Re: [R] command help

2015-04-03 Thread Jeff Newmiller
I cannot make sense of your email. This is partly due to your use of HTML, 
which the Posting Guidelines warn you not to do. 

The symbolic language used on this mailing list is R. The linear algebra 
operations available in R are rather straightforward. Please read the 
documentation and convey your request using R do we can understand you.
---
Jeff NewmillerThe .   .  Go Live...
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--- 
Sent from my phone. Please excuse my brevity.

On April 3, 2015 11:16:11 AM PDT, ali alRubaiee  
wrote:
>Dear colleagues,
>I wanted to run a command to do iteration for the following equation:
>Pj+1=TT11+FtransposeTT22F-FtransposeTT21-TT12
>F+y[(Atranspose-FtransposeBtranspose)P(A-BF)]
>
>where TT11, TT22,TT12, TT21, F transpose, A, B and F are matrices and y
>is
>a scalar
>we want to find the iteration for P using loop. Can you send me the
>right
>command for the loop for the above equation?
>
>
>Thank you in advance,
>Ali
>
>   [[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
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>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.

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Re: [R] glmnet: converting coefficients back to original scale

2015-04-03 Thread Suzen, Mehmet
This is interesting, can you post your lm.ridge solution as well?  I
suspect in glmnet, you need to use model.matrix with intercept, that
could be the reason.

-m

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Re: [R] Kruskal-Wallace power calculations.

2015-04-03 Thread Greg Snow
Here is some sample code:

## Simulation function to create data, analyze it using
## kruskal.test, and return the p-value
## change rexp to change the simulation distribution

simfun <- function(means, k=length(means), n=rep(50,k)) {
  mydata <- lapply( seq_len(k), function(i) {
rexp(n[i], 1) - 1 + means[i]
  })
  kruskal.test(mydata)$p.value
}

# simulate under the null to check proper sizing
B <- 1
out1 <- replicate(B, simfun(rep(3,4)))
hist(out1)
mean( out1 <= 0.05 )
binom.test( sum(out1 <= 0.05), B, p=0.05)

### Now simulate for power

B <- 1
out2 <- replicate(B, simfun( c(3,3,3.2,3.3)))
hist(out2)
mean( out2 <= 0.05 )
binom.test( sum(out2 <= 0.05), B, p=0.05 )

This simulates from a continuous exponential (skewed) and shifts to
get the means (shifted location is a common assumption, though not
required for the actual test).

On Thu, Apr 2, 2015 at 8:19 PM, Collin Lynch  wrote:
> Thank you Jim, I did see those (though not my typo :) and am still
> pondering the warning about post-hoc analyses.
>
> The situation that I am in is that I have a set of individuals who
> have been assigned a course grade.  We have then clustered these
> individuals into about 50 communities using standard community
> detection algorithms with the goal of determining whether community
> membership affects one of their grades.  We are using the KW test as
> the grade data is strongly non-normal and my coauthors preferred KW as
> an alternative.
>
> The two issues that I am struggling with are: 1) whether the post-hoc
> power analysis would be useful; and 2) how to code the simulation
> studies that are described in:
> http://onlinelibrary.wiley.com/doi/10.1002/bimj.4710380510/abstract
>
>
> Problem #1 is of course beyond the scope of this e-mail list though I
> would welcome anyone's suggestions on that point.  I am not sure that
> I buy the arguments against it offered here:
>
> http://graphpad.com/support/faq/why-it-is-not-helpful-to-compute-the-power-of-an-experiment-to-detect-the-difference-actually-observed-why-is-post-hoc-power-analysis-futile/
>
> It seems that the rationale boils down to "you didn't find it so you
> couldn't find it" but that does not tell me how far off I was from the
> goal.  I am still perusing the articles the author cites however.
>
>
> With respect to question #2 I am trying to lay my hands on the article
> and did find this old r-help discussion:
> http://r.789695.n4.nabble.com/Power-of-Kruskal-Wallis-Test-td4671188.html
> however I am not sure how to adapt the simulation studies that it
> links to to my current problem.  The links it leads to focus on
> mixed-effects models.  This may be more of a pure stats question and
> not suited for this list but I thought I'd ask in the hopes that
> anyone had any more specific KW code or knew of a good tutorial for
> the right kinds of simulation studies.
>
> Thank you,
> Collin.
>
>
>
>
> On Thu, Apr 2, 2015 at 6:35 PM, Jim Lemon  wrote:
>> Hi Collin,
>> Have a look at this:
>>
>> http://stats.stackexchange.com/questions/70643/power-analysis-for-kruskal-wallis-or-mann-whitney-u-test-using-r
>>
>> Although, thinking about it, this might have constituted your "perusal of
>> the literature".
>>
>> Plus it always looks better when you spell the names properly
>>
>> Jim
>>
>>
>> On Fri, Apr 3, 2015 at 2:23 AM, Jeff Newmiller 
>> wrote:
>>>
>>> Please stop... you are acting like a broken record, and are also posting
>>> in HTML format. Please read the Posting Guide and demonstrate that you have
>>> used a search engine on this topic before posting again.
>>>
>>> ---
>>> Jeff NewmillerThe .   .  Go
>>> Live...
>>> DCN:Basics: ##.#.   ##.#.  Live
>>> Go...
>>>   Live:   OO#.. Dead: OO#..  Playing
>>> Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
>>> /Software/Embedded Controllers)   .OO#.   .OO#.
>>> rocks...1k
>>>
>>> ---
>>> Sent from my phone. Please excuse my brevity.
>>>
>>> On April 2, 2015 7:25:20 AM PDT, Collin Lynch  wrote:
>>> >Greetings, I am working on a project where we are applying the
>>> >Kruskal-Wallace test to some factor data to evaluate their correlation
>>> >with
>>> >existing grade data.  I know that the grade data is nonnormal therefore
>>> >we
>>> >cannot rely on ANOVA or a similar parametric test.  What I would like
>>> >to
>>> >find is a mechanism for making power calculations for the KW test given
>>> >the
>>> >nonparametric assumptions.  My perusal of the literature has suggested
>>> >that
>>> >a simulation would be the best method.
>>> >
>>> >Can anyone point me to good examples of such simulations for KW in R?
>>> >And
>>> >does anyone have a favourite package for generating simulated data or
>>> >conducting such tests?
>>> >
>>> >Thank you

[R] command help

2015-04-03 Thread ali alRubaiee
Dear colleagues,
I wanted to run a command to do iteration for the following equation:
Pj+1=TT11+FtransposeTT22F-FtransposeTT21-TT12
F+y[(Atranspose-FtransposeBtranspose)P(A-BF)]

where TT11, TT22,TT12, TT21, F transpose, A, B and F are matrices and y is
a scalar
we want to find the iteration for P using loop. Can you send me the right
command for the loop for the above equation?


Thank you in advance,
Ali

[[alternative HTML version deleted]]

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Re: [R] open xlsx file using read.xls function of gdata package

2015-04-03 Thread Hadley Wickham
You might try the readxl package - it's only available on github but it
reads both xlsx and xls. All going well, it should be on its way to CRAN
next week.

