date:20120522

On Tue, May 22, 2012 at 9:01 AM, jiangxijixzy jiangxiji...@163.com wrote:
 The function I wrote can run well with the small data, but with the large
 data, the function runs very very slowly. How can I correct it? Thank you
 very much. My function as below:

I guess this is a classic loops vs vectorization problem.

 fortune('treat')

Contrary to popular belief the speed of R's interpreter is rarely the limiting
factor to R's speed. People treating R like C is typically the limiting factor.
You have vector operations, USE THEM.
   -- Byron Ellis
  R-help (October 2005)

I would suggest that you read up on vectorization in R to understand
why it is often preferred to loops. Two obvious places are an Rnews
article by John Fox (if I remember well) and his Companion to Applied
Regression book.

Moreover, please sign your messages with your real name. Regards
Liviu

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[R] What's wrong with MEAN?

2012-05-22 Thread Vincy Pyne


Dear R helpers,

I have recently installed R version 2.15.0

I just wanted to calculate 

mean(16, 18)

Surprisingly I got answer as 

 mean(16, 18)
[1] 16


 mean(18, 16)

[1] 18

 mean(14, 11, 17, 9, 5, 18)
[1] 14


So instead of calculating simple Arithmetic average, mean command is generating 
first element as average. I restarted the machine, changed the machine, but 
still the reply is same. I have been using this mean function ever since I 
strated learning R, but this has never happened.

Kindly guide

Vincy




[[alternative HTML version deleted]]

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Re: [R] barplot: legend: two rows

2012-05-22 Thread Marion Wenty

Hello Uwe,

thanks for drawing this to my attention! I tried it again without adding
another empty legend element and it worked - yes, the second row not being
visible was my problem. Now it works!

Thanks again,
Marion

2012/5/15 Uwe Ligges lig...@statistik.tu-dortmund.de



 On 15.05.2012 17:38, Marion Wenty wrote:

 Hello,

 I solved my problem with the uneven numbers of legend elements, which
 should be put into more than one rows, in the following way:

 I added another legend element which is empty () and is filled with
 white:

 legend(0,-1.4,xjust=0,ncol=3,**legend=c(colnames(dat)[-1],)**
 ,fill=c(navyblue,**steelblue4,steelblue2,**lightsteelblue3,**
 lightsteelblue1,white),bty=**n,border=white,cex=1.2)

 Not quite such a nice way, but it works, anyway.



 The previous version already worked for me, you just had to increase the
 bottom margin to make the second row of labels visible:

 For example

 par(mar=c(10,4.5+0.3529412*(**max(nchar(colnames(dat1)))-1),**1,1))

 worked for me.

 Uwe



  Marion

 2012/5/11 Marion Wentymarion.we...@gmail.com

  Hello,

 thank you very much for your answers. Using the command ncol worked in my
 special case, if I have got a legend with 4 elements, but with 5 it
 doesn't
 work. I am using two functions which I had created and checked what might
 be the difference, so that the second one doesn't work, but I couldn't
 find
 out what the problem is. So I am including the text of my .csv files and
 my
 functions:


 My .csv file abb8a looks like this:

 ;sehr gut;eher gut;eher nicht gut;nicht gut
 Kontakt mit der Schule;46;49;4;1
 Die richtige Wahl;32;58;8;2


 mw_g_4stap-function(pfad=F:/**04 Archiv/04 Programme/19
 R_Syntaxe/Grafiken_Funktionen**,abb=abb8a)
 {
   setwd(pfad)
   postscript(file=paste(abb,.**eps,sep=))
   dat-read.csv2(paste(abb,.**csv,sep=),header=T,check.**names=F)
   dat1-t(as.matrix(dat[,2:5],**nrow=2))
   colnames(dat1)-dat[,1]
   zehn-seq(10,100,10)
   xmax-max(dat1)
   ind-min(which(zehn=xmax))
   ticks-seq(0,100,10)
   par(las=1)
   par(mar=c(5.9,4.5+0.3529412*(**max(nchar(colnames(dat1)))-1),**1,1))

 barplot(dat1,width=0.61,horiz=**T,col=c(steelblue4,**
 steelblue2,lightsteelblue3,**lightsteelblue1),border=NA**
 ,axes=F,beside=F,xlim=c(0,100)**,cex.names=1.2,ylim=c(0,10))
   par(xpd=TRUE)
   abline(v = seq(10, 100, by = 10), col = white)
   par(xpd=F)
   axis(1,at=ticks,las=1,labels=**paste(ticks,%,sep=))
   par(xpd=TRUE)

 legend(0,-1.4,xjust=0,ncol=2,**legend=colnames(dat)[-1],fill=**
 c(steelblue4,steelblue2,**lightsteelblue3,**
 lightsteelblue1),bty=n,**border=white,cex=1.2)
   par(xpd=FALSE)
   dev.off()
 }
 mw_g_4stap()

 ###

 My .csv file probe8_5 looks like this:

 ;stimme zu;stimme zu;stimme zu;stimme zu;stimme zu
 Eltern konnten mir beim ler;23;70;1;1;5
 leichte Ent;20;10;10;10;50
 leichte Ent;20;10;10;10;50
 leichte Ent;20;10;10;10;50
 leichte Ent;20;10;10;10;50
 leichte Ent;20;10;10;10;50
 leichte Ent;20;10;10;10;50


 mw_g_5stap-function(pfad=F:/**04 Archiv/04 Programme/19
 R_Syntaxe/Grafiken_Funktionen**,abb=probe8_5)
 {
   setwd(pfad)
   postscript(file=paste(abb,.**eps,sep=))
   dat-read.csv2(paste(abb,.**csv,sep=),header=T,check.**names=F)
   dat1-t(as.matrix(dat[,2:6],**nrow=2))
   dat1
   colnames(dat1)-dat[,1]
   dat1
   zehn-seq(10,100,10)
   xmax-max(dat1)
   ind-min(which(zehn=xmax))
   ticks-seq(0,100,10)
   par(las=1)
   par(mar=c(5,4.5+0.3529412*(**max(nchar(colnames(dat1)))-1),**1,1))

 barplot(dat1,width=0.61,horiz=**T,col=c(navyblue,**
 steelblue4,steelblue2,**lightsteelblue3,**
 lightsteelblue1),border=NA,**axes=F,beside=F,xlim=c(0,100),**
 cex.names=1.2,ylim=c(0,10))
   par(xpd=TRUE)
   abline(v = seq(10, 100, by = 10), col = white)
   par(xpd=F)
   axis(1,at=ticks,las=1,labels=**paste(ticks,%,sep=))
   par(xpd=TRUE)

 legend(0,-1.4,xjust=0,ncol=3,**legend=colnames(dat)[-1],fill=**
 c(navyblue,steelblue4,**steelblue2,lightsteelblue3,**
 lightsteelblue1),bty=n,**border=white,cex=1.2)
   dev.off()
 }
 mw_g_5stap()

 ###

 The first function works but the second one doesn't.

 Does anyone know why?

 Thank you very much for your help in advance!!

 Marion



 2012/5/9 Uwe 
 Liggeslig...@statistik.tu-**dortmund.delig...@statistik.tu-dortmund.de
 



 On 09.05.2012 13:23, Marion Wenty wrote:

  dear r-helpers,

 i have got another question:

 i am using the functions

 par(xpd=T)
 legend

 to create a legend below the x-axis. i used the parameter horiz=T.

 now i would like to put the elements of the legend in two rows:

 e.g. if my legend has got 5 elements, i would like 3 elements in one
 row
 and the last two elements in the next row.

 does anyone know how to do that?


 Example:

 bp- barplot(1)
 par(xpd = TRUE)
 legend(bp, 0, xjust=0.5, legend=letters[1:5], lwd=1:5, ncol=3)

 Uwe Ligges


  thank you very much for your help in advance!


 marion

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Re: [R] What's wrong with MEAN?

2012-05-22 Thread ONKELINX, Thierry

You'll need to pass the data as a vector.

mean(16, 18) is asking the mean of 16. 18 is passed to the second argument 
which is trim. So you are doing mean(16, trim = 18)

What you want is

mean(c(16, 18))

Best regards,

Thierry

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and 
Forest
team Biometrie  Kwaliteitszorg / team Biometrics  Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more than 
asking him to perform a post-mortem examination: he may be able to say what the 
experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not ensure 
that a reasonable answer can be extracted from a given body of data.
~ John Tukey


-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens 
Vincy Pyne
Verzonden: dinsdag 22 mei 2012 11:10
Aan: r-help@r-project.org
Onderwerp: [R] What's wrong with MEAN?


Dear R helpers,

I have recently installed R version 2.15.0

I just wanted to calculate

mean(16, 18)

Surprisingly I got answer as

 mean(16, 18)
[1] 16


 mean(18, 16)

[1] 18

 mean(14, 11, 17, 9, 5, 18)
[1] 14


So instead of calculating simple Arithmetic average, mean command is generating 
first element as average. I restarted the machine, changed the machine, but 
still the reply is same. I have been using this mean function ever since I 
strated learning R, but this has never happened.

Kindly guide

Vincy




[[alternative HTML version deleted]]

__
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and provide commented, minimal, self-contained, reproducible code.
* * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * *
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en 
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The views expressed in this message and any annex are purely those of the 
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Re: [R] What's wrong with MEAN?

2012-05-22 Thread Rainer Schuermann

 mean( 16, 18 )
[1] 16
 mean( c( 16, 18 ) )
[1] 17



On Tuesday 22 May 2012 02:10:27 Vincy Pyne wrote:
 
 Dear R helpers,
 
 I have recently installed R version 2.15.0
 
 I just wanted to calculate 
 
 mean(16, 18)
 
 Surprisingly I got answer as 
 
  mean(16, 18)
 [1] 16
 
 
  mean(18, 16)
 
 [1] 18
 
  mean(14, 11, 17, 9, 5, 18)
 [1] 14
 
 
 So instead of calculating simple Arithmetic average, mean command is 
 generating first element as average. I restarted the machine, changed the 
 machine, but still the reply is same. I have been using this mean function 
 ever since I strated learning R, but this has never happened.
 
 Kindly guide
 
 Vincy
 
 
 
 
   [[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] What's wrong with MEAN?

2012-05-22 Thread Jim Holtman

you need to provide a vector:

mean(c(16,18))

Sent from my iPad

On May 22, 2012, at 5:10, Vincy Pyne vincy_p...@yahoo.ca wrote:

 
 Dear R helpers,
 
 I have recently installed R version 2.15.0
 
 I just wanted to calculate 
 
 mean(16, 18)
 
 Surprisingly I got answer as 
 
 mean(16, 18)
 [1] 16
 
 
 mean(18, 16)
 
 [1] 18
 
 mean(14, 11, 17, 9, 5, 18)
 [1] 14
 
 
 So instead of calculating simple Arithmetic average, mean command is 
 generating first element as average. I restarted the machine, changed the 
 machine, but still the reply is same. I have been using this mean function 
 ever since I strated learning R, but this has never happened.
 
 Kindly guide
 
 Vincy
 
 
 
 
[[alternative HTML version deleted]]
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

__
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Re: [R] What's wrong with MEAN?

2012-05-22 Thread Vincy Pyne

Dear Mr. Thierry,

Thanks a lot for pointing out such a silly mistake from my side. I was simply 
wondering how come I am not getting such a simple mean. 

Thanks again.

Vincy

--- On Tue, 5/22/12, ONKELINX, Thierry thierry.onkel...@inbo.be wrote:

From: ONKELINX, Thierry thierry.onkel...@inbo.be
Subject: RE: [R] What's wrong with MEAN?
To: Vincy Pyne vincy_p...@yahoo.ca, r-help@r-project.org 
r-help@r-project.org
Received: Tuesday, May 22, 2012, 9:17 AM

You'll need to pass the data as a vector.

mean(16, 18) is asking the mean of 16. 18 is passed to the second argument 
which is trim. So you are doing mean(16, trim = 18)

What you want is

mean(c(16, 18))

Best regards,

Thierry

ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and 
Forest
team Biometrie  Kwaliteitszorg / team Biometrics  Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more than 
asking him to perform a post-mortem examination: he may be able to say what the 
experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not ensure 
that a reasonable answer can be extracted from a given body of data.
~ John Tukey


-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens 
Vincy Pyne
Verzonden: dinsdag 22 mei 2012 11:10
Aan: r-help@r-project.org
Onderwerp: [R] What's wrong with MEAN?


Dear R helpers,

I have recently installed R version 2.15.0

I just wanted to calculate

mean(16, 18)

Surprisingly I got answer as

 mean(16, 18)
[1] 16


 mean(18, 16)

[1] 18

 mean(14, 11, 17, 9, 5, 18)
[1] 14


So instead of calculating simple Arithmetic average, mean command is generating 
first element as average. I restarted the machine, changed the machine, but 
still the reply is same. I have been using this mean function ever since I 
strated learning R, but this has never happened.

Kindly guide

Vincy




        [[alternative HTML version deleted]]

__
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.
* * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * *
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en 
binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is 
door een geldig ondertekend document.
The views expressed in this message and any annex are purely those of the 
writer and may not be regarded as stating an official position of INBO, as long 
as the message is not confirmed by a duly signed document.

[[alternative HTML version deleted]]

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

[R] iterate over vector with range vector inside

2012-05-22 Thread chschulz


   Hello,
   Have anybody a hint,  how I could iterate over the vector below.
   So I'll wish only 8 iterations. Put the range vector in c() didn't help.
   for(i in c(1,2,3,4,5,11:20,21:50,51:100)) {
   sum(dfc[i,cnt]) # or similar actions
   }
   Thanks in advance
   Christian
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Re: [R] iterate over vector with range vector inside

2012-05-22 Thread Jim Holtman

use 'list' instead of 'c'



Sent from my iPad

On May 22, 2012, at 5:23, chsch...@email.de wrote:

 
   Hello,
   Have anybody a hint,  how I could iterate over the vector below.
   So I'll wish only 8 iterations. Put the range vector in c() didn't help.
   for(i in 
 c(1,2,3,4,5,11:20,21:50,51:http://spinroot.com/spin/Doc/Book91_PDF/x20v_1991.pdf))
  {
   sum(dfc[i,cnt]) # or similar actions
   }
   Thanks in advance
   Christian
 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] What's wrong with MEAN?

On Tue, May 22, 2012 at 11:22 AM, Vincy Pyne vincy_p...@yahoo.ca wrote:
 Thanks a lot for pointing out such a silly mistake from my side. I was simply 
 wondering how come I am not getting such a simple mean.

To avoid such mistakes it would help to first store your data in a vector.

 x-c(16,18)
 mean(x)
[1] 17

If you did otherwise:
 x- 16,18
Error: unexpected ',' in x- 16,

then R would complain.

Liviu

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Re: [R] confused with indexing

2012-05-22 Thread Jim Holtman

use is.null for the test

if (is.null(indexSkipped))

Sent from my iPad

On May 22, 2012, at 2:10, Alaios ala...@yahoo.com wrote:

 Dear all,
 I have a code that looks like the following (I am sorry that this is not a 
 reproducible example)
 
 
 indexSkipped-NULL
 
 
 
  code Skipped that might alter indexSkipped
 
 
 if (length(indexSkipped)==0)
 
 spatial_structure-spatial_structures_from_measurements(DataList[[i]]$Lon,DataList[[i]]$Lat,meanVector)
  
 else
 
 spatial_structure-spatial_structures_from_measurements(DataList[[i]]$Lon[-indexSkipped],DataList[[i]]$Lat[-indexSkipped],meanVector)
  # 
 
 
 
 What I am doing here is that I am processing files. Every files has a 
 measurement table and Longtitude and Latitude fields. If one file is marked 
 as invalid I keep a number of of the skipped index so to remove the element 
 of the Longtitude and Latitide vectors.
 
 1) That works correct, I was just wondering if it would be possible to remove 
 some how the given if statement and initialize the indexSkipped in such a way 
 so 
 
 the DataList[[i]]$Lon[-indexSkipped],DataList[[i]]$Lat[-indexSkipped] do 
 nothing, aka remove no element, in case the indexSkipped remains unchanged 
 (in its initial value).
 
 2) When u define a variable as empty, I usually use NULL, how I can check 
 afterwords if that holds or not. If I use  the 
 
  (indexSkipped==NULL)
 logical(0)
 
 this does not return true or false. How I can do that check?
 
 
 Iwould like to thank you in advance for your help
 
 B.R
 Alex
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Re: [R] sweave tables as images?

On Mon, May 21, 2012 at 10:29 PM, Alexander Shenkin ashen...@ufl.edu wrote:
 So, I think it will be better if I can somehow generate the tables as
 images.  Is there any way to generate tables as images in separate files
 in Sweave?  Or, is there another tree up which I should be barking?

Hmm, there are Sweave equivalents for LibreOffice [1] and MS Word [2].
Have you looked into those? You may find other suggestions in the
ReproducibleResearch Task View [3].

Liviu

[1] http://crantastic.org/packages/odfWeave
[2] http://crantastic.org/packages/SWordInstaller
[3] http://crantastic.org/task_views/ReproducibleResearch

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Re: [R] “For” calculation is so slow

Hi

 
 Dear All,
 
 The function I wrote can run well with the small data, but with the 
large
 data, the function runs very very slowly. How can I correct it? Thank 
you

Your function does not run slowly, it does not run at all.

 l(10,20)
Error in l(10, 20) : object 's' not found

No s object found. I presume that also gx and gy will not be found.

