date:20130610

Unless I completely misunderstand what you are doing you don't need to 
aggregate, just drop the one column and rename things

newtemp  -  temp_df[, c(1,3)]
names(newtemp) -  c(names, w)
newtemp


John Kane
Kingston ON Canada


 -Original Message-
 From: abhishek@gmail.com
 Sent: Sun, 9 Jun 2013 23:15:48 -0700
 To: r-help@r-project.org
 Subject: [R] reshaping a data frame
 
 Hi Guys
 
 I am trying to cast a data frame but not aggregate the rows for the
 same variable.
 
 here is a contrived example.
 
 **input**
 temp_df  -
 data.frame(names=c('foo','foo','foo'),variable=c('w','w','w'),value=c(34,65,12))
 temp_df
   names variable value
 1   foow34
 2   foow65
 3   foow12
 
 
 ###
 **Want this**
 
 names  w
 foo 34
 foo 65
 foo 12
 
 
 ##
 **getting this***
 ##
 cast(temp_df)
 Aggregation requires fun.aggregate: length used as default
   names w
 1   foo 3
 
 
 In real dataset  the categorical column 'variable' will have many more
 categorical variable.
 
 Thanks!
 -Abhi
 
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Re: [R] All against all correlation matrix with GGPLOT Facet

No image.  The R-help list tends to strip out a lot of files. A pdf or txt 
usually gets through.  In any case I understand what you want this may do it.

library(ggplot2)
dat1  -  data.frame( v = rnorm(13),
w = rnorm(13),
x = rnorm(13),
y = rnorm(13),
z = rnorm(13))
plotmatrix(dat1)

John Kane
Kingston ON Canada


 -Original Message-
 From: gunda...@gmail.com
 Sent: Mon, 10 Jun 2013 12:26:44 +0900
 To: r-h...@stat.math.ethz.ch
 Subject: [R] All against all correlation matrix with GGPLOT Facet
 
 I have the following data:
 
 v - rnorm(13)
 w - rnorm(13)
 x - rnorm(13)
 y - rnorm(13)
 z - rnorm(13)
 
 
 Using GGPLOT facet, what I want to do is to create a 5*5 matrix,
 where each cells plot the correlation between
 each pair of the above data. E.g. v-v,v-w; v-x,...,z-z
 
 
 What's the way to do it?
 Attached is the image.
 
 GV.
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Re: [R] reshaping a data frame

Hi,If your dataset is similar to the one below:
set.seed(24)
temp1_df- 
data.frame(names=rep(c('foo','foo1'),each=6),variable=rep(c('w','x'),times=6),value=sample(25:40,12,replace=TRUE),stringsAsFactors=FALSE)

library(reshape2)
 
res-dcast(within(temp1_df,{Seq1-ave(value,names,variable,FUN=seq_along)}),names+Seq1~variable,value.var=value)[,-2]
res
#  names  w  x
#1   foo 29 28
#2   foo 36 33
#3   foo 35 39
#4  foo1 29 37
#5  foo1 37 29
#6  foo1 34 30
A.K.


- Original Message -
From: Abhishek Pratap abhishek@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc: 
Sent: Monday, June 10, 2013 2:15 AM
Subject: [R] reshaping a data frame

Hi Guys

I am trying to cast a data frame but not aggregate the rows for the
same variable.

here is a contrived example.

**input**
temp_df  - 
data.frame(names=c('foo','foo','foo'),variable=c('w','w','w'),value=c(34,65,12))
 temp_df
  names variable value
1   foo        w    34
2   foo        w    65
3   foo        w    12


###
**Want this**

names  w
foo         34
foo         65
foo         12


##
**getting this***
##
 cast(temp_df)
Aggregation requires fun.aggregate: length used as default
  names w
1   foo 3


In real dataset  the categorical column 'variable' will have many more
categorical variable.

Thanks!
-Abhi

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Re: [R] please check this

Hi,
Try this:
which(duplicated(res10Percent))
# [1] 117 125 157 189 213 235 267 275 278 293 301 327 331 335 339 367 369 371 
379
#[20] 413 415 417 441 459 461 477 479 505
res10PercentSub1-subset(res10Percent[which(duplicated(res10Percent)),],dummy==1)
  #most of the duplicated are dummy==1
res10PercentSub0-subset(res10Percent[which(duplicated(res10Percent)),],dummy==0)
 indx1-as.numeric(row.names(res10PercentSub1))
indx11-sort(c(indx1,indx1+1))
indx0- as.numeric(row.names(res10PercentSub0))
 indx00- sort(c(indx0,indx0-1))
indx10- sort(c(indx11,indx00))

 nrow(res10Percent[-indx10,])
#[1] 452
 res10PercentNew-res10Percent[-indx10,]
 nrow(subset(res10PercentNew,dummy==1))
#[1] 226
 nrow(subset(res10PercentNew,dummy==0))
#[1] 226
 nrow(unique(res10PercentNew))
#[1] 452
A.K.



- Original Message -
From: Cecilia Carmo cecilia.ca...@ua.pt
To: arun smartpink...@yahoo.com
Cc: 
Sent: Monday, June 10, 2013 10:19 AM
Subject: RE: please check this

But I don't want it like this. 
Once a firm is paired with another, these two firms should not be paired again.
Could you solve this?
Thanks,
Cecília



De: arun [smartpink...@yahoo.com]
Enviado: segunda-feira, 10 de Junho de 2013 15:12
Para: Cecilia Carmo
Assunto: Re: please check this

I did look into that.
If you look for the nrow() in each category, then it will be different.  It 
means that the duplicates are not pairwise, but in the whole `result`.  The 
explanation is again with the multiple matches.  So, here we selected the one 
with dummy==0 that closely matches the dimension of one dummy==1.  Suppose, the 
value of dimension with dummy==1` is `2554` and it got a match with dummy==0 
with `2580`.  Now, consider another case with dimension as `2570` with dummy==1 
(which also comes within the same split group).  Then it got a match with 
`2580' with dummy==0.  I guess it was based on the way in which it was tested.







From: Cecilia Carmo cecilia.ca...@ua.pt
To: arun smartpink...@yahoo.com
Sent: Monday, June 10, 2013 10:02 AM
Subject: please check this




When I do

res10Percent- fun1(final3New,0.1,200)
dim(res10Percent)
[1] 508   5
#[1] 508   5
nrow(subset(res10Percent,dummy==0))
#[1] 254
nrow(subset(res10Percent,dummy==1))
#[1] 254


testingDuplicates-unique(res10Percent)
nrow(testingDuplicates)
[1] 480 #this should be 508, if not there are duplicated rows, or not?


Thanks
Cecilia

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Re: [R] recode: how to avoid nested ifelse

2013-06-10 Thread Paul Johnson

Thanks, guys.


On Sat, Jun 8, 2013 at 2:17 PM, Neal Fultz nfu...@gmail.com wrote:

 rowSums and Reduce will have the same problems with bad data you alluded
 to earlier, eg
 cg = 1, hs = 0

 But that's something to check for with crosstabs anyway.


This wrong data thing is a distraction here.  I guess I'd have to craft 2
solutions, depending on what the researcher says. (We can't assume es = 0
or es = NA and cg = 1 is bad data. There are some people who finish college
without doing elementary school (wasn't Albert Einstein one of those?) or
high school. I once went to an eye doctor who didn't finish high school,
but nonetheless was admitted to optometrist school.)

I did not know about the Reduce function before this. If we enforce the
ordering and clean up the data in the way you imagine, it would work.

I think the pmax is the most teachable and dependably not-getting-wrongable
approach if the data is not wrong.


 Side note: you should check out the microbenchmark pkg, it's quite handy.


Perhaps the working example of microbenchmark is the best thing in this
thread! I understand the idea behind it, but it seems like I can never get
it to work right. It helps to see how you do it.


 Rrequire(microbenchmark)
 Rmicrobenchmark(
 +   f1(cg,hs,es),
 +   f2(cg,hs,es),
 +   f3(cg,hs,es),
 +   f4(cg,hs,es)
 + )
 Unit: microseconds
expr   min lq median uq   max neval
  f1(cg, hs, es) 23029.848 25279.9660 27024.9640 29996.6810 55444.112   100
  f2(cg, hs, es)   730.665   755.5750   811.7445   934.3320  6179.798   100
  f3(cg, hs, es)85.029   101.6785   129.8605   196.2835  2820.187   100
  f4(cg, hs, es)   762.232   804.4850   843.7170  1079.0800 24869.548   100

 On Fri, Jun 07, 2013 at 08:03:26PM -0700, Joshua Wiley wrote:
  I still argue for na.rm=FALSE, but that is cute, also substantially
 faster
 
  f1 - function(x1, x2, x3) do.call(paste0, list(x1, x2, x3))
  f2 - function(x1, x2, x3) pmax(3*x3, 2*x2, es, 0, na.rm=FALSE)
  f3 - function(x1, x2, x3) Reduce(`+`, list(x1, x2, x3))
  f4 - function(x1, x2, x3) rowSums(cbind(x1, x2, x3))
 
  es - rep(c(0, 0, 1, 0, 1, 0, 1, 1, NA, NA), 1000)
  hs - rep(c(0, 0, 1, 0, 1, 0, 1, 0, 1, NA), 1000)
  cg - rep(c(0, 0, 0, 0, 1, 0, 1, 0, NA, NA), 1000)
 
  system.time(replicate(1000, f1(cg, hs, es)))
  system.time(replicate(1000, f2(cg, hs, es)))
  system.time(replicate(1000, f3(cg, hs, es)))
  system.time(replicate(1000, f4(cg, hs, es)))
 
   system.time(replicate(1000, f1(cg, hs, es)))
 user  system elapsed
22.730.03   22.76
   system.time(replicate(1000, f2(cg, hs, es)))
 user  system elapsed
 0.920.040.95
   system.time(replicate(1000, f3(cg, hs, es)))
 user  system elapsed
 0.190.020.20
system.time(replicate(1000, f4(cg, hs, es)))
 user  system elapsed
 0.950.030.98
 
 
  R version 3.0.0 (2013-04-03)
  Platform: x86_64-w64-mingw32/x64 (64-bit)
 
 
 
 
  On Fri, Jun 7, 2013 at 7:25 PM, Neal Fultz nfu...@gmail.com wrote:
   I would do this to get the highest non-missing level:
  
   x - pmax(3*cg, 2*hs, es, 0, na.rm=TRUE)
  
   rock chalk...
  
