Re: [R] Suggestions for vectorizing/double loop

[PIKAL Petr]

Well, your second post is rather confusing too, as is your not reproducible 
code.

If I understand correctly, you just want to change names of files in a 
directory according to some rules, which you did not clerly specify.

First I am not sure if R is the best tool for it. Nevertheless, you do not 
change a contents of any file, just change the name.

I found this with simple answer questioning internet.
http://stackoverflow.com/questions/10758965/how-do-i-rename-files-using-r

Is this what you want?

Here is another possibility
https://www.r-bloggers.com/operating-on-files-with-r-copy-and-rename/

Anyway the concept I would use is.

#read old names

old_names <- list.files(old_path)

#setting new names (you did not explain how do you want to do this)
new_names <- some vector of new names

If order of names is correct
file.rename(old_names, new_names)

othervise it would be needed correctly order both vectors before file.rename.

The crucial information is how do you want set the new names vector. You shold 
specify it by reproducible way if you want more precise answer.

Copying files is in this case best done by OS itself.

Cheers
Petr

> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Luanna
> Dixson
> Sent: Monday, January 23, 2017 6:39 PM
> To: r-help@r-project.org
> Subject: Re: [R] Suggestions for vectorizing/double loop
>
> Hi all,
>
> Thanks very much for your help! You are correct in thinking the list is the
> same as before, actually, my question was more about how to do the next
> steps, where I needed to match the filenames of the files in my directory
> with old (i.e current) and new file name prefixes in my list. For each match I
> then wanted to rename the original file using the corresponding new
> filename prefix from the list.
>
> Sorry for being a bit confusing, I didn't post my first attempt to do this at 
> first
> as my code just didn't work at all, but I have had another go using a matrix
> instead and I think this does the job.
>
> old_path="./old/"
> new_path="./new/"
>
> old_names <- list.files(old_path)
> head(old_names)
>
> oldnames=c('1002', '1003')
> newnames=c('1002_new', '1003_new')
> mapping=cbind(oldnames,newnames)
> head(mapping)
>
> for (i in old_names){
> temp <- unlist(strsplit(i, "[.]"))[1]
> n <- which(is.element(mapping,temp))
> if(length(n)>0) {
> #copy the file to the new folder
> #re name it and the name is paste(mapping[n,2], '.xls', sep="")
> print(paste(mapping[n,2], '.xls', sep=""))
> newnames <- paste(mapping[n,2], '.xls', sep="")
> file.copy(from = paste(old_path, i, sep=""),
>   to = paste(new_path, newnames))
> }
> }
>
>
> On 23 January 2017 at 13:28, Luanna Dixson
> 
> wrote:
>
> > I need to rename a bunch of files, by searching for string matches in
> > a list. Each list element containings a character with the old
> > filename that I want to match to, and the new file name that I want to
> rename by.
> >
> > For instance, here filename '1001.xls' should match to
> > list[[1]]$oldname and I want to rename it to 'newa.xls' from
> list[[1]]$newname.
> >
> > The actual new file names I have will feature a random alphanumeric
> > number and the list will have a length of ~600.
> >
> >
> > # files I want to rename
> >
> > files with old file name =c('1001.xls', '1002.xls')
> >
> > # list with old file names and new file names
> >
> > oldnames=c('1001', '1002', '1003')
> >
> > newnames=c('newa', 'newb', 'newc')
> >
> > df=data.frame(oldnames,newnames)
> >
> > list <- split(df, rownames(df))
> >
> >
> > # turn list elements in character
> >
> > for(i in 1:length(list)) list[[i]]$oldnames=as.
> > character(list[[i]]$oldnames)
> >
> > for(i in 1:length(list)) list[[i]]$newnames=as.
> > character(list[[i]]$newnames)
> >
> >
> > I heard that it would be better to vectorize this than trying to do a
> > double loop so if someone could give me a hint about how to do this I
> > would be very grateful!
> >
>
>   [[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.


Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny 
pouze jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně 
jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze 
svého systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email 
jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či 
zpožděním přenosu e-mailu.

V případě, že je tento 

Re: [R] separate mixture of gamma and normal (with mixtools or ??)

[Thomas Petzoldt]

Dear Bert,

thank you very much for your suggestion. You are right, ill-conditioning 
was sometimes a problem for 3 components, but not in the two-component 
case. The modes are well separated, and the sample size is high.


My main problem is (1) the shape of the distributions and (2) the 
diversity of available packages and approaches to this topic.


In the mean time I made some progress in this matter by treating the 
data as a mixture of gamma distributions (package mixdist, see below), 
so what I want to know is the purely R technical question ;-)


Has someone else has ever stumbled across something like this and can 
make a suggestion which package to use?


Thanks, Thomas


## Approximate an Exponential+Gaussian mixture with
##   a mixture of Gammas

library("mixdist")
set.seed(123)
lambda <- c(0.25, 0.75)
N <- 2000
x <- c(rexp(lambda[1]*N, 1), rnorm(lambda[2]*N, 20, 4))

xx <- mixgroup(x, breaks=0:40)
pp <- mixparam(mu=c(1, 8), sigma=c(1, 3), pi=c(0.2, 0.5))
mix <-  mix(xx, pp, dist="gamma", emsteps=10)

summary(mix)
p <- coef(mix)
beta <- with(p, sigma^2/mu)
alpha <- with(p, mu /beta)
lambda <- p$pi

plot(mix, xlim=c(0, 35))
x1 <- seq(0, 35, 0.1)
lines(x1, lambda[1]*dgamma(x1, alpha[1], 1/beta[1]),
  col="orange", lwd=2)
lines(x1, lambda[2]*dgamma(x1, alpha[2], 1/beta[2]),
  col="magenta", lwd=2)



Am 24.01.2017 um 00:34 schrieb Bert Gunter:

Fitting multicomponent mixtures distributions -- and 3 is already a
lot of components -- is inherently ill-conditioned. You may need to
reassess your strategy. You might wish to post on stackexchange
instead to discuss such statistical issues.

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 Mon, Jan 23, 2017 at 2:32 PM, Thomas Petzoldt  wrote:

Dear friends,

I am trying to separate bi- (and sometimes tri-) modal univariate mixtures
of biological data, where the first component is left bounded (e.g.
exponential or gamma) and the other(s) approximately Gaussian.

After checking several packages, I'm not really clear what to do. Here is an
example with "mixtools" that already works quite good, however, the left
component is not Gaussian (and not symmetric).

Any idea about a more adequate function or package for this problem?

Thanks a lot!

Thomas



library(mixtools)
set.seed(123)

lambda <- c(0.25, 0.75)
N  <- 200

## dist1 ~ gamma (or exponential as a special case)
#dist1 <- rexp(lambda[1]*N, 1)
dist1 <- rgamma(lambda[1]*N, 1, 1)

## dist2 ~ normal
dist2 <- rnorm(lambda[2]*N, 12, 2)

## mixture
x <- c(dist1, dist2)

mix <-  spEMsymloc(x, mu0=2, eps=1e-3, verbose=TRUE)
plot(mix, xlim=c(0, 25))
summary(mix)


--
Thomas Petzoldt
TU Dresden, Institute of Hydrobiology
http://www.tu-dresden.de/Members/thomas.petzoldt

__
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.


__
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.

Re: [R] Extracting first number after * in a character vector

[Michael Hannon]

Elegant I don't know, but I think the appended does the trick.

