date:20120301



On Mar 1, 2012, at 12:52 AM, Bernardo Powaga wrote:


Hello R users,

I would like to fit a generalized linear model for the binomial  
family with some non standard link functions. For instance, this is  
the Aranda-Ordaz link:


η = ln( ( (1 - π)^-α - 1 )/α)

I know how to define a new link function to use with glm(), but I my  
problem is that I have an extra parameter to estimate and I have no  
clue how to incorporate that. Is there any way to tell glm() to add  
this parameter in the estimation or do I have to write my own  
estimator with optim()?


Bernardo;

Let me introduce you to a new friend. You very own link to a search  
engine to r-help. That way you won't need to repeat questions that  
have been asked answered and even incorported into announced packages:


http://www.googlesyndicatedsearch.com/u/newcastlemaths?q=Aranda-Ordaz+linksa=Google+Search

The Newcastle site is just one of several that you could have used.  
MarkMail is another popular one and the venerable Baron maintained  
finzi site to which RSiteSearch has in the past directed you to. It,  
however, no longer is being maintained for rhelp or rdevel searches.


Enjoy the company of your new friends.

--

David Winsemius, MD
West Hartford, CT

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Re: [R] Cleaning up messy Excel data

But there are some important reasons to use Excel.  In my work there
are a lot of people that I have to send the equivalent of a data.frame
to who want to look at the data and possibly slice/dice the data
differently and then send back to me updates.  These folks do not know
how to use R, but do have Microsoft Office installed on their
computers and know how to use the different products.

I have been very successful in conveying what I am doing for them by
communicating via Excel spreadsheets.  It is also an important medium
in dealing with some international companies who provide data via
Excel and expect responses back via Excel.

When dealing with data in a tabular form, Excel does provide a way for
a majority of the people I work with to understand the data.  Yes,
there are problems with some of the ways that people use Excel, and
yes I have had to invest time in scrubbing some of the data that I get
from them, but if I did not, then I would probably not have a job
working for them.  I use R exclusively for the analysis that I do, but
find it convenient to use Excel to provide a communication mechanism
to the majority of the non-R users that I have to deal with.  It is a
convenient work-around because I would never get them to invest the
time to learn R.

So in the real world these is a need to Excel and we are not going to
cause it to go away; we have to learn how to live with it, and from my
standpoint, it has definitely benefited me in being able to
communicate with my users and continuing to provide them with results
that they are happy with.  They refer to letting me work my magic on
the data; all they know is they see the result via Excel and in the
background R is doing the heavy lifting that they do not have to know
about.

On Wed, Feb 29, 2012 at 4:41 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
 On 01/03/12 04:43, John Kane wrote:

 (mydata- as.factor(c(1,2,3, 2, 5, 2)))
 str(mydata)

 newdata- as.character(mydata)

 newdata[newdata==2]- 0
 newdata- as.numeric(newdata)
 str(newdata)

 We really need to keep Excel (and other spreadsheets) out of peoples
 hands.


 Amen, bro'!!!

    cheers,

        Rolf Turner

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-- 
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Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] convert list to text file

Perhaps something like

sink(outtext.txt)
lapply(LIST, print)
sink()

You could replace print with cat and friends if you wanted more
detailed control over the look of the output.

Michael

On Thu, Mar 1, 2012 at 5:28 AM,  t.galesl...@ebh.umcn.nl wrote:
 Dear R users,

 Is it possible to write the following list to a text-file?

 List:

 [[1]]
 [1] 500

 [[2]]
 [1] 1

 [[3]]
    [,1] [,2] [,3] [,4] [,5]
 FID    1    2    3    4    5
 Var    2    0    2    1    1

 I would like to have the textfile look like this:

 500
 1
 FID 1 2 3 4 5
 Var 2 0 2 1 1

 Thank you very much in advance for your help!

 Kind regards,

 Tessel Galesloot
 Department of Epidemiology, Biostatistics and HTA (133)
 Radboud University Nijmegen Medical Centre



 Het UMC St Radboud staat geregistreerd bij de Kamer van Koophandel in het 
 handelsregister onder nummer 41055629.
 The Radboud University Nijmegen Medical Centre is listed in the Commercial 
 Register of the Chamber of Commerce under file number 41055629.

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Re: [R] Re : Create a function automatically from lm formula and coefficients?

2012-03-01 Thread Frank Harrell

yes, e.g.
require(rms)
f - ols( ) # ols is a wrapper for lm
g - Function(f)
g(age=30) # get predicted mean at age=30 and defaults for other variables
(medians  modes)
Frank



Pascal Oettli-2 wrote
 
 Hi Keith,
 
 Do you mean as predict.lm can do?
 
 Regards,
 Pascal
 
 
 
 De : Keith Weintraub lt;kw1958@gt;
 À : r-help@ 
 Envoyé le : Jeudi 1 mars 2012 11h41
 Objet : [R] Create a function automatically from lm formula and
 coefficients?
 
 I hope the subject says it all.
 
 I want to be able to use an lm object and the associated coefficients to
 create  function that can produce expected y values given inputs.
 
 Thanks,
 KW
 
 --
 
 
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-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Re: [R] linear regression by column

2012-03-01 Thread Filoche

Thank everyone for your help.

Problem solved. I'm getting more used with vectorization with your help,

Regards,
Phil

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[R] 'break' function in loop

2012-03-01 Thread Tsai, Pei-Chien

Dear R helpers,

I have some difficulties in using 'break' function with loop, and the 
followings are my script. What I try to do is (1) permute 'or' first; (2) doing 
t-test if this 'or' pass criteria 1 (k=1); (3) end the loop when I get 10 
permutations; (4) redo everything again but this time use criteria 2 (k=2) (I 
have more criteria 1:n).

Somehow using my script, the final dataset (results1) only contains the result 
from criteria 1 (twice!) but not the result from criteria 2. I guess probably I 
put the break function under the wrong loop but I cannot fix it. Sorry if the 
whole script looks quite messy, I will be very appreciate if someone can help 
me fix this problem or probably give me some advices to write it in a smart way.


My attempt:

controlall - rbeta(1,1.5,6)
caseall - rbeta(1,1.6,6)

results - NULL
results1 - NULL
or -vector(list,length=10)
criteria - matrix(data=c(1.05,1.15,1.15,1.25),ncol=2,nrow=2)

for (k in 1:2) {

for (i in 1:1000)
 {
 control - sample(controlall,100)
 case - sample(caseall,100)
 or[i] - 
round(mean(case)*(1-mean(control))/(mean(control)*(1-mean(case))),digit=2)

 if (or[i]criteria[k,2]or[i]=criteria[k,1])
 {
 group - c(rep(1,100),rep(0,100))
 value - c(case,control)
 ttest - (t.test(case,control,alternative=two.sided,paired=F))$p.value
 all - c(or[i],ttest)
 results - rbind(results,all)
 }
 ifelse(nrow(results)==5,break,1)
 }
results1 - rbind(results1,results)
 }


Thank you so much in advance,
Amber

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[R] Execution of Rprofile.site

2012-03-01 Thread Ross Bowden

Hi everyone. I have recently installed R 2.14.1 on my 64-bit Windows 7 
laptop. I am attempting to include some favourite functions in the 
Rprofile.site file to run at R start-up as I did with my previous 32-bit 
XP machine. I have edited the Rprofile.site file in C:\Program 
Files\R\R-2.14.1\etc\ but the added code doesn't seem to be executed 
when I run R, whether through Tinn-R or independently. Any ideas please 
on what's going wrong? I've searched my laptop and there is only one 
Rprofile.site file. Also Sys.getenv(R_HOME) returns: [1] 
C:/PROGRA~1/R/R-214~1.1 so R should be able to find the Rprofile.site 
file. Note that Rprofile.site has been modified by Tinn-R.


Many thanks

Ross Bowden
PhD Candidate, Maths Department, Murdoch University.

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Re: [R] 'break' function in loop

I have never seen 'break' used in an 'ifelse'; you probably meant to
use an 'if' statement there.

On Thu, Mar 1, 2012 at 8:24 AM, Tsai, Pei-Chien
pei-chien.t...@kcl.ac.uk wrote:
 Dear R helpers,

 I have some difficulties in using 'break' function with loop, and the 
 followings are my script. What I try to do is (1) permute 'or' first; (2) 
 doing t-test if this 'or' pass criteria 1 (k=1); (3) end the loop when I get 
 10 permutations; (4) redo everything again but this time use criteria 2 (k=2) 
 (I have more criteria 1:n).

 Somehow using my script, the final dataset (results1) only contains the 
 result from criteria 1 (twice!) but not the result from criteria 2. I guess 
 probably I put the break function under the wrong loop but I cannot fix it. 
 Sorry if the whole script looks quite messy, I will be very appreciate if 
 someone can help me fix this problem or probably give me some advices to 
 write it in a smart way.


 My attempt:

 controlall - rbeta(1,1.5,6)
 caseall - rbeta(1,1.6,6)

 results - NULL
 results1 - NULL
 or -vector(list,length=10)
 criteria - matrix(data=c(1.05,1.15,1.15,1.25),ncol=2,nrow=2)

 for (k in 1:2) {

 for (i in 1:1000)
     {
         control - sample(controlall,100)
         case - sample(caseall,100)
         or[i] - 
 round(mean(case)*(1-mean(control))/(mean(control)*(1-mean(case))),digit=2)

  if (or[i]criteria[k,2]or[i]=criteria[k,1])
  {
     group - c(rep(1,100),rep(0,100))
     value - c(case,control)
     ttest - (t.test(case,control,alternative=two.sided,paired=F))$p.value
     all - c(or[i],ttest)
     results - rbind(results,all)
  }
         ifelse(nrow(results)==5,break,1)
  }
    results1 - rbind(results1,results)
  }


 Thank you so much in advance,
 Amber

        [[alternative HTML version deleted]]

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-- 
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Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] update.formula has 512 char buffer?

2012-03-01 Thread Duncan Murdoch


On 12-02-28 2:11 PM, Chris Hane wrote:

Hello,

I am trying to paste together a formula to use in the mob function of
party. This means the formula will be of the form y ~ x1+ ...+xM | z1+..zN.

I am doing some preliminary fits of y ~ x1+ ...+xM, then want to add the
conditional part of the equation using update().

Here's the test code:
var1- 1:78
x1- paste(x, var1, sep=)
f1- paste(f, var1[1:10], sep=)

# use first 77 variables
fmla- as.formula( paste(y ~ , paste(x1[1:77], collapse= + , sep=),
sep=))
fmla2- update(fmla, paste(. ~ . | , paste(f1, collapse=  + ), sep=))

# CHANGE x to all 78 variables
fmla- as.formula( paste(y ~ , paste(x1, collapse= + , sep=),
sep=))
fmla2- update(fmla, paste(. ~ . | , paste(f1, collapse=  + ), sep=))

I have run this in Windows and Linux (64 bit) and both fail when using all
78 terms (and anything more than 78 terms). The error message contains
Error in parse(text = x) : :1:514: unexpected ')'.

Changing the length of the names of the x variables will break the update()
with fewer variables, but always with an error referring to just more than
512 characters.  There is nothing special about 77 or 78 variables here; I
want to do this with hundreds of variables.

Is there a workaround to this?


I've just committed code to R-devel (to become 2.15.0 at the end of the 
month) to remove the truncation.  I don't know any workaround that will 
address this in earlier versions, other than don't use very long formulas.


It also affects as.character() applied to a formula object, and some 
other cases where very long language objects are displayed.


Please test R-devel or one of the alpha/beta versions once they are 
built; this fix went into revision 58544.


Duncan Murdoch



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Re: [R] 'break' function in loop

2012-03-01 Thread Duncan Murdoch


On 12-03-01 8:24 AM, Tsai, Pei-Chien wrote:

Dear R helpers,

I have some difficulties in using 'break' function with loop, and the 
followings are my script. What I try to do is (1) permute 'or' first; (2) doing 
t-test if this 'or' pass criteria 1 (k=1); (3) end the loop when I get 10 
permutations; (4) redo everything again but this time use criteria 2 (k=2) (I 
have more criteria 1:n).

Somehow using my script, the final dataset (results1) only contains the result 
from criteria 1 (twice!) but not the result from criteria 2. I guess probably I 
put the break function under the wrong loop but I cannot fix it. Sorry if the 
whole script looks quite messy, I will be very appreciate if someone can help 
me fix this problem or probably give me some advices to write it in a smart way.


My attempt:

controlall- rbeta(1,1.5,6)
caseall- rbeta(1,1.6,6)

results- NULL
results1- NULL
or-vector(list,length=10)
criteria- matrix(data=c(1.05,1.15,1.15,1.25),ncol=2,nrow=2)

for (k in 1:2) {

for (i in 1:1000)
  {
  control- sample(controlall,100)
  case- sample(caseall,100)
  or[i]- 
round(mean(case)*(1-mean(control))/(mean(control)*(1-mean(case))),digit=2)

  if (or[i]criteria[k,2]or[i]=criteria[k,1])
  {
  group- c(rep(1,100),rep(0,100))
  value- c(case,control)
  ttest- (t.test(case,control,alternative=two.sided,paired=F))$p.value
  all- c(or[i],ttest)
  results- rbind(results,all)
  }
  ifelse(nrow(results)==5,break,1)


Jim was right, this line doesn't make sense.  Use

if(nrow(results)==5) break

instead.

