[R] Factored ARMA models
I am trying to replicate ARIMA models in R from SAS. Specifically - Factored ARMA models, which I can do very easily in SAS ( http://v8doc.sas.com/sashtml/ets/chap7/sect11.htm). I would like to know if there is a way to do this in any of the R packages. Thanks in advance, Indrajit [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Factored ARMA models
Hello, Please have a look at: http://cran.r-project.org/web/views/TimeSeries.html Regards, Pascal 2013/9/20 Indrajit Sengupta indrajitsg2...@gmail.com I am trying to replicate ARIMA models in R from SAS. Specifically - Factored ARMA models, which I can do very easily in SAS ( http://v8doc.sas.com/sashtml/ets/chap7/sect11.htm). I would like to know if there is a way to do this in any of the R packages. Thanks in advance, Indrajit [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-3.0.1 g77 errors
On 19/09/2013 15:28, Prigot, Jonathan wrote: Sadly, I am limited to the Solaris 10 system. I wish that I could use Linux, the world uses it. What does that have to do with this? The CRAN check farm uses gfortran 4.8 on a Solaris 10 system. g77 is not the native compiler there, nor is that used for most binary software. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] [R-pkgs] depmixS4 version 1.3-0 on CRAN
Package news (see below for general description of functionality) depmixS4 version 1.3-0 has been released on CRAN. See the NEWS file for an overview of all changes. The most important user-visible changes are: 1) more compact pretty-printing of parameters in print/summary of (dep)mix objects (following lm/glm style of presenting results) 2) some speed improvements in the EM algorithm, most notable in large data/models 3) EM has an optional argument to use the classification likelihood instead of the usual likelihood; this can be useful as a means of starting value generation; use with caution as results are often unstable. Best, happy mixing, Ingmar Maarten Package general information depmixS4 is a framework for specifying and fitting dependent mixture models, otherwise known as hidden or latent Markov models. Optimization is done with the EM algorithm or optionally with Rdonlp2 when (general linear (in-)equality) constraints on the parameters need to be incorporated. Models can be fitted on (multiple) sets of observations. The response densities for each state may be chosen from the GLM family, or a multinomial. User defined response densities are easy to add; for the latter an example is given for the ex-gauss distribution as well as the multivariate normal distribution. Mixture or latent class (regression) models can also be fitted; these are the limit case in which the length of observed time series is 1 for all cases. ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need help to find out the name of my columns and rows in a data file
Hi -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of David Winsemius Sent: Friday, September 20, 2013 6:39 AM To: ivanc010 Cc: r-help@r-project.org Subject: Re: [R] Need help to find out the name of my columns and rows in a data file On Sep 19, 2013, at 3:59 PM, ivanc010 wrote: I've been assigned homework to analyze a file. The R package is car. The specific data file is Florida. So, I did the usual stuff: library(car) data(Florida) summary(Florida) My specific assignment is to run a t-test between GORE and BUSH. (This file has information on the 2000 election.) To run my t-test, my code must be something analogues to: t.test(case0102$Salary[case0102$Sex==Female],case0102$Salary[case010 2$Sex==Male]) Unfortunately, for my Florida data I can't find the analogues titles of the rows and columns (i.e. Sex and Salary). Well, can not resist. I believe both are Male. see ?str Regards Petr Please read the Posting Guide. Many of the participants in R-Help are academics and they are asking _their_ students not to post howmework questions here. The Posting Guide lays out the reasoning. Please read it, ... and don't expect replies. -- View this message in context: http://r.789695.n4.nabble.com/Need-help-to-find-out-the-name-of-my- col umns-and-rows-in-a-data-file-tp4676534.html Sent from the R help mailing list archive at Nabble.com. Nabble is NOT the archive for R-help. -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Hope u have some time for 2 more questions
Hi,
as far as I know, there is no limitation on data size in regard to foreach. You
should reserve though enough memory for your application on the cluster (via
ulimit -s unlimited and ulimit -v unlimited).
Furthermore I would check the following:
Check if there are two versions of R on the cluster/your home directory on the
frontend (LSF loads this frontend environment and uses the R version installed
there). If you have two R executables (R and R64) make sure you use the 64bit
version.
Run R and call memory.limit() to see what are the limits of memory in your
system.
If this is limited to sizes below your needed sizes, increase it by calling R
in the LSF script with the options --max-mem-size=YourSize and if you get
errors of kind cannot allocate vector of size you should also use
--max-vsize=YourVSize.
Then, check if there is a memory leak in your application: If you compiled R
with the --enable-memory-profiling you can use Rprof to do this otherwise you
must rely on profiling instruments given by the cluster environment (I think
you work there as well with modules, so type in the shell 'module avail' for
listing available modules).
If you detect a memory leak or if you see, that at certain points in your
algorithms some objects are not used anymore call rm(ObjectName) and gc() for
garbage collection.
To your nested loop using foreach: That is a highly delicate issue in parallel
computing and for the foreach syntax I refer to the must-read
http://cran.r-project.org/web/packages/foreach/vignettes/nested.pdf.
Using nested loops should be considered carefully in regard to organizing the
nesting. In C++ you have the ability to determine how many cores should work on
which loop. In the foreach environment using doMC this seems to me not
possible.
And, please keep the discussion to the r-help mailing list, so others can learn
from it and researchers with more experience can also leave comments.
Best
Simon
On Sep 19, 2013, at 9:24 PM, pko...@bgc-jena.mpg.de wrote:
Hi again,
if you have some time I would like to bother you again with 2 more questions.
After your response the parallel code is working perfect but when I implement
that to the real case (big matrices) I get an error for not numeric dimension
and i guess that again it returns NULL or something. Are you aware if foreach
loop can handle only a certain size objects? the equation that I am using
includes 3 objects with 2Gb size each.
The second question has to deal with the cores that foreach uses. Although I
am asking to our cluster (LSF) to give me certain number of cpus, and also i
am specifing that with
library(doMC)
registerDoMC(n)
it seems from the top command that I am using all the cores. I am using 2
foreach as nest foreach(i in 1:16){
foreach(j in 1:10) etc etc..
maybe i should do something with this kind of nest? I am not aware about that.