   -nfultz
  
   On Fri, Jun 07, 2013 at 06:24:50PM -0700, Joshua Wiley wrote:
   Hi Paul,
  
   Unless you have truly offended the data generating oracle*, the
   pattern: NA, 1, NA, should be a data entry error --- graduating HS
   implies graduating ES, no?  I would argue fringe cases like that
   should be corrected in the data, not through coding work arounds.
   Then you can just do:
  
   x - do.call(paste0, list(es, hs, cg))
  
table(factor(x, levels = c(000, 100, 110, 111), labels =
 c(none, es,hs, cg)))
   none   es   hs   cg
  4112
  
   Cheers,
  
   Josh
  
   *Drawn from comments by Judea Pearl one lively session.
  
  
   On Fri, Jun 7, 2013 at 6:13 PM, Paul Johnson pauljoh...@gmail.com
 wrote:
In our Summer Stats Institute, I was asked a question that amounts
 to
reversing the effect of the contrasts function (reconstruct an
 ordinal
predictor from a set of binary columns). The best I could think of
 was to
link together several ifelse functions, and I don't think I want to
 do this
if the example became any more complicated.
   
I'm unable to remember a less error prone method :). But I expect
 you might.
   
Here's my working example code
   
## Paul Johnson pauljohn at ku.edu
## 2013-06-07
   
## We need to create an ordinal factor from these indicators
## completed elementary school
es - c(0, 0, 1, 0, 1, 0, 1, 1)
## completed high school
hs - c(0, 0, 1, 0, 1, 0, 1, 0)
## completed college graduate
cg - c(0, 0, 0, 0, 1, 0, 1, 0)
   
ed - ifelse(cg == 1, 3,
 ifelse(hs == 1, 2,
ifelse(es == 1, 1, 0)))
   
edf - factor(ed, levels = 0:3,  labels = c(none, es, hs,
 cg))
data.frame(es, hs, cg, ed, edf)
   
## Looks OK, but what if there are missings?
es - c(0, 0, 1,

Re: [R] How to expand.grid with string elements (the half!)

If you can explain why those particular six combinations out of the
complete set of nine, then perhaps someone can tell you how.

-Don

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/9/13 10:25 PM, Gundala Viswanath gunda...@gmail.com wrote:

I have the following result of expand grid:

 d - expand.grid(c(x,y,z),c(x,y,z))

What I want is to create a combination of strings
but only the half of the all combinations:

  Var1 Var2
1xx
2yx
3   yy
4   zy
5   xz
6zz


What's the way to do it?

G.V.

   [[alternative HTML version deleted]]

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[R] Substituting the values on the y-axis

2013-06-10 Thread diddle1990

Hello, 

I plotted a graph on R showing how salinity (in ‰, y-axis) changes with time(in 
years, x-axis). However, right from the beginning on the Excel spreadsheet the v
alues for salinity appeared as, for example, 35000‰ instead of 35‰, which I gues
sed must have been a typing error for the website from which I extracted the dat
a (NOAA).Thus, I now would like to substitute these values with the correspondin
g smaller value, as it follows: 

25000 35000- 25, 35   and so on.

Is there any way I can change this on R or do I have to modify these numbers bef
ore inputting the data on R (for example on Excel)? If so, can anybody tell me h
ow to do either of these? 

Many thanks! 

Emanuela 

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[R] twoby2 (Odds Ratio) for variables with 3 or more levels

2013-06-10 Thread Vlatka Matkovic Puljic

Dear all,

I am using Epi package to calculate Odds ratio in my bivariate analysis.
How can I make *twoby2 *in variables that have 3 or more levels.

For example:
I have 4 level var (Age)
m=matrix(c(290, 100,232, 201, 136, 99, 182, 240), nrow=4, ncol=2)
twoby2(m)

R gives me only
Comparing : Row 1 vs. Row 2

While I would like to have reference value in Row 1, and compare Row 2, Row
3 and Row 4 with it.


Thanks for your help!

[[alternative HTML version deleted]]

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[R] Rcmdr seit heute nicht mehr ladbar

2013-06-10 Thread Bastian Wimmer

Wenn man ihn mal braucht ist er tot. folgende Fehlermeldung ereilt mich seit 
heute beim starten des R Commander auf dem Mac:

 library(Rcmdr)
Lade nötiges Paket: car
Lade nötiges Paket: MASS
Lade nötiges Paket: nnet
Error : .onAttach in attachNamespace() für 'Rcmdr' fehlgeschlagen, Details:
  Aufruf: structure(.External(.C_dotTclObjv, objv), class = tclObj)
  Fehler: [tcl] invalid command name image.

Zusätzlich: Warnmeldung:
In fun(libname, pkgname) :
  couldn't connect to display /tmp/launch-K8nELf/org.macosforge.xquartz:0
Fehler: Laden von Paket oder Namensraum für 'Rcmdr' fehlgeschlagen

Ich bin ziemlich angefressen. Folgende erfolglose Versuche:
- R neu installiert
- x11 neu installiert
- alle Ordner dabei gelöscht
- Pakete neu installiert
Nichts. Rcmdr will nicht mehr. 


--  
Beste Grüße,
Yours,
Bastian Wimmer M.A.

Research Associate at the Chair of Educational Psychology
University of Erlangen-Nuremberg
Dutzendteichstraße 24
90478 Nuremberg
Germany

Phone: +49 (0) 9171 83924 84
Fax: +49 (0) 3222 64968 14
Email: bastian.wim...@fau.de
Web: http://j.mp/Umkf4U (Chair of educational Psychology)

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Re: [R] Substituting the values on the y-axis

2013-06-10 Thread Bert Gunter

Sounds like you have made no effort to learn R, e.g. by reading the
Intro to R tutorial packaged with R or other online tutorial (there
are many).

Don't you think you need to do some homework first?

-- Bert

On Mon, Jun 10, 2013 at 7:26 AM,  diddle1...@fastwebnet.it wrote:
 Hello,

 I plotted a graph on R showing how salinity (in ‰, y-axis) changes with 
 time(in
 years, x-axis). However, right from the beginning on the Excel spreadsheet 
 the v
 alues for salinity appeared as, for example, 35000‰ instead of 35‰, which I 
 gues
 sed must have been a typing error for the website from which I extracted the 
 dat
 a (NOAA).Thus, I now would like to substitute these values with the 
 correspondin
 g smaller value, as it follows:

 25000 35000- 25, 35   and so on.

 Is there any way I can change this on R or do I have to modify these numbers 
 bef
 ore inputting the data on R (for example on Excel)? If so, can anybody tell 
 me h
 ow to do either of these?

 Many thanks!

 Emanuela

 __
 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.



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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

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Re: [R] Substituting the values on the y-axis


Just calculate a new sequence if those percentages are in an orderly sequence. 
See ?seq
 v  -  seq(25, 200, by = 10)
or perhaps the values are actually  text
?substr
x  -  substr(v, 1,2)

John Kane
Kingston ON Canada


 -Original Message-
 From: diddle1...@fastwebnet.it
 Sent: Mon, 10 Jun 2013 16:26:54 +0200 (CEST)
 To: r-help@r-project.org
 Subject: [R] Substituting the values on the y-axis
 
 Hello,
 
 I plotted a graph on R showing how salinity (in ‰, y-axis) changes with
 time(in
 years, x-axis). However, right from the beginning on the Excel
 spreadsheet the v
 alues for salinity appeared as, for example, 35000‰ instead of 35‰, which
 I gues
 sed must have been a typing error for the website from which I extracted
 the dat
 a (NOAA).Thus, I now would like to substitute these values with the
 correspondin
 g smaller value, as it follows:
 
 25000 35000- 25, 35   and so on.
 
 Is there any way I can change this on R or do I have to modify these
 numbers bef
 ore inputting the data on R (for example on Excel)? If so, can anybody
 tell me h
 ow to do either of these?
 
 Many thanks!
 
 Emanuela
 
 __
 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
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Re: [R] Substituting the values on the y-axis

2013-06-10 Thread Emanuela

I did look into tutorials but I could not find the exact request I am looking
for. I just started using R so I am still a beginner.  If you then know
where I can find it, can you please redirect me to it 




--
View this message in context: 
http://r.789695.n4.nabble.com/Substituting-the-values-on-the-y-axis-tp4669165p4669171.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Rcmdr seit heute nicht mehr ladbar

2013-06-10 Thread John Fox

Dear Bastian,

I'm afraid that I don't read German, but (as near as I can tell) since you say 
that you're using the most recent version of R and have X11 installed, you 
should have the software you need. Just in case, you might check the Rcmdr 
installation notes for Mac users at 
http://socserv.socsci.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html. 
Apparently, R is having difficulty connecting to X11. I'm copying this response 
to Rob Goedman, who has often been able to help with Rcmdr issues under Mac OS 
X.

Best,
 John

---
John Fox
Senator McMaster Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada




 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Bastian Wimmer
 Sent: Monday, June 10, 2013 5:27 AM
 To: r-help@r-project.org
 Subject: [R] Rcmdr seit heute nicht mehr ladbar
 
 Wenn man ihn mal braucht ist er tot. folgende Fehlermeldung ereilt mich
 seit heute beim starten des R Commander auf dem Mac:
 
  library(Rcmdr)
 Lade nötiges Paket: car
 Lade nötiges Paket: MASS
 Lade nötiges Paket: nnet
 Error : .onAttach in attachNamespace() für 'Rcmdr' fehlgeschlagen,
 Details:
   Aufruf: structure(.External(.C_dotTclObjv, objv), class = tclObj)
   Fehler: [tcl] invalid command name image.
 