Duncan Murdoch


  }
 results1- rbind(results1,results)
  }


Thank you so much in advance,
Amber

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Re: [R] Error in solve.default(res$hessian * n.used) :Lapack routine dgesv: system is exactly singular

2012-03-01 Thread VictorDelgado

Vinicius,


Vinicius Magalhães wrote
 
 
 + n - length(*x.ts*)
 + for (p in 0:maxord[1]) for (d in 0:maxord[2]) for (q in 0:maxord[3])
 +  for (P in 0:maxord[4]) for (D in 0:maxord[5]) for (Q in 0:maxord[6])
 +  {
 + fit - arima(*x.ts*, order=c(p,d,q),
 + seas = list(order=c(P,D,Q),
 + frequency(x.ts)), method = CSS)
 + fit.aic - -2 * fit$loglik + (log(n) + 1) * length(fit$coef)
 
 

I think you are using 'x.ts' instead 's.ts'. Is it correct?! 
Maybe you can also put here objects likely 'x', 'x.ts' to help us give
feedback to you routine.

Best regards. 


-
Victor Delgado
cedeplar.ufmg.br P.H.D. student
www.fjp.mg.gov.br reseacher
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[R] barplots of several variables with different number of categories

2012-03-01 Thread jon waterhouse

If I have two factors, v1 and v2 and I want to have a stacked bar graph of
the two variables side by side I could do

barplot(cbind(table(v1),table(v2)))

if v1 and v2 have the same number of categories.

If they don't have the same number of categories this won't work.

I'm sure there's a simple solution?

Thanks,

Jon

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Re: [R] 'break' function in loop

Hello,


jholtman wrote
 
 I have never seen 'break' used in an 'ifelse'; you probably meant to
 use an 'if' statement there.
 
 On Thu, Mar 1, 2012 at 8:24 AM, Tsai, Pei-Chien
 lt;pei-chien.tsai@.acgt; wrote:
 Dear R helpers,

 I have some difficulties in using 'break' function with loop, and the
 followings are my script. What I try to do is (1) permute 'or' first; (2)
 doing t-test if this 'or' pass criteria 1 (k=1); (3) end the loop when I
 get 10 permutations; (4) redo everything again but this time use criteria
 2 (k=2) (I have more criteria 1:n).

 Somehow using my script, the final dataset (results1) only contains the
 result from criteria 1 (twice!) but not the result from criteria 2. I
 guess probably I put the break function under the wrong loop but I cannot
 fix it. Sorry if the whole script looks quite messy, I will be very
 appreciate if someone can help me fix this problem or probably give me
 some advices to write it in a smart way.


 My attempt:

 controlall - rbeta(1,1.5,6)
 caseall - rbeta(1,1.6,6)

 results - NULL
 results1 - NULL
 or -vector(list,length=10)
 criteria - matrix(data=c(1.05,1.15,1.15,1.25),ncol=2,nrow=2)

 for (k in 1:2) {

 for (i in 1:1000)
     {
         control - sample(controlall,100)
         case - sample(caseall,100)
         or[i] -
 round(mean(case)*(1-mean(control))/(mean(control)*(1-mean(case))),digit=2)

  if (or[i]lt;criteria[k,2]amp;or[i]gt;=criteria[k,1])
  {
     group - c(rep(1,100),rep(0,100))
     value - c(case,control)
     ttest -
 (t.test(case,control,alternative=two.sided,paired=F))$p.value
     all - c(or[i],ttest)
     results - rbind(results,all)
  }
         ifelse(nrow(results)==5,break,1)
  }
    results1 - rbind(results1,results)
  }


 Thank you so much in advance,
 Amber

        [[alternative HTML version deleted]]

 __
 R-help@ mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.
 
 
 
 -- 
 Jim Holtman
 Data Munger Guru
 
 What is the problem that you are trying to solve?
 Tell me what you want to do, not how you want to do it.
 
 __
 R-help@ mailing list
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 and provide commented, minimal, self-contained, reproducible code.
 


Right.
There are other things I'd like to note:
1. When debugging, if you use random number generators, also use 'set.seed'.
And get reproducible
conditions for the code tests.

2. Your first 'if' has the short conjunction ''. This is a vectorized
version, you only want 1 result.
See

?.

(Jim's statement is another 'vectorized' vs. '1 result' case. You want
the latter on both cases.)

3. Now the bug. You don't reinitialize 'results' to NULL and it therefore
reuses the first one.
It's a good pratice to always put 'var - NULL' as the instruction just
before the loop that forms it.

Is this what you want?


set.seed(1)   # Note 1 above
results1 - NULL
for (k in 1:2) {
results - NULL   # Note 3 above.
for (i in 1:1000)
{
control - sample(controlall, 100)
case - sample(caseall, 100)
or[i] -
round(mean(case)*(1-mean(control))/(mean(control)*(1-mean(case))), digit=2)

if (or[i]  criteria[k,2]  or[i] = criteria[k, 1])   # Note 
2 above
{
group - c(rep(1,100),rep(0,100))
value - c(case,control)
ttest - 
(t.test(case,control,alternative=two.sided,paired=F))$p.value
all - c(or[i], ttest)
results - rbind(results, all)
}
if(!is.null(results)  nrow(results) == 5) break  # Note 2 
above
}
results1 - rbind(results1, results)
}
results
results1


Hope this helps

Rui Barradas


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[R] Problems downloading file

2012-03-01 Thread Pete Brecknock

I am running the following line to download data from the US Energy
Information Administration. This function has worked successfully for me in
the past but yesterday gave the error/warning messages below. 

If I simply type http://ir.eia.gov/wpsr/psw09.xls; (no quotes) into a
browser, the file is available to download. 

I am running R 2.13.0 on Windows XP. 

# Download File Attempt
download.file(http://ir.eia.gov/wpsr/psw09.xls,c:\\temp\\psw09.xls,mode=wb;)

# Error  Warning Messages
trying URL 'http://ir.eia.gov/wpsr/psw09.xls'
Error in download.file(http://ir.eia.gov/wpsr/psw09.xls;,
c:\\temp\\psw09.xls,  : 
  cannot open URL 'http://ir.eia.gov/wpsr/psw09.xls'
In addition: Warning message:
In download.file(http://ir.eia.gov/wpsr/psw09.xls;, c:\\temp\\psw09.xls, 
:
  InternetOpenUrl failed: 'The operation timed out'

Thanks for any insights.

Pete Brecknock


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Re: [R] Problems downloading file

It might be a local network or OS issue: this works fine for me on my
personal Mac

download.file(url = http://ir.eia.gov/wpsr/psw09.xls;, destfile =
~/herewego.xls, mode = wb)

Someone with more Windows knowledge may have to help you out, but
often using IE settings (activated by the setInternet2 command if I
recall correctly) is a good first move.

Michael

On Thu, Mar 1, 2012 at 9:47 AM, Pete Brecknock peter.breckn...@bp.com wrote:
 I am running the following line to download data from the US Energy
 Information Administration. This function has worked successfully for me in
 the past but yesterday gave the error/warning messages below.

 If I simply type http://ir.eia.gov/wpsr/psw09.xls; (no quotes) into a
 browser, the file is available to download.

 I am running R 2.13.0 on Windows XP.

 # Download File Attempt
 download.file(http://ir.eia.gov/wpsr/psw09.xls,c:\\temp\\psw09.xls,mode=wb;)

 # Error  Warning Messages
 trying URL 'http://ir.eia.gov/wpsr/psw09.xls'
 Error in download.file(http://ir.eia.gov/wpsr/psw09.xls;,
 c:\\temp\\psw09.xls,  :
  cannot open URL 'http://ir.eia.gov/wpsr/psw09.xls'
 In addition: Warning message:
 In download.file(http://ir.eia.gov/wpsr/psw09.xls;, c:\\temp\\psw09.xls,
 :
  InternetOpenUrl failed: 'The operation timed out'

 Thanks for any insights.

 Pete Brecknock


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Re: [R] How to remove an object that is loaded every time R is opened?

2012-03-01 Thread Jason Love

Many thanks Jim and Chris for your helpful answers!
J

On Wed, Feb 29, 2012 at 11:48 AM, Chris Campbell 
ccampb...@mango-solutions.com wrote:

 Hi Jason

 If you close an R session and save without choosing a filename, a file
 called .RData will be created. Open a new session and type getwd(). Then
 have a look in the named file with your folder options set to Show Hidden
 Files. Deleting or renaming this file should remove the imported object.

 Hope this helps

 Chris

 Chris Campbell
 MANGO SOLUTIONS
 Data Analysis that Delivers
 +44 1249 705450


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
 On Behalf Of Jason Love
 Sent: 29 February 2012 15:10
 To: r-help@r-project.org
 Subject: [R] How to remove an object that is loaded every time R is opened?

 Dear R users,
 I'm a newbie and have another basic question that you guys can answer for
 me.
 So, I've been noticing that an object (data frame) called FossilFuel is
 loaded as default when I first open up the R (see below).
 I created this data frame a while ago which is a data set for tutorial and
 I don't need this any more.
 I'm not sure why this happens and wonder if there is any way to remove
 this object.
 Let me know.
 -Jason


 ---
 R version 2.14.1 (2011-12-22)
 Copyright (C) 2011 The R Foundation for Statistical Computing ISBN
 3-900051-07-0
 Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

 R is free software and comes with ABSOLUTELY NO WARRANTY.
 You are welcome to redistribute it under certain conditions.
 Type 'license()' or 'licence()' for distribution details.

  Natural language support but running in an English locale

 R is a collaborative project with many contributors.
 Type 'contributors()' for more information and 'citation()' on how to cite
 R or R packages in publications.

 Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()'
 for an HTML browser interface to help.
 Type 'q()' to quit R.

 [R.app GUI 1.43 (5989) x86_64-apple-darwin9.8.0]

 [Workspace restored from /R/.RData]
 [History restored from /R/.Rapp.history]

  ls()
 [1] fossilfuel
 

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Re: [R] 'break' function in loop

2012-03-01 Thread pei-chien tsai

Thank you so much Jim, Duncan, and special thanks to Rui!!

The modified script from Rui works perfectly and I really learned a lot from
all your suggestions.

Regards,
Amber



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Re: [R] barplots of several variables with different number of categories

2012-03-01 Thread ilai

Jon,
You could create new variables with the combined levels just for the
purpose of plotting.
Assume I have data.frame bpt
str(bpt)
'data.frame':   12 obs. of  2 variables:
 $ V1: Factor w/ 3 levels low,med,high: 1 1 1 1 2 2 2 2 3 3 ...
 $ V2: Factor w/ 6 levels 1,2,3,4,..: 1 1 2 2 3 3 4 4 5 5 ...

 bpt$V3 - factor(bpt$V1,c(levels(bpt$V1),levels(bpt$V2)))
 bpt$V4 - factor(bpt$V2,c(levels(bpt$V1),levels(bpt$V2)))

 with(bpt,barplot(cbind(table(V3),table(V4

Hope this helps

Elai

On Thu, Mar 1, 2012 at 7:19 AM, jon waterhouse jonwaterho...@gov.nl.ca wrote:
 If I have two factors, v1 and v2 and I want to have a stacked bar graph of
 the two variables side by side I could do

 barplot(cbind(table(v1),table(v2)))

 if v1 and v2 have the same number of categories.

 If they don't have the same number of categories this won't work.

 I'm sure there's a simple solution?

 Thanks,

 Jon

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Re: [R] select rows by criteria

On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
 Hello,
 
 I am stuck with selecting the right rows from a data frame. I think the
 problem is rather how to select them
 then how to implement the R code.
 
 Consider the following data frame:
 
 df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
 c(34,12,23,25,34,42,48,29,30,27))
 
 What I want to achieve is to select 7 rows (values) so that the mean value
 of those rows are closest
 to the value of 35 and the remaining 3 rows (values) are closest to 45.
 However, each value is only
 allowed to be sampled once!

Hi.

If some 3 rows have mean close to 45, then they have sum close
to 3*45, so the remaining 7 rows have sum close to

  sum(df$value) - 3*45 # [1] 169

and they have mean close to 169/7 = 24.14286. In other words,
the two criteria cannot be optimized together.

For this reason, let me choose the criterion on 3 rows.
The closest solution may be found as follows.