I am sorry for the long text , and thank you for your nice solution
_
Sent from http://r.789695.n4.nabble.com
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[R] Subcripting matrix
Hello again, I have one question on subscripting matrix. Let say I have following matrix: Mat - matrix(1:9, 3) colnames(Mat) - c(a, b, a) Mat a b a [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 Now I want to fetch data for colnames 'a'.I did following: Mat[, a] [1] 1 2 3 However it is not taking second 'a' colume. Basically I expected to get 1st and 3rd columns Can somebody tell me how to achieve that? Thanks and regards, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subcripting matrix
On 20-09-2013, at 12:52, Christofer Bogaso bogaso.christo...@gmail.com wrote: Hello again, I have one question on subscripting matrix. Let say I have following matrix: Mat - matrix(1:9, 3) colnames(Mat) - c(a, b, a) Mat a b a [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 Now I want to fetch data for colnames 'a'.I did following: Mat[, a] [1] 1 2 3 However it is not taking second 'a' colume. Basically I expected to get 1st and 3rd columns Can somebody tell me how to achieve that? I think this is just silly. How about Mat[,which(colnames(Mat)==a)] Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subcripting matrix
Hi, it seems the following works as needed... Mat[,colnames(Mat)==a] Olivier. On Fri, 20 Sep 2013 16:22:37 +0530 Christofer Bogaso bogaso.christo...@gmail.com wrote: Hello again, I have one question on subscripting matrix. Let say I have following matrix: Mat - matrix(1:9, 3) colnames(Mat) - c(a, b, a) Mat a b a [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 Now I want to fetch data for colnames 'a'.I did following: Mat[, a] [1] 1 2 3 However it is not taking second 'a' colume. Basically I expected to get 1st and 3rd columns Can somebody tell me how to achieve that? Thanks and regards, __ 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. -- Olivier Crouzet, PhD Laboratoire de Linguistique -- EA3827 Université de Nantes Chemin de la Censive du Tertre - BP 81227 44312 Nantes cedex 3 France phone:(+33) 02 40 14 14 05 (lab.) (+33) 02 40 14 14 36 (office) fax: (+33) 02 40 14 13 27 e-mail: olivier.crou...@univ-nantes.fr http://www.lling.univ-nantes.fr/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Factored ARMA models
Can you point to any function in specific? On Fri, Sep 20, 2013 at 12:10 PM, Pascal Oettli kri...@ymail.com wrote: Hello, Please have a look at: http://cran.r-project.org/web/views/TimeSeries.html Regards, Pascal 2013/9/20 Indrajit Sengupta indrajitsg2...@gmail.com I am trying to replicate ARIMA models in R from SAS. Specifically - Factored ARMA models, which I can do very easily in SAS ( http://v8doc.sas.com/sashtml/ets/chap7/sect11.htm). I would like to know if there is a way to do this in any of the R packages. Thanks in advance, Indrajit [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] binary symmetric matrix combination
Hi Elio,
Try this:
library(stringr)
lines1-str_trim(gsub(\t, ,readLines(elio.txt)))
lst1-lapply(split(lines1,cumsum(lines1==)),function(x) x[x!=])
lst2- lapply(lst1[lapply(lst1,length)0],function(x)
as.matrix(read.table(text=x,row.names=1)))
names(lst2)- paste0(m,seq_along(lst2))
dat- do.call(rbind,lapply(names(lst2),function(x) {x1- lst2[[x]];
cbind(expand.grid(rep(list(colnames(x1)),2),stringsAsFactors=FALSE),value=as.vector(x1))}))
library(reshape2)
res- dcast(dat,Var1~Var2,value.var=value,sum)
row.names(res)- res[,1]
res- as.matrix(res[,-1])
dim(res)
#[1] 14 14
A.K.
From: Elio Shijaku sel...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, September 20, 2013 3:38 AM
Subject: Re: binary symmetric matrix combination
Hi Arun,
Let me explain my point. With your great help, I was able to create matrices
such as m1, m2, etc. My initial objective was to combine the binary matrices
and I succeeded in that again with your help. The only problem I have left is
that since I have about 350 matrices to combine for a single giant symmetric
binary matrix of 470x470 size and I have to repeat this 22 times (got 22 giant
matrices to build), I was hoping to not have to manually input each matrix but
to use your list2 command to get them all from Excel. Now the problem is how to
get the elements from list2 to then apply the commands:
Out1-
as.matrix(read.table(text=y1 g24 c1 c2 l17 h4 s2 s30 e5
l15,sep=,header=TRUE))
names1-unique(c(colnames(m1),colnames(m2),colnames(m3),colnames(m4),colnames(m5)))
Out3-matrix(0,length(names1),length(names1),dimnames=list(names1,names1))
vecOut-paste0(colnames(Out3)[col(Out3)],rownames(Out3)[row(Out3)])
lst1-sapply(paste0(m,1:5),function(x)
{x1- get(x); x2-paste0(colnames(x1)[col(x1)],rownames(x1)[row(x1)]);
match(x2,vecOut)})
lst2- list(m1,m2,m3,m4,m5)
N- length(lst1)
fn1- function(N,Out){
i=1
while(i=N){
Out[lst1[[i]]]-lst2[[i]]
i-i+1
}
Out
}
fn1(N,Out3)
Thanks a lot!!
Best,
Elio
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Re: [R] Averate memory usage trend
Replying with some code. input- readLines(textConnection( 6.2 httpd (18) 4.0 httpd (11) 4.0 httpd (11) 3.3 httpd (9) 4.2 httpd (12) 4.2httpd (12) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) )) data-input[input!=] data [1] 6.2 httpd (18) 4.0 httpd (11) 4.0 httpd (11) 3.3 httpd (9) [5] 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) [9] 4.2 httpd (12) m - t(sapply(1:1,function(x) unlist(strsplit(data, + [,1] [,2] [,3][,4] [,5] [,6] [,7][,8] [,9] [,10] [,11] [,12] [,13] [,14] [1,]6.2 httpd (18)4.0 httpd (11)4.0 httpd (11) 3.3 [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [1,] httpd (9) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [1,] (12) 4.2 httpd (12) 4.2 httpd (12) What I want to do is this ? 6.2 httpd 18 4.0 httpd 11 4.0 httpd 11 3.3 httpd9) 4.2 httpd 12 4.2httpd12 4.2 httpd 12 4.2 httpd 12 4.2 httpd 12 Thanks, Mohan From: mohan.radhakrish...@polarisft.com To: r-help@r-project.org Date: 09/20/2013 10:44 AM Subject:[R] Averate memory usage trend Sent by:r-help-boun...@r-project.org Hi, I would like to understand how to draw a graph to find out the average memory used by a single httpd process given these details collected over a period of several days. I code R but this is about a method to find an average memory utilization. I believe I have enough data. How should I statistically calculate this ? The figures inside braces are the total httpd processes at that time. Will there be an error margin if I have combined totals like this ? Sample data : Private Shared RAM used Program 4.0 MiB3.3 MiB7.2 MiB httpd (11) 3.3 MiB3.0 MiB 6.3 MiB httpd (9) 4.2 MiB3.3 MiB 7.5 MiB httpd (12) Thanks, Mohan This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averate memory usage trend