 Zusätzlich: Warnmeldung:
 In fun(libname, pkgname) :
   couldn't connect to display /tmp/launch-
 K8nELf/org.macosforge.xquartz:0
 Fehler: Laden von Paket oder Namensraum für 'Rcmdr' fehlgeschlagen
 
 Ich bin ziemlich angefressen. Folgende erfolglose Versuche:
 - R neu installiert
 - x11 neu installiert
 - alle Ordner dabei gelöscht
 - Pakete neu installiert
 Nichts. Rcmdr will nicht mehr.
 
 
 --
 Beste Grüße,
 Yours,
 Bastian Wimmer M.A.
 
 Research Associate at the Chair of Educational Psychology
 University of Erlangen-Nuremberg
 Dutzendteichstraße 24
 90478 Nuremberg
 Germany
 
 Phone: +49 (0) 9171 83924 84
 Fax: +49 (0) 3222 64968 14
 Email: bastian.wim...@fau.de
 Web: http://j.mp/Umkf4U (Chair of educational Psychology)
 
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[R] woby2 (Odds Ratio) for variables with 3 or more levels

2013-06-10 Thread Vlatka Matkovic Puljic

Dear all,

I am using Epi package to calculate Odds ratio in my bivariate analysis.
How can I make *twoby2 *in variables that have 3 or more levels.

For example:
I have 4 level var (Age)
m=matrix(c(290, 100,232, 201, 136, 99, 182, 240), nrow=4, ncol=2)
library (Epi)
twoby2(m)

R gives me only
Comparing : Row 1 vs. Row 2

While I would like to have reference value in Row 1, and compare Row 2, Row
3 and Row 4 with it.


Thanks for your help!

[[alternative HTML version deleted]]

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

Re: [R] Substituting the values on the y-axis

Hi Emanuela,

Welcome to R

It can be hard finding even relatively simple things when you are just
starting. You might want to have a look at
http://www.unt.edu/rss/class/Jon/R_SC/ or
http://www.burns-stat.com/documents/tutorials/impatient-r/ if ou have not
already seen them. Patrick Burn's site
http://www.introductoryr.co.uk/R_Resources_for_Beginners.html has some useful
links

If you are a refugee from SAS or SPSS, this paper by Bob Muenchen is very
useful www.et.bs.ehu.es/~etptupaf/pub/R/RforSASSPSSusers.pdf

Some tricks for asking a good question in the R help list is here:
https://github.com/hadley/devtools/wiki/Reproducibility or
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

In most cases it is very useful to provide some data. See ?dput in the last two
links. A small bit of sample data in your original post would definately have
helped.

Many or most R-help readers do not use nabble and really hate to have to go
there to see the context of a message. You should always leave the important
parts of earlier messages to let the R-help reader see what the problems and
other suggested solutions may be.

John Kane
Kingston ON Canada

-Original Message-
From: diddle1...@fastwebnet.it
Sent: Mon, 10 Jun 2013 09:08:59 -0700 (PDT)
To: r-help@r-project.org
Subject: Re: [R] Substituting the values on the y-axis

I did look into tutorials but I could not find the exact request I am
looking
for. I just started using R so I am still a beginner. If you then know
where I can find it, can you please redirect me to it

--
View this message in context:
http://r.789695.n4.nabble.com/Substituting-the-values-on-the-y-axis-tp4669165p4669171.html
Sent from the R help mailing list archive at Nabble.com.

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FREE ONLINE PHOTOSHARING - Share your photos online with your friends and
family!
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Re: [R] How to expand.grid with string elements (the half!)

2013-06-10 Thread William Dunlap

Perhaps the OP wants the unique combinations of V1 and V2, as in
  R d - expand.grid(V1=c(x,y,z),V2=c(x,y,z))
  R d[ as.numeric(d$V1) = as.numeric(d$V2), ]
V1 V2
  1  x  x
  4  x  y
  5  y  y
  7  x  z
  8  y  z
  9  z  z
or
  R V - letters[24:26]
  R rbind(t(combn(V,m=2)), cbind(V,V))
   V   V  
  [1,] x y
  [2,] x z
  [3,] y z
  [4,] x x
  [5,] y y
  [6,] z z

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Rolf Turner
 Sent: Monday, June 10, 2013 2:20 AM
 To: Gundala Viswanath
 Cc: r-h...@stat.math.ethz.ch
 Subject: Re: [R] How to expand.grid with string elements (the half!)
 
 
 Your question makes no sense at all.  The grid expansion
 has 9 rows.  In case you hadn't noticed, 9 is an odd number
 (i.e. not divisible by 2).  There are no halves.
 
 Do not expect the list to read your mind.  Instead, ask a
 meaningful question.
 
  cheers,
 
  Rolf Turner
 
 On 10/06/13 17:25, Gundala Viswanath wrote:
  I have the following result of expand grid:
 
  d - expand.grid(c(x,y,z),c(x,y,z))
  What I want is to create a combination of strings
  but only the half of the all combinations:
 
 Var1 Var2
  1xx
  2yx
  3   yy
  4   zy
  5   xz
  6zz
 
 
  What's the way to do it?
 
 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Selecting divergent colors

2013-06-10 Thread Brian Smith

Hi,

I was trying to make a density plot with 13 samples. To distinguish each
sample, it would be good if each color is as different as possible from the
other colors. I could use the built in function, but that does not do more
than 8 colors and then goes back to recycling the cols. If I use a palette,
then it is really difficult to distinguish between the colors.

So, is there a way that I can select a large number of colors (i.e. perhaps
20) that are as different from each other as possible?

Here is my example code using the palette:

**
mat - matrix(sample(1:1000,1000,replace=T),nrow=20,ncol=20)
snames - paste('Sample_',1:ncol(mat),sep='')
colnames(mat) - snames

mycols - palette(rainbow(ncol(mat)))

for(k in 1:ncol(mat)){
  plot(density(mat[,k]),col=mycols[k],xlab='',ylab='',axes=F,main=F)
  par(new=T)
}

legend(x='topright',legend=snames,fill=mycols)



thanks!

[[alternative HTML version deleted]]

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Re: [R] read.csv timing

2013-06-10 Thread ivo welch

here are some small benchmarks on an i7-2600k with an SSD:

input file: 104,126 rows with 76 columns.  all numeric.

linux time bzcat bzfile.csv.bz2  /dev/null  -- 1.8 seconds

R d - read.csv( pipe( bzfile ) )   -- 6.3 seconds
R d - read.csv( pipe( bzfile ), colClasses=numeric)  -- 4.2 seconds

R more than doubles the time it takes to load the file to convert it
into an R data structure.  if the colClasses are not specified, then
it takes another 50% longer.


some more experiments: save in R format (gzip format) --- this
increases file size from 15MB to 20MB.  how fast is the filesystem?

linux time gzcat file.Rdata  /dev/null  -- 0.4 seconds


the linux file system and CPU can decompress the 15MB .bz2 file in 1.8
seconds and decompress the 20MB .gz file in 0.4 seconds.  this is
surprising.  let's make sure that this is due to the .gz format.
indeed:

linux bunzip bzfile.csv.bz2 ; gzip bzfile.csv
linux time gzcat bzfile.csv.gz  /dev/null  -- 0.4 seconds


reading .gz files is much faster on my linux system than reading bz
files.  this surprises me.  I would have thought my CPU is so fast at
decompressing even bzip2 that it is almost zero, so I thought the disk
space was the primary determinant of speed, and bzip2 should have been
faster.  well, ok, maybe slower, but not by a factor of 4.


now I am thinking that maybe I should use .gz files to store my data.
but the advantages are surprisingly not as great:

R d - read.csv( pipe( gzfile ) )   -- 5.7 seconds
R d - read.csv( pipe( gzfile ), colClasses=numeric)  -- 2.6 seconds
R d - read.csv( gzfile( gzfile ), colClasses=numeric) -- 4.5
seconds   (surprisingly slower)

(the first and second versions are using R's gzfile, but literally
gzcat .. | in a pipe here.)


conclusion: a .gz file can be read from file to memory about four
times faster than a .bz file by the linux file system (outside R).
the conversion from strings in memory nto R doubles takes about as
much time as the .bz file system decompression read.  bzip2 is a more
efficient storage method than .gz, but its decompression is
considerably slower (the fact that there is less to read from disk
does not make up for the CPU decompression overhead).

saving the data in native R format essentially has no decompression
penalty and becomes close to native fast reading of .gz data.  chances
are this is because it has .gz support baked in.  gzfile does not help
with read.csv, however.

/iaw

Ivo Welch (ivo.we...@gmail.com)


On Mon, Jun 10, 2013 at 10:09 AM, ivo welch ivo.we...@gmail.com wrote:
 Surely you know the types of the columns?  If you specify it in advance,
 read.table and relatives will be much faster.

 Duncan Murdoch

 thx, duncan.  yes, I do know the types of columns, but I did not
 realize how much faster these functions become.  on my SSD-based
 system, the speedup is about a factor of 2.  that is, read.csv on a
 bzip2 file that takes 10 seconds without colClasses takes 5 seconds
 with colClasses.  I don't know how to benchmark intermittent memory
 usage, but my guess is that with colClasses, it requires less memory,
 too.  in fact, my naive and incorrect assumption had been that
 read.csv would just read ithe file nto a dynamic string array and then
 convert each string, and this would not take much longer than if it
 converted as it went along.  so, I had thought more memory use but
 not more time.  wrong.

 I would add to the man (.Rd) page the sentence Specifying colClasses
 can speed up read.csv where it describes the option.)


 once I will figure out how to bake C into R, I may try to write a fast
 filter function for myself, but share it for others wanting to use it.

 regards,

 /iaw

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Re: [R] woby2 (Odds Ratio) for variables with 3 or more levels

2013-06-10 Thread David Winsemius


On Jun 10, 2013, at 9:27 AM, Vlatka Matkovic Puljic wrote:

 Dear all,
 
 I am using Epi package to calculate Odds ratio in my bivariate analysis.
 How can I make *twoby2 *in variables that have 3 or more levels.

