date:20110407

I've just published a new book for R beginners in Portuguese: 
Conhecendo o R: uma visão estatística (something like Knowing R: an 
statistical approach). I'd like to include it on the list Books at 
R-project.org. How can I do it?

More informations about the book at
http://www.editoraufv.com.br/produtos/conhecendo-o-r

--
Marcio Pupin Mello

Survey Engineer
Ph.D student in Remote Sensing
National Institute for Space Research (INPE) - Brazil
Laboratory of Remote Sensing in Agriculture and Forestry (LAF)
www.dsr.inpe.br/~mello

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Re: [R] force output dimension of table function


You can do it using factor:

s0-factor(0:5)
s1-sample(s0,5,re=T)
s1
  [1] 0 2 2 4 0
  Levels: 0 1 2 3 4 5

table(s1)
  s1
  0 1 2 3 4 5
  2 0 2 0 1 0

Good luck!

--
Marcio Pupin Mello

Survey Engineer
Ph.D student in Remote Sensing
National Institute for Space Research (INPE) - Brazil
Laboratory of Remote Sensing in Agriculture and Forestry (LAF)
www.dsr.inpe.br/~mello


Em 4/7/2011 12:37 AM, fisken escreveu:

I have a small annoying problem.

When I use the 'table' function on a simple vector it counts the
number of occurences.
So depending on the values of my input vector the function returns a
class of type table with different lengths.

Is there an easy way to tell the table function, the values to expect?

That is
#

set.seed(0)
s-sample(0:5,5,rep=T)
s

[1] 5 1 2 3 5

ts-table(s)
ts

s
1 2 3 5
1 1 1 2

##

And what I wanted was

0 1 2 3 4 5
0 1 1 1 0 2


Thanks



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[R] R licence

2011-04-07 Thread Stanislav Bek

Hi,

is it possible to use some statistic computing by R in proprietary software?
Our software is written in c#, and we intend to use
http://rdotnet.codeplex.com/
to get R work there. Especially we want to use loess function.

Thanks,

Best regards,
Stanislav

[[alternative HTML version deleted]]

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[R] Two questions about metacharacter in regexprs and function return

2011-04-07 Thread Tuexy

for the script, please kindly see the script below. At line 10 and line 13,
my problems occurs. 
The first one is I try to retrieve the gene official name from a column of a
table. The pattern of official name is something  starting with gene_name.
For detail problems, please see the according lines. 
Any suggestions are appreciated

example of matching source (extract the Nnat, sometime it would be the
character N/A): 
AB004048|MM8;NCBI Build 36|transcript|chr2|157251580|157253958|ExemplarFor
'AB004048'; gene_id '18111'; transcript_id 'AB004048'; gene_name 'Nnat'; alt
'5730414I02Rik|AW107673|Peg5';
neuronatin|http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?db=genecmd=Retrievedopt=full_reportlist_uids=18111;

#obtain the exprs matrix for cluster analysis
#ask questions
DEG_files - grep(bak, dir());#pay attention to the filenames
exprs_files - grep(copy, dir());
protein - c();
assign_exprs - function(files, protein) { #use to find the DEGs or exprs
for cmeans clustering
for(i in 1:length(files)) {
microarray_data - read.csv(file = files[i], header = T, sep =
\t);
microarray_data[, 7] - gsub(([\\s\\S]+gene_name '(\\w*)';.+),
\\2, microarray_data[, 7], perl = T);#why [\\w]* cannot workable? also the
[(\\w*)(N/A)] cannot be workable.
assign(files[i], microarray_data, envir=.GlobalEnv); #get(dir()[i]()
can obtain the data of interest.`variable_names` can also work
protein - c(protein, get(files[i])[, 7]);  #used for obtain all
the DEGs only
}   
#return protein; #why this line is not workable?
assign(all_protein, protein, envir=.GlobalEnv);
}
exist_to_cluster_exprs - function(x, cluster_exprs, all) {
if(exists(all, x[1])){ #exists function
cluster_exprs - cbind(cluster_exprs, x); 
}   
#return cluster_exprs;
}
assign_exprs(dir()[DEG_files], protein);
all_protein - unique(all_protein);
assign_exprs(dir()[exprs_files], protein);
for(i in 1:2) {
apply(get(dir()[exprs_files[i]]), 1, exist_to_cluster_exprs,
cluster_exprs, all);
#assign(paste(exprs_files()[i], exprs_data), cluster_exprs[, c(2,
3, 5, 7)];
exprs_data - cbind(exprs_data, cluster_exprs[, 3]);
}
exprs_data;

--
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Re: [R] force output dimension of table function

2011-04-07 Thread Philipp Pagel

On Thu, Apr 07, 2011 at 05:37:08AM +0200, fisken wrote:
 When I use the 'table' function on a simple vector it counts the
 number of occurences.
 So depending on the values of my input vector the function returns a
 class of type table with different lengths.
 
 Is there an easy way to tell the table function, the values to expect?
 
 And what I wanted was
 
 0 1 2 3 4 5
 0 1 1 1 0 2

The solution using factos has already been posted. if you are really
interested in integers only you could also use tabulate():

 tabulate(s)
[1] 1 1 1 0 2

Note that this excludes zero, though.

cu
Philipp

-- 
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
Maximus-von-Imhof-Forum 3
85354 Freising, Germany
http://webclu.bio.wzw.tum.de/~pagel/

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[R] set locale information in R

2011-04-07 Thread Raji

Hi R-helpers,

  Is it possible to localise the error messages/warnings that comes from
R.My application takes in a locale information.I used the following command
to set the locale in R.But in RGui, i still get the error messages in
English only.Can you please help me out with this?

Sys.setlocale(LC_ALL, French)  

Regards,
Raji

--
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http://r.789695.n4.nabble.com/set-locale-information-in-R-tp3432760p3432760.html
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Re: [R] R licence

2011-04-07 Thread Liviu Andronic

On Thu, Apr 7, 2011 at 9:45 AM, Stanislav Bek
stanislav.pavel@gmail.com wrote:
 Hi,

 is it possible to use some statistic computing by R in proprietary software?

I don't know if this covers your case, but SAS and SPSS provide
interfaces to R. Regards
Liviu


 Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.

 Thanks,

 Best regards,
 Stanislav

        [[alternative HTML version deleted]]

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-- 
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail

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[R] plot layout with several plots ON plot area of previous plot

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Hi

I would like to have a plot layout, on which line graphs are on the plot
area of an underlying bar graph.
How can I achieve this? I know about layout() and par() to create a
layout, but as far as I know, they assign non-overlapping plot areas.

Effectively, it should look as below:

||
||
|  |---|  |---|  |
|  |   |  |   |  |
|  |   |  |   |  |
|  |-L-|  |-L-|  |
||
||
||
||
||
||
|-the bargraph---|

L: the line graphs

Any help appreciated,

Rainer



- -- 
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main Campus, Merriman Avenue
Stellenbosch
South Africa

Tel:+33 - (0)9 53 10 27 44
Cell:   +27 - (0)8 39 47 90 42
Fax (SA):   +27 - (0)8 65 16 27 82
Fax (D) :   +49 - (0)3 21 21 25 22 44
Fax (FR):   +33 - (0)9 58 10 27 44
email:  rai...@krugs.de

Skype:  RMkrug
-BEGIN PGP SIGNATURE-
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Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/

iEYEARECAAYFAk2dg3UACgkQoYgNqgF2egpegwCfbqQP19qZqyof21+rA9Cbn+Vf
UKcAnAgl6AoUNcygx/sSaMcYV+zoum/S
=9/P0
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Re: [R] plot layout with several plots ON plot area of previous plot

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 07/04/11 11:27, Rainer M Krug wrote:
 Hi
 
 I would like to have a plot layout, on which line graphs are on the plot
 area of an underlying bar graph.
 How can I achieve this? I know about layout() and par() to create a
 layout, but as far as I know, they assign non-overlapping plot areas.

Sorry - did not do my homework and forgot to google - found the solution
(http://tolstoy.newcastle.edu.au/R/help/04/11/6449.html).

But an additional question - uis there a way to position the plots in
relation to the co-ordinates of the first plot, i.e. the x axis of the
first plot (ehich covers the whole plot area) is from 1 to 6 - can I
position the second overlapping plot over the x-coordinate 3 of the
first plot?

Rainer


 
 Effectively, it should look as below:
 
 ||
 ||
 |  |---|  |---|  |
 |  |   |  |   |  |
 |  |   |  |   |  |
 |  |-L-|  |-L-|  |
 ||
 ||
 ||
 ||
 ||
 ||
 |-the bargraph---|
 
 L: the line graphs
 
 Any help appreciated,
 
 Rainer
 
 
 

- -- 
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main Campus, Merriman Avenue
Stellenbosch
South Africa

Tel:+33 - (0)9 53 10 27 44
Cell:   +27 - (0)8 39 47 90 42
Fax (SA):   +27 - (0)8 65 16 27 82
Fax (D) :   +49 - (0)3 21 21 25 22 44
Fax (FR):   +33 - (0)9 58 10 27 44
email:  rai...@krugs.de

Skype:  RMkrug
-BEGIN PGP SIGNATURE-
Version: GnuPG v1.4.10 (GNU/Linux)
Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/

iEYEARECAAYFAk2dhmMACgkQoYgNqgF2egqysACeMJ0nl81wSBlr8w4s7sM2Zo2k
gvUAnjTlzoFoOnDHPzbziRAbS/8MbV0p
=LYr+
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Re: [R] R licence

2011-04-07 Thread peter dalgaard


On Apr 7, 2011, at 09:45 , Stanislav Bek wrote:

 Hi,
 
 is it possible to use some statistic computing by R in proprietary software?
 Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.

You need to take legal advice to be certain, but offhand I would say that this 
kind of circumvention of the GPL is _not_ allowed. 

It all depends on whether the end product is a derivative work, in which 
case, the whole must be distributed under a GPL-compatible licence. The 
situation around GPL-incompatible plug-ins or plug-ins interfacing to R in GPL 
-incompatible software is legally murky, but using R as a subroutine library 
for proprietary code is clearly crossing the line, as far as I can tell.  

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

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Re: [R] plot layout with several plots ON plot area of previous plot

2011-04-07 Thread Jim Lemon


On 04/07/2011 07:39 PM, Rainer M Krug wrote:

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 07/04/11 11:27, Rainer M Krug wrote:

Hi

I would like to have a plot layout, on which line graphs are on the plot
area of an underlying bar graph.
How can I achieve this? I know about layout() and par() to create a
layout, but as far as I know, they assign non-overlapping plot areas.


Sorry - did not do my homework and forgot to google - found the solution
(http://tolstoy.newcastle.edu.au/R/help/04/11/6449.html).

But an additional question - uis there a way to position the plots in
relation to the co-ordinates of the first plot, i.e. the x axis of the
first plot (ehich covers the whole plot area) is from 1 to 6 - can I
position the second overlapping plot over the x-coordinate 3 of the
first plot?

Rainer




Effectively, it should look as below:

||
||
|  |---|  |---|  |
|  |   |  |   |  |
|  |   |  |   |  |
|  |-L-|  |-L-|  |
||
||
||
||
||
||
|-the bargraph---|

L: the line graphs


Hi Rainer,
I think you want the subplot function in the TeachingDemos package.

Jim

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[R] [R-pkgs] [R-package] PredictABEL 1.1: Assessment of risk prediction models in R

2011-04-07 Thread Suman Kundu


Dear R users,

A new R package (PredictABEL, 
http://cran.at.r-project.org/web/packages/PredictABEL/ ) has been published 
recently that includes functions to assess the performance of (genetic) risk 
models.


The package provides the following measures: 
 

Univariate and multivariate odds ratios (OR) of the predictors
The c-statistic (or area under the receiver operating characteristic (ROC) 
curve (AUC))
Hosmer-Lemeshow goodness of fit test
Reclassification table
Net reclassification improvement (NRI)
Integrated discrimination improvement (IDI)


You can find more information in our paper [1] and on PredictABEL website
http://www.genabel.org/packages/PredictABEL
 
Hope you'll find it useful!

Suman Kundu

--- 
References:
[1] http://www.ncbi.nlm.nih.gov/pubmed/21431839

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Re: [R] plot layout with several plots ON plot area of previous plot

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 07/04/11 12:53, Jim Lemon wrote:
 On 04/07/2011 07:39 PM, Rainer M Krug wrote:
 -BEGIN PGP SIGNED MESSAGE-
 Hash: SHA1

 On 07/04/11 11:27, Rainer M Krug wrote:
 Hi

 I would like to have a plot layout, on which line graphs are on the plot
 area of an underlying bar graph.
 How can I achieve this? I know about layout() and par() to create a
 layout, but as far as I know, they assign non-overlapping plot areas.

 Sorry - did not do my homework and forgot to google - found the solution
 (http://tolstoy.newcastle.edu.au/R/help/04/11/6449.html).

 But an additional question - uis there a way to position the plots in
 relation to the co-ordinates of the first plot, i.e. the x axis of the
 first plot (ehich covers the whole plot area) is from 1 to 6 - can I
 position the second overlapping plot over the x-coordinate 3 of the
 first plot?

 Rainer



 Effectively, it should look as below:

 ||
 ||
 |  |---|  |---|  |
 |  |   |  |   |  |
 |  |   |  |   |  |
 |  |-L-|  |-L-|  |
 ||
 ||
 ||
 ||
 ||
 ||
 |-the bargraph---|

 L: the line graphs

 Hi Rainer,
 I think you want the subplot function in the TeachingDemos package.

Exactly - why is such a gem of a function hiding in a package with Demos
in the name? I would never have thought about that.

Thanks a lot,

Rainer

 
 Jim
 


- -- 
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main Campus, Merriman Avenue
Stellenbosch
South Africa

Tel:+33 - (0)9 53 10 27 44
Cell:   +27 - (0)8 39 47 90 42
Fax (SA):   +27 - (0)8 65 16 27 82
Fax (D) :   +49 - (0)3 21 21 25 22 44
Fax (FR):   +33 - (0)9 58 10 27 44
email:  rai...@krugs.de

Skype:  RMkrug
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uesAn3emEYA7679U4Z7RfZO80T0p0pZ/
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[R] understanding randomForest results

2011-04-07 Thread divya karnad

How do I read/interpret the output of varImpPlot() for a randomForest
object?

Regards,

Divya


[[alternative HTML version deleted]]

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Re: [R] plot layout with several plots ON plot area of previous plot

2011-04-07 Thread Jim Lemon


On 04/07/2011 08:03 PM, Rainer M Krug wrote:

...
Exactly - why is such a gem of a function hiding in a package with Demos
in the name? I would never have thought about that.


Maybe Greg thought it was a package for the people.

Jim

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[R] predict fGarch doubt

2011-04-07 Thread Luis Felipe Parra

Hello I am using the predict method in fGarch. I tried to replicate what I
supposed it was doing when I estimated an ARMA(2,2)+GARCH(1,1) by doing:

(temp is my ouput form garchFit)
fit = temp@fit$par
Nobs = length(temp@data)
# Predecir media
  fit[mu]+sum(fit[2:(2+2-1)]*temp@data[length(temp@data):(length(temp@data
)-2+1)]),0)+
sum(fit[(2+2):(2+2+o2-1)]*temp@residuals[length(temp@residuals
):(length(temp@residuals)-2+1)]),0)

and I got -0.005585084 while the output from the predict method gave me
-0.00752829. Does anybody know what might be going on? am I using the wrong
slots in the output?


Thank you

Felipe Parra

[[alternative HTML version deleted]]

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Re: [R] plot layout with several plots ON plot area of previous plot

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

On 07/04/11 13:09, Jim Lemon wrote:
 On 04/07/2011 08:03 PM, Rainer M Krug wrote:
 ...
 Exactly - why is such a gem of a function hiding in a package with Demos
 in the name? I would never have thought about that.

 Maybe Greg thought it was a package for the people.