library(stringr) read.table(text=gsub([()],,str_trim(input[input!=])),sep=,header=FALSE,stringsAsFactors=FALSE) # V1 V2 V3 #1 6.2 httpd 18 #2 4.0 httpd 11 #3 4.0 httpd 11 #4 3.3 httpd 9 #5 4.2 httpd 12 #6 4.2 httpd 12 #7 4.2 httpd 12 #8 4.2 httpd 12 #9 4.2 httpd 12 A.K. - Original Message - From: mohan.radhakrish...@polarisft.com mohan.radhakrish...@polarisft.com To: r-help@r-project.org Cc: Sent: Friday, September 20, 2013 9:15 AM Subject: Re: [R] Averate memory usage trend Replying with some code. input- readLines(textConnection( 6.2 httpd (18) 4.0 httpd (11) 4.0 httpd (11) 3.3 httpd (9) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) )) data-input[input!=] data [1] 6.2 httpd (18) 4.0 httpd (11) 4.0 httpd (11) 3.3 httpd (9) [5] 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd (12) [9] 4.2 httpd (12) m - t(sapply(1:1,function(x) unlist(strsplit(data, + [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [1,] 6.2 httpd (18) 4.0 httpd (11) 4.0 httpd (11) 3.3 [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [1,] httpd (9) 4.2 httpd (12) 4.2 httpd (12) 4.2 httpd [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [1,] (12) 4.2 httpd (12) 4.2 httpd (12) What I want to do is this ? 6.2 httpd 18 4.0 httpd 11 4.0 httpd 11 3.3 httpd 9) 4.2 httpd 12 4.2 httpd 12 4.2 httpd 12 4.2 httpd 12 4.2 httpd 12 Thanks, Mohan From: mohan.radhakrish...@polarisft.com To: r-help@r-project.org Date: 09/20/2013 10:44 AM Subject: [R] Averate memory usage trend Sent by: r-help-boun...@r-project.org Hi, I would like to understand how to draw a graph to find out the average memory used by a single httpd process given these details collected over a period of several days. I code R but this is about a method to find an average memory utilization. I believe I have enough data. How should I statistically calculate this ? The figures inside braces are the total httpd processes at that time. Will there be an error margin if I have combined totals like this ? Sample data : Private Shared RAM used Program 4.0 MiB 3.3 MiB 7.2 MiB httpd (11) 3.3 MiB 3.0 MiB 6.3 MiB httpd (9) 4.2 MiB 3.3 MiB 7.5 MiB httpd (12) Thanks, Mohan This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Renaming variables
Hi, I guess this is pretty basic. I have 25 variables in the data file (name: score), i.e. X1,X2,.,X25. I dont want to use score$X1, score$X2 everytime I use these variables. Is there a way I can rename all these variables as simply X1,X2,.X25 without writing 25 lines of code, one line for renaming each variable (eg: X1=score.X1 X2=score.X2 and so on) ? Thanks for your help. Regards, Preetam -- Preetam Pal (+91)-9432212774 M-Stat 2nd Year, Room No. N-114 Statistics Division, C.V.Raman Hall Indian Statistical Institute, B.H.O.S. Kolkata. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need help to find out the name of my columns and rows in a data file
You are close--think names, not titles, as in rownames or colnames (no reason to completely spell out column). Summary already gave you the column names, so type ?rownames to learn more. Clint BowmanINTERNET: cl...@ecy.wa.gov Air Quality Modeler INTERNET: cl...@math.utah.edu Department of Ecology VOICE: (360) 407-6815 PO Box 47600FAX:(360) 407-7534 Olympia, WA 98504-7600 USPS: PO Box 47600, Olympia, WA 98504-7600 Parcels:300 Desmond Drive, Lacey, WA 98503-1274 On Thu, 19 Sep 2013, ivanc010 wrote: I've been assigned homework to analyze a file. The R package is car. The specific data file is Florida. So, I did the usual stuff: library(car) data(Florida) summary(Florida) My specific assignment is to run a t-test between GORE and BUSH. (This file has information on the 2000 election.) To run my t-test, my code must be something analogues to: t.test(case0102$Salary[case0102$Sex==Female],case0102$Salary[case0102$Sex==Male]) Unfortunately, for my Florida data I can't find the analogues titles of the rows and columns (i.e. Sex and Salary). Please help. -- View this message in context: http://r.789695.n4.nabble.com/Need-help-to-find-out-the-name-of-my-columns-and-rows-in-a-data-file-tp4676534.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Renaming variables
You haven't said yet, what object your 'data file' is. If you mean a data.frame I would use colnames(dataName) - c(Col1Name, col2Name, ….) Best Simon On Sep 20, 2013, at 4:10 PM, Preetam Pal lordpree...@gmail.com wrote: Hi, I guess this is pretty basic. I have 25 variables in the data file (name: score), i.e. X1,X2,.,X25. I dont want to use score$X1, score$X2 everytime I use these variables. Is there a way I can rename all these variables as simply X1,X2,.X25 without writing 25 lines of code, one line for renaming each variable (eg: X1=score.X1 X2=score.X2 and so on) ? Thanks for your help. Regards, Preetam -- Preetam Pal (+91)-9432212774 M-Stat 2nd Year, Room No. N-114 Statistics Division, C.V.Raman Hall Indian Statistical Institute, B.H.O.S. Kolkata. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Clustering of data set documentation files in package description
Dear R help list, I was just wondering whether there is a way to cluster the documentation files of data sets in the package documentation index file, so that common prefixes such as dat... are not necessary. Best wishes, Alrik Dr. Alrik Thiem Post-Doctoral Researcher Department of Humanities, Social and Political Sciences Swiss Federal Institute of Technology Zurich (ETHZ) Building IFW, Office C 29.2 Haldeneggsteig 4 CH-8092 Zurich +41 44 63 20937 (landline) +41 76 52 78083 (mobile) http://www.alrik-thiem.net http://www.compasss.org Dr. Alrik Thiem Post-Doctoral Researcher Department of Humanities, Social and Political Sciences Swiss Federal Institute of Technology Zurich (ETHZ) Building IFW, Office C 29.2 Haldeneggsteig 4 CH-8092 Zurich +41 44 63 20937 (landline) +41 76 52 78083 (mobile) http://www.alrik-thiem.net http://www.compasss.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Renaming variables
On Fri, Sep 20, 2013 at 10:10 AM, Preetam Pal lordpree...@gmail.com wrote: I have 25 variables in the data file (name: score), i.e. X1,X2,.,X25. I dont want to use score$X1, score$X2 everytime I use these variables. attach(score) plot(X1, X2) # etc. etc. -Aaron [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating dummy vars with contrasts - why does the returned identity matrix contain all levels (and not n-1 levels) ?
