[R] reading and parsing gzipped files
Hi all,I have large compressed text tab delimited files, I am trying to write efficient function to read them, I am using gzfile() and readLines() zz - gzfile(exampl.txt.gz, r) # compressed file system.time(temp1-readLines(zz )) close(zz) which work fast, and create vector of strings. The problem is to parse the result, if I use strsplit it takes longer then decompress file manually , read it with scan and erase it. Can anybody recommend an efficient way of parsing large vector ~200,000 entries Dmitriy [[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] question on FARIMA innovationsIn-Reply-To=[EMAIL PROTECTED]
Hi Ferebee,
I am also working on a similar problem. Have you solved it? I am trying to
write a function to to determine the innovations.
However, I think that the Farma package may do this. I haven't succeeded in
getting Farma to work. It requires OX.
Can you let me know if you have made any progress
Tony Cricenti
Regards
A.L. Cricenti,
Academic Leader Telecommunications
Faculty of Information and Communication Technologies
Swinburne University of Technology,
PO Box 218, Hawthorn, Vic 3122, Australia.
email: [EMAIL PROTECTED]
Phone: (voice) +61-3-9214 5506 (fax) +61-3-9819 6443
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Re: [R] isoMDS and dist
On Thu, 4 Sep 2008, Carolyn A. Pucko wrote: I am starting with a matrix in which rows are vegetation plots and columns are various characteristics including ID# and elevation. I removed elevation and ID columns to avoid having those characteristics influence the distances between points which I calculated using the dist command. The resulting distance file was then used in isoMDS. What I want to know is whether I can reattach the ID and elevation onto the point location information given from isoMDS? or is it possible that the order given in the isoMDS output is not the same as the order of the original data used to calculate distances? The order is the same. From the help page Value: Two components: points: A k-column vector of the fitted configuration. ... and see how it describes 'the fitted configuration'. BTW, it is regarded as basic courtesy to credit the work of others: isoMDS is from package MASS. Thanks, Carrie Pucko University of Vermont __ 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. -- Brian D. Ripley, [EMAIL PROTECTED] 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.
Re: [R] how to deal with NaN in boot object
On Fri, 5 Sep 2008, Jinsong Zhao wrote:
Hi there,
I use boot() to do bootstrap simulation on my statistic, I get a boot
object, named obj. For certain reasons, there are some NaN in obj$t.
Now, I hope to get confidence interval using boot.ci(), it give the
following error:
Error in if (const(t, min(1e-08, mean(t)/1e+06))) { :
missing value where TRUE/FALSE needed
Now, what I could do is I have to edit the obj by hand.
Any suggestions? Thanks in advanced!
Correct your 'statistic' function not to return NaN.
--
Brian D. Ripley, [EMAIL PROTECTED]
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
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Re: [R] algorithm to create unique identifiers
For a much simpler solution that does always work for numbers, see unique's methods for matrices and data frames. On Thu, 4 Sep 2008, Henrik Bengtsson wrote: On Thu, Sep 4, 2008 at 8:44 PM, Ralph S. [EMAIL PROTECTED] wrote: Hi all, I am trying to create a unique identifier for each row, combining numbers from three columns. Do you know if there is a general formula to do this (or some manual where I can read about this)? I figure I can use the numeric entries of the columns as coordinates and multiply them with different coefficients (different magnitudes) to get the unique ID - but it would be nice to read about such algorithms in general. What are you numbers? Are they in a fixed range? Integers or reals? If fixed range integers, it is easy. Think regular numerical representation, e.g. binary, octadecimal, decimal and hexadecimal. For a more generic solution that works with any data types, see e.g. MD5 [http://en.wikipedia.org/wiki/MD5]. It is not guaranteed to generated unique codes, but it is extremely rare that two different inputs gives the same MD5 code. MD5 (and others) are implemented in the 'digest' packages, e.g. library(digest) digest(list(a=1, b=list(1:10, c=letters))) [1] 73e0ae066a97bfff7f79d41c65b55fde My $.02 /Henrik Any links/input would be great - Ralph -- Brian D. Ripley, [EMAIL PROTECTED] 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.
Re: [R] Trouble with R CMD check: I can't seem to get dependencies right (maybe I'm using R_LIBS incorrectly?)
'Writing R Extensions' does say
@quotation Note
@code{R CMD check} and @code{R CMD build} run @R{} with
@option{--vanilla}, so none of the user's startup files are read. If
you need @env{R_LIBS} set (to find packages in a non-standard library)
you will need to set it in the environment.