Nothing to add :-)

Rainer

 
 Jim
 


- -- 
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
Biology, UCT), Dipl. Phys. (Germany)

Centre of Excellence for Invasion Biology
Natural Sciences Building
Office Suite 2039
Stellenbosch University
Main Campus, Merriman Avenue
Stellenbosch
South Africa

Tel:+33 - (0)9 53 10 27 44
Cell:   +27 - (0)8 39 47 90 42
Fax (SA):   +27 - (0)8 65 16 27 82
Fax (D) :   +49 - (0)3 21 21 25 22 44
Fax (FR):   +33 - (0)9 58 10 27 44
email:  rai...@krugs.de

Skype:  RMkrug
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Re: [R] How to speed up grouping time series, help please

2011-04-07 Thread Den Alpin

I found a faster implementation (by an order of magnitude from my
tests) than the one using xts, split, merge (from Joshua).
I report the two fastest solution below with code to generate a test
case; some work still to be done for columns order and naming,
Test case has grown from my previous post to get a more realistic timing.

Any comment or idea to further speed up multivariate time series
creation with classes xts or timeSeries starting from a data.frame
like the one reported here is welcome.

Best regards,
Den


a data.frame example (code below to generate it)

  IDDATE VALUE
14  3 2000-01-01 00:00:03 0.5726334
4   1 2000-01-01 00:00:03 0.8830174
1   1 2000-01-01 00:00:00 0.2875775
15  3 2000-01-01 00:00:04 0.1029247
11  3 2000-01-01 00:00:00 0.9568333
9   2 2000-01-01 00:00:03 0.5514350
7   2 2000-01-01 00:00:01 0.5281055
6   2 2000-01-01 00:00:00 0.0455565
12  3 2000-01-01 00:00:01 0.4533342
8   2 2000-01-01 00:00:02 0.8924190
3   1 2000-01-01 00:00:02 0.4089769
13  3 2000-01-01 00:00:02 0.6775706

And I want to get a timeSeries object or xts object like this:

   1 2 3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04NANA 0.1029247

# CODE:

set.seed(123)
# set N to 5 to reproduce above data.frame
N - 1000
# set K to 3 to reproduce above data.frame
K - 10
X - data.frame(
  ID = rep(1:K, each = N),
  DATE = as.character(rep(as.POSIXct(2000-01-01, tz = GMT)+ 0:(N-1), K)),
  VALUE = runif(N*K), stringsAsFactors = FALSE)
X - X[sample(1:(N*K), N*K),]
X - X[-(sample(1:nrow(X), floor(nrow(X)*0.2))),]
str(X)


xtsSplit - function(x)
{
  library(xts)
  x - xts(x[,c(ID,VALUE)], as.POSIXct(x[,DATE]))
  return(do.call(merge, split(x$VALUE,x$ID)))
}
xtsSplitTime - replicate(50,
  system.time(xtsSplit(X))[[1]])
median(xtsSplitTime)

xtsReshape - function(x)
{
  library(xts)
  x - reshape(x, idvar = DATE, timevar = ID, direction = wide)
  x - xts(x[,-1], as.POSIXct(x[,1]))
  return(x)
}
xtsReshapeTime - replicate(50,
  system.time(xtsReshape(X))[[1]])
median(xtsReshapeTime)

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[R] Hypercluster Finder Function in R

2011-04-07 Thread Vijayan Padmanabhan


Dear R Users
I am looking at carrying out Multivariate Supervised Discretization on a 
set of continuous attributes to determine cut off values on each attribute 
which would allow best explanation of a binary outcome variable.
I understand that there is an algorithm for doing this (The HyperCluster 
Finder Algorithm from the paper -Muhlenbach F.,  Rakotomalala R. (2002). 
Multivariate Supervised Discretization, a Neighborhood Graph Approach. 
Proceedings of the 1st IEEE International Conference on Data Mining 
(ICDM-2002), 314-321.)
I donot know if this implementation is available in R already
Does someone know if there is any package that exists for running this 
algorithm? or does anybody have a function developed on this basis?

Thanks

Regards
Vijayan Padmanabhan


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[R] extract variance for random effects from mer object

2011-04-07 Thread Maas James Dr (MED)

I get an output from a glmer package, but would like to extract the variance of 
the Random effects, in this case called trtpair and the value is 0.

Can I extract it directly or do I need to do some further calculations on the 
mer object to get it?

Thanks

Jim

 class(fednmaout)
[1] mer
attr(,package)
[1] lme4
 fednmaout
Generalized linear mixed model fit by the Laplace approximation 
Formula: cbind(fednmadat$events, fednmadat$patnums - fednmadat$events) ~  
treat + tn + (1 | trtpair) 
   AIC   BIC logLik deviance
 84.01 153.1 -9.00618.01
Random effects:
 Groups  NameVariance Std.Dev.
 trtpair (Intercept)  00  
Number of obs: 60, groups: trtpair, 3

===
Dr. Jim Maas
University of East Anglia

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[R] Time series of spatial data

2011-04-07 Thread idham

Hi guys,

I'm really new in R.

Trying to analyze series of spatial datasets (365 satellite images) in order
to find the best model that fit the data.

Any suggestion which package that could help me?

Thanks in advance.

Cheers

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Re: [R] Ports for communicating with R-mirros sites?


On 11-04-06 10:55 PM, Debraj GuhaThakurta wrote:

Hi,



I have been having an error since past couple of days to connect to several of 
the mirror sites. I wonder if anyone knows the port/s for communicating with 
the R-mirror sites for downloading packages. If we know this we may be able to 
figure the problem at our end.


Just the standard http port 80, unless they advertise something else 
like ftp, etc.


Duncan Murdoch




Debraj


Debraj GuhaThakurta
Email: d_guhathaku...@hotmail.com

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Re: [R] set locale information in R

2011-04-07 Thread Raji

Hi Rhelpers,

  Do we need to install some font packages in R for the locale information
to take effect.
Alternatively, i found that we can use the following command for the
messages

Sys.setlocale(LC_MESSAGES,French)

Should this be done for getting error messages in French? The help for
setLocale says that the LC_MESSAGES cannot be set for Windows.

Can you please let me know how i can get the error messages in the locale
set?
Please do apologize if i am missing something obvious here.

Thanks,
Raji

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Re: [R] extract variance for random effects from mer object

2011-04-07 Thread Ben Bolker

Maas James Dr (MED J.Maas at uea.ac.uk writes:

 I get an output from a glmer package, but would like 
 to extract the variance of the Random effects, in this
 case called trtpair and the value is 0.
[snip]
  fednmaout
 Generalized linear mixed model fit by the Laplace approximation 
 Formula: cbind(fednmadat$events, fednmadat$patnums - fednmadat$events) ~ 
treat + tn + (1 | trtpair) 
AIC   BIC logLik deviance
  84.01 153.1 -9.00618.01
 Random effects:
  Groups  NameVariance Std.Dev.
  trtpair (Intercept)  00  
 Number of obs: 60, groups: trtpair, 3
 

  VarCorr(fednmaout)[[1]]

(the [[1]] means to extract the variance for the
first (and only in this case) random effect.)

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[R] comparing ARIMA model to data

2011-04-07 Thread Andrew Collier

hi,

i am trying to teach myself about ARIMA models. i have followed examples
from a number of sources and have more or less got the hang of how it
works. i would like to compare the output from the fitted model to the
original data. is this possible? or even a meaningful thing to do?

to be clear, for example, having generated a fit to some data using

 fit - arima(LakeHuron, order = c(1, 0, 1))

and then plotting the data with

 plot(LakeHuron)

is it possible to overlay the output of the model on the original data
to compare how well it captures the variations in the data? i know that
predict can be used to extrapolate beyond the end of the data series,
but i want to evaluate the model within (not beyond) the original data.

best regards,
andrew.

-- 
Andrew B. Collier

Physicist
Waves and Space Plasmas Group
Hermanus Magnetic Observatory

Honorary Senior Lecturer tel: +27 31 2601157
Space Physics Research Institute fax: +27 31 2607795
University of KwaZulu-Natal, Durban, South Africagsm: +27 83 3813655

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Re: [R] How can I include a new book on the

2011-04-07 Thread Ben Bolker

Marcio Pupin Mello mello at ieee.org writes:

 I've just published a new book for R beginners in Portuguese: 
 Conhecendo o R: uma visão estatística (something like Knowing R: an 
 statistical approach). I'd like to include it on the list Books at 
 R-project.org. How can I do it?
 More informations about the book at
 http://www.editoraufv.com.br/produtos/conhecendo-o-r

  I think you should try contacting Kurt Hornik at r-project.org

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[R] Two functions as parametrs of a function.

2011-04-07 Thread Kenneth Roy Cabrera Torres

Hi R users:

I'm trying to make a function where two of the parameters are
functions, but I don't know how to put each set of parameters for
each function.

What am I missing?

I try this code:

f2-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){
force(fun1)
force(fun2)
force(n)
p1-unlist(par1)
p2-unlist(par2)
force(p1)
force(p2)
localfun1 - function() fun1(n, p1)
localfun2 - function() fun2(n, p2)
vp-replicate(nsim,t.test(localfun1(), localfun2())$p.value)
return(vp)
}

f2(fun1=rbeta,par1=list(shape1=2,shape2=2),fun2=rbeta,par2=list(shape1=1,shape2=2))

Thank you for your help.

Kenneth

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Re: [R] Scrap java scripts and styles from an html document

2011-04-07 Thread antujsrv

Hi ,

I am working on developing a web crawler.
Removing javascripts and styles is a part of the cleaning of the html
document. 
What I want is a cleaned html document with only the html tags and textual
information, 
so that i can figure out the pattern of the web page. This is being done to
extract relevant 
information from the webpage like comments for a particular product.

For e.g the amazon.com has all such comments within the 
 and tags, 
with regular 
 occuring for breaks. So tags which appear the most help us in
 locating the required information. Different websites have different
patterns, 
but its more likely that tags that will occur the most will have the
relevant information enclosed in them. 

So, once the html page is cleaned, it would be easy to role up the tags and
knowing their frequency of occurrence, we can target the information. 

Should there be any suggestions to help, please let me know. I would be more
than pleased.

Regards,
Antuj

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[R] Quasipoisson with geeglm

2011-04-07 Thread JANSEN, Ivy

Dear all,

I am trying to use the GEE methodology to fit a trend for the number of 
butterflies observed at several sites. In total, there are 66 sites, and 19 
years for which observations might be available. However, only 326 observations 
are available (instead of 1254). For the time being, I ignore the large number 
of missing values, and the fact that GEE is only valid under MCAR. When I run 
the following code

geeglm(SumOfButterflies ~ RES_YEAR, family = poisson, data = ManijurtNoNA, id = 
RES_ROTE_ID, corstr = ar1)

I obtain normal output. Not surprisingly, overdispersion is present 
(Estimated Scale Parameters:  [1] 185.8571), so changing to quasipoisson is 
needed. However, the code below

geeglm(SumOfButterflies ~ RES_YEAR, family = quasipoisson, data = ManijurtNoNA, 
id = RES_ROTE_ID, corstr = ar1)

produces the following error

Error in geese.fit(xx, yy, id, offset, soffset, w, waves = waves, zsca,  :   
variance invalid.

Other correlation structures encounter the same problem. I also tried adding 
waves = RES_YEAR (although I am not sure how waves should be used correctly), 
but it does not change anything.

Any suggestions what might be wrong?

Regards,
Ivy



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Re: [R] R licence

2011-04-07 Thread Matthew Dowle

Peter,

If the proprietary part of REvolution's product is ok, then surely 
Stanislav's suggestion is too. No?

Matthew


peter dalgaard pda...@gmail.com wrote in message 
news:be157cf5-9b4b-45a0-a7d4-363b774f1...@gmail.com...

 On Apr 7, 2011, at 09:45 , Stanislav Bek wrote:

 Hi,

 is it possible to use some statistic computing by R in proprietary 
 software?
 Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.

 You need to take legal advice to be certain, but offhand I would say that 
 this kind of circumvention of the GPL is _not_ allowed.

 It all depends on whether the end product is a derivative work, in which 
 case, the whole must be distributed under a GPL-compatible licence. The 
 situation around GPL-incompatible plug-ins or plug-ins interfacing to R in 
 GPL -incompatible software is legally murky, but using R as a subroutine 
 library for proprietary code is clearly crossing the line, as far as I can 
 tell.

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


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Re: [R] Two functions as parametrs of a function.


On 07/04/2011 7:45 AM, Kenneth Roy Cabrera Torres wrote:

Hi R users:

I'm trying to make a function where two of the parameters are
functions, but I don't know how to put each set of parameters for
each function.

What am I missing?

I try this code:

f2-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){
 force(fun1)
 force(fun2)
 force(n)
 p1-unlist(par1)
 p2-unlist(par2)


You don't want to unlist those.  Leave them as lists, and use the 
do.call() function for the calls.  For example,


localfun1 - function() do.call(fun1, c(list(n), par1))



 force(p1)
 force(p2)


Calling force() is only useful for arguments to your function:  it kicks 
the evaluator, so it's not so lazy.   Since p1 and p2 were calculated 
earlier, there's no need to force them.  You might need to force n, 
since it is only used in the localfun1 and localfun2 calls, and if the 
argument to f2 changes before you make those calls, the wrong value will 
be used.  That's pretty unlikely with your code, but since fun1 and fun2 
can do anything, it's safest to fix the value of n before you call 
them.  The same argument applies to fun1 and fun2 themselves.


Duncan Murdoch


 localfun1- function() fun1(n, p1)
 localfun2- function() fun2(n, p2)
 vp-replicate(nsim,t.test(localfun1(), localfun2())$p.value)
 return(vp)
}

f2(fun1=rbeta,par1=list(shape1=2,shape2=2),fun2=rbeta,par2=list(shape1=1,shape2=2))

Thank you for your help.

Kenneth

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Re: [R] Two functions as parametrs of a function.

2011-04-07 Thread Henrique Dallazuanna

Try this:

f2 - function(n, nsim, fun1, fun2) {
vp - replicate(nsim, do.call(fun1$name, c(n, fun1$args)),
do.call(fun1$name, c(n, fun1$args)))
vp

}

f2(2, 100, fun1 = list(name = rbeta, args = list(shape1 = 2, shape2 =
2), fun1 = list(name = rbeta, args = list(shape1 = 2, shape2 = 2

On Thu, Apr 7, 2011 at 8:45 AM, Kenneth Roy Cabrera Torres
krcab...@une.net.co wrote:
 Hi R users:

 I'm trying to make a function where two of the parameters are
 functions, but I don't know how to put each set of parameters for
 each function.

 What am I missing?

 I try this code:

 f2-function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){
    force(fun1)
    force(fun2)
    force(n)
    p1-unlist(par1)
    p2-unlist(par2)
    force(p1)
    force(p2)
    localfun1 - function() fun1(n, p1)
    localfun2 - function() fun2(n, p2)
    vp-replicate(nsim,t.test(localfun1(), localfun2())$p.value)
    return(vp)
 }

 f2(fun1=rbeta,par1=list(shape1=2,shape2=2),fun2=rbeta,par2=list(shape1=1,shape2=2))

 Thank you for your help.

 Kenneth

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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] R licence

2011-04-07 Thread Allan Engelhardt

The licences are available on the web site and you really should have yor 
lawyers look at them and give you professional advise.

The GPL2+ is probably the relevant one for your purposes and essentially 
require you to provide the source for the parts of R that you distribute.

However, the R licencing is a mess with core packages not on a GPL licence. We 
looked into it and it is a nightmare and I don't even think the UK CRAN mirrors 
are strictly speaking legal. So look carefully at what you use and get advice 
from somewhere other than a mailing list.

happy to discuss the UK side of things offline if you want, but I never looked 
at other jurisdictions.

Allan

- Original message -
 Hi,
 
 is it possible to use some statistic computing by R in proprietary
 software? Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.
 