-- Mike

> foo <- c(" 1 X[0,SMITH]   *  0 0 1 ",
+  " 2 X[0,JOHNSON] *  0 0 1 ",
+  " 3 X[0,WILLIAMS]  *  1 0 1 ",
+  " 4 X[0,JONES]   *  0 0 1 ",
+   [TRUNCATED]

> as.numeric(gsub("^[^*]+[*][^0-9]+([01]).*$", "\\1", foo))
[1] 0 0 1 0 0 0 0 0 0
>

On Mon, Jan 23, 2017 at 1:27 PM, Jim Lemon  wrote:
> Hi Abhinaba,
> I'm sure that someone will post a terrifyingly elegant regular
> expression that does this, but:
>
>  ardat<-
>  c([1] " 1 X[0,SMITH]   *  0 0 1 ",
>  ...
> numpoststar<-function(x) {
>  xsplit<-unlist(strsplit(x,""))
>  starpos<-which(xsplit=="*")
>  # watch out for a missing asterisk, they cause an infinite loop
>  if(length(starpos)) {
>   digits<-c("0","1","2","3","4","5","6","7","8","9")
>   while(!any(digits %in% xsplit[starpos])) starpos<-starpos+1
>   return(as.numeric(xsplit[starpos]))
>  }
>  return(NA)
> }
>
> for(i in 1:length(ardat)) print(numpoststar(ardat[i]))
>
> The observant will wonder why I didn't use sapply. Because for some
> reason it returned the original strings rather than the numbers. I
> dunno.
>
> Jim
>
> On Mon, Jan 23, 2017 at 11:29 PM, Abhinaba Roy  
> wrote:
>> Hi,
>>
>> How do I extract the first number after '*' in a vector?
>>
>> The vector is given below
>>
>>> dput(out[1:10])
>> c(" 1 X[0,SMITH]   *  0 0 1 ",
>> " 2 X[0,JOHNSON] *  0 0 1 ",
>> " 3 X[0,WILLIAMS]", "*  1 0
>> 1 ",
>> " 4 X[0,JONES]   *  0 0 1 ",
>> " 5 X[0,BROWN]   *  0 0 1 ",
>> " 6 X[0,DAVIS]   *  0 0 1 ",
>> " 7 X[0,MILLER]  *  0 0 1 ",
>> " 8 X[0,WILSON]  *  0 0 1 ",
>> " 9 X[0,MOORE]   *  0 0 1 "
>> )
>>
>> I want a vector with the first number after the asterisk.
>>
>> So the output would give me, a vector (0,0,1,0,0,0,0,0,0,0)
>>
>> How can I do it in R?
>>
>> Best,
>> Abhinaba
>>
>> [[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.
>
> __
> 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.

__
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.

Re: [R-es] Eliminación de filas en data frame según versión del fichero de origen

[Freddy Omar López Quintero]

​Hola.​

2017-01-23 14:27 GMT-03:00 Rubén Coca :

> El caso es que para un mismo id y date debo quedarme con la observación que
> tenga la versión más alta (descartando el resto).
>

Si es válido apoyarse en SQL, yo usaría algo como:

> library(sqldf)

> sqldf("select rowid, id, max(value) from df group by id")

  rowid   id max(value)

1 4 0001   19.57054

2 5 0002   14.70713

3 7 0003   19.34788


donde se reporta el número de la fila por si se quiere hacer otro join.​

​Saludos.​

-- 
«Pídeles sus títulos a los que te persiguen, pregúntales
cuándo nacieron, diles que te demuestren su existencia.»

Rafael Cadenas

[[alternative HTML version deleted]]

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

Re: [R] Can R markdown do beamer palo alto theme?

[C W]

Actually, I figured out that you can, but the syntax is different.

You don't declare using \usedocument{beamer} \usetheme{PaloAlto}, instead,
title and output are the correct syntax.

Is there a full conversion list from Latex Beamer to Markdown Beamer?

Thanks!

On Mon, Jan 23, 2017 at 7:20 PM, C W  wrote:

> Hi R list,
>
> Is it possible to use R markdown with beamer palo alto theme?  Are they
> compatible?
>
> I copy pasted my LaTex code over to R, I get the following error message:
>
> ! LaTeX Error: Can be used only in preamble.
>
> See the LaTeX manual or LaTeX Companion for explanation.
> Type  H   for immediate help.
>  ...
>
> l.89 \end{frame}
>
> pandoc: Error producing PDF
> Error: pandoc document conversion failed with error 43
> Execution halted
>
> Thanks for your help in advance!
>

[[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.

[R] Can R markdown do beamer palo alto theme?

[C W]

Hi R list,

Is it possible to use R markdown with beamer palo alto theme?  Are they
compatible?

I copy pasted my LaTex code over to R, I get the following error message:

! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H   for immediate help.
 ...

l.89 \end{frame}

pandoc: Error producing PDF
Error: pandoc document conversion failed with error 43
Execution halted

Thanks for your help in advance!

[[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.

Re: [R] separate mixture of gamma and normal (with mixtools or ??)

[Bert Gunter]

Fitting multicomponent mixtures distributions -- and 3 is already a
lot of components -- is inherently ill-conditioned. You may need to
reassess your strategy. You might wish to post on stackexchange
instead to discuss such statistical issues.

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 Mon, Jan 23, 2017 at 2:32 PM, Thomas Petzoldt  wrote:
> Dear friends,
>
> I am trying to separate bi- (and sometimes tri-) modal univariate mixtures
> of biological data, where the first component is left bounded (e.g.
> exponential or gamma) and the other(s) approximately Gaussian.
>
> After checking several packages, I'm not really clear what to do. Here is an
> example with "mixtools" that already works quite good, however, the left
> component is not Gaussian (and not symmetric).
>
> Any idea about a more adequate function or package for this problem?
>
> Thanks a lot!
>
> Thomas
>
>
>
> library(mixtools)
> set.seed(123)
>
> lambda <- c(0.25, 0.75)
> N  <- 200
>
> ## dist1 ~ gamma (or exponential as a special case)
> #dist1 <- rexp(lambda[1]*N, 1)
> dist1 <- rgamma(lambda[1]*N, 1, 1)
>
> ## dist2 ~ normal
> dist2 <- rnorm(lambda[2]*N, 12, 2)
>
> ## mixture
> x <- c(dist1, dist2)
>
> mix <-  spEMsymloc(x, mu0=2, eps=1e-3, verbose=TRUE)
> plot(mix, xlim=c(0, 25))
> summary(mix)
>
>
> --
> Thomas Petzoldt
> TU Dresden, Institute of Hydrobiology
> http://www.tu-dresden.de/Members/thomas.petzoldt
>
> __
> 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.

__
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.

[R] separate mixture of gamma and normal (with mixtools or ??)

[Thomas Petzoldt]

Dear friends,

I am trying to separate bi- (and sometimes tri-) modal univariate 
mixtures of biological data, where the first component is left bounded 
(e.g. exponential or gamma) and the other(s) approximately Gaussian.


After checking several packages, I'm not really clear what to do. Here 
is an example with "mixtools" that already works quite good, however, 
the left component is not Gaussian (and not symmetric).


Any idea about a more adequate function or package for this problem?

Thanks a lot!

Thomas



library(mixtools)
set.seed(123)

lambda <- c(0.25, 0.75)
N  <- 200

## dist1 ~ gamma (or exponential as a special case)
#dist1 <- rexp(lambda[1]*N, 1)
dist1 <- rgamma(lambda[1]*N, 1, 1)

## dist2 ~ normal
dist2 <- rnorm(lambda[2]*N, 12, 2)

## mixture
x <- c(dist1, dist2)

mix <-  spEMsymloc(x, mu0=2, eps=1e-3, verbose=TRUE)
plot(mix, xlim=c(0, 25))
summary(mix)


--
Thomas Petzoldt
TU Dresden, Institute of Hydrobiology
http://www.tu-dresden.de/Members/thomas.petzoldt

__
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.

Re: [R] removing dropouts from survival analysis

[Bert Gunter]

Sorry. You may get private replies, but this *is* way OT on this list.
Try stats.stackexchange.com instead for statistical queries. Or,
better yet, find local consulting help. Non-random dropouts are a
difficult issue.