On Sep 13, 2013, at 11:21 PM, E Joffe wrote: Hi David, First I ordered the levels of each factor in a descending order based on frequency. Then, I used the following code to generate a matrix from the dataframe with dummy variables and subsequently run the glmnet (coxnet) ## tranform categorical variables into binary variables with dummy for trainSet predict_matrix - model.matrix(~ ., data=trainSet, contrasts.arg = lapply (trainSet[,sapply(trainSet, is.factor)], contrasts)) ## remove the status/time variables from the predictor matrix (x) for glmnet predict_matrix - subset (predict_matrix, select=c(-time,-status)) ## create a glmnet cox object using lasso regularization and cross validation glmnet.cv - cv.glmnet (predict_matrix, surv_obj, family=cox) I hope I did not do anything wrong . Can't thank you enough for your advice and interest. Thank you for outlining the process that you used. It looks from the outside as though it respects the constraints on the first two argument imposed by the more constrained input requirements of cv.glmnet. I didn't realize that subset could accept a `-`sign as an operator inside a c() expression, but if you are getting success then I guess it must. -- David. Erel -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Friday, September 13, 2013 8:51 PM To: E Joffe Cc: r-help@r-project.org Subject: Re: [R] Creating dummy vars with contrasts - why does the returned identity matrix contain all levels (and not n-1 levels) ? On Sep 13, 2013, at 9:33 AM, E Joffe wrote: Thank you so much for your answer ! As far as I understand, glmnet doesn't accept categorical variables only binary factors - so I had to create dummy variables for all categorical variables. I was rather puzzled by your question. The conventions used by glmnet should prevent constrasts from being pre-specified. Only matrices are accepted as data objects and one cannot assign contrast attributes to matrix columns. It worked perfectly. Erel Erel Joffe MD MSc School of Biomedical Informatics University of Texas - Health Science Center in Houston 832.287.0829 (c) -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Friday, September 13, 2013 3:05 PM To: E Joffe Cc: r-help@r-project.org Subject: Re: [R] Creating dummy vars with contrasts - why does the returned identity matrix contain all levels (and not n-1 levels) ? On Sep 13, 2013, at 4:15 AM, E Joffe wrote: Hello, I have a problem with creating an identity matrix for glmnet by using the contrasts function. Why do you want to do this? I have a factor with 4 levels. When I create dummy variables I think there should be n-1 variables (in this case 3) - so that the contrasts would be against the baseline level. This is also what is written in the help file for 'contrasts'. The problem is that the function creates a matrix with n variables (i.e. the same as the number of levels) and not n-1 (where I would have a baseline level for comparison). Only if you specify contrasts=FALSE does it do so and this is documented in that help file. My questions are: 1. How can I create a matrix with n-1 dummy vars ? See below. was I supposed to define explicitly that I want contr.treatment (contrasts) ? No need to do so. 2. If it is not possible, how should I interpret the hazard ratios in the Cox regression I am generating (I use glmnet for variable selection and then generate a Cox regression) - That is, if I get an HR of 3 for the variable 300mg what does it mean ? the hazard is 3 times higher of what ? Relative hazards are generally referenced to the baseline hazard, i.e. the hazard for a group with the omitted level for treatment constrasts and the mean value for any numeric predictors. Here is some code to reproduce the issue: # Create a 4 level example factor trt - factor( sample( c(PLACEBO, 300 MG, 600 MG, 1200 MG), 100, replace=TRUE ) ) # If your intent is to use constrasts different than the defaults used by # regression functions, these factor contrasts need to be assigned, either # within the construction of the factor or after the fact. contrasts(trt) 300 MG 600 MG PLACEBO 1200 MG 0 0 0 300 MG 1 0 0 600 MG 0 1 0 PLACEBO 0 0 1 # the default value for the contrasts parameter is TRUE and the default type is treatement # That did not cause any change to the 'trt'-object: trt #To make a change you need to use the `contrasts-` function: contrasts (trt) - contrasts(trt) trt # Use contrasts to get the identity matrix of dummy variables to be used in glmnet trt2 - contrasts (trt,contrasts=FALSE) Results (as you can see all levels are represented in the identity matrix): levels (trt) [1] 1200 MG 300 MG 600 MG PLACEBO print (trt2) 1200 MG 300 MG 600 MG
Re: [R] Renaming variables
or with(score.plot(X1, X2)) Clint BowmanINTERNET: cl...@ecy.wa.gov Air Quality Modeler INTERNET: cl...@math.utah.edu Department of Ecology VOICE: (360) 407-6815 PO Box 47600FAX:(360) 407-7534 Olympia, WA 98504-7600 USPS: PO Box 47600, Olympia, WA 98504-7600 Parcels:300 Desmond Drive, Lacey, WA 98503-1274 On Fri, 20 Sep 2013, Aaron Mackey wrote: On Fri, Sep 20, 2013 at 10:10 AM, Preetam Pal lordpree...@gmail.com wrote: I have 25 variables in the data file (name: score), i.e. X1,X2,.,X25. I dont want to use score$X1, score$X2 everytime I use these variables. attach(score) plot(X1, X2) # etc. etc. -Aaron [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Is there a way to change x and y axis tick mark inside plot()?
Dear R community, I am having trouble changing the tick marks on y-axis to every 5 units? I have the following: x - c(12, 16, 6, 23, 27, 8, 5, 19, 23, 13, 16, 8) y - c(29, 29, 23, 34, 38, 24, 22, 34, 36, 27, 33, 27) plot(x, y, pch=19) Should I change ylim=c(0,40), and then use axis()? I kept on thinking I can do everything inside plot(), but there is actually axis(), par(), ..., and so on. Could someone tell me why is there so many functions outside plot(). I'm sure there is a reason, but I don't seem to understand why. Thanks, Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Rmpfr question
Hello everyone, R beginner, I am confronted with the need to use Rmpf. In my first scripts I made use of X=read.table(file.choose(), header=FALSE, sep=,,dec=.) X=as.matrix(X) to load into a matrix data from file before matrix use. How can I do to load the same data in a mpfrMatrix. Is it possible to use with mpfrMatrix the same as operations M1%*%M2 scale(M1,TRUE,FALSE) Sorry but I'm a newbe Thanks in advance Michel [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] time zones from longitude, latitude, and date
I have data that provide longitude, latitude, and local date and time but no information about the corresponding time zone. How to identify the time zone so they can be converted to a common date/time? Thanks, Carlisle [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Access of odfWeave function to variables that are defined inside a function
Hi everyone!
I have now been using odfWeave() for a while in order to make some word
document reports of my data. I have encountered a problem when I try to
organize my program script into some functions. It seems that if I use the
odfWeave() within a function, then odfWeave() is not able to access the
variables that are defined inside the body of the same function that runs
odfWeave(). Here is a simple example of what I mean:
report.year - function(){
year=2011
odfWeave(sourcefile,outputfile)
}
report.year()
When I run report.year(), I get the error message: object 'year' not found
However, if I define the year out of the body of the function where I run
report.year(), then I don't get such error and I can produce my report.
I am wondering if this is something expected when using odfWeave due to its
nature of being able to only access the variables on the first level, or
should I somehow give access to the variables that are defined in the higher
levels (I mean within the body of my functions)?
I would actually like to report some of the variables that can only be
defined within this function, is there any way of doing it inside the body
of my report.year() function?
I appreciate it a lot if anyone can give me some hint about what I can do in
order to correct for this.
Best regards,
Sahar
--
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Re: [R] gam and optim
please ignore this, I see the error.
greg
hi
probably a silly mistake, but I expected gam to minimise the penalised
deviance.
thanks
greg
set.seed(1)
library(mgcv)
x-runif(100)
lp-exp(-2*x)*sin(8*x)
y-rpois(100,exp(lp))
plot(x,y)
m1-gam(y~s(x),poisson)
points(x,exp(lp),pch=16,col=green3)
points(x,fitted(m1),pch=16,cex=0.5,col=blue)
W-diag(fitted(m1))
X-predict(m1,type=lpmatrix)
S-m1$smooth[[1]]$S[[1]]
S-rbind(0,cbind(0,S))
A-X%*%solve(t(X)%*%W%*%X+m1$sp*S)%*%t(X)%*%W
sum(diag(A))
sum(m1$edf)
fit-fitted(m1)
b-m1$coef
range(exp(X%*%b)-fit)
z-y/fit-1+X%*%b
range(A%*%z-X%*%b)
dv-function(t)
{
f-exp(X%*%t)
-2*sum(y*log(f)-f-ifelse(y==0,0,y*log(y))+y)+t%*%S%*%t
}
dv(b)
m1$dev+b%*%S%*%b
#so far, so good
t1-optim(rep(0,10),dv)
t1$p
b
#different
dv(t1$p)
dv(b)
#different, and dv(t1$p) is lower!
fit1-exp(X%*%t1$p)
points(x,fit1,pch=16,cex=0.5,col=red)
# different
# gam found b which does approximate the true curve, but does not minimise
the penalised deviance, by a long shot.