 Thanks,
 
 Best regards,
 Stanislav
 
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Re: [R] Lars package



On Apr 7, 2011, at 3:43 AM, ac...@cse.iitb.ac.in wrote:


Hello,

I am a student of IIT Bombay and I am using lars for one of my  
projects. I

wanted to enquire if lars is using C under the hood. Is it true? Can
someone point me to the corresponding C function? Thank You very much.



If you just type lars at the console (after loading the package) , you  
will see the uncommented code.


Most of R has C (or Fortran) under the hood, so there may be a  
semantic problem in answering that direct question.


You can get the full source on any CRAN mirror.
--

David Winsemius, MD
West Hartford, CT

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Re: [R] R licence


On 07/04/2011 7:47 AM, Matthew Dowle wrote:

Peter,

If the proprietary part of REvolution's product is ok, then surely
Stanislav's suggestion is too. No?


Revolution has said that they believe they follow the GPL, and they 
haven't been challenged on that.   If you think that they don't, you 
could let an R copyright holder know what they're doing that's a license 
violation.


My opinion of Stanislav's question is that he doesn't give enough 
information to answer.  If he is planning to distribute R as part of his 
product, he needs to follow the GPL.  If not, I don't think any R 
copyright holder has anything to complain about.


Duncan Murdoch


Matthew


peter dalgaardpda...@gmail.com  wrote in message
news:be157cf5-9b4b-45a0-a7d4-363b774f1...@gmail.com...

  On Apr 7, 2011, at 09:45 , Stanislav Bek wrote:

  Hi,

  is it possible to use some statistic computing by R in proprietary
  software?
  Our software is written in c#, and we intend to use
  http://rdotnet.codeplex.com/
  to get R work there. Especially we want to use loess function.

  You need to take legal advice to be certain, but offhand I would say that
  this kind of circumvention of the GPL is _not_ allowed.

  It all depends on whether the end product is a derivative work, in which
  case, the whole must be distributed under a GPL-compatible licence. The
  situation around GPL-incompatible plug-ins or plug-ins interfacing to R in
  GPL -incompatible software is legally murky, but using R as a subroutine
  library for proprietary code is clearly crossing the line, as far as I can
  tell.

  -- 
  Peter Dalgaard

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


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Re: [R] comparing ARIMA model to data

2011-04-07 Thread nandan amar

Hello Andrew,
There are some parameters associated with the o/p of arima(),
In your case fit.
If one search for help(arima) ,
Values like sigma2,var.coef,loglik, aic will give you some information
regarding how good the model is.

If you want to overlay the output, you have to produce the output and you
already know predict().
regards.

On 7 April 2011 17:04, Andrew Collier collie...@gmail.com wrote:

 hi,

 i am trying to teach myself about ARIMA models. i have followed examples
 from a number of sources and have more or less got the hang of how it
 works. i would like to compare the output from the fitted model to the
 original data. is this possible? or even a meaningful thing to do?

 to be clear, for example, having generated a fit to some data using

  fit - arima(LakeHuron, order = c(1, 0, 1))

 and then plotting the data with

  plot(LakeHuron)

 is it possible to overlay the output of the model on the original data
 to compare how well it captures the variations in the data? i know that
 predict can be used to extrapolate beyond the end of the data series,
 but i want to evaluate the model within (not beyond) the original data.

 best regards,
 andrew.

 --
 Andrew B. Collier

 Physicist
 Waves and Space Plasmas Group
 Hermanus Magnetic Observatory

 Honorary Senior Lecturer tel: +27 31 2601157
 Space Physics Research Institute fax: +27 31 2607795
 University of KwaZulu-Natal, Durban, South Africagsm: +27 83 3813655

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-- 
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Re: [R] R licence


On 07/04/2011 8:07 AM, Allan Engelhardt wrote:

The licences are available on the web site and you really should have yor 
lawyers look at them and give you professional advise.

The GPL2+ is probably the relevant one for your purposes and essentially 
require you to provide the source for the parts of R that you distribute.


I think it's stronger than that (requiring the source for everything if 
you include any GPL'd bits), but your first paragraph is good advice.


Duncan Murdoch


However, the R licencing is a mess with core packages not on a GPL licence. We 
looked into it and it is a nightmare and I don't even think the UK CRAN mirrors 
are strictly speaking legal. So look carefully at what you use and get advice 
from somewhere other than a mailing list.

happy to discuss the UK side of things offline if you want, but I never looked 
at other jurisdictions.

Allan

- Original message -
  Hi,

  is it possible to use some statistic computing by R in proprietary
  software? Our software is written in c#, and we intend to use
  http://rdotnet.codeplex.com/
  to get R work there. Especially we want to use loess function.

  Thanks,

  Best regards,
  Stanislav

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[R] Automated Fixed Order Stepwise Regression Function

2011-04-07 Thread Tyler Rinker


Greetings,
 
I am interested in creating a stepwise fixed order regression function.  
There's a function for this already called add1( ).  The F statistics are 
calculated using type 2 anova (the SS and the F changes don't match SPSS's).  
You can see my use of this at the very end of the email.
 
What I want: a function to make an anova table with f changes and delt R^2.  
 
I ran into 10 snags to making this a fully automated function using the full 
linear model (order matters here).  Each snag is marked with a Comment #.  Some 
snags are repeated because I couldn't do the first time and certainly couldn't 
do it after that. Help with any/all snags would be very appreciated. 
 
I'm a 2 1/2 month [R] user who's reading everything online (incl. manuals)  
ordering every book I can (looking at Dalgaard, Crawly's and Teetor's very 
helpful books right now).  Loops and their usage is a foreign thing to me, 
despite studying them, and unfortunately I think that my function may call for 
them.  I also realize that beyond 10 predictors may make this function way to 
bulky.
 
I'm running the latest version of R (2.12.2)on a windows 7 machine.

DATASET
 
mtcars
full.model-lm(mpg~cyl+disp+hp+drat, data=mtcars)
 
CODE

stepFO-function(model)
{
m-data.frame(model.frame(model))
num.of.var-length(colnames(m))
mod1-lm(m[,1]~m[,2])
mod2-lm(m[,1]~m[,2]+m[,3])
mod3-lm(m[,1]~m[,2]+m[,3]+m[,4])
mod4-lm(m[,1]~m[,2]+m[,3]+m[,4]+m[,5])
#Comment 1--I don't know how to automated this process(above) of adding 
#...additional variables.  Probably a loop is needed but I don't understand 
#...how to apply it here.  Maybe update.model [1:num.ofvar]?
a1-anova(mod1)
a2-anova(mod2)
a3-anova(mod3)
a4-anova(mod4)
#Comment 2--SAME AS COMMENT 1 except applied to the anova tables.  How do I make
#...[R] add a5, a6...an   as necessary?

rb-rbind(a1,a2,a3,a4)
#Comment 3--again I can't automate this to make the  addition of a's automated

anova1-rbind(rb[1,],rb[4,],rb[8,],rb[13,],rb[14,])
#Comment 4--the rb's follow a pattern of 1+3+4+5...+n variables
#then I row bind these starting with 1 and rowbind one more after the last 
#...rb to include the bottom piece of the anova table that tells 
#...about residuals (how do I aoutomate this?)

anova-anova1[,1:num.of.var]
anova.table-data.frame(anova)
#Comment 5--Something that bugs me here is that I have to turn it into a 
dataframe to 
#...add the totals row and the delta R^2 (tried playing w/ tkinsert to no avail)
#...I miss the stuff that's at the bottom of the anova table (sig values)
#Comment 6--I'd love to turn the place value to round to 2 after the decimal.
#...I've worked this many ways including changing my options but this does 
#...not seem to apply to a data frame

Total-c(sum(anova[,1]),sum(anova[,2]), ,  ,  )
anova.table-rbind(anova.table,Total)
R1-summary(mod1)[[8]][[1]]
R2-summary(mod2)[[8]][[1]]
R3-summary(mod3)[[8]][[1]]
R4-summary(mod4)[[8]][[1]]
#Comment 7--SAME AS COMMENT 2.  How do I make
#...[R] add R5, R6...Rn   as necessary?

deltaR.1-R1
deltaR.2-R2-R1
deltaR.3-R3-R2
deltaR.4-R4-R3
#Comment 8--SAME AS COMMENT 7.  How would I aoutomate this process?

Delta.R.Squared-c(deltaR.1,deltaR.2,deltaR.3,deltaR.4, ,)
#Comment 9--I need a way to add as many deltaR's as 
#...necessary(n of R = n of predictors)

anova.table-cbind(anova.table, Delta.R.Squared)
colnames(anova.table)-c(df,Sum Sq,Mean Sq,F-change,
P-value,Delta.R.Squared)
rownames(anova.table)-c(X1, X2 elminating for X1, 
X3 eliminating for X2  X3, X4 eliminating for X1,X2,  X3,Residuals,
 Total)
anova.table
}
#Comment 10--Again I would need [R] to automate the list for row names as we 
#...add more predictors.
#See the final product below I'm after (with two places after the decimal)
 anova.table
df   Sum Sq  Mean Sq 
F-changeP-value Delta.R.Squared
X1   1 817.712952354614 817.712952354614 
79.5610275293349 6.112687142581e-10   0.726180005093805
X2 elminating for X1 1 37.5939529324851 37.5939529324851  
4.0268283172755 0.0541857201845285  0.0333857704630917
X3 eliminating for X2  X3   1 9.37092926438942 9.37092926438942 
1.00388976918267  0.324951851250774 0.00832196853596723
X4 eliminating for X1,Xx2,  X3  1 16.4674349222492 16.4674349222492 
1.81550535203668  0.189048514740917  0.0146241073243205
Residuals   27 244.901918026262 9.07044140838007
 NA   NA
Total   31 1126.0471875  
 
 
USING THE ADD1() FUNCTION  NOT WHAT I WANT 
 
mod1-lm(mpg~cyl, data=mtcars)
add1(mod1,~cyl+disp+hp+drat, data=mtcars, test=F)

Model:
mpg ~ cyl
   Df Sum of SqRSSAIC F value   Pr(F)  
none  308.33 76.494  
disp137.594 270.74 74.334  4.0268 0.05419 .
hp  116.360 291.98 76.750  1.6249 0.21253  
drat115.841 292.49 76.807  1.5706 0.22012  
---
Signif. codes:  0 *** 0.001 **

Re: [R] Getting number of students with zeroes in long format

2011-04-07 Thread Christopher Desjardins

Hi Jorge,
I want to make sure this does what I want.

So I want to get a count of students that never get a suspension. Once a
student has a non-zero I don't want to count that student. Each id_r is may
be associated with multiple sus. Are these commands doing this? Because ...

 suslm[175953:nrow(suslm),c(id_r,sus)]
   id_r  sus
999881.5 999881   1
999881.6 999881   7
999881.7 999881   0
999881.8 999881   0
999886.5 999886   0
999886.6 999886   0
999886.7 999886   0
999886.8 999886   0
999890.5 999890   0
999890.6 999890   0
999890.7 999890   0
999890.8 999890   0
999892.5 999892   0
999892.6 999892   0
999892.7 999892   0
999892.8 999892   0
999896.5 999896   0
999896.6 999896   4
999896.7 999896   3
999896.8 999896   0
999897.5 999897   0
999897.6 999897   0
999897.7 999897   0

 tail(with(suslm,tapply(sus,id_r,function(x) any(x==0
999881 999886 999890 999892 999896 999897
  TRUE   TRUE   TRUE   TRUE   TRUE   TRUE
 r - with(suslm, tapply(sus, id_r, function(x) any(x  0))
 tail(with(suslm, tapply(sus, id_r, function(x) any(x  0
999881 999886 999890 999892 999896 999897
  TRUE  FALSE  FALSE  FALSE   TRUE  FALSE

Based on this 999881 and 999896 should be FALSE not TRUE

I would expect if they were true for the first command they should be false
for the second command right?

 tail(names(r[ r == TRUE ]))
[1] 999752 999767 999806 999807 999881 999896
 tail(names(r[ r == FALSE ]))
[1] 999869 999870 999886 999890 999892 999897

This command seems to do the right thing. Is that right?


On Wed, Apr 6, 2011 at 10:25 PM, Jorge Ivan Velez
jorgeivanve...@gmail.comwrote:

 Hi Chris,

 Sorry I did not see your email before ;-)   Here is one option:

   r - with(d, tapply(sus, id_r, function(x) any(x  0)))
  r
111516181920212224252630
  3132
 FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE
 FALSE FALSE
33
 FALSE
  names(r[ r == TRUE ])
 [1] 15 25

 Regards,
 Jorge


 On Wed, Apr 6, 2011 at 5:03 PM, Christopher Desjardins  wrote:

 Thanks. And how many could I find that have greater than 0?
 Chris


 On Wed, Apr 6, 2011 at 3:58 PM, Jorge Ivan Velez  wrote:

 Hi Chris,

 Is this what you have in mind?

  sum(with(yourdata, tapply(sus, id_r, function(x) any(x==0
 [1] 13

 HTH,
 Jorge


 On Wed, Apr 6, 2011 at 4:44 PM, Christopher Desjardins  wrote:

 Hi,
 I have longitudinal school suspension data on students. I would like to
 figure out how many students (id_r) have no suspensions (sus), i.e. have
 a
 code of '0'. My data is in long format and the first 20 records look
 like
 the following:

  suslm[1:20,c(1,7)]
   id_r sus
   11   0
   15  10
   16   0
   18   0
   19   0
   19   0
   20   0
   21   0
   21   0
   22   0
   24   0
   24   0
   25   3
   26   0
   26   0
   30   0
   30   0
   31   0
   32   0
   33   0

 Each id_r is unique and I'd like to know the number of id_r that have a
 0
 for sus not the total number of 0. Does that make sense?
 Thanks!
 Chris

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 http://www.R-project.org/posting-guide.html
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Re: [R] Getting number of students with zeroes in long format

2011-04-07 Thread Christopher Desjardins

On Thu, Apr 7, 2011 at 8:07 AM, Christopher Desjardins 
cddesjard...@gmail.com wrote:

 Hi Jorge,
 I want to make sure this does what I want.

 So I want to get a count of students that never get a suspension. Once a
 student has a non-zero I don't want to count that student. Each id_r is may
 be associated with multiple sus. Are these commands doing this? Because ...


Also sus = suspension


  suslm[175953:nrow(suslm),c(id_r,sus)]
id_r  sus
 999881.5 999881   1
 999881.6 999881   7
 999881.7 999881   0
 999881.8 999881   0
 999886.5 999886   0
 999886.6 999886   0
 999886.7 999886   0
 999886.8 999886   0
 999890.5 999890   0
 999890.6 999890   0
 999890.7 999890   0
 999890.8 999890   0
 999892.5 999892   0
 999892.6 999892   0
 999892.7 999892   0
 999892.8 999892   0
 999896.5 999896   0
 999896.6 999896   4
 999896.7 999896   3
 999896.8 999896   0
 999897.5 999897   0
 999897.6 999897   0
 999897.7 999897   0
 
  tail(with(suslm,tapply(sus,id_r,function(x) any(x==0
 999881 999886 999890 999892 999896 999897
   TRUE   TRUE   TRUE   TRUE   TRUE   TRUE
  r - with(suslm, tapply(sus, id_r, function(x) any(x  0))
  tail(with(suslm, tapply(sus, id_r, function(x) any(x  0
 999881 999886 999890 999892 999896 999897
   TRUE  FALSE  FALSE  FALSE   TRUE  FALSE

 Based on this 999881 and 999896 should be FALSE not TRUE


for tail(with(suslm,tapply(sus,id_r,function(x) any(x==0


 I would expect if they were true for the first command they should be false
 for the second command right?

  tail(names(r[ r == TRUE ]))
 [1] 999752 999767 999806 999807 999881 999896
  tail(names(r[ r == FALSE ]))
 [1] 999869 999870 999886 999890 999892 999897

 This command seems to do the right thing. Is that right?