@end quotation
~/.Renviron is one of those startup files (and so is ~/.Rprofile).
On Thu, 4 Sep 2008, Dylan Arena wrote:
Hi there,
I'm in the following directory:
~/Documents/Rstuff/diceFiles/dice_1.1
The directory dice is in this directory, with all the usual build
files (DESCRIPTION, NAMESPACE, etc). I'm trying to run the following
command:
R CMD check dice
(where dice is the name of the package I'm checking), and I get the following:
* checking package dependencies ... ERROR
Packages required but not available:
gtools
My DESCRIPTION file does (correctly) list gtools as a dependency, so
I'm happy this check is ensuring its availability. What I can't
figure out is how to get the check command to see that gtools is in
fact installed. I've created a file called .Renviron in my home
directory that has the following line in it:
R_LIBS=~/Documents/Rstuff/library
And in the directory specified by R_LIBS above is the gtools package.
(I've also got a file called .Rprofile in my home directory, with the
line .libPaths(~/Documents/Rstuff/library)
in it.) My reading of the R-intro manual is that setting R_LIBS in
this way should allow the R CMD check command to search in the
directory specified by R_LIBS when looking for gtools. I'm pretty
sure at this point, though, that my understanding is incorrect.
At this point I'm in over my depth. I'm writing because I'm hopeful
that to someone on this list my problem is trivially solvable. My
basic question is: what must I do to get R CMD check to find the
gtools package?
Please let me know any ideas you might have for me,
Dylan
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--
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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
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Re: [R] how to write a escape sequences in to a file
A newline is \n. So hai \nhru. On Fri, 5 Sep 2008, Kurapati, Ravichandra (Ravichandra) wrote: Hi Str1-hai Str2-hru I want to write these 2 strings in a file separated by newline. How can I get this. Thanks K.Ravichandra [[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. -- Brian D. Ripley, [EMAIL PROTECTED] 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.
Re: [R] reading and parsing gzipped files
On Thu, 4 Sep 2008, Dmitriy Skvortsov wrote: Hi all,I have large compressed text tab delimited files, I am trying to write efficient function to read them, I am using gzfile() and readLines() zz - gzfile(exampl.txt.gz, r) # compressed file system.time(temp1-readLines(zz )) close(zz) which work fast, and create vector of strings. The problem is to parse the result, if I use strsplit it takes longer then decompress file manually , read it with scan and erase it. Can anybody recommend an efficient way of parsing large vector ~200,000 entries 'parse'? What is wrong with using read.delim (reading 'tab delimited files' is its job)? It (and scan) work with gzfile connections, so there is no need to decompress manually. See the 'R Data Import/Export Manual' for how to use read.delim efficiently. Dmitriy [[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. PLEASE do. -- Brian D. Ripley, [EMAIL PROTECTED] 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] Odp: Help for par(mfrow)
Hi
see from plot.boot help page
Side Effects
All screens are closed and cleared and a number of plots are produced on
the current graphics device. Screens are closed but not cleared at
termination of this function.
You need to change a behaviour of this function what shall not be simple.
Or maybe you can produce set of graphs in one pdf file or in several png
files and then to add them together by external means
Regards
Petr
[EMAIL PROTECTED] napsal dne 03.09.2008 12:03:03:
Hi, R users
I have a question with par(mfrow). I try to histograms and qqplots
form boot output for 5 statistics but par(mfrow=c(5,2)) or
par(mfrow=c(5,1)) does not work. R still display each figure
separately. What did I do wrong? (I check ?par)
c1.boot-boot(c1data,c1.fun,R=999)
c1.boot
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = c1data, statistic = c1.fun, R = 999)
Bootstrap Statistics :
originalbiasstd. error
t1* 0.30447696 0.0014228306 0.01026153
t2* 0.05967183 -0.0014505285 0.01274977
t3* 1.11852318 -0.0009339349 0.31586852
t4* 4.12856813 -0.0310221125 0.39602210
t5* -1.29958196 0.0222876889 0.67625299
par(mfrow=c(5,2))
for (i in 1:5) {plot(c1.boot, index=i, col=i)}
Many Thanks
Chunhao
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Re: [R] restricted bootstrap