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[R] gam and optim
hi
probably a silly mistake, but I expected gam to minimise the penalised
deviance.
thanks
greg
set.seed(1)
library(mgcv)
x-runif(100)
lp-exp(-2*x)*sin(8*x)
y-rpois(100,exp(lp))
plot(x,y)
m1-gam(y~s(x),poisson)
points(x,exp(lp),pch=16,col=green3)
points(x,fitted(m1),pch=16,cex=0.5,col=blue)
W-diag(fitted(m1))
X-predict(m1,type=lpmatrix)
S-m1$smooth[[1]]$S[[1]]
S-rbind(0,cbind(0,S))
A-X%*%solve(t(X)%*%W%*%X+m1$sp*S)%*%t(X)%*%W
sum(diag(A))
sum(m1$edf)
fit-fitted(m1)
b-m1$coef
range(exp(X%*%b)-fit)
z-y/fit-1+X%*%b
range(A%*%z-X%*%b)
dv-function(t)
{
f-exp(X%*%t)
-2*sum(y*log(f)-f-ifelse(y==0,0,y*log(y))+y)+t%*%S%*%t
}
dv(b)
m1$dev+b%*%S%*%b
#so far, so good
t1-optim(rep(0,10),dv)
t1$p
b
#different
dv(t1$p)
dv(b)
#different, and dv(t1$p) is lower!
fit1-exp(X%*%t1$p)
points(x,fit1,pch=16,cex=0.5,col=red)
# different
# gam found b which does approximate the true curve, but does not minimise
the penalised deviance, by a long shot.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there a way to change x and y axis tick mark inside plot()?
On 20/09/2013 11:52 AM, C W wrote: Dear R community, I am having trouble changing the tick marks on y-axis to every 5 units? I have the following: x - c(12, 16, 6, 23, 27, 8, 5, 19, 23, 13, 16, 8) y - c(29, 29, 23, 34, 38, 24, 22, 34, 36, 27, 33, 27) plot(x, y, pch=19) Should I change ylim=c(0,40), and then use axis()? Don't use ylim, use yaxt=n, then use axis(). I kept on thinking I can do everything inside plot(), but there is actually axis(), par(), ..., and so on. Could someone tell me why is there so many functions outside plot(). I'm sure there is a reason, but I don't seem to understand why. R is designed to be flexible. If you create giant functions that can do everything, you end up with a design like SAS, which is extremely inflexible. It's good at what it can do, but it's very hard to get it to do something the designers didn't think of. Duncan Murdoch __ 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] Best way to specify a mixed ANCOVA in R?
I initially posted this question to one of the StackExchange sites, and they suggested that I repost my problem here. After using ezANOVA as my primary way of specifying mixed ANOVAs, I've hit a stumbling block when it come to adding a covariate to the model. I am using an ANCOVA in order to determine if there is a developmental trajectory in my data; namely, I need to be able to see the F-statistic and p-values for interactions with the covariate (see p.466 onwards here [http://www.psyc.bbk.ac.uk/research/DNL/personalpages/annaz_etal_2009.pdf] if you want an example). Using ezANOVA, I can include covariates but the output does not show the F-statistic and p-values for interactions with the covariate - the main effect of the covariate is also not tested using this method. My ezANOVA model is as follows: aov.model-ezANOVA( data=textureView.child.outliersRemoved , dv=.(x) , wid=.(ID) , within=.(Texture,View) , between=.(TNOGroup) , between_covariates=.(Age) , type=3 , return_aov=TRUE ) Another option is to use lm or Anova, but I don't know how to specify the error terms properly for either and I'm limited because I want to use Type-III sums of squares (drop1 doesn't work in the cases where I've tried to use the aov wrapper for lm; it fails while reporting 'Error in formula.default(object, env = baseenv()) : invalid formula'). Finally, I've heard about using the nlme package to specify my ANCOVA as a mixed model instead, but I don't know where to begin here (despite spending a while reading about it). To give a summary, I'm trying to do a 2 (between; TNOGroup) x 2 (within, Texture) x2 (within, View) mixed ANCOVA, with age as a covariate. I want to use Type-III sums of squares, and see the F-statistic and p-values for interactions with the covariate, as well as for the main effect of the covariate. Any advice on the best way to do this would be much appreciated. Thanks, Danielle __ 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] saving as TIFF - problem with compression
Hi,
I am struggling to save my figures as TIFF files (yes, the yournal only accept
TIFF and not any other format).
I can manage to save a simple plot as TIFF at the correct dpi (I need at least
600dpi) and compression works. Only not for my own plot. Everything looks
distorted. I have read Stack Overflow and tried to change the point size to 6,
but with no apparent effect. Why is this and what would be the solution? The
TIFF-stuff is in the end of the script.
thank you for your time!
Ndata - data.frame(
Ncellpercent = rnorm(400, mean = rep(c(14, 18, 65), each = 40),
sd = rep(c(1, 3, 6), each = 40)),
fyear = rep(c('2007', '2008'), each = 100*2),
Station = sample(c('B1', 'H2', 'H3', 'H4'), 400, replace = TRUE),
Week = sample(c('19', '21', '23', '25'), 400, replace = TRUE))
Pdata - data.frame(
Ppercentcell = rnorm(400, mean = rep(c(4, 17, 22), each = 40),
sd = rep(c(0.1, 0.2, 0.4), each = 40)),
fyear = rep(c('2007', '2008'), each = 100*2),
Station = sample(c('B1', 'H2', 'H3', 'H4'), 400, replace = TRUE),
Week = sample(c('19', '21', '23', '25'), 400, replace = TRUE))
SummNdata - ddply(Ndata, .(Week, fyear, Station), summarise,
mean = mean(Ncellpercent),
sd = sd(Ncellpercent))
names(Pdata)
SummPdata - ddply(Pdata, .(Week, fyear, Station), summarise,
mean = mean(Ppercentcell),
sd = sd(Ppercentcell))
SummPdata
library(lattice)
library(latticeExtra)
library(HH)
font.settings - list( font = 1, cex = 1.2, fontfamily = serif)
my.theme - list(
par.xlab.text = font.settings,
par.ylab.text = font.settings,
axis.text = font.settings,
par.sub=font.settings)
plotN - xyplot(mean ~ Week | Station*fyear,
col=black,
pch=1,
cex=1.1,
lty=1,
strip = strip.custom(bg = 'white', style=1), # why can I not
use fontfamily=serif here ???
key=list(text=list(c(),
col=c(black)),
points=list(pch=1, lty=1, cex=1.5,
col=c(black)),
columns=1, border=F,
x = 0.02, y = 0.55, corner = c(2, 2),
title=, cex.title=1.3),
ylab = (Nc),
xlab=Week,
data= SummNdata,type=o,
par.settings = my.theme)
plotP - xyplot(mean ~ Week | Station*fyear,
col=black,
pch=2,
cex=1.1,
lty=2,
strip = strip.custom(bg = 'white', style=1),
key=list(text=list(c(),
col=c(black)),
points=list(pch=1, lty=1, cex=1.5,
col=c(black)),
columns=1, border=F,
x = 0.2, y = 0.2, corner = c(2, 2),
title=, cex.title=1.3),
ylab = (Pc),
xlab=Week,
data= SummPdata,type=o,
par.settings = my.theme)
tiff(file=myplot.tiff, bg=white, res=800,
width=13.3, height=9.45, units=cm, pointsize=6,
compression = lzw)
doubleYScale(plotN, plotP, add.ylab2 = TRUE)
dev.off()
this works:
tiff(file=myplot8.tiff, bg=white, res=800,
width=13.3, height=9.45, units=cm, pointsize=6,
compression = lzw)
plot(1:100)
dev.off()
Anna Zakrisson Braeunlich
PhD student
Department of Ecology, Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden/Sverige
Lives in Berlin.