 On Wed, Apr 6, 2011 at 10:25 PM, Jorge Ivan Velez 
 jorgeivanve...@gmail.com wrote:

 Hi Chris,

 Sorry I did not see your email before ;-)   Here is one option:

   r - with(d, tapply(sus, id_r, function(x) any(x  0)))
  r
111516181920212224252630
  3132
 FALSE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE
 FALSE FALSE
33
 FALSE
  names(r[ r == TRUE ])
 [1] 15 25

 Regards,
 Jorge


 On Wed, Apr 6, 2011 at 5:03 PM, Christopher Desjardins  wrote:

  Thanks. And how many could I find that have greater than 0?
 Chris


 On Wed, Apr 6, 2011 at 3:58 PM, Jorge Ivan Velez  wrote:

 Hi Chris,

 Is this what you have in mind?

  sum(with(yourdata, tapply(sus, id_r, function(x) any(x==0
 [1] 13

 HTH,
 Jorge


 On Wed, Apr 6, 2011 at 4:44 PM, Christopher Desjardins  wrote:

 Hi,
 I have longitudinal school suspension data on students. I would like to
 figure out how many students (id_r) have no suspensions (sus), i.e.
 have a
 code of '0'. My data is in long format and the first 20 records look
 like
 the following:

  suslm[1:20,c(1,7)]
   id_r sus
   11   0
   15  10
   16   0
   18   0
   19   0
   19   0
   20   0
   21   0
   21   0
   22   0
   24   0
   24   0
   25   3
   26   0
   26   0
   30   0
   30   0
   31   0
   32   0
   33   0

 Each id_r is unique and I'd like to know the number of id_r that have a
 0
 for sus not the total number of 0. Does that make sense?
 Thanks!
 Chris

[[alternative HTML version deleted]]

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 http://www.R-project.org/posting-guide.html
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[R] How to speed up grouping time series, help please

2011-04-07 Thread Den Alpin

I found a faster implementation (by an order of magnitude from my
tests) than the one using xts, split, merge (from Joshua).
I report the two fastest solution below with code to generate a test
case; some work still to be done for columns order and naming,
Test case has grown from my previous post to get a more realistic timing.

Any comment or idea to further speed up multivariate time series
creation with classes xts or timeSeries starting from a data.frame
like the one reported here is welcome.

Best regards,
Den


a data.frame example (code below to generate it)

 ID                DATE     VALUE
14  3 2000-01-01 00:00:03 0.5726334
4   1 2000-01-01 00:00:03 0.8830174
1   1 2000-01-01 00:00:00 0.2875775
15  3 2000-01-01 00:00:04 0.1029247
11  3 2000-01-01 00:00:00 0.9568333
9   2 2000-01-01 00:00:03 0.5514350
7   2 2000-01-01 00:00:01 0.5281055
6   2 2000-01-01 00:00:00 0.0455565
12  3 2000-01-01 00:00:01 0.4533342
8   2 2000-01-01 00:00:02 0.8924190
3   1 2000-01-01 00:00:02 0.4089769
13  3 2000-01-01 00:00:02 0.6775706

And I want to get a timeSeries object or xts object like this:

                          1         2         3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01        NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04        NA        NA 0.1029247

# CODE:

set.seed(123)
# set N to 5 to reproduce above data.frame
N - 1000
# set K to 3 to reproduce above data.frame
K - 10
X - data.frame(
 ID = rep(1:K, each = N),
 DATE = as.character(rep(as.POSIXct(2000-01-01, tz = GMT)+ 0:(N-1), K)),
 VALUE = runif(N*K), stringsAsFactors = FALSE)
X - X[sample(1:(N*K), N*K),]
X - X[-(sample(1:nrow(X), floor(nrow(X)*0.2))),]
str(X)


xtsSplit - function(x)
{
 library(xts)
 x - xts(x[,c(ID,VALUE)], as.POSIXct(x[,DATE]))
 return(do.call(merge, split(x$VALUE,x$ID)))
}
xtsSplitTime - replicate(50,
 system.time(xtsSplit(X))[[1]])
median(xtsSplitTime)

xtsReshape - function(x)
{
 library(xts)
 x - reshape(x, idvar = DATE, timevar = ID, direction = wide)
 x - xts(x[,-1], as.POSIXct(x[,1]))
 return(x)
}
xtsReshapeTime - replicate(50,
 system.time(xtsReshape(X))[[1]])
median(xtsReshapeTime)

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Re: [R] unexpected sort order with merge

2011-04-07 Thread Johann Hibschman

B77S bps0...@auburn.edu writes:

 That is odd, I noticed some weird sorting with merge() a while back too and
 always am careful with it now.  Fortunately, sort=FALSE seems to work the
 way one would think most of the time.

Thanks for checking.  Is this on a more recent version of R than 2.10.1?
(I'm half-hoping this is something fixed in a newer R, so I can use it
as an excuse to demand an upgrade.)

 Although, the following results seem weird too!  (adding by=date makes it
 not sort oddly, regardless of sort=TRUE or FALSE)

[...]

 merge(d1, d2, by=date, sort=T)
   date icpn.x foo icpn.y bar
 1 Jan 2000500   1500  10
 2 Feb 2000500   2500  20
 3 Mar 2000500   3500  30
 4 Apr 2000500   4500  40
 5 May 2000500   5500  50
 6 Jun 2000500   6500  60

I think this is equivalent to the single column version.  For yearmon
objects, when `merge` only has one column to sort by, it seems to do the
right thing.  It only uses alphabetical order when there are more than
one column.  For ordered factors, though, even the single-column merge
was giving me strange sorts.

-Johann


P.S. Just in case people have bad threading on their mail/news reader, here's
the bad sort example:
 merge(d1, d2)
   date icpn foo bar
 1 Apr 2000  500   4  40
 2 Feb 2000  500   2  20
 3 Jan 2000  500   1  10
 4 Jun 2000  500   6  60
 5 Mar 2000  500   3  30
 6 May 2000  500   5  50

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Re: [R] help on pspline in coxph

2011-04-07 Thread Terry Therneau

 Hi there,

 I have a question on how to extract the linear term in the penalized 
 spline in coxph. Here is a sample code:

Here is a slightly more general example: 
   fit - coxph(Surv(time, status) ~ ph.ecog + pspline(age) + sex, lung)
(The lung data set is part of the survival package.)
There are 14 coefficients, of which 2-13 are the for the spline.  

The slope logic is subtle. Luckily I left comments on this in the source
code for pspline, because I remember it taking a long time for me to
figure it out.  
  slope = (X'VX)^{-1} X'V c
A weighted regression on the coefficients where c= spline coefficients =
fit$coef[2:13] in this case, V= fit$var[2:13,2:13], and X is a two
column matrix with column 1=1 (intercept) and column 2 = centers of the
pspline basis functions.  These centers are saved away in the print
function:
get(cbase, environment(fit$printfun[[1]]))
If there were multiple psline terms in the fit use [[2]], [[3]] etc as
appropriate.  No, you won't find this without reading the source code.

  Note that this is the linear slope from a fit with only a linear term,
i.e., what would have resulted from a fit with age instead of
pspline(age).  I thought it quite interesting that this quantity could
be derived at all.  I leave the details as a homework exercise for the
reader.

Terry Therneau

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Re: [R] R licence

2011-04-07 Thread Hadley Wickham

If all you need is loess, I suspect it would be cheaper to re-write it
in C# than to get a considered legal opinion on the matter.

Hadley

On Thu, Apr 7, 2011 at 2:45 AM, Stanislav Bek
stanislav.pavel@gmail.com wrote:
 Hi,

 is it possible to use some statistic computing by R in proprietary software?
 Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.

 Thanks,

 Best regards,
 Stanislav

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-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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[R] transform() on selective names. Is it possible?

2011-04-07 Thread Juan Carlos Borrás

Hi all,
I am whitening my data:

# code begins
N - 300
M - 2
x - matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N)
y - matrix(c(1,-2,-2,1), ncol=M, nrow=M)
z - data.frame(x %*% y)
colnames(z) - c('x','y')
par(mfrow=c(1,3))
plot(z, pch=5, col=blue)

whiten - function(x) { (x-mean(x))/sd(x) }

zz - transform(z, x=whiten(x), y=whiten(y))
plot(zz, pch=3, col=red)

#code ends

And everything looks fine enough.
But now I want to withen just one of the columns and I won't know
which one until my script is running, hence I can't hard code it in
the script.
Then I though, well maybe if I define some convenient f...

#begin code

f - function(a) { paste(a,=withen(,a,), sep='') }
a - 'x' # or a - 'y' depending on user input.
f(a)
 [1] x=withen(x)
# so I could try
zzz - transform(z, eval(f('x')))
# which of course doesn't work
plot(zz, pch=3, col=green)

head(z, n=2)
  x y
1 17.167380  6.884402
2  8.234507 13.940932
head(zzz, n=2)
  x y
1 17.167380  6.884402
2  8.234507 13.940932

#end code

Could someone provide me with some hint on whether the attempted trick
above is possible and how to proceed further?
Thanks in advance.
jcb!

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[R] package

2011-04-07 Thread Nelson Doki

I'd like to get the package about Potts and, Swenson-Wang models.
Haw can I do that?
There is a package about Wolff model, too? If yes, how to get it?
Thanks.
Nelson Doki.

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Re: [R] metaplot

2011-04-07 Thread cheba meier

Many thanks for your reply!  the Package ggplot was removed from the CRAN
repository. Formerly available versions can be obtained from the
archivehttp://cran.r-project.org/src/contrib/Archive/ggplot(Unfortunately,
I am a windows user).  The metafor and meta.summaries()
still need the (a,b,c,d) or standard errors for the effect estimates! I have
the following data

rr - c(0.61, 0.35, 1.16, 0.94, 1.16, 0.46, 0.99, 1.05, 1.15, 0.27)
lci - c(-1.69, -1.27, -0.24, -0.54, -1.01, -1.8,-0.89, -1.38, -1.05,-2.04)
uci - c(0.70, -0.82,0.54,0.41,1.30,0.24,0.87,1.48,1.33,-0.57)
rrdata - data.frame(rr,lci,uci)

I can compute standard errors from my data? but the results show different
confidence interval?

Many thanks in advance,
Cheba


2011/4/6 Scott Chamberlain scttchamberla...@gmail.com

  What about the metafor package?


 Or just create your own plot.

 For example, using ggplot2 package:

 limits - aes(ymax = OR + (OR - 95%LCI), ymin = OR - (OR - 95%LCI))
 ggplot(dataframe, aes(x = Study.Name, y = OR)) + geom_point() +
 geom_errobar(limits)

 Best,
 Scott

 On Wednesday, April 6, 2011 at 11:53 AM, cheba meier wrote:

 Dear all,

 I have a four variable: Stuy.Name, OR, 95%LCI and 95%UCI and I would like
 to
 create a meta analysis plot. I can't use meta.MH function in metaplot
 because I do not have
 n.trt, n.ctrl, col.trt, col.ctrl are not available! Is there an alternative
 way to do it?

 Many thanks in advance,
 Cheba

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Re: [R] transform() on selective names. Is it possible?



On Apr 7, 2011, at 9:56 AM, Juan Carlos Borrás wrote:


Hi all,
I am whitening my data:

# code begins
N - 300
M - 2
x - matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N)
y - matrix(c(1,-2,-2,1), ncol=M, nrow=M)
z - data.frame(x %*% y)
colnames(z) - c('x','y')
par(mfrow=c(1,3))
plot(z, pch=5, col=blue)

whiten - function(x) { (x-mean(x))/sd(x) }


Consider:

 whiten - scale  # no need to re-invent the wheel
 fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[coln]))
 colxy - x
 z - fc(z, colxy)  # the [ function will interpret colxy
 z



zz - transform(z, x=whiten(x), y=whiten(y))
plot(zz, pch=3, col=red)

#code ends

And everything looks fine enough.
But now I want to withen just one of the columns and I won't know
which one until my script is running, hence I can't hard code it in
the script.
Then I though, well maybe if I define some convenient f...

#begin code

f - function(a) { paste(a,=withen(,a,), sep='') }
a - 'x' # or a - 'y' depending on user input.
f(a)

[1] x=withen(x)

# so I could try
zzz - transform(z, eval(f('x')))
# which of course doesn't work
plot(zz, pch=3, col=green)

head(z, n=2)

x y
1 17.167380  6.884402
2  8.234507 13.940932

head(zzz, n=2)

x y
1 17.167380  6.884402
2  8.234507 13.940932


#end code

Could someone provide me with some hint on whether the attempted trick
above is possible and how to proceed further?
Thanks in advance.
jcb!

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


David Winsemius, MD
West Hartford, CT

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[R] [SOLVED] Two functions as parametrs of a function.

2011-04-07 Thread KENNETH R CABRERA

Dr Dallazuanna:

Thank you for your help.

It works very good!


- Mensaje original -
De: Henrique Dallazuanna www...@gmail.com
Fecha: Jueves, 7 de Abril de 2011, 7:05 am
Asunto: Re: [R] Two functions as parametrs of a function.
A: Kenneth Roy Cabrera Torres krcab...@une.net.co
CC: r-help r-help@r-project.org

 Try this:
 
 f2 - function(n, nsim, fun1, fun2) {
   vp - replicate(nsim, do.call(fun1$name, c(n, fun1$args)),
 do.call(fun1$name, c(n, fun1$args)))
   vp
 
 }
 
 f2(2, 100, fun1 = list(name = rbeta, args = list(shape1 = 2, 
 shape2 =
 2), fun1 = list(name = rbeta, args = list(shape1 = 2, shape2 = 2
 
 On Thu, Apr 7, 2011 at 8:45 AM, Kenneth Roy Cabrera Torres
 krcab...@une.net.co wrote:
  Hi R users:
 
  I'm trying to make a function where two of the parameters are
  functions, but I don't know how to put each set of parameters for
  each function.
 
  What am I missing?
 
  I try this code:
 
  f2-
 function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){
 force(fun1)
 force(fun2)
 force(n)
 p1-unlist(par1)
 p2-unlist(par2)
 force(p1)
 force(p2)
 localfun1 - function() fun1(n, p1)
 localfun2 - function() fun2(n, p2)
 vp-replicate(nsim,t.test(localfun1(), localfun2())$p.value)
 return(vp)
  }
 
  
 f2(fun1=rbeta,par1=list(shape1=2,shape2=2),fun2=rbeta,par2=list(shape1=1,shape2=2))
  Thank you for your help.
 
  Kenneth
 
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  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.
 
 
 
 
 -- 
 Henrique Dallazuanna
 Curitiba-Paraná-Brasil
 25° 25' 40 S 49° 16' 22 O

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[R] [SOLVED] Two functions as parametrs of a function.

2011-04-07 Thread KENNETH R CABRERA

Dr. Murdoch:

Thank you very much for your help.

Where can I find a more systematic documentation about these topics?

I mean, examples of do.call, environments, as.function, the
scoping and the way understand when to use force, etc.

Again, thank you.

Kenneth

- Mensaje original -
De: Duncan Murdoch murdoch.dun...@gmail.com
Fecha: Jueves, 7 de Abril de 2011, 7:04 am
Asunto: Re: [R] Two functions as parametrs of a function.
A: Kenneth Roy Cabrera Torres krcab...@une.net.co
CC: r-help r-help@r-project.org

 On 07/04/2011 7:45 AM, Kenneth Roy Cabrera Torres wrote:
 Hi R users:
 
 I'm trying to make a function where two of the parameters are
 functions, but I don't know how to put each set of parameters for
 each function.
 
 What am I missing?
 