And besides I am pretty sure that you probably does not need any loop at 
all.

 very much. My function as below:
 
 a-c(1:240)
 b-c(1:240)
 l=function(a,b){
 v=0
 u=0
 uv=0
 v[1]=0

Why?

v-0
v
[1] 0
v[1]-0
v
[1] 0

So there is no difference between first v and second v


 u[1]=0
 uv[1]=0
 for (i in 1:(length(s)-1)){
 
v[i]-((gx[[i]][b,(gx[[i]][a,1]+1)])-(gx[[i]][a,gx[[i]][a,1]+1]))/(gx[[i]]
 [a,gx[[i]][a,1]+1])
 
u[i]-((gy[[i]][a,(gy[[i]][a,1]+1)])-(gy[[i]][b,gy[[i]][a,1]+1]))/(gy[[i]]
 [a,gy[[i]][a,1]+1])
 uv[i]-v[i]+u[i]
 }
 w=0
 w=mean(uv)

Your function does not return anything. Here you can test it by yourself 
on shortened version

l=function(a,b){
v=0
u=0
uv=0
v-10
u-20
uv-v+u
w=0
w=mean(uv)
}

Regards
Petr





 }
 kk-data.frame()
 for (a in 1:240){
 for (b in 1:240){
 if (ab)
 kk[a,b]=l(a,b)
 else kk[a,b]=0
 }}
 max(kk)
 
 
 --
 View this message in context: http://r.789695.n4.nabble.com/For-
 calculation-is-so-slow-tp4630830.html
 Sent from the R help mailing list archive at Nabble.com.
 
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http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] Quantmod, Xts, TTR and Postgresql

2012-05-22 Thread Joshua Ulrich

On Tue, May 22, 2012 at 2:15 AM, R-type Studios rtypestud...@gmail.com wrote:
 Hi Everyone,

 I'm currently using the latest build of R and R-Studio server (both are
 amazing products)

 I'm still very new to this but I came across this issue:

 I'm trying to do a select from postgres and put the data into and xts
 object like so:

 # Libs
 library('RPostgreSQL') # http://code.google.com/p/rpostgresql/
 library('quantmod')
 library('TTR')
 library('xts') #
 http://cran.r-project.org/web/packages/xts/vignettes/xts.pdf

 # Connect and get data
 drv - dbDriver('PostgreSQL')
 db - dbConnect(drv, host='localhost', user='postgres', dbname='technica',
 password='password')
 fr - dbGetQuery(db, 'SELECT date as Date, open as Open, high as
 High, low as Low, close as Close, volume as Volume, random() as
 Adjusted FROM stocks s INNER JOIN historical_prices hp ON s.id =
 hp.stock_id WHERE s.symbol = \'SDL\' ORDER BY date DESC limit 50')

 # copied from the mysql code in the quantmod source
 fr - data.frame(fr[,-1],row.names=fr[,1])
 fr - xts(as.matrix(fr[,-1]), order.by=as.Date(fr[,1],
 origin='1970-01-01'), src='technica', updated=Sys.time())
 colnames(fr) - paste('SDL',
 c('Open','High','Low','Close','Volume','Adjusted'), sep='.')

 dvi - DVI(Cl(fr))
 print(dvi)

 When the code is executed I receive the error for the line dvi -
 DVI(Cl(fr)): Error in runSum(x, n) : Invalid 'n'

The defaults for DVI's magnitude and stretch arguments require there
be at least 100 observations.  My guess is that your 'fr' object
doesn't have the required 100 observations.  If it doesn't, then using
the default value of n = 252 may also be an issue (giving misleading
results, if not an error).

 But if I do this (fetch the data from yahoo):

 getSymbols(SDL.AX)
 dvi - DVI(Cl(SDL.AX))
 print(dvi)

 All seems to work fine, but I cant see any difference when I print the two
 data sets out, except the data set size.

 Any ideas what I'm doing wrong? Id really like to be able to import from my
 postgres database.

 Also i would be happy to write a postgres routine and submit it to quantmod
 if i get this working

 Thanks!
 Max


Best,
--
Joshua Ulrich  |  FOSS Trading: www.fosstrading.com

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Re: [R] Need to help to get value for bigger calculation

2012-05-22 Thread Hans W Borchers

Rehena Sultana hena_4567 at yahoo.com writes:

 I want to calculate values like 15^200 or 17^300 in R. In normal case it can 
 calculate the small values of b (a^b).
 I have fixed width = 1 and digits = 22 but still answers are Inf.
 
 How to deal the cases like these? Thanks in advance.

library(Rmpfr)
m15 - mpfr(15, precBits= 1024)
m15^200
165291991078820803015600259355571011187461128806050897708002963982861165
279305672605355515846488679511983197774726563035424119280679882914553697
406070345089013507994161512883544043567583004683422057135011584705353016
03376865386962890625

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Re: [R] Quantmod, Xts, TTR and Postgresql

2012-05-22 Thread Joshua Ulrich

On Tue, May 22, 2012 at 5:54 AM, R-type Studios rtypestud...@gmail.com wrote:
 Hi Joshua,

 Oh snap, awesome to have the author of the blog your reading at that
 moment to reply to your message.

 I updated the query to be 200 sessions, the function now prints, but appears
 to get the dates all wrong.  (I actually was following along with the post
 here:
 http://blog.fosstrading.com/2011/03/how-to-backtest-strategy-in-r.html great
 blog by the way.)

 What I have now (simplified it a bit):

 # Libs
 library('RPostgreSQL') # http://code.google.com/p/rpostgresql/
 library('quantmod')
 library('TTR')

 # Connect and get data


 drv - dbDriver('PostgreSQL')
 db - dbConnect(drv, host='localhost', user='postgres', dbname='technica',
 password='password')
 fr - dbGetQuery(db, 'SELECT date as Date, open as Open, high as High,
 low as Low, close as Close, volume as Volume, 0 as Adjusted FROM
 stocks s INNER JOIN historical_prices hp ON s.id = hp.stock_id WHERE
 s.symbol = \'SDL\' ORDER BY date DESC limit 200')

 fr - data.frame(fr[,-1], row.names=fr[,1])
 fr - xts(as.matrix(fr[,-1]), order.by=as.Date(fr[,1]), updated=Sys.time())

Your problem is likely here and has nothing to do with DVI.  Take a
step back and look at your intermediate objects.

The result of dbGetQuery is a data.frame with columns Date, OHLC,
Volume, and Adjusted.  Then you overwrite 'fr' with another data.frame
that drops the Date column after setting them as the row names.  Then
you overwrite 'fr' again with an xts object that contains HLC, Volume,
and Adjusted columns, ordered by the Open prices converted to Dates.

 dvi - DVI(Cl(fr))
 print(dvi)

 # Clean up.
 dbDisconnect(db)
 dbUnloadDriver(drv)

 Then as you can see below the dates are all out of whack, My guess was it
 was the date format that postgres returns (being mm/dd/) but no luck,
 any ideas?

See above.  If my assumptions aren't correct, then please provide a
small sample of the result from dbGetQuery.

 Thanks!
 Max

   dvi.mag dvi.str    dvi
 1970-01-01  NA  NA NA
 (Repeated a few hundread times)
 Then this:
 1970-01-01 0.134920635 0.349206349 0.1778
snip
 1970-01-01 0.734126984 0.996031746 0.78650794


 On Tue, May 22, 2012 at 6:26 PM, Joshua Ulrich josh.m.ulr...@gmail.com
 wrote:

 On Tue, May 22, 2012 at 2:15 AM, R-type Studios rtypestud...@gmail.com
 wrote:
  Hi Everyone,
 
  I'm currently using the latest build of R and R-Studio server (both are
  amazing products)
 
  I'm still very new to this but I came across this issue:
 
  I'm trying to do a select from postgres and put the data into and xts
  object like so:
 
  # Libs
  library('RPostgreSQL') # http://code.google.com/p/rpostgresql/
  library('quantmod')
  library('TTR')
  library('xts') #
  http://cran.r-project.org/web/packages/xts/vignettes/xts.pdf
 
  # Connect and get data
  drv - dbDriver('PostgreSQL')
  db - dbConnect(drv, host='localhost', user='postgres',
  dbname='technica',
  password='password')
  fr - dbGetQuery(db, 'SELECT date as Date, open as Open, high as
  High, low as Low, close as Close, volume as Volume, random() as
  Adjusted FROM stocks s INNER JOIN historical_prices hp ON s.id =
  hp.stock_id WHERE s.symbol = \'SDL\' ORDER BY date DESC limit 50')
 
  # copied from the mysql code in the quantmod source
  fr - data.frame(fr[,-1],row.names=fr[,1])
  fr - xts(as.matrix(fr[,-1]), order.by=as.Date(fr[,1],
  origin='1970-01-01'), src='technica', updated=Sys.time())
  colnames(fr) - paste('SDL',
  c('Open','High','Low','Close','Volume','Adjusted'), sep='.')
 
  dvi - DVI(Cl(fr))
  print(dvi)
 
  When the code is executed I receive the error for the line dvi -
  DVI(Cl(fr)): Error in runSum(x, n) : Invalid 'n'
 
 The defaults for DVI's magnitude and stretch arguments require there
 be at least 100 observations.  My guess is that your 'fr' object
 doesn't have the required 100 observations.  If it doesn't, then using
 the default value of n = 252 may also be an issue (giving misleading
 results, if not an error).

  But if I do this (fetch the data from yahoo):
 
  getSymbols(SDL.AX)
  dvi - DVI(Cl(SDL.AX))
  print(dvi)
 
  All seems to work fine, but I cant see any difference when I print the
  two
  data sets out, except the data set size.
 
  Any ideas what I'm doing wrong? Id really like to be able to import from
  my
  postgres database.
 
  Also i would be happy to write a postgres routine and submit it to
  quantmod
  if i get this working
 
  Thanks!
  Max
 

 Best,
 --
 Joshua Ulrich  |  FOSS Trading: www.fosstrading.com



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[R] getting a Likert plot from a data frame

2012-05-22 Thread Gavin Rudge

I'm creating a stacked bar chart using the likert command in the HH package.  
My data are in a data frame, with two numeric variables and a categorical 
variable, I can't get likert to use the column containing the categorical 
variable as a my y axis label.

Here is a quick example:

library(HH)
#my data are:
df-data.frame(as.character(cat=c(group1,group2,group3,group4, 
group5)),males=c(20,30,45,12,5),females=c(35,23,32,8,5))
#make a pyramid Likert chart
p-likert(df)
as.pyramidLikert(p)

It tries to plot three variables here when I just want two. I think I 
understand what is happening - my categorical variable is treated as a factor 
and I think it gets inserted as an integer into the matrix which the command 
derives from my data fame, to make the plot with(?) It's then used as a 
variable to be plotted just like the other two variables. what I don't get is 
how the example given in the package does something differently, which is how I 
want mine to work.

## Population Pyramid
data(USAge.table)
USA79 - USAge.table[75:1, 2:1, 1979]/100
PL - likert(USA79,
 main=Population of United States 1979 (ages 0-74),
 xlab=Count in Millions,
 ylab=Age,
 scales=list(
   y=list(
 limits=c(0,77),
 at=seq(1,76,5),
 labels=seq(0,75,5),
 tck=.5))
 )
PL
as.pyramidLikert(PL)

This does exactly what I'm trying to achieve.  here the two population counts 
are plotted in the likert plot and the age groups in the first columns are used 
as labels.

I can't work out why in my example the age group variable is not used in the 
same way as the in my plot in the same way as the agegroups in this example, 
other than the example takes it's data from a table and mine is coming from a 
data frame.  The end point I want is a stacked Likert bar chart based on a data 
frame where the column containing the description of my groups is used as the y 
axis labels and the other two columns are used to draw the bars. I'm sure I'm 
missing a simple solution.  any help gratefully received.

Gavin.

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[R] package grid: mirror grob objects along an axis

2012-05-22 Thread Thomas Zumbrunn

Hi everyone

I'd like to flip grobs (grid graphical objects) along an axis, e.g. flip grobs 
horizontally or vertically. I couldn't find any hints, neither in the 
documentation nor by searching the web. Does anybody know how to achieve this?

Cheers
/thomas

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Re: [R] variate generation

2012-05-22 Thread Mohan Radhakrishnan

I think part of my question about using R facilities from a program is
answered by http://binfalse.de/2011/02/talking-r-through-java/



Thanks.

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mohan Radhakrishnan
Sent: Tuesday, May 22, 2012 10:33 AM
To: R. Michael Weylandt
Cc: r-help@r-project.org
Subject: Re: [R] variate generation

I was trying to understand how to use R to generate distributions of
data, for example, uniform, and use the data in a program. I send raw
bytes to the server. Is there a recommended way or book that I should
read to understand this ?

I use R but this is a beginner question. When I plot the no: of bytes
against time I would like to ensure that over a period of 5 minutes the
data fits a uniform distribution or an exponential one. This is to
stress the SUT.


Mohan

-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] 
Sent: Monday, May 21, 2012 8:39 PM
To: Mohan Radhakrishnan
Cc: r-help@r-project.org
Subject: Re: [R] variate generation

And what distribution would that be 

R provides many built in distributions, but if those aren't enough for
you, you can check:
http://cran.r-project.org/web/views/Distributions.html

Best,
Michael

On Mon, May 21, 2012 at 7:26 AM, Mohan Radhakrishnan moh...@fss.co.in
wrote:
 Hi,

   I plot no: of bytes against time and find the distribution
 curve using R. These bytes are sent from the client to the server.

 Is there a way to generate bytes randomly using R according to a
 distribution ? I would like to send these bytes to the server. Hope I
am
 not misguided here. My goal is to simulate a certain distribution of
 bytes.



 Thanks,

 Mohan



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Re: [R] include a dataset in my package

2012-05-22 Thread Uwe Ligges




On 22.05.2012 06:24, Richard M. Heiberger wrote:

you do have a dataset x. it is probably inside the test.rda file.

start a fresh R session and library(yourPackage)

then
ls()
data(test)


No, you meant

data(x)

Note that the name of the filename is still the original mname it had 
hen creating the rda file, the filename of the rda file is irrelevant 
(and it may contain more than just one object).


Best,
Uwe Ligges




ls()  ## you will probably have now have x.

Should you need to use load, then use
load(/full/path/to/test.rda)  ## in quotes
ls()

The idiom for saving a dataset is
save(mydataset, file=mydataset.rda)

On Mon, May 21, 2012 at 8:45 PM, di jianingjianin...@gmail.com  wrote:


Hey R-users,

I think I followed the steps but still couldn't figure this out.. I am
creating a personal package and I want to include several datasets in
the package. I created a subdirectory 'data' in the package, save a
dataset 'test.rda' there, built the package, checked it, installed it.
Then I loaded the package and tried load(test), data(test),
attach(test), none of them gave me the actual data. Another thing,
which I am not sure if it is relevant, is that when I was checking the
package before installation, I got a warning (not an error) that I
have a dataset 'x' without document. I actually don't have any dataset
with the name 'x'.

Any thoughts?

thanks!

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[R] Interpreting output from the val.prob function in the rms package

2012-05-22 Thread Gustav Nilsonne

Dear mail-list,

I have used the val.prob function from the rms package to validate a logistic 
regression model. It is however not clear to me how to interpret the two 
different curves that are generated (logistic vs. nonparametric). The 
documentation doesn't enter into any detail on this point, and to a 
non-statistician like myself it isn't obvious.

If I compare the output to that of the val.prob.ci function by E. Steyerberg, 
it looks like the non-parametric curve is a smoothed line between the means of 
quantiles. But what is then the logistic calibration line?

Most grateful for any help to understand this.

Best, Gustav


Gustav Nilsonne, MD, PhD
Postdoc
+46 (0) 736-798 743

Stockholm University
Stress Research Institute
106 91 Stockholm
gustav.nilso...@stressforskning.su.se

Karolinska Institutet
Department of Clinical Neuroscience and Osher Center for Integrative Medicine
Retzius väg 8 A2:3
171 77 Stockholm
gustav.nilso...@ki.se

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[R] How to remove square brackets, etc. from address strings?

2012-05-22 Thread Sabina Arndt


Hello,



I'd like to remove the individual pairs of square brackets along with 
their content - plus the space directly behind it - from address strings
 such as this:


  [Swidsinski, Alexander; Loening-Baucke, Vera; Lochs, Herbert] Charite 
Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; [Hale, Laura P.] 
Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA



 I'd like get the following result:



  Charite Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; Duke Univ, Med 
Ctr, Dept Pathol, Durham, NC 27710 USA



I tried


  address = gsub((.*)[(.*)], \\2, address)



But this deletes everything from the first opening bracket to the last closing 
bracket and leaves only the very last address:


  Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA



How can I remove only the individual pairs of square brackets along with their 
content?



Thank you very much in advance!   
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Re: [R] Quantmod, Xts, TTR and Postgresql

2012-05-22 Thread R-type Studios

Hi Joshua,

Oh snap, awesome to have the author of the blog your reading at that
moment to reply to your message.