  # generate all triples and compute their means
  tripleMeans - colMeans(combn(df$value, 3))

  # select the index of the triple with mean closest to 35
  indClosest - which.min(abs(tripleMeans - 35))

  # generate the indices, which form the closest triple in df$value
  tripleInd - combn(1:length(df$value), 3)[, indClosest]
  tripleInd # [1] 1 3 7

  # check the mean of the triple
  mean(df$value[tripleInd]) # [1] 35

This code constructs all triples. If it is used for k-tuples
for a larger k and for a set of n values, its complexity
will be proportional to choose(n, k), so it will be large
even for moderate n, k. It is hard to provide a significant
speed up, since some variants of knapsack problem, which
is NP-complete, may be reduced to your question. Consequently,
it is, in general, NP-complete.

Hope this helps.

Petr Savicky.

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Re: [R] select rows by criteria

On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
 On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
  Hello,
  
  I am stuck with selecting the right rows from a data frame. I think the
  problem is rather how to select them
  then how to implement the R code.
  
  Consider the following data frame:
  
  df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
  c(34,12,23,25,34,42,48,29,30,27))
  
  What I want to achieve is to select 7 rows (values) so that the mean value
  of those rows are closest
  to the value of 35 and the remaining 3 rows (values) are closest to 45.
  However, each value is only
  allowed to be sampled once!
 
 Hi.
 
 If some 3 rows have mean close to 45, then they have sum close
 to 3*45, so the remaining 7 rows have sum close to
 
   sum(df$value) - 3*45 # [1] 169
 
 and they have mean close to 169/7 = 24.14286. In other words,
 the two criteria cannot be optimized together.
 
 For this reason, let me choose the criterion on 3 rows.
 The closest solution may be found as follows.
 
   # generate all triples and compute their means
   tripleMeans - colMeans(combn(df$value, 3))
 
   # select the index of the triple with mean closest to 35
   indClosest - which.min(abs(tripleMeans - 35))

I am sorry. There should be 45 and not 35.

  indClosest - which.min(abs(tripleMeans - 45))

  # generate the indices, which form the closest triple in df$value
  tripleInd - combn(1:length(df$value), 3)[, indClosest]
  tripleInd # [1] 1 6 7

  # check the mean of the triple
  mean(df$value[tripleInd]) # [1] 41.3

Petr Savicky.

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[R] Parameterization of Inverse Wishart distribution available in MCMCpack and bayesm libraries

2012-03-01 Thread Shantanu MULLICK

Hello Everyone

Both the MCMCpack and the bayesm libraries allow us to make draws from the
Inverse Wishart distribution.

But I wanted to find out how exactly is the Inverse Wishart distribution
parameterized in these libraries.

The reason I ask is the following:

Now its generally standard to express Inverse Wishart  as IW(0.5 * DOF,0.5*
Scale).(DOF- Degree of freedom, Scale - Scale parameter).

If we follow standard usage when we refer to the Degree of Freedom of the
above IW distribution it is =  DOF (and not 0.5* DOF).

Similarly the Scale parameters of the above IW it is= Scale (and not
0.5*Scale).

For the MCMCpack the IW draws are made by riwish(v,S).

*Question:* Does this compute IW( v,S) or IW(0.5*v,0.5*S) ?

This is the reason I want to find out the way these libraries parameterize
the Inverted Wishart distribution.

Best
Shantanu

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Re: [R] TM reader with text

2012-03-01 Thread Mickael R problem

Hi Richard,
clearly there is a problem with latin ligature because the word resulting
from my ask with  findFreqTerms give me some wordsU+FB01n  

U+FB01nancement
 U+FB01nancier U+FB01nanciÃ¨reU+FB01nanciÃ¨res
 U+FB01nanciersU+FB01xe 
 where U+FB01 is a code for latin ligature. The problem is well identified
ok.

Now, how can I tretaed it. The package TAU seems to offer a solution for
text but not for corpus. 

quoation TAU  translate Translate Unicode Latin Ligatures Description
Translate Unicode “Latin ligature” characters to their respective
constituents. Usage translate_Unicode_latin_ligatures(x) Arguments
x a character vector in UTF-8 encoding. 
Details In typography, a ligature occurs where two or more graphemes are
joined as a single glyph. (See
http://en.wikipedia.org/wiki/Typographic_ligature for more information.)
Unicode (http://www.unicode.org/) lists the following “Latin” ligatures:
Code Name
0132 LATIN CAPITAL LIGATURE IJ
0133 LATIN SMALL LIGATURE IJ
0152 LATIN CAPITAL LIGATURE OE
0153 LATIN SMALL LIGATURE OE
FB00 LATIN SMALL LIGATURE FF
util 9
FB01 LATIN SMALL LIGATURE FI
FB02 LATIN SMALL LIGATURE FL
FB03 LATIN SMALL LIGATURE FFI
FB04 LATIN SMALL LIGATURE FFL
FB05 LATIN SMALL LIGATURE LONG S T
FB06 LATIN SMALL LIGATURE ST

translate_Unicode_latin_ligatures translates these to their respective
constituent characters.

I need this king of fonction for corpus not only text or characters. Any
ideas ?
Thank's for comments and your answers. We are in progress!
Mickaël

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Re: [R] select rows by criteria

Hello,



syrvn wrote
 
 Hello,
 
 I am stuck with selecting the right rows from a data frame. I think the
 problem is rather how to select them
 then how to implement the R code.
 
 Consider the following data frame:
 
 df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
 c(34,12,23,25,34,42,48,29,30,27))
 
 What I want to achieve is to select 7 rows (values) so that the mean value
 of those rows are closest
 to the value of 35 and the remaining 3 rows (values) are closest to 45.
 However, each value is only
 allowed to be sampled once!
 
 Any ideas, how to achieve that?
 
 
 Cheers
 

See

?combn

It gives all possible combinations as a matrix (default) or list. Then,
'apply'.


#---
# Name changed to 'DF',
# 'df' is the R function for the F distribution density
# (and a frequent choice for example data in R-help!)
#
DF - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10),
value = c(34,12,23,25,34,42,48,29,30,27)) 

f - function(j, v, const) abs(mean(v[j]) - const)

inxmat - with(DF, combn(ID, 7))
meansDist1 - apply(inxmat, 2, function(jnx) f(jnx, DF$value, 35))
(i1 - which(meansDist1 == min(meansDist1)))

inxmat - with(DF, combn(ID, 3))
meansDist2 - apply(inxmat, 2, function(jnx) f(jnx, DF$value, 45))
(i2 - which(meansDist2 == min(meansDist2)))

meansDist3 - meansDist1 + meansDist2  # Compromise of both criteria?
(i3 - which(meansDist3 == min(meansDist3)))

Maybe it's combn(1:10, 3)[, 101] you want, or maybe there's another way to
compromise the two criteria.

Hope this helps,

Rui Barradas


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[R] Connecting points on a line with arcs/curves

2012-03-01 Thread hendersi


Hello,

I have a spreadsheet of pairs of coordinates and I would like to plot a line
along which curves/arcs connect each pair of coordinates. The aim is to
visualise the pattern of point connections. 

Thanks! Ian

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[R] how to change or copy to another the names of models

2012-03-01 Thread Waldir de Carvalho Junior

Hi
I would like to know how I can change the name of a model for each
trainning cycle of a model.
I work with the RSNNS package and to build a neural network, I used :
for (i in 5:30) 
model_ANN - mlp(X, Y, size=n,) # where size is the number of neurons
in the hidden layer
but I need to save each time that the model that is build (the end of each
cycle), e.g., when i = 5, I need to save the model with a especific name,
when i = 6, also I need to save the model with another name
How i am doing, i am saving only the last model with n = 30 and with the
name model_ANN
Question
how can I change the name of the model (model_ANN) at each end of cycle of
i values?
I have already tryied copy, save, rename,.. but unsuccessfully
thanks for any answer

-- 

Waldir de Carvalho Junior
Pedologia/Pedologue
Pesquisador/Chercheur
Embrapa Solos/INRA - UMR- LISAH

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Re: [R] select rows by criteria

Sorry, correction:

The second index matrix is the matrix of elements not in the first,
not another combination, this time 3 out of 10.

Change this in my first post

 
 inxmat - with(DF, combn(ID, 3))
 meansDist2 - apply(inxmat, 2, function(jnx) f(jnx, DF$value, 45))
 (i2 - which(meansDist2 == min(meansDist2)))
 

to this

inxmat2 - with(DF, apply(inxmat, 2, function(x) setdiff(ID, x)))
meansDist2 - apply(inxmat2, 2, function(jnx) f(jnx, DF$value, 45))
(i2 - which(meansDist2 == min(meansDist2)))

Rui Barradas



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[R] Help with segmented package

2012-03-01 Thread Filoche

Hi everyone.

I'm using segmented package to find break point in a bi-linear relationship.
In a particular case, I find 1 pointcut (so 2 slopes).

I would like to know if it is possible to retrieve information in the
segmented object that could let me to plot 1 particular segment with a
different color.

For example, in that folowing example, I would like to plot the second
segment in red.

library(segmented)
set.seed(1234)
z-runif(100)
y-rpois(100,exp(2+1.8*pmax(z-.6,0)))
o-glm(y~z,family=poisson)
o.seg-segmented(o,seg.Z=~z,psi=list(z=.5))
plot(o.seg)

Thank in advance,
Phil



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[R] Delete rows from data.frame matching a certain criteria

2012-03-01 Thread mails

Hello,


consider the following data.frame:

test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
c(1,1,NA,1,NA))

 test
  n v pattern
1  1 6   1
2  2 5   1
3  3 7  NA
4  4 5   1
5  5 3  NA


I tried to use apply and the adply function to set v to NA where pattern = 1
and v to v where pattern = 1


So basically the result should look like this:
 test
  n v pattern
1  1 NA   1
2  2 NA  1
3  3 7  NA
4  4 NA   1
5  5 3  NA

So far, I solved it by creating subsets and using merge but it turns out to
be super slow. Is there a way to do that
with the apply function?

Any help/hint is appreciated

Thanks


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[R] Need help using Melt and cast to compute correlation for a cross tabulation

2012-03-01 Thread Walter Anderson

I have a data frame with a number of observed and predicted values by 
classification as shown below:


Count   Volume  FCLASS
1  55000   6   Grade Separated
2  43000   39000   Grade Separated
3  26000   26500   Major Arterial
4  19500   2   Major Arterial
...

There are four classes here: Grade Separated, Major Arterial, Minor 
Arterial, and Collector


I am looking to compute the following information

FCLASS   cor(Count,Volume)
Grade Separated   0.999
Major Arterial 0.999
Minor Arterial 0.999
Collector 0.999

I am attempting to use the following commands to achieve this:

library(reshape)
tmp - melt(dataframe)
cast(tmp, FCLASS ~ variable, function(Count, Volume) cor(Count, Volume))

but this is generating:  Error in is.data.frame(y) : argument Volume 
is missing, with no default


Any suggestions?

Walter Anderson

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Re: [R] select rows by criteria

On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
 On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
  Hello,
  
  I am stuck with selecting the right rows from a data frame. I think the
  problem is rather how to select them
  then how to implement the R code.
  
  Consider the following data frame:
  
  df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
  c(34,12,23,25,34,42,48,29,30,27))
  
  What I want to achieve is to select 7 rows (values) so that the mean value
  of those rows are closest
  to the value of 35 and the remaining 3 rows (values) are closest to 45.
  However, each value is only
  allowed to be sampled once!
 
 Hi.
 
 If some 3 rows have mean close to 45, then they have sum close
 to 3*45, so the remaining 7 rows have sum close to
 
   sum(df$value) - 3*45 # [1] 169
 
 and they have mean close to 169/7 = 24.14286. In other words,
 the two criteria cannot be optimized together.
 
 For this reason, let me choose the criterion on 3 rows.
 The closest solution may be found as follows.
 
   # generate all triples and compute their means
   tripleMeans - colMeans(combn(df$value, 3))
 
   # select the index of the triple with mean closest to 35
   indClosest - which.min(abs(tripleMeans - 35))
 
   # generate the indices, which form the closest triple in df$value
   tripleInd - combn(1:length(df$value), 3)[, indClosest]
   tripleInd # [1] 1 3 7
 
   # check the mean of the triple
   mean(df$value[tripleInd]) # [1] 35
 
 This code constructs all triples. If it is used for k-tuples
 for a larger k and for a set of n values, its complexity
 will be proportional to choose(n, k), so it will be large
 even for moderate n, k. It is hard to provide a significant
 speed up, since some variants of knapsack problem, which
 is NP-complete, may be reduced to your question. Consequently,
 it is, in general, NP-complete.

Hi.

Also this statement requires a correction. It applies to the
search of an exact optimum if the numbers in df$value are large.
There are efficient algorithms, which find an approximate solution.
Also, if the numbers in df$value are integers (or may be rounded
to integers after an appropriate scaling), then there is an
algorithm, whose complexity is O(k*n*max(df$value)). This
may be significantly less than choose(n, k).

CRAN task view Optimization and Mathematical Programming

  http://cran.at.r-project.org/web/views/Optimization.html

may suggest also other solutions.

Petr Savicky.

