer es coger un dia al azar de todo el periodo, y a partuir de 
ese dia, coger por ejemplo los 20 dias siguientes.

Recuerdo que para cogerlos al azar hacia lo siguiente:

set.seed(121)
final<-0
nuevo<-0
for(i in 1:10){
nuevo<-sample(datos$pedidos,replace=T)
final[i]<-sum(nuevo[1:20])
}

donde aqui estoy cogiendo los 20 dias al azar. 

�Como haria para coger estos 20 dias seguidos??

Gracias
Jes�s 



  
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Re: [R] ts or xts with high-frequency data within a year

[R] function to carry out Bootstrap LRT with poLCA results

[R-es] Bootstrap de días seguidos

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