I hope looking at that again you will see how odd it sounds to be requesting 
advice about how to use a program for 2 x 2 tables on data that doesn't meet 
those requirements. If you want to stay within the Epi package world, you can 
probably use the 'mh' function since it says it can handle multi-way tables (or 
you can learn to use 'glm' in the regular stats package to do either logistic 
regression or Poisson regression.)
 
 For example:
 I have 4 level var (Age)
 m=matrix(c(290, 100,232, 201, 136, 99, 182, 240), nrow=4, ncol=2)
 library (Epi)
 twoby2(m)
 
 R gives me only
 Comparing : Row 1 vs. Row 2
 
 While I would like to have reference value in Row 1, and compare Row 2, Row
 3 and Row 4 with it.

That is the default set of contrasts for 'glm' (and probably for 'mh' although 
it's not clear from the help page.)

(Epi does have its own mailing list.)

-- 
David Winsemius
Alameda, CA, USA

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[R] Where Query in SQL

2013-06-10 Thread Sneha Bishnoi

Hey all

I am trying to use where in clause in sql query in R
here is my code:

sql.select-paste(select PERSON_NAME from UNITS where UNIT_ID in
(',cathree,'),sep=)

where cathree is 1 variable with 16 observations as follows

UNIT_ID
1 205
2 209
3 213
4 217
5 228
6 232
7 236
8 240
9 245
10 249
11 253
12 257
13 268
14 272
15 276
16 280

but when i run this code, 0 rows are selected eventhough there exist 3 rows
which satisfy the above query


Thanks
Sneha

[[alternative HTML version deleted]]

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Re: [R] woby2 (Odds Ratio) for variables with 3 or more levels

2013-06-10 Thread James C. Whanger

You may want to consider a cumulative logit model which effectively
bifurcates an ordinal variable by utilizing the odds of being in a given
level or below (depending on your coding).


On Mon, Jun 10, 2013 at 12:27 PM, Vlatka Matkovic Puljic vlatk...@gmail.com
 wrote:

 Dear all,

 I am using Epi package to calculate Odds ratio in my bivariate analysis.
 How can I make *twoby2 *in variables that have 3 or more levels.

 For example:
 I have 4 level var (Age)
 m=matrix(c(290, 100,232, 201, 136, 99, 182, 240), nrow=4, ncol=2)
 library (Epi)
 twoby2(m)

 R gives me only
 Comparing : Row 1 vs. Row 2

 While I would like to have reference value in Row 1, and compare Row 2, Row
 3 and Row 4 with it.


 Thanks for your help!

 [[alternative HTML version deleted]]

 __
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 http://www.R-project.org/posting-guide.html
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-- 
*James C. Whanger*
*
*

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Re: [R] Where Query in SQL

Do this

cat(sql.select,'\n')

and then decide whether the query is what it should be according to
standard SQL syntax.
(If it is not, then fix it.)

-Don

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/10/13 11:47 AM, Sneha Bishnoi sneha.bish...@gmail.com wrote:

Hey all

I am trying to use where in clause in sql query in R
here is my code:

sql.select-paste(select PERSON_NAME from UNITS where UNIT_ID in
(',cathree,'),sep=)

where cathree is 1 variable with 16 observations as follows

UNIT_ID
1 205
2 209
3 213
4 217
5 228
6 232
7 236
8 240
9 245
10 249
11 253
12 257
13 268
14 272
15 276
16 280

but when i run this code, 0 rows are selected eventhough there exist 3
rows
which satisfy the above query


Thanks
Sneha

   [[alternative HTML version deleted]]

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Re: [R] Selecting divergent colors

2013-06-10 Thread Adams, Jean

It will be hard to come up with 20 clearly distinguishable colors.  Check
out the website http://colorbrewer2.org/ and the R package RColorBrewer.
 It does not have a 20-color palette, but it does have some 8- to 12-color
palettes that are very nice.

library(RColorBrewer)
display.brewer.all(n=NULL, type=all, select=NULL, exact.n=TRUE)

You could use these colors in combination with line type to build up to 72
unique combinations.  For example ...

nuniq - ncol(mat)
mycols - rep(brewer.pal(12, Set3), length=nuniq)
myltys - rep(1:6, rep(12, 6))[1:nuniq]

for(k in 1:nuniq){
 plot(density(mat[,k]), col=mycols[k], xlab='', ylab='', axes=F, main=F,
lwd=3, lty=myltys[k])
par(new=TRUE)
 }
legend('topright', legend=snames, col=mycols, lty=myltys, lwd=3)

Jean



On Mon, Jun 10, 2013 at 12:33 PM, Brian Smith bsmith030...@gmail.comwrote:

 Hi,

 I was trying to make a density plot with 13 samples. To distinguish each
 sample, it would be good if each color is as different as possible from the
 other colors. I could use the built in function, but that does not do more
 than 8 colors and then goes back to recycling the cols. If I use a palette,
 then it is really difficult to distinguish between the colors.

 So, is there a way that I can select a large number of colors (i.e. perhaps
 20) that are as different from each other as possible?

 Here is my example code using the palette:

 **
 mat - matrix(sample(1:1000,1000,replace=T),nrow=20,ncol=20)
 snames - paste('Sample_',1:ncol(mat),sep='')
 colnames(mat) - snames

 mycols - palette(rainbow(ncol(mat)))

 for(k in 1:ncol(mat)){
   plot(density(mat[,k]),col=mycols[k],xlab='',ylab='',axes=F,main=F)
   par(new=T)
 }

 legend(x='topright',legend=snames,fill=mycols)

 

 thanks!

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[[alternative HTML version deleted]]

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Re: [R] reshaping a data frame

2013-06-10 Thread Abhishek Pratap

Thanks everyone for your quick reply. I think my contrived example hid
the complexity I wanted to show by using only one variable.

@Arun: I think your example is exactly what I was looking for. Very
cool trick with 'ave' and 'seq_along'...just dint occur to me.

Best,
-Abh


 On Mon, Jun 10, 2013 at 7:13 AM, arun smartpink...@yahoo.com wrote:
 Hi,If your dataset is similar to the one below:
 set.seed(24)
 temp1_df- 
 data.frame(names=rep(c('foo','foo1'),each=6),variable=rep(c('w','x'),times=6),value=sample(25:40,12,replace=TRUE),stringsAsFactors=FALSE)

 library(reshape2)
  
 res-dcast(within(temp1_df,{Seq1-ave(value,names,variable,FUN=seq_along)}),names+Seq1~variable,value.var=value)[,-2]
 res
 #  names  w  x
 #1   foo 29 28
 #2   foo 36 33
 #3   foo 35 39
 #4  foo1 29 37
 #5  foo1 37 29
 #6  foo1 34 30
 A.K.


 - Original Message -
 From: Abhishek Pratap abhishek@gmail.com
 To: r-help@r-project.org r-help@r-project.org
 Cc:
 Sent: Monday, June 10, 2013 2:15 AM
 Subject: [R] reshaping a data frame

 Hi Guys

 I am trying to cast a data frame but not aggregate the rows for the
 same variable.

 here is a contrived example.

 **input**
 temp_df  - 
 data.frame(names=c('foo','foo','foo'),variable=c('w','w','w'),value=c(34,65,12))
 temp_df
   names variable value
 1   foow34
 2   foow65
 3   foow12


 ###
 **Want this**
 
 names  w
 foo 34
 foo 65
 foo 12


 ##
 **getting this***
 ##
 cast(temp_df)
 Aggregation requires fun.aggregate: length used as default
   names w
 1   foo 3


 In real dataset  the categorical column 'variable' will have many more
 categorical variable.

 Thanks!
 -Abhi

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[R] parameters estimation of a normal-lognormal multivariate model

2013-06-10 Thread Hertzog Gladys

Dear all,

I have to create a model which is a mixture of a normal and log-normal 
distribution. To create it, I need to estimate the 2 covariance matrixes and 
the mixing parameter (total =7 parameters) by maximizing the log-likelihood 
function. This maximization has to be performed by the nlm routine.
As I use relative data, the means are known and equal to 1.

I’ve already tried to do it in 1 dimension (with 1 set of relative data) and it 
works well. However, when I introduce the 2nd set of relative data I get 
illogical results for the correlation and a lot of warnings messages (at all 
25).

To estimates the parameters I defined first the log-likelihood function with 
the 2 commands dmvnorm and dlnorm.plus. Then I assign starting values of the 
parameters and finally I use the nlm routine to estimate the parameters (see 
script below).

# Importing and reading the grid files. Output are 2048x2048 matrixes
 
P - read.ascii.grid(d:/Documents/JOINT_FREQUENCY/grid_E727_P-3000.asc, 
return.header= FALSE ); 
V - read.ascii.grid(d:/Documents/JOINT_FREQUENCY/grid_E727_V-3000.asc, 
return.header= FALSE ); 
 
p - c(P); # tranform matrix into a vector
v - c(V);
 
p- p[!is.na(p)] # removing NA values
v- v[!is.na(v)]
 
p_rel - p/mean(p) #Transforming the data to relative values
v_rel - v/mean(v) 
PV - cbind(p_rel, v_rel) # create a matrix of vectors
 
L - function(par,p_rel,v_rel) {
 
return (-sum(log( (1- par[7])*dmvnorm(PV, mean=c(1,1), sigma= 
matrix(c(par[1]^2, par[1]*par[2]*par[3],par[1]*par[2]*par[3], par[2]^2 
),nrow=2, ncol=2))+
par[7]*dlnorm.rplus(PV, meanlog=c(1,1), varlog= 
matrix(c(par[4]^2,par[4]*par[5]*par[6],par[4]*par[5]*par[6],par[5]^2), 
nrow=2,ncol=2)))))
 
}
par.start- c(0.74, 0.66 ,0.40, 1.4, 1.2, 0.4, 0.5) # log-likelihood estimators
 
result-nlm(L,par.start,v_rel=v_rel,p_rel=p_rel, hessian=TRUE, iterlim=200, 
check.analyticals= TRUE)
Messages d'avis :
1: In log(eigen(sigma, symmetric = TRUE, only.values = TRUE)$values) :
  production de NaN
2: In sqrt(2 * pi * det(varlog)) : production de NaN
3: In nlm(L, par.start, p_rel = p_rel, v_rel = v_rel, hessian = TRUE) :
  NA/Inf replaced by maximum positive value
4: In log(eigen(sigma, symmetric = TRUE, only.values = TRUE)$values) :
  production de NaN
…. Until 25.

par.hat - result$estimate
 
cat(sigN_p =, par[1],\n,sigN_v =, par[2],\n,rhoN =, 
par[3],\n,sigLN_p =, par[4],\n,sigLN_v =, par[5],\n,rhoLN =, 
par[6],\n,mixing parameter =, par[7],\n)
 
sigN_p = 0.5403361 
 sigN_v = 0.6667375 
 rhoN = 0.6260181 
 sigLN_p = 1.705626 
 sigLN_v = 1.592832 
 rhoLN = 0.9735974 
 mixing parameter = 0.8113369
 
Does someone know what is wrong in my model or how should I do to find these 
parameters in 2 dimensions?