Hadley

On Friday, April 3, 2015, Luigi Marongiu  wrote:

> Dear all,
> I am trying to open excel files using the gdata package. I can do that
> using a .xls file, but the same file, containing the same data,
> formatted in .xlsx gives error (R does not recognize the pattern from
> where to start reading the data).
> Doen anybody knows whether it is possible to read .xlslx with this package?
> Am I missing another package to implement the reading of the .xlsx?
> Thank you
> Luigi
>
> PS: this is the error I get:
> > my.file <- "array.xlsx"
> > my.data<-read.xls(
> +   my.file,
> +   sheet="sheet x",
> +   verbose=FALSE,
> +   pattern="row name",
> +   na.strings=c("NA","#DIV/0!"),
> +   method="tab",
> +   perl="perl"
> + )
> > Warning message:
> In read.xls(my.file, sheet = "sheet x", verbose = FALSE,  :
>   pattern not found
>
>
> The verbose version runs like this:
> “array.xlsx”
> to tab  file
> “/tmp/Rtmp2tAjzz/filef06102dd018.tab”
> ...
>
> Executing ' '/usr/bin/perl'
> '/home/gigiux/R/x86_64-pc-linux-gnu-library/3.0/gdata/perl/xls2tab.pl'
>  'array.xlsx' '/tmp/Rtmp2tAjzz/filef06102dd018.tab' 'sheet x' '...
>
> Loading 'array.xlsx'...
> Done.
>
> Orignal Filename: array.xlsx
> Number of Sheets: 2
>
> Writing sheet 'sheet x' to file '/tmp/Rtmp2tAjzz/filef06102dd018.tab'
> Minrow=31 Maxrow=17310 Mincol=0 Maxcol=4
>   (Ignored 0 blank lines.)
>
> 0
>
> Done.
>
> Searching for lines tfntaining pattern  row name ...
> Warning message:
> In read.xls(my.file, sheet = "sheet x", verbose = TRUE,  :
>   pattern not found
> >
>
> __
> R-help@r-project.org  mailing list -- To UNSUBSCRIBE and
> more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
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> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



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

[[alternative HTML version deleted]]

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[R] species names on a RDA plot

2015-04-03 Thread Antonio Silva
Dear R users

I'm trying to do a RDA analysis based on Borcard et al. 2011 Numerical
Ecology with R examples.

What I cannot understand is why when I run the script (RDA.R) using the
dataset from book (dataset1.rar) RDA triplot shows species names and when I
use my dataset (dataset2.rar) species names are not shown.

Data and script can be downloaded at
https://app.box.com/s/oayq7tglbmdsu72fj05h83dlzclsiglg

Does anyone know why this happens? Thanks for any clue.

Best regards

Antônio Olinto Ávila da Silva
Fisheries Institute
São Paulo, Brasil

[[alternative HTML version deleted]]

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Re: [R] Kruskal-Wallace power calculations.

2015-04-03 Thread Collin Lynch
Thank you very much Greg, I will give that a try.

Best,
Collin.

On Fri, Apr 3, 2015 at 1:43 PM, Greg Snow <538...@gmail.com> wrote:
> Here is some sample code:
>
> ## Simulation function to create data, analyze it using
> ## kruskal.test, and return the p-value
> ## change rexp to change the simulation distribution
>
> simfun <- function(means, k=length(means), n=rep(50,k)) {
>   mydata <- lapply( seq_len(k), function(i) {
> rexp(n[i], 1) - 1 + means[i]
>   })
>   kruskal.test(mydata)$p.value
> }
>
> # simulate under the null to check proper sizing
> B <- 1
> out1 <- replicate(B, simfun(rep(3,4)))
> hist(out1)
> mean( out1 <= 0.05 )
> binom.test( sum(out1 <= 0.05), B, p=0.05)
>
> ### Now simulate for power
>
> B <- 1
> out2 <- replicate(B, simfun( c(3,3,3.2,3.3)))
> hist(out2)
> mean( out2 <= 0.05 )
> binom.test( sum(out2 <= 0.05), B, p=0.05 )
>
> This simulates from a continuous exponential (skewed) and shifts to
> get the means (shifted location is a common assumption, though not
> required for the actual test).
>
> On Thu, Apr 2, 2015 at 8:19 PM, Collin Lynch  wrote:
>> Thank you Jim, I did see those (though not my typo :) and am still
>> pondering the warning about post-hoc analyses.
>>
>> The situation that I am in is that I have a set of individuals who
>> have been assigned a course grade.  We have then clustered these
>> individuals into about 50 communities using standard community
>> detection algorithms with the goal of determining whether community
>> membership affects one of their grades.  We are using the KW test as
>> the grade data is strongly non-normal and my coauthors preferred KW as
>> an alternative.
>>
>> The two issues that I am struggling with are: 1) whether the post-hoc
>> power analysis would be useful; and 2) how to code the simulation
>> studies that are described in:
>> http://onlinelibrary.wiley.com/doi/10.1002/bimj.4710380510/abstract
>>
>>
>> Problem #1 is of course beyond the scope of this e-mail list though I
>> would welcome anyone's suggestions on that point.  I am not sure that
>> I buy the arguments against it offered here:
>>
>> http://graphpad.com/support/faq/why-it-is-not-helpful-to-compute-the-power-of-an-experiment-to-detect-the-difference-actually-observed-why-is-post-hoc-power-analysis-futile/
>>
>> It seems that the rationale boils down to "you didn't find it so you
>> couldn't find it" but that does not tell me how far off I was from the
>> goal.  I am still perusing the articles the author cites however.
>>
>>
>> With respect to question #2 I am trying to lay my hands on the article
>> and did find this old r-help discussion:
>> http://r.789695.n4.nabble.com/Power-of-Kruskal-Wallis-Test-td4671188.html
>> however I am not sure how to adapt the simulation studies that it
>> links to to my current problem.  The links it leads to focus on
>> mixed-effects models.  This may be more of a pure stats question and
>> not suited for this list but I thought I'd ask in the hopes that
>> anyone had any more specific KW code or knew of a good tutorial for
>> the right kinds of simulation studies.
>>
>> Thank you,
>> Collin.
>>
>>
>>
>>
>> On Thu, Apr 2, 2015 at 6:35 PM, Jim Lemon  wrote:
>>> Hi Collin,
>>> Have a look at this:
>>>
>>> http://stats.stackexchange.com/questions/70643/power-analysis-for-kruskal-wallis-or-mann-whitney-u-test-using-r
>>>
>>> Although, thinking about it, this might have constituted your "perusal of
>>> the literature".
>>>
>>> Plus it always looks better when you spell the names properly
>>>
>>> Jim
>>>
>>>
>>> On Fri, Apr 3, 2015 at 2:23 AM, Jeff Newmiller 
>>> wrote:

 Please stop... you are acting like a broken record, and are also posting
 in HTML format. Please read the Posting Guide and demonstrate that you have
 used a search engine on this topic before posting again.

 ---
 Jeff NewmillerThe .   .  Go
 Live...
 DCN:Basics: ##.#.   ##.#.  Live
 Go...
   Live:   OO#.. Dead: OO#..  Playing
 Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
 /Software/Embedded Controllers)   .OO#.   .OO#.
 rocks...1k

 ---
 Sent from my phone. Please excuse my brevity.

 On April 2, 2015 7:25:20 AM PDT, Collin Lynch  wrote:
 >Greetings, I am working on a project where we are applying the
 >Kruskal-Wallace test to some factor data to evaluate their correlation
 >with
 >existing grade data.  I know that the grade data is nonnormal therefore
 >we
 >cannot rely on ANOVA or a similar parametric test.  What I would like
 >to
 >find is a mechanism for making power calculations for the KW test given
 >the
 >nonparametric assumptions.  My perusal of the litera

Re: [R] Repeated failures to install "caret" package (of Max Kuhn)

2015-04-03 Thread Uwe Ligges

What is the error message?