 I try this code:
 
 f2-
 function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){ 
 force(fun1)
  force(fun2)
  force(n)
  p1-unlist(par1)
  p2-unlist(par2)
 
 You don't want to unlist those.  Leave them as lists, and 
 use the do.call() function for the calls.  For example,
 
 localfun1 - function() do.call(fun1, c(list(n), par1))
 
 
  force(p1)
  force(p2)
 
 Calling force() is only useful for arguments to your 
 function:  it kicks the evaluator, so it's not so 
 lazy.   Since p1 and p2 were calculated earlier, 
 there's no need to force them.  You might need to force n, 
 since it is only used in the localfun1 and localfun2 calls, and 
 if the argument to f2 changes before you make those calls, the 
 wrong value will be used.  That's pretty unlikely with your 
 code, but since fun1 and fun2 can do anything, it's safest to 
 fix the value of n before you call them.  The same argument 
 applies to fun1 and fun2 themselves.
 
 Duncan Murdoch
 
  localfun1- function() fun1(n, p1)
  localfun2- function() fun2(n, p2)
  vp-
 replicate(nsim,t.test(localfun1(), localfun2())$p.value)
  return(vp)
 }
 
 f2(fun1=rbeta,par1=list(shape1=2,shape2=2),fun2=rbeta,par2=list(shape1=1,shape2=2))
 
 Thank you for your help.
 
 Kenneth
 
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Re: [R] transform() on selective names. Is it possible?

2011-04-07 Thread Juan Carlos Borrás

Wonderful, or the closest to heaven I've been the whole afternoon, but
not quite there:

# begin code
N - 300
M - 2
x - matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N)
y - matrix(c(1,-2,-2,1), ncol=M, nrow=M)
z - data.frame(x %*% y)
colnames(z) - c('x','y')
par(mfrow=c(1,3))
plot(z, pch=5, col=blue)

whiten - scale  # no need to re-invent the wheel, I agree
fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[coln]))
colxy - x
z1 - fc(z, colxy)  # the [ function will interpret colxy
colnames(z1)
 [1] x   y   x.1

fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[,coln]))
colxy - x
z2 - fc(z, colxy)  # the [ function will interpret colxy
colnames(z2)
 [1] xycoln

#end code

What I want is to know whether I can customize the column name of the
result of the transform() call.
Your hint is fantastic, thanks there, but I keep getting into that
particular pattern of computation over and over and I wonder if it's
possible to skip a column clean-up after applying your trick.

2011/4/7 David Winsemius dwinsem...@comcast.net:

 On Apr 7, 2011, at 9:56 AM, Juan Carlos Borrás wrote:

 Hi all,
 I am whitening my data:

 # code begins
 N - 300
 M - 2
 x - matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N)
 y - matrix(c(1,-2,-2,1), ncol=M, nrow=M)
 z - data.frame(x %*% y)
 colnames(z) - c('x','y')
 par(mfrow=c(1,3))
 plot(z, pch=5, col=blue)

 whiten - function(x) { (x-mean(x))/sd(x) }

 Consider:

  whiten - scale  # no need to re-invent the wheel
  fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[coln]))
  colxy - x
  z - fc(z, colxy)  # the [ function will interpret colxy
  z


 zz - transform(z, x=whiten(x), y=whiten(y))
 plot(zz, pch=3, col=red)

 #code ends

 And everything looks fine enough.
 But now I want to withen just one of the columns and I won't know
 which one until my script is running, hence I can't hard code it in
 the script.
 Then I though, well maybe if I define some convenient f...

 #begin code

 f - function(a) { paste(a,=withen(,a,), sep='') }
 a - 'x' # or a - 'y' depending on user input.
 f(a)

 [1] x=withen(x)

 # so I could try
 zzz - transform(z, eval(f('x')))
 # which of course doesn't work
 plot(zz, pch=3, col=green)

 head(z, n=2)

        x         y
 1 17.167380  6.884402
 2  8.234507 13.940932

 head(zzz, n=2)

        x         y
 1 17.167380  6.884402
 2  8.234507 13.940932

 #end code

 Could someone provide me with some hint on whether the attempted trick
 above is possible and how to proceed further?
 Thanks in advance.
 jcb!

 __
 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.

 David Winsemius, MD
 West Hartford, CT




Cheers,
jcb!

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Re: [R] R licence

2011-04-07 Thread Matthew Dowle

Duncan,

Letting you know then that I just don't see how the first paragraph here :

http://www.revolutionanalytics.com/downloads/gpl-sources.php

is compatible with clause 2(b) here :

http://www.gnu.org/licenses/gpl-2.0.html

Perhaps somebody could explain why it is?

Matthew


Duncan Murdoch murdoch.dun...@gmail.com wrote in message 
news:4d9da9ff.9020...@gmail.com...
 On 07/04/2011 7:47 AM, Matthew Dowle wrote:
 Peter,

 If the proprietary part of REvolution's product is ok, then surely
 Stanislav's suggestion is too. No?

 Revolution has said that they believe they follow the GPL, and they 
 haven't been challenged on that.   If you think that they don't, you could 
 let an R copyright holder know what they're doing that's a license 
 violation.

 My opinion of Stanislav's question is that he doesn't give enough 
 information to answer.  If he is planning to distribute R as part of his 
 product, he needs to follow the GPL.  If not, I don't think any R 
 copyright holder has anything to complain about.

 Duncan Murdoch

 Matthew


 peter dalgaardpda...@gmail.com  wrote in message
 news:be157cf5-9b4b-45a0-a7d4-363b774f1...@gmail.com...
 
   On Apr 7, 2011, at 09:45 , Stanislav Bek wrote:
 
   Hi,
 
   is it possible to use some statistic computing by R in proprietary
   software?
   Our software is written in c#, and we intend to use
   http://rdotnet.codeplex.com/
   to get R work there. Especially we want to use loess function.
 
   You need to take legal advice to be certain, but offhand I would say 
  that
   this kind of circumvention of the GPL is _not_ allowed.
 
   It all depends on whether the end product is a derivative work, in 
  which
   case, the whole must be distributed under a GPL-compatible licence. 
  The
   situation around GPL-incompatible plug-ins or plug-ins interfacing to 
  R in
   GPL -incompatible software is legally murky, but using R as a 
  subroutine
   library for proprietary code is clearly crossing the line, as far as I 
  can
   tell.
 
   -- 
   Peter Dalgaard
   Center for Statistics, Copenhagen Business School
   Solbjerg Plads 3, 2000 Frederiksberg, Denmark
   Phone: (+45)38153501
   Email: pd@cbs.dk  Priv: pda...@gmail.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.


__
R-help@r-project.org mailing list
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Re: [R] R licence

2011-04-07 Thread Ista Zahn

I don't think R-help is the appropriate place for this question.
Probably you will have more luck at http://www.theattorneysforum.com/
or some such.

Best,
Ista
On Thu, Apr 7, 2011 at 10:36 AM, Matthew Dowle mdo...@mdowle.plus.com wrote:
 Duncan,

 Letting you know then that I just don't see how the first paragraph here :

 http://www.revolutionanalytics.com/downloads/gpl-sources.php

 is compatible with clause 2(b) here :

 http://www.gnu.org/licenses/gpl-2.0.html

 Perhaps somebody could explain why it is?

 Matthew


 Duncan Murdoch murdoch.dun...@gmail.com wrote in message
 news:4d9da9ff.9020...@gmail.com...
 On 07/04/2011 7:47 AM, Matthew Dowle wrote:
 Peter,

 If the proprietary part of REvolution's product is ok, then surely
 Stanislav's suggestion is too. No?

 Revolution has said that they believe they follow the GPL, and they
 haven't been challenged on that.   If you think that they don't, you could
 let an R copyright holder know what they're doing that's a license
 violation.

 My opinion of Stanislav's question is that he doesn't give enough
 information to answer.  If he is planning to distribute R as part of his
 product, he needs to follow the GPL.  If not, I don't think any R
 copyright holder has anything to complain about.

 Duncan Murdoch

 Matthew


 peter dalgaardpda...@gmail.com  wrote in message
 news:be157cf5-9b4b-45a0-a7d4-363b774f1...@gmail.com...
 
   On Apr 7, 2011, at 09:45 , Stanislav Bek wrote:
 
   Hi,
 
   is it possible to use some statistic computing by R in proprietary
   software?
   Our software is written in c#, and we intend to use
   http://rdotnet.codeplex.com/
   to get R work there. Especially we want to use loess function.
 
   You need to take legal advice to be certain, but offhand I would say
  that
   this kind of circumvention of the GPL is _not_ allowed.
 
   It all depends on whether the end product is a derivative work, in
  which
   case, the whole must be distributed under a GPL-compatible licence.
  The
   situation around GPL-incompatible plug-ins or plug-ins interfacing to
  R in
   GPL -incompatible software is legally murky, but using R as a
  subroutine
   library for proprietary code is clearly crossing the line, as far as I
  can
   tell.
 
   --
   Peter Dalgaard
   Center for Statistics, Copenhagen Business School
   Solbjerg Plads 3, 2000 Frederiksberg, Denmark
   Phone: (+45)38153501
   Email: pd@cbs.dk  Priv: pda...@gmail.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.


 __
 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.




-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] [SOLVED] Two functions as parametrs of a function.


On 07/04/2011 10:12 AM, KENNETH R CABRERA wrote:

Dr. Murdoch:

Thank you very much for your help.

Where can I find a more systematic documentation about these topics?

I mean, examples of do.call, environments, as.function, the
scoping and the way understand when to use force, etc.


For details of the concepts, you should look in the R Language 
Definition manual (and perhaps An Introduction to R for some more 
examples).


For individual functions, the help page for each usually includes some 
examples.


And then there's the source code, but if the manuals are unclear, it may 
be no clearer.


Duncan Murdoch



Again, thank you.

Kenneth

- Mensaje original -
De: Duncan Murdochmurdoch.dun...@gmail.com
Fecha: Jueves, 7 de Abril de 2011, 7:04 am
Asunto: Re: [R] Two functions as parametrs of a function.
A: Kenneth Roy Cabrera Torreskrcab...@une.net.co
CC: r-helpr-help@r-project.org

  On 07/04/2011 7:45 AM, Kenneth Roy Cabrera Torres wrote:
  Hi R users:
  
  I'm trying to make a function where two of the parameters are
  functions, but I don't know how to put each set of parameters for
  each function.
  
  What am I missing?
  
  I try this code:
  
  f2-
  function(n=2,nsim=100,fun1=rnorm,par1=list(),fun2=rnorm,par2=list()){  
force(fun1)
force(fun2)
force(n)
p1-unlist(par1)
p2-unlist(par2)

  You don't want to unlist those.  Leave them as lists, and
  use the do.call() function for the calls.  For example,

  localfun1- function() do.call(fun1, c(list(n), par1))


force(p1)
force(p2)

  Calling force() is only useful for arguments to your
  function:  it kicks the evaluator, so it's not so
  lazy.   Since p1 and p2 were calculated earlier,
  there's no need to force them.  You might need to force n,
  since it is only used in the localfun1 and localfun2 calls, and
  if the argument to f2 changes before you make those calls, the
  wrong value will be used.  That's pretty unlikely with your
  code, but since fun1 and fun2 can do anything, it's safest to
  fix the value of n before you call them.  The same argument
  applies to fun1 and fun2 themselves.

  Duncan Murdoch

localfun1- function() fun1(n, p1)
localfun2- function() fun2(n, p2)
vp-
  replicate(nsim,t.test(localfun1(), localfun2())$p.value)
return(vp)
  }
  
  
f2(fun1=rbeta,par1=list(shape1=2,shape2=2),fun2=rbeta,par2=list(shape1=1,shape2=2))
  
  Thank you for your help.
  
  Kenneth
  
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Re: [R] R licence


On 07/04/2011 10:36 AM, Matthew Dowle wrote:

Duncan,

Letting you know then that I just don't see how the first paragraph here :

http://www.revolutionanalytics.com/downloads/gpl-sources.php

is compatible with clause 2(b) here :

http://www.gnu.org/licenses/gpl-2.0.html

Perhaps somebody could explain why it is?


No, I certainly wouldn't attempt that, but thanks for the information.

Duncan Murdoch


Matthew


Duncan Murdochmurdoch.dun...@gmail.com  wrote in message
news:4d9da9ff.9020...@gmail.com...
  On 07/04/2011 7:47 AM, Matthew Dowle wrote:
  Peter,

  If the proprietary part of REvolution's product is ok, then surely
  Stanislav's suggestion is too. No?

  Revolution has said that they believe they follow the GPL, and they
  haven't been challenged on that.   If you think that they don't, you could
  let an R copyright holder know what they're doing that's a license
  violation.

  My opinion of Stanislav's question is that he doesn't give enough
  information to answer.  If he is planning to distribute R as part of his
  product, he needs to follow the GPL.  If not, I don't think any R
  copyright holder has anything to complain about.

  Duncan Murdoch

  Matthew


  peter dalgaardpda...@gmail.com   wrote in message
  news:be157cf5-9b4b-45a0-a7d4-363b774f1...@gmail.com...
  
 On Apr 7, 2011, at 09:45 , Stanislav Bek wrote:
  
 Hi,
  
 is it possible to use some statistic computing by R in proprietary
 software?
 Our software is written in c#, and we intend to use
 http://rdotnet.codeplex.com/
 to get R work there. Especially we want to use loess function.
  
 You need to take legal advice to be certain, but offhand I would say
that
 this kind of circumvention of the GPL is _not_ allowed.
  
 It all depends on whether the end product is a derivative work, in
which
 case, the whole must be distributed under a GPL-compatible licence.
The
 situation around GPL-incompatible plug-ins or plug-ins interfacing to
R in
 GPL -incompatible software is legally murky, but using R as a
subroutine
 library for proprietary code is clearly crossing the line, as far as I
can
 tell.
  
 -- 
 Peter Dalgaard

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

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Re: [R] R licence

2011-04-07 Thread peter dalgaard


On Apr 7, 2011, at 16:36 , Matthew Dowle wrote:

 Duncan,
 
 Letting you know then that I just don't see how the first paragraph here :
 
 http://www.revolutionanalytics.com/downloads/gpl-sources.php
 
 is compatible with clause 2(b) here :
 
 http://www.gnu.org/licenses/gpl-2.0.html
 
 Perhaps somebody could explain why it is?


As I understand it, they believe that it falls under the mere aggregation 
exception at the end of sec.2, and they believe that their plugins are not 
derived works, but add-ons. The legal hair-splitting is rather painful... 

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

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Re: [R] Quasipoisson with geeglm

2011-04-07 Thread Søren Højsgaard

Dear Ivy,

In gee there is no quasipossion, because gee is in a way already quasi.

With GEE we do not fit a poisson glm, but use in the construction of the 
sandwich covariance matrix the variance function of the poisson family. In Gee 
always an 'overdispersion' is estimated.

Regards
Søren


Fra: r-help-boun...@r-project.org [r-help-boun...@r-project.org] P#229; vegne 
af JANSEN, Ivy [ivy.jan...@inbo.be]
Sendt: 7. april 2011 13:32
Til: r-help@r-project.org
Emne: [R] Quasipoisson with geeglm

Dear all,

I am trying to use the GEE methodology to fit a trend for the number of 
butterflies observed at several sites. In total, there are 66 sites, and 19 
years for which observations might be available. However, only 326 observations 
are available (instead of 1254). For the time being, I ignore the large number 
of missing values, and the fact that GEE is only valid under MCAR. When I run 
the following code

geeglm(SumOfButterflies ~ RES_YEAR, family = poisson, data = ManijurtNoNA, id = 
RES_ROTE_ID, corstr = ar1)

I obtain normal output. Not surprisingly, overdispersion is present 
(Estimated Scale Parameters:  [1] 185.8571), so changing to quasipoisson is 
needed. However, the code below

geeglm(SumOfButterflies ~ RES_YEAR, family = quasipoisson, data = ManijurtNoNA, 
id = RES_ROTE_ID, corstr = ar1)

produces the following error

Error in geese.fit(xx, yy, id, offset, soffset, w, waves = waves, zsca,  :   
variance invalid.