I updated the query to be 200 sessions, the function now prints, but
appears to get the dates all wrong.  (I actually was following along with
the post here:
http://blog.fosstrading.com/2011/03/how-to-backtest-strategy-in-r.html great
blog by the way.)

What I have now (simplified it a bit):

# Libs
library('RPostgreSQL') # http://code.google.com/p/rpostgresql/
library('quantmod')
library('TTR')

# Connect and get data
drv - dbDriver('PostgreSQL')
db - dbConnect(drv, host='localhost', user='postgres', dbname='technica',
password='password')
fr - dbGetQuery(db, 'SELECT date as Date, open as Open, high as
High, low as Low, close as Close, volume as Volume, 0 as Adjusted
FROM stocks s INNER JOIN historical_prices hp ON s.id = hp.stock_id WHERE
s.symbol = \'SDL\' ORDER BY date DESC limit 200')

fr - data.frame(fr[,-1], row.names=fr[,1])
fr - xts(as.matrix(fr[,-1]), order.by=as.Date(fr[,1]), updated=Sys.time())

dvi - DVI(Cl(fr))
print(dvi)
# Clean up.
dbDisconnect(db)
dbUnloadDriver(drv)

Then as you can see below the dates are all out of whack, My guess was it
was the date format that postgres returns (being mm/dd/) but no luck,
any ideas?

Thanks!
Max

  dvi.mag dvi.strdvi
1970-01-01  NA  NA NA
(Repeated a few hundread times)
Then this:
1970-01-01 0.134920635 0.349206349 0.1778
1970-01-01 0.079365079 0.369047619 0.13730159
1970-01-01 0.071428571 0.261904762 0.10952381
1970-01-01 0.067460317 0.281746032 0.11031746
1970-01-01 0.091269841 0.38889 0.15079365
1970-01-01 0.123015873 0.289682540 0.15634921
1970-01-01 0.170634921 0.289682540 0.1944
1970-01-01 0.075396825 0.289682540 0.11825397
1970-01-01 0.023809524 0.289682540 0.07698413
1970-01-01 0.023809524 0.492063492 0.11746032
1970-01-01 0.02778 0.492063492 0.12063492
1970-01-01 0.039682540 0.492063492 0.13015873
1970-01-01 0.285714286 0.575396825 0.34365079
1970-01-01 0.531746032 0.575396825 0.54047619
1970-01-01 0.559523810 0.615079365 0.57063492
1970-01-01 0.587301587 0.650793651 0.6000
1970-01-01 0.531746032 0.650793651 0.5556
1970-01-01 0.313492063 0.619047619 0.37460317
1970-01-01 0.202380952 0.52778 0.26746032
1970-01-01 0.198412698 0.452380952 0.24920635
1970-01-01 0.146825397 0.563492063 0.23015873
1970-01-01 0.198412698 0.6 0.26984127
1970-01-01 0.230158730 0.4 0.27301587
1970-01-01 0.309523810 0.456349206 0.3389
1970-01-01 0.329365079 0.341269841 0.33174603
1970-01-01 0.182539683 0.119047619 0.16984127
1970-01-01 0.063492063 0.130952381 0.07698413
1970-01-01 0.095238095 0.373015873 0.15079365
1970-01-01 0.043650794 0.488095238 0.13253968
1970-01-01 0.02778 0.353174603 0.09285714
1970-01-01 0.285714286 0.234126984 0.27539683
1970-01-01 0.599206349 0.234126984 0.52619048
1970-01-01 0.420634921 0.234126984 0.3833
1970-01-01 0.396825397 0.142857143 0.34603175
1970-01-01 0.515873016 0.142857143 0.44126984
1970-01-01 0.3 0.242063492 0.31507937
1970-01-01 0.186507937 0.142857143 0.1778
1970-01-01 0.234126984 0.007936508 0.1889
1970-01-01 0.261904762 0.003968254 0.21031746
1970-01-01 0.19444 0.003968254 0.15634921
1970-01-01 0.178571429 0.003968254 0.14365079
1970-01-01 0.226190476 0.007936508 0.18253968
1970-01-01 0.273809524 0.019841270 0.22301587
1970-01-01 0.329365079 0.011904762 0.26587302
1970-01-01 0.369047619 0.003968254 0.29603175
1970-01-01 0.380952381 0.003968254 0.3056
1970-01-01 0.337301587 0.007936508 0.27142857
1970-01-01 0.242063492 0.023809524 0.19841270
1970-01-01 0.206349206 0.023809524 0.16984127
1970-01-01 0.186507937 0.023809524 0.15396825
1970-01-01 0.226190476 0.05556 0.19206349
1970-01-01 0.182539683 0.05556 0.15714286
1970-01-01 0.178571429 0.05556 0.15396825
1970-01-01 0.190476190 0.099206349 0.1722
1970-01-01 0.238095238 0.091269841 0.20873016
1970-01-01 0.178571429 0.087301587 0.16031746
1970-01-01 0.269841270 0.079365079 0.23174603
1970-01-01 0.218253968 0.079365079 0.19047619
1970-01-01 0.289682540 0.150793651 0.26190476
1970-01-01 0.234126984 0.27778 0.24285714
1970-01-01 0.265873016 0.285714286 0.26984127
1970-01-01 0.309523810 0.448412698 0.33730159
1970-01-01 0.488095238 0.448412698 0.48015873
1970-01-01 0.3 0.297619048 0.32619048
1970-01-01 0.376984127 0.297619048 0.3611
1970-01-01 0.41667 0.452380952 0.42380952
1970-01-01 0.313492063 0.58333 0.36746032
1970-01-01 0.202380952 0.58333 0.27857143
1970-01-01 0.289682540 0.452380952 0.3222
1970-01-01 0.36111 0.452380952 0.37936508
1970-01-01 0.349206349 0.452380952 0.36984127
1970-01-01 0.309523810 0.297619048 0.30714286
1970-01-01 0.297619048 0.297619048 0.29761905
1970-01-01 0.503968254 0.297619048 0.46269841
1970-01-01 0.511904762 0.484126984 0.50634921
1970-01-01 0.757936508 0.642857143 0.73492063
1970-01-01 0.845238095 0.507936508 0.7778
1970-01-01 0.853174603 0.515873016

[R] Re : “For” calculation is so slow

2012-05-22 Thread Pascal Oettli

Hello,

's' is missing in your code.

Regards



- Mail original -
De : jiangxijixzy jiangxiji...@163.com
À : r-help@r-project.org
Cc : 
Envoyé le : Mardi 22 mai 2012 16h01
Objet : [R] “For” calculation is so slow

Dear All,

The function I wrote can run well with the small data, but with the large
data, the function runs very very slowly. How can I correct it? Thank you
very much. My function as below:

a-c(1:240)
b-c(1:240)
l=function(a,b){
v=0
u=0
uv=0
v[1]=0
u[1]=0
uv[1]=0
for (i in 1:(length(s)-1)){
v[i]-((gx[[i]][b,(gx[[i]][a,1]+1)])-(gx[[i]][a,gx[[i]][a,1]+1]))/(gx[[i]][a,gx[[i]][a,1]+1])
u[i]-((gy[[i]][a,(gy[[i]][a,1]+1)])-(gy[[i]][b,gy[[i]][a,1]+1]))/(gy[[i]][a,gy[[i]][a,1]+1])
uv[i]-v[i]+u[i]
}
w=0
w=mean(uv)
}
kk-data.frame()
for (a in 1:240){
for (b in 1:240){
if (ab)
kk[a,b]=l(a,b)
else kk[a,b]=0
}}
max(kk)


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[R] ReName

2012-05-22 Thread HAOLONG HOU

Dear list,

The name of R-language is too short and is not friendly to search engines.
Do you think it can be renamed to something like Rsio or Radio ?
Thank you so much for this useful software!

Best wishes.

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[R] ¨Time series and variables with different lengths

2012-05-22 Thread Isabel Scheckenbach

Hey,
we are currently analyzing the liquidty in the Danish mortgage bond market.
For this project we have several irregular time series variables as Bond 
prices, interest rates etc. We declared all the variables as irregular time 
series, and created the first differences of them to make them stationary. 
Now we are trying to run a linear regression on the price of the bon including 
dummy variables. The dummy variables are already included in the data table. Do 
we have to declare them in R in addition as dummy variables or is the command 
as.factor sufficient?
Moreover we always get the error message that the variable lengths differ, 
since after taking the first differences the main variables have on observation 
less than the dummy variables.. How can I solve for this?
Thanks a lot,
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[R] writing my own logistic regression function

2012-05-22 Thread mael

Hi,

I have problems writing a script for a logistic model. My data looks
something like this:

PI  A   N1  N2  N3
0   9   0   1   0   
1   4   1   0   1
0   8   0   0   0
1   6   0   1   0

And so on…

The model I want to fit looks like this: 

PI=exp^f(sw)/1+exp^f(sw) 
where
 f(sw)=c+(a1*N1/A^z)+(a2*N2/A^z)+(a3*N3/A^z)

So I want to estimate c,a1,a2,a3 and z…
Is there someone who can help me on this problem, because I don’t even know
how to start.

Best,
Anna


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Re: [R] How to remove square brackets, etc. from address strings?

2012-05-22 Thread Sarah Goslee

Hi Sabina,

You've run into two characteristics of regular expressions:
[ ] are special characters
* is a greedy match
Reading an intro regular expression document will help with both of those.

Meanwhile:

 x - [Swidsinski, Alexander; Loening-Baucke, Vera; Lochs, Herbert] Charite 
 Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; [Hale, Laura P.] Duke 
 Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA
 x
[1] [Swidsinski, Alexander; Loening-Baucke, Vera; Lochs, Herbert]
Charite Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; [Hale,
Laura P.] Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA

 gsub(\\[.*?\\] , , x) # escape [ and ] and make * lazy instead of greedy
[1] Charite Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; Duke
Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA

Sarah

On Tue, May 22, 2012 at 6:08 AM, Sabina Arndt sabina.ar...@hotmail.de wrote:

 Hello,



 I'd like to remove the individual pairs of square brackets along with
 their content - plus the space directly behind it - from address strings
  such as this:


  [Swidsinski, Alexander; Loening-Baucke, Vera; Lochs, Herbert] Charite
 Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; [Hale, Laura P.]
 Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA



  I'd like get the following result:



  Charite Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; Duke Univ, Med 
 Ctr, Dept Pathol, Durham, NC 27710 USA



 I tried


  address = gsub((.*)[(.*)], \\2, address)



 But this deletes everything from the first opening bracket to the last 
 closing bracket and leaves only the very last address:


  Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA



 How can I remove only the individual pairs of square brackets along with 
 their content?



 Thank you very much in advance!
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-- 
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http://www.stringpage.com
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Re: [R] “For” calculation is so slow

2012-05-22 Thread Zhou Fang

For loops are really, really slow in R. In general, you want to avoid them
like the plague. If you absolutely must insist on using them in large,
computationally intense and complex code, consider implementing the relevant
parts in C, say, and calling that from R.

Staying within R, you can probably considerably speed up that code by
storing gx and gy as a multi-dimensional arrays. (e.g. for sample data,
something like

rawGy = sample( 1:240, 240^2* 241, replace = T)
rawGx = sample( 1:240, 240^2 *241, replace = T)
gx = array(rawGx, dim = c(length(s) - 1, 240,  max(rawGx)+1  ) )
gy = array(rawGy, dim = c(length(s) - 1, 240,  max(rawGy)+1 ) )

 ), in which case, you can easily do the computation without loops by

gxa = (gx[ ,a,1]+ 1)
gya =(gy[ ,a, 1] +1)
uv = gx[cbind(1:(length(s) - 1) , b, gxa)] / gx[cbind(1:(length(s) - 1) , a,
gxa)]  - gy[cbind(1:(length(s) - 1) ,b, gya)]/gy[cbind(1:(length(s) - 1) ,a,
gya)]

or similar, which will be enormously faster (on my computer, there's an over
30x speed up). With a bit of thought, I'm sure you can also figure out how
to let it vectorise in a, as well...

Zhou

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Re: [R] as.function parameters

2012-05-22 Thread jackl

Hi there,

~ the 'problem' or rather the task I'm trying to solve is to implement
an algorithm to compute the ask/bid price of american options in a 
close to R related program language. 

because i'm not really using R but just it's basic functionalities I cannot
rely on different packages included in the R space.

is it usual for R to compute nested functions with such a workload?

Best thanks

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Re: [R] package grid: mirror grob objects along an axis

2012-05-22 Thread Rui Barradas

Hello,

Just flip 'xlim' or 'ylim'. Or both.
Using the iris example in help(grid), make the following changes:


op - par(mfcol = c(2,2))  # Two columns, first is the original, second
flipped.

with(iris,

  [... etc ...]
 # row 1, col 2: flip x axis
 plot(Sepal.Length, Sepal.Width, col = as.integer(Species),
  xlim = c(8, 4), ylim = c(2, 4.5), panel.first = grid(),
  main = with(iris,  plot(, panel.first = grid(), ..) ))
 # row 2, col 2: flip y axis
 plot(Sepal.Length, Sepal.Width, col = as.integer(Species),
  xlim = c(4, 8), ylim = c(4.5, 2), 
  panel.first = grid(3, lty=1,lwd=2),
  main = ... panel.first = grid(3, lty=1,lwd=2), ..)

 [... etc ...]
)
par(op)


Hope this helps,

Rui Barradas

Thomas Zumbrunn-3 wrote
 
 Hi everyone
 
 I'd like to flip grobs (grid graphical objects) along an axis, e.g. flip
 grobs 
 horizontally or vertically. I couldn't find any hints, neither in the 
 documentation nor by searching the web. Does anybody know how to achieve
 this?
 
 Cheers
 /thomas
 
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[R] Adding Text to a Plot

2012-05-22 Thread Canto Casasola, Vicente David

Hi, all!

 

I'm pretty sure I'm missing something about this.

Is there a smart way of typping hat(R)^2 and it's value from a linear
regression?

 

I've just found this tricky one:

 

# Sample data

x - sample(1:100,10)

y - 2+3*x+rnorm(10)

 

# Run the regression

lm1 - lm(y~x)

 

# Plotting

plot(x,y, main=Linear Regression, col=red)

abline(lm1, col=blue)

 

placex - par(usr)[1]+.1*(par(usr)[2]-par(usr)[1])

placey1 - par(usr)[3]+.9*(par(usr)[4]-par(usr)[3])

placey2 - par(usr)[3]+.8*(par(usr)[4]-par(usr)[3])

 

# HERE: Is this the right way?

text(x=placex, y=placey1,

 bquote(R^2 == .(summary(lm1)$r.squared)), adj=c(0,0))

text(x=placex, y=placey2,

 bquote(R^2 == .(summary(lm1)$adj.r.squared)), adj=c(0,0))

text(x=placex,y=placey2,

 expression(hat(R)), adj=c(0,0))

 

In addition, when I save the plot as PDF, the expression hat(R) seems to be
somewhat displaced.

 

Any advice?

 

Thanks in advance,

 

Vicente

 

Aviso de confidencialidad

Este correo electrónico y, en su caso, cualquier fichero anexo, contiene
información confidencial exclusivamente dirigida a su(s) destinatario(s).
Toda copia o divulgación deberá ser autorizada por la Subdirección de
Análisis y Control Interno. Si ha recibido este mensaje por error, le
rogamos que nos lo comunique inmediatamente por esta misma vía y proceda a
su eliminación.
Antes de imprimir este mensaje, asegúrese de que es necesario. El medio
ambiente está en nuestras manos.

 


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Re: [R] getting a Likert plot from a data frame

2012-05-22 Thread Richard M. Heiberger

Gavin,

thank you for using likert()

There are several problems in the use of a data.frame.
1. df is a bad name to use because df is the name of a base function (that
isn't an error, it is bad style).
2. the as.character() made the line you sent not work.
3. you indeed have the variable cat as a factor.  you need to make it the
row.names of the data.frame.

mydata - data.frame(
row.names=c(group1, group2, group3, group4, group5),
males=c(20,30,45,12,5),
females=c(35,23,32,8,5))
## make a pyramid Likert chart
as.pyramidLikert(likert(mydata), panel.width=.46)

Rich

On Tue, May 22, 2012 at 7:29 AM, Gavin Rudge g.ru...@bham.ac.uk wrote:

 I'm creating a stacked bar chart using the likert command in the HH
 package.  My data are in a data frame, with two numeric variables and a
 categorical variable, I can't get likert to use the column containing the
 categorical variable as a my y axis label.

 Here is a quick example:

 library(HH)
 #my data are:
 df-data.frame(as.character(cat=c(group1,group2,group3,group4,
 group5)),males=c(20,30,45,12,5),females=c(35,23,32,8,5))
 #make a pyramid Likert chart
 p-likert(df)
 as.pyramidLikert(p)

 It tries to plot three variables here when I just want two. I think I
 understand what is happening - my categorical variable is treated as a
 factor and I think it gets inserted as an integer into the matrix which the
 command derives from my data fame, to make the plot with(?) It's then used
 as a variable to be plotted just like the other two variables. what I don't
 get is how the example given in the package does something differently,
 which is how I want mine to work.