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 Mon, Jan 23, 2017 at 12:48 PM, Damjan Krstajic  wrote:
> Dear All.
>
>
> Apologies for posting a question regarding survival analysis, and not R, to 
> the R-help list. In the past I received the best advices from the R community.
>
>
> The random censorship model (the censoring times independent of the failure 
> times and vice versa) is one of the fundamental assumptions in the survival 
> analysis. In the medical studies we have random entry to study and study end 
> which is a censoring mechanism independent of the failure times. However, in 
> reality we may have dropout subjects, lost to follow-up, which are censored 
> by a different mechanism which may not be independent of the failure times. 
> The inclusion of dropout subjects in the survival analysis may break the 
> random censorship model and include bias in our estimates of survival with 
> KM. I have studied papers on this subject (e.g. double sampling, copula 
> approach for dependent censoring), but I have not found any research paper 
> which examines the removal of dropout subjects from the survival analysis.
>
>
> I am alone in my research and would be grateful to hear thoughts on this 
> subject. Thank you in advance and apologies for using the R-help list for my 
> research question.
>
>
> DK
>
> [[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.

__
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.

Re: [R] [FORGED] Fitting arima Models with Exogenous Variables

[Rolf Turner]

This should have been sent to the R-help mailing list, not to me 
personally.  I am not an expert on this sort of time series modelling
and cannot thereby provide any useful advice.  My reply to you was of a 
"generic" nature --- when making an enquiry, provide a reproducible 
example!!!


I am cc-ing this email to the R-help list, since someone on that list 
*may* be able to answer your question.  I have (re-) attached the data 
sets that you sent to me.


cheers,

Rolf Turner

--
Technical Editor ANZJS
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276

On 24/01/17 04:36, Paul Bernal wrote:

Hello Rolf,

Thank you for your kind reply. I am attaching two datasets, one with the
historical data that I used to train the model, and the other one with
the exogenous variables.

The R code that I used is as follows:


library(forecast)
library(tseries)
library(TSA)
library(stats)
library(stats4)
TrainingDat<-read.csv("Training Data.csv")

ExogVars<-read.csv("ExogenousVariables5.csv")
#The file ExogVars contains 5 columns, one column for each regressor
Model1<-auto.arima(TrainingDat[,5], xreg=ExogVars)
#In Model1 I was able to incorporate xreg without any trouble
#The problem comes when trying to incorporate newxreg
Model2<-auto.arima(ExoVars[1:5])

Error in as.ts(x) : object 'ExoVars' not found


Model2<-auto.arima(ExogVars[1:5])


Error in auto.arima(ExogVars[1:5]) : No suitable ARIMA model found


Model2<-auto.arima(ExogVars[,1])

NewXReg<-forecast(Model2, h=12)

Forec<-forecast(Model1, newxreg=NewXReg)

Error in forecast.Arima(Model1, newxreg = NewXReg) :
  No regressors provided
In addition: Warning message:
In forecast.Arima(Model1, newxreg = NewXReg) :
  The non-existent newxreg arguments will be ignored.


Forec<-forecast(Model1, newxreg=NewXReg$mean)

Error in forecast.Arima(Model1, newxreg = NewXReg$mean) :
  No regressors provided
In addition: Warning message:
In forecast.Arima(Model1, newxreg = NewXReg$mean) :
  The non-existent newxreg arguments will be ignored.

I would like to generate the forecasts for all 4 variables included in
the Training set, along with all 5 regressors, but it seems like I can
only chose one training variable at a time, and one regressor at a time.

Please let me know if you can work this out,



__
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.

Re: [R] Extracting first number after * in a character vector

[Jim Lemon]

Hi Abhinaba,
I'm sure that someone will post a terrifyingly elegant regular
expression that does this, but:

 ardat<-
 c([1] " 1 X[0,SMITH]   *  0 0 1 ",
 ...
numpoststar<-function(x) {
 xsplit<-unlist(strsplit(x,""))
 starpos<-which(xsplit=="*")
 # watch out for a missing asterisk, they cause an infinite loop
 if(length(starpos)) {
  digits<-c("0","1","2","3","4","5","6","7","8","9")
  while(!any(digits %in% xsplit[starpos])) starpos<-starpos+1
  return(as.numeric(xsplit[starpos]))
 }
 return(NA)
}

for(i in 1:length(ardat)) print(numpoststar(ardat[i]))

The observant will wonder why I didn't use sapply. Because for some
reason it returned the original strings rather than the numbers. I
dunno.

Jim

On Mon, Jan 23, 2017 at 11:29 PM, Abhinaba Roy  wrote:
> Hi,
>
> How do I extract the first number after '*' in a vector?
>
> The vector is given below
>
>> dput(out[1:10])
> c(" 1 X[0,SMITH]   *  0 0 1 ",
> " 2 X[0,JOHNSON] *  0 0 1 ",
> " 3 X[0,WILLIAMS]", "*  1 0
> 1 ",
> " 4 X[0,JONES]   *  0 0 1 ",
> " 5 X[0,BROWN]   *  0 0 1 ",
> " 6 X[0,DAVIS]   *  0 0 1 ",
> " 7 X[0,MILLER]  *  0 0 1 ",
> " 8 X[0,WILSON]  *  0 0 1 ",
> " 9 X[0,MOORE]   *  0 0 1 "
> )
>
> I want a vector with the first number after the asterisk.
>
> So the output would give me, a vector (0,0,1,0,0,0,0,0,0,0)
>
> How can I do it in R?
>
> Best,
> Abhinaba
>
> [[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.

__
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.

Re: [R] R Graphics: Device 2 (Active)

[Greg Snow]

What is the result of running:

getOption("device")

?

It should be something like: "RStudioGD".  It sounds like this has
been changed to something else, if that is the case it is a matter of
either changing it back, or figuring out where the change is being
made and fixing that.

On Mon, Jan 23, 2017 at 11:19 AM, Jackson Rodrigues
 wrote:
> Hi,
>
> after updating R and RStudio I am no longer able to see my plots in the
> plot pane. Instead a new window opens called: R Graphics: Device 2 (Active).
>
> I tried to use dev.off(), reinstalling, update again and etc but it does
> not help.
>
> I've checked this discussion list but I could not find any solution.
>
> The following are some info about my R version
>
>> sessionInfo()
>
> R version 3.3.2 (2016-10-31)
>
> Platform: x86_64-w64-mingw32/x64 (64-bit)
>
> Running under: Windows >= 8 x64 (build 9200)
>
> locale:
>
> [1] LC_COLLATE=Portuguese_Brazil.1252  LC_CTYPE=Portuguese_Brazil.1252
>
> [3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C
>
> [5] LC_TIME=Portuguese_Brazil.1252
>
>
> attached base packages:
>
> [1] stats graphics  grDevices utils datasets  methods   base
>
> other attached packages:
>
>  [1] RColorBrewer_1.1-2 labdsv_1.8-0   cluster_2.0.5  MASS_7.3-45
>
>
>  [5] mgcv_1.8-15nlme_3.1-128   analogue_0.17-0vegan_2.4-1
>
>
>  [9] lattice_0.20-34permute_0.9-4
>
> loaded via a namespace (and not attached):
>
> [1] Matrix_1.2-7.1   parallel_3.3.2   tools_3.3.2  brglm_0.5-9
>
> [5] grid_3.3.2   princurve_1.1-12
>
>
> $version
>
> [1] ‘0.99.441’
>
>
> Any help are very welcome.
>
> Thanks
>
> Jackson
>
> [[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.



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

__
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.

[R] removing dropouts from survival analysis

[Damjan Krstajic]

Dear All.


Apologies for posting a question regarding survival analysis, and not R, to the 
R-help list. In the past I received the best advices from the R community.


The random censorship model (the censoring times independent of the failure 
times and vice versa) is one of the fundamental assumptions in the survival 
analysis. In the medical studies we have random entry to study and study end 
which is a censoring mechanism independent of the failure times. However, in 
reality we may have dropout subjects, lost to follow-up, which are censored by 
a different mechanism which may not be independent of the failure times. The 
inclusion of dropout subjects in the survival analysis may break the random 
censorship model and include bias in our estimates of survival with KM. I have 
studied papers on this subject (e.g. double sampling, copula approach for 
dependent censoring), but I have not found any research paper which examines 
the removal of dropout subjects from the survival analysis.