For paper mail:
Katzbachstr. 21
D-10965, Berlin - Kreuzberg
Germany/Deutschland
E-mail: anna.zakris...@su.se
Tel work: +49-(0)3091541281
Mobile: +49-(0)15777374888
LinkedIn: http://se.linkedin.com/pub/anna-zakrisson-braeunlich/33/5a2/51b
º`. . `. . `. . º`. . `. . `. .º`. . `. .
`. .º
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Re: [R] Compare two subsequent rows based on specific values of a string
Hi,
May be this helps:
dat1- read.table(text=x1 x2 x3 x4
1 xz ab cd ef
2 ab fz cd ef
3 ab cd dy dx,sep=,header=TRUE,stringsAsFactors=FALSE)
dat1$changes_to_row_above- sapply(seq_len(nrow(dat1)),function(i)
{x1-dat1[,i]%in% dat1[,i-1];if(any(x1)) sum(x1,na.rm=TRUE) else NA})
dat1
# x1 x2 x3 x4 changes_to_row_above
#1 xz ab cd ef NA
#2 ab fz cd ef 1
#3 ab cd dy dx 2
A.K.
Dear all,
I would like to compare two rows and check whether something
changed. The only thing is, it is not always the same column that needs
to be compared. It is rather a matrix of values. The outcome would be a
number of changes in these, for instance: of row 2, there are 3 exactly
same values in row 3. I will make up an example:
x1 x2 x3 x4 changes_to_row_above
1 xz ab cd ef NA
2 ab fz cd ef 1
3 ab cd dy dx 2
and so forth...
Do you think that would be possible? Is there an easy function
to do so? It would be great help guys, thanks a lot in advance, sorry
for my bad data manipulation knowledge...
Tobi
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[R] Comparing two GAMs using anova (mgcv)
Hey all, I've fitted two GAMs to some data using mgcv. The only difference between the two models is that one includes an additional smooth term (the smooth terms are s(x), s(y) and s(log(y)), the difference being that one model contains s(y) as additional term whereas the other one only contains s(x) and s(log(y)) - x and y being my explanatory variables). I'm now trying to decide between those two models. There's no difference in deviance explained or R^2 and the diagnostic plots returned by gam.check() look fairly similar although the one of the fuller model looks slightly more satisfactory as far as the histogram of the residuals is concerned. I'm wondering whether it is appropriate to conduct an approximate F test using the anova function. I'm not 100% clear I've understood the documentation on that completely. Is it appropriate to conduct such a test if the only difference between models is the inclusion/exclusion of a smooth term? Conducting the test, I get the result that there's no reason to reject the null hypothesis that the simpler model (without s(y)) is correct. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] time zones from longitude, latitude, and date
If you make no further assumptions then this question is not solvable. For example we use standard time in our data collection systems even though legal time here applies daylight savings offset in the summer. In some cases I have seen data collected from sites in multiple time zones recorded in one data base with a single time zone. Even if you do assume local legal time applies in all cases, the boundaries of the time zones have changed over time. There exist time zone maps (e.g. http://efele.net/maps/tz/world/) that you could assume apply but I am not aware of any in CRAN. Someone on r-sig-geo might be able to help. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. carlisle thacker carlisle.thac...@gmail.com wrote: I have data that provide longitude, latitude, and local date and time but no information about the corresponding time zone. How to identify the time zone so they can be converted to a common date/time? Thanks, Carlisle [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there a way to change x and y axis tick mark inside plot()?
On Sep 20, 2013, at 11:17 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 20/09/2013 11:52 AM, C W wrote: Dear R community, I am having trouble changing the tick marks on y-axis to every 5 units? I have the following: x - c(12, 16, 6, 23, 27, 8, 5, 19, 23, 13, 16, 8) y - c(29, 29, 23, 34, 38, 24, 22, 34, 36, 27, 33, 27) plot(x, y, pch=19) Should I change ylim=c(0,40), and then use axis()? Don't use ylim, use yaxt=n, then use axis(). I kept on thinking I can do everything inside plot(), but there is actually axis(), par(), ..., and so on. Could someone tell me why is there so many functions outside plot(). I'm sure there is a reason, but I don't seem to understand why. R is designed to be flexible. If you create giant functions that can do everything, you end up with a design like SAS, which is extremely inflexible. It's good at what it can do, but it's very hard to get it to do something the designers didn't think of. Duncan Murdoch Fortune candidate, with cc: to Z. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compare two subsequent rows based on specific values of a string
Thanks I'm lookin for yur example
-Message d'origine-
De : r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] De
la part de arun
Envoyé : vendredi 20 septembre 2013 18:11
À : R help
Objet : Re: [R] Compare two subsequent rows based on specific values of a
string
Hi,
May be this helps:
dat1- read.table(text=x1 x2 x3 x4
1 xz ab cd ef
2 ab fz cd ef
3 ab cd dy dx,sep=,header=TRUE,stringsAsFactors=FALSE)
dat1$changes_to_row_above- sapply(seq_len(nrow(dat1)),function(i)
{x1-dat1[,i]%in% dat1[,i-1];if(any(x1)) sum(x1,na.rm=TRUE) else NA})
dat1
# x1 x2 x3 x4 changes_to_row_above
#1 xz ab cd ef NA
#2 ab fz cd ef 1
#3 ab cd dy dx 2
A.K.
Dear all,
I would like to compare two rows and check whether something changed. The
only thing is, it is not always the same column that needs to be compared.
It is rather a matrix of values. The outcome would be a number of changes in
these, for instance: of row 2, there are 3 exactly same values in row 3.
I will make up an example:
x1 x2 x3 x4 changes_to_row_above
1 xz ab cd ef NA
2 ab fz cd ef 1
3 ab cd dy dx 2
and so forth...
Do you think that would be possible? Is there an easy function to do so? It
would be great help guys, thanks a lot in advance, sorry for my bad data
manipulation knowledge...
Tobi
__
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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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and provide commented, minimal, self-contained, reproducible code.
[R] SPATIO-TEMPORAL ANALYSIS AND BIG DATA PROCESSING USING FREE AND OPEN SOURCE SW
HI, we are pleased to announce 2 weeks intensive training in: *SPATIO-TEMPORAL ANALYSIS AND BIG DATA PROCESSING USING FREE AND OPEN SOURCE SOFTWARE* 16-20 December 2013 20-24 January 2014 University of Twente - Enschede - The Netherlands 4 ECTS Over the few decades there has been an explosion of available data for environmental spatio-temporal research. This big data allows us to address a number of old and new important questions with unprecedented rigor and generality. Leveraging these new data streams requires new tools and increasingly complex workflows. This 2-week course introduces a set of free and open source software (GRASS, R, Python, AWK, BASH, GDAL) tools to perform spatio-temporal analysis and modelling of environmental data under Linux environment. It consists of a set of lectures and practical sessions where participants use this software packages to perform typical Geographic Information System (GIS) and Remote Sensing (RS) data analysis tasks. In this course, attention is paid to the use of command line rather than the graphical user interface. Yet no programming experience is required to register for this course as basic principles are introduced. More info: Objectives and program of the course http://www.spatial-ecology.net/ost4sem_dokuwiki/doku.php?id=wiki:enschede2013 Registration http://www.sense.nl/?module=coursesfunc=displayplannedplanningid=1332 **Please forward to interested persons. Best Regards * Staff* Dr. Raul Zurita-Milla (University of Twente, NL) Dr. Giuseppe Amatulli (Yale University, USA) Dr. Stefano Casalegno (University of Exeter, UK) Dr. Pieter Kempeneers (VITO, BE) -- Giuseppe Amatulli Web: www.spatial-ecology.net [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] time zones from longitude, latitude, and date
On Sep 20, 2013, at 8:02 AM, carlisle thacker wrote: I have data that provide longitude, latitude, and local date and time but no information about the corresponding time zone. How to identify the time zone so they can be converted to a common date/time? Perhaps as an approximation you could divide the longitude by 360/24, truncate and subtract to get an estimated GMT. It really depends on your purposes, the encoding of local time, need for accuracy, and perhaps further details regarding the data collection methods, none of which you have provided. -- David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] saving as TIFF - problem with compression
Hi,
Are you using Linux? If so, you may use ImageMagick and try your luck
using
convert filename.whatever.format filename.tiff
There are lots of options in ImageMagick. (Read the manual.)