Best,
Uwe Ligges

On 03.04.2015 23:07, Ronald Wyllys wrote:

For an edx course, MIT's "The Analtics Edge", I need to install the
"caret" package that was originated and is maintained by Dr. Max Kuhn of
Pfizer. So far, every effort I've made to try to
install.packages("caret") has failed.  (I'm using R v. 3.1.3 and RStudio
v. 0.98.1103 in LinuxMint 17.1)

Here are some of the things I've tried unsuccessfully:
install.packages("caret", repos=c("http://rstudio.org/_packages";,
"http://cran.rstudio.com";))
install.packages("caret", dependencies=TRUE)
install.packages("caret", repos=c("http://rstudio.org/_packages";,
"http://cran.rstudio.com";), dependencies=TRUE)
install.packages("caret", dependencies = c("Depends", "Suggests"))
install.packages("caret", repos="http://cran.rstudio.com/";)

I've changed my CRAN mirror from UCLA to Revolution Analytics in Dallas,
and tried the above installs again, unsuccessfully.

I've succeeded in individually installing a number of packages on which
"caret" appears to be dependent.  Specifically, I've been able to
install  "nloptr", "minqa", "Rcpp", "reshape2", "stringr", and
"scales".  But I've had no success with trying to do individual installs
of "BradleyTerry2", "car", "lme4", "quantreg", and "RcppEigen".

Any suggestions will be very gratefully received (and tried out quickly).

Thanks in advance.

Ron Wyllys

__
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Re: [R] Repeated failures to install "caret" package (of Max Kuhn)

2015-04-03 Thread Sarah Goslee
Hi Ron,

Without knowing more it's really hard to help. What error messages are
you getting?
"failed" and "no success" are utterly uninformative - there are many
things that could be wrong.

The first thing, though, is always to check whether you have the
necessary dependencies installed, eg devel libraries.

Sarah

On Fri, Apr 3, 2015 at 5:07 PM, Ronald Wyllys  wrote:
> For an edx course, MIT's "The Analtics Edge", I need to install the "caret"
> package that was originated and is maintained by Dr. Max Kuhn of Pfizer. So
> far, every effort I've made to try to install.packages("caret") has failed.
> (I'm using R v. 3.1.3 and RStudio v. 0.98.1103 in LinuxMint 17.1)
>
> Here are some of the things I've tried unsuccessfully:
> install.packages("caret", repos=c("http://rstudio.org/_packages";,
> "http://cran.rstudio.com";))
> install.packages("caret", dependencies=TRUE)
> install.packages("caret", repos=c("http://rstudio.org/_packages";,
> "http://cran.rstudio.com";), dependencies=TRUE)
> install.packages("caret", dependencies = c("Depends", "Suggests"))
> install.packages("caret", repos="http://cran.rstudio.com/";)
>
> I've changed my CRAN mirror from UCLA to Revolution Analytics in Dallas, and
> tried the above installs again, unsuccessfully.
>
> I've succeeded in individually installing a number of packages on which
> "caret" appears to be dependent.  Specifically, I've been able to install
> "nloptr", "minqa", "Rcpp", "reshape2", "stringr", and "scales".  But I've
> had no success with trying to do individual installs of "BradleyTerry2",
> "car", "lme4", "quantreg", and "RcppEigen".
>
> Any suggestions will be very gratefully received (and tried out quickly).
>
> Thanks in advance.
>
> Ron Wyllys
>

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

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[R] Repeated failures to install "caret" package (of Max Kuhn)

2015-04-03 Thread Ronald Wyllys
For an edx course, MIT's "The Analtics Edge", I need to install the 
"caret" package that was originated and is maintained by Dr. Max Kuhn of 
Pfizer. So far, every effort I've made to try to 
install.packages("caret") has failed.  (I'm using R v. 3.1.3 and RStudio 
v. 0.98.1103 in LinuxMint 17.1)


Here are some of the things I've tried unsuccessfully:
install.packages("caret", repos=c("http://rstudio.org/_packages";, 
"http://cran.rstudio.com";))

install.packages("caret", dependencies=TRUE)
install.packages("caret", repos=c("http://rstudio.org/_packages";, 
"http://cran.rstudio.com";), dependencies=TRUE)

install.packages("caret", dependencies = c("Depends", "Suggests"))
install.packages("caret", repos="http://cran.rstudio.com/";)

I've changed my CRAN mirror from UCLA to Revolution Analytics in Dallas, 
and tried the above installs again, unsuccessfully.


I've succeeded in individually installing a number of packages on which 
"caret" appears to be dependent.  Specifically, I've been able to 
install  "nloptr", "minqa", "Rcpp", "reshape2", "stringr", and 
"scales".  But I've had no success with trying to do individual installs 
of "BradleyTerry2", "car", "lme4", "quantreg", and "RcppEigen".


Any suggestions will be very gratefully received (and tried out quickly).

Thanks in advance.

Ron Wyllys

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Re: [R] WEIBULL or EXPONENTIAL?

2015-04-03 Thread varin sacha
Hi Xavier,

I use the fitdistrplus and logspline packages to know which distribution fits 
better my data.


Here is an example :

install.packages("fitdistrplus")
library(fitdistrplus)
instal.packages("logspline")
library(logspline)
x=c(44986,18288,56147,44488,41018,40631,27301,39025,45688,47172,12300,21558,16103,48874,67245,36119,10398,42630,12879,34058,84443,30639)
descdist(x,discrete=FALSE) 

Cheers,
S.




De : CHIRIBOGA Xavier 
À : "r-help@r-project.org"  
Envoyé le : Vendredi 3 avril 2015 16h33
Objet : [R] WEIBULL or EXPONENTIAL?


Dear members,



I am doing a survival analysis wiith the function coxph...however I am 
wondering how can I know if my data follows a EXPONENTIAL or WEIBULL 
distribution?

I have 3 censored datum. Using R studio.



Thanks for the suggestions,



Xavier

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Re: [R] open xlsx file using read.xls function of gdata package

2015-04-03 Thread Jeff Newmiller
I had poor luck with gdata. I have had better luck with XLConnect. There is no 
single best package for this, since each seems to leverage efforts made in 
other languages (so there are non-R configuration requirements to keep working) 
and Excel is a proprietary moving target. In general YMMV when it comes to 
Excel data. In most cases I just export the data to CSV and avoid the issue.
---
Jeff NewmillerThe .   .  Go Live...
DCN:Basics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

On April 3, 2015 11:24:29 AM PDT, Luigi Marongiu  
wrote:
>Dear all,
>I am trying to open excel files using the gdata package. I can do that
>using a .xls file, but the same file, containing the same data,
>formatted in .xlsx gives error (R does not recognize the pattern from
>where to start reading the data).
>Doen anybody knows whether it is possible to read .xlslx with this
>package?
>Am I missing another package to implement the reading of the .xlsx?
>Thank you
>Luigi
>
>PS: this is the error I get:
>> my.file <- "array.xlsx"
>> my.data<-read.xls(
>+   my.file,
>+   sheet="sheet x",
>+   verbose=FALSE,
>+   pattern="row name",
>+   na.strings=c("NA","#DIV/0!"),
>+   method="tab",
>+   perl="perl"
>+ )
>> Warning message:
>In read.xls(my.file, sheet = "sheet x", verbose = FALSE,  :
>  pattern not found
>
>
>The verbose version runs like this:
>“array.xlsx”
>to tab  file
>“/tmp/Rtmp2tAjzz/filef06102dd018.tab”
>...
>
>Executing ' '/usr/bin/perl'
>'/home/gigiux/R/x86_64-pc-linux-gnu-library/3.0/gdata/perl/xls2tab.pl'
> 'array.xlsx' '/tmp/Rtmp2tAjzz/filef06102dd018.tab' 'sheet x' '...
>
>Loading 'array.xlsx'...
>Done.
>
>Orignal Filename: array.xlsx
>Number of Sheets: 2
>
>Writing sheet 'sheet x' to file '/tmp/Rtmp2tAjzz/filef06102dd018.tab'
>Minrow=31 Maxrow=17310 Mincol=0 Maxcol=4
>  (Ignored 0 blank lines.)
>
>0
>
>Done.
>
>Searching for lines tfntaining pattern  row name ...
>Warning message:
>In read.xls(my.file, sheet = "sheet x", verbose = TRUE,  :
>  pattern not found
>>
>
>__
>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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[R] open xlsx file using read.xls function of gdata package