I am alone in my research and would be grateful to hear thoughts on this 
subject. Thank you in advance and apologies for using the R-help list for my 
research question.


DK

[[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.

Re: [R] Replicate Christensen, Dib (2008) using BMR package

[Jeff Newmiller]

Only your pdf attachment made it through. You would need to follow the Posting 
Guide regarding file types to achieve successful transmission. 

I will say that open-ended debugging of your code does not fit the Posting 
Guide recommendations either... this is an R language mailing list, not a 
research assistance mailing list. It might fit in better at 
stats.stackexchange.com, but even they want a focused question rather than 
general assistance requests. 
-- 
Sent from my phone. Please excuse my brevity.

On January 23, 2017 7:09:30 AM PST, Xue Li  wrote:
>Dear all,
>
>
>I am trying to replicate Christensen, Dib (2008)'s estimation results
>using "EDSGE" function in BMR package. Attached are my R code, the data
>I collected and preprocessed, and the paper.
>
>
>I set the priors following the literature (e.g., Smets and Wouters,
>2007). However, when I ran the code, R reported error messages like:
>
>
>Beginning MCMC run, Mon Jan 23 22:42:03 2017.
>Error in chol.default(CovM) :
>  the leading minor of order 2 is not positive definite
>In addition: Warning message:
>In sqrt(parModeHessian) : NaNs produced
>
>I also found that the estimation is sensitive to initial values and/or
>prior settings. I am struggling with debugging the code and hope
>someone could help me out. Thanks!
>
>
>Best,
>
>April

__
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.

Re: [R] R Graphics: Device 2 (Active)

[Bert Gunter]

This sounds like an RStudio issue to me (just a not-so-informed guess,
though). Did you reinstall that, too? Did you look/post on RStudio's
site?

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 Mon, Jan 23, 2017 at 10:19 AM, Jackson Rodrigues
 wrote:
> Hi,
>
> after updating R and RStudio I am no longer able to see my plots in the
> plot pane. Instead a new window opens called: R Graphics: Device 2 (Active).
>
> I tried to use dev.off(), reinstalling, update again and etc but it does
> not help.
>
> I've checked this discussion list but I could not find any solution.
>
> The following are some info about my R version
>
>> sessionInfo()
>
> R version 3.3.2 (2016-10-31)
>
> Platform: x86_64-w64-mingw32/x64 (64-bit)
>
> Running under: Windows >= 8 x64 (build 9200)
>
> locale:
>
> [1] LC_COLLATE=Portuguese_Brazil.1252  LC_CTYPE=Portuguese_Brazil.1252
>
> [3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C
>
> [5] LC_TIME=Portuguese_Brazil.1252
>
>
> attached base packages:
>
> [1] stats graphics  grDevices utils datasets  methods   base
>
> other attached packages:
>
>  [1] RColorBrewer_1.1-2 labdsv_1.8-0   cluster_2.0.5  MASS_7.3-45
>
>
>  [5] mgcv_1.8-15nlme_3.1-128   analogue_0.17-0vegan_2.4-1
>
>
>  [9] lattice_0.20-34permute_0.9-4
>
> loaded via a namespace (and not attached):
>
> [1] Matrix_1.2-7.1   parallel_3.3.2   tools_3.3.2  brglm_0.5-9
>
> [5] grid_3.3.2   princurve_1.1-12
>
>
> $version
>
> [1] ‘0.99.441’
>
>
> Any help are very welcome.
>
> Thanks
>
> Jackson
>
> [[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.

__
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.

[R] [R-pkgs] Announcing caesar

[Jacob Kaplan]

Dear R users,

I am happy to announce that the package caesar is now on CRAN (
https://cran.r-project.org/web/packages/caesar/index.html).

The caesar package lets you encrypt and decrypt strings using the common
Caesar cipher or a more secure pseudorandom number generator method.

If you would like to know more please visit the repository on GitHub (
https://github.com/jacobkap/caesar).

Best,
Jacob

[[alternative HTML version deleted]]

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

__
R-help@r-project.org mailing list -- 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.

Re: [R] R Graphics: Device 2 (Active)

[Jeff Newmiller]

You need to be clear in your own mind about the distinction between RStudio and 
R before you can communicate clearly about it.  Specifically, we don't know how 
to debug problems with RStudio here, and what you describe sounds completely 
normal for the R Gui program that ships with R.

It sounds to me like you need to communicate with RStudio support, but 
providing a reproducible example here, run from R Gui rather than RStudio, 
could make the issue more clear.
-- 
Sent from my phone. Please excuse my brevity.

On January 23, 2017 10:19:02 AM PST, Jackson Rodrigues 
 wrote:
>Hi,
>
>after updating R and RStudio I am no longer able to see my plots in the
>plot pane. Instead a new window opens called: R Graphics: Device 2
>(Active).
>
>I tried to use dev.off(), reinstalling, update again and etc but it
>does
>not help.
>
>I've checked this discussion list but I could not find any solution.
>
>The following are some info about my R version
>
>> sessionInfo()
>
>R version 3.3.2 (2016-10-31)
>
>Platform: x86_64-w64-mingw32/x64 (64-bit)
>
>Running under: Windows >= 8 x64 (build 9200)
>
>locale:
>
>[1] LC_COLLATE=Portuguese_Brazil.1252  LC_CTYPE=Portuguese_Brazil.1252
>
>[3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C
>
>[5] LC_TIME=Portuguese_Brazil.1252
>
>
>attached base packages:
>
>[1] stats graphics  grDevices utils datasets  methods   base
>
>other attached packages:
>
>[1] RColorBrewer_1.1-2 labdsv_1.8-0   cluster_2.0.5 
>MASS_7.3-45
>
>
>[5] mgcv_1.8-15nlme_3.1-128   analogue_0.17-0   
>vegan_2.4-1
>
>
> [9] lattice_0.20-34permute_0.9-4
>
>loaded via a namespace (and not attached):
>
>[1] Matrix_1.2-7.1   parallel_3.3.2   tools_3.3.2  brglm_0.5-9
>
>[5] grid_3.3.2   princurve_1.1-12
>
>
>$version
>
>[1] ‘0.99.441’
>
>
>Any help are very welcome.
>
>Thanks
>
>Jackson
>
>   [[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.

__
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.

Re: [R] R Graphics: Device 2 (Active)

[Rui Barradas]

Hello,

That's a question for R Studio, ask here: https://support.rstudio.com

Hope this helps,

Rui Barradas


Em 23-01-2017 18:19, Jackson Rodrigues escreveu:

Hi,

after updating R and RStudio I am no longer able to see my plots in the
plot pane. Instead a new window opens called: R Graphics: Device 2 (Active).

I tried to use dev.off(), reinstalling, update again and etc but it does
not help.

I've checked this discussion list but I could not find any solution.

The following are some info about my R version


sessionInfo()


R version 3.3.2 (2016-10-31)

Platform: x86_64-w64-mingw32/x64 (64-bit)

Running under: Windows >= 8 x64 (build 9200)

locale:

[1] LC_COLLATE=Portuguese_Brazil.1252  LC_CTYPE=Portuguese_Brazil.1252

[3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C

[5] LC_TIME=Portuguese_Brazil.1252


attached base packages:

[1] stats graphics  grDevices utils datasets  methods   base

other attached packages:

  [1] RColorBrewer_1.1-2 labdsv_1.8-0   cluster_2.0.5  MASS_7.3-45


  [5] mgcv_1.8-15nlme_3.1-128   analogue_0.17-0vegan_2.4-1


  [9] lattice_0.20-34permute_0.9-4

loaded via a namespace (and not attached):

[1] Matrix_1.2-7.1   parallel_3.3.2   tools_3.3.2  brglm_0.5-9

[5] grid_3.3.2   princurve_1.1-12


$version

[1] ‘0.99.441’


Any help are very welcome.

Thanks

Jackson

[[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.