 ## Population Pyramid
 data(USAge.table)
 USA79 - USAge.table[75:1, 2:1, 1979]/100
 PL - likert(USA79,
 main=Population of United States 1979 (ages 0-74),
 xlab=Count in Millions,
 ylab=Age,
 scales=list(
   y=list(
 limits=c(0,77),
 at=seq(1,76,5),
 labels=seq(0,75,5),
 tck=.5))
 )
 PL
 as.pyramidLikert(PL)

 This does exactly what I'm trying to achieve.  here the two population
 counts are plotted in the likert plot and the age groups in the first
 columns are used as labels.

 I can't work out why in my example the age group variable is not used in
 the same way as the in my plot in the same way as the agegroups in this
 example, other than the example takes it's data from a table and mine is
 coming from a data frame.  The end point I want is a stacked Likert bar
 chart based on a data frame where the column containing the description of
 my groups is used as the y axis labels and the other two columns are used
 to draw the bars. I'm sure I'm missing a simple solution.  any help
 gratefully received.

 Gavin.

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Re: [R] sweave tables as images?

2012-05-22 Thread Alexander Shenkin

Thanks Liviu,

I've looked into SWord, and while it's an impressive project, I'm
concerned that it doesn't quite give me the fine-grain control I'd like
over the R environment where it's being executed.  It's still in the
back of my mind though, and I may indeed go with it at some point.

If I understand it correctly, odfWeave doesn't have a path backwards
from the odf doc back to the original odfWeave doc (which is the main
restriction of all these Sweave/knitr/etc solutions in my use case). 
Please correct me if I'm wrong about that!

Thanks,
Allie

On 5/22/2012 4:36 AM, Liviu Andronic wrote:
 On Mon, May 21, 2012 at 10:29 PM, Alexander Shenkin ashen...@ufl.edu wrote:
 So, I think it will be better if I can somehow generate the tables as
 images.  Is there any way to generate tables as images in separate files
 in Sweave?  Or, is there another tree up which I should be barking?

 Hmm, there are Sweave equivalents for LibreOffice [1] and MS Word [2].
 Have you looked into those? You may find other suggestions in the
 ReproducibleResearch Task View [3].

 Liviu

 [1] http://crantastic.org/packages/odfWeave
 [2] http://crantastic.org/packages/SWordInstaller
 [3] http://crantastic.org/task_views/ReproducibleResearch

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Re: [R] sweave tables as images?

On Tue, May 22, 2012 at 3:24 PM, Alexander Shenkin ashen...@ufl.edu wrote:
 If I understand it correctly, odfWeave doesn't have a path backwards
 from the odf doc back to the original odfWeave doc (which is the main
 restriction of all these Sweave/knitr/etc solutions in my use case).
 Please correct me if I'm wrong about that!

Unfortunately I cannot help you with this specific question, not least
because I've never used odfWeave.

From a quick glance to the docs, it seems that you always have two
documents: source odfWeave .odt (equivalent to .Rnw) and output .odt
(equivalent to .tex). Then using the .odt (.tex) you can generate a
PDF. I would be surprised if it were possible to go back from .odt
(.tex) to the source .odt (.Rnw).

I would suggest that you contact the authors of the respective
packages to make sure that the functionality is (not) available.
Regards
Liviu

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Re: [R] getting a Likert plot from a data frame

Dear Richard


On Tue, May 22, 2012 at 3:20 PM, Richard M. Heiberger r...@temple.edu wrote:
 mydata - data.frame(
    row.names=c(group1, group2, group3, group4, group5),
    males=c(20,30,45,12,5),
    females=c(35,23,32,8,5))
 ## make a pyramid Likert chart
 as.pyramidLikert(likert(mydata), panel.width=.46)


Loading library(HH) and running the code above fails with this error:
  as.pyramidLikert(likert(mydata), panel.width=.46)
Error in inherits(x, trellis) : could not find function likert

Am I doing something wrong?
Liviu

 sessionInfo()
R version 2.14.2 (2012-02-29)
Platform: x86_64-pc-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C
 [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
 [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8
 [7] LC_PAPER=C LC_NAME=C
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

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

other attached packages:
 [1] HH_2.2-23   latticeExtra_0.6-19 RColorBrewer_1.0-5
 [4] leaps_2.9   multcomp_1.2-12 survival_2.36-12
 [7] mvtnorm_0.9-9992lattice_0.20-6  plyr_1.7.1
[10] Defaults_1.1-1  fortunes_1.5-0  sos_1.3-1
[13] brew_1.0-6

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Re: [R] getting a Likert plot from a data frame

2012-05-22 Thread Richard M. Heiberger

out-of-phase problem

Update to HH_2.3-15 that was published on CRAN yesterday.

In HH_2.2-23 you needed to use a slightly longer name
plot.likert() instead of likert()

On Tue, May 22, 2012 at 9:44 AM, Liviu Andronic landronim...@gmail.comwrote:

 Dear Richard


 On Tue, May 22, 2012 at 3:20 PM, Richard M. Heiberger r...@temple.edu
 wrote:
  mydata - data.frame(
 row.names=c(group1, group2, group3, group4, group5),
 males=c(20,30,45,12,5),
 females=c(35,23,32,8,5))
  ## make a pyramid Likert chart
  as.pyramidLikert(likert(mydata), panel.width=.46)
 

 Loading library(HH) and running the code above fails with this error:
   as.pyramidLikert(likert(mydata), panel.width=.46)
 Error in inherits(x, trellis) : could not find function likert

 Am I doing something wrong?
 Liviu

  sessionInfo()
 R version 2.14.2 (2012-02-29)
 Platform: x86_64-pc-linux-gnu (64-bit)

 locale:
  [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C
  [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
  [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8
  [7] LC_PAPER=C LC_NAME=C
  [9] LC_ADDRESS=C   LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

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

 other attached packages:
  [1] HH_2.2-23   latticeExtra_0.6-19 RColorBrewer_1.0-5
  [4] leaps_2.9   multcomp_1.2-12 survival_2.36-12
  [7] mvtnorm_0.9-9992lattice_0.20-6  plyr_1.7.1
 [10] Defaults_1.1-1  fortunes_1.5-0  sos_1.3-1
 [13] brew_1.0-6


[[alternative HTML version deleted]]

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Re: [R] “For” calculation is so slow

Hi

 
 For loops are really, really slow in R. In general, you want to avoid 
them

I strongly disagree. ***Proper*** use of looping is quite convenient and 
reasonably fast.

Consider

 system.time( {
+ a=0
+ for (i in 1:1000) {
+ a -a+i
+ }
+ a
+ })
   user  system elapsed 
  10.220.02   10.28 
 
 system.time(b-sum(as.numeric(1:1000)))
   user  system elapsed 
   0.090.010.11 
 identical(a,b)
[1] TRUE

It is usually implementing C habits into R code what makes looping slow. 
The slowest part of a program is usually programming and it is influenced 
mainly by programmer.

Regards
Petr

 like the plague. If you absolutely must insist on using them in large,
 computationally intense and complex code, consider implementing the 
relevant
 parts in C, say, and calling that from R.
 
 Staying within R, you can probably considerably speed up that code by
 storing gx and gy as a multi-dimensional arrays. (e.g. for sample data,
 something like
 
 rawGy = sample( 1:240, 240^2* 241, replace = T)
 rawGx = sample( 1:240, 240^2 *241, replace = T)
 gx = array(rawGx, dim = c(length(s) - 1, 240,  max(rawGx)+1  ) )
 gy = array(rawGy, dim = c(length(s) - 1, 240,  max(rawGy)+1 ) )
 
  ), in which case, you can easily do the computation without loops by
 
 gxa = (gx[ ,a,1]+ 1)
 gya =(gy[ ,a, 1] +1)
 uv = gx[cbind(1:(length(s) - 1) , b, gxa)] / gx[cbind(1:(length(s) - 1) 
, a,
 gxa)]  - gy[cbind(1:(length(s) - 1) ,b, gya)]/gy[cbind(1:(length(s) - 1) 
,a,
 gya)]
 
 or similar, which will be enormously faster (on my computer, there's an 
over
 30x speed up). With a bit of thought, I'm sure you can also figure out 
how
 to let it vectorise in a, as well...
 
 Zhou
 
 --
 View this message in context: http://r.789695.n4.nabble.com/For-
 calculation-is-so-slow-tp4630830p4630855.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.

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Re: [R] “For” calculation is so slow

And as followup

 system.time(d-1000*1001/2)
   user  system elapsed 
   0.020.000.02 
 identical(a,b,d)
[1] TRUE


Regards
Petr

 
 Hi
 
  
  For loops are really, really slow in R. In general, you want to avoid 
 them
 
 I strongly disagree. ***Proper*** use of looping is quite convenient and 

 reasonably fast.
 
 Consider
 
  system.time( {
 + a=0
 + for (i in 1:1000) {
 + a -a+i
 + }
 + a
 + })
user  system elapsed 
   10.220.02   10.28 
  
  system.time(b-sum(as.numeric(1:1000)))
user  system elapsed 
0.090.010.11 
  identical(a,b)
 [1] TRUE
 
 It is usually implementing C habits into R code what makes looping slow. 

 The slowest part of a program is usually programming and it is 
influenced 
 mainly by programmer.
 
 Regards
 Petr
 
  like the plague. If you absolutely must insist on using them in large,
  computationally intense and complex code, consider implementing the 
 relevant
  parts in C, say, and calling that from R.
  
  Staying within R, you can probably considerably speed up that code by
  storing gx and gy as a multi-dimensional arrays. (e.g. for sample 
data,
  something like
  
  rawGy = sample( 1:240, 240^2* 241, replace = T)
  rawGx = sample( 1:240, 240^2 *241, replace = T)
  gx = array(rawGx, dim = c(length(s) - 1, 240,  max(rawGx)+1  ) )
  gy = array(rawGy, dim = c(length(s) - 1, 240,  max(rawGy)+1 ) )
  
   ), in which case, you can easily do the computation without loops by
  
  gxa = (gx[ ,a,1]+ 1)
  gya =(gy[ ,a, 1] +1)
  uv = gx[cbind(1:(length(s) - 1) , b, gxa)] / gx[cbind(1:(length(s) - 
1) 
 , a,
  gxa)]  - gy[cbind(1:(length(s) - 1) ,b, gya)]/gy[cbind(1:(length(s) - 
1) 
 ,a,
  gya)]
  
  or similar, which will be enormously faster (on my computer, there's 
an 
 over
  30x speed up). With a bit of thought, I'm sure you can also figure out 

 how
  to let it vectorise in a, as well...
  
  Zhou
  
  --
  View this message in context: http://r.789695.n4.nabble.com/For-
  calculation-is-so-slow-tp4630830p4630855.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.
 
 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
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Re: [R] ReName

2012-05-22 Thread David Winsemius



On May 22, 2012, at 4:08 AM, HAOLONG HOU wrote:


Dear list,

The name of R-language is too short and is not friendly to search  
engines.

Do you think it can be renamed to something like Rsio or Radio ?
Thank you so much for this useful software!


The notion of renaming R to 'radio' seems at best humorous. Try  
searching with:


r-project
CRAN

Or use:Rseek.org

Or use:allinurl:r-project in a google search

Or: http://code.google.com/hosting/search?q=label:R+

At one point I used `language:r` but I can find no support in the  
google search documentation that currently supports that strategy.


--

David Winsemius, MD
West Hartford, CT

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[R] Creating functions with a loop.

2012-05-22 Thread Etienne Larrivée-Hardy

Hi

I am trying to create n functions where each function is defined in function 
one step before, i.e. something like

ff.k(x) = ff.j(x) - sum(1:j),  for j=k-1

Is it possible? If it isn't and I manually create each function then is their a 
way to call them through a loop?  My objective is to calculate something like

result.k = ff.k(x1)/ff.k(x2)  for k in 2:n

Thank you for your time,
Etienne
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Re: [R] ReName

2012-05-22 Thread Jeff Newmiller

No. There are search tools that work. The RSiteSearch() function and the 
rseek.org website are two such options.
---
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.

HAOLONG HOU hihaol...@gmail.com wrote:

Dear list,

The name of R-language is too short and is not friendly to search
engines.
Do you think it can be renamed to something like Rsio or Radio ?
Thank you so much for this useful software!

Best wishes.

   [[alternative HTML version deleted]]

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

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Re: [R] ReName

Hi

 
 On May 22, 2012, at 4:08 AM, HAOLONG HOU wrote:
 
  Dear list,
 
  The name of R-language is too short and is not friendly to search 
  engines.

Did you try it? Even with plain Google

something R will usually lead to quite good hit. Try e.g.

linear model R

Regards
Petr

  Do you think it can be renamed to something like Rsio or Radio ?
  Thank you so much for this useful software!
 
 The notion of renaming R to 'radio' seems at best humorous. Try 
 searching with:
 
 r-project
 CRAN
 
 Or use:Rseek.org
 
 Or use:allinurl:r-project in a google search
 
 Or: http://code.google.com/hosting/search?q=label:R+
 
 At one point I used `language:r` but I can find no support in the 
 google search documentation that currently supports that strategy.
 
 -- 
 
 David Winsemius, MD
 West Hartford, CT
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] “For” calculation is so slow

2012-05-22 Thread Jeff Newmiller

Thanks, I will take that factor of 100 anytime rather than keep some syntactic 
familiarity from other languages.

Not to say I don't ever use loops, but I always try to vectorize the inner 
loop, if not the inner two loops.
---
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.

Petr PIKAL petr.pi...@precheza.cz wrote:

Hi

 
 For loops are really, really slow in R. In general, you want to avoid

them

I strongly disagree. ***Proper*** use of looping is quite convenient
and 
reasonably fast.

Consider

 system.time( {
+ a=0
+ for (i in 1:1000) {
+ a -a+i
+ }
+ a
+ })
   user  system elapsed 
  10.220.02   10.28 
 
 system.time(b-sum(as.numeric(1:1000)))
   user  system elapsed 
   0.090.010.11 
 identical(a,b)
[1] TRUE

It is usually implementing C habits into R code what makes looping
slow. 
The slowest part of a program is usually programming and it is
influenced 
mainly by programmer.

Regards
Petr

 like the plague. If you absolutely must insist on using them in
large,
 computationally intense and complex code, consider implementing the 
relevant
 parts in C, say, and calling that from R.
 
 Staying within R, you can probably considerably speed up that code by
 storing gx and gy as a multi-dimensional arrays. (e.g. for sample
data,
 something like
 
 rawGy = sample( 1:240, 240^2* 241, replace = T)
 rawGx = sample( 1:240, 240^2 *241, replace = T)
 gx = array(rawGx, dim = c(length(s) - 1, 240,  max(rawGx)+1  ) )
 gy = array(rawGy, dim = c(length(s) - 1, 240,  max(rawGy)+1 ) )
 
  ), in which case, you can easily do the computation without loops by
 
 gxa = (gx[ ,a,1]+ 1)
 gya =(gy[ ,a, 1] +1)
 uv = gx[cbind(1:(length(s) - 1) , b, gxa)] / gx[cbind(1:(length(s) -
1) 
, a,
 gxa)]  - gy[cbind(1:(length(s) - 1) ,b, gya)]/gy[cbind(1:(length(s) -
1) 
,a,
 gya)]
 
 or similar, which will be enormously faster (on my computer, there's
an 
over
 30x speed up). With a bit of thought, I'm sure you can also figure
out 
how
 to let it vectorise in a, as well...
 
 Zhou
 
 --
 View this message in context: http://r.789695.n4.nabble.com/For-
 calculation-is-so-slow-tp4630830p4630855.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.

__
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Re: [R] “For” calculation is so slow

2012-05-22 Thread Zhou Fang

I'm not sure what you are trying to prove with that example - the loopless
versions are massively faster, no?

I don't disagree that loops are sometimes unavoidable, and I suppose
sometimes loops can be faster when the non-loop version e.g. breaks your
memory budget, or performs tons of needless computations. But I think
avoiding for loops whenever you can is a good rule of thumb in R coding.

--
View this message in context: 
http://r.789695.n4.nabble.com/For-calculation-is-so-slow-tp4630830p4630897.html
Sent from the R help mailing list archive at Nabble.com.

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

Re: [R] getting a Likert plot from a data frame

On Tue, May 22, 2012 at 4:00 PM, Richard M. Heiberger r...@temple.edu wrote:
 out-of-phase problem

 Update to HH_2.3-15 that was published on CRAN yesterday.

Thanks! It works nicely now.