2015-04-03 Thread Luigi Marongiu
Dear all,
I am trying to open excel files using the gdata package. I can do that
using a .xls file, but the same file, containing the same data,
formatted in .xlsx gives error (R does not recognize the pattern from
where to start reading the data).
Doen anybody knows whether it is possible to read .xlslx with this package?
Am I missing another package to implement the reading of the .xlsx?
Thank you
Luigi

PS: this is the error I get:
> my.file <- "array.xlsx"
> my.data<-read.xls(
+   my.file,
+   sheet="sheet x",
+   verbose=FALSE,
+   pattern="row name",
+   na.strings=c("NA","#DIV/0!"),
+   method="tab",
+   perl="perl"
+ )
> Warning message:
In read.xls(my.file, sheet = "sheet x", verbose = FALSE,  :
  pattern not found


The verbose version runs like this:
“array.xlsx”
to tab  file
“/tmp/Rtmp2tAjzz/filef06102dd018.tab”
...

Executing ' '/usr/bin/perl'
'/home/gigiux/R/x86_64-pc-linux-gnu-library/3.0/gdata/perl/xls2tab.pl'
 'array.xlsx' '/tmp/Rtmp2tAjzz/filef06102dd018.tab' 'sheet x' '...

Loading 'array.xlsx'...
Done.

Orignal Filename: array.xlsx
Number of Sheets: 2

Writing sheet 'sheet x' to file '/tmp/Rtmp2tAjzz/filef06102dd018.tab'
Minrow=31 Maxrow=17310 Mincol=0 Maxcol=4
  (Ignored 0 blank lines.)

0

Done.

Searching for lines tfntaining pattern  row name ...
Warning message:
In read.xls(my.file, sheet = "sheet x", verbose = TRUE,  :
  pattern not found
>

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

Re: [R] Color US counties on US map using a numeric variable for color intensity

2015-04-03 Thread Adams, Jean
Dimitri,

To answer your questions:
The colorRamp() function creates a new function, newpal().
The value returned by newpal() is a numeric matrix of RGB color values.
The rgb() function is then used to convert this numeric matrix to colors,
with the argument maxColorValue
giving the maximum of the color values range.  Typically either 255 or 1.
See the help files for more information

?colorRamp
?rgb


I think Jim Lemon's suggestion to use color.scale() function is a handier
solution.

Jean

On Thu, Apr 2, 2015 at 6:05 PM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.com> wrote:

> This is really cool, Jim - thanks a lot!
>
> On Thu, Apr 2, 2015 at 6:18 PM, Jim Lemon  wrote:
> > Hi Dimitri,
> > You can also try the color.scale function in plotrix, which allows you to
> > specify the NA color in the call.
> >
> >
> newcol<-color.scale(mydata.final$Mean.Wait,extremes=c("yellow","red"),na.color="white")
> >
> > Jim
> >
> >
> > On Fri, Apr 3, 2015 at 8:08 AM, Dimitri Liakhovitski
> >  wrote:
> >>
> >> Jean, I think I fixed it:
> >>
> >> newpal <- colorRamp(c("yellow", "red"))
> >> missing <- is.na(mydata.final$Mean.Wait)
> >> newcol <- ifelse(missing, "white",
> >>
> >> rgb(newpal(mydata.final$Mean.Wait[!is.na(mydata.final$Mean.Wait)]/
> >>   max(mydata.final$Mean.Wait,
> >> na.rm=T)), maxColorValue=255))
> >> map('county', fill=TRUE, col=newcol,
> >> resolution=0, lty=0, bg="transparent")
> >> map('state', lwd=1, add=TRUE)
> >>
> >> One understanding question: what exactly does this rgb line do and why
> >> do we have to say "maxColorValue=255"?
> >> Thank you!
> >>
> >> On Thu, Apr 2, 2015 at 5:02 PM, Dimitri Liakhovitski
> >>  wrote:
> >> > Thank you, Jean, but I think this newcol line is not working. I am
> >> > running:
> >> >
> >> > newcol <- ifelse(missing, "white",
> >> >
> >> > rgb(newpal(mydata.final$Mean.Wait/max(mydata.final$Mean.Wait,
> >> > na.rm=T)),
> >> >  maxColorValue=255))
> >> >
> >> > # And I am getting:
> >> > Error in rgb(newpal(mydata.final$Mean.Wait/max(mydata.final$Mean.Wait,
> >> > :
> >> >   color intensity NA, not in 0:255
> >> >
> >> > I think it's not liking the NAs - despite the ifelse...
> >> >
> >> > On Thu, Apr 2, 2015 at 4:26 PM, Adams, Jean  wrote:
> >> >> Dimitri,
> >> >>
> >> >> You could use colorRamp() and rgb() to get more continuous colors.
> >> >> For example
> >> >>
> >> >> newpal <- colorRamp(c("yellow", "red"))
> >> >> missing <- is.na(mydata.final$Mean.Wait)
> >> >> newcol <- ifelse(missing, "white",
> >> >>   rgb(newpal(mydat$Mean.Wait/max(mydat$Mean.Wait)),
> maxColorValue=255))
> >> >> map('county', fill=TRUE, col=newcol,
> >> >> resolution=0, lty=0, bg="transparent")
> >> >> map('state', lwd=1, add=TRUE)
> >> >>
> >> >> Jean
> >> >>
> >> >>
> >> >> On Thu, Apr 2, 2015 at 12:03 PM, Dimitri Liakhovitski
> >> >>  wrote:
> >> >>>
> >> >>> I have a data frame 'mydata.final' (see below) that contains US
> >> >>> counties and a continuous numeric variable 'Mean.Wait' that ranges
> >> >>> from zero to 10 or so. I also created variable 'wait' that is based
> on
> >> >>> the 'Mean.Wait' and takes on discrete values from 1 (lowest values
> on
> >> >>> 'Mean.Wait') to 5 (highest values on 'Mean.Wait').
> >> >>>
> >> >>> I can create a map of the US with the counties colored based on the
> >> >>> values of 'wait' using R package 'maps':
> >> >>>
> >> >>> #
> >> >>> ### Generating an artificial data file:
> >> >>> #
> >> >>> library(maps)
> >> >>> mydata.final <- data.frame(county = (map('county', plot =
> >> >>> FALSE)$names),
> >> >>>  stringsAsFactors = F)
> >> >>>
> >> >>> ### My numeric variable:
> >> >>> set.seed(123)
> >> >>> mydata.final$Mean.Wait <- runif(nrow(mydata.final)) * 10
> >> >>>
> >> >>> ### Introducing NAs to mimic my real data set:
> >> >>> set.seed(1234)
> >> >>> mydata.final$Mean.Wait[sample(1:nrow(mydata.final), 1500)] <- NA
> >> >>>
> >> >>> ### Cutting the original numeric variable into categories
> >> >>> ### because I don't know how to color based on 'Mean.Wait':
> >> >>> mydata.final$wait <- cut(mydata.final$Mean.Wait, breaks = 5)
> >> >>> levels(mydata.final$wait) <- 1:5
> >> >>> mydata.final$wait <- as.numeric(as.character(mydata.final$wait))
> >> >>>
> >> >>> 
> >> >>> Building a US map based on 'wait' (5 categories)
> >> >>> #
> >> >>>
> >> >>> ### Creating my 5 colors:
> >> >>> pal <- colorRampPalette(c("yellow", "red"))
> >> >>> allcolors <- pal(5)
> >> >>>
> >> >>> ### Looking at my 5 colors:
> >> >>> barplot(1:5, rep(1,5), col = allcolors, horiz = T)
> >> >>>
> >> >>> ### Builiding the US map using 5 categories in 'wait':
> >> >>> map('county', fill = TRUE, col = allcolors[mydata.final$wait],