Other correlation structures encounter the same problem. I also tried adding 
waves = RES_YEAR (although I am not sure how waves should be used correctly), 
but it does not change anything.

Any suggestions what might be wrong?

Regards,
Ivy



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Re: [R] transform() on selective names. Is it possible?



On Apr 7, 2011, at 10:27 AM, Juan Carlos Borrás wrote:


Wonderful, or the closest to heaven I've been the whole afternoon, but
not quite there:

# begin code
N - 300
M - 2
x - matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N)
y - matrix(c(1,-2,-2,1), ncol=M, nrow=M)
z - data.frame(x %*% y)
colnames(z) - c('x','y')
par(mfrow=c(1,3))
plot(z, pch=5, col=blue)

whiten - scale  # no need to re-invent the wheel, I agree
fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[coln]))
colxy - x
z1 - fc(z, colxy)  # the [ function will interpret colxy
colnames(z1)

[1] x   y   x.1


fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[,coln]))
colxy - x
z2 - fc(z, colxy)  # the [ function will interpret colxy
colnames(z2)

[1] xycoln


#end code

What I want is to know whether I can customize the column name of the
result of the transform() call.


I haven't stumbled on a solution to that task. I am wondering if you  
could use something like:


inpnames - names(dfrm) # before the transform step
outdfrm - transform(...)
names(outdfrm) - c(inpnames, paste(colxy, new, sep=_) )

 kludgy to be sure.


Your hint is fantastic, thanks there, but I keep getting into that
particular pattern of computation over and over and I wonder if it's
possible to skip a column clean-up after applying your trick.

2011/4/7 David Winsemius dwinsem...@comcast.net:


On Apr 7, 2011, at 9:56 AM, Juan Carlos Borrás wrote:


Hi all,
I am whitening my data:

# code begins
N - 300
M - 2
x - matrix(data=rnorm(N*M, 0, 3)-10, ncol=M, nrow=N)
y - matrix(c(1,-2,-2,1), ncol=M, nrow=M)
z - data.frame(x %*% y)
colnames(z) - c('x','y')
par(mfrow=c(1,3))
plot(z, pch=5, col=blue)

whiten - function(x) { (x-mean(x))/sd(x) }


Consider:

 whiten - scale  # no need to re-invent the wheel
 fc - function(dfrm, coln) transform(dfrm, coln=whiten(dfrm[coln]))
 colxy - x
 z - fc(z, colxy)  # the [ function will interpret colxy
 z



zz - transform(z, x=whiten(x), y=whiten(y))
plot(zz, pch=3, col=red)

#code ends

And everything looks fine enough.
But now I want to withen just one of the columns and I won't know
which one until my script is running, hence I can't hard code it in
the script.
Then I though, well maybe if I define some convenient f...

#begin code

f - function(a) { paste(a,=withen(,a,), sep='') }
a - 'x' # or a - 'y' depending on user input.
f(a)


[1] x=withen(x)


# so I could try
zzz - transform(z, eval(f('x')))
# which of course doesn't work
plot(zz, pch=3, col=green)

head(z, n=2)


   x y
1 17.167380  6.884402
2  8.234507 13.940932


head(zzz, n=2)


   x y
1 17.167380  6.884402
2  8.234507 13.940932


#end code

Could someone provide me with some hint on whether the attempted  
trick

above is possible and how to proceed further?
Thanks in advance.
jcb!

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David Winsemius, MD
West Hartford, CT





Cheers,
jcb!


David Winsemius, MD
West Hartford, CT

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[R] dotplot as a background for multiple barchart plots (with Lattice)

2011-04-07 Thread Christian Charles


Hi R users,

I'm kind stuck in my R experience...
I want to have a multiple barchart with n windows, but with the same 
background for each window, and I want that background to be a recorded 
dotplot.

First, is it possible ?
If so, I guess I have to call a record plot in the panel function but how ?
As there is no panel.plot and I can't use the recorded plot as an 
argument for the panel.dotplot() call, I'm stuck.

I think I close but I need some help.

Many thanks and regards,

Christian


### dummy data building
test - data.frame(Subject=rep(c(A, B, C, D), each=6),
   Type=rep(c(EEE, FFF, GGG, HHH, III, 
JJJ), times=4),

   MeanX=abs(rnorm(24)),
   MeanZ=abs(rnorm(24)))

### data for regrouping all means
all - data.frame(Type=levels(factor(test$Type)), MeanX = c(0), MeanZ = 
c(0))


### building means...
for(s in levels(factor(test$Subject)))
{
tmp - data.frame(Type = all$Type)
for(t in tmp$Type)
{
tmp$MeanX[tmp$Type == t] - mean(test$MeanX[test$Type==t  
test$Subject==s])
tmp$MeanZ[tmp$Type == t] - mean(test$MeanZ[test$Type==t  
test$Subject==s])

}
all$MeanX = all$MeanX + tmp$MeanX
all$MeanZ = all$MeanZ + tmp$MeanZ
}
l - length(levels(factor(test$Subject)))
all$MeanX = all$MeanX/l
all$MeanX = all$MeanZ/l

### plot for means
meanplot - dotplot(MeanX+MeanZ ~ Type, data=all, cex = 1.2, xlab=, 
ylab=,

panel = panel.superpose,
panel.groups = function(x, ..., group.number){
panel.dotplot(x + (group.number - 1.5)/3, ... )
})

### test plot in order to see if there is something wrong.
plot(meanplot)

### final plot
barchart(MeanX + MeanZ ~ Type|Subject,
 data=test ,
 scales = list(x = list(rot = 60)),
 layout=c(1, l),
 panel = function(...) {
panel.superpose
#panel.dotplot(meanplot) # so this is want I want, my 
recorded dotplot as a background...

panel.abline(h=0)
panel.grid(v = -length(levels(factor(test$Type))), h=0)
 panel.barchart(...)
})


BTW, this R code might not be perfect and surely, there are better ways 
to build my data frames, but that's my newbie way to do it.


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[R] K means algorithm C code

2011-04-07 Thread Jean-Eudes Dazard

Dear R user,



 

How can I get the C or C++ source code of the R_kmeans_MacQueen or
R_kmeans_Lloyd subroutines implemented in the R kmeans function of the R
stats-package {stats}? Would these be available as a C header file (*.h)
somewhere from the R installation???

Any help to get a C or C++ implementation of these clustering  algorithms
would be very welcome.

 

Thanks for your help.

Johnny

 

~~~

Jean-Eudes DAZARD, PhD.

Assistant Professor

Bioinformatics Division

Center for Proteomics and Bioinformatics

School Of Medicine, BRB 936

Case Western Reserve University

10900 Euclid Avenue

CLEVELAND, OHIO 44106 - 4988

Tel : 1-216-368-3157

Fax : 1-216-368-6846

e-mail : jxd...@case.edu

~~~

 


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Re: [R] Scrap java scripts and styles from an html document

2011-04-07 Thread Mike Marchywka

Date: Thu, 7 Apr 2011 04:15:50 -0700
From: antuj...@gmail.com
To: r-help@r-project.org
Subject: Re: [R] Scrap java scripts and styles from an html document

Hi ,

I am working on developing a web crawler.

Comments like this come up on the list every few weeks or so and I
keep suggesting that someone ( other than me of course LOL) investigates
an R interface to webkit for any efforts that require mimic of
large parts of a browser function. Perhaps just make
a debug build or custom build of webkit to dump whatever it is you want
into a structured text file
( I've actually done this for what would amount to a crawler, I modified
maybe one or two classes to output the links being fetched to stdout but
I think there are ways to dump a DOM or other stuff in a format usable by R).
For valid pages, you can just parse html as xml and get what you want in this
case but usually people are looking for information only apparent after
large pieces of js are executed. If you want comments only, these
may be easy to isolate yourself.If you google CRAN HTML parser some
hits do come up, for example

http://cran.r-project.org/web/packages/scrapeR/scrapeR.pdf

http://r.789695.n4.nabble.com/How-to-import-HTML-and-SQL-files-td879480.html

Removing javascripts and styles is a part of the cleaning of the html
document.
What I want is a cleaned html document with only the html tags and textual
information,
so that i can figure out the pattern of the web page. This is being done to
extract relevant
information from the webpage like comments for a particular product.

For e.g the amazon.com has all such comments within the
and tags,
with regular
occuring for breaks. So tags which appear the most help us in
locating the required information. Different websites have different
patterns,
but its more likely that tags that will occur the most will have the
relevant information enclosed in them.

So, once the html page is cleaned, it would be easy to role up the tags and
knowing their frequency of occurrence, we can target the information.

Should there be any suggestions to help, please let me know. I would be more
than pleased.

Regards,
Antuj

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Re: [R] How can I include a new book on the


Thanks Ben! I will!

Em 4/7/2011 8:32 AM, Ben Bolker escreveu:

Marcio Pupin Mellomelloat  ieee.org  writes:


I've just published a new book for R beginners in Portuguese:
Conhecendo o R: uma visão estatística (something like Knowing R: an
statistical approach). I'd like to include it on the list Books at
R-project.org. How can I do it?
More informations about the book at
http://www.editoraufv.com.br/produtos/conhecendo-o-r


   I think you should try contacting Kurt Hornik at r-project.org

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--
Marcio Pupin Mello

Survey Engineer
Ph.D student in Remote Sensing
National Institute for Space Research (INPE) - Brazil
Laboratory of Remote Sensing in Agriculture and Forestry (LAF)
www.dsr.inpe.br/~mello

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[R] Two questions about metacharacter in regexprs and function return

2011-04-07 Thread Tuexy

for the script, please kindly see the script below. At line 10 and line 13,
my problems occurs. 
The first one is I try to retrieve the gene official name from a column of a
table. The pattern of official name is something  starting with gene_name.
For detail problems, please see the according lines. 
Any suggestions are appreciated

example of matching source (extract the Nnat, sometime it would be the
character N/A): 
AB004048|MM8;NCBI Build 36|transcript|chr2|157251580|157253958|ExemplarFor
'AB004048'; gene_id '18111'; transcript_id 'AB004048'; gene_name 'Nnat'; alt
'5730414I02Rik|AW107673|Peg5';
neuronatin|http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?db=genecmd=Retrievedopt=full_reportlist_uids=18111;

#obtain the exprs matrix for cluster analysis
#ask questions
DEG_files - grep(bak, dir());#pay attention to the filenames
exprs_files - grep(copy, dir());
protein - c();
assign_exprs - function(files, protein) { #use to find the DEGs or exprs
for cmeans clustering
for(i in 1:length(files)) {
microarray_data - read.csv(file = files[i], header = T, sep =
\t);
microarray_data[, 7] - gsub(([\\s\\S]+gene_name '(\\w*)';.+),
\\2, microarray_data[, 7], perl = T);#why [\\w]* cannot workable? also the
[(\\w*)(N/A)] cannot be workable.
assign(files[i], microarray_data, envir=.GlobalEnv); #get(dir()[i]()
can obtain the data of interest.`variable_names` can also work
protein - c(protein, get(files[i])[, 7]);  #used for obtain all
the DEGs only
}   
#return protein; #why this line is not workable?
assign(all_protein, protein, envir=.GlobalEnv);
}
exist_to_cluster_exprs - function(x, cluster_exprs, all) {
if(exists(all, x[1])){ #exists function
cluster_exprs - cbind(cluster_exprs, x); 
}   
#return cluster_exprs;
}
assign_exprs(dir()[DEG_files], protein);
all_protein - unique(all_protein);
assign_exprs(dir()[exprs_files], protein);
for(i in 1:2) {
apply(get(dir()[exprs_files[i]]), 1, exist_to_cluster_exprs,
cluster_exprs, all);
#assign(paste(exprs_files()[i], exprs_data), cluster_exprs[, c(2,
3, 5, 7)];
exprs_data - cbind(exprs_data, cluster_exprs[, 3]);
}
exprs_data;

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Re: [R] Time series of spatial data


Hi idham,
	Welcome to the R! It will depend on what you want to do with your data. 
But you can take a look at the packages sp, spdep. And I also recomend 
the book:
R.S. Bivand, E.J. Pebesma, and V. Gómez-Rubio, Applied spatial data 
analysis with R, New York, USA: Springer, 2008.

Good luck!
--
Marcio Pupin Mello

Survey Engineer
Ph.D student in Remote Sensing
National Institute for Space Research (INPE) - Brazil
Laboratory of Remote Sensing in Agriculture and Forestry (LAF)
www.dsr.inpe.br/~mello


Em 4/7/2011 7:38 AM, idham escreveu:

Hi guys,

I'm really new in R.

Trying to analyze series of spatial datasets (365 satellite images) in order
to find the best model that fit the data.

Any suggestion which package that could help me?

Thanks in advance.

Cheers

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[R] Problem Installing rJava on MS Win 7 Platform

2011-04-07 Thread Stephen P Molnar

I'm a new R user and, frankly, this problem has got me stumped.

 

I'm using the 64 bit version of RGui and followed the installation
instructions on http://www.rforge.net/rJava/

 

Here is the result:

 

package 'rJava' successfully unpacked and MD5 sums checked

 library(rJava)

Error in utils::readRegistry(key, HLM, 2) : 

  Registry key 'Software\JavaSoft\Java Runtime Environment' not found

Error in utils::readRegistry(key, HLM, 2) : 

  Registry key 'Software\JavaSoft\Java Development Kit' not found

Error : .onLoad failed in loadNamespace() for 'rJava', details:

  call: fun(...)

  error: JAVA_HOME cannot be found from the Registry

Error: package/namespace load failed for 'rJava'

 

I don't Have the faintest idea as t what I've done, or failed to do.

 

Assistance will be greatly appreciated.

 

Thanks in advance.

 

Stephen P. Molnar, Ph.D.  Life is a
fuzzy set

Foundation for Chemistry Stochastic
and multivriate

http://www.FoundationForChemistry.com

 


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Re: [R] package


See
http://cran.opensourceresources.org/web/packages/index.html

--
Marcio Pupin Mello

Survey Engineer
Ph.D student in Remote Sensing
National Institute for Space Research (INPE) - Brazil
Laboratory of Remote Sensing in Agriculture and Forestry (LAF)
www.dsr.inpe.br/~mello


Em 4/7/2011 10:59 AM, Nelson Doki escreveu:

I'd like to get the package about Potts and, Swenson-Wang models.
Haw can I do that?
There is a package about Wolff model, too? If yes, how to get it?
Thanks.
Nelson Doki.

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[R] Regrouping data

2011-04-07 Thread Sandeepa Ramakrishnan










Dear R forum

I have just started my venture with R. While I am trying to learn R through the 
tutorials, I think the current problem I need to address to is beyond my 
knowledge about R.

I have a dataframe as defined below -


rat_dat = data.frame(name = c(A, A, A, B, B, B, C, C, C, D, 
D, D),
  period = c(31-Mar-11, 30-Mar-11, 29-Mar-11, 
31-Mar-11, 30-Mar-11, 29-Mar-11,  31-Mar-11, 30-Mar-11, 29-Mar-11, 
31-Mar-11, 30-Mar-11, 29-Mar-11),
cy_date = c(31-May-11,31-May-11, 31-May-11, 31-May-11, 31-May-11, 
31-May-11, 31-May-11, 31-May-11, 31-May-11, 31-May-11, 31-May-11, 
31-May-11),  
values = c(10.42, 10.46, 10.37, 12.62, 12.65, 12.61, 14.10, 14.12, 14.08, 
11.11, 11.15, 11.17))  


 rat_dat
   name    period   cy_date values
1 A 31-Mar-11 31-May-11  10.42
2 A 30-Mar-11 31-May-11  10.46
3 A 29-Mar-11 31-May-11  10.37
4 B 31-Mar-11 31-May-11  12.62
5 B 30-Mar-11 31-May-11  12.65
6 B 29-Mar-11 31-May-11  12.61
7 C 31-Mar-11 31-May-11  14.10
8 C 30-Mar-11 31-May-11  14.12
9 C 29-Mar-11 31-May-11  14.08
10    D 31-Mar-11 31-May-11  11.11
11    D 30-Mar-11 31-May-11  11.15
12    D 29-Mar-11 31-May-11  11.17

My actual data is too large. I need to 

(1) rearrange this input as 









period  
  cy_date
  A
  B
    C
    D

31-Mar-11
 31-May-11 
10.42
  12.62
14.10
11.11

30-Mar-11
 31-May-11 
10.46
  12.65
14.12
11.15

29-Mar-11
 31-May-11 
10.37
  12.61
14.08
11.17

(2) Also, I need to find the difference between (rat_dat$cy_date) - 
(rat_dat$period)

I hope I am able to put forward my requirement properly. If not, please forgive 
me.