Liviu


 In HH_2.2-23 you needed to use a slightly longer name
 plot.likert() instead of likert()

 On Tue, May 22, 2012 at 9:44 AM, Liviu Andronic landronim...@gmail.com
 wrote:

 Dear Richard


 On Tue, May 22, 2012 at 3:20 PM, Richard M. Heiberger r...@temple.edu
 wrote:
  mydata - data.frame(
     row.names=c(group1, group2, group3, group4, group5),
     males=c(20,30,45,12,5),
     females=c(35,23,32,8,5))
  ## make a pyramid Likert chart
  as.pyramidLikert(likert(mydata), panel.width=.46)
 

 Loading library(HH) and running the code above fails with this error:
   as.pyramidLikert(likert(mydata), panel.width=.46)
 Error in inherits(x, trellis) : could not find function likert

 Am I doing something wrong?
 Liviu

  sessionInfo()
 R version 2.14.2 (2012-02-29)
 Platform: x86_64-pc-linux-gnu (64-bit)

 locale:
  [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C
  [3] LC_TIME=en_US.UTF-8        LC_COLLATE=en_US.UTF-8
  [5] LC_MONETARY=en_US.UTF-8    LC_MESSAGES=en_US.UTF-8
  [7] LC_PAPER=C                 LC_NAME=C
  [9] LC_ADDRESS=C               LC_TELEPHONE=C
 [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

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

 other attached packages:
  [1] HH_2.2-23           latticeExtra_0.6-19 RColorBrewer_1.0-5
  [4] leaps_2.9           multcomp_1.2-12     survival_2.36-12
  [7] mvtnorm_0.9-9992    lattice_0.20-6      plyr_1.7.1
 [10] Defaults_1.1-1      fortunes_1.5-0      sos_1.3-1
 [13] brew_1.0-6





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Re: [R] Adding Text to a Plot

2012-05-22 Thread David Winsemius



On May 22, 2012, at 9:10 AM, Canto Casasola, Vicente David wrote:


I'm pretty sure I'm missing something about this.

Is there a smart way of typping hat(R)^2 and it's value from a linear
regression?

I've just found this tricky one:

# Sample data
x - sample(1:100,10)
y - 2+3*x+rnorm(10)

# Run the regression
lm1 - lm(y~x)
# Plotting
plot(x,y, main=Linear Regression, col=red)
abline(lm1, col=blue)
placex - par(usr)[1]+.1*(par(usr)[2]-par(usr)[1])
placey1 - par(usr)[3]+.9*(par(usr)[4]-par(usr)[3])
placey2 - par(usr)[3]+.8*(par(usr)[4]-par(usr)[3])

# HERE: Is this the right way?


# To do   what? I see that you over-plotted the R, presumably  
because you did not like the way that:

  bquote(hat(R)^2 == .(summary(lm1)$adj.r.squared))

... ended up looking (with the exponent higher than in the non-hatted  
version.)




text(x=placex, y=placey1,
bquote(R^2 == .(summary(lm1)$r.squared)), adj=c(0,0))
text(x=placex, y=placey2,
bquote(R^2 == .(summary(lm1)$adj.r.squared)), adj=c(0,0))
text(x=placex,y=placey2,
expression(hat(R)), adj=c(0,0))

In addition, when I save the plot as PDF, the expression hat(R)  
seems to be

somewhat displaced.

Any advice?

Are you sure? It looks  to me that the lower R is slightly shifted  
to the left, but when I actually measure it, there does not appear to  
be a shift. If it is real and not just an optical illusion, this may  
have something to do with your unstated version of R or your also  
unstated OS.


--
David Winsemius, MD
West Hartford, CT

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Re: [R] Creating functions with a loop.

2012-05-22 Thread David Winsemius



On May 22, 2012, at 10:06 AM, Etienne Larrivée-Hardy wrote:


Hi

I am trying to create n functions where each function is defined in  
function one step before, i.e. something like


ff.k(x) = ff.j(x) - sum(1:j),  for j=k-1


There is a cumsum function:

 cumsum(1:10)
 [1]  1  3  6 10 15 21 28 36 45 55

Did you mean something like?:

 ff.k[j] - ff.j[j] - sum(1:j),  for j=k-1

In R the paren, (, indicates that you are calling a function whereas  
you appear interested in making an indexed assignment to a vector.




Is it possible? If it isn't and I manually create each function then  
is their a way to call them through a loop?  My objective is to  
calculate something like


result.k = ff.k(x1)/ff.k(x2)  for k in 2:n


You may have clear idea about which k-indices should go where in that  
expression, but I surely do not. What are 'x1' and 'x2'?


--

David Winsemius, MD
West Hartford, CT

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Re: [R] as.function parameters

Jumping in from a finance perspective...

You really don't want to actually use nested lists for this -- the
overhead and index-book-keeping will quickly become quite annoying.

Instead use a 2D array (I assume you're using the CRR Binomial Tree
model for an American Option) where the column number less one is how
many steps into the tree you are and the row is how many ups/downs
you've had (your call which one) -- then just roll your option valuing
formula with loops over the array.

This is implemented in the fOptions package in the CRRBinomialTree
function, but if you can't use real R that might/might not help you.
It is open source however...

Michael

On Tue, May 22, 2012 at 6:22 AM, jackl jacks...@hotmail.de wrote:
 Hi there,

 ~ the 'problem' or rather the task I'm trying to solve is to implement
 an algorithm to compute the ask/bid price of american options in a
 close to R related program language.

 because i'm not really using R but just it's basic functionalities I cannot
 rely on different packages included in the R space.

 is it usual for R to compute nested functions with such a workload?

 Best thanks

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/as-function-parameters-tp4620390p4630858.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] RE : Creating functions with a loop.

2012-05-22 Thread Etienne Larrivée-Hardy

I must say I'm new here and I don't quite know how things work so you will have 
to excuse me.

The first part of my problem is that it would help me greatly to have a loop 
creating all the functions.  Here is an example of how the 10th function would 
need to look like:

ff10 - function (q, alph, lam) 
{
ff9(q, alph, lam) - sum(v.pp[1:(10 - 1)] * exp(-v.beta[1:(10 - 1)] * cc1 * 
10^(1:(10 - 1) - 1)))
}

where v.pp and v.beta are vector for which values are known up to 9 in this 
case. cc1, bb, alpha, lambda are fixed.

The second part of my problem is that with those functions I need to calculate 
the next value of v.beta and v.pp. 

v.beta[10] - 
1/((bb-1)*cc1*10^-(10-1))*log(ff2(cc1*10^-(10-1),alpha,lambda)/ff10(cc1*10^-(10-1)*bb,alpha,lambda))
and 
v.pp[10] - ...

I don't know if that made it any clearer...

De : David Winsemius [dwinsem...@comcast.net]
Date d'envoi : 22 mai 2012 11:08
À : Etienne Larrivée-Hardy
Cc : r-help@r-project.org
Objet : Re: [R] Creating functions with a loop.

On May 22, 2012, at 10:06 AM, Etienne Larrivée-Hardy wrote:

 Hi

 I am trying to create n functions where each function is defined in
 function one step before, i.e. something like

 ff.k(x) = ff.j(x) - sum(1:j),  for j=k-1

There is a cumsum function:

  cumsum(1:10)
  [1]  1  3  6 10 15 21 28 36 45 55

Did you mean something like?:

  ff.k[j] - ff.j[j] - sum(1:j),  for j=k-1

In R the paren, (, indicates that you are calling a function whereas
you appear interested in making an indexed assignment to a vector.


 Is it possible? If it isn't and I manually create each function then
 is their a way to call them through a loop?  My objective is to
 calculate something like

 result.k = ff.k(x1)/ff.k(x2)  for k in 2:n

You may have clear idea about which k-indices should go where in that
expression, but I surely do not. What are 'x1' and 'x2'?

--

David Winsemius, MD
West Hartford, CT

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[R] RNORM matrix based on CSV file values for MEAN and SD

2012-05-22 Thread dcoakley

This should (hopefully) be a pretty simple task. What I'd like to do is read
in a csv file containing means and standard deviations for a large number of
'n' parameters (up to 2000). The list would be in the following format (see
attached read.csv):

Paramter(1), mean, standard dev.,
Paramter(2), mean, standard dev.,
Paramter(3), mean, standard dev.,
...
Paramter(n), mean, standard dev.,


Based on the above csv file, I would then like to generate a large sample
matrix for 's' samples, using the rnorm function. The matrix will be in the
following format:

1,0,0, P1(1), P2(1), P3(1), ... Pn(1)
2,0,0, P1(2), P2(2), P3(2), ... Pn(2)

s,0,0, P1(s), P2(s), P3(s), ... Pn(s)

The first column contains the Row number. Taking s=3, we would have rows
numbered 1 to 30,000.

The second and third column are fixed values - 0

The forth and subsequent columns contain values from the rnorm distribution
for each parameter. P1(1) is the first value generated for the first
parameter, P1(2) is the second value generated and so forth. P2(1) is the
first value generated for the second parameter, P2(2) is the second value
generated and so forth.  Pn(1) is the first value generated for the n'th
parameter, Pn(2) is the second value generated and so forth.

Again the number of rows depends on 's', the number of samples.

Therefore, I will be generating a fairly large matrix. This could be a
1,000,000 x 2,000 matrix. However, due to memory constraints, it may be
necessary to break this down into smaller sub-matrices where I limit the
number of rows. Firstly, is this possible in r, and secondly, can anyone
help suggest a method for creating such a matrix.

I'd really appreciate any help on this. Thank you.



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[R] Graph to visualize paired t test

2012-05-22 Thread jhartsho

http://r.789695.n4.nabble.com/file/n4630909/pfaff_fig1.gif 

I have run the statistics and found no significance in my pairwise t test. 
I want to create a graph similar to the one I included showing similar
slopes/lines for my data points.  For my data a correlation graph is not
appropriate and looks very confusing.  Is the easiest way to create several
lines between points I enter manually or is there a way to create this graph
from entering my pairwise.t.test data?

Thanks in advance

--
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[R] How to evaluate R things from Visual Studio?

2012-05-22 Thread Giannis Mamalikidis

Hello all, Iâm new here and this is actually my very first post.
My name is Giannis and I am an undergraduate student. I am programming in 
Microsoftâs Visual Studio 2010 (VB.NET mostly) and I want to create a program 
for the research group I am in.

I merely want to evaluate actions and use simple statistical things such as ks 
test (Kolmogorov-Smirnov), and some plots.
The reason I need to call R from my program is that the users of my program 
will NOT know any programming language, and by extension they will not know R 
either. So everything will happen by the click of a button.
For me to do that, I need to be able to use interoperability between R and 
VS.NET 2010, so that I can use Râs evaluator.

Iâve already tried Baierâs StatConnDCOM (http://rcom.univie.ac.at) and I 
did succeed in creating a program that runs perfectly but came to see the hard 
way that â for my case, is kind of useless for the sole reason of 
StatconnDCOMâs redistribution restriction. 
As Mr. Baier told me himself, there is no guarantee that StatconnDCOM will be 
available free of charge for ever. Hence I cannot create a freeware program 
now, and have my users uncertain of where they will be able to use this program 
for free in the future.

Iâve already tried R.NET (http://rdotnet.codeplex.com/ ) Version 1.4 (stable) 
and 1.4.1 (Change Set: 6d2c3f161801 ).
This does seem wonderful and it doesnât have StaconnDCOMâs ridiculous 
restrictions, BUT it refuses to work on me!

So, now that you have all the necessary background info, it is time I presented 
my questions:

1) Is there another way of using R from Visual Studio that you can point me to?
2) Is anyone here using R.NET (any version) successfully? If so, can you help 
me to get it working? Unfortunately they canât help me on R.NETâs official 
discussion forum

I mean, I just need to evaluate the basic R stuff.. Why is it THAT difficult?

Last but not least, let me give you info that might be relevant in your 
deciding how to aid me.
OS: Windows 7 Professional x64 English (Genuine, it is purchased)
R: Version 2.15.0
Programming Environment: Microsoft Visual Studio .net 2010

For any additional Info, please do not hesitate to ask me either here or 
directly at giannis_mamaliki...@msn.com
I hope we can make this work. thanks in advance.
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Re: [R] Creating functions with a loop.

Interpreting your question slightly differently, where ff.k(x) is a
function call.

Suppose you want to get a list of functions like

x^2
x^2 -1
x^2 - 3
x^2 - 6

It's perfectly possible with something like this:

NextFunc - function(f, i) {
   # Takes in a function f and returns
   # a different function that gives you
   # f(x) - i

   force(i) # probably unnecessary, but good practice as this just
might bite you in a loop


  function(x) f(x) - i
}

Then

f - function(x) x^2

f2 - NextFunc(f, 1)

f3 - NextFunc(f2, 2)

f3(3) # 6 = 3^2 - 1 - 2

Which should be wrappable in a loop.

Michael

On Tue, May 22, 2012 at 11:08 AM, David Winsemius
dwinsem...@comcast.net wrote:

 On May 22, 2012, at 10:06 AM, Etienne Larrivée-Hardy wrote:

 Hi

 I am trying to create n functions where each function is defined in
 function one step before, i.e. something like

 ff.k(x) = ff.j(x) - sum(1:j),      for j=k-1


 There is a cumsum function:

 cumsum(1:10)
  [1]  1  3  6 10 15 21 28 36 45 55

 Did you mean something like?:

  ff.k[j] - ff.j[j] - sum(1:j),      for j=k-1

 In R the paren, (, indicates that you are calling a function whereas you
 appear interested in making an indexed assignment to a vector.



 Is it possible? If it isn't and I manually create each function then is
 their a way to call them through a loop?  My objective is to calculate
 something like

 result.k = ff.k(x1)/ff.k(x2)      for k in 2:n


 You may have clear idea about which k-indices should go where in that
 expression, but I surely do not. What are 'x1' and 'x2'?

 --

 David Winsemius, MD
 West Hartford, CT


 __
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Re: [R] LM with summation function

2012-05-22 Thread Robbie Edwards

Hi all,

Thanks for the replies, but I realize I've done a bad job explaining my
problem.  To help, I've created some sample data to explain the problem.

df - data.frame(x=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), y=c(109, 232,
363, 496, 625, 744, 847, 928, 981, 1000, 979, 912), s=c(109, 341, 704,
1200, 1825, 2569, 3416, 4344, 5325, 6325, 7304, 8216))

In this data frame, y results from y = x * b1 + x^2 * b2 + x^3 * b3 and s
is sum of the current y and all previous y (s3 = y1 + y2 + y3).

I know I can find b1, b2 and b3 using:
lm(y ~ 0 + x + I(x^2) + I(x^3), data=df)

yielding...
Coefficients:
 x  I(x^2)  I(x^3)
   100  10  -1

However, I need to find b1, b2 and b3 using the s column.  The reason
being, I don't actually know the values of y in the actual data set.  And
in the actual data, I only have a few of the values.  Imagine this data is
being used a reward schedule for like a loyalty points program.  y
represents the number of points needed for each level while s is the total
number of points to reach that level.  In the real problem, my data looks
more like this:

d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))

Where I need to use a few sample points to help define the parameters of
the curve.

thanks again and hopefully this makes the problem a bit clearer.

robbie



On Fri, May 18, 2012 at 7:40 PM, David Winsemius dwinsem...@comcast.netwrote:


 On May 18, 2012, at 1:44 PM, Robbie Edwards wrote:

  Hi all,

 I'm trying to model some data where the y is defined by

 y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3

 Hopefully that reads clearly for email.


 cumsum( rowSums( cbind(B1 * x,  B2 * x^2, B3 * x^3)))



  Anyway, if it wasn't for the summation, I know I would do it like this

 lm(y ~ x + x2 + x3)

 Where x2 and x3 are x^2 and x^3.

 However, since each value of x is related to the previous values of x, I
 don't know how to do this.  Any help is greatly appreciated.




 David Winsemius, MD
 West Hartford, CT



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Re: [R] RNORM matrix based on CSV file values for MEAN and SD

No CSV came through so I'll just assume you get in a data.frame from
read.csv() that looks something like this

params - data.frame(mean = c(1,4,7), sd = c(2,2,5))

and you want 10 samples from each. If you're on memory constraints,
you can simply loop over rows and append to a growing CSV.

for(i in NROW(params)){
write.table(c(i, 0, 0, rnorm(10, params$mean[i], params$sd[i])),
out.csv, append = TRUE, sep =,, row.names = FALSE, col.names =
FALSE)
}

Note that we have to set the names to false or the appending gets messy.

It's probably faster (though more work) to do a few rows at a time and
to use textConnections so you aren't constantly opening and closing
the file, but this should get you started.

See the examples of ?textConnection for how to do that bit properly.

Best,

Michael


On Tue, May 22, 2012 at 10:43 AM, dcoakley danielcoakl...@gmail.com wrote:
 This should (hopefully) be a pretty simple task. What I'd like to do is read
 in a csv file containing means and standard deviations for a large number of
 'n' parameters (up to 2000). The list would be in the following format (see
 attached read.csv):

 Paramter(1), mean, standard dev.,
 Paramter(2), mean, standard dev.,
 Paramter(3), mean, standard dev.,
 ...
 Paramter(n), mean, standard dev.,


 Based on the above csv file, I would then like to generate a large sample
 matrix for 's' samples, using the rnorm function. The matrix will be in the
 following format:

 1,0,0, P1(1), P2(1), P3(1), ... Pn(1)
 2,0,0, P1(2), P2(2), P3(2), ... Pn(2)
 
 s,0,0, P1(s), P2(s), P3(s), ... Pn(s)

 The first column contains the Row number. Taking s=3, we would have rows
 numbered 1 to 30,000.

 The second and third column are fixed values - 0

 The forth and subsequent columns contain values from the rnorm distribution
 for each parameter. P1(1) is the first value generated for the first
 parameter, P1(2) is the second value generated and so forth. P2(1) is the
 first value generated for the second parameter, P2(2) is the second value
 generated and so forth.  Pn(1) is the first value generated for the n'th
 parameter, Pn(2) is the second value generated and so forth.

 Again the number of rows depends on 's', the number of samples.