I have done this in the past, and it has worked. However, your exact
situation may be different, so sorry if this information is not
particularly helpful.
Ranjan
On Fri, 20 Sep 2013 10:55:03 + Anna Zakrisson Braeunlich
anna.zakris...@su.se wrote:
Hi,
I am struggling to save my figures as TIFF files (yes, the yournal only
accept TIFF and not any other format).
I can manage to save a simple plot as TIFF at the correct dpi (I need at
least 600dpi) and compression works. Only not for my own plot. Everything
looks distorted. I have read Stack Overflow and tried to change the point
size to 6, but with no apparent effect. Why is this and what would be the
solution? The TIFF-stuff is in the end of the script.
thank you for your time!
Ndata - data.frame(
Ncellpercent = rnorm(400, mean = rep(c(14, 18, 65), each = 40),
sd = rep(c(1, 3, 6), each = 40)),
fyear = rep(c('2007', '2008'), each = 100*2),
Station = sample(c('B1', 'H2', 'H3', 'H4'), 400, replace = TRUE),
Week = sample(c('19', '21', '23', '25'), 400, replace = TRUE))
Pdata - data.frame(
Ppercentcell = rnorm(400, mean = rep(c(4, 17, 22), each = 40),
sd = rep(c(0.1, 0.2, 0.4), each = 40)),
fyear = rep(c('2007', '2008'), each = 100*2),
Station = sample(c('B1', 'H2', 'H3', 'H4'), 400, replace = TRUE),
Week = sample(c('19', '21', '23', '25'), 400, replace = TRUE))
SummNdata - ddply(Ndata, .(Week, fyear, Station), summarise,
mean = mean(Ncellpercent),
sd = sd(Ncellpercent))
names(Pdata)
SummPdata - ddply(Pdata, .(Week, fyear, Station), summarise,
mean = mean(Ppercentcell),
sd = sd(Ppercentcell))
SummPdata
library(lattice)
library(latticeExtra)
library(HH)
font.settings - list( font = 1, cex = 1.2, fontfamily = serif)
my.theme - list(
par.xlab.text = font.settings,
par.ylab.text = font.settings,
axis.text = font.settings,
par.sub=font.settings)
plotN - xyplot(mean ~ Week | Station*fyear,
col=black,
pch=1,
cex=1.1,
lty=1,
strip = strip.custom(bg = 'white', style=1), # why can I not
use fontfamily=serif here ???
key=list(text=list(c(),
col=c(black)),
points=list(pch=1, lty=1, cex=1.5,
col=c(black)),
columns=1, border=F,
x = 0.02, y = 0.55, corner = c(2, 2),
title=, cex.title=1.3),
ylab = (Nc),
xlab=Week,
data= SummNdata,type=o,
par.settings = my.theme)
plotP - xyplot(mean ~ Week | Station*fyear,
col=black,
pch=2,
cex=1.1,
lty=2,
strip = strip.custom(bg = 'white', style=1),
key=list(text=list(c(),
col=c(black)),
points=list(pch=1, lty=1, cex=1.5,
col=c(black)),
columns=1, border=F,
x = 0.2, y = 0.2, corner = c(2, 2),
title=, cex.title=1.3),
ylab = (Pc),
xlab=Week,
data= SummPdata,type=o,
par.settings = my.theme)
tiff(file=myplot.tiff, bg=white, res=800,
width=13.3, height=9.45, units=cm, pointsize=6,
compression = lzw)
doubleYScale(plotN, plotP, add.ylab2 = TRUE)
dev.off()
this works:
tiff(file=myplot8.tiff, bg=white, res=800,
width=13.3, height=9.45, units=cm, pointsize=6,
compression = lzw)
plot(1:100)
dev.off()
Anna Zakrisson Braeunlich
PhD student
Department of Ecology, Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden/Sverige
Lives in Berlin.
For paper mail:
Katzbachstr. 21
D-10965, Berlin - Kreuzberg
Germany/Deutschland
E-mail: anna.zakris...@su.se
Tel work: +49-(0)3091541281
Mobile: +49-(0)15777374888
LinkedIn: http://se.linkedin.com/pub/anna-zakrisson-braeunlich/33/5a2/51b
_`_. . _ `_. ._ `_. . _`_. . _ `_. ._ `_. ._`_. . _ `_.
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Re: [R] Is there a way to change x and y axis tick mark inside plot()?