Re: [R] applying cumsum within groups

2015-04-03 Thread peter dalgaard
ave() is your friend (unfortunately named as it may be):

> ave(dat$seq, dat$ts, FUN=cumsum)
 [1] 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 1
[39] 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3


> On 03 Apr 2015, at 14:17 , Morway, Eric  wrote:
> 
> This small example will be applied to a problem with 1.4e6 lines of data.
> First, here is the dataset and a few lines of R script, followed by an
> explanation of what I'd like to get:
> 
> dat <- read.table(textConnection("ISEG  IRCH  val
> 11   265
> 12   260
> 13   234
> 54   39   467
> 54   40   468
> 54   41   460
> 54   42   489
> 11   265
> 12   276
> 13   217
> 54   39   456
> 54   40   507
> 54   41   483
> 54   42   457
> 11   265
> 12   287
> 13   224
> 54   39   473
> 54   40   502
> 54   41   497
> 54   42   447
> 11   230
> 12   251
> 13   199
> 54   39   439
> 54   40   474
> 54   41   477
> 54   42   413
> 11   230
> 12   262
> 13   217
> 54   39   455
> 54   40   493
> 54   41   489
> 54   42   431
> 11   1002
> 12   1222
> 13   1198
> 54   39   1876
> 54   40   1565
> 54   41   1455
> 54   42   1427
> 11   1002
> 12   1246
> 13   1153
> 54   39   1813
> 54   40   1490
> 54   41   1518
> 54   42   1486
> 11   1002
> 12   1229
> 13   1142
> 54   39   1797
> 54   40   1517
> 54   41   1527
> 54   42   1514"),header=TRUE)
> 
> dat$seq <- ifelse(dat$ISEG==1 & dat$IRCH==1, 1, 0)
> tmp <- diff(dat[dat$seq==1,]$val)!=0
> dat$idx <- 0
> dat[dat$seq==1,][c(TRUE,tmp),]$idx <- 1
> dat$ts <- cumsum(dat$idx)
> 
> At this point, I'd like to add one more column called "iter" that counts up
> by 1 based on "seq", but within each "ts".  So, the result would look like
> this (undoubtedly this is a simple problem with something like ddply, but
> I've been unable to construct the R for it):
> 
> dat
> ISEG IRCH  val seq idx ts iter
>11  265   1   1  11
>12  260   0   0  11
>13  234   0   0  11
>   54   39  467   0   0  11
>   54   40  468   0   0  11
>   54   41  460   0   0  11
>   54   42  489   0   0  11
>11  265   1   0  12
>12  276   0   0  12
>13  217   0   0  12
>   54   39  456   0   0  12
>   54   40  507   0   0  12
>   54   41  483   0   0  12
>   54   42  457   0   0  12
>11  265   1   0  13
>12  287   0   0  13
>13  224   0   0  13
>   54   39  473   0   0  13
>   54   40  502   0   0  13
>   54   41  497   0   0  13
>   54   42  447   0   0  13
>11  230   1   1  21
>12  251   0   0  21
>13  199   0   0  21
>   54   39  439   0   0  21
>   54   40  474   0   0  21
>   54   41  477   0   0  21
>   54   42  413   0   0  21
>11  230   1   0  22
>12  262   0   0  22
>13  217   0   0  22
>   54   39  455   0   0  22
>   54   40  493   0   0  22
>   54   41  489   0   0  22
>   54   42  431   0   0  22
>11 1002   1   1  31
>12 1222   0   0  31
>13 1198   0   0  31
>   54   39 1876   0   0  31
>   54   40 1565   0   0  31
>   54   41 1455   0   0  31
>   54   42 1427   0   0  31
>11 1002   1   0  32
>12 1246   0   0  32
>13 1153   0   0  32
>   54   39 1813   0   0  32
>   54   40 1490   0   0  32
>   54   41 1518   0   0  32
>   54   42 1486   0   0  32
>11 1002   1   0  33
>12 1229   0   0  33
>13 1142   0   0  33
>   54   39 1797   0   0  33
>   54   40 1517   0   0  33
>   54   41 1527   0   0  33
>   54   42 1514   0   0  33
> 
>   [[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.

-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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


Re: [R] applying cumsum within groups

2015-04-03 Thread David Winsemius

On Apr 3, 2015, at 5:17 AM, Morway, Eric wrote:

> This small example will be applied to a problem with 1.4e6 lines of data.
> First, here is the dataset and a few lines of R script, followed by an
> explanation of what I'd like to get:
> 
> dat <- read.table(textConnection("ISEG  IRCH  val
> 11   265
> 12   260
> 13   234
> 54   39   467
> 54   40   468
> 54   41   460
> 54   42   489
> 11   265
> 12   276
> 13   217
> 54   39   456
> 54   40   507
> 54   41   483
> 54   42   457
> 11   265
> 12   287
> 13   224
> 54   39   473
> 54   40   502
> 54   41   497
> 54   42   447
> 11   230
> 12   251
> 13   199
> 54   39   439
> 54   40   474
> 54   41   477
> 54   42   413
> 11   230
> 12   262
> 13   217
> 54   39   455
> 54   40   493
> 54   41   489
> 54   42   431
> 11   1002
> 12   1222
> 13   1198
> 54   39   1876
> 54   40   1565
> 54   41   1455
> 54   42   1427
> 11   1002
> 12   1246
> 13   1153
> 54   39   1813
> 54   40   1490
> 54   41   1518
> 54   42   1486
> 11   1002
> 12   1229
> 13   1142
> 54   39   1797
> 54   40   1517
> 54   41   1527
> 54   42   1514"),header=TRUE)
> 
> dat$seq <- ifelse(dat$ISEG==1 & dat$IRCH==1, 1, 0)
> tmp <- diff(dat[dat$seq==1,]$val)!=0
> dat$idx <- 0
> dat[dat$seq==1,][c(TRUE,tmp),]$idx <- 1
> dat$ts <- cumsum(dat$idx)
> 
> At this point, I'd like to add one more column called "iter" that counts up
> by 1 based on "seq", but within each "ts".  So, the result would look like
> this (undoubtedly this is a simple problem with something like ddply, but
> I've been unable to construct the R for it):