 Therefore, I will be generating a fairly large matrix. This could be a
 1,000,000 x 2,000 matrix. However, due to memory constraints, it may be
 necessary to break this down into smaller sub-matrices where I limit the
 number of rows. Firstly, is this possible in r, and secondly, can anyone
 help suggest a method for creating such a matrix.

 I'd really appreciate any help on this. Thank you.



 --
 View this message in context: 
 http://r.789695.n4.nabble.com/RNORM-matrix-based-on-CSV-file-values-for-MEAN-and-SD-tp4630901.html
 Sent from the R help mailing list archive at Nabble.com.

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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Graph to visualize paired t test

If I understand what you are looking for, this should get you started:

https://www.stat.math.ethz.ch/pipermail/r-help/2012-May/312287.html

Michael

On Tue, May 22, 2012 at 11:14 AM, jhartsho jhart...@uark.edu wrote:
 http://r.789695.n4.nabble.com/file/n4630909/pfaff_fig1.gif

 I have run the statistics and found no significance in my pairwise t test.
 I want to create a graph similar to the one I included showing similar
 slopes/lines for my data points.  For my data a correlation graph is not
 appropriate and looks very confusing.  Is the easiest way to create several
 lines between points I enter manually or is there a way to create this graph
 from entering my pairwise.t.test data?

 Thanks in advance

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/Graph-to-visualize-paired-t-test-tp4630909.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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Re: [R] Graph to visualize paired t test

2012-05-22 Thread Richard M. Heiberger

tmp - data.frame(A=sample(20,10), B=sample(20, 10))
with(tmp, t.test(A, B))
matplot(t(tmp), type=b)

This does what you asked for.  I don't understand the legend on your plot.

For future queries,
PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.

The nabble interface is not encouraged.  Please write directly to the list
R-help@r-project.org

Rich

On Tue, May 22, 2012 at 11:14 AM, jhartsho jhart...@uark.edu wrote:

 http://r.789695.n4.nabble.com/file/n4630909/pfaff_fig1.gif

 I have run the statistics and found no significance in my pairwise t test.
 I want to create a graph similar to the one I included showing similar
 slopes/lines for my data points.  For my data a correlation graph is not
 appropriate and looks very confusing.  Is the easiest way to create several
 lines between points I enter manually or is there a way to create this
 graph
 from entering my pairwise.t.test data?

 Thanks in advance

 --
 View this message in context:
 http://r.789695.n4.nabble.com/Graph-to-visualize-paired-t-test-tp4630909.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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 http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
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Re: [R] Syntax for lme function to model random factors and interactions

2012-05-22 Thread Joshua Wiley

See inline

On Mon, May 21, 2012 at 11:17 AM, i_like_macs dk...@mac.com wrote:
 Hello Joshua,

 Many thanks for your help, especially from a fellow Bruin (I went there as
 an undergrad!).

 I understand that there is another forum for mixed models. If my problem
 can't be solved within this thread, I'll have to go there. I do understand
 some theory about mixed models, but obviously am far from an expert.

 My question is not so much statistical advice, as it concerns the correct
 syntax to include random factors and interactions (which include these
 random factors) for the lme function. Maybe it's because I'm used to SPSS,
 but I find R very difficult to use, even after looking up its built-in help.

 I could run your neat code suggestion:

 lme(Y ~ (A + B + C + D)^3, data = myData, random = ~ 1 | C, method = ML)

 but would like to know how to also include D as a random factor. My
 understanding is that the random argument for the lme function is coded
 as:

 ~ x1 + ... + xn | g1 / ... / gm

so in this case x1 ... xn are random effects where the effects are
allowed to vary across levels of g1 .. gm.


 where the left side describes the model for random effects, and the right
 side describes the grouping structure. Reading other posts, I learned that I
 need both sides for the code to run without errors. However, it's not clear

that is correct, the left side is the random effect (random intercepts
and/or random slopes), the right side is whatever variable codes the
levels that the random effect can vary across.


 to me what both sides represent. The left side appears to be where the
 random factors are specified, perhaps like this:

 random = ~ C + D

so this would indicate that C and D are random effects, but you will
need something to indicate what they get to vary across.


 But then this results in errors. Does this mean I have to somehow join the
 two following lines of code to specify both random factors?

 random = ~ 1 | C
 random = ~ 1 | D

I do not think that is what you want (and besides I do not think that
lme() allows multiple random arguments though I could be wrong because
I work with lmer more than lme).


 It's not clear what the ~ 1 represents here, as I would have guessed that
 this is where the random factors would be specified. Is this related to an
 intercept-only model?

yes, ~ 1 | C  means that there is a random intercept for each level of
C.  If you do this, you will get an estimate of the average intercept
for the intercept in your model, but you will also get an estimate of
the variance in intercepts (technically the intercepts are assumed to
come from a normal distribution, you will get the mean and variance
(or standard deviation) of the maximum likelihood estimate of that
distribution).


 I'm sorry for sounding so lost. This is because I am. Perhaps I need to know
 more theory of mixed models, but this seems to be possible only if I

Understanding more theory would probably help.  The Pinheiro and Bates
text as wonderful as it is, may not be the easiest place to start.  I
have not seen you mention anything about a grouping or nesting
structure in your data.  This may be part of the confusion too.  Many
uses of mixed models are for that case.  A classical example would be
students nested within classrooms.  In that case, the research
question could be does number of hours spent on homework predict
grades.  The model could look like:

grades ~ 1 + homework

or in ordinary regression notation instead of R's formula:

grades = Xb + e

where X is a design matrix the first few rows of which might look like:

1  2
1  2.5
1  3
1  2

the first column being the intercept adn the second the number of
hours spent on homework for each student.  b will be a vector of
coefficients, the first coefficient being the estimated intercept and
the second the slope of grades on homework.  e is a vector of
residuals, that part of grades which cannot be explained by the
intercept and homework.  The assumption in ordinary regression is that
e is identically and independently distributed, but students are
within classrooms, and we might guess that in fact, each student was
not really an indepedent observation---there is some similarity
because they share a classroom.  Mixed models address this by adding
random effects.  Following the above example, we might do:

grades ~ 1 + homework
random = ~ 1 | ClassroomID

this allows the intercepts to randomly vary by classroom, which is
sensible---some classes may have more or less skilled students so
given that everyone did 0 hours of homework, we still might expect
some classrooms to have higher or lower grades.  This models that.
Now lets say that further, you think that the effects of homework
might vary across classrooms.  Perhaps for students in very low
performing classes, they get an enormous benefit from spending time on
homework, whereas in the very high performing classes, their grades
only marginally improve for every additional hour of homework.

[R] how to remove the 'promise' attribute of an R object (.Random.seed)?

2012-05-22 Thread Yihui Xie

Hi,

The problem arises when I lazyLoad() the .Random.seed from a
previously saved database. To simplify the process of reproducing the
problem, see the example below:

## this assignment may not really make sense, but illustrates the problem
delayedAssign('.Random.seed', 1L)

typeof(.Random.seed)
# [1] integer

rnorm(1)
# Error in rnorm(1) :
#  .Random.seed is not an integer vector but of type 'promise'

typeof(.Random.seed)
# [1] integer

So there must be an attribute promise somewhere attached to
.Random.seed, and I cannot find it. The R function typeof() does not
reveal it, but the TYPEOF() function in src/main/RNG.c says it is a
'promise'.

My question is, how to make R use the real value of .Random.seed
instead of complaining about the promise? Thanks!

Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA

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Re: [R] LM with summation function

But if I understand your problem correctly, you can get the y values
from the s values. I'm relying on your statement that s is sum of the
current y and all previous y (s3 = y1 + y2 + y3). E.g.,

y - c(1, 4, 6, 9, 3, 7)

s1 = 1
s2 = 4 + s1 = 5
s3 = 6 + s2 = 11

more generally

s - cumsum(y)

Then if we only see s, we can get back the y vector by doing

c(s[1], diff(s))

which is identical to y.

So for your data, the underlying y must have been c(109, 1091, 4125,
2891) right?

Or have I completely misunderstood your problem?

Michael

On Tue, May 22, 2012 at 12:25 PM, Robbie Edwards
robbie.edwa...@gmail.com wrote:
 Actually, I can't.  I don't know the y values.  Only the s and only for a
 subset of the data.

 Like this.

 d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))



 On Tue, May 22, 2012 at 11:57 AM, R. Michael Weylandt
 michael.weyla...@gmail.com wrote:

 You can reconstruct the y values by taking first-differences of the s
 vector, no? Then it sounds like you're good to go

 Best, Michael

 On Tue, May 22, 2012 at 11:40 AM, Robbie Edwards
 robbie.edwa...@gmail.com wrote:
  Hi all,
 
  Thanks for the replies, but I realize I've done a bad job explaining my
  problem.  To help, I've created some sample data to explain the problem.
 
  df - data.frame(x=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), y=c(109,
  232,
  363, 496, 625, 744, 847, 928, 981, 1000, 979, 912), s=c(109, 341, 704,
  1200, 1825, 2569, 3416, 4344, 5325, 6325, 7304, 8216))
 
  In this data frame, y results from y = x * b1 + x^2 * b2 + x^3 * b3 and
  s
  is sum of the current y and all previous y (s3 = y1 + y2 + y3).
 
  I know I can find b1, b2 and b3 using:
  lm(y ~ 0 + x + I(x^2) + I(x^3), data=df)
 
  yielding...
  Coefficients:
      x  I(x^2)  I(x^3)
    100      10      -1
 
  However, I need to find b1, b2 and b3 using the s column.  The reason
  being, I don't actually know the values of y in the actual data set.
   And
  in the actual data, I only have a few of the values.  Imagine this data
  is
  being used a reward schedule for like a loyalty points program.  y
  represents the number of points needed for each level while s is the
  total
  number of points to reach that level.  In the real problem, my data
  looks
  more like this:
 
  d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
 
  Where I need to use a few sample points to help define the parameters of
  the curve.
 
  thanks again and hopefully this makes the problem a bit clearer.
 
  robbie
 
 
 
  On Fri, May 18, 2012 at 7:40 PM, David Winsemius
  dwinsem...@comcast.netwrote:
 
 
  On May 18, 2012, at 1:44 PM, Robbie Edwards wrote:
 
   Hi all,
 
  I'm trying to model some data where the y is defined by
 
  y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3
 
  Hopefully that reads clearly for email.
 
 
  cumsum( rowSums( cbind(B1 * x,  B2 * x^2, B3 * x^3)))
 
 
 
   Anyway, if it wasn't for the summation, I know I would do it like this
 
  lm(y ~ x + x2 + x3)
 
  Where x2 and x3 are x^2 and x^3.
 
  However, since each value of x is related to the previous values of x,
  I
  don't know how to do this.  Any help is greatly appreciated.
 
 
 
 
  David Winsemius, MD
  West Hartford, CT
 
 
 
         [[alternative HTML version deleted]]
 
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Re: [R] LM with summation function

2012-05-22 Thread Robbie Edwards

I don't think I can.

For the sample data

 d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))

when x = 4, s = 1200.  However, that s4 is sum of y1 + y2 + y3 + y4.
 Wouldn't I have to know the y for x = 2 and x = 3 to get the value of y
for x = 4?

In the previous message, I created two sample data frames.  d is what I'm
trying to use to create df.  I only know what's in d, df is just used to
illustrate what I'm trying to get from d.

robbie





On Tue, May 22, 2012 at 12:30 PM, R. Michael Weylandt 
michael.weyla...@gmail.com wrote:

 But if I understand your problem correctly, you can get the y values
 from the s values. I'm relying on your statement that s is sum of the
 current y and all previous y (s3 = y1 + y2 + y3). E.g.,

 y - c(1, 4, 6, 9, 3, 7)

 s1 = 1
 s2 = 4 + s1 = 5
 s3 = 6 + s2 = 11

 more generally

 s - cumsum(y)

 Then if we only see s, we can get back the y vector by doing

 c(s[1], diff(s))

 which is identical to y.

 So for your data, the underlying y must have been c(109, 1091, 4125,
 2891) right?

 Or have I completely misunderstood your problem?

 Michael

 On Tue, May 22, 2012 at 12:25 PM, Robbie Edwards
 robbie.edwa...@gmail.com wrote:
  Actually, I can't.  I don't know the y values.  Only the s and only for a
  subset of the data.
 
  Like this.
 
  d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
 
 
 
  On Tue, May 22, 2012 at 11:57 AM, R. Michael Weylandt
  michael.weyla...@gmail.com wrote:
 
  You can reconstruct the y values by taking first-differences of the s
  vector, no? Then it sounds like you're good to go
 
  Best, Michael
 
  On Tue, May 22, 2012 at 11:40 AM, Robbie Edwards
  robbie.edwa...@gmail.com wrote:
   Hi all,
  
   Thanks for the replies, but I realize I've done a bad job explaining
 my
   problem.  To help, I've created some sample data to explain the
 problem.
  
   df - data.frame(x=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), y=c(109,
   232,
   363, 496, 625, 744, 847, 928, 981, 1000, 979, 912), s=c(109, 341, 704,
   1200, 1825, 2569, 3416, 4344, 5325, 6325, 7304, 8216))
  
   In this data frame, y results from y = x * b1 + x^2 * b2 + x^3 * b3
 and
   s
   is sum of the current y and all previous y (s3 = y1 + y2 + y3).
  
   I know I can find b1, b2 and b3 using:
   lm(y ~ 0 + x + I(x^2) + I(x^3), data=df)
  
   yielding...
   Coefficients:
   x  I(x^2)  I(x^3)
 100  10  -1
  
   However, I need to find b1, b2 and b3 using the s column.  The reason
   being, I don't actually know the values of y in the actual data set.
And
   in the actual data, I only have a few of the values.  Imagine this
 data
   is
   being used a reward schedule for like a loyalty points program.  y
   represents the number of points needed for each level while s is the
   total
   number of points to reach that level.  In the real problem, my data
   looks
   more like this:
  
   d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
  
   Where I need to use a few sample points to help define the parameters
 of
   the curve.
  
   thanks again and hopefully this makes the problem a bit clearer.
  
   robbie
  
  
  
   On Fri, May 18, 2012 at 7:40 PM, David Winsemius
   dwinsem...@comcast.netwrote:
  
  
   On May 18, 2012, at 1:44 PM, Robbie Edwards wrote:
  
Hi all,
  
   I'm trying to model some data where the y is defined by
  
   y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3
  
   Hopefully that reads clearly for email.
  
  
   cumsum( rowSums( cbind(B1 * x,  B2 * x^2, B3 * x^3)))
  
  
  
Anyway, if it wasn't for the summation, I know I would do it like
 this
  
   lm(y ~ x + x2 + x3)
  
   Where x2 and x3 are x^2 and x^3.
  
   However, since each value of x is related to the previous values of
 x,
   I
   don't know how to do this.  Any help is greatly appreciated.
  
  
  
  
   David Winsemius, MD
   West Hartford, CT
  
  
  
  [[alternative HTML version deleted]]
  
   __
   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.
 
 


[[alternative HTML version deleted]]

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Re: [R] LM with summation function

Ahh sorry -- I didn't understand that x was supposed to be an
index so I was using the row number an index for the summation -- yes,
my proposal probably won't work without further assumptions[I.e.,
you could assume linear growth between observations, but that will
bias something some direction...(not sure which)]

I'll ponder it some more and get back to you if I come up with anything

Michael

On Tue, May 22, 2012 at 12:43 PM, Robbie Edwards
robbie.edwa...@gmail.com wrote:
 I don't think I can.

 For the sample data

  d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))

 when x = 4, s = 1200.  However, that s4 is sum of y1 + y2 + y3 + y4.
  Wouldn't I have to know the y for x = 2 and x = 3 to get the value of y
 for x = 4?

 In the previous message, I created two sample data frames.  d is what I'm
 trying to use to create df.  I only know what's in d, df is just used to
 illustrate what I'm trying to get from d.

 robbie





 On Tue, May 22, 2012 at 12:30 PM, R. Michael Weylandt 
 michael.weyla...@gmail.com wrote:

 But if I understand your problem correctly, you can get the y values
 from the s values. I'm relying on your statement that s is sum of the
 current y and all previous y (s3 = y1 + y2 + y3). E.g.,

 y - c(1, 4, 6, 9, 3, 7)

 s1 = 1
 s2 = 4 + s1 = 5
 s3 = 6 + s2 = 11

 more generally

 s - cumsum(y)

 Then if we only see s, we can get back the y vector by doing

 c(s[1], diff(s))

 which is identical to y.

 So for your data, the underlying y must have been c(109, 1091, 4125,
 2891) right?

 Or have I completely misunderstood your problem?

 Michael

 On Tue, May 22, 2012 at 12:25 PM, Robbie Edwards
 robbie.edwa...@gmail.com wrote:
  Actually, I can't.  I don't know the y values.  Only the s and only for a
  subset of the data.
 