Sandeepa

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Re: [R] Problem Installing rJava on MS Win 7 Platform

2011-04-07 Thread Uwe Ligges


You failed to install an appropriate Java version.

Best,
Uwe Ligges


On 07.04.2011 16:30, Stephen P Molnar wrote:

I'm a new R user and, frankly, this problem has got me stumped.



I'm using the 64 bit version of RGui and followed the installation
instructions on http://www.rforge.net/rJava/



Here is the result:



package 'rJava' successfully unpacked and MD5 sums checked


library(rJava)


Error in utils::readRegistry(key, HLM, 2) :

   Registry key 'Software\JavaSoft\Java Runtime Environment' not found

Error in utils::readRegistry(key, HLM, 2) :

   Registry key 'Software\JavaSoft\Java Development Kit' not found

Error : .onLoad failed in loadNamespace() for 'rJava', details:

   call: fun(...)

   error: JAVA_HOME cannot be found from the Registry

Error: package/namespace load failed for 'rJava'



I don't Have the faintest idea as t what I've done, or failed to do.



Assistance will be greatly appreciated.



Thanks in advance.



Stephen P. Molnar, Ph.D.  Life is a
fuzzy set

Foundation for Chemistry Stochastic
and multivriate

http://www.FoundationForChemistry.com




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Re: [R] [SOLVED] Two functions as parametrs of a function.



On Apr 7, 2011, at 10:47 AM, Duncan Murdoch wrote:

And then there's the source code, but if the manuals are unclear, it  
may be no clearer.



-- Duncan Murdoch ... after citing the the usual suspects.
   R-help (April 2011)

Fortune candidate?

--
David Winsemius, MD
West Hartford, CT

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Re: [R] K means algorithm C code

2011-04-07 Thread Uwe Ligges


See
Ligges, U. (2006): R Help Desk: Accessing the Sources. R News 6 (4), 43-45.

Best,
Uwe Ligges


On 07.04.2011 16:05, Jean-Eudes Dazard wrote:

Dear R user,





How can I get the C or C++ source code of the R_kmeans_MacQueen or
R_kmeans_Lloyd subroutines implemented in the R kmeans function of the R
stats-package {stats}? Would these be available as a C header file (*.h)
somewhere from the R installation???

Any help to get a C or C++ implementation of these clustering  algorithms
would be very welcome.



Thanks for your help.

Johnny



~~~

Jean-Eudes DAZARD, PhD.

Assistant Professor

Bioinformatics Division

Center for Proteomics and Bioinformatics

School Of Medicine, BRB 936

Case Western Reserve University

10900 Euclid Avenue

CLEVELAND, OHIO 44106 - 4988

Tel : 1-216-368-3157

Fax : 1-216-368-6846

e-mail : jxd...@case.edu

~~~




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[R] Assigning a larger number of levels to a factor that has fewer levels

2011-04-07 Thread Dimitri Liakhovitski

Hello!

I have larger and a smaller data frame with 1 factor in each - it's
the same factor:

large.frame-data.frame(myfactor=LETTERS[1:10])
small.frame-data.frame(myfactor=LETTERS[c(9,7,5,3,1)])
levels(large.frame$myfactor)
levels(small.frame$myfactor)
table(large.frame$myfactor)
table(small.frame$myfactor)

myfactor has 10 levels in large.frame and 5 levels in small.frame. All
5 levels in small.frame are present in large.frame.
How could I make levels(small.frame$myfactor) contain all the same
levels as levels(large.frame$myfactor)?
In other words, I want table(small.frame$myfactor) to have the same
number of entries as table(small.frame$myfactor) but with 5 zeros in
it:
A B C D E F G J I J
1 0 1 0 1 0 1 0 1 0

Thank you very much for your suggestions!
-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] Assigning a larger number of levels to a factor that has fewer levels

2011-04-07 Thread Ista Zahn

Hi Dimitri,
The factor() function allows you to specify the levels. So

small.frame$myfactor - factor(small.frame$myfactor, levels =
levels(large.frame$myfactor))

should do it.

Best,
Ista

On Thu, Apr 7, 2011 at 11:47 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have larger and a smaller data frame with 1 factor in each - it's
 the same factor:

 large.frame-data.frame(myfactor=LETTERS[1:10])
 small.frame-data.frame(myfactor=LETTERS[c(9,7,5,3,1)])
 levels(large.frame$myfactor)
 levels(small.frame$myfactor)
 table(large.frame$myfactor)
 table(small.frame$myfactor)

 myfactor has 10 levels in large.frame and 5 levels in small.frame. All
 5 levels in small.frame are present in large.frame.
 How could I make levels(small.frame$myfactor) contain all the same
 levels as levels(large.frame$myfactor)?
 In other words, I want table(small.frame$myfactor) to have the same
 number of entries as table(small.frame$myfactor) but with 5 zeros in
 it:
 A B C D E F G J I J
 1 0 1 0 1 0 1 0 1 0

 Thank you very much for your suggestions!
 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.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.




-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] R licence

2011-04-07 Thread Jeff Newmiller

If the discussion on the GPL Wikipedia page 
(http://en.wikipedia.org/wiki/GNU_General_Public_License ) is representative, 
you are not alone but nevertheless it is not clear-cut either. The debate 
revolves around the definition of derived work for software. If you fall into 
the camp that believes any call to a Linux-specific system call makes the 
calling program a derivative of Linux, then your example is clearly a 
violation. I think the majority view is that there are categories of linkage 
that remove the derived work designation even when the caller is useless 
without the callee. The GPLv3 implicitly acknowledges this as it addresses 
situations where the proprietary elements of a mixed system of proprietary and 
GPLed software impose constraints on substituting modified versions of the free 
software components (tivoization).

Short answer: the proprietary parts are asserted not to be derived works. You 
are free to disagree, but only the judges' opinions count.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Matthew Dowle mdo...@mdowle.plus.com wrote:

Duncan, Letting you know then that I just don't see how the first paragraph 
here : http://www.revolutionanalytics.com/downloads/gpl-sources.php is 
compatible with clause 2(b) here : http://www.gnu.org/licenses/gpl-2.0.html 
Perhaps somebody could explain why it is? Matthew Duncan Murdoch 
murdoch.dun...@gmail.com wrote in message news:4d9da9ff.9020...@gmail.com... 
 On 07/04/2011 7:47 AM, Matthew Dowle wrote:  Peter,   If the 
proprietary part of REvolution's product is ok, then surely  Stanislav's 
suggestion is too. No?   Revolution has said that they believe they follow 
the GPL, and they  haven't been challenged on that. If you think that they 
don't, you could  let an R copyright holder know what they're doing that's a 
license  violation.   My opinion of Stanislav's question is that he doesn't 
give enough  information to answer. If he is planning to distribute R as part 
of his  product, he needs to follow the GPL. If not, I don't think any R  
copyright ho
 lder
has anything to complain about.   Duncan Murdoch   Matthewpeter 
dalgaardpda...@gmail.com wrote in message  
news:be157cf5-9b4b-45a0-a7d4-363b774f1...@gmail.com... On Apr 7, 
2011, at 09:45 , Stanislav Bek wrote: Hi, is it possible 
to use some statistic computing by R in proprietary   software?   Our 
software is written in c#, and we intend to use   
http://rdotnet.codeplex.com/   to get R work there. Especially we want to 
use loess function. You need to take legal advice to be certain, but 
offhand I would say   that   this kind of circumvention of the GPL is 
_not_ allowed. It all depends on whether the end product is a 
derivative work, in   which   case, the whole must be distributed under 
a GPL-compatible licence.   The   situation around GPL-incompatible 
plug-ins or plug-ins interfacing to   R in   GPL -incompatible software 
is legally murky, but using R as a  
 
subroutine   library for proprietary code is clearly crossing the line, as 
far as I   can   tell. --   Peter Dalgaard   Center for 
Statistics, Copenhagen Business School   Solbjerg Plads 3, 2000 
Frederiksberg, Denmark   Phone: (+45)38153501   Email: pd@cbs.dk 
Priv: pda...@gmail.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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html 
and provide commented, minimal, self-contained, reproducible code. 


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[R] Adding text to page margin with lattice graphics

2011-04-07 Thread Dennis Fisher

R: 2.12.1
OS X

Colleagues

I am working with lattice graphics for the first time and I am confused by some 
aspects of controlling these graphics.  The most pressing issue is the 
following:

I can print the graphics without problems with the command:
print(MYFUNCTION(SOMEOBJECT)
However, I would like to add margin text (this terminology might not apply in 
this setting).  I am able to do so with the following command:
print(MYFUNCTION(SOMEOBJECT, page=function(page) grid.text(SOMETEXT, 
x=0.5, y=0.01)))

Two issues:
1.  The added text overlaps text in SOMEOBJECT.  How do I reserve space at the 
margins of the page in which I can print the added text? (the equivalent of 
par(mar=c(1,1,1,1)))

2.  I was able to construct the command with some assistance from a colleague.  
But, I remain at a loss to understand the syntax:
page=function(page), 
i.e., page is a function of itself?
Any help in interpreting the function would be appreciated.

Dennis

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

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[R] plyr workaround to converting by() to a data frame

2011-04-07 Thread Liviu Andronic

Dear all
Is there a clean plyr version of the following by() and do.call(rbind,
...) construct:
 df-data.frame(a=1:10,b=11:20,c=21:30,grp1=c(x,y),grp2=c(x,y),grp3=c(x,y))
 dfsum-by(df[c(a,b,c)], df[c(grp1,grp2,grp3)], range)
 as.data.frame(dfsum)
Error in as.data.frame.default(dfsum) :
  cannot coerce class 'by' into a data.frame
 do.call(rbind, dfsum)
 [,1] [,2]
[1,]1   29
[2,]2   30


Thank you
Liviu


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Re: [R] Assigning a larger number of levels to a factor that has fewer levels

2011-04-07 Thread Dimitri Liakhovitski

Thanks a lot, Ista!

On Thu, Apr 7, 2011 at 11:52 AM, Ista Zahn iz...@psych.rochester.edu wrote:
 Hi Dimitri,
 The factor() function allows you to specify the levels. So

 small.frame$myfactor - factor(small.frame$myfactor, levels =
 levels(large.frame$myfactor))

 should do it.

 Best,
 Ista

 On Thu, Apr 7, 2011 at 11:47 AM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!

 I have larger and a smaller data frame with 1 factor in each - it's
 the same factor:

 large.frame-data.frame(myfactor=LETTERS[1:10])
 small.frame-data.frame(myfactor=LETTERS[c(9,7,5,3,1)])
 levels(large.frame$myfactor)
 levels(small.frame$myfactor)
 table(large.frame$myfactor)
 table(small.frame$myfactor)

 myfactor has 10 levels in large.frame and 5 levels in small.frame. All
 5 levels in small.frame are present in large.frame.
 How could I make levels(small.frame$myfactor) contain all the same
 levels as levels(large.frame$myfactor)?
 In other words, I want table(small.frame$myfactor) to have the same
 number of entries as table(small.frame$myfactor) but with 5 zeros in
 it:
 A B C D E F G J I J
 1 0 1 0 1 0 1 0 1 0

 Thank you very much for your suggestions!
 --
 Dimitri Liakhovitski
 Ninah Consulting
 www.ninah.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.




 --
 Ista Zahn
 Graduate student
 University of Rochester
 Department of Clinical and Social Psychology
 http://yourpsyche.org




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] Regrouping data

2011-04-07 Thread Henrique Dallazuanna

Try this:

1)
xtabs(values ~ name + period + cy_date, rat_dat)


2)
as.Date(rat_dat$cy_date, '%d-%b-%y') - as.Date(rat_dat$period, '%d-%b-%y')

On Thu, Apr 7, 2011 at 9:03 AM, Sandeepa Ramakrishnan
sandeepa_ramakrish...@yahoo.com wrote:









 Dear R forum

 I have just started my venture with R. While I am trying to learn R through 
 the tutorials, I think the current problem I need to address to is beyond my 
 knowledge about R.

 I have a dataframe as defined below -


 rat_dat = data.frame(name = c(A, A, A, B, B, B, C, C, C, 
 D, D, D),
   period = c(31-Mar-11, 30-Mar-11, 
 29-Mar-11, 31-Mar-11, 30-Mar-11, 29-Mar-11,  31-Mar-11, 
 30-Mar-11, 29-Mar-11, 31-Mar-11, 30-Mar-11, 29-Mar-11),
 cy_date = c(31-May-11,31-May-11, 31-May-11, 31-May-11, 31-May-11, 
 31-May-11, 31-May-11, 31-May-11, 31-May-11, 31-May-11, 31-May-11, 
 31-May-11),
 values = c(10.42, 10.46, 10.37, 12.62, 12.65, 12.61, 14.10, 14.12, 14.08, 
 11.11, 11.15, 11.17))


 rat_dat
    name    period   cy_date values
 1 A 31-Mar-11 31-May-11  10.42
 2 A 30-Mar-11 31-May-11  10.46
 3 A 29-Mar-11 31-May-11  10.37
 4 B 31-Mar-11 31-May-11  12.62
 5 B 30-Mar-11 31-May-11  12.65
 6 B 29-Mar-11 31-May-11  12.61
 7 C 31-Mar-11 31-May-11  14.10
 8 C 30-Mar-11 31-May-11  14.12
 9 C 29-Mar-11 31-May-11  14.08
 10    D 31-Mar-11 31-May-11  11.11
 11    D 30-Mar-11 31-May-11  11.15
 12    D 29-Mar-11 31-May-11  11.17

 My actual data is too large. I need to

 (1) rearrange this input as









 period
   cy_date
   A
   B
     C
     D

 31-Mar-11
  31-May-11
 10.42
   12.62
 14.10
 11.11

 30-Mar-11
  31-May-11
 10.46
   12.65
 14.12
 11.15

 29-Mar-11
  31-May-11
 10.37
   12.61
 14.08
 11.17

 (2) Also, I need to find the difference between (rat_dat$cy_date) - 
 (rat_dat$period)

 I hope I am able to put forward my requirement properly. If not, please 
 forgive me.

 Sandeepa

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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] plyr workaround to converting by() to a data frame

2011-04-07 Thread rmailbox

Is this what you are looking for?

ddply ( .variables = c(grp1, grp2, grp3 ), .data = df, .fun = 
plyr::summarize, abmin = min(c(a, b, c) ), abmax = max ( c ( a, b, c)  ) )

- Original message -
From: Liviu Andronic landronim...@gmail.com
To: r-help@r-project.org Help r-help@r-project.org
Date: Thu, 7 Apr 2011 18:39:30 +0200
Subject: [R] plyr workaround to converting by() to a data frame

Dear all
Is there a clean plyr version of the following by() and do.call(rbind,
...) construct:
 df-data.frame(a=1:10,b=11:20,c=21:30,grp1=c(x,y),grp2=c(x,y),grp3=c(x,y))
 dfsum-by(df[c(a,b,c)], df[c(grp1,grp2,grp3)], range)
 as.data.frame(dfsum)
Error in as.data.frame.default(dfsum) :
  cannot coerce class 'by' into a data.frame
 do.call(rbind, dfsum)
 [,1] [,2]
[1,]1   29
[2,]2   30

Thank you
Liviu

-- 
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[R] Randomisation tests

2011-04-07 Thread Ben Ward

Hi,

I want to ue a randomisation test, to compare two sample means, for a 
small set of data, I've looked at towt.permutation, perm.test and 
permtest, as well as permTS(), which is the best one to use for such a 
simple purpose? I know perm.test package isn't used anymore and coins 
recommended, but reading the manual I couldnt find the function most 
like it, ore more likely I didn't understand the lingo used to know it's 
what I wanted.