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[R] Standard variance / devistion clarification

2012-03-01 Thread Suranga Kasthurirathne

Dear gurus,

Im a newbie, and I want to ask a very general question.
Assume that I have a set of numbers as follows,

1, 1, 2, 10, 100, 10,1

From these, I need to identify which number is the most different as
compared to others. (in this case, it will be 100, since its way larger
than the other numbers). It doesnt have to be specifically this way, but I
need to identify which number(s) are most different compared to the others.

Any idea as to what I need to do this ? Im a math noob, so I'm also going
to need to ask it this is called 'standard deviation' or 'variance' :-)

-- 
Best Regards,

Suranga

[[alternative HTML version deleted]]

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Re: [R] Delete rows from data.frame matching a certain criteria

2012-03-01 Thread Sarah Goslee

Hi,

On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:
 Hello,


 consider the following data.frame:

 test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
 c(1,1,NA,1,NA))

 test
  n v pattern
 1  1     6       1
 2  2     5       1
 3  3     7      NA
 4  4     5       1
 5  5     3      NA


Thanks for the reproducible example.

 I tried to use apply and the adply function to set v to NA where pattern = 1
 and v to v where pattern = 1

Presumably that should be set v to NA where pattern == 1 and v to v where
pattern != 1


 So basically the result should look like this:
 test
  n v pattern
 1  1     NA       1
 2  2     NA      1
 3  3     7      NA
 4  4     NA       1
 5  5     3      NA

 So far, I solved it by creating subsets and using merge but it turns out to
 be super slow. Is there a way to do that
 with the apply function?

Far too much work. What about:

 test$v - ifelse(test$pattern == 1, NA, v)
 test
  n  v pattern
1 1 NA   1
2 2 NA   1
3 3 NA  NA
4 4 NA   1
5 5 NA  NA


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

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Re: [R] Delete rows from data.frame matching a certain criteria

Your criteria did not make sense since in both cases pattern == 1, so
I chose to set to NA if pattern == 1

 test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
+ c(1,1,NA,1,NA))
 test
  n v pattern
1 1 6   1
2 2 5   1
3 3 7  NA
4 4 5   1
5 5 3  NA
 # set v to NA when pattern = 1
 test$v[!is.na(test$pattern)  (test$pattern == 1)] - NA
 test
  n  v pattern
1 1 NA   1
2 2 NA   1
3 3  7  NA
4 4 NA   1
5 5  3  NA




On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:

 Hello,


 consider the following data.frame:

 test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
 c(1,1,NA,1,NA))

  test
  n v pattern
 1  1     6       1
 2  2     5       1
 3  3     7      NA
 4  4     5       1
 5  5     3      NA


 I tried to use apply and the adply function to set v to NA where pattern =
 1
 and v to v where pattern = 1


 So basically the result should look like this:
  test
  n v pattern
 1  1     NA       1
 2  2     NA      1
 3  3     7      NA
 4  4     NA       1
 5  5     3      NA

 So far, I solved it by creating subsets and using merge but it turns out
 to
 be super slow. Is there a way to do that
 with the apply function?

 Any help/hint is appreciated

 Thanks


 --
 View this message in context:
 http://r.789695.n4.nabble.com/Delete-rows-from-data-frame-matching-a-certain-criteria-tp4435414p4435414.html
 Sent from the R help mailing list archive at Nabble.com.

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




--
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

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Re: [R] Need help using Melt and cast to compute correlation for a cross tabulation

Here is another way of doing it:

 x - read.table(text =Count   Volume  FCLASS
+ 1  55000   6   'Grade Separated'
+ 2  43000   39000   'Grade Separated'
+ 3  26000   26500   'Major Arterial'
+ 4  19500   2   'Major Arterial', as.is = TRUE)
 result - sapply(split(x, x$FCLASS), function(.grp) cor(.grp$Count,
.grp$Volume))

 result
Grade Separated  Major Arterial
  1   1



On Thu, Mar 1, 2012 at 12:23 PM, Walter Anderson wandrso...@gmail.comwrote:

 I have a data frame with a number of observed and predicted values by
 classification as shown below:

Count   Volume  FCLASS
 1  55000   6   Grade Separated
 2  43000   39000   Grade Separated
 3  26000   26500   Major Arterial
 4  19500   2   Major Arterial
 ...

 There are four classes here: Grade Separated, Major Arterial, Minor
 Arterial, and Collector

 I am looking to compute the following information

 FCLASS   cor(Count,Volume)
 Grade Separated   0.999
 Major Arterial 0.999
 Minor Arterial 0.999
 Collector 0.999

 I am attempting to use the following commands to achieve this:

 library(reshape)
 tmp - melt(dataframe)
 cast(tmp, FCLASS ~ variable, function(Count, Volume) cor(Count, Volume))

 but this is generating:  Error in is.data.frame(y) : argument Volume is
 missing, with no default

 Any suggestions?

 Walter Anderson

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




-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

[[alternative HTML version deleted]]

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Re: [R] Standard variance / devistion clarification



On Mar 1, 2012, at 12:30 PM, Suranga Kasthurirathne wrote:


Dear gurus,

Im a newbie, and I want to ask a very general question.
Assume that I have a set of numbers as follows,

1, 1, 2, 10, 100, 10,1


From these, I need to identify which number is the most different as
compared to others. (in this case, it will be 100, since its way  
larger
than the other numbers). It doesnt have to be specifically this way,  
but I
need to identify which number(s) are most different compared to the  
others.


Any idea as to what I need to do this ? Im a math noob, so I'm also  
going
to need to ask it this is called 'standard deviation' or  
'variance' :-)


You are requested not to use rhelp as a statistics QA list. It was  
not set up to remediate gaps in statistics education. There are other  
websites and perhaps mailing list whose purposes would be a better fit  
for this question.


--

David Winsemius, MD
West Hartford, CT

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Re: [R] GLM with regularization

2012-03-01 Thread Bert Gunter

Google is your friend! -- as usual.

If you had searched on glm with regularization you would have bumped
into the glmnet R package, which I think is what you're looking for.

-- Bert

On Wed, Feb 29, 2012 at 6:22 PM, Dmitriy Lyubimov dlie...@gmail.com wrote:
Hello,

Thank you for probably not so new question, but i am new to R.

Does any of packages have something like glm+regularization? So far i
see probably something close to that as a ridge regression in MASS but
I think i need something like GLM, in particular binomial regularized
versions of polynomial regression.

Also I am not sure how some of the K-fold crossvalidation helpers out
there (cv.glm) could be used to adjust reg rate as there seems to be
no way to apply them over data not used for training (or i am not
seeing a solution here as training is completely separated from
crossvalidation error computation here) .

The example here in cv.glm doesn't look right to me since it computes
cv error over model trained on 100% of data. (e.g. wikipedia
crossvalidation article lists this as an example of misuse of K-fold
CV).

- doc quote
# leave-one-out and 6-fold cross-validation prediction error for
# the mammals data set.
data(mammals, package=MASS)
mammals.glm - glm(log(brain)~log(body),data=mammals)
cv.err - cv.glm(mammals,mammals.glm)
cv.err.6 - cv.glm(mammals, mammals.glm, K=6)
end of quote ---

Those seem to be pretty common techniques, any poniter in the right
direction (package) will be greatly appreciated.

thank you very much.
-Dmitriy

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

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Re: [R] Delete rows from data.frame matching a certain criteria

2012-03-01 Thread Nordlund, Dan (DSHS/RDA)

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of mails
 Sent: Thursday, March 01, 2012 8:11 AM
 To: r-help@r-project.org
 Subject: [R] Delete rows from data.frame matching a certain criteria

 Hello,

 consider the following data.frame:

 test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
 c(1,1,NA,1,NA))

  test
   n v pattern
 1  1 6   1
 2  2 5   1
 3  3 7  NA
 4  4 5   1
 5  5 3  NA

 I tried to use apply and the adply function to set v to NA where
 pattern = 1
 and v to v where pattern = 1

 So basically the result should look like this:
  test
   n v pattern
 1  1 NA   1
 2  2 NA  1
 3  3 7  NA
 4  4 NA   1
 5  5 3  NA

 So far, I solved it by creating subsets and using merge but it turns
 out to
 be super slow. Is there a way to do that
 with the apply function?

 Any help/hint is appreciated

 Thanks

There is no need for apply here, this is a simple indexing problem.  Something 
like this should work

test$v - ifelse(is.na(test$pattern), test$v, NA)

Hope this is helpful,

Dan

Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204

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Re: [R] Delete rows from data.frame matching a certain criteria

2012-03-01 Thread Ista Zahn

Hi,

On Mar 1, 2012, at 12:38 PM, Sarah Goslee wrote:

 Hi,
 
 On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:
 Hello,
 
 
 consider the following data.frame:
 
 test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
 c(1,1,NA,1,NA))
 

 snip 

 So basically the result should look like this:
 test
  n v pattern
 1  1 NA   1
 2  2 NA  1
 3  3 7  NA
 4  4 NA   1
 5  5 3  NA
 
 So far, I solved it by creating subsets and using merge but it turns out to
 be super slow. Is there a way to do that
 with the apply function?
 
 Far too much work. What about:
 
 test$v - ifelse(test$pattern == 1, NA, v)
 test
  n  v pattern
 1 1 NA   1
 2 2 NA   1
 3 3 NA  NA
 4 4 NA   1
 5 5 NA  NA

Actually that doesn't work because of those pesky missing values. You need

test -  transform(test, v = ifelse(pattern == 1  !is.na(pattern), NA, v))

Best,
Ista

 
 
 -- 
 Sarah Goslee
 http://www.functionaldiversity.org
 
 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Robust ARMA Fitting in R?

2012-03-01 Thread isabella

 

Hello,   BODY { font-family:Arial, Helvetica,
sans-serif;font-size:12px; }  

Does any one know if there are any functions/packages available in R
for robust fitting of ARMA time series models (e.g., similar to the
function arima.rob() in S-PLUS)?   

Many thanks and kind regards,  

Isabella  

Isabella R. Ghement, Ph.D. 
 Ghement Statistical Consulting Company 
 301-7031 Blundell Road, Richmond, B.C., Canada, V6Y 1J5 
 Tel: 604-767-1250 
 Fax: 604-270-3922 
 E-mail: isabe...@ghement.ca 
 Web: www.ghement.ca 
 
[[alternative HTML version deleted]]

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Re: [R] Delete rows from data.frame matching a certain criteria

2012-03-01 Thread Sarah Goslee

You're all correct: I copied in the wrong thing. My apologies!

On Thu, Mar 1, 2012 at 1:00 PM, Ista Zahn istaz...@gmail.com wrote:
 Hi,

 On Mar 1, 2012, at 12:38 PM, Sarah Goslee wrote:

 Hi,

 On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:
 Hello,


 consider the following data.frame:

 test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
 c(1,1,NA,1,NA))


  snip 

 So basically the result should look like this:
 test
  n v pattern
 1  1     NA       1
 2  2     NA      1
 3  3     7      NA
 4  4     NA       1
 5  5     3      NA

 So far, I solved it by creating subsets and using merge but it turns out to
 be super slow. Is there a way to do that
 with the apply function?

 Far too much work. What about:

 test$v - ifelse(test$pattern == 1, NA, v)
 test
  n  v pattern
 1 1 NA       1
 2 2 NA       1
 3 3 NA      NA
 4 4 NA       1
 5 5 NA      NA

 Actually that doesn't work because of those pesky missing values. You need

 test -  transform(test, v = ifelse(pattern == 1  !is.na(pattern), NA, v))

 Best,
 Ista



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Re: [R] GLM with regularization

2012-03-01 Thread Dmitriy Lyubimov

Thank you.

On Thu, Mar 1, 2012 at 9:58 AM, Bert Gunter gunter.ber...@gene.com wrote:
Google is your friend! -- as usual.

If you had searched on glm with regularization you would have bumped
into the glmnet R package, which I think is what you're looking for.

-- Bert

On Wed, Feb 29, 2012 at 6:22 PM, Dmitriy Lyubimov dlie...@gmail.com wrote:
Hello,

Thank you for probably not so new question, but i am new to R.

Those seem to be pretty common techniques, any poniter in the right
direction (package) will be greatly appreciated.

thank you very much.
-Dmitriy

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

Re: [R] Delete rows from data.frame matching a certain criteria



On Mar 1, 2012, at 1:02 PM, Sarah Goslee wrote:


You're all correct: I copied in the wrong thing. My apologies!

On Thu, Mar 1, 2012 at 1:00 PM, Ista Zahn istaz...@gmail.com wrote:

Hi,

On Mar 1, 2012, at 12:38 PM, Sarah Goslee wrote:


Hi,

On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:

Hello,


consider the following data.frame:

test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
c(1,1,NA,1,NA))



 snip 


So basically the result should look like this:

test

 n v pattern
1  1 NA   1
2  2 NA  1
3  3 7  NA
4  4 NA   1
5  5 3  NA


So far, I solved it by creating subsets and using merge but it  
turns out to

be super slow. Is there a way to do that
with the apply function?