Thank you very much for taking time to look at my questions.

Regards,

Gladys Hertzog
Master student in environmental engineering, ETH Zurich
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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.

Re: [R] Selecting divergent colors

2013-06-10 Thread Ben Tupper

Hi,

On Jun 10, 2013, at 3:46 PM, Adams, Jean wrote:

 It will be hard to come up with 20 clearly distinguishable colors.  Check
 out the website http://colorbrewer2.org/ and the R package RColorBrewer.
 It does not have a 20-color palette, but it does have some 8- to 12-color
 palettes that are very nice.
 
 library(RColorBrewer)
 display.brewer.all(n=NULL, type=all, select=NULL, exact.n=TRUE)
 

It sounds like Brian is looking for categorical coloring rather than divergent 
coloring.  The Glasbey LUT works really well in image processing for just such 
purposes.  It would be easy to use that within R for your lines.

http://www.bioss.ac.uk/people/chris/colorpaper.pdf

You might be able to snag the color table out of this collection of Java 
plugins for ImageJ software. 

http://www.dentistry.bham.ac.uk/landinig/software/morphology.zip

Within that archive is a text file called glasbey.lut which is a simple text 
file of RGB color values.

Cheers,
Ben




 You could use these colors in combination with line type to build up to 72
 unique combinations.  For example ...
 
 nuniq - ncol(mat)
 mycols - rep(brewer.pal(12, Set3), length=nuniq)
 myltys - rep(1:6, rep(12, 6))[1:nuniq]
 
 for(k in 1:nuniq){
 plot(density(mat[,k]), col=mycols[k], xlab='', ylab='', axes=F, main=F,
 lwd=3, lty=myltys[k])
 par(new=TRUE)
 }
 legend('topright', legend=snames, col=mycols, lty=myltys, lwd=3)
 
 Jean
 
 
 
 On Mon, Jun 10, 2013 at 12:33 PM, Brian Smith bsmith030...@gmail.comwrote:
 
 Hi,
 
 I was trying to make a density plot with 13 samples. To distinguish each
 sample, it would be good if each color is as different as possible from the
 other colors. I could use the built in function, but that does not do more
 than 8 colors and then goes back to recycling the cols. If I use a palette,
 then it is really difficult to distinguish between the colors.
 
 So, is there a way that I can select a large number of colors (i.e. perhaps
 20) that are as different from each other as possible?
 
 Here is my example code using the palette:
 
 **
 mat - matrix(sample(1:1000,1000,replace=T),nrow=20,ncol=20)
 snames - paste('Sample_',1:ncol(mat),sep='')
 colnames(mat) - snames
 
 mycols - palette(rainbow(ncol(mat)))
 
 for(k in 1:ncol(mat)){
  plot(density(mat[,k]),col=mycols[k],xlab='',ylab='',axes=F,main=F)
  par(new=T)
 }
 
 legend(x='topright',legend=snames,fill=mycols)
 
 
 
 thanks!
 
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Ben Tupper
Bigelow Laboratory for Ocean Sciences
60 Bigelow Drive, P.O. Box 380
East Boothbay, Maine 04544
http://www.bigelow.org

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[R] Fwd: Problem with ODBC connection

2013-06-10 Thread Christofer Bogaso

Any response please? Was my question not clear to the list? Please let me
know.

Thanks and regards,

-- Forwarded message --
From: Christofer Bogaso bogaso.christo...@gmail.com
Date: Sat, Jun 8, 2013 at 9:39 PM
Subject: Re: Problem with ODBC connection
To: r-help r-help@r-project.org

Hello All,

My previous post remains unanswered probably because the attachment was not
working properly.

So I am re-posting it again.

My problem is in reading an Excel-2003 file through ODBC connection using
RODBC package. Let say I have this Excel file:

http://www.2shared.com/document/HS3JeFyW/MyFile.html

I saved it in my F: drive and tried reading the contents using RODBC
connection:

 library(RODBC)
 MyData - sqlFetch(odbcConnectExcel(f:/MyFile.xls), )
 head(MyData, 30)

However it looks that the second column (with header 's') is not read
properly.

Can somebody here explain this bizarre thing? Did I do something wrong in
reading that?

Really appreciate if someone could point out anything what might go wrong.

Thanks and regards,

On Fri, Jun 7, 2013 at 4:46 PM, Christofer Bogaso 
bogaso.christo...@gmail.com wrote:

 Hello again,

 I am having problem with ODBC connection using the RODBC package.

 I am basically trying to read the attached Excel-2003 file using RODBC
 package. Here is my code:

  head(sqlFetch(odbcConnectExcel(d:/1.xls), ), 30);
 odbcCloseAll()
Criteria  s  d fd  ffd1
 f1fd2f2 fd3 f3 F12 F13 F14 F15 F16 F17
 F18 F19 F20
 1 a NA NA NA NA 0.
 0.27755576 -0.00040332321NA  NA NA  NA
  NA  NA  NA  NA  NA  NA  NA  NA
 2 s NA  0 NA NA 0.
 0.  0.000NA  NA NA  NA
  NA  NA  NA  NA  NA  NA  NA  NA
 3 d NA  0 NA NA 0.01734723
 0.06938894  0.2775558  5.00  NA NA  NA
  NA  NA  NA  NA  NA  NA  NA  NA
 4 f NA NA NA NA NA
 NA NA -4.25  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 5 f NA  0 NA NA 0.
 0.  0.000 -1.53  NA NA  NA
  NA  NA  NA  NA  NA  NA  NA  NA
 6 f NA NA NA NA NA
 NA  0.000  0.00  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 7 f NA NA NA NA NA
 NA  0.000NA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 8 f NA  0 NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 9 f NA  0 NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 10f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 11f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 12f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA
 13f NA NA NA NA NA
 NA NANA  NA NA  NA  NA  NA  NA  NA  NA
  NA  NA  NA

 Here you see the data in second column could not read at all.

 Can somebody point me if I did something wrong?

 Thanks and regards,

[[alternative HTML version deleted]]

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[R] Apply a PCA to other datasets

2013-06-10 Thread edelance

I have run a PCA on one data set.  I need the standard deviation of the first
two bands for my analysis.  I now want to apply the same PCA rotation I used
in the first one to all my other data sets.  Is there any way to do this in
r?  Thanks.




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[R] Speed up or alternative to 'For' loop

2013-06-10 Thread Trevor Walker

I have a For loop that is quite slow and am wondering if there is a faster
option:

df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
df$Height - exp(-0.1 + 0.2*df$Age)
df$HeightGrowth - NA   #intialize with NA
for (i in 2:nrow(df))
 {if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
 }

Trevor Walker
Email: trevordaviswal...@gmail.com

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[R] Combining CSV data

2013-06-10 Thread Shreya Rawal

Hello R community,

I am trying to combine two CSV files that look like this:

File A

Row_ID_CR,   Data1,Data2,Data3
1,   aa,  bb,  cc
2,   dd,  ee,  ff


File B

Row_ID_N,   Src_Row_ID,   DataN1
1a,   1,   This is comment 1
2a,   1,   This is comment 2
3a,   2,   This is comment 1
4a,   1,   This is comment 3

And the output I am looking for is, comparing the values of Row_ID_CR and
Src_Row_ID

Output

ROW_ID_CR,Data1,Data2,Data3,DataComment1,
 DataComment2,  DataComment3
1,  aa, bb, cc,This is
comment1,This is comment2, This is comment 3
2,  dd,  ee, ff,  This is
comment1


I am a novice R user, I am able to replicate a left join but I need a bit
more in the final result.


Thanks!!

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Re: [R] How sum all possible combinations of rows, given 4 matrices

2013-06-10 Thread Estigarribia, Bruno

It works, Arun. Thanks!
(FYI, a couple a the matrices I am dealing with have 1000+ rows, so I had
to do in on a supercomputer at work. For the curious, I am trying to find
all possible scores in a model f language mixing described in:
Title: Structured Variation in Codeswitching: Towards an Empirically Based
Typology of Bilingual Speech Patterns
Authors: Deuchar, Margaret; Muysken, Pieter; Wang, Sung-Lan
Publication Date: 2007
Journal Name: International Journal of Bilingual Education and
Bilingualism)


Bruno Estigarribia
Assistant Professor of Spanish, Department of Romance Languages and
Literatures
Research Assistant Professor of Psychology, Cognitive Science Program
Affiliate Faculty, Global Studies
Dey Hall, Room 332, CB# 3170
University of North Carolina at Chapel Hill
estig...@email.unc.edu
917-348-8162





On 5/27/13 1:54 PM, arun smartpink...@yahoo.com wrote:

Hi,
Not sure if this is what you expected:

set.seed(24)
mat1- matrix(sample(1:20,3*4,replace=TRUE),ncol=3)
set.seed(28)
mat2- matrix(sample(1:25,3*6,replace=TRUE),ncol=3)
set.seed(30)
mat3- matrix(sample(1:35,3*8,replace=TRUE),ncol=3)
set.seed(35)
mat4- matrix(sample(1:40,3*10,replace=TRUE),ncol=3)
 
dat1-expand.grid(seq(dim(mat1)[1]),seq(dim(mat2)[1]),seq(dim(mat3)[1]),se
q(dim(mat4)[1]))
vec1-paste0(mat,1:4)
matNew-do.call(cbind,lapply(seq_len(ncol(dat1)),function(i)
get(vec1[i])[dat1[,i],]))
colnames(matNew)- (seq(12)-1)%%3+1
datNew-data.frame(matNew)
res-sapply(split(colnames(datNew),gsub(\\..*,,colnames(datNew))),func
tion(x) rowSums(datNew[,x]))

dim(res)
#[1] 19203
 head(res)
# X1 X2 X3
#[1,] 46 63 70
#[2,] 45 68 59
#[3,] 55 55 66
#[4,] 51 65 61
#[5,] 48 84 75
#[6,] 47 89 64

A.K.