Thanks,
Ben.

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[R] An extention of outer() ?

2011-04-07 Thread Bogaso Christofer

Dear all, as per my understanding, outer() function is to facilitate
2-dimensional function evaluation, like to evaluate f(x,y) for different
values of x and y. However I have slightly modified version of that, where x
is a matrix with, say, 5 rows and 2 columns and y is a vector. Let take
following example:

 

fn1 - function(x1, x2, y) return (x1+x2+y)

 

x - matrix(1:10, nc = 2)

y - 1:5

 

with this input, I want to construct a 5x5 matrix whose (1,1)th element will
be : fn1(x[1,1], x[1,2], y[1]), (1,2)th element will be : fn1(x[1,1],
x[1,2], y[2]). (1,5)th element will be : fn1(x[1,1], x[1,2],
y[5]).similarly (2,1)th element will be : fn1(x[2,1], x[2,2],
y[1])etc

 

If my 'x' object is a vector then I can easily use outer() function
effectively. However for my present case, I am looking for some ***outer()
like function*** to get the result effectively. Can somebody help me on how
can I do that without using any for loop?

 

Thanks and regards,

 


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Re: [R] An extention of outer() ?

2011-04-07 Thread Henrique Dallazuanna

Try this:

aperm(outer(x, y, fn1, y), c(3, 1, 2))[,,2]


On Thu, Apr 7, 2011 at 3:21 PM, Bogaso Christofer
bogaso.christo...@gmail.com wrote:
 Dear all, as per my understanding, outer() function is to facilitate
 2-dimensional function evaluation, like to evaluate f(x,y) for different
 values of x and y. However I have slightly modified version of that, where x
 is a matrix with, say, 5 rows and 2 columns and y is a vector. Let take
 following example:



 fn1 - function(x1, x2, y) return (x1+x2+y)



 x - matrix(1:10, nc = 2)

 y - 1:5



 with this input, I want to construct a 5x5 matrix whose (1,1)th element will
 be : fn1(x[1,1], x[1,2], y[1]), (1,2)th element will be : fn1(x[1,1],
 x[1,2], y[2]). (1,5)th element will be : fn1(x[1,1], x[1,2],
 y[5]).similarly (2,1)th element will be : fn1(x[2,1], x[2,2],
 y[1])etc



 If my 'x' object is a vector then I can easily use outer() function
 effectively. However for my present case, I am looking for some ***outer()
 like function*** to get the result effectively. Can somebody help me on how
 can I do that without using any for loop?



 Thanks and regards,




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-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O

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Re: [R] Grid on Map

2011-04-07 Thread Jaimin Dave

Hi,
I tried it using abline but it is not showing any o/p(i.e no grids are
formed on map). I did following :

 map('state','new york',proj='azequalarea',orient=c(42.5,-74.56,0))
 abline(h=seq(41, 44, length=3))
 abline(v=seq(-79, -76, length=3))

I also tried using map.grid() but in some cases it is dividing into exact 4
quadrants.

On Wed, Apr 6, 2011 at 2:49 PM, MacQueen, Don macque...@llnl.gov wrote:

 Possibly something similar to

  abline(v=seq(long.min, long.max, length=3)
  abline(h=seq(lat.min, lat.max, length=3)

 ?

 The above will add vertical and horizontal lines to an existing plot, and
 assumes that the plot is in long/lat coordinates. Of course, this ignores
 the fact that long/lat is not a cartesian coordinate system.

 (can't provide more detail without more information)

 -Don

 --
 Don MacQueen

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





 -Original Message-
 From: Jaimin Dave davejaim...@gmail.com
 Date: Mon, 4 Apr 2011 18:39:45 -0700
 To: r-help@r-project.org r-help@r-project.org
 Subject: [R] Grid on Map

 I am new to R.I want to draw grid from a csv file which contains latitude
 minimum ,latitude maximum ,longitude minimum ,longitude maximum.The grid
 should be divided into exactly 4 quadrants. The map is of NY state of
 USA. I
 want to know how can I do it.
 Help would be appreciated.
 
 Thanks
 Jaimin
 
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[R] classification

2011-04-07 Thread array chip

Dear all, this is not a pure R question, but really about how to set up a 
multinomial logistic regression model to do a multi-class classification. I 
would really appreciate if any of you would give me some of your thoughts and 
recommendation.

Let's say we have 3-class classification problem: A, B and C. I have certain 
number of samples, with each sample, I have 3 variables (Xa, Xb and Xc). The 
trick here is that these 3 variables measures the extent of the likelihood of 
the samples being class A, B and C, i.e., Xa for class A, Xb for class B and Xc 
for class C. For a given sample i, we can simply make a rough prediction based 
on the values of Xa, Xb and Xc. For example:

 for sample 1, Xa=10, Xb=50, Xc=15, then most likely I would predict sample 1 
as 
class B.

Then I have another set of variables Ya, Yb and Yc doing similar things.

I can construct a dataset as below:
   Xa   XbXc Ya   Yb   Yc  class
sample 1   10   50150.2  0.8  0.1   B
sample 2   84 6 0.7  0.5  0.3   A
:
:


and then make a model fit-multinom(class~Xa+Xb+Xc+Ya+Yb+Yc)

But my understanding is that this model is not working in a way of by simply 
looking at each row of the data and pick the class that has the best Xs and/or 
Ys. In leave-one-out, sometimes it picks up a class that apparently is not a 
winner if I compare across Xs and Ys.

Greatly appreciate if anyone can suggest a more sensible way to construct the 
data and/or a different way of thinking of the problem at all.

John
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[R] Panel data - replicating Stata's xtpcse in R

2011-04-07 Thread Florian Markowetz

Dear list,
 
I am trying to replicate an econometrics study that was orginally done in 
Stata. (Blanton and Blanton. 2009. A Sectoral Analysis of Human Rights and FDI: 
Does Industry Type Matter?  International Studies Quarterley 53 (2):469 - 493.) 
The model I try to replicate is in Stata given as

xtpcse total_FDI lag_total ciri human_cap worker_rts polity_4 market income 
econ_growth log_trade fix_dollar fixed_xr xr_fluct lab_growth english, pairwise 
corr(ar1)

According to the paper, this is an OLS regression with panel corrected standard 
errors including a lagged dependent variable (lag_total is total_FDI t-1) and 
controlling first order correlations within each panel (corr(ar1)).

The BIG QUESTION is how to replicate this line in R.

Econometrics is a new field to me, but a bit of searching showed that  packages 
like plm, nlme, pcse should be able to handle this kind of problem. In 
particular, function gls() uses auto-correlation structure and pcse() corrects 
the standard errors of the fitted model. Below is some code to show what I have 
done, and some problems I ran into.

## setup and load data from web
library(foreign)
library(nlme)
library(pcse)
D - 
read.dta(http://umdrive.memphis.edu/rblanton/public/ISQ_data/blanton_isq08_data.dta;)
D[544,year] - 2005 ## fixing an unexpected NA in the year column

## Model formula
form - total_FDI ~ lag_total + ciri + human_cap + worker_rts + polity_4 + 
market_size + income + econ_growth + log_trade + fixed_xr + fix_dollar + 
xr_fluct + english + lab_growth

## Model 1: no auto-correlation
res1  - gls(model=form, data=D,correlation=NULL,na.action=na.omit)
coefficients(res1)

## Model 2: with auto-correlation
corr - corAR1(.1,~1|c_name)
res2  - gls(model=form, data=D,correlation=corr,na.action=na.omit)
coefficients(res2)

Now, I know from the paper how the Stata coefficients looked like.  For 
example, for log_total it should be .852 and for market_size .21 (these were 
the two significant ones). The result of Model1 is closer to this than the 
result of Model 2, but there is still quite a gap.

The goal is to do OLS on panel data with AR(1) and PCSE - am I on the right 
track here? More specifically:

Question 1: Auto-correlation
- how to specify the parameter 'value' in corAR1 (the .1 above is completely 
arbitrary) 
- Any other ideas how to translate Stata's corr(AR1) into R? (I'm not even 
completely sure what Stata does there and didn't find any details in the online 
manuals)

Question 2: PCSE
- the pcse function seems to work on objects of class 'lm' only. Any way to use 
it for gls-objects?

Any help is greatly appreciated!
Florian

--
Florian Markowetz

Cancer Research UK
Cambridge Research Institute
Li Ka Shing Centre
Robinson Way, Cambridge, CB2 0RE, UK

phone: +44 (0) 1223 40 4315
email: florian.markow...@cancer.org.uk
web  : http://www.markowetzlab.org
skype: florian.markowetz

This communication is from Cancer Research UK. Our website is at 
www.cancerresearchuk.org. We are a registered charity in England and Wales 
(1089464) and in Scotland (SC041666) and a company limited by guarantee 
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Re: [R] Problem Installing rJava on MS Win 7 Platform

2011-04-07 Thread Stephen P Molnar

Thanks for the reply.

Yes, you are quite correct, but:

 install.packages('rJava')
Warning in install.packages(rJava) :
  argument 'lib' is missing: using
'C:\Users\Stephen\Documents/R/win64-library/2.11'
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository
http://www.stats.ox.ac.uk/pub/RWin/bin/windows64/contrib/2.11
trying URL 'http://cran.case.edu/bin/windows64/contrib/2.11/rJava_0.8-5.zip'
Content type 'application/zip' length 658913 bytes (643 Kb)
opened URL
downloaded 643 Kb

package 'rJava' successfully unpacked and MD5 sums checked

The downloaded packages are in
C:\Users\Stephen\AppData\Local\Temp\Rtmpcy0hAb\downloaded_packages

install.packages('C;\Users\Stephen\AppData\Local/Temp/Rtmpcy0hAb\downloaded_
$
Error: '\U' used without hex digits in character string starting C;\U


So, obviously, at this point I am attempting something that R does not like.

Stephen P. Molnar, Ph.D.  Life is a
fuzzy set
Foundation for Chemistry Stochastic
and multivriate
http://www.FoundationForChemistry.com


-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] 
Sent: Thursday, April 07, 2011 11:34 AM
To: Stephen P Molnar
Cc: r-help@r-project.org
Subject: Re: [R] Problem Installing rJava on MS Win 7 Platform

You failed to install an appropriate Java version.

Best,
Uwe Ligges


On 07.04.2011 16:30, Stephen P Molnar wrote:
 I'm a new R user and, frankly, this problem has got me stumped.



 I'm using the 64 bit version of RGui and followed the installation
 instructions on http://www.rforge.net/rJava/



 Here is the result:



 package 'rJava' successfully unpacked and MD5 sums checked

 library(rJava)

 Error in utils::readRegistry(key, HLM, 2) :

Registry key 'Software\JavaSoft\Java Runtime Environment' not found

 Error in utils::readRegistry(key, HLM, 2) :

Registry key 'Software\JavaSoft\Java Development Kit' not found

 Error : .onLoad failed in loadNamespace() for 'rJava', details:

call: fun(...)

error: JAVA_HOME cannot be found from the Registry

 Error: package/namespace load failed for 'rJava'



 I don't Have the faintest idea as t what I've done, or failed to do.



 Assistance will be greatly appreciated.



 Thanks in advance.



 Stephen P. Molnar, Ph.D.  Life is
a
 fuzzy set

 Foundation for Chemistry
Stochastic
 and multivriate

 http://www.FoundationForChemistry.com




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[R] error reading data help please

2011-04-07 Thread Ouattara

Dear Readers,

Sorry to bother you.

Have been try to read my data into R (something I have done in the past) but
having an error message with this one.

Any possible help please?

Thanks.

Oz

 

This is the error I am getting (please see bold):

 setwd(D:/Replication/)

library(gmp)

 dataFrame2=read.table(data.txt,header=T)

Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,
: 

  line 1 did not have 58 elements


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[R] df with max function applied to 6 lags of a variable?!?

2011-04-07 Thread Rita Carreira


Hello everyone!
I have a data frame of 136 variables with 270 observations. I would like to 
create a new data frame such that each element of that data frame contains the 
maximum value of the 6 prior lags of the initial data frame. So for example, if 
my original data frame, A, were
A1=c(7.72, 7.94, 7.56, 7.54, 0.93, 0.59, 7.21, 8.00, 7.19, 7.57)A2=c(4.27, 
3.70, 3.80, 3.67, 3.83, 3.95, 4.02, 2.06, 3.28, 2.17)A3=c(4.16, 2.91, 3.89, 
3.73, 4.00, 3.81, 3.86, 1.88, 3.74, 3.75)A4=c(3.56, 3.27, 3.42, 3.32, 3.48, 
3.54, 3.60, 2.04, 3.38, 3.37)A5=c(8.24, 8.67, 8.49, 0.00, 9.39, 0.00, 8.49, 
8.94, 0.00, 0.00)A6=c(4.28, 4.18, 4.20, 4.14, 4.40, 4.20, 4.20, 4.15, 4.17, 
4.16)  
A - cbind(A1, A2, A3, A4, A5, A6)
My new data frame, Amax, where each element is amax(i,j)=max[a(i-1,j), 
a(i-2,j), a(i-3,j), a(i-4,j), a(1-5,j), a(i-6,j)] would be:NA   NA   NA   NA   
NANA NA   NA   NA   NA   NANA NA   NA   NA   NA   NANA NA   NA   NA 
  NA   NANA NA   NA   NA   NA   NANA NA   NA   NA   NA   NANA 7.94
  4.27
  4.16
  3.56
  9.39
  4.40 7.94
  4.02
  4.00
  3.60
  9.39
  4.40 8.00
  4.02
  4.00
  3.60
  9.39
  4.40 8.00
  4.02
  4.00
  3.60
  9.39
  4.40 8.00
  4.02
  4.00
  3.60
  9.39
  4.40  
So the question is: what is the most efficient way of doing this in R?
Thanks!Rita  If you think education is 
expensive, try ignorance--Derek Bok


  
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Re: [R] error reading data help please

2011-04-07 Thread Joshua Wiley

Hi Oz,

On Thu, Apr 7, 2011 at 10:47 AM, Ouattara b.ouatt...@swansea.ac.uk wrote:
 Dear Readers,

 Sorry to bother you.

 Have been try to read my data into R (something I have done in the past) but
 having an error message with this one.

 Any possible help please?

 Thanks.

 Oz

 This is the error I am getting (please see bold):

This listserv only uses plain text emails (which is how the posting
guide respectfully requests you submit all queries).  HTML emails are
converted to plain text, so bold does not come through to us.


 setwd(D:/Replication/)

library(gmp)

 dataFrame2=read.table(data.txt,header=T)

 Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,
 :
  line 1 did not have 58 elements

Assuming the data file is not malformed, try using a different
delimitter.  read.table() defaults to using space (sep =  ) I think,
but your data may be tab delimitted (sep = \t), or comma (sep =
,), or ...

If you are not certain, open your data in a regular text editor, and
see what sort of stuff is in between each cell (i.e., how can you tell
where one cell stops and another starts).  I think space, commas, and
tabs, are the most common, but there are many others.  More generally,
look through the documentation for ?read.table to get ideas for what
different arguments you might try to properly read in your data.

Hope this helps,

Josh



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sends us the plain text.

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^^ This is where you could have read that the list only sends plain text.

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


-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] Problem Installing rJava on MS Win 7 Platform