Far too much work. What about:


test$v - ifelse(test$pattern == 1, NA, v)
test

 n  v pattern
1 1 NA   1
2 2 NA   1
3 3 NA  NA
4 4 NA   1
5 5 NA  NA


Actually that doesn't work because of those pesky missing values.  
You need


test -  transform(test, v = ifelse(pattern == 1  !is.na(pattern),  
NA, v))



What about just using `is.na-`:

is.na(test$v) - test$pattern==1



--
David Winsemius, MD
West Hartford, CT

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[R] 6 different errors while using glm.nb

2012-03-01 Thread Lucas

Hello to everyone.
I need your help. I´m trying to fit the same *glm.nb* to a different data
set and i am getting these errors in some of the data. Sometimes, one data
set has two of these errors when fitting the model.

1.- Error en while ((it - it + 1)  limit  abs(del)  eps) { :

  valor ausente donde TRUE/FALSE es necesario


2.- Mensajes de aviso perdidos

1: In sqrt(1/i) : Se han producido NaNs2: In sqrt(1/i) : Se han producido NaNs


3.- glm.fit: fitted rates numerically 0 occurred


4.- glm.fit : algorithm did not converge


5.- Error: no valid set of coefficients has been found: please supply
starting values


6.- Error en glm.fitter(x = X, y = Y, w = w, etastart = eta, offset =
offset,  :

  NA/NaN/Inf en llamada a una función externa (arg 1)


I don´t know why i have this errors. If someone could help me to
understand what´s going on i would appreciate.


thank you.

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Re: [R] Robust ARMA Fitting in R?

2012-03-01 Thread Berend Hasselman


On 01-03-2012, at 19:03, isabe...@ghement.ca wrote:

 
 
   Hello,   BODY { font-family:Arial, Helvetica,
 sans-serif;font-size:12px; }  
 
   Does any one know if there are any functions/packages available in R
 for robust fitting of ARMA time series models (e.g., similar to the
 function arima.rob() in S-PLUS)?   

CRAN:  http://cran.r-project.org/

In Task Views (left side)  goto TimeSeries.
And search for arima on that page.

Berend

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Re: [R] Robust ARMA Fitting in R?

2012-03-01 Thread isabella

 

  BODY { font-family:Arial, Helvetica, sans-serif;font-size:12px; }
Hi Berend,  

Many thanks for your prompt reply.  I followed your instructions but
couldn't find what I was looking for.  I was hoping that someone who's
already worked with such a function could be able to point it out to
me.  My Google searches came up empty handed.   

Kind regards,  

Isabella 
 Isabella R. Ghement, Ph.D. 
 Ghement Statistical Consulting Company 
 301-7031 Blundell Road, Richmond, B.C., Canada, V6Y 1J5 
 Tel: 604-767-1250 
 Fax: 604-270-3922 
 E-mail: isabe...@ghement.ca 
 Web: www.ghement.ca 
 On Thu 01/03/12 10:29 AM , Berend Hasselman b...@xs4all.nl sent:
 On 01-03-2012, at 19:03,  wrote:
  
  
  Hello, BODY { font-family:Arial, Helvetica,
  sans-serif;font-size:12px; } 
  
  Does any one know if there are any functions/packages available in
R
  for robust fitting of ARMA time series models (e.g., similar to
the
  function arima.rob() in S-PLUS)? 
 CRAN: http://cran.r-project.org/ [2]
 In Task Views (left side) goto TimeSeries.
 And search for arima on that page.
 Berend
 

Links:
--
[2]
http://sitemail.netnation.com/parse.php?redirect=http%3A%2F%2Fcran.r-project.org%2F

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Re: [R] Execution of Rprofile.site

2012-03-01 Thread Uwe Ligges




On 01.03.2012 14:39, Ross Bowden wrote:

Hi everyone. I have recently installed R 2.14.1 on my 64-bit Windows 7
laptop. I am attempting to include some favourite functions in the
Rprofile.site file to run at R start-up as I did with my previous 32-bit
XP machine. I have edited the Rprofile.site file in C:\Program
Files\R\R-2.14.1\etc\ but the added code doesn't seem to be executed
when I run R, whether through Tinn-R or independently. Any ideas please
on what's going wrong? I've searched my laptop and there is only one
Rprofile.site file. Also Sys.getenv(R_HOME) returns: [1]
C:/PROGRA~1/R/R-214~1.1 so R should be able to find the Rprofile.site
file. Note that Rprofile.site has been modified by Tinn-R.


Have you disabled the Rprofile.site?
What makes you think code is not executed?

Uwe Ligges



Many thanks

Ross Bowden
PhD Candidate, Maths Department, Murdoch University.

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Re: [R] Robust ARMA Fitting in R?

2012-03-01 Thread Berend Hasselman


On 01-03-2012, at 19:33, isabe...@ghement.ca wrote:

 Hi Berend,
 
 Many thanks for your prompt reply.  I followed your instructions but couldn't 
 find what I was looking for.  I was hoping that someone who's already worked 
 with such a function could be able to point it out to me.  My Google searches 
 came up empty handed. 

I just tried searching in Google for arima R (without the quotes) and got

http://stat.ethz.ch/R-manual/R-patched/library/stats/html/arima.html

and a lot more.

And then searched for arima R robust  
It appears there is a package TSA which might bw of use to you.

Berend

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Re: [R] Connecting points on a line with arcs/curves

2012-03-01 Thread Uwe Ligges


?plot


On 01.03.2012 16:15, hendersi wrote:


Hello,

I have a spreadsheet of pairs of coordinates and I would like to plot a line
along which curves/arcs connect each pair of coordinates. The aim is to
visualise the pattern of point connections.

Thanks! Ian

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Re: [R] TM reader with text

2012-03-01 Thread Milan Bouchet-Valat

Le jeudi 01 mars 2012 à 07:07 -0800, Mickael R problem a écrit :
 Hi Richard,
 clearly there is a problem with latin ligature because the word resulting
 from my ask with  findFreqTerms give me some wordsU+FB01n
   
 U+FB01nancement
  U+FB01nancier U+FB01nanciÃ¨reU+FB01nanciÃ¨res
  U+FB01nanciersU+FB01xe 
  where U+FB01 is a code for latin ligature. The problem is well identified
 ok.
 
 Now, how can I tretaed it. The package TAU seems to offer a solution for
 text but not for corpus. 
 
 quoation TAU  translate Translate Unicode Latin Ligatures Description
 Translate Unicode “Latin ligature” characters to their respective
 constituents. Usage translate_Unicode_latin_ligatures(x) Arguments
 x a character vector in UTF-8 encoding. 
 Details In typography, a ligature occurs where two or more graphemes are
 joined as a single glyph. (See
 http://en.wikipedia.org/wiki/Typographic_ligature for more information.)
 Unicode (http://www.unicode.org/) lists the following “Latin” ligatures:
 Code Name
 0132 LATIN CAPITAL LIGATURE IJ
 0133 LATIN SMALL LIGATURE IJ
 0152 LATIN CAPITAL LIGATURE OE
 0153 LATIN SMALL LIGATURE OE
 FB00 LATIN SMALL LIGATURE FF
 util 9
 FB01 LATIN SMALL LIGATURE FI
 FB02 LATIN SMALL LIGATURE FL
 FB03 LATIN SMALL LIGATURE FFI
 FB04 LATIN SMALL LIGATURE FFL
 FB05 LATIN SMALL LIGATURE LONG S T
 FB06 LATIN SMALL LIGATURE ST
 
 translate_Unicode_latin_ligatures translates these to their respective
 constituent characters.
 
 I need this king of fonction for corpus not only text or characters. Any
 ideas ?
Try:
corpus - tm_map(corpus, translate_Unicode_latin_ligatures)
(with 'corpus' your corpus, of course ;-)

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Re: [R] Plotting a registered sign

2012-03-01 Thread Dennis Fisher

R 2.14.0
OS X

Marc's proposed solution (appearing at the end of this email) is perfect -- 
thanks so much
However, some questions remain:

1.  The following works:
plot(1, type=n)
text(1, 1, expression(symbol(\342)))

but this does not work (TEXT appears in a symbol font)
plot(1, type=n)
TEXT- \342
text(1, 1, expression(symbol(TEXT)))

nor does this work (it yields:  Error: could not find function symbol)
plot(1, type=n)
TEXT- symbol(\342)
text(1, 1, expression(TEXT))
Clearly, I do not understand the subtleties of plotmath.  Could someone explain?

2.  How would one learn that \342 corresponds to the copyright sign in the 
symbol font? (i.e., is there a list of all possible characters)



Dennis Fisher MD
P  (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com

On Feb 29, 2012, at 6:55 PM, Marc Schwartz wrote:

 Dennis,
 
 Depending upon some additional fine tuning, here is a generic example:
 
  plot(1)
  legend(topright, pch = 1, legend = expression(SOMETEXT ^ symbol(\342)))
 
 
 See ?plotmath for more information. The above works on my MBP running Lion.
 
 HTH,
 
 Marc Schwartz
 


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Re: [R] Quantile scores as dependent variables.. an R and general method question

2012-03-01 Thread Thomas Lumley

On Thu, Mar 1, 2012 at 12:07 PM, Doran, Harold hdo...@air.org wrote:
 Typically this list doesn't support general statistical questions and 
 unfortunately I don't have a better recommendation. It may be more helpful 
 for you to work with a statistician than seek help here.

 My point is simply that quantile regression is not for modeling outcomes that 
 are quantiles. The fact that you have a dependent variables expressed as a 
 percentile doesn't mean that quantile regression is the appropriate approach.


It's fairly widely true that percentiles or ranks behave
probabilistically as if they were independent uniform random variables
(though unfortunately it can be quite hard to prove in individual
cases) so that might be where to start.


  -thomas

-- 
Thomas Lumley
Professor of Biostatistics
University of Auckland

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Re: [R] select rows by criteria

Hello, again.


Petr Savicky wrote
 
 On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
 On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
  Hello,
  
  I am stuck with selecting the right rows from a data frame. I think the
  problem is rather how to select them
  then how to implement the R code.
  
  Consider the following data frame:
  
  df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
  c(34,12,23,25,34,42,48,29,30,27))
  
  What I want to achieve is to select 7 rows (values) so that the mean
 value
  of those rows are closest
  to the value of 35 and the remaining 3 rows (values) are closest to 45.
  However, each value is only
  allowed to be sampled once!
 
 Hi.
 
 If some 3 rows have mean close to 45, then they have sum close
 to 3*45, so the remaining 7 rows have sum close to
 
   sum(df$value) - 3*45 # [1] 169
 
 and they have mean close to 169/7 = 24.14286. In other words,
 the two criteria cannot be optimized together.
 
 For this reason, let me choose the criterion on 3 rows.
 The closest solution may be found as follows.
 
   # generate all triples and compute their means
   tripleMeans - colMeans(combn(df$value, 3))
 
   # select the index of the triple with mean closest to 35
   indClosest - which.min(abs(tripleMeans - 35))
 
 I am sorry. There should be 45 and not 35.
 
   indClosest - which.min(abs(tripleMeans - 45))
 
   # generate the indices, which form the closest triple in df$value
   tripleInd - combn(1:length(df$value), 3)[, indClosest]
   tripleInd # [1] 1 6 7
 
   # check the mean of the triple
   mean(df$value[tripleInd]) # [1] 41.3
 
 Petr Savicky.
 
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There are two solutions for the 3 rows criterion, 'which.min' only finds
one, the first in the order given by 'combn'.
(And I've corrected my first post but still with an error)

# Forgot to change the index matrix
meansDist2 - apply(inxmat2, 2, function(jnx) f(jnx, DF$value, 45))

# Two solutions
(i2 - which(meansDist2 == min(meansDist2)))
inxmat2[, i2]

mean(DF$value[inxmat2[, i2][, 1]])
[1] 41.3

Petr's solution and mine give the same mean value.
But use for small values of (n, k) only.

Rui Barradas



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Re: [R] How to deal with missing values when using Random Forrest

2012-03-01 Thread Kevin Corry

Hi,

Thanks for your help,

This worked very well:

na.action=na.roughfix

Kevin

On Sun, Feb 26, 2012 at 3:10 PM, Weidong Gu anopheles...@gmail.com wrote:

 Hi,

 You can set na.action=na.roughfix which fills NAs with the mean or
 mode of the missing variable.

 Other option is to impute missing values using rfImpute, then run
 randomForest on the complete data set.