Thanks Duncan. I have no knowledge of SAS, though many in industry use it. I hope R would expand its usage to more at the industry level. Mike On Fri, Sep 20, 2013 at 12:17 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote: On 20/09/2013 11:52 AM, C W wrote: Dear R community, I am having trouble changing the tick marks on y-axis to every 5 units? I have the following: x - c(12, 16, 6, 23, 27, 8, 5, 19, 23, 13, 16, 8) y - c(29, 29, 23, 34, 38, 24, 22, 34, 36, 27, 33, 27) plot(x, y, pch=19) Should I change ylim=c(0,40), and then use axis()? Don't use ylim, use yaxt=n, then use axis(). I kept on thinking I can do everything inside plot(), but there is actually axis(), par(), ..., and so on. Could someone tell me why is there so many functions outside plot(). I'm sure there is a reason, but I don't seem to understand why. R is designed to be flexible. If you create giant functions that can do everything, you end up with a design like SAS, which is extremely inflexible. It's good at what it can do, but it's very hard to get it to do something the designers didn't think of. Duncan Murdoch [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] search species with all absence in a presence-absence matrix
Dear list I have a matrix composed of islandID as rows and speciesID as columns. IslandID: Island A, B, C .O (15 islands in total) SpeciesID: D0001, D0002, D0003 .D0100 (100 species in total) The cell of the matrix describes presence (1) or absence (0) of the species in an island. Now I would like to search the species with absence (0) in all the islands (Island A to Island O.) Please kindly advise the R code for the search purpose. Thank you. Elaine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] search species with all absence in a presence-absence matrix
Once you learn to use dput() I am sure someone will be happy to help you. http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example John Kane Kingston ON Canada -Original Message- From: elaine.kuo...@gmail.com Sent: Sat, 21 Sep 2013 07:14:38 +0800 To: r-help@r-project.org Subject: [R] search species with all absence in a presence-absence matrix Dear list I have a matrix composed of islandID as rows and speciesID as columns. IslandID: Island A, B, C?.O (15 islands in total) SpeciesID: D0001, D0002, D0003?.D0100 (100 species in total) The cell of the matrix describes presence (1) or absence (0) of the species in an island. Now I would like to search the species with absence (0) in all the islands (Island A to Island O.) Please kindly advise the R code for the search purpose. Thank you. Elaine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Renaming variables
Depending on what your doing perhaps with() could help? Or assuming a data.frame or matrix, score[ , 25] will give you Score$X25 If you would supply a bit more information (and code) about what you are doing we probably can help more. John Kane Kingston ON Canada -Original Message- From: lordpree...@gmail.com Sent: Fri, 20 Sep 2013 19:40:23 +0530 To: r-help@r-project.org Subject: [R] Renaming variables Hi, I guess this is pretty basic. I have 25 variables in the data file (name: score), i.e. X1,X2,.,X25. I dont want to use score$X1, score$X2 everytime I use these variables. Is there a way I can rename all these variables as simply X1,X2,.X25 without writing 25 lines of code, one line for renaming each variable (eg: X1=score.X1 X2=score.X2 and so on) ? Thanks for your help. Regards, Preetam -- Preetam Pal (+91)-9432212774 M-Stat 2nd Year, Room No. N-114 Statistics Division, C.V.Raman Hall Indian Statistical Institute, B.H.O.S. Kolkata. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Receive Notifications of Incoming Messages Easily monitor multiple email accounts access them with a click. Visit http://www.inbox.com/notifier and check it out! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] search species with all absence in a presence-absence matrix
Hi, Try this: set.seed(248) lst1- lapply(1:1000,function(i) matrix( sample(0:1,15*100,replace=TRUE),ncol=100,dimnames=list(paste(Island,LETTERS[1:15]), paste0(D,sprintf(%04d,1:100) lst2-lst1[sapply(lst1,function(x) any(colSums(x)==0))] ##The above steps are just to create some matrices with zeros in all the islands mat1-lst2[[1]] mat1[,colSums(mat1)==0,drop=FALSE] # D0038 #Island A 0 #Island B 0 #Island C 0 #Island D 0 #Island E 0 #Island F 0 #Island G 0 #Island H 0 #Island I 0 #Island J 0 #Island K 0 #Island L 0 #Island M 0 #Island N 0 #Island O 0 colnames(mat1)[colSums(mat1)==0] #[1] D0038 mat2-lst2[[3]] colnames(mat2)[colSums(mat2)==0] #[1] D0086 A.K. - Original Message - From: Elaine Kuo elaine.kuo...@gmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Friday, September 20, 2013 7:14 PM Subject: [R] search species with all absence in a presence-absence matrix Dear list I have a matrix composed of islandID as rows and speciesID as columns. IslandID: Island A, B, C….O (15 islands in total) SpeciesID: D0001, D0002, D0003….D0100 (100 species in total) The cell of the matrix describes presence (1) or absence (0) of the species in an island. Now I would like to search the species with absence (0) in all the islands (Island A to Island O.) Please kindly advise the R code for the search purpose. Thank you. Elaine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Grouping variables by a irregular time interval
Hello all, I´m have a very large data frame (more than 5 million lines) as below (dput example at the end of mail): Station Antenna TagDateTime Power Events 1 1 2 999 22/07/2013 11:00:2117 1 2 1 2 999 22/07/2013 11:33:4731 1 3 1 2 999 22/07/2013 11:34:0019 1 4 1 2 999 22/07/2013 11:34:1653 1 5 1 2 999 22/07/2013 11:43:2015 1 6 1 2 999 22/07/2013 11:43:3517 1 To each Tag, in each Antenna, in each Station, I need to create a 10 min interval and sum the number of Events and mean of Power in the time interval, as below (complete wanted output at the end of mail). Station Antenna Tag StartDateTime EndDateTime Power Events 1 1 2 999 22/07/2013 11:00:21 22/07/2013 11:00:2117 1 2 1 2 999 22/07/2013 11:34:16 22/07/2013 11:43:3527 5 3 1 2 999 22/07/2013 11:44:35 22/07/2013 11:45:4017 14 4 2 1 1 25/07/2013 14:19:45 25/07/2013 14:20:3965 4 5 2 1 2 25/07/2013 14:20:13 25/07/2013 14:25:1421 3 6 2 1 4 25/07/2013 14:20:46 25/07/2013 14:20:4628 1 Show start and end points of each interval is optional, not necessary. I put both to show the irregular time interval (look at tag 999) First I tried a for-loop, without success. After that, I tried this code: require (plyr) ddply (data, .(Station, Antenna, Tag, cut(data$DateTime, 10 min)), summarise, Power = round (mean(Power), 0), Events = sum (Events)) Is almost what I want, because cut() divided in regular time intervals, but in some cases I do not have this, and it split a unique observation in two. Any ideas to solve this issue? R version 3.0.1 (2013-05-16) -- Good Sport Platform: x86_64-w64-mingw32/x64 (64-bit) Windows 7 Professional Thanks in advanced, Raoni -- Raoni Rosa Rodrigues Research Associate of Fish Transposition Center CTPeixes Universidade Federal de Minas Gerais - UFMG Brasil rodrigues.ra...@gmail.com ##complete data dput structure(list(Station = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Antenna = c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Tag = c(999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 999L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 4L, 18L, 18L, 18L, 21L, 22L, 36L, 36L, 36L, 36L, 36L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L, 48L), DateTime = structure(c(3L, 4L, 5L, 5L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 18L, 19L, 19L, 19L, 19L, 20L, 23L, 19L, 17L, 17L, 17L, 23L, 18L, 1L, 1L, 1L, 2L, 2L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 12L, 12L, 13L, 13L, 13L, 13L, 14L, 14L, 14L, 14L, 14L, 14L, 15L, 15L, 15L, 15L, 15L, 15L, 15L, 16L, 16L, 16L, 16L, 18L, 19L, 21L, 21L, 21L, 21L, 21L, 22L, 22L, 22L, 22L, 22L, 23L, 24L, 24L, 24L, 24L, 24L, 24L, 25L, 25L, 25L, 25L, 25L, 26L, 26L, 26L, 26L, 26L, 26L, 26L, 27L, 27L, 27L, 27L, 27L, 27L, 28L, 28L, 28L, 28L, 28L), .Label = c(19/06/2013 22:15, 19/06/2013 22:16, 22/07/2013 11:00, 22/07/2013 11:33, 22/07/2013 11:34, 22/07/2013 11:43, 22/07/2013 11:44, 22/07/2013 11:45, 25/07/2013 14:10, 25/07/2013 14:11, 25/07/2013 14:12, 25/07/2013 14:13, 25/07/2013 14:14, 25/07/2013 14:15, 25/07/2013 14:16, 25/07/2013 14:17, 25/07/2013 14:18, 25/07/2013 14:19, 25/07/2013 14:20, 25/07/2013 14:21, 25/07/2013 14:23, 25/07/2013 14:24, 25/07/2013 14:25, 25/07/2013 14:26, 25/07/2013 14:27, 25/07/2013 14:28, 25/07/2013 14:29, 25/07/2013 14:30), class = factor), Power = c(17L, 31L, 19L, 53L, 15L, 17L, 21L, 12L, 15L, 22L, 19L, 15L, 13L, 14L, 15L, 12L, 23L, 19L, 16L, 20L, 30L, 37L, 25L, 167L, 24L, 14L,