- Original Message -
From: Estigarribia, Bruno estig...@email.unc.edu
To: r-help@R-project.org r-help@r-project.org
Cc: 
Sent: Monday, May 27, 2013 11:24 AM
Subject: [R] How sum all possible combinations of rows, given 4 matrices

Hello all,

I have 4 matrices with 3 columns each (different number of rows though). I
want to find a function that returns all possible 3-place vectors
corresponding to the sum by columns of picking one row from matrix 1, one
from matrix 2, one from matrix 3, and one from matrix 4. So basically, all
possible ways of picking one row from each matrix and then sum their
columns to obtain a 3-place vector.
Is there a way to use expand.grid and reduce to obtain this result? Or am
I on the wrong track?
Thank you,
Bruno
PS:I believe I have given all relevant info. I apologize in advance if my
question is ill-posed or ambiguous.

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Re: [R] please check this

Hi,
Try this:
res10Percent- fun1(final3New,0.1,200)

res10PercentSub1-subset(res10Percent[duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE),],dummy==1)
indx1-as.numeric(row.names(res10PercentSub1))

res10PercentSub2-res10PercentSub1[order(res10PercentSub1$dimension),]
indx11-as.numeric(row.names(res10PercentSub2))
names(indx11)-(seq_along(indx11)-1)%/%2+1
res10PercentSub3-res10Percent[c(indx11,indx11+1),]
res10PercentSub3$id- names(c(indx11,indx11+1))
 
res10PercentSub4-do.call(rbind,lapply(split(res10PercentSub3,res10PercentSub3$id),function(x)
 
{x1-x[-1,];x2-x1[which.max(abs(x1$dimension[1]-x1$dimension[-1]))+1,];x3-x[x$dummy==1,][which.min(abs(as.numeric(row.names(x[x$dummy==1,]))-as.numeric(row.names(x2,];rbind(x3,x2)}))

res10PercentSub0-subset(res10Percent[duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE),],dummy==0)
indx0-as.numeric(row.names(res10PercentSub0))

res10PercentSub20-res10PercentSub0[order(res10PercentSub0$dimension),]
indx00-as.numeric(row.names(res10PercentSub20))
names(indx00)-(seq_along(indx00)-1)%/%2+1
res10PercentSub30- res10Percent[c(indx00-1,indx00),]
res10PercentSub30$id- names(c(indx00-1,indx00))
res10PercentSub40- 
do.call(rbind,lapply(split(res10PercentSub30,res10PercentSub30$id),function(x){x1-subset(x,dummy==1);
 
x2-subset(x,dummy==0);x3-x1[which.max(abs(x1$dimension-unique(x2$dimension))),];x4-x2[which.min(abs(as.numeric(row.names(x3))-as.numeric(row.names(x2,];rbind(x3,x4)}))

row.names(res10PercentSub40)-gsub(.*\\.,,row.names(res10PercentSub40))
indxNew- 
sort(as.numeric(c(row.names(res10PercentSub5),row.names(res10PercentSub40
res10PercentFinal-res10Percent[-indxNew,]
 dim(res10PercentFinal)
#[1] 454   5
 nrow(subset(res10PercentFinal,dummy==0))
#[1] 227
 nrow(subset(res10PercentFinal,dummy==1))
#[1] 227

nrow(unique(res10PercentFinal))
#[1] 454
which(duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE))
# [1] 113 117 123 125 153 157 187 189 207 213 223 235 265 267 269 275 276 278 
279
#[20] 283 293 301 303 305 309 317 327 331 335 339 341 343 347 351 367 369 371 
379
#[39] 385 399 407 413 415 417 429 437 441 453 459 461 471 473 477 479 501 505
 res10Percent[c(113:114,117:118),]
# firm year industry dummy dimension
#113 500221723 2005   26 1  3147
#114 500601429 2005   26 0  3076
#117 500221723 2005   26 1  3147
#118 502668920 2005   26 0  3249
 
res10PercentFinal[c(113:114,117:118),]  #deleted the duplicated row and the 
accompanying pair with the maximum difference
# firm year industry dummy dimension
#113 500221723 2005   26 1  3147
#114 500601429 2005   26 0  3076
#119 500115362 2006   26 1  6239
#120 500060223 2006   26 0  6208

A.K.

row.names(res10PercentSub4)-gsub(.*\\.,,row.names(res10PercentSub4))
res10PercentSub5-res10PercentSub4[order(as.numeric(res10PercentSub4$id)),]

- Original Message -
From: Cecilia Carmo cecilia.ca...@ua.pt
To: arun smartpink...@yahoo.com
Cc: 
Sent: Monday, June 10, 2013 1:41 PM
Subject: RE: please check this

I think it could be better to eliminate that one.
If you could do it I appreciate.

Cecília


De: arun [smartpink...@yahoo.com]
Enviado: segunda-feira, 10 de Junho de 2013 18:14
Para: Cecilia Carmo
Assunto: Re: please check this

If you wanted to eliminate the duplicate rows that have the pair with the 
maximum difference, it is possible.
Just informing you.




- Original Message -
From: Cecilia Carmo cecilia.ca...@ua.pt
To: arun smartpink...@yahoo.com
Cc:
Sent: Monday, June 10, 2013 10:51 AM
Subject: RE: please check this

I think it is ok now.

Thanks
Cecília


De: arun [smartpink...@yahoo.com]
Enviado: segunda-feira, 10 de Junho de 2013 15:39
Para: Cecilia Carmo
Cc: R help
Assunto: Re: please check this

Hi,
Try this:
which(duplicated(res10Percent))
# [1] 117 125 157 189 213 235 267 275 278 293 301 327 331 335 339 367 369 371 
379
#[20] 413 415 417 441 459 461 477 479 505
res10PercentSub1-subset(res10Percent[which(duplicated(res10Percent)),],dummy==1)
  #most of the duplicated are dummy==1
res10PercentSub0-subset(res10Percent[which(duplicated(res10Percent)),],dummy==0)
indx1-as.numeric(row.names(res10PercentSub1))
indx11-sort(c(indx1,indx1+1))
indx0- as.numeric(row.names(res10PercentSub0))
indx00- sort(c(indx0,indx0-1))
indx10- sort(c(indx11,indx00))

nrow(res10Percent[-indx10,])
#[1] 452
res10PercentNew-res10Percent[-indx10,]
nrow(subset(res10PercentNew,dummy==1))
#[1] 226
nrow(subset(res10PercentNew,dummy==0))
#[1] 226
nrow(unique(res10PercentNew))
#[1] 452
A.K.



- Original Message -
From: Cecilia Carmo cecilia.ca...@ua.pt
To: arun smartpink...@yahoo.com
Cc:
Sent: Monday, June 10, 2013 10:19 AM
Subject: RE: please check this

But I don't want it like this.
Once a firm is paired with another, these two firms

Re: [R] Speed up or alternative to 'For' loop

2013-06-10 Thread Rui Barradas


Hello,

One way to speed it up is to use a matrix instead of a data.frame. Since 
data.frames can hold data of all classes, the access to their elements 
is slow. And your data is all numeric so it can be hold in a matrix. The 
second way below gave me a speed up by a factor of 50.



system.time({
for (i in 2:nrow(df))
 {if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
 }
})

system.time({
df2 - data.matrix(df)
for(i in seq_len(nrow(df2))[-1]){
if(df2[i, TreeID] == df2[i - 1, TreeID])
df2[i, HeightGrowth] - df2[i, Height] - df2[i - 1, 
Height]
}
})

all.equal(df, as.data.frame(df2))  # TRUE


Hope this helps,

Rui Barradas

Em 10-06-2013 18:28, Trevor Walker escreveu:

I have a For loop that is quite slow and am wondering if there is a faster
option:

df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
df$Height - exp(-0.1 + 0.2*df$Age)
df$HeightGrowth - NA   #intialize with NA
for (i in 2:nrow(df))
  {if(df$TreeID[i]==df$TreeID[i-1])
   {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
   }
  }

Trevor Walker
Email: trevordaviswal...@gmail.com

[[alternative HTML version deleted]]

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Re: [R] Combining CSV data

2013-06-10 Thread jim holtman

try this:

 fileA - read.csv(text = Row_ID_CR,   Data1,Data2,Data3
+ 1,   aa,  bb,  cc
+ 2,   dd,  ee,  ff, as.is = TRUE)

 fileB - read.csv(text = Row_ID_N,   Src_Row_ID,   DataN1
+ 1a,   1,   This is comment 1
+ 2a,   1,   This is comment 2
+ 3a,   2,   This is comment 1
+ 4a,   1,   This is comment 3, as.is = TRUE)

 # get rid of leading/trailing blanks on comments
 fileB$DataN1 - gsub(^ *| *$, , fileB$DataN1)

 # merge together
 result - merge(fileA, fileB, by.x = 'Row_ID_CR', by.y = Src_Row_ID)

 # now partition by Row_ID_CR and aggregate the comments
 result2 - do.call(rbind,
+ lapply(split(result, result$Row_ID_CR), function(.grp){
+ cbind(.grp[1L, -c(5,6)], comment = paste(.grp$DataN1, collapse =
', '))
+ })
+ )
 result2
  Row_ID_CR Data1Data2Data3
comment
1 1aa   bb   cc This is comment
1, This is comment 2, This is comment 3
2 2dd   ee   ff
  This is comment 1




On Mon, Jun 10, 2013 at 4:38 PM, Shreya Rawal rawal.shr...@gmail.comwrote:

 Hello R community,

 I am trying to combine two CSV files that look like this:

 File A

 Row_ID_CR,   Data1,Data2,Data3
 1,   aa,  bb,  cc
 2,   dd,  ee,  ff


 File B

 Row_ID_N,   Src_Row_ID,   DataN1
 1a,   1,   This is comment 1
 2a,   1,   This is comment 2
 3a,   2,   This is comment 1
 4a,   1,   This is comment 3

 And the output I am looking for is, comparing the values of Row_ID_CR and
 Src_Row_ID

 Output

 ROW_ID_CR,Data1,Data2,Data3,DataComment1,
  DataComment2,  DataComment3
 1,  aa, bb, cc,This is
 comment1,This is comment2, This is comment 3
 2,  dd,  ee, ff,  This is
 comment1


 I am a novice R user, I am able to replicate a left join but I need a bit
 more in the final result.