> dat$iter2 <- ave(dat$seq, dat$ts,FUN=cumsum)
> dat
   ISEG IRCH  val seq idx ts iter iter2
1 11  265   1   1  1  1_1 1
2 12  260   0   0  1  1_1 1
3 13  234   0   0  1  1_1 1
454   39  467   0   0  1  1_1 1
554   40  468   0   0  1  1_1 1
654   41  460   0   0  1  1_1 1
754   42  489   0   0  1  1_1 1
8 11  265   1   0  1  1_2 2
9 12  276   0   0  1  1_2 2
1013  217   0   0  1  1_2 2
11   54   39  456   0   0  1  1_2 2
12   54   40  507   0   0  1  1_2 2
13   54   41  483   0   0  1  1_2 2
14   54   42  457   0   0  1  1_2 2
1511  265   1   0  1  1_3 3
1612  287   0   0  1  1_3 3
1713  224   0   0  1  1_3 3
18   54   39  473   0   0  1  1_3 3
19   54   40  502   0   0  1  1_3 3
20   54   41  497   0   0  1  1_3 3
21   54   42  447   0   0  1  1_3 3
2211  230   1   1  2  2_4 1
2312  251   0   0  2  2_4 1
snipped->

-- 
David
> 
> dat
> ISEG IRCH  val seq idx ts iter
>11  265   1   1  11
>12  260   0   0  11
>13  234   0   0  11
>   54   39  467   0   0  11
>   54   40  468   0   0  11
>   54   41  460   0   0  11
>   54   42  489   0   0  11
>11  265   1   0  12
>12  276   0   0  12
>13  217   0   0  12
>   54   39  456   0   0  12
>   54   40  507   0   0  12
>   54   41  483   0   0  12
>   54   42  457   0   0  12
>11  265   1   0  13
>12  287   0   0  13
>13  224   0   0  13
>   54   39  473   0   0  13
>   54   40  502   0   0  13
>   54   41  497   0   0  13
>   54   42  447   0   0  13
>11  230   1   1  21
>12  251   0   0  21
>13  199   0   0  21
>   54   39  439   0   0  21
>   54   40  474   0   0  21
>   54   41  477   0   0  21
>   54   42  413   0   0  21
>11  230   1   0  22
>12  262   0   0  22
>13  217   0   0  22
>   54   39  455   0   0  22
>   54   40  493   0   0  22
>   54   41  489   0   0  22
>   54   42  431   0   0  22
>11 1002   1   1  31
>12 1222   0   0  31
>13 1198   0   0  31
>   54   39 1876   0   0  31
>   54   40 1565   0   0  31
>   54   41 1455   0   0  31
>   54   42 1427   0   0  31
>11 1002   1   0  32
>12 1246   0   0  32
>13 1153   0   0  32
>   54   39 1813   0   0  32
>   54   40 1490   0   0  32
>   54   41 1518   0   0  32
>   54   42 1486   0   0  32
>11 1002   1   0  33
>12 1229   0   0  33
>13 1142   0   0  33
>   54   39 1797   0   0  33
>   54   40 1517   0   0  33
>   54   41 1527   0   0  33
>   54   42 1514   0   0  33
> 
>   [[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.

David Winsemius
Alameda, CA, USA

__

Re: [R] Cut breaks in descending order

2015-04-03 Thread David Winsemius

On Apr 3, 2015, at 5:09 AM, Wing Keong Lew wrote:

> Hi,
> 
> Is it a requirement to provide the break intervals of the cut function in 
> ascending order?

Apparently not. I get teh sam splits even with random permutations. It is 
apparently a "requirement" to make sure you labels match the sorted order of 
the breaks.

The findInterval function does require that its `vec` argument be 
non-decreasing, but I do not see a discussion of break order in the help page. 
Looking at cut.default the first think it does to the breaks is sort them.

  #... snipped code that deals with length(breaks)==1
   else nb <- length(breaks <- sort.int(as.double(breaks)))

-- 
David.

> The help documentation for cut didn't specify this but the labels returned 
> are reversed if I indicate the break intervals in a descending order. Here is 
> an example
> 
> tbl<-data.frame(x=c(0:10))
> tbl$ascending<-cut(tbl$x, breaks=c(0,3,5,9,99), 
> labels=c('<3','4-5','6-9','>9'), include.lowest=T)
> tbl$descending<-cut(tbl$x, breaks=c(99,9,5,3,0), 
> labels=c('>9','6-8','4-5','<3'), include.lowest=T)
> tbl
> x ascending descending
> 1   0<3>9
> 2   1<3>9
> 3   2<3>9
> 4   3<3>9
> 5   4   4-5   6-8
> 6   5   4-5   6-8
> 7   6   6-9   4-5
> 8   7   6-9   4-5
> 9   8   6-9   4-5
> 10  9   6-9   4-5
> 11 10>9<3
> 
> Appreciate any guidance on this.
> Regards
> Wing Keong
> 
> 
> __
> 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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[R] Cut breaks in descending order

2015-04-03 Thread Wing Keong Lew
Hi,

Is it a requirement to provide the break intervals of the cut function in 
ascending order? The help documentation for cut didn't specify this but the 
labels returned are reversed if I indicate the break intervals in a descending 
order. Here is an example

tbl<-data.frame(x=c(0:10))
tbl$ascending<-cut(tbl$x, breaks=c(0,3,5,9,99), 
labels=c('<3','4-5','6-9','>9'), include.lowest=T)
tbl$descending<-cut(tbl$x, breaks=c(99,9,5,3,0), 
labels=c('>9','6-8','4-5','<3'), include.lowest=T)
tbl
    x ascending descending
1   0        <3        >9
2   1        <3        >9
3   2        <3        >9
4   3        <3        >9
5   4       4-5       6-8
6   5       4-5       6-8
7   6       6-9       4-5
8   7       6-9       4-5
9   8       6-9       4-5
10  9       6-9       4-5
11 10        >9        <3

Appreciate any guidance on this.
Regards
Wing Keong

  
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[R] question on waveletcomp plot image

2015-04-03 Thread Sudheer Joseph
Dear R experts,
 I have used waveletcomp package of R  and was
trying to get the dates formatted as month & year but get below error while
trying it with the example provided  in
http://www.hs-stat.com/projects/WaveletComp/WaveletComp_guided_tour.pdf
Kindly help me with the dateformat so that I can get %Y-%m as xaxis label
data(FXtrade.transactions)
my.data.a = FXtrade.transactions[FXtrade.transactions$active == T, ]
my.w.a = analyze.wavelet(
+ my.data.a, "transactions",
+ loess.span = 0.0, # no detrending required
+ dt = 1/(12*24),
+ # one day has 12*24 5-minute time slots
+ dj = 1/250,
+ # resolution along period axis
+ lowerPeriod = 1/8, # lowest period of interest: 3 hours
+ make.pval = T,
+ # draws white lines indicating significance
+ n.sim = 10)


wt.image(my.w.a, n.levels = 250, periodlab = "periods (days)",
+  legend.params = list(lab = "wavelet power levels"),
+  show.date = T, date.format = "%Y %m", timelab = "")
Error in plot.window(...) : need finite 'xlim' values
In addition: Warning messages:
1: In min(x) : no non-missing arguments to min; returning Inf
2: In max(x) : no non-missing arguments to max; returning -In
with best regards

Sudheer

**
Dr. Sudheer Joseph

Scientist,

INCOIS, MoES, Govt. of India.
"OCEAN VALLEY" , Pragathi Nagar (BO), Nizampet SO,  Telangana, India. PIN-
500 090.
Tel:+91-9440832534(Mobile) Tel:+91-40-23886047(O),Fax:+91-40-23892910(O)
E-mail: sjo.in...@gmail.com;  s...@incois.gov.in.
---* 
"The ultimate measure of a man is not where he stands in moments of
comfort and convenience, but where he stands at times of challenge and
controversy."
Martin Luther King, Jr.
***

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Re: [R] double-axis labels function of each other

2015-04-03 Thread John Kane
Notepad should be fine but you will, I think, still have to correct the 
apostrophes if you are copying from Word.  