 Weidong Gu

 On Sat, Feb 25, 2012 at 6:24 PM, kevin123 kevincorry...@gmail.com wrote:
  I am using the package Random Forrest to test and train a model,
  I aim to predict (LengthOfStay.days),:
 
  library(randomForest)
  model - randomForest( LengthOfStay.days~.,data = training,
  + importance=TRUE,
  + keep.forest=TRUE
  + )
 
 
  *This is a small portion of the data frame:   *
 
  *data(training)*
 
  LengthOfStay.days CharlsonIndex.numeric DSFS.months
  1  0   0.0 8.5
  6  0   0.0 3.5
  7  0   0.0 0.5
  8  0   0.0 0.5
  9  0   0.0 1.5
  11 0   1.5 NaN
 
 
 
  *Error message*
 
  Error in na.fail.default(list(LengthOfStay.days = c(0, 0, 0, 0, 0, 0,  :
   missing values in object,
 
  I would greatly appreciate any help
 
  Thanks
 
  Kevin
 
 
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[R] Rscript example

2012-03-01 Thread statquant2

Hi there,
I am trying to find an example how to use Rscript
Let's suppose I want to pass 3 arguments (I don't want [options] and -e
[expressions] as described in help) 

*on the command line
myRscript.R -arg1=value1 -arg2=value2 -arg3=value3

*In the script
#! /path/to/Rscript
args = commandArgs(TRUE);

From what I see args is just a string, do I do things correctly ?

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[R] fridays date to date

Hello,

How do I get the dates of all Fridays between two dates?

thanks,

Ben

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Re: [R] Plotting a registered sign

2012-03-01 Thread Marc Schwartz

Hi Dennis,

There are some subtleties in the way that 'symbol' is handled in plotmath. 
There is no symbol() function, per se, though there is an as.symbol() function, 
which is used differently.

'symbol' as used in plotmath, is telling R to plot the character using a symbol 
font, which is the same as 'font = 5'.

If you look at some of the examples in ?plotmath and ?bquote, you can see that 
there is the use of the syntax:

  .(VarName)

where VarName is the name of a scalar that contains the value/character that is 
desired to be plotted.

Thus, to do the sort of thing you desire:

  TEXT - \342

  plot(1, type = n)
  text(1, 1, bquote(symbol(.(TEXT

or:

  plot(1, type = n)
  text(1, 1, bquote(.(TEXT)), font = 5)

Also, with respect to a listing of the symbols, see the References section of 
?plotmath, which leads you to pages such as:

  http://www.stat.auckland.ac.nz/~paul/R/CM/AdobeSym.html

HTH,

Marc

On Mar 1, 2012, at 1:40 PM, Dennis Fisher wrote:

 R 2.14.0
 OS X
 
 Marc's proposed solution (appearing at the end of this email) is perfect -- 
 thanks so much
 However, some questions remain:
 
 1.  The following works:
   plot(1, type=n)
   text(1, 1, expression(symbol(\342)))
 
 but this does not work (TEXT appears in a symbol font)
   plot(1, type=n)
   TEXT- \342
   text(1, 1, expression(symbol(TEXT)))
 
 nor does this work (it yields:Error: could not find function symbol)
   plot(1, type=n)
   TEXT- symbol(\342)
   text(1, 1, expression(TEXT))
 Clearly, I do not understand the subtleties of plotmath.  Could someone 
 explain?
 
 2.  How would one learn that \342 corresponds to the copyright sign in the 
 symbol font? (i.e., is there a list of all possible characters)
 
 
 
 Dennis Fisher MD
 P  (The P Less Than Company)
 Phone: 1-866-PLessThan (1-866-753-7784)
 Fax: 1-866-PLessThan (1-866-753-7784)
 www.PLessThan.com
 
 On Feb 29, 2012, at 6:55 PM, Marc Schwartz wrote:
 
 Dennis,
 
 Depending upon some additional fine tuning, here is a generic example:
 
  plot(1)
  legend(topright, pch = 1, legend = expression(SOMETEXT ^ symbol(\342)))
 
 
 See ?plotmath for more information. The above works on my MBP running Lion.
 
 HTH,
 
 Marc Schwartz
 
 


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Re: [R] function for filtering and deleting vector entries

2012-03-01 Thread babyluck

Thank you Rui, that helped a lot. The correct values show up when I'm using
the following code. Now  fun(Temp,v)  returns a matrix, and  Temp  and   v  
stay the same. But I'd like to use the reduced vectors in some
calculations..can they be extracted  in some way so that I have them
separately again? 



fun - function(Temp, v){ 
   unwanted - Temp = 16 | Temp = 38.5 
Temp - Temp[!unwanted] 
v - v[!unwanted] 
  list(Temp=Temp, v=v) 
  
} 

fun(Temp,v)

PS: thank you too,andrija..but it's not what I'm looking for


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Re: [R] function for filtering and deleting vector entries

Of course, just use

x - fun(Temp, v)

x$Temp # To get back temp
x[[Temp]]
x$v # To get back v
x[[v]]

Michael


On Thu, Mar 1, 2012 at 3:15 PM, babyluck madr...@gmx.ch wrote:
 Thank you Rui, that helped a lot. The correct values show up when I'm using
 the following code. Now  fun(Temp,v)  returns a matrix, and  Temp  and   v
 stay the same. But I'd like to use the reduced vectors in some
 calculations..can they be extracted  in some way so that I have them
 separately again?



 fun - function(Temp, v){
       unwanted - Temp = 16 | Temp = 38.5
 Temp - Temp[!unwanted]
    v - v[!unwanted]
  list(Temp=Temp, v=v)

 }

 fun(Temp,v)

 PS: thank you too,andrija..but it's not what I'm looking for


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Re: [R] fridays date to date

Inelegant, but here's one way:

d1 - Sys.Date()
d2 - Sys.Date() + 100

library(lubridate)

d - seq(d1, d2, by = day)
d[wday(d)==6]

Michael

On Thu, Mar 1, 2012 at 3:02 PM, Ben quant ccqu...@gmail.com wrote:
 Hello,

 How do I get the dates of all Fridays between two dates?

 thanks,

 Ben

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[R] Converting a string vector with names to a numeric vector with names

2012-03-01 Thread John C Nash

Not paying close attention to detail, I entered the equivalent of

pstr-c(b1=200, b2=50, b3=0.3)

when what I wanted was

pnum-c(b1=200, b2=50, b3=0.3)

There was a list thread in 2010 that shows how to deal with un-named vectors, 
but the same
lapply solution doesn't seem to work here i.e.,

pnum-lapply(pstr, as.numeric)

or similar vapply version. The names and = signs seem to mess things up.

This is clearly not critical, but it would be nice to know an appropriate 
transformation.

JN

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Re: [R] fridays date to date

2012-03-01 Thread Marc Schwartz

On Mar 1, 2012, at 2:02 PM, Ben quant wrote:

 Hello,
 
 How do I get the dates of all Fridays between two dates?
 
 thanks,
 
 Ben


Days - seq(from = as.Date(2012-03-01), 
to = as.Date(2012-07-31), 
by = day)

 str(Days)
 Date[1:153], format: 2012-03-01 2012-03-02 2012-03-03 2012-03-04 ...

# See ?weekdays

 Days[weekdays(Days) == Friday]
 [1] 2012-03-02 2012-03-09 2012-03-16 2012-03-23 2012-03-30
 [6] 2012-04-06 2012-04-13 2012-04-20 2012-04-27 2012-05-04
[11] 2012-05-11 2012-05-18 2012-05-25 2012-06-01 2012-06-08
[16] 2012-06-15 2012-06-22 2012-06-29 2012-07-06 2012-07-13
[21] 2012-07-20 2012-07-27

HTH,

Marc Schwartz

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Re: [R] function for filtering and deleting vector entries

2012-03-01 Thread babyluck

Thank you very much!! Exactly how I wanted it :)

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Re: [R] fridays date to date

Great thanks!

ben

On Thu, Mar 1, 2012 at 1:30 PM, Marc Schwartz marc_schwa...@me.com wrote:

 On Mar 1, 2012, at 2:02 PM, Ben quant wrote:

  Hello,
 
  How do I get the dates of all Fridays between two dates?
 
  thanks,
 
  Ben


 Days - seq(from = as.Date(2012-03-01),
to = as.Date(2012-07-31),
by = day)

  str(Days)
  Date[1:153], format: 2012-03-01 2012-03-02 2012-03-03 2012-03-04
 ...

 # See ?weekdays

  Days[weekdays(Days) == Friday]
  [1] 2012-03-02 2012-03-09 2012-03-16 2012-03-23 2012-03-30
  [6] 2012-04-06 2012-04-13 2012-04-20 2012-04-27 2012-05-04
 [11] 2012-05-11 2012-05-18 2012-05-25 2012-06-01 2012-06-08
 [16] 2012-06-15 2012-06-22 2012-06-29 2012-07-06 2012-07-13
 [21] 2012-07-20 2012-07-27

 HTH,

 Marc Schwartz



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Re: [R] Plotting a registered sign



On Mar 1, 2012, at 2:40 PM, Dennis Fisher wrote:


R 2.14.0
OS X

Marc's proposed solution (appearing at the end of this email) is  
perfect -- thanks so much

However, some questions remain:

1.  The following works:
plot(1, type=n)
text(1, 1, expression(symbol(\342)))

but this does not work (TEXT appears in a symbol font)
plot(1, type=n)
TEXT- \342
text(1, 1, expression(symbol(TEXT)))


?substitute
  text(1, 1, substitute( symbol(TEXT), list(TEXT=TEXT) )  )

?bquote

text(1, 1, bquote( symbol( .(TEXT )  ) ) )



nor does this work (it yields:	Error: could not find function  
symbol)

plot(1, type=n)
TEXT- symbol(\342)
text(1, 1, expression(TEXT))
Clearly, I do not understand the subtleties of plotmath.


expression() does not evaluate names within its arguments. Similar  
issues arise with the use of $ as a function. That's why it's often  
the answer to programming questions that one should use obj[[colnam]]  
than it is to use obj$colnam. [[ does evaluate its argument.



 Could someone explain?


Try instead:

plot(1, type=n)
TEXT- expression(symbol(\342))
text(1, 1, TEXT)

If you want to build up from character-type components then you can  
use paste and wrap the whole think in as.expression(). bquote() is  
much better, though.


--
David.



2.  How would one learn that \342 corresponds to the copyright sign  
in the symbol font? (i.e., is there a list of all possible characters)




Dennis Fisher MD
P  (The P Less Than Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com

On Feb 29, 2012, at 6:55 PM, Marc Schwartz wrote:


Dennis,

Depending upon some additional fine tuning, here is a generic  
example:


plot(1)
legend(topright, pch = 1, legend = expression(SOMETEXT ^  
symbol(\342)))



See ?plotmath for more information. The above works on my MBP  
running Lion.


HTH,

Marc Schwartz




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David Winsemius, MD
West Hartford, CT

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Re: [R] Rscript example

On Thu, Mar 01, 2012 at 11:53:00AM -0800, statquant2 wrote:
 Hi there,
 I am trying to find an example how to use Rscript
 Let's suppose I want to pass 3 arguments (I don't want [options] and -e
 [expressions] as described in help) 
 
 *on the command line
 myRscript.R -arg1=value1 -arg2=value2 -arg3=value3
 
 *In the script
 #! /path/to/Rscript
 args = commandArgs(TRUE);
 
 From what I see args is just a string, do I do things correctly ?

Hi.

Extending your example with print(args), i obtained

  ./myRscript.R -arg1=value1 -arg2=value2 -arg3=value3
  [1] -arg1=value1 -arg2=value2 -arg3=value3

If you can use a fixed order of the arguments, then you
need not use the -arg[i]= parts. The arguments are
strings on the command line, so they are also passed
to R as strings, but you can convert them inside the
script. For example, with the above script, i get

  ./myRscript.R 23 45 78
  [1] 23 45 78

However, if the script contains

   args - as.numeric(args)

, then i get

  ./myRscript.R 23 45 78
  [1] 23 45 78

which is a numeric vector.

If you want to keep the arg[i]= structure, you can have in
the script

  x - strsplit(args, =)
  print(x)

and a call produces

  ./myRscript.R -arg1=value1 -arg2=value2 -arg3=value3
  [[1]]
  [1] -arg1  value1
  
  [[2]]
  [1] -arg2  value2
  
  [[3]]
  [1] -arg3  value3

The list x may then be further analyzed.

Hope this helps.

Petr Savicky.

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Re: [R] Converting a string vector with names to a numeric vector with names

2012-03-01 Thread Peter Langfelder

On Thu, Mar 1, 2012 at 12:28 PM, John C Nash nas...@uottawa.ca wrote:
 Not paying close attention to detail, I entered the equivalent of

 pstr-c(b1=200, b2=50, b3=0.3)

 when what I wanted was

 pnum-c(b1=200, b2=50, b3=0.3)

 There was a list thread in 2010 that shows how to deal with un-named vectors, 
 but the same
 lapply solution doesn't seem to work here i.e.,

 pnum-lapply(pstr, as.numeric)

 or similar vapply version. The names and = signs seem to mess things up.

 This is clearly not critical, but it would be nice to know an appropriate 
 transformation.

Here's one:

split = sapply(strsplit(pstr, split = =), I);

pnum = as.numeric(split[2, ]);
names(pnum) = split[1, ];

 pnum
   b1b2b3
200.0  50.0   0.3

HTH

Peter

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[R] fill data forward in data frame.