__
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.

[R] R Graphics: Device 2 (Active)

[Jackson Rodrigues]

Hi,

after updating R and RStudio I am no longer able to see my plots in the
plot pane. Instead a new window opens called: R Graphics: Device 2 (Active).

I tried to use dev.off(), reinstalling, update again and etc but it does
not help.

I've checked this discussion list but I could not find any solution.

The following are some info about my R version

> sessionInfo()

R version 3.3.2 (2016-10-31)

Platform: x86_64-w64-mingw32/x64 (64-bit)

Running under: Windows >= 8 x64 (build 9200)

locale:

[1] LC_COLLATE=Portuguese_Brazil.1252  LC_CTYPE=Portuguese_Brazil.1252

[3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C

[5] LC_TIME=Portuguese_Brazil.1252


attached base packages:

[1] stats graphics  grDevices utils datasets  methods   base

other attached packages:

 [1] RColorBrewer_1.1-2 labdsv_1.8-0   cluster_2.0.5  MASS_7.3-45


 [5] mgcv_1.8-15nlme_3.1-128   analogue_0.17-0vegan_2.4-1


 [9] lattice_0.20-34permute_0.9-4

loaded via a namespace (and not attached):

[1] Matrix_1.2-7.1   parallel_3.3.2   tools_3.3.2  brglm_0.5-9

[5] grid_3.3.2   princurve_1.1-12


$version

[1] ‘0.99.441’


Any help are very welcome.

Thanks

Jackson

[[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.

Re: [R] spatial analysis using quickPCNM

[Jeff Newmiller]

I don't know if I will be able to help solve your problem, but failure to 
follow the recommendations in the Posting Guide was probably getting you off in 
the wrong foot when you posted on the other (more appropriate?) mailing list, 
as it is here. 

A) Code is garbled. Post using plain text format... this is a setting you have 
to choose in your email program. 

B) Code is not reproducible. We cannot run your code to get your error due to 
missing library statements and no data. The Posting Guide and Google have 
suggestions for how you can make your example reproducible. 

C) Assuming you are using the PCNM package, the documentation found by Google 
(which you can probably read using ?quickPCNM) says quickPCNM expects a matrix, 
but it looks like you are giving it a data frame. Read your introductory R 
training materials (e.g. the Introduction to R document that ships with R) to 
understand the distinction.  But I have never used this function so this could 
be a red herring. 

Please go read the Posting Guide. 
-- 
Sent from my phone. Please excuse my brevity.

On January 23, 2017 6:42:28 AM PST, Andrew Halford  
wrote:
>Hi Listers,
>
>I posted this message to the R-sig-ecology group last Friday but have
>not
>had a response hence my post here.
>
>I have been trying to run spatial analyses on a fish community dataset.
>
>My fish dataset has 114 species(variables) x 45 sites
>My spatial dataset has the Lat and Long values for each site, converted
>to
>cartesian coordinates
>
>
>
>> fish <- read.table(file.choose())> latlong <-
>read.table(file.choose()) > fish.h <- decostand (fish, "hellinger")>
>fish.PCNM.quick <- quickPCNM(fish.h,latlong)
>
>Truncation level = 639.5348
>Time to compute PCNMs = 0.82  sec Error in if (temp2.test[1, 5] <=
>alpha) { : argument is of length zeroTiming stopped at: 1.06 0.05 1.19
>
>I do not understand the error message coming up and would appreciate
>some advice.
>
>Andy

__
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.

Re: [R] Suggestions for vectorizing/double loop

[Luanna Dixson]

Hi all,

Thanks very much for your help! You are correct in thinking the list is the
same as before, actually, my question was more about how to do the next
steps, where I needed to match the filenames of the files in my directory
with old (i.e current) and new file name prefixes in my list. For each
match I then wanted to rename the original file using the corresponding new
filename prefix from the list.

Sorry for being a bit confusing, I didn't post my first attempt to do this
at first as my code just didn't work at all, but I have had another go
using a matrix instead and I think this does the job.

old_path="./old/"
new_path="./new/"

old_names <- list.files(old_path)
head(old_names)

oldnames=c('1002', '1003')
newnames=c('1002_new', '1003_new')
mapping=cbind(oldnames,newnames)
head(mapping)

for (i in old_names){
temp <- unlist(strsplit(i, "[.]"))[1]
n <- which(is.element(mapping,temp))
if(length(n)>0) {
#copy the file to the new folder
#re name it and the name is paste(mapping[n,2], '.xls', sep="")
print(paste(mapping[n,2], '.xls', sep=""))
newnames <- paste(mapping[n,2], '.xls', sep="")
file.copy(from = paste(old_path, i, sep=""),
  to = paste(new_path, newnames))
}
}


On 23 January 2017 at 13:28, Luanna Dixson 
wrote:

> I need to rename a bunch of files, by searching for string matches in a
> list. Each list element containings a character with the old filename that
> I want to match to, and the new file name that I want to rename by.
>
> For instance, here filename '1001.xls' should match to list[[1]]$oldname
> and I want to rename it to 'newa.xls' from list[[1]]$newname.
>
> The actual new file names I have will feature a random alphanumeric number
> and the list will have a length of ~600.
>
>
> # files I want to rename
>
> files with old file name =c('1001.xls', '1002.xls')
>
> # list with old file names and new file names
>
> oldnames=c('1001', '1002', '1003')
>
> newnames=c('newa', 'newb', 'newc')
>
> df=data.frame(oldnames,newnames)
>
> list <- split(df, rownames(df))
>
>
> # turn list elements in character
>
> for(i in 1:length(list)) list[[i]]$oldnames=as.
> character(list[[i]]$oldnames)
>
> for(i in 1:length(list)) list[[i]]$newnames=as.
> character(list[[i]]$newnames)
>
>
> I heard that it would be better to vectorize this than trying to do a
> double loop so if someone could give me a hint about how to do this I would
> be very grateful!
>

[[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.

Re: [R] Granger-causality test using vars package

[Pfaff, Bernhard Dr.]

Dear T.Riedle,

it is a 'combined' test, see ?causality for a formal description of the test 
statistic. 
If you would like results on an 'equation' by equation' approach, you could 
employ anova() on restricted and unrestricted lm-objects.

Best wishes,
Bernhard


-Ursprüngliche Nachricht-
Von: R-help [mailto:r-help-boun...@r-project.org] Im Auftrag von T.Riedle
Gesendet: Montag, 23. Januar 2017 13:26
An: R-help@r-project.org
Betreff: [EXT] Re: [R] Granger-causality test using vars package

Thank you for your reply. The code follows the example in the vignette and I 
changed it only a little as shown below.

library(vars)
data(Canada)
summary(Canada)
stat.desc(Canada,basic=FALSE)
plot(Canada, nc=2, xlab="")

# Testing for unit roots using ADF
adf1<-adf.test(Canada[,"prod"])
adf1
adf2<-adf.test(Canada[,"e"])
adf2
adf3<-adf.test(Canada[,"U"])
adf3
adf4<-adf.test(Canada[,"rw"])
adf4

# Use VAR to create a list of class varest Canada<-Canada[, c("prod", "e", "U", 
"rw")] p1ct<-VAR(Canada, p=1, type = "both") p1ct summary(p1ct, equation="e") 
plot(p1ct, names = "e")

#Run Granger-causality test
causality(p1ct)

The Granger-causality test returns following output $Granger

Granger causality H0: prod do not Granger-cause e U rw

data:  VAR object p1ct
F-Test = 11.956, df1 = 3, df2 = 308, p-value = 1.998e-07


$Instant

H0: No instantaneous causality between: prod and e U rw

data:  VAR object p1ct
Chi-squared = 3.7351, df = 3, p-value = 0.2915


Warning message:
In causality(p1ct) : 
Argument 'cause' has not been specified; using first variable in 'x$y' (prod) 
as cause variable.

I am struggling with the result as it is not clear to me whether the variable 
prod Granger-causes e or U or rw. H0 is that prod does not Granger-cause e U 
rw. What does that mean? How can I find out if prod Granger-causes e, U and rw, 
respectively i.e. how can I determine that prod Granger-causes e, U and rw?