Any code written in Notepad should be fine.  

There are several dedicated editors or IDE's for use with R that you might want 
to look into.  RStudio, Tinn-R and EMACs with ESS are some that come to mind.  
Among other things, they usually offer code highlighting which can be very 
useful for finding where there are typos, missing commas, and so on.

I run Linux (Ubuntu) and find gedit with an R plug-in to be very good also.


John Kane
Kingston ON Canada


> -Original Message-
> From: hill0...@umn.edu
> Sent: Thu, 2 Apr 2015 23:59:17 -0700 (PDT)
> To: r-help@r-project.org
> Subject: Re: [R] double-axis labels function of each other
> 
> Yes, I keep a copy in MS Word.
> Would Notepad be OK?
> I need sleep now, will work tomorrow.
> 
> 
> 
> 
> --
> View this message in context:
> http://r.789695.n4.nabble.com/double-axis-labels-function-of-each-other-tp4705457p4705463.html
> Sent from the R help mailing list archive at Nabble.com.
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


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Use Password manager! It stores your passwords & protects your account.

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Re: [R] idiom for constructing data frame

2015-04-03 Thread peter dalgaard

> On 03 Apr 2015, at 16:46 , William Dunlap  wrote:
> 
> > 
> > df <- as.data.frame(rep(list(rep(NA_real_, 10)),3))
> > names(df) <- names
> 
> As a matter of personal style (and functional programming
> sensibility), I prefer not to make named objects and then modify them.
> Also, the names coming out of that as.data.frame call are exceedingly
> ugly and I'd rather not generate them at all.
> 

Ah, yes, I missed the generation of intermediate names. You can name the list 
before as.data.frame, though:

l <- rep(list(rep(NA_real_, 10)),3)
names(l) <- names
as.data.frame(l)

or as a one-liner:

as.data.frame(structure(rep(list(rep(NA_real_, 10)), 3) , .Names=names))

-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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Re: [R] WEIBULL or EXPONENTIAL?

2015-04-03 Thread JLucke
As a start you can use an exploratory approach.  Standard survival 
analysis texts show you how to use a log-log plot to assess whether a 
distribution is Weibull.   Of course, the exponential is a special case of 
the Weibull. 



CHIRIBOGA Xavier  
Sent by: "R-help" 
04/03/2015 10:33 AM

To
"r-help@r-project.org" , 
cc

Subject
[R] WEIBULL or EXPONENTIAL?






Dear members,



I am doing a survival analysis wiith the function coxph...however I am 
wondering how can I know if my data follows a EXPONENTIAL or WEIBULL 
distribution?

I have 3 censored datum. Using R studio.



Thanks for the suggestions,



Xavier

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Re: [R] idiom for constructing data frame

2015-04-03 Thread William Dunlap
> but wouldn't it be more to the point to do
>
> df <- as.data.frame(rep(list(rep(NA_real_, 10)),3))
> names(df) <- names

As a matter of personal style (and functional programming
sensibility), I prefer not to make named objects and then modify them.
Also, the names coming out of that as.data.frame call are exceedingly
ugly and I'd rather not generate them at all.

Also adding the names after calling data.frame means can give
different results than passing them into data.frame(), which can
mangle nonsyntactic names like "Second Name" into "Second.Name".
It is often preferable, but it is different.



Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Fri, Apr 3, 2015 at 5:51 AM, peter dalgaard  wrote:

>
> > On 31 Mar 2015, at 20:55 , William Dunlap  wrote:
> >
> > You can use structure() to attach the names to a list that is input to
> > data.frame.
> > E.g.,
> >
> > dfNames <- c("First", "Second Name")
> > data.frame(lapply(structure(dfNames, names=dfNames),
> > function(name)rep(NA_real_, 5)))
> >
>
> Yes, I cooked up something similar:
>
> names <- c("foo","bar","baz")
> names(names) <- names # confuse 'em
> as.data.frame(lapply(names, function(x) rep(NA_real_,10)))
>
> but wouldn't it be more to the point to do
>
> df <- as.data.frame(rep(list(rep(NA_real_, 10)),3))
> names(df) <- names
>
> ?
>
> The lapply() approach could be generalized to a vector of column classes,
> though.
>
> A general solution looks impracticable; once you start considering how to
> specify factor columns with each their own level set, things get a bit out
> of hand.
>
> -pd
>
> >
> > Bill Dunlap
> > TIBCO Software
> > wdunlap tibco.com
> >
> > On Tue, Mar 31, 2015 at 11:37 AM, Sarah Goslee 
> > wrote:
> >
> >> Hi,
> >>
> >> Duncan Murdoch suggested:
> >>
> >>> The matrix() function has a dimnames argument, so you could do this:
> >>>
> >>> names <- c("strat", "id", "pid")
> >>> data.frame(matrix(NA, nrow=10, ncol=3, dimnames=list(NULL, names)))
> >>
> >> That's a definite improvement, thanks. But no way to skip matrix()? It
> >> just seems unRlike, although since it's only full of NA values there
> >> are no coercion issues with column types or anything, so it doesn't
> >> hurt. It's just inelegant. :)
> >>
> >> Sarah
> >> --
> >> Sarah Goslee
> >> http://www.functionaldiversity.org
> >>
> >> __
> >> 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.
>
> --
> Peter Dalgaard, Professor,
> Center for Statistics, Copenhagen Business School
> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
> Phone: (+45)38153501
> Email: pd@cbs.dk  Priv: pda...@gmail.com
>
>
>
>
>
>
>
>
>

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[R] WEIBULL or EXPONENTIAL?

2015-04-03 Thread CHIRIBOGA Xavier
Dear members,



I am doing a survival analysis wiith the function coxph...however I am 
wondering how can I know if my data follows a EXPONENTIAL or WEIBULL 
distribution?

I have 3 censored datum. Using R studio.



Thanks for the suggestions,



Xavier

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


Re: [R] Package build system adds line break in DESCRIPTION URL

2015-04-03 Thread Ben Bolker
Daniel Lewandowski  nextpagesoft.net> writes:

> 
> Has anybody noticed that if field URL in DESCRIPTION contains a uri with 
> 66 or more characters, then file DESCRIPTION in the resulting package 
> includes a line break at the beginning?
> 
> So this (source DESCRIPTION):
> 
> URL: 
> http://ecdc.europa.eu/en/data-tools/
   seroincidence-calculator-tool/Pages/default.aspx

> 
> becomes (again file DESCRIPTION, but inside the package)
> 
> URL:
> http://ecdc.europa.eu/en/data-tools/seroincidence-calculator-tool/
   Pages/default.aspx
> 
> This has been tested with R on Windows 8.1 (devel 01/04/2015 and 3.1.3) 
> and Linux Mint (3.1.3). It has many sad implications including not 
> acceptance of such packages to CRAN.
> 

  (links line-broken above to make gmane happy).