  Like this.
 
  d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
 
 
 
  On Tue, May 22, 2012 at 11:57 AM, R. Michael Weylandt
  michael.weyla...@gmail.com wrote:
 
  You can reconstruct the y values by taking first-differences of the s
  vector, no? Then it sounds like you're good to go
 
  Best, Michael
 
  On Tue, May 22, 2012 at 11:40 AM, Robbie Edwards
  robbie.edwa...@gmail.com wrote:
   Hi all,
  
   Thanks for the replies, but I realize I've done a bad job explaining
 my
   problem.  To help, I've created some sample data to explain the
 problem.
  
   df - data.frame(x=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), y=c(109,
   232,
   363, 496, 625, 744, 847, 928, 981, 1000, 979, 912), s=c(109, 341, 704,
   1200, 1825, 2569, 3416, 4344, 5325, 6325, 7304, 8216))
  
   In this data frame, y results from y = x * b1 + x^2 * b2 + x^3 * b3
 and
   s
   is sum of the current y and all previous y (s3 = y1 + y2 + y3).
  
   I know I can find b1, b2 and b3 using:
   lm(y ~ 0 + x + I(x^2) + I(x^3), data=df)
  
   yielding...
   Coefficients:
       x  I(x^2)  I(x^3)
     100      10      -1
  
   However, I need to find b1, b2 and b3 using the s column.  The reason
   being, I don't actually know the values of y in the actual data set.
    And
   in the actual data, I only have a few of the values.  Imagine this
 data
   is
   being used a reward schedule for like a loyalty points program.  y
   represents the number of points needed for each level while s is the
   total
   number of points to reach that level.  In the real problem, my data
   looks
   more like this:
  
   d - data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
  
   Where I need to use a few sample points to help define the parameters
 of
   the curve.
  
   thanks again and hopefully this makes the problem a bit clearer.
  
   robbie
  
  
  
   On Fri, May 18, 2012 at 7:40 PM, David Winsemius
   dwinsem...@comcast.netwrote:
  
  
   On May 18, 2012, at 1:44 PM, Robbie Edwards wrote:
  
    Hi all,
  
   I'm trying to model some data where the y is defined by
  
   y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3
  
   Hopefully that reads clearly for email.
  
  
   cumsum( rowSums( cbind(B1 * x,  B2 * x^2, B3 * x^3)))
  
  
  
    Anyway, if it wasn't for the summation, I know I would do it like
 this
  
   lm(y ~ x + x2 + x3)
  
   Where x2 and x3 are x^2 and x^3.
  
   However, since each value of x is related to the previous values of
 x,
   I
   don't know how to do this.  Any help is greatly appreciated.
  
  
  
  
   David Winsemius, MD
   West Hartford, CT
  
  
  
          [[alternative HTML version deleted]]
  
   __
   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.
 
 


        [[alternative HTML version deleted]]

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[R] Could incomplete final line found be more serious than a warning?

2012-05-22 Thread Michael Bärtl


Dear all,

I've been successfully reading Web of Science-data from tab-delimited 
text files into a data.frame using an R-script based on readLines().

With new data I just downloaded I suddenly get this warning:

  incomplete final line found

I know this warning has already been discussed numerous times but none 
of the previously suggested solutions worked for me, unfortunately; so 
please bear with me:


I shut the warning down using warn = FALSE, but the data still won't 
get read so this seems to be more serious than a warning.


Adding a blank line or two at the end of the file did NOT help, i.e. R 
still does not read the file.


But my old files still work properly, though.
So I opened the text files using Notepad++ and saw that the last lines 
of both old text files (i.e. working) as well as new text files (i.e. 
the ones that don't work for some reason) always end with a tab stop 
followed by a line break. Personally I couldn't tell any difference 
between the ways these files ended. Their endings looked identical to me.


I was using R 2.14.0 (64 bit) on Windows when I dioscovered the problem. 
So I upgraded to 2.15.0 (64-bit) but the problem persists.


You can see small examples of an old and new file at 
https://www.dropbox.com/s/2joadjo9ce86rij/WoS-old.txt and 
https://www.dropbox.com/s/lp9l1exx4mfws1s/WoS-new.txt, respectively.


Does anybody happen to have an idea of what could cause these problems 
for me?


Thank you very much for your consideration!

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[R] utils:::menuInstallLocal()

2012-05-22 Thread Emiliano Zapata

Hello R,

I'm trying to install a package (class) locally; in windows 7, 64 bits
machine. The only massage I see on the R Console is:
utils:::menuInstallLocal()
 nothing else. What does this means, shouldn't I got some source of massage
on the Console.

EZ

[[alternative HTML version deleted]]

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

Re: [R] R Memory Issues

2012-05-22 Thread Emiliano Zapata

As a continuation to my original question, here is the massage that I get:

Error in glm.fit(x = structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,  :
  cannot allocate memory block of size 2.1 Gb

The model glm.fit is a logistic type (in the family of GLM) model. Maybe
this is not enough information; again!, but some feedback will be
appreciated. To me the issues appears to be associated with manipulation of
large dataset. Howeverl the algorithm runs fine in Unix; but not in Windows
(64 bits windows 7).

EZ

On Sun, May 20, 2012 at 4:09 PM, Emiliano Zapata ezapata...@gmail.comwrote:

 Already then, thank you everyone. This information was extremly useful,
 and I'll do a better job on the web next time.

 On Sun, May 20, 2012 at 2:10 PM, Prof Brian Ripley 
 rip...@stats.ox.ac.ukwrote:

 On 20/05/2012 18:42, jim holtman wrote:

 At the point in time that you get the error message, how big are the
 objects that you have in memory?  What does 'memory.size()' show as
 being used?  What does 'memory.limit()' show?  Have you tried using
 'gc()' periodically to do some garbage collection?  It might be that
 you memory is fragmented.  You need to supply some additional
 information.


 Either this is a 32-bit version of R in which case the wrong version is
 being used, or your advice is wrong: there are no credible fragmentation
 issues (and no need to use gc()) on a 64-bit build of R.

 But, we have a posting guide, we require 'at a minimum information', and
 the OP failed to give it to us so we are all guessing, completely
 unnecessarily.



 On Sun, May 20, 2012 at 12:09 PM, Emiliano Zapataezapata...@gmail.com
  wrote:

 -- Forwarded message --
 From: Emiliano Zapataezapata...@gmail.com
 Date: Sun, May 20, 2012 at 12:09 PM
 Subject:
 To: R-help@r-project.org


 Hi,

 I have a 64 bits machine (Windows) with a total of 192GB of physical
 memory
 (RAM), and total of 8 CPU. I wanted to ask how can I make R make use of
 all
 the memory. I recently ran a script requiring approximately 92 GB of
 memory
 to run, and got the massage:

  cannot allocate memory block of size 2.1 Gb



 I read on the web that if you increase the memory you have to reinstall
 R;
 would that be enough. Could I just increase the memory manually.


 Take you for any comments, or links on the web.


 EZ

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 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.






 --
 Brian D. Ripley,  rip...@stats.ox.ac.uk
 Professor of Applied Statistics,  
 http://www.stats.ox.ac.uk/~**ripley/http://www.stats.ox.ac.uk/%7Eripley/
 University of Oxford, Tel:  +44 1865 272861 (self)
 1 South Parks Road, +44 1865 272866 (PA)
 Oxford OX1 3TG, UKFax:  +44 1865 272595




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

Re: [R] How to remove square brackets, etc. from address strings?

2012-05-22 Thread arun

Hi,

 text - [Swidsinski, Alexander; Loening-Baucke, Vera; Lochs, Herbert] Charite 
Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; [Hale, Laura P.] Duke 
Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA


 gsub(\\[.+?],,text)
A.K.



- Original Message -
From: Sabina Arndt sabina.ar...@hotmail.de
To: r-help@r-project.org
Cc: 
Sent: Tuesday, May 22, 2012 6:08 AM
Subject: [R] How to remove square brackets, etc. from address strings?


Hello,



I'd like to remove the individual pairs of square brackets along with 
their content - plus the space directly behind it - from address strings
such as this:


  [Swidsinski, Alexander; Loening-Baucke, Vera; Lochs, Herbert] Charite 
Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; [Hale, Laura P.] 
Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA



I'd like get the following result:



  Charite Humboldt Univ, Innere Klin, D-10098 Berlin, Germany; Duke Univ, Med 
Ctr, Dept Pathol, Durham, NC 27710 USA



I tried


  address = gsub((.*)[(.*)], \\2, address)



But this deletes everything from the first opening bracket to the last closing 
bracket and leaves only the very last address:


  Duke Univ, Med Ctr, Dept Pathol, Durham, NC 27710 USA



How can I remove only the individual pairs of square brackets along with their 
content?



Thank you very much in advance!                           
    [[alternative HTML version deleted]]

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Re: [R] RNORM matrix based on CSV file values for MEAN and SD

2012-05-22 Thread dcoakley

Thanks Michael,

This seems to get me half-way towards a solution. However, I am still having
some difficulty in getting the out.csv exactly as I would like. I should
probably explain the issue a bit further.

I am reading in a csv file containing a list of 'n' parameters. I have
attached a csv ( http://r.789695.n4.nabble.com/file/n4630928/read.csv
read.csv ) which contains 10 parameters as an example. However, the final
problem may have up to 2000 parameters. Currently, these have string titles
in the first column. However, I can change these to numbers if it causes
problems in R.

As you said correctly, this will give me a data.frame with mean and sd
values.

The problem I have is in the creation of the matrix of rnorm values. Using
the for loop specified, I am getting a single column with the rnorm values
for the last parameter specified (see attached
http://r.789695.n4.nabble.com/file/n4630928/out.csv out.csv ). Re-running
the for loop appends another column to the csv file, below the original
column.

I have attached an example of the output I am looking for (
http://r.789695.n4.nabble.com/file/n4630928/OutputSample.csv
OutputSample.csv ). I believe, the approach here is correct, I just need to
make a few modifications to the for loop. This also needs to be scalable to
allow for the inclusion of multiple paramters (up to 2000) and samples (up
to 1 million). I will look into the dataConnections process you mentioned to
address this task.

Also, just in case the above attachments don't work, here is a link to the
shared files on Google Drive:
https://docs.google.com/open?id=0Bwwrh1DeZXteSFBXNVI1NGZfQjg

Thanks again for the help. Very much appreciated.

--
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Re: [R] RNORM matrix based on CSV file values for MEAN and SD

On Tue, May 22, 2012 at 12:40 PM, dcoakley danielcoakl...@gmail.com wrote:
Thanks Michael,

This seems to get me half-way towards a solution. However, I am still having
some difficulty in getting the out.csv exactly as I would like. I should
probably explain the issue a bit further.

It looks like

read.csv(http://r.789695.n4.nabble.com/file/n4630928/read.csv;)

works just fine for me here -- can you provide the code you're using
to read things in? It might be that you need header = TRUE if you're
using read.table() instead of read.csv() [The only reason I'm having
you use write.table() is that write.csv() doesn't append nicely]

As you said correctly, this will give me a data.frame with mean and sd
values.

Ahh, my apologies -- you probably need to transpose your data to a row
before writing:

e.g.,

write.table(1:10, temp.csv, sep = ,, row.names = FALSE, col.names = FALSE)
read.csv(temp.csv, header = FALSE)
# Gives a single column

write.table(t(1:10), temp.csv, sep = ,, row.names = FALSE,
col.names = FALSE)
read.csv(temp.csv, header = FALSE)
# Gives a single row

Best,
Michael

Also, just in case the above attachments don't work, here is a link to the
shared files on Google Drive:
https://docs.google.com/open?id=0Bwwrh1DeZXteSFBXNVI1NGZfQjg

Thanks again for the help. Very much appreciated.

--
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[R] In R, how do I execute a script that sets environment variables within the same shell session?

2012-05-22 Thread Michael

In R, how do I execute a script that sets environment variables within the
same shell session?

Hi all,

Could you please shed some lights on how to do this?

In a shell, I launched the R session.

But then in R, I realized that some environment variables need to be set up.

Of course I can use Sys.setenv()...

But I have so many of them...

And for some special reason, I have to first launch R and then run the
script from within R to set up the environment variables...

i.e.

I cannot do the other way around, which is first execute the script in the
shell,

and then launch the R session(which is the usual sequenc of ordering)...

So my question is:

I have to launch R session first, and then run the scrip from within R.

But if I naively use shell or system, that's going to not affect the
same shell that the R is using...

I would like the R session to be able to access those global environment
variables...

How do I do that?

Thank you!

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Re: [R] Could incomplete final line found be more serious than a warning?

2012-05-22 Thread peter dalgaard

(Original below)

Looks like someone had the bright idea of changing it to 16-bit UTF, so every 
2nd byte is NUL. It works for me with 

x - readLines(file(~/Downloads/WoS-new.txt, encoding=UTF-16))

(except that for some reason, x won't print properly although each individual 
line prints fine. Never mind, who cares as long as it reads...)

-pd

PS: The reason the printing is wacky is that one line has 148934 characters in 
it and the print routines pad all lines to the maximum length. Not sure what 
the point is in that.

On May 22, 2012, at 18:26 , Michael Bärtl wrote:

 Dear all,
 
 I've been successfully reading Web of Science-data from tab-delimited text 
 files into a data.frame using an R-script based on readLines().
 With new data I just downloaded I suddenly get this warning:
 
  incomplete final line found
 
 I know this warning has already been discussed numerous times but none of the 
 previously suggested solutions worked for me, unfortunately; so please bear 
 with me:
 
 I shut the warning down using warn = FALSE, but the data still won't get 
 read so this seems to be more serious than a warning.
 
 Adding a blank line or two at the end of the file did NOT help, i.e. R still 
 does not read the file.
 
 But my old files still work properly, though.
 So I opened the text files using Notepad++ and saw that the last lines of 
 both old text files (i.e. working) as well as new text files (i.e. the ones 
 that don't work for some reason) always end with a tab stop followed by a 
 line break. Personally I couldn't tell any difference between the ways these 
 files ended. Their endings looked identical to me.
 
 I was using R 2.14.0 (64 bit) on Windows when I dioscovered the problem. So I 
 upgraded to 2.15.0 (64-bit) but the problem persists.
 
 You can see small examples of an old and new file at 
 https://www.dropbox.com/s/2joadjo9ce86rij/WoS-old.txt and 
 https://www.dropbox.com/s/lp9l1exx4mfws1s/WoS-new.txt, respectively.
 
 Does anybody happen to have an idea of what could cause these problems for me?
 
 Thank you very much for your consideration!
 
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Re: [R] package grid: mirror grob objects along an axis

2012-05-22 Thread Thomas Zumbrunn

Maybe my question was not concise enough. I was referring to objects created 
with the package grid (also called grobs), not to the function grid 
from package graphics. For instance, let's say I have a polygon createad 
with grid::polygonGrob and want to mirror it along a specified axis. Of course 
one could transform the polygon's coordinates, but I was hoping that there is 
some more generic way of doing this (and that this could e.g. also be done 
with text grobs).

Cheers
/thomas



On Tuesday 22 May 2012, Rui Barradas wrote:
 Hello,
 
 Just flip 'xlim' or 'ylim'. Or both.
 Using the iris example in help(grid), make the following changes:
 
 
 op - par(mfcol = c(2,2))  # Two columns, first is the original, second
 flipped.
 
 with(iris,
 
   [... etc ...]
  # row 1, col 2: flip x axis
  plot(Sepal.Length, Sepal.Width, col = as.integer(Species),
   xlim = c(8, 4), ylim = c(2, 4.5), panel.first = grid(),
   main = with(iris,  plot(, panel.first = grid(), ..) ))
  # row 2, col 2: flip y axis
  plot(Sepal.Length, Sepal.Width, col = as.integer(Species),
   xlim = c(4, 8), ylim = c(4.5, 2),
   panel.first = grid(3, lty=1,lwd=2),
   main = ... panel.first = grid(3, lty=1,lwd=2), ..)
 
  [... etc ...]
 )
 par(op)
 
 
 Hope this helps,
 
 Rui Barradas
 
 Thomas Zumbrunn-3 wrote
 
  Hi everyone
  
  I'd like to flip grobs (grid graphical objects) along an axis, e.g. flip
  grobs
  horizontally or vertically. I couldn't find any hints, neither in the
  documentation nor by searching the web. Does anybody know how to achieve
  this?
  
  Cheers
  /thomas
  
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 is-tp4630866p4630870.html Sent from the R help mailing list archive at
 Nabble.com.
 
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Clinical Trial Unit (CTU)
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Schanzenstr. 55, CH-4031 Basel
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Re: [R] how to remove the 'promise' attribute of an R object (.Random.seed)?

2012-05-22 Thread luke-tierney


On Tue, 22 May 2012, Yihui Xie wrote:


Hi,

The problem arises when I lazyLoad() the .Random.seed from a
previously saved database. To simplify the process of reproducing the
problem, see the example below:

## this assignment may not really make sense, but illustrates the problem
delayedAssign('.Random.seed', 1L)

typeof(.Random.seed)
# [1] integer

rnorm(1)
# Error in rnorm(1) :
#  .Random.seed is not an integer vector but of type 'promise'

typeof(.Random.seed)
# [1] integer

So there must be an attribute promise somewhere attached to
.Random.seed, and I cannot find it. The R function typeof() does not
reveal it, but the TYPEOF() function in src/main/RNG.c says it is a
'promise'.

My question is, how to make R use the real value of .Random.seed
instead of complaining about the promise? Thanks!


Siple answer: Don't creat the promise in the first place, i.e. don't
use delayedAssign.

What is the real context where this arises? Knowing that may help us
decide whether the internals should address this possibility.