Thanks for your support in advance.

From: Pfaff, Bernhard Dr. 
Sent: 23 January 2017 09:12
To: T.Riedle; R-help@r-project.org
Subject: AW:  [R] Granger-causality test using vars package

Dear T.Riedle,

you cannot assign *all* variables as a cause at once. Incidentally, in your 
example, you missed a 'data(Canada)'.
Having said this, you can loop over the variables names and extract the 
statistic/p-values. These are contained as named list elements 'statistic' and 
'p.value' in the returned list object 'Granger' which is of informal class 
'htest'.

Best wishes,
Bernhard

-Ursprüngliche Nachricht-
Von: R-help [mailto:r-help-boun...@r-project.org] Im Auftrag von T.Riedle
Gesendet: Sonntag, 22. Januar 2017 14:11
An: R-help@r-project.org
Betreff: [EXT] [R] Granger-causality test using vars package

Dear R-users,

I am trying to compute the test statistics for Granger-causality for a VAR(p) 
model using the "vars" package. I simply used the example proposed by the vars 
vignette and added the code for the Granger-causality. The code looks as follows



library(vars)

Canada<-Canada[, c("prod", "e", "U", "rw")] p1ct<-VAR(Canada, p=1, type = 
"both")

causality(p1ct, cause = c("prod","e","U","rw"))



Unfortunately I get the error

Error in `[<-`(`*tmp*`, i, w[i], value = 1) : subscript out of bounds



Does any body know what is wrong with the code? I would like to create a matrix 
containing the F-values and the corresponding significance. How can I do that?

[[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.
*
Confidentiality Note: The information contained in this ...{{dropped:9}}

__
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.

[R] Can mice() handle crr()? Fine-Gray model

[Therneau, Terry M., Ph.D.]

Look at the finegray command within the survival package; the competing risks vignette has 
coverage of it.  The command creates an expanded data set with case weights, such that 
coxph() on the new data set = the Fine Gray model for the original data.  Anything that 
works with coxph is valid on the new data.  Caveat -- I don't know mice() well at all: the 
dat1 data set below has multiple observations per subject, should the mice() command be 
cognizant of this?


Terry Therneau


library(survival)
library(mice)

test1 <- data.frame (time=c(4,3,1,1,2,2,3,5,2,4,5,1,
4,3,1,1,2,2,3,5,2,4,5,1),
 status=c(1,1,1,0,2,2,0,0,1,1,2,0,
  1,1,1,0,2,2,0,0,1,1,2,0),
 x=c(0,2,1,1,NA,NA,0,1,1,2,0,1,
 0,2,1,1,NA,NA,0,1,1,2,0,1),
 sex=c(0,0,0,NA,1,1,1,1,NA,1,0,0,
   0,0,0,NA,1,1,1,1,NA,1,0,0)))

# Endpoint 1, simple, then with mice
fit0 <- copxh(Surv(time, status==1) ~ sex + x, test1)
dat0 <- mice(test1, m=10)
mfit0 <- with(dat0, coxph(Surv(time, status==1) ~ sex + x))
summary(pool(mfit0))

# Endpoint 1, Fine-Gray model
fg1 <- finegray(Surv(time, factor(status)) ~ ., data=test1, etype=1)
fit1 <- coxph(Surv(fgstart, fgstop, fgstatus) ~ sex + x, data=fg1,
  weight= fgwt)

dat1 <- mice(fg1, m=10)
mfit1 <- with(dat1, coxph(Surv(fgstart, fgstop, fgstatus) ~ sex + x,
 weight=fgwt))


On 01/23/2017 05:00 AM, r-help-requ...@r-project.org wrote:

Here is an example:


# example

library(survival)
library(mice)
library(cmprsk)

test1 <- as.data.frame(list(time=c(4,3,1,1,2,2,3,5,2,4,5,1,
4,3,1,1,2,2,3,5,2,4,5,1),
 status=c(1,1,1,0,2,2,0,0,1,1,2,0,
1,1,1,0,2,2,0,0,1,1,2,0),
 x=c(0,2,1,1,NA,NA,0,1,1,2,0,1,
0,2,1,1,NA,NA,0,1,1,2,0,1),
 sex=c(0,0,0,NA,1,1,1,1,NA,1,0,0,
0,0,0,NA,1,1,1,1,NA,1,0,0)))

dat <- mice(test1,m=10)

#Cox regression: cause 1

models.cox1 <- with(dat,coxph(Surv(time, status==1) ~ x +sex
))

summary(pool(models.cox1))

#Cox regression: cause 1 or 2

models.cox <- with(dat,coxph(Surv(time, status==1 | status==2) ~ x +sex
))
models.cox
summary(pool(models.cox))


# crr()

#Fine-Gray model

models.FG<- with(dat,crr(ftime=time, fstatus=status,  cov1=test1[,c(
"x","sex")], failcode=1, cencode=0, variance=TRUE))

summary(pool(models.FG))

#Error in pool(models.FG) : Object has no vcov() method.

models.FG




-- Andreu Ferrero Gregori


__
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.

[R] spatial analysis using quickPCNM

[Andrew Halford]

Hi Listers,

I posted this message to the R-sig-ecology group last Friday but have not
had a response hence my post here.

I have been trying to run spatial analyses on a fish community dataset.

My fish dataset has 114 species(variables) x 45 sites
My spatial dataset has the Lat and Long values for each site, converted to
cartesian coordinates



> fish <- read.table(file.choose())> latlong <- read.table(file.choose()) > 
> fish.h <- decostand (fish, "hellinger")> fish.PCNM.quick <- 
> quickPCNM(fish.h,latlong)

Truncation level = 639.5348
Time to compute PCNMs = 0.82  sec Error in if (temp2.test[1, 5] <=
alpha) { : argument is of length zeroTiming stopped at: 1.06 0.05 1.19

I do not understand the error message coming up and would appreciate
some advice.

Andy


-- 
Andrew Halford Ph.D
Research Scientist (Kimberley Marine Parks)
Dept. Parks and Wildlife
Western Australia

Ph: +61 8 9219 9795
Mobile: +61 (0) 468 419 473

[[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.

Re: [R] Suggestions for vectorizing/double loop

[PIKAL Petr]

Hi

In your case if you want to rename abot 600 items and your doble loop works it 
is not worth to vectorize it. However if you want to repeat such task often and 
you expect much bigger number of files vectorizing can speed things up and make 
them cleaner.

Although I must admit that I do not understand what your commands should 
actually do. AFAICU the list is the same before and after your "double" loop.

Cheers
Petr

> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Luanna
> Dixson
> Sent: Monday, January 23, 2017 1:28 PM
> To: r-help@r-project.org
> Subject: [R] Suggestions for vectorizing/double loop
>
> I need to rename a bunch of files, by searching for string matches in a list.
> Each list element containings a character with the old filename that I want to
> match to, and the new file name that I want to rename by.
>
> For instance, here filename '1001.xls' should match to list[[1]]$oldname and I
> want to rename it to 'newa.xls' from list[[1]]$newname.
>
> The actual new file names I have will feature a random alphanumeric number
> and the list will have a length of ~600.
>
>
> # files I want to rename
>
> files with old file name =c('1001.xls', '1002.xls')
>
> # list with old file names and new file names
>
> oldnames=c('1001', '1002', '1003')
>
> newnames=c('newa', 'newb', 'newc')
>
> df=data.frame(oldnames,newnames)
>
> list <- split(df, rownames(df))
>
>
> # turn list elements in character
>
> for(i in 1:length(list)) list[[i]]$oldnames=as.character(list[[i]]$oldnames)
>
> for(i in 1:length(list)) list[[i]]$newnames=as.character(list[[i]]$newnames)
>
>
> I heard that it would be better to vectorize this than trying to do a double
> loop so if someone could give me a hint about how to do this I would be very
> grateful!
>
>   [[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.


Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny 
pouze jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně 
jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze 
svého systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email 
jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či 
zpožděním přenosu e-mailu.

V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a 
to z jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; 
Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce 
s dodatkem či odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným 
dosažením shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost 
žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně 
pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně 
osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi 
či osobě jím zastoupené známá.

This e-mail and any documents attached to it may be confidential and are 
intended only for its intended recipients.
If you received this e-mail by mistake, please immediately inform its sender. 
Delete the contents of this e-mail with all attachments and its copies from 
your system.
If you are not the intended recipient of this e-mail, you are not authorized to 
use, disseminate, copy or disclose this e-mail in any manner.
The sender of this e-mail shall not be liable for any possible damage caused by 
modifications of the e-mail or by delay with transfer of the email.

In case that this e-mail forms part of business dealings:
- the sender reserves the right to end negotiations about entering into a 
contract in any time, for any reason, and without stating any reasoning.
- if the e-mail contains an offer, the recipient is entitled to immediately 
accept such offer; The sender of this e-mail (offer) excludes any acceptance of 
the offer on the part of the recipient containing any amendment or variation.
- the sender insists on that the respective contract is concluded only upon an 
express mutual agreement on all its aspects.
- the sender of this e-mail informs that he/she is not authorized to enter into 
any contracts on behalf of the company except for cases in which he/she is 
expressly authorized to do so in writing, and such authorization or power of 
attorney is submitted to the recipient or the person 

Re: [R] Suggestions for vectorizing/double loop

[Michael Dewey]

Dear Luanna

It is not compulsory to avoid for loops but see below

On 23/01/2017 12:28, Luanna Dixson wrote:

I need to rename a bunch of files, by searching for string matches in a
list. Each list element containings a character with the old filename that
I want to match to, and the new file name that I want to rename by.

For instance, here filename '1001.xls' should match to list[[1]]$oldname
and I want to rename it to 'newa.xls' from list[[1]]$newname.

The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.


# files I want to rename

files with old file name =c('1001.xls', '1002.xls')

# list with old file names and new file names

oldnames=c('1001', '1002', '1003')

newnames=c('newa', 'newb', 'newc')

df=data.frame(oldnames,newnames)

list <- split(df, rownames(df))



Both df and list are already in existience and you are overwriting them. 
It is best practice not to do that.


# turn list elements in character

for(i in 1:length(list)) list[[i]]$oldnames=as.character(list[[i]]$oldnames)



Something like

lapply(list, function(x) x$oldnames <- as.character(x$oldnames))

would probably work.
You might want sapply instead of lapply


for(i in 1:length(list)) list[[i]]$newnames=as.character(list[[i]]$newnames)


I heard that it would be better to vectorize this than trying to do a
double loop so if someone could give me a hint about how to do this I would
be very grateful!

[[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.



--
Michael
http://www.dewey.myzen.co.uk/home.html

__
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.

Re: [R] Suggestions for vectorizing/double loop

[Thierry Onkelinx]

Dear Luanna,

Assuming that oldnames and newnames are character (and not factor), the
just use stringsAsFactors = FALSE. That will save you from having to
convert the factors back to character.

data.frame(oldnames, newnames, stringsAsFactor = FALSE)

Best regards,

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to say
what the experiment died of. ~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data. ~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of data.
~ John Tukey

2017-01-23 13:28 GMT+01:00 Luanna Dixson :

> I need to rename a bunch of files, by searching for string matches in a
> list. Each list element containings a character with the old filename that
> I want to match to, and the new file name that I want to rename by.
>
> For instance, here filename '1001.xls' should match to list[[1]]$oldname
> and I want to rename it to 'newa.xls' from list[[1]]$newname.
>
> The actual new file names I have will feature a random alphanumeric number
> and the list will have a length of ~600.
>
>
> # files I want to rename
>
> files with old file name =c('1001.xls', '1002.xls')
>
> # list with old file names and new file names
>
> oldnames=c('1001', '1002', '1003')
>
> newnames=c('newa', 'newb', 'newc')
>
> df=data.frame(oldnames,newnames)
>
> list <- split(df, rownames(df))
>
>
> # turn list elements in character
>
> for(i in 1:length(list)) list[[i]]$oldnames=as.
> character(list[[i]]$oldnames)
>
> for(i in 1:length(list)) list[[i]]$newnames=as.
> character(list[[i]]$newnames)
>
>
> I heard that it would be better to vectorize this than trying to do a
> double loop so if someone could give me a hint about how to do this I would
> be very grateful!
>
> [[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.
>

[[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.

Re: [R] Chi-square test

[Sergio Ferreira Cardoso]

Dear David and John,

Thank you for your replies. Indeed I'm using ape and nlme packages. Here it is:

> fit<-gls(fcl~mass+activity+agility,correlation=corBrownian(phy=tree),data=df,method="ML",weights=varFixed(~vf))
> Anova(fit)
Analysis of Deviance Table (Type II tests)

Response: fcl
 Df  Chisq Pr(>Chisq)
mass  1 0.1756 0.6752
activity  2 0.5549 0.7577
agility   4 3.2903 0.5105

Anyway, I have the help I was looking for. Thank you vey much.

Best regards,
Sérgio.

- Mensagem original -
> De: "Fox, John" 
> Para: "Sergio Ferreira Cardoso" 
> Cc: "R-help list" 
> Enviadas: Sábado, 21 De Janeiro de 2017 6:09:22
> Assunto: Re: [R] Chi-square test

> Dear Sergio,
> 
> You appear to have asked this question twice on r-help.
> 
> Anova() has no specific method for “gls” models (I assume, though you don’t 
> say
> so, that the model is fit by gls() in the nlme package), but the default 
> method
> works and provides Wald chi-square tests for terms in the model. I don’t
> understand the model formula x ~ 1 + 2 + 3 + x, however, and so I have no idea
> what gls() would do with this model, other than report an error. Perhaps you
> can show us the output — or, better yet, provide a reproducible example.
> 
> As a general matter, for 1-df terms in an additive model, the 1-df chi-square
> values reported by Anova() will simply be the squares of the corresponding 
> Wald
> statistics (labelled “t” I believe) reported in the summary of the model.
> Although the p-value is from the upper tail of the chi-square distribution, 
> the
> test is inherently two-sided.
> 
> Best,
> John
> 
> -
> John Fox, Professor
> McMaster University
> Hamilton, Ontario, Canada
> Web: http::/socserv.mcmaster.ca/jfox
> 
>> On Jan 20, 2017, at 8:36 AM, Sergio Ferreira Cardoso
>>  wrote:
>> 
>> Dear all,
>> 
>> Anova() for .car package retrieves Chi-square statistics when I'm testing a
>> model the significance of a multivariate .gls model
>> gls(x~1+2+3+x,corBrownian(phy=tree), ...).
>> Is this Chi-square a two-sided test?
>> 
>> Thank you.
>> 
>> Best,
>> Sérgio.
>> 
>>  [[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.

__
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.

[R] Suggestions for vectorizing/double loop

[Luanna Dixson]

I need to rename a bunch of files, by searching for string matches in a
list. Each list element containings a character with the old filename that
I want to match to, and the new file name that I want to rename by.

For instance, here filename '1001.xls' should match to list[[1]]$oldname
and I want to rename it to 'newa.xls' from list[[1]]$newname.

The actual new file names I have will feature a random alphanumeric number
and the list will have a length of ~600.


# files I want to rename

files with old file name =c('1001.xls', '1002.xls')

# list with old file names and new file names

oldnames=c('1001', '1002', '1003')

newnames=c('newa', 'newb', 'newc')

df=data.frame(oldnames,newnames)

list <- split(df, rownames(df))


# turn list elements in character

for(i in 1:length(list)) list[[i]]$oldnames=as.character(list[[i]]$oldnames)

for(i in 1:length(list)) list[[i]]$newnames=as.character(list[[i]]$newnames)


I heard that it would be better to vectorize this than trying to do a
double loop so if someone could give me a hint about how to do this I would
be very grateful!