 Thanks!!

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-- 
Jim Holtman
Data Munger Guru

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

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Re: [R] Speed up or alternative to 'For' loop

How about

for (ir in unique(df$TreeID)) {
  in.ir - df$TreeID == ir
  df$HeightGrowth[in.ir] - cumsum(df$Height[in.ir])
}

Seemed fast enough to me.

In R, it is generally good to look for ways to operate on entire vectors
or arrays, rather than element by element within them. The cumsum()
function does that in this example.

-Don


-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/10/13 10:28 AM, Trevor Walker trevordaviswal...@gmail.com wrote:

I have a For loop that is quite slow and am wondering if there is a faster
option:

df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
df$Height - exp(-0.1 + 0.2*df$Age)
df$HeightGrowth - NA   #intialize with NA
for (i in 2:nrow(df))
 {if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
 }

Trevor Walker
Email: trevordaviswal...@gmail.com

   [[alternative HTML version deleted]]

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Re: [R] please check this

Sorry, I forgot to paste some lines and change the names:


res10Percent- fun1(final3New,0.1,200)

res10PercentSub1-subset(res10Percent[duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE),],dummy==1)
indx1-as.numeric(row.names(res10PercentSub1))

res10PercentSub2-res10PercentSub1[order(res10PercentSub1$dimension),]
indx11-as.numeric(row.names(res10PercentSub2))
names(indx11)-(seq_along(indx11)-1)%/%2+1
res10PercentSub3-res10Percent[c(indx11,indx11+1),]
res10PercentSub3$id- names(c(indx11,indx11+1))
res10PercentSub4-do.call(rbind,lapply(split(res10PercentSub3,res10PercentSub3$id),function(x)
 
{x1-x[-1,];x2-x1[which.max(abs(x1$dimension[1]-x1$dimension[-1]))+1,];x3-x[x$dummy==1,][which.min(abs(as.numeric(row.names(x[x$dummy==1,]))-as.numeric(row.names(x2,];rbind(x3,x2)}))
row.names(res10PercentSub4)-gsub(.*\\.,,row.names(res10PercentSub4)) 
#forgot

res10PercentSub0-subset(res10Percent[duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE),],dummy==0)
indx0-as.numeric(row.names(res10PercentSub0))

res10PercentSub20-res10PercentSub0[order(res10PercentSub0$dimension),]
indx00-as.numeric(row.names(res10PercentSub20))
names(indx00)-(seq_along(indx00)-1)%/%2+1
res10PercentSub30- res10Percent[c(indx00-1,indx00),]
res10PercentSub30$id- names(c(indx00-1,indx00))
res10PercentSub40- 
do.call(rbind,lapply(split(res10PercentSub30,res10PercentSub30$id),function(x){x1-subset(x,dummy==1);
 
x2-subset(x,dummy==0);x3-x1[which.max(abs(x1$dimension-unique(x2$dimension))),];x4-x2[which.min(abs(as.numeric(row.names(x3))-as.numeric(row.names(x2,];rbind(x3,x4)}))

row.names(res10PercentSub40)-gsub(.*\\.,,row.names(res10PercentSub40))
indxNew- 
sort(as.numeric(c(row.names(res10PercentSub4),row.names(res10PercentSub40 
#res10PercentSub4
res10PercentFinal-res10Percent[-indxNew,]
dim(res10PercentFinal)
#[1] 454  5
nrow(subset(res10PercentFinal,dummy==0))
#[1] 227
nrow(subset(res10PercentFinal,dummy==1))
#[1] 227

nrow(unique(res10PercentFinal))

A.K.

- Original Message -
From: Cecilia Carmo cecilia.ca...@ua.pt
To: arun smartpink...@yahoo.com
Cc: 
Sent: Monday, June 10, 2013 5:48 PM
Subject: RE: please check this

Error message:

Error in row.names(res10PercentSub5) : 
  object 'res10PercentSub5' not found



De: arun [smartpink...@yahoo.com]
Enviado: segunda-feira, 10 de Junho de 2013 22:05
Para: Cecilia Carmo
Cc: R help
Assunto: Re: please check this

Hi,
Try this:
res10Percent- fun1(final3New,0.1,200)

res10PercentSub1-subset(res10Percent[duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE),],dummy==1)
indx1-as.numeric(row.names(res10PercentSub1))

res10PercentSub2-res10PercentSub1[order(res10PercentSub1$dimension),]
indx11-as.numeric(row.names(res10PercentSub2))
names(indx11)-(seq_along(indx11)-1)%/%2+1
res10PercentSub3-res10Percent[c(indx11,indx11+1),]
res10PercentSub3$id- names(c(indx11,indx11+1))
res10PercentSub4-do.call(rbind,lapply(split(res10PercentSub3,res10PercentSub3$id),function(x)
 
{x1-x[-1,];x2-x1[which.max(abs(x1$dimension[1]-x1$dimension[-1]))+1,];x3-x[x$dummy==1,][which.min(abs(as.numeric(row.names(x[x$dummy==1,]))-as.numeric(row.names(x2,];rbind(x3,x2)}))

res10PercentSub0-subset(res10Percent[duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE),],dummy==0)
indx0-as.numeric(row.names(res10PercentSub0))

res10PercentSub20-res10PercentSub0[order(res10PercentSub0$dimension),]
indx00-as.numeric(row.names(res10PercentSub20))
names(indx00)-(seq_along(indx00)-1)%/%2+1
res10PercentSub30- res10Percent[c(indx00-1,indx00),]
res10PercentSub30$id- names(c(indx00-1,indx00))
res10PercentSub40- 
do.call(rbind,lapply(split(res10PercentSub30,res10PercentSub30$id),function(x){x1-subset(x,dummy==1);
 
x2-subset(x,dummy==0);x3-x1[which.max(abs(x1$dimension-unique(x2$dimension))),];x4-x2[which.min(abs(as.numeric(row.names(x3))-as.numeric(row.names(x2,];rbind(x3,x4)}))

row.names(res10PercentSub40)-gsub(.*\\.,,row.names(res10PercentSub40))
indxNew- 
sort(as.numeric(c(row.names(res10PercentSub5),row.names(res10PercentSub40
res10PercentFinal-res10Percent[-indxNew,]
dim(res10PercentFinal)
#[1] 454   5
nrow(subset(res10PercentFinal,dummy==0))
#[1] 227
nrow(subset(res10PercentFinal,dummy==1))
#[1] 227

nrow(unique(res10PercentFinal))
#[1] 454
which(duplicated(res10Percent)|duplicated(res10Percent,fromLast=TRUE))
# [1] 113 117 123 125 153 157 187 189 207 213 223 235 265 267 269 275 276 278 
279
#[20] 283 293 301 303 305 309 317 327 331 335 339 341 343 347 351 367 369 371 
379
#[39] 385 399 407 413 415 417 429 437 441 453 459 461 471 473 477 479 501 505
res10Percent[c(113:114,117:118),]
#         firm year industry dummy dimension
#113 500221723 2005       26     1      3147
#114 500601429 2005       26     0      3076
#117 500221723 2005       26     1      3147
#118 502668920 2005       26     0      3249

Re: [R] help needed! RMSE

2013-06-10 Thread Ben Bolker

mansor nad nadsim88 at hotmail.com writes:

 
 i need HELPPP!! how do i calculate the RMSE value for two GEV
 models?first GEV is where the three parameters are constant.2nd GEV
 model a 4 parameter model with the location parameter is allowed to
 vary linearly with respect to time while holding the other
 parameters at constant.  is there any programming code for this?  i
 really really need help. please reply to me as soon as
 possible. thanks in advance.

  Have you read the posting guide (URL/link at the bottom of
every posting at this list)?  Can you provide a reproducible example?
It may seem perverse, but urgency (I need HELP! ... I really really
need help ... please reply to me as soon as possible ...) doesn't
actually generally improve your chances of getting help here -- it
comes across as shouting.  Providing reproducible examples not only
makes it easier for people to answer, and improving the chances
that the answers you get will be ones you really need, it also 
demonstrates evidence that you have invested some effort.

  You might want to start with this example:

  library(fExtremes)
  g1 - gevFit(gevSim())
  sqrt(sum(g1@residuals^2))
  ?gevFit

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Re: [R] Speed up or alternative to 'For' loop

2013-06-10 Thread David Winsemius


On Jun 10, 2013, at 10:28 AM, Trevor Walker wrote:

 I have a For loop that is quite slow and am wondering if there is a faster
 option:
 
 df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
 df$Height - exp(-0.1 + 0.2*df$Age)
 df$HeightGrowth - NA   #intialize with NA
 for (i in 2:nrow(df))
 {if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
 }
 
Ivoid tests with if(){}e;se(). Use vectorized code, possibly with 'ifelse' but 
in this case you need a function that does calcualtions within groups.