Hello,

My direct desire is a good (fast) way to fill values forward until there is
another value then fill that value foward in the data xx (at the bottom of
this email).  For example, from row 1 to row 45 should be NA (no change),
but from row 46 row 136 the value should be 12649, and from row 137 to the
next value should be 13039.00.  The last line of code is all you need for
this part.

If you are so inclined, my goal is this: I want to create a weekly time
series out of some data based on the report date. The report date is 'rd'
below, and is the correct date for the time series. My idea (in part seen
below) is to align rd and ua via the incorrect date (the time series date),
then merge that using the report date (rd) and a daily series (so I capture
all of the dates) of dates (dt). That gets the data in the right start
period. I've done all of this so far below and it looks fine. Then I plan
to roll all of those values forward to the next value (see question above),
then I'll do something like this:

xx[weekdays(xx[,1]) == Friday,]

...to get a weekly series of Friday values. I'm thinking someone probably
has a faster way of doing this. I have to do this many times, so speed is
important. Thanks!

Here is what I have done so far:

dt - seq(from =as.Date(2009-06-01), to = Sys.Date(), by = day)

 nms
[1] 2009-06-30 2009-09-30 2009-12-31 2010-03-31 2010-06-30
2010-09-30 2010-12-31 2011-03-31 2011-06-30 2011-09-30
[11] 2011-12-31

 rd
2009-06-30   2009-09-30   2009-12-31   2010-03-31   2010-06-30
2010-09-30   2010-12-31   2011-03-31   2011-06-30   2011-09-30
2009-07-16 2009-10-15 2010-01-19 2010-04-19 2010-07-19
2010-10-18 2011-01-18 2011-04-19 2011-07-18 2011-10-17
2011-12-31
2012-01-19

 ua
2009-06-30 2009-09-30 2009-12-31 2010-03-31 2010-06-30 2010-09-30
2010-12-31 2011-03-31 2011-06-30 2011-09-30 2011-12-31
12649.00   13039.00   13425.00   13731.00   14014.00   14389.00
14833.00   15095.00   15481.43   15846.43   16186.43

 x = merge(ua,rd,by='row.names')
 names(x) = c('z.date','val','rt_date')
 xx = merge(dt,x,by.y= 'rt_date',by.x=1,all.x=T)
 xx
x  z.date   val
1   2009-06-01   NANA
2   2009-06-02   NANA
3   2009-06-03   NANA
4   2009-06-04   NANA
5   2009-06-05   NANA

...ect

36  2009-07-06   NANA
37  2009-07-07   NANA
38  2009-07-08   NANA
39  2009-07-09   NANA
40  2009-07-10   NANA
41  2009-07-11   NANA
42  2009-07-12   NANA
43  2009-07-13   NANA
44  2009-07-14   NANA
45  2009-07-15   NANA
46  2009-07-16 2009-06-30 12649
47  2009-07-17   NANA
48  2009-07-18   NANA
49  2009-07-19   NANA
50  2009-07-20   NANA
51  2009-07-21   NANA
52  2009-07-22   NANA
53  2009-07-23   NANA
54  2009-07-24   NANA
55  2009-07-25   NANA
56  2009-07-26   NANA
57  2009-07-27   NANA
58  2009-07-28   NANA

...ect

129  2009-10-07   NA   NA
130  2009-10-08   NA   NA
131  2009-10-09   NA   NA
132  2009-10-10   NA   NA
133  2009-10-11   NA   NA
134  2009-10-12   NA   NA
135  2009-10-13   NA   NA
136  2009-10-14   NA   NA
137  2009-10-15 2009-09-30 13039.00
138  2009-10-16   NA   NA
139  2009-10-17   NA   NA
140  2009-10-18   NA   NA
141  2009-10-19   NA   NA
142  2009-10-20   NA   NA
143  2009-10-21   NA   NA

...ect

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Re: [R] Converting a string vector with names to a numeric vector with names

On Thu, Mar 01, 2012 at 03:28:31PM -0500, John C Nash wrote:
 Not paying close attention to detail, I entered the equivalent of
 
 pstr-c(b1=200, b2=50, b3=0.3)
 
 when what I wanted was
 
 pnum-c(b1=200, b2=50, b3=0.3)
 
 There was a list thread in 2010 that shows how to deal with un-named vectors, 
 but the same
 lapply solution doesn't seem to work here i.e.,
 
 pnum-lapply(pstr, as.numeric)
 
 or similar vapply version. The names and = signs seem to mess things up.
 
 This is clearly not critical, but it would be nice to know an appropriate 
 transformation.

Hi.

Try the following.

  mat - sapply(strsplit(pstr, =), identity)
  x - as.numeric(mat[2, ])
  names(x) - mat[1, ]
  x

 b1b2b3 
  200.0  50.0   0.3 
 
Hope this helps.

Petr Savicky.

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Re: [R] Converting a string vector with names to a numeric vector with names

2012-03-01 Thread John C Nash


Gotta love R.

Thanks to Bill Dunlap, Peter Langfelder and Jim Holtman for no less than 3 different 
solutions.


JN


On 12-03-01 04:25 PM, Peter Langfelder wrote:

pstr-c(b1=200, b2=50, b3=0.3)



split = sapply(strsplit(pstr, split = =), I);

pnum = as.numeric(split[2, ]);
names(pnum) = split[1, ];



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Re: [R] Connecting points on a line with arcs/curves

2012-03-01 Thread John Kane

That's nice. 

Please read the posting guidelines and get back to us with some information on 
what the data looks like an what you are doing.

For example do you just want lines or do you want a smoother, etc?

John Kane
Kingston ON Canada


 -Original Message-
 From: ir...@cam.ac.uk
 Sent: Thu, 1 Mar 2012 07:15:52 -0800 (PST)
 To: r-help@r-project.org
 Subject: [R] Connecting points on a line with arcs/curves
 
 
 Hello,
 
 I have a spreadsheet of pairs of coordinates and I would like to plot a
 line
 along which curves/arcs connect each pair of coordinates. The aim is to
 visualise the pattern of point connections.
 
 Thanks! Ian
 
 --
 View this message in context:
 http://r.789695.n4.nabble.com/Connecting-points-on-a-line-with-arcs-curves-tp4435247p4435247.html
 Sent from the R help mailing list archive at Nabble.com.
 
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Re: [R] Help with segmented package

2012-03-01 Thread vito.muggeo

dear Phil,
plot.segmented() accepts vectorized 'col', 'lty' and 'lwd' arguments. Then,

par(mfrow=c(1,2))
plot(o.seg,col=2:3,lty=2:3,lwd=c(1,2)) 
plot(z,y)
plot(o.seg,col=2:3,lty=1,linkinv=T,add=T,lwd=2) 
 

hope this helps you,
vito


On Thu, 1 Mar 2012 08:57:39 -0800 (PST), Filoche wrote
 Hi everyone.
 
 I'm using segmented package to find break point in a bi-linear relationship.
 In a particular case, I find 1 pointcut (so 2 slopes).
 
 I would like to know if it is possible to retrieve information in the
 segmented object that could let me to plot 1 particular segment with a
 different color.
 
 For example, in that folowing example, I would like to plot the second
 segment in red.
 
 library(segmented)
 set.seed(1234)
 z-runif(100)
 y-rpois(100,exp(2+1.8*pmax(z-.6,0)))
 o-glm(y~z,family=poisson)
 o.seg-segmented(o,seg.Z=~z,psi=list(z=.5))
 plot(o.seg)
 
 Thank in advance,
 Phil
 
 --
 View this message in context: http://r.789695.n4.nabble.com/Help-with-
 segmented-package-tp4435550p4435550.html Sent from the R help mailing list 
 archive at Nabble.com.
 
 __
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[R] problem with sum function

2012-03-01 Thread Mark A. Albins


Hi!

I'm running R version 2.13.0 (2011-04-13)
Platform: i386-pc-mingw32/i386 (32-bit)

When i type in the command:

sum(c(-0.2, 0.8, 0.8, -3.2, 1.8))

R returns the value:

-5.551115e-17

Why doesn't R return zero in this case?  There shouldn't be any rounding 
error in a simple sum.


Thanks,

Mark

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Re: [R] Standard variance / devistion clarification

2012-03-01 Thread John Kane

No it's an outlier problem, I think.

If you have a fairly small number of sets of these numbers simple visual 
inspection of a boxplot for each set would probably acomplish what you want.

Try this in R for an example. Just paste the next two lines into R 

xx - c(1, 1, 2, 10, 100, 10,1)
boxplot(xx)

After this it gets more complicated, but it you're new here let's take it one 
step at a time


John Kane
Kingston ON Canada


 -Original Message-
 From: suranga...@gmail.com
 Sent: Thu, 1 Mar 2012 09:30:59 -0800
 To: r-help@r-project.org
 Subject: [R] Standard variance / devistion clarification
 
 Dear gurus,
 
 Im a newbie, and I want to ask a very general question.
 Assume that I have a set of numbers as follows,
 
 1, 1, 2, 10, 100, 10,1
 
 From these, I need to identify which number is the most different as
 compared to others. (in this case, it will be 100, since its way larger
 than the other numbers). It doesnt have to be specifically this way, but
 I
 need to identify which number(s) are most different compared to the
 others.
 
 Any idea as to what I need to do this ? Im a math noob, so I'm also going
 to need to ask it this is called 'standard deviation' or 'variance' :-)
 
 --
 Best Regards,
 
 Suranga
 
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Re: [R] problem with sum function

2012-03-01 Thread Sarah Goslee

Of course there's rounding error: your computer can't
store those decimal numbers precisely. See R FAQ 7.31 for
details.

See also:
sum(10*c(-0.2, 0.8, 0.8, -3.2, 1.8)) / 10

Sarah

On Thu, Mar 1, 2012 at 4:49 PM, Mark A. Albins kamoko...@gmail.com wrote:
 Hi!

 I'm running R version 2.13.0 (2011-04-13)
 Platform: i386-pc-mingw32/i386 (32-bit)

 When i type in the command:

 sum(c(-0.2, 0.8, 0.8, -3.2, 1.8))

 R returns the value:

 -5.551115e-17

 Why doesn't R return zero in this case?  There shouldn't be any rounding
 error in a simple sum.

 Thanks,

 Mark

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

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Re: [R] fill data forward in data frame.

On Thu, Mar 01, 2012 at 02:31:01PM -0700, Ben quant wrote:
 Hello,
 
 My direct desire is a good (fast) way to fill values forward until there is
 another value then fill that value foward in the data xx (at the bottom of
 this email).  For example, from row 1 to row 45 should be NA (no change),
 but from row 46 row 136 the value should be 12649, and from row 137 to the
 next value should be 13039.00.  The last line of code is all you need for
 this part.
 
 If you are so inclined, my goal is this: I want to create a weekly time
 series out of some data based on the report date. The report date is 'rd'
 below, and is the correct date for the time series. My idea (in part seen
 below) is to align rd and ua via the incorrect date (the time series date),
 then merge that using the report date (rd) and a daily series (so I capture
 all of the dates) of dates (dt). That gets the data in the right start
 period. I've done all of this so far below and it looks fine. Then I plan
 to roll all of those values forward to the next value (see question above),
 then I'll do something like this:
 
 xx[weekdays(xx[,1]) == Friday,]
 
 ...to get a weekly series of Friday values. I'm thinking someone probably
 has a faster way of doing this. I have to do this many times, so speed is
 important. Thanks!
 
 Here is what I have done so far:
 
 dt - seq(from =as.Date(2009-06-01), to = Sys.Date(), by = day)
 
  nms
 [1] 2009-06-30 2009-09-30 2009-12-31 2010-03-31 2010-06-30
 2010-09-30 2010-12-31 2011-03-31 2011-06-30 2011-09-30
 [11] 2011-12-31
 
  rd
 2009-06-30   2009-09-30   2009-12-31   2010-03-31   2010-06-30
 2010-09-30   2010-12-31   2011-03-31   2011-06-30   2011-09-30
 2009-07-16 2009-10-15 2010-01-19 2010-04-19 2010-07-19
 2010-10-18 2011-01-18 2011-04-19 2011-07-18 2011-10-17
 2011-12-31
 2012-01-19
 
  ua
 2009-06-30 2009-09-30 2009-12-31 2010-03-31 2010-06-30 2010-09-30
 2010-12-31 2011-03-31 2011-06-30 2011-09-30 2011-12-31
 12649.00   13039.00   13425.00   13731.00   14014.00   14389.00
 14833.00   15095.00   15481.43   15846.43   16186.43
 
  x = merge(ua,rd,by='row.names')
  names(x) = c('z.date','val','rt_date')
  xx = merge(dt,x,by.y= 'rt_date',by.x=1,all.x=T)
  xx
 x  z.date   val
 1   2009-06-01   NANA
 2   2009-06-02   NANA
 3   2009-06-03   NANA
 4   2009-06-04   NANA
 5   2009-06-05   NANA
 
 ...ect
 
 36  2009-07-06   NANA
 37  2009-07-07   NANA
 38  2009-07-08   NANA
 39  2009-07-09   NANA
 40  2009-07-10   NANA
 41  2009-07-11   NANA
 42  2009-07-12   NANA
 43  2009-07-13   NANA
 44  2009-07-14   NANA
 45  2009-07-15   NANA
 46  2009-07-16 2009-06-30 12649
 47  2009-07-17   NANA
 48  2009-07-18   NANA
 49  2009-07-19   NANA
 50  2009-07-20   NANA
 51  2009-07-21   NANA
 52  2009-07-22   NANA
 53  2009-07-23   NANA
 54  2009-07-24   NANA
 55  2009-07-25   NANA
 56  2009-07-26   NANA
 57  2009-07-27   NANA
 58  2009-07-28   NANA
 
 ...ect
 
 129  2009-10-07   NA   NA
 130  2009-10-08   NA   NA
 131  2009-10-09   NA   NA
 132  2009-10-10   NA   NA
 133  2009-10-11   NA   NA
 134  2009-10-12   NA   NA
 135  2009-10-13   NA   NA
 136  2009-10-14   NA   NA
 137  2009-10-15 2009-09-30 13039.00
 138  2009-10-16   NA   NA
 139  2009-10-17   NA   NA
 140  2009-10-18   NA   NA
 141  2009-10-19   NA   NA
 142  2009-10-20   NA   NA
 143  2009-10-21   NA   NA

Hi.