[[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.

Re: [R] Extracting first number after * in a character vector

[Uwe Ligges]

On 23.01.2017 13:29, Abhinaba Roy wrote:

Hi,

How do I extract the first number after '*' in a vector?

The vector is given below


dput(out[1:10])

c(" 1 X[0,SMITH]   *  0 0 1 ",
" 2 X[0,JOHNSON] *  0 0 1 ",
" 3 X[0,WILLIAMS]", "*  1 0
1 ",
" 4 X[0,JONES]   *  0 0 1 ",
" 5 X[0,BROWN]   *  0 0 1 ",
" 6 X[0,DAVIS]   *  0 0 1 ",
" 7 X[0,MILLER]  *  0 0 1 ",
" 8 X[0,WILSON]  *  0 0 1 ",
" 9 X[0,MOORE]   *  0 0 1 "
)

I want a vector with the first number after the asterisk.

So the output would give me, a vector (0,0,1,0,0,0,0,0,0,0)

How can I do it in R?


You know that your vector (called x below) contains an element without 
an asterisk?

If that happened by accident, use
 gsub(".+\\* *([[:digit:]]+).*", "\\1", x)
and if it could happen to have elements without an asterisk or number 
that follows, you can set these results to NA in a seperate step.


Best,
Uwe Ligges









Best,
Abhinaba

[[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.



__
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.

[R] Extracting first number after * in a character vector

[Abhinaba Roy]

Hi,

How do I extract the first number after '*' in a vector?

The vector is given below

> dput(out[1:10])
c(" 1 X[0,SMITH]   *  0 0 1 ",
" 2 X[0,JOHNSON] *  0 0 1 ",
" 3 X[0,WILLIAMS]", "*  1 0
1 ",
" 4 X[0,JONES]   *  0 0 1 ",
" 5 X[0,BROWN]   *  0 0 1 ",
" 6 X[0,DAVIS]   *  0 0 1 ",
" 7 X[0,MILLER]  *  0 0 1 ",
" 8 X[0,WILSON]  *  0 0 1 ",
" 9 X[0,MOORE]   *  0 0 1 "
)

I want a vector with the first number after the asterisk.

So the output would give me, a vector (0,0,1,0,0,0,0,0,0,0)

How can I do it in R?

Best,
Abhinaba

[[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.

Re: [R] Granger-causality test using vars package

[T.Riedle]

Thank you for your reply. The code follows the example in the vignette and I 
changed it only a little as shown below.

library(vars)
data(Canada)
summary(Canada)
stat.desc(Canada,basic=FALSE)
plot(Canada, nc=2, xlab="")

# Testing for unit roots using ADF
adf1<-adf.test(Canada[,"prod"])
adf1
adf2<-adf.test(Canada[,"e"])
adf2
adf3<-adf.test(Canada[,"U"])
adf3
adf4<-adf.test(Canada[,"rw"])
adf4

# Use VAR to create a list of class varest
Canada<-Canada[, c("prod", "e", "U", "rw")]
p1ct<-VAR(Canada, p=1, type = "both")
p1ct
summary(p1ct, equation="e")
plot(p1ct, names = "e")

#Run Granger-causality test
causality(p1ct)

The Granger-causality test returns following output
$Granger

Granger causality H0: prod do not Granger-cause e U rw

data:  VAR object p1ct
F-Test = 11.956, df1 = 3, df2 = 308, p-value = 1.998e-07


$Instant

H0: No instantaneous causality between: prod and e U rw

data:  VAR object p1ct
Chi-squared = 3.7351, df = 3, p-value = 0.2915


Warning message:
In causality(p1ct) : 
Argument 'cause' has not been specified;
using first variable in 'x$y' (prod) as cause variable.

I am struggling with the result as it is not clear to me whether the variable 
prod Granger-causes e or U or rw. H0 is that prod does not Granger-cause e U 
rw. What does that mean? How can I find out if prod Granger-causes e, U and rw, 
respectively i.e. how can I determine that prod Granger-causes e, U and rw?

Thanks for your support in advance.

From: Pfaff, Bernhard Dr. 
Sent: 23 January 2017 09:12
To: T.Riedle; R-help@r-project.org
Subject: AW:  [R] Granger-causality test using vars package

Dear T.Riedle,

you cannot assign *all* variables as a cause at once. Incidentally, in your 
example, you missed a 'data(Canada)'.
Having said this, you can loop over the variables names and extract the 
statistic/p-values. These are contained as named list elements 'statistic' and 
'p.value' in the returned list object 'Granger' which is of informal class 
'htest'.

Best wishes,
Bernhard

-Ursprüngliche Nachricht-
Von: R-help [mailto:r-help-boun...@r-project.org] Im Auftrag von T.Riedle
Gesendet: Sonntag, 22. Januar 2017 14:11
An: R-help@r-project.org
Betreff: [EXT] [R] Granger-causality test using vars package

Dear R-users,

I am trying to compute the test statistics for Granger-causality for a VAR(p) 
model using the "vars" package. I simply used the example proposed by the vars 
vignette and added the code for the Granger-causality. The code looks as follows



library(vars)

Canada<-Canada[, c("prod", "e", "U", "rw")] p1ct<-VAR(Canada, p=1, type = 
"both")

causality(p1ct, cause = c("prod","e","U","rw"))



Unfortunately I get the error

Error in `[<-`(`*tmp*`, i, w[i], value = 1) : subscript out of bounds



Does any body know what is wrong with the code? I would like to create a matrix 
containing the F-values and the corresponding significance. How can I do that?

[[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.
*
Confidentiality Note: The information contained in this ...{{dropped:11}}

__
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.

Re: [R] Granger-causality test using vars package

[Pfaff, Bernhard Dr.]

Dear T.Riedle,

you cannot assign *all* variables as a cause at once. Incidentally, in your 
example, you missed a 'data(Canada)'.
Having said this, you can loop over the variables names and extract the 
statistic/p-values. These are contained as named list elements 'statistic' and 
'p.value' in the returned list object 'Granger' which is of informal class 
'htest'.

Best wishes,
Bernhard

-Ursprüngliche Nachricht-
Von: R-help [mailto:r-help-boun...@r-project.org] Im Auftrag von T.Riedle
Gesendet: Sonntag, 22. Januar 2017 14:11
An: R-help@r-project.org
Betreff: [EXT] [R] Granger-causality test using vars package

Dear R-users,

I am trying to compute the test statistics for Granger-causality for a VAR(p) 
model using the "vars" package. I simply used the example proposed by the vars 
vignette and added the code for the Granger-causality. The code looks as follows



library(vars)

Canada<-Canada[, c("prod", "e", "U", "rw")] p1ct<-VAR(Canada, p=1, type = 
"both")

causality(p1ct, cause = c("prod","e","U","rw"))



Unfortunately I get the error

Error in `[<-`(`*tmp*`, i, w[i], value = 1) : subscript out of bounds



Does any body know what is wrong with the code? I would like to create a matrix 
containing the F-values and the corresponding significance. How can I do that?

[[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.
*
Confidentiality Note: The information contained in this ...{{dropped:10}}

__
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.

Re: [R] runmed {stat}

[Martin Maechler]

>   on Sun, 22 Jan 2017 10:26:24 -0600 writes:

> Dear Dr Mächler, I am using runmed from R's stat
> package. I understand that you are the author of this package.

not of the package - but of function runmed().

I'm reply to R-help, so this answer maybe available to future
web searches.

> I am using the function with even length k=40 and the
> function forces it to be odd as k=41. I am sure there must
> be reason behind this overriding behavior. May I ask if
> you could enlighten me on this? Thank you in advance.

The help page - which one should really read (!) - says that k must be odd.

Why?  The median of an *odd* number of observations is "the
middle".  That's not the case with an even number, but a very
desirable property, which is kept even when iterating running
medians.
(Further, mathematically there are as many odd numbers as integers, so
 odd numbers should be sufficient ;-) ;-))

Martin Maechler
ETH Zurich

__
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.