The ave() function with diff() will do it compactly and efficiently:

 df - data.frame(TreeID=rep(1:5,each=4), Age=rep(seq(1,4,1),5))
 df$Height - exp(-0.1 + 0.2*df$Age)
 df$HeightGrowth - NA   #intialize with NA

 df$HeightGrowth - ave(df$Height, df$TreeID, FUN= function(vec) c(NA, 
 diff(vec)))
 df
   TreeID Age   Height HeightGrowth
1   1   1 1.105171   NA
2   1   2 1.3498590.2446879
3   1   3 1.6487210.2988625
4   1   4 2.0137530.3650314
5   2   1 1.105171   NA
6   2   2 1.3498590.2446879
7   2   3 1.6487210.2988625
8   2   4 2.0137530.3650314
9   3   1 1.105171   NA
10  3   2 1.3498590.2446879
11  3   3 1.6487210.2988625
12  3   4 2.0137530.3650314
13  4   1 1.105171   NA
14  4   2 1.3498590.2446879
15  4   3 1.6487210.2988625
16  4   4 2.0137530.3650314
17  5   1 1.105171   NA
18  5   2 1.3498590.2446879
19  5   3 1.6487210.2988625
20  5   4 2.0137530.3650314

(On my machine it was over six times as fast as the if-based code from Arun. )

-- 

David Winsemius
Alameda, CA, USA

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Re: [R] Speed up or alternative to 'For' loop

Sorry, it looks like I was hasty.
Absent another dumb mistake, the following should do it.

The request was for differences, i.e., the amount of growth from one
period to the next, separately for each tree.

for (ir in unique(df$TreeID)) {
  in.ir - df$TreeID == ir
  df$HeightGrowth[in.ir] - c(NA, diff(df$Height[in.ir]))
}



And this gives the same result as Rui Barradas' previous response.

-Don

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/10/13 2:51 PM, MacQueen, Don macque...@llnl.gov wrote:

How about

for (ir in unique(df$TreeID)) {
  in.ir - df$TreeID == ir
  df$HeightGrowth[in.ir] - cumsum(df$Height[in.ir])
}

Seemed fast enough to me.

In R, it is generally good to look for ways to operate on entire vectors
or arrays, rather than element by element within them. The cumsum()
function does that in this example.

-Don


-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/10/13 10:28 AM, Trevor Walker trevordaviswal...@gmail.com wrote:

I have a For loop that is quite slow and am wondering if there is a
faster
option:

df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
df$Height - exp(-0.1 + 0.2*df$Age)
df$HeightGrowth - NA   #intialize with NA
for (i in 2:nrow(df))
 {if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
 }

Trevor Walker
Email: trevordaviswal...@gmail.com

  [[alternative HTML version deleted]]

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

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

Re: [R] Combining CSV data

Hi,
Try this:

dat1-read.table(text=
Row_ID_CR,  Data1,    Data2,    Data3
1,  aa,  bb,  cc
2,  dd,  ee,  ff
,sep=,,header=TRUE,stringsAsFactors=FALSE)

dat2-read.table(text=
Row_ID_N,  Src_Row_ID,  DataN1
1a,  1,  This is comment 1
2a,  1,  This is comment 2
3a,  2,  This is comment 1
4a,  1,  This is comment 3
,sep=,,header=TRUE,stringsAsFactors=FALSE)
library(stringr)
dat2$DataN1-str_trim(dat2$DataN1)
res- merge(dat1,dat2,by.x=1,by.y=2)
 res1-res[,-5]
library(plyr)
 res2-ddply(res1,.(Row_ID_CR,Data1,Data2,Data3),summarize, DataN1=list(DataN1))
 res2
 # Row_ID_CR    Data1    Data2    Data3
#1 1   aa   bb   cc
#2 2   dd   ee   ff
#   DataN1
#1 This is comment 1, This is comment 2, This is comment 3
#2   This is comment 1



res3-data.frame(res2[,-5],t(apply(do.call(rbind,res2[,5]),1,function(x) 
{x[duplicated(x)]-NA;x})))
 colnames(res3)[grep(X,colnames(res3))]- 
paste0(DataComment,gsub([[:alpha:]],,colnames(res3)[grep(X,colnames(res3))]))
res3
#  Row_ID_CR    Data1    Data2    Data3  DataComment1
#1 1   aa   bb   cc This is comment 1
#2 2   dd   ee   ff This is comment 1
#   DataComment2  DataComment3
#1 This is comment 2 This is comment 3
#2  NA  NA

A.K.


- Original Message -
From: Shreya Rawal rawal.shr...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Monday, June 10, 2013 4:38 PM
Subject: [R] Combining CSV data

Hello R community,

I am trying to combine two CSV files that look like this:

File A

Row_ID_CR,   Data1,    Data2,    Data3
1,                   aa,          bb,          cc
2,                   dd,          ee,          ff


File B

Row_ID_N,   Src_Row_ID,   DataN1
1a,               1,                   This is comment 1
2a,               1,                   This is comment 2
3a,               2,                   This is comment 1
4a,               1,                   This is comment 3

And the output I am looking for is, comparing the values of Row_ID_CR and
Src_Row_ID

Output

ROW_ID_CR,    Data1,    Data2,    Data3,    DataComment1,
DataComment2,          DataComment3
1,                      aa,         bb,         cc,        This is
comment1,    This is comment2,     This is comment 3
2,                      dd,          ee,         ff,          This is
comment1


I am a novice R user, I am able to replicate a left join but I need a bit
more in the final result.


Thanks!!

    [[alternative HTML version deleted]]

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


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

Re: [R] Speed up or alternative to 'For' loop

Well, speaking of hasty...

This will also do it, provided that each tree's initial height is less
than the previous tree's final height. In principle, not a safe
assumption, but might be ok depending on where the data came from.

df$delta - c(NA,diff(df$Height))
df$delta[df$delta  0] - NA

-Don



-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/10/13 2:51 PM, MacQueen, Don macque...@llnl.gov wrote:

How about

for (ir in unique(df$TreeID)) {
  in.ir - df$TreeID == ir
  df$HeightGrowth[in.ir] - cumsum(df$Height[in.ir])
}

Seemed fast enough to me.

In R, it is generally good to look for ways to operate on entire vectors
or arrays, rather than element by element within them. The cumsum()
function does that in this example.

-Don


-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 6/10/13 10:28 AM, Trevor Walker trevordaviswal...@gmail.com wrote:

I have a For loop that is quite slow and am wondering if there is a
faster
option:

df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
df$Height - exp(-0.1 + 0.2*df$Age)
df$HeightGrowth - NA   #intialize with NA
for (i in 2:nrow(df))
 {if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
 }

Trevor Walker
Email: trevordaviswal...@gmail.com

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

Re: [R] Speed up or alternative to 'For' loop

Hi,
Some speed comparisons:


df - data.frame(TreeID=rep(1:6000,each=20), Age=rep(seq(1,20,1),6000))
df$Height - exp(-0.1 + 0.2*df$Age)
df1- df
df3-df
library(data.table)
dt1- data.table(df)
df$HeightGrowth - NA 


system.time({  #Rui's 2nd function
df2 - data.matrix(df)
for(i in seq_len(nrow(df2))[-1]){
    if(df2[i, TreeID] == df2[i - 1, TreeID])
        df2[i, HeightGrowth] - df2[i, Height] - df2[i - 1, Height]
}
})
# user  system elapsed 
 # 1.108   0.000   1.109 


system.time({for (ir in unique(df$TreeID)) {   #Don's first function
  in.ir - df$TreeID == ir
  df$HeightGrowth[in.ir] - c(NA, diff(df$Height[in.ir]))
}})
#  user  system elapsed 
#100.004   0.704 100.903 

system.time({df3$delta - c(NA,diff(df3$Height)) ##Don's 2nd function
df3$delta[df3$delta  0] - NA}) #winner 
#   user  system elapsed 
 # 0.016   0.000   0.014 

system.time(df1$HeightGrowth - ave(df1$Height, df1$TreeID, FUN= function(vec) 
c(NA, diff(vec #David's
 #user  system elapsed 
 # 0.136   0.000   0.137 
 system.time(dt1[,HeightGrowth:=c(NA,diff(Height)),by=TreeID])
#  user  system elapsed 
 # 0.076   0.000   0.079 


 identical(df1,as.data.frame(dt1))
#[1] TRUE
 identical(df1,df)
#[1] TRUE


head(df1,2)
#  TreeID Age   Height HeightGrowth
#1  1   1 1.105171   NA
#2  1   2 1.349859    0.2446879
head(df2,2)
# TreeID Age   Height HeightGrowth
#[1,]  1   1 1.105171   NA
#[2,]  1   2 1.349859    0.2446879

A.K.



- Original Message -
From: Trevor Walker trevordaviswal...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Monday, June 10, 2013 1:28 PM
Subject: [R] Speed up or alternative to 'For' loop

I have a For loop that is quite slow and am wondering if there is a faster
option:

df - data.frame(TreeID=rep(1:500,each=20), Age=rep(seq(1,20,1),500))
df$Height - exp(-0.1 + 0.2*df$Age)
df$HeightGrowth - NA   #intialize with NA
for (i in 2:nrow(df))
{if(df$TreeID[i]==df$TreeID[i-1])
  {df$HeightGrowth[i] - df$Height[i]-df$Height[i-1]
  }
}

Trevor Walker
Email: trevordaviswal...@gmail.com

    [[alternative HTML version deleted]]

__
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Apply a PCA to other datasets

2013-06-10 Thread Thomas Stewart

Short answer: Yes.

Long answer: Your question does not provide specific information;
therefore, I cannot provide a specific answer.


On Mon, Jun 10, 2013 at 1:23 PM, edelance delanceye...@gmail.com wrote:

 I have run a PCA on one data set.  I need the standard deviation of the
 first
 two bands for my analysis.  I now want to apply the same PCA rotation I
 used
 in the first one to all my other data sets.  Is there any way to do this in
 r?  Thanks.




 --
 View this message in context:
 http://r.789695.n4.nabble.com/Apply-a-PCA-to-other-datasets-tp4669182.html
 Sent from the R help mailing list archive at Nabble.com.

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[[alternative HTML version deleted]]

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Re: [R] Combining CSV data