Best,

luke



Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA

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--
Luke Tierney
Chair, Statistics and Actuarial Science
Ralph E. Wareham Professor of Mathematical Sciences
University of Iowa  Phone: 319-335-3386
Department of Statistics andFax:   319-335-3017
   Actuarial Science
241 Schaeffer Hall  email:   luke-tier...@uiowa.edu
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[R] Is there a way to save EVERYTHING in R?

2012-05-22 Thread Michael

Is there a way to save EVERYTHING in R?

Hi all,

Could you please shed some lights on me about this?

I am trying to see if there is a way to save EVERYTHING in R to a file,
something like save.image...

But save.image doesn't save environment variables such as those
properties that are obtained using Sys.getenv() function...

So is there a way to save EVERYTHING?

Thanks a lot!

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Re: [R] Adding Text to a Plot

2012-05-22 Thread Peter Ehlers


On 2012-05-22 07:51, David Winsemius wrote:


On May 22, 2012, at 9:10 AM, Canto Casasola, Vicente David wrote:


I'm pretty sure I'm missing something about this.

Is there a smart way of typping hat(R)^2 and it's value from a linear
regression?

I've just found this tricky one:

# Sample data
x- sample(1:100,10)
y- 2+3*x+rnorm(10)

# Run the regression
lm1- lm(y~x)
# Plotting
plot(x,y, main=Linear Regression, col=red)
abline(lm1, col=blue)
placex- par(usr)[1]+.1*(par(usr)[2]-par(usr)[1])
placey1- par(usr)[3]+.9*(par(usr)[4]-par(usr)[3])
placey2- par(usr)[3]+.8*(par(usr)[4]-par(usr)[3])

# HERE: Is this the right way?


# To do   what? I see that you over-plotted the R, presumably
because you did not like the way that:
bquote(hat(R)^2 == .(summary(lm1)$adj.r.squared))

... ended up looking (with the exponent higher than in the non-hatted
version.)



text(x=placex, y=placey1,
 bquote(R^2 == .(summary(lm1)$r.squared)), adj=c(0,0))
text(x=placex, y=placey2,
 bquote(R^2 == .(summary(lm1)$adj.r.squared)), adj=c(0,0))
text(x=placex,y=placey2,
 expression(hat(R)), adj=c(0,0))

In addition, when I save the plot as PDF, the expression hat(R)
seems to be
somewhat displaced.

Any advice?


Are you sure? It looks  to me that the lower R is slightly shifted
to the left, but when I actually measure it, there does not appear to
be a shift. If it is real and not just an optical illusion, this may
have something to do with your unstated version of R or your also
unstated OS.


I don't see any shift either (Windows), easily checked with insert of an 
appropriate vertical line, but I would recommend not

overwriting the 'R' by using the phantom() construct:

  text(x=placex, y=placey2,
bquote(phantom(R)^2 == .(summary(lm1)$adj.r.squared)), adj=c(0,0))
  text(x=placex, y=placey2,
expression(hat(R)), adj=c(0,0))


Peter Ehlers



--
David Winsemius, MD
West Hartford, CT


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[R] Naming dimnames in an array using the results of an expression

2012-05-22 Thread Dr Bob Phillips


dear all

i'm struggling with naming in an array

diag.data is one of a series of 2x2 diagnostic testing arrays, with
'Outcome' columns (true/false) and 'Test' rows (High-risk, Low-risk),
drawn from a larger list object of 'results'

i can hard-code the names of the array using dimnames;

diag.data-array(c(19,2,125,50),c(2,2)) #example to run - actually comes
out with rownames/colnames already attached

dimnames(diag.data)-list(Rule=c(LR,HR),Outcome=c(FALSE,TRUE))

but i would really love to be able to soft-code it, using the different
'rule' names and slightly different 'outcomes' as extracted from the
main list of results, rather than just 'rule' and 'outcome' e.g.

dimnames(diag.data)-list(names(result[1])=rownames(diag.data),names(result[[1]][2])=colnames(diag.data))

this doesn't work

i understand that the dimnames command is NOT executing the function
'names' and giving me the result of this, and have struggled and failed
to use the 'eval' expression with alternative environments (which i
don't understand ...)

can anyone assist?

cheers - bob


-- 


Dr Bob Phillips
MRC Research Training Fellow
CRD
University of York
York
YO10 5DD
t:  +44 (0)1904 321099
f:  +44 (0)1904 321041
e:  bob.phill...@york.ac.uk
www.york.ac.uk/inst/crd
www.crd.york.ac.uk/prospero

CRD is part of the National Institute for Health Research and is a
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Re: [R] Could incomplete final line found be more serious than a warning?

2012-05-22 Thread Zhou Fang

If you look at the new file in raw mode, you'll see that it's chock full of
ASCII nuls, while the old file has none. This is probably what's giving you
the problems, because R does not allow strings containing embedded nul
characters. (I believe this is because Nul in strings is pretty dangerous in
programming, because they are often used to delimit the end of strings, and
so allowing you to read it in directly can be used for various code
injection exploits.)

To read the new data files, you need some way of dealing with the file as a
raw stream, and stripping out all the nul characters before converting back
to character. Investigate ?readBin...

Zhou

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Re: [R] Creating functions with a loop.

2012-05-22 Thread Etienne Larrivée-Hardy

Michael's method seems to be working but I still can't get to wrap it in a
loop since I cannot get the loop to dynamically change the functions' name,
i.e. ff1, ff2, ff3, ... In other words, I would need the first iteration to
create the function ff1, the second to create the function ff2, ...
Furthermore, does anyone know of a way to call said functions from a loop in
a way that the first step of the loop calls ff1, the second calls ff2 and so
on?

The 2 problems should be closely related I think.

Thanks for the time

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[R] scatterplot x axis specifications

2012-05-22 Thread jhartsho

I have created a scatter plot that has come out okay but I am having trouble
with the x axis.  My data consists of 4 treatments but these treatments are
days so R keeps reading them as numeric and making my x axis continuous. 
Here is what I have so far:


plot(pair$MC~pair$Day, pch=c(19,24)[as.factor(Cookie)],
main='Paired t Test',
xlab='Days in Field',
ylab='Moisture Content (Percent)')

I have also done: as.factor(Day) to change it to 4 levels and when I do
is.factor(Day) it returns with FALSE.

I'm assuming this is the reason that I get the message that my x and y
lengths differ when trying to convert my x axis using:

axis(side=1, at=c(0,15,30,45)

When I switch my numbers to words (i.e. zero, fifteen...) it only brings up
a boxplot and I specifically want the points so I can assign different pch
by group. Am I totally missing something here?

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Re: [R] Is there a way to save EVERYTHING in R?

2012-05-22 Thread Uwe Ligges




On 22.05.2012 19:50, Michael wrote:

Is there a way to save EVERYTHING in R?

Hi all,

Could you please shed some lights on me about this?

I am trying to see if there is a way to save EVERYTHING in R to a file,
something like save.image...

But save.image doesn't save environment variables such as those
properties that are obtained using Sys.getenv() function...

So is there a way to save EVERYTHING?


Well, the answer is yes, but not as you expect: If you have a virtual 
machine running only accessing a local filesystem, you can create a 
snapshot.


So, for other practical purposes and in the sense but not wording of you 
question:

No, environment variables are part of the OS (or the shell).
If you want to save EVERYTHING this way, you'd need an image of both, 
the RAM of you machine as well as the current state of all filesystems etc.


Best,
Uwe Ligges






Thanks a lot!

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Re: [R] Is there a way to save EVERYTHING in R?

2012-05-22 Thread Albert-Jan Roskam

Hi Michael,

check ?save

 
Regards,
Albert-Jan


~~
All right, but apart from the sanitation, the medicine, education, wine, public 
order, irrigation, roads, a 
fresh water system, and public health, what have the Romans ever done for us?
~~ 



 From: Michael comtech@gmail.com
To: r-help r-h...@stat.math.ethz.ch 
Sent: Tuesday, May 22, 2012 7:50 PM
Subject: [R] Is there a way to save EVERYTHING in R?
 
Is there a way to save EVERYTHING in R?

Hi all,

Could you please shed some lights on me about this?

I am trying to see if there is a way to save EVERYTHING in R to a file,
something like save.image...

But save.image doesn't save environment variables such as those
properties that are obtained using Sys.getenv() function...

So is there a way to save EVERYTHING?

Thanks a lot!

    [[alternative HTML version deleted]]

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[[alternative HTML version deleted]]

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Re: [R] scatterplot x axis specifications

2012-05-22 Thread Uwe Ligges




On 22.05.2012 19:59, jhartsho wrote:

I have created a scatter plot that has come out okay but I am having trouble
with the x axis.  My data consists of 4 treatments but these treatments are
days so R keeps reading them as numeric and making my x axis continuous.
Here is what I have so far:


plot(pair$MC~pair$Day, pch=c(19,24)[as.factor(Cookie)],
main='Paired t Test',
xlab='Days in Field',
ylab='Moisture Content (Percent)')

I have also done: as.factor(Day) to change it to 4 levels and when I do
is.factor(Day) it returns with FALSE.

I'm assuming this is the reason that I get the message that my x and y
lengths differ when trying to convert my x axis using:

axis(side=1, at=c(0,15,30,45)

When I switch my numbers to words (i.e. zero, fifteen...) it only brings up
a boxplot and I specifically want the points so I can assign different pch
by group. Am I totally missing something here?



Yes, missing a reproducible example so that we could easily show how it 
works.


Best,
Uwe Ligges




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Re: [R] How to evaluate R things from Visual Studio?

2012-05-22 Thread Peter Ehlers


Sorry, I have no solution for your problem, but a comment;
see inline below.

On 2012-05-22 08:02, Giannis Mamalikidis wrote:

Hello all, I’m new here and this is actually my very first post.
My name is Giannis and I am an undergraduate student. I am programming in 
Microsoft’s Visual Studio 2010 (VB.NET mostly) and I want to create a program 
for the research group I am in.

I merely want to evaluate actions and use simple statistical things such as ks 
test (Kolmogorov-Smirnov), and some plots.
The reason I need to call R from my program is that the users of my program 
will NOT know any programming language, and by extension they will not know R 
either. So everything will happen by the click of a button.
For me to do that, I need to be able to use interoperability between R and 
VS.NET 2010, so that I can use R’s evaluator.

I’ve already tried Baier’s StatConnDCOM (http://rcom.univie.ac.at) and I did 
succeed in creating a program that runs perfectly but came to see the hard way 
that – for my case, is kind of useless for the sole reason of StatconnDCOM’s 
redistribution restriction.
As Mr. Baier told me himself, there is no guarantee that StatconnDCOM will be 
available free of charge for ever. Hence I cannot create a freeware program 
now, and have my users uncertain of where they will be able to use this program 
for free in the future.

I’ve already tried R.NET (http://rdotnet.codeplex.com/ ) Version 1.4 (stable) 
and 1.4.1 (Change Set: 6d2c3f161801 ).
This does seem wonderful and it doesn’t have StaconnDCOM’s ridiculous 
restrictions, BUT it refuses to work on me!

So, now that you have all the necessary background info, it is time I presented 
my questions:

1) Is there another way of using R from Visual Studio that you can point me to?
2) Is anyone here using R.NET (any version) successfully? If so, can you help 
me to get it working? Unfortunately they can’t help me on R.NET’s official 
discussion forum

I mean, I just need to evaluate the basic R stuff.. Why is it THAT difficult?


This appears to imply that you consider it to be R's fault that you
can't easily interface it with Visual Studio. Isn't that rather the
fault of VS? If VS were open source, as R is, you could probably
find an easy solution yourself.

If you don't get a solution from one of those much more knowledgeable
than I, perhaps you might contact Microsoft.

Peter Ehlers



Last but not least, let me give you info that might be relevant in your 
deciding how to aid me.
OS: Windows 7 Professional x64 English (Genuine, it is purchased)
R: Version 2.15.0
Programming Environment: Microsoft Visual Studio .net 2010

For any additional Info, please do not hesitate to ask me either here or 
directly at giannis_mamaliki...@msn.com
I hope we can make this work. thanks in advance.
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[R] pad leading zeros in front of strings

2012-05-22 Thread Hui Du

Dear All,

This question sounds very simple but I don't know where I am wrong. I just want 
to pad leading zeros in some string, for example, 123 becomes 00123. What 
is wrong if I do following?

 sprintf(%05s, 123)
[1]   123


It didn't return 00123, instead it padded with 'blank'.


Thank you for your help in advance.

HXD

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[R] Is there an inverse function of Sys.getenv()

2012-05-22 Thread Michael

Using Sys.getenv(),

I got a list of 380 environment variables...

Now I open a new R session,

how do I set those environment variables in my new R session using
Sys.setenv()?

There is no batch function for Sys.setenv()?

Is there an import/export environment variables function?

Basically I want to export all the environment variables from one R
session and import them into another R session...

Thank you!

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Re: [R] pad leading zeros in front of strings

I think once upon a time this was found to be OS-dependent since it
calls the system's C  sprintf()  -- I get the leading zeros on Mac. I
presume you're on Windows?

Michael

On Tue, May 22, 2012 at 2:41 PM, Hui Du hui...@dataventures.com wrote:
 Dear All,

 This question sounds very simple but I don't know where I am wrong. I just 
 want to pad leading zeros in some string, for example, 123 becomes 00123. 
 What is wrong if I do following?

 sprintf(%05s, 123)
 [1]   123


 It didn't return 00123, instead it padded with 'blank'.


 Thank you for your help in advance.

 HXD

        [[alternative HTML version deleted]]

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Re: [R] pad leading zeros in front of strings

2012-05-22 Thread Sarah Goslee

Hi,

I think it's because   is used to pad strings, while 0 is used to
pad numbers*. If your values are always numeric, but stored as
strings, you could use:

 x - 123
 sprintf(%05d, as.numeric(x))
[1] 00123


* From ?sprintf:
‘0’ For numbers, pad to the field width with leading zeros.

I think some language implementations allow for specifying different
pad characters, but R's doesn't seem to.

Sarah

On Tue, May 22, 2012 at 2:41 PM, Hui Du hui...@dataventures.com wrote:
 Dear All,

 This question sounds very simple but I don't know where I am wrong. I just 
 want to pad leading zeros in some string, for example, 123 becomes 00123. 
 What is wrong if I do following?

 sprintf(%05s, 123)
 [1]   123


 It didn't return 00123, instead it padded with 'blank'.


 Thank you for your help in advance.

 HXD



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

__
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.

Re: [R] pad leading zeros in front of strings

Please do reply to the list -- I'm not on Windows so someone else will
have to pick the question up to help you out.

It's not great, but you could do something like

zeroPad - function(str, len.out, num.zeros = len.out[1] - nchar(str)){
   paste0(paste(rep(0, num.zeros), collapse = ), str)
}

as a temporary work-around. Probably possible to vectorize that pretty
easily as well.

Best,
Michael

On Tue, May 22, 2012 at 2:50 PM, Hui Du hui...@dataventures.com wrote:
 Thank you for your replay. Yes, I am on windows.

 Best Regards,
 Hui Du

 Data Ventures Inc

 -Original Message-
 From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
 Sent: Tuesday, May 22, 2012 11:49 AM
 To: Hui Du
 Cc: r-help@r-project.org
 Subject: Re: [R] pad leading zeros in front of strings

 I think once upon a time this was found to be OS-dependent since it
 calls the system's C  sprintf()  -- I get the leading zeros on Mac. I
 presume you're on Windows?

 Michael

 On Tue, May 22, 2012 at 2:41 PM, Hui Du hui...@dataventures.com wrote:
 Dear All,

 This question sounds very simple but I don't know where I am wrong. I just 
 want to pad leading zeros in some string, for example, 123 becomes 
 00123. What is wrong if I do following?

 sprintf(%05s, 123)
 [1]   123


 It didn't return 00123, instead it padded with 'blank'.


 Thank you for your help in advance.

 HXD

        [[alternative HTML version deleted]]

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

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

Re: [R] pad leading zeros in front of strings

2012-05-22 Thread Sarah Goslee

Michael, I'm curious: if you pass sprintf() a string, it still pads
with zeros? What's the output of:

sprintf(%05s, 123)
sprintf(%05s, abc)

On linux, sprintf() pads strings with spaces, as you'd expect. Padding
strings with zeros is... odd.

Sarah

 sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: x86_64-redhat-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_US.UTF-8   LC_NUMERIC=C
 [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
 [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8
 [7] LC_PAPER=C LC_NAME=C
 [9] LC_ADDRESS=C   LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C

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

loaded via a namespace (and not attached):
[1] tools_2.15.0

On Tue, May 22, 2012 at 2:49 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 I think once upon a time this was found to be OS-dependent since it
 calls the system's C  sprintf()  -- I get the leading zeros on Mac. I
 presume you're on Windows?

 Michael

 On Tue, May 22, 2012 at 2:41 PM, Hui Du hui...@dataventures.com wrote:
 Dear All,

 This question sounds very simple but I don't know where I am wrong. I just 
 want to pad leading zeros in some string, for example, 123 becomes 
 00123. What is wrong if I do following?

 sprintf(%05s, 123)
 [1]   123


 It didn't return 00123, instead it padded with 'blank'.


 Thank you for your help in advance.

 HXD


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

__
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.

Re: [R] pad leading zeros in front of strings