  Haven't tested, but seems it would not be *too* hard to trace
through the code to see what's happening in the package-building
process.  Two thoughts:

 (1) as a workaround, could you use a URL-shortener such as tinyurl?
tinyurl allows you to specify a somewhat meaningful name for the
shortened URL, e.g. http://tinyurl.com/reproducible-000

 (2) this feels like it is more suitable for r-de...@r-project.org

  Ben Bolker

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Re: [R] idiom for constructing data frame

2015-04-03 Thread Hadley Wickham
On Tue, Mar 31, 2015 at 6:42 PM, Sarah Goslee  wrote:
> On Tue, Mar 31, 2015 at 6:35 PM, Richard M. Heiberger  wrote:
>> I got rid of the extra column.
>>
>> data.frame(r=seq(8), foo=NA, bar=NA, row.names="r")
>
> Brilliant!
>
> After much fussing, including a disturbing detour into nested lapply
> statements from which I barely emerged with my sanity (arguable, I
> suppose), here is a one-liner that creates a data frame of arbitrary
> number of rows given an existing data frame as template for column
> number and name:
>
>
> n <- 8
> df1 <- data.frame(A=runif(9), B=runif(9))
>
> do.call(data.frame, setNames(c(list(seq(n), "r"), as.list(rep(NA,
> ncol(df1, c("r", "row.names", colnames(df1
>
> It's not elegant, but it is fairly R-ish. I should probably stop
> hunting for an elegant solution now.

Given a template df, you can create a new df with subsetting:

df2 <- df1[rep(NA_real_, 8), ]
rownames(df2) <- NULL
df2

This has the added benefit of preserving the types.

Hadley

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

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Re: [R] idiom for constructing data frame

2015-04-03 Thread peter dalgaard

> On 31 Mar 2015, at 20:55 , William Dunlap  wrote:
> 
> You can use structure() to attach the names to a list that is input to
> data.frame.
> E.g.,
> 
> dfNames <- c("First", "Second Name")
> data.frame(lapply(structure(dfNames, names=dfNames),
> function(name)rep(NA_real_, 5)))
> 

Yes, I cooked up something similar:

names <- c("foo","bar","baz")
names(names) <- names # confuse 'em
as.data.frame(lapply(names, function(x) rep(NA_real_,10)))

but wouldn't it be more to the point to do

df <- as.data.frame(rep(list(rep(NA_real_, 10)),3))
names(df) <- names

?

The lapply() approach could be generalized to a vector of column classes, 
though. 

A general solution looks impracticable; once you start considering how to 
specify factor columns with each their own level set, things get a bit out of 
hand. 

-pd

> 
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
> 
> On Tue, Mar 31, 2015 at 11:37 AM, Sarah Goslee 
> wrote:
> 
>> Hi,
>> 
>> Duncan Murdoch suggested:
>> 
>>> The matrix() function has a dimnames argument, so you could do this:
>>> 
>>> names <- c("strat", "id", "pid")
>>> data.frame(matrix(NA, nrow=10, ncol=3, dimnames=list(NULL, names)))
>> 
>> That's a definite improvement, thanks. But no way to skip matrix()? It
>> just seems unRlike, although since it's only full of NA values there
>> are no coercion issues with column types or anything, so it doesn't
>> hurt. It's just inelegant. :)
>> 
>> Sarah
>> --
>> Sarah Goslee
>> http://www.functionaldiversity.org
>> 
>> __
>> 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.

-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk  Priv: pda...@gmail.com

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[R] applying cumsum within groups

2015-04-03 Thread Morway, Eric
This small example will be applied to a problem with 1.4e6 lines of data.
First, here is the dataset and a few lines of R script, followed by an
explanation of what I'd like to get:

dat <- read.table(textConnection("ISEG  IRCH  val
 11   265
 12   260
 13   234
54   39   467
54   40   468
54   41   460
54   42   489
 11   265
 12   276
 13   217
54   39   456
54   40   507
54   41   483
54   42   457
 11   265
 12   287
 13   224
54   39   473
54   40   502
54   41   497
54   42   447
 11   230
 12   251
 13   199
54   39   439
54   40   474
54   41   477
54   42   413
 11   230
 12   262
 13   217
54   39   455
54   40   493
54   41   489
54   42   431
 11   1002
 12   1222
 13   1198
54   39   1876
54   40   1565
54   41   1455
54   42   1427
 11   1002
 12   1246
 13   1153
54   39   1813
54   40   1490
54   41   1518
54   42   1486
 11   1002
 12   1229
 13   1142
54   39   1797
54   40   1517
54   41   1527
54   42   1514"),header=TRUE)

dat$seq <- ifelse(dat$ISEG==1 & dat$IRCH==1, 1, 0)
tmp <- diff(dat[dat$seq==1,]$val)!=0
dat$idx <- 0
dat[dat$seq==1,][c(TRUE,tmp),]$idx <- 1
dat$ts <- cumsum(dat$idx)

At this point, I'd like to add one more column called "iter" that counts up
by 1 based on "seq", but within each "ts".  So, the result would look like
this (undoubtedly this is a simple problem with something like ddply, but
I've been unable to construct the R for it):

dat
 ISEG IRCH  val seq idx ts iter
11  265   1   1  11
12  260   0   0  11
13  234   0   0  11
   54   39  467   0   0  11
   54   40  468   0   0  11
   54   41  460   0   0  11
   54   42  489   0   0  11
11  265   1   0  12
12  276   0   0  12
13  217   0   0  12
   54   39  456   0   0  12
   54   40  507   0   0  12
   54   41  483   0   0  12
   54   42  457   0   0  12
11  265   1   0  13
12  287   0   0  13
13  224   0   0  13
   54   39  473   0   0  13
   54   40  502   0   0  13
   54   41  497   0   0  13
   54   42  447   0   0  13
11  230   1   1  21
12  251   0   0  21
13  199   0   0  21
   54   39  439   0   0  21
   54   40  474   0   0  21
   54   41  477   0   0  21
   54   42  413   0   0  21
11  230   1   0  22
12  262   0   0  22
13  217   0   0  22
   54   39  455   0   0  22
   54   40  493   0   0  22
   54   41  489   0   0  22
   54   42  431   0   0  22
11 1002   1   1  31
12 1222   0   0  31
13 1198   0   0  31
   54   39 1876   0   0  31
   54   40 1565   0   0  31
   54   41 1455   0   0  31
   54   42 1427   0   0  31
11 1002   1   0  32
12 1246   0   0  32
13 1153   0   0  32
   54   39 1813   0   0  32
   54   40 1490   0   0  32
   54   41 1518   0   0  32
   54   42 1486   0   0  32
11 1002   1   0  33
12 1229   0   0  33
13 1142   0   0  33
   54   39 1797   0   0  33
   54   40 1517   0   0  33
   54   41 1527   0   0  33
   54   42 1514   0   0  33

[[alternative HTML version deleted]]

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[R] RInside

2015-04-03 Thread Michel
help

Hello
I'm newbee with R and RInside
My question is about stderr in R. Is there a way to collect R stderr in C++
program embedding R Thanks in advance

Michel


---
L'absence de virus dans ce courrier électronique a été vérifiée par le logiciel 
antivirus Avast.

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[R] Package build system adds line break in DESCRIPTION URL

2015-04-03 Thread Daniel Lewandowski
Has anybody noticed that if field URL in DESCRIPTION contains a uri with 
66 or more characters, then file DESCRIPTION in the resulting package 
includes a line break at the beginning?


So this (source DESCRIPTION):

URL: 
http://ecdc.europa.eu/en/data-tools/seroincidence-calculator-tool/Pages/default.aspx


becomes (again file DESCRIPTION, but inside the package)

URL:
http://ecdc.europa.eu/en/data-tools/seroincidence-calculator-tool/Pages/default.aspx

This has been tested with R on Windows 8.1 (devel 01/04/2015 and 3.1.3) 
and Linux Mint (3.1.3). It has many sad implications including not 
acceptance of such packages to CRAN.


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Re: [R] double-axis labels function of each other

2015-04-03 Thread Hurr
Yes, I keep a copy in MS Word. 
Would Notepad be OK?
I need sleep now, will work tomorrow.




--
View this message in context: 
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Sent from the R help mailing list archive at Nabble.com.

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