Try first the following simpler version.

  # an input vector
  x - rep(NA, times=20)
  x[4] - A
  x[9] - B
  x[17] - C

  # extending the values forward 
  values - c(NA, x[!is.na(x)])
  ind - cumsum(!is.na(x)) + 1
  y - values[ind]

  # compare with the original
  cbind(x, y)

x   y  
   [1,] NA  NA 
   [2,] NA  NA 
   [3,] NA  NA 
   [4,] A A
   [5,] NA  A
   [6,] NA  A
   [7,] NA  A
   [8,] NA  A
   [9,] B B
  [10,] NA  B
  [11,] NA  B
  [12,] NA  B
  [13,] NA  B
  [14,] NA  B
  [15,] NA  B
  [16,] NA  B
  [17,] C C
  [18,] NA  C
  [19,] NA  C
  [20,] NA  C

This could be applied directly to the last two columns of your
data frame xx. However, it may be more natural to obtain the
vector values from the input data and not from their sparse
form, which is the data frame. Also, the logical vector !is.na(x)
is the same for the last two columns of your data frame, so
it may be computed only once.

Hope this helps.

Petr Savicky.

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Re: [R] problem with sum function

2012-03-01 Thread William Dunlap

In base ten, using any fixed number of digits, compute
   1/3 + 1/3 + 1/3
(doing the divisions before the additions).
Why isn't it 1?

1/5 has the same sort of problem in base two.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf Of Mark A. Albins
 Sent: Thursday, March 01, 2012 1:50 PM
 To: r-help@r-project.org
 Subject: [R] problem with sum function
 
 Hi!
 
 I'm running R version 2.13.0 (2011-04-13)
 Platform: i386-pc-mingw32/i386 (32-bit)
 
 When i type in the command:
 
 sum(c(-0.2, 0.8, 0.8, -3.2, 1.8))
 
 R returns the value:
 
 -5.551115e-17
 
 Why doesn't R return zero in this case?  There shouldn't be any rounding
 error in a simple sum.
 
 Thanks,
 
 Mark
 
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Re: [R] problem with sum function

On Thu, Mar 01, 2012 at 01:49:44PM -0800, Mark A. Albins wrote:
 Hi!
 
 I'm running R version 2.13.0 (2011-04-13)
 Platform: i386-pc-mingw32/i386 (32-bit)
 
 When i type in the command:
 
 sum(c(-0.2, 0.8, 0.8, -3.2, 1.8))
 
 R returns the value:
 
 -5.551115e-17
 
 Why doesn't R return zero in this case?  There shouldn't be any rounding 
 error in a simple sum.

Hi.

There is a rounding error, since numerical values are represented
in binary system and, for example, 0.2 = 1/5 cannot be represented
in binary exactly. A simpler version is

  0.1 + 0.2 - 0.3

  [1] 5.551115e-17

Use round(), for example

  round(sum(c(-0.2, 0.8, 0.8, -3.2, 1.8)), digits=7)

  [1] 0

See FAQ 7.31 and/or 

  http://rwiki.sciviews.org/doku.php?id=misc:r_accuracy

for further hints.

Hope this helps.

Petr Savicky.

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[R] Memory problem in R

2012-03-01 Thread saqlain raza

Hi all, I am running an -MNP- multinomial probit model package using R. It 
gives me the following objection instead of giving me the results:
Erreur : impossible d'allouer un vecteur de taille 137.9 Mo (in english: cannot 
allocate a 137.9 Mb vector memory). 

I have already increased the memory size upto 2047Mb. This problem has been 
discussed in 2008 (archives) but no profitable answers were delievered. I am 
sending the programming and bit of data. Thanks in advance for your help.

model-mnp(choice~1+Asso+FP+BEV+IAA+CER+FrtVeg+Meat+Others+lnADH+lnTO+Emp+LT50inReg+lnEXinEU+lnEXoutEU+GDist+RET+WS+LT50SC+Others_hotel,
base=4, n.draws=1, burnin=2000, thin=3, verbose=TRUE, trace=FALSE,
p.scale=1, coef.start=0, invcdf=FALSE, cov.start=1)

AssoFPCGIAABEVCERFrtVegMeatMilkOilOtherslnADHlnTOEmpLT50inReglnEXinEUlnEXoutEUGDistRETWSLT50SCOthers_hotelchoice
0001103,68897,55331,,0,010004
0001104,86759,01681,,0,14
0001104,90538,98720,,0,010004
0001104,41888,03530,,0,000104
0001105,634815,19545,,69310,000101
10100010003,401214,59170,0010,0,001001
0001104,499815,12016,,0,14
1101103,784213,49204,,0,6931001001
1001101,38638,86052,,0,02
 
Saqlain RAZA
PhD Student
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Re: [R] fill data forward in data frame.

That is great! Thank you very much.

Ben

On Thu, Mar 1, 2012 at 2:57 PM, Petr Savicky savi...@cs.cas.cz wrote:

 On Thu, Mar 01, 2012 at 02:31:01PM -0700, Ben quant wrote:
  Hello,
 
  My direct desire is a good (fast) way to fill values forward until there
 is
  another value then fill that value foward in the data xx (at the bottom
 of
  this email).  For example, from row 1 to row 45 should be NA (no change),
  but from row 46 row 136 the value should be 12649, and from row 137 to
 the
  next value should be 13039.00.  The last line of code is all you need for
  this part.
 
  If you are so inclined, my goal is this: I want to create a weekly time
  series out of some data based on the report date. The report date is 'rd'
  below, and is the correct date for the time series. My idea (in part seen
  below) is to align rd and ua via the incorrect date (the time series
 date),
  then merge that using the report date (rd) and a daily series (so I
 capture
  all of the dates) of dates (dt). That gets the data in the right start
  period. I've done all of this so far below and it looks fine. Then I plan
  to roll all of those values forward to the next value (see question
 above),
  then I'll do something like this:
 
  xx[weekdays(xx[,1]) == Friday,]
 
  ...to get a weekly series of Friday values. I'm thinking someone probably
  has a faster way of doing this. I have to do this many times, so speed is
  important. Thanks!
 
  Here is what I have done so far:
 
  dt - seq(from =as.Date(2009-06-01), to = Sys.Date(), by = day)
 
   nms
  [1] 2009-06-30 2009-09-30 2009-12-31 2010-03-31 2010-06-30
  2010-09-30 2010-12-31 2011-03-31 2011-06-30 2011-09-30
  [11] 2011-12-31
 
   rd
  2009-06-30   2009-09-30   2009-12-31   2010-03-31   2010-06-30
  2010-09-30   2010-12-31   2011-03-31   2011-06-30   2011-09-30
  2009-07-16 2009-10-15 2010-01-19 2010-04-19 2010-07-19
  2010-10-18 2011-01-18 2011-04-19 2011-07-18 2011-10-17
  2011-12-31
  2012-01-19
 
   ua
  2009-06-30 2009-09-30 2009-12-31 2010-03-31 2010-06-30 2010-09-30
  2010-12-31 2011-03-31 2011-06-30 2011-09-30 2011-12-31
  12649.00   13039.00   13425.00   13731.00   14014.00   14389.00
  14833.00   15095.00   15481.43   15846.43   16186.43
 
   x = merge(ua,rd,by='row.names')
   names(x) = c('z.date','val','rt_date')
   xx = merge(dt,x,by.y= 'rt_date',by.x=1,all.x=T)
   xx
  x  z.date   val
  1   2009-06-01   NANA
  2   2009-06-02   NANA
  3   2009-06-03   NANA
  4   2009-06-04   NANA
  5   2009-06-05   NANA
 
  ...ect
 
  36  2009-07-06   NANA
  37  2009-07-07   NANA
  38  2009-07-08   NANA
  39  2009-07-09   NANA
  40  2009-07-10   NANA
  41  2009-07-11   NANA
  42  2009-07-12   NANA
  43  2009-07-13   NANA
  44  2009-07-14   NANA
  45  2009-07-15   NANA
  46  2009-07-16 2009-06-30 12649
  47  2009-07-17   NANA
  48  2009-07-18   NANA
  49  2009-07-19   NANA
  50  2009-07-20   NANA
  51  2009-07-21   NANA
  52  2009-07-22   NANA
  53  2009-07-23   NANA
  54  2009-07-24   NANA
  55  2009-07-25   NANA
  56  2009-07-26   NANA
  57  2009-07-27   NANA
  58  2009-07-28   NANA
 
  ...ect
 
  129  2009-10-07   NA   NA
  130  2009-10-08   NA   NA
  131  2009-10-09   NA   NA
  132  2009-10-10   NA   NA
  133  2009-10-11   NA   NA
  134  2009-10-12   NA   NA
  135  2009-10-13   NA   NA
  136  2009-10-14   NA   NA
  137  2009-10-15 2009-09-30 13039.00
  138  2009-10-16   NA   NA
  139  2009-10-17   NA   NA
  140  2009-10-18   NA   NA
  141  2009-10-19   NA   NA
  142  2009-10-20   NA   NA
  143  2009-10-21   NA   NA

 Hi.

 Try first the following simpler version.

  # an input vector
  x - rep(NA, times=20)
  x[4] - A
  x[9] - B
  x[17] - C

  # extending the values forward
  values - c(NA, x[!is.na(x)])
  ind - cumsum(!is.na(x)) + 1
  y - values[ind]

  # compare with the original
  cbind(x, y)

x   y
   [1,] NA  NA
   [2,] NA  NA
   [3,] NA  NA
   [4,] A A
   [5,] NA  A
   [6,] NA  A
   [7,] NA  A
   [8,] NA  A
   [9,] B B
  [10,] NA  B
  [11,] NA  B
  [12,] NA  B
  [13,] NA  B
  [14,] NA  B
  [15,] NA  B
  [16,] NA  B
  [17,] C C
  [18,] NA  C
  [19,] NA  C
  [20,] NA  C

 This could be applied directly to the last two columns of your
 data frame xx. However, it may be more natural to obtain the
 vector values from the input data and not from their sparse
 form, which is the data frame. Also, the logical vector !is.na(x)
 is the same for the last two columns of your data frame, so
 it may be computed only once.

 Hope this helps.

 Petr Savicky.

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Re: [R] how to change or copy to another the names of models

2012-03-01 Thread Jeff Newmiller

A) you will generally get a better response when your question includes 
reproducible code/sample data, and a clear identification of the desired final 
result.

B) in most cases like this, a proliferation of names is not as useful as the OP 
(you) thinks it is. Much better is to build a list of results that can be 
indexed by position or by name.

mymodel - vector( list, 30 )
for (i in 5:30) {
  mymodel[[i]] - mlp(X, Y, size=n,) 
 }
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  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.

Waldir de Carvalho Junior waldi...@gmail.com wrote:

Hi
I would like to know how I can change the name of a model for each
trainning cycle of a model.
I work with the RSNNS package and to build a neural network, I used :
for (i in 5:30) 
model_ANN - mlp(X, Y, size=n,) # where size is the number of
neurons
in the hidden layer
but I need to save each time that the model that is build (the end of
each
cycle), e.g., when i = 5, I need to save the model with a especific
name,
when i = 6, also I need to save the model with another name
How i am doing, i am saving only the last model with n = 30 and with
the
name model_ANN
Question
how can I change the name of the model (model_ANN) at each end of cycle
of
i values?
I have already tryied copy, save, rename,.. but
unsuccessfully
thanks for any answer

-- 

Waldir de Carvalho Junior
Pedologia/Pedologue
Pesquisador/Chercheur
Embrapa Solos/INRA - UMR- LISAH

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Re: [R] problem with sum function