Grant, Have you considered a gls model instead of a lm model? In a gls model one can model the correlation between the measures. So you won't need to select a subset of your data. You can kind gls in the nlme package. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Grant Gillis Verzonden: donderdag 4 september 2008 14:57 Aan: r-help@r-project.org Onderwerp: Re: [R] restricted bootstrap Hello Professor Ripely, Sorry for not being clear. I posted after a long day of struggling. Also my toy distance matrix should have been symmetrical. Simply put I have spatially autocorrelated data collected from many points. I would like to do a linear regression on these data. To deal with the autocrrelation I want to resample a subset of my data with replacement but I need to restrict subsets such that no two locations where data was collected are closer than Xm apart (further apart than the autocrrelation in the data). Thanks for having a look at this for me. I will look up the hard-core spatial point process. Grant 2008/9/4 Prof Brian Ripley [EMAIL PROTECTED] I see nothing here to do with the 'bootstrap', which is sampling with replacement. Do you know what you mean exactly by 'randomly sample'? In general the way to so this is to sample randomly (uniformly, whatever) and reject samples that do not meet your restriction. For some restrictions there are more efficient algorithms, but I don't understand yours. (What are the 'rows'? Do you want to sample rows in space or xy locations? How come 'dist' is not symmetric?) For some restrictions, an MCMC sampling scheme is needed, the hard-core spatial point process being a related example. On Wed, 3 Sep 2008, Grant Gillis wrote: Hello List, I am not sure that I have the correct terminology here (restricted bootstrap) which may be hampering my archive searches. I have quite a large spatially autocorrelated data set. I have xy coordinates and the corresponding pairwise distance matrix (metres) for each row. I would like to randomly sample some number of rows but restricting samples such that the distance between them is larger than the threshold of autocorrelation. I have been been unsuccessfully trying to link the 'sample' function to values in the distance matrix. My end goal is to randomly sample M thousand rows of data N thousand times calculating linear regression coefficients for each sample but am stuck on taking the initial sample. I believe I can figure out the rest. Example Question I would like to radomly sample 3 rows further but withe the restriction that they are greater than 100m apart example data: main data: y- c(1, 2, 9, 5, 6) x-c( 1, 3, 5, 7, 9) z-c(2, 4, 6, 8, 10) a-c(3, 9, 6, 4 ,4) maindata-cbind(y, x, z, a) y x x a [1,] 1 1 1 3 [2,] 2 3 3 9 [3,] 9 5 5 6 [4,] 5 7 7 4 [5,] 6 9 9 4 distance matrix: row1-c(0, 123, 567, 89) row2-c(98, 0, 345, 543) row3-c(765, 90, 0, 987) row4-c(654, 8, 99, 0) dist-rbind(row1, row2, row3, row4) [,1] [,2] [,3] [,4] row10 123 567 89 row2 980 345 543 row3 765 900 987 row4 6548 990 Thanks for all of the help in the past and now Cheers Grant [[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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/http://www.stats.ox.ac.uk/%7Eripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[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
[R] Add marks to a point pattern
My name is Vanessa and I am from Spain.I have in Excel a list of children that go and don´t go to the dentist, and I want to add these categorical marks to the point pattern in package spatstat.How can I do it? Thanks for your help. [[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] Package for Tidying-up R-code
Dear all, Is there any such package? I am thinking of something equivalent to Perl::Tidy. For example with VI editor one can visual highlight a portion of Perlcode and then issue the command: !perltidy then the code will be automatically arranged. - Gundala Viswanath Jakarta - Indonesia __ 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] Package for Tidying-up R-code
check Section 3.1 of the `Writing R Extensions' manual. I hope it helps. Best, Dimitris Gundala Viswanath wrote: Dear all, Is there any such package? I am thinking of something equivalent to Perl::Tidy. For example with VI editor one can visual highlight a portion of Perlcode and then issue the command: !perltidy then the code will be automatically arranged. - Gundala Viswanath Jakarta - Indonesia __ 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. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043399 Fax: +31/(0)10/7044657 __ 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] Upgrade to R 2.7.2 from Debian Repository
I am running R 2.7.1 on an eeepc with the standard Xandros Linux. I install/upgrade R binaries using synaptic from the http://cran.r-project.org/bin/linux/debian/etch-cran/ repository. R 2.7.2 does not appear to be available from this repository. Will it be available soon or is there another repository I should be using? Tom -- View this message in context: http://www.nabble.com/Upgrade-to-R-2.7.2-from-Debian-Repository-tp19328325p19328325.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] controlling lattice plot ticks with relation=free
How do you persuade lattice to draw tick marks on both the left and right
side of the y-axis, when relation=free in the scales component?
#Ticks appear on both sides
histogram(~height|voice.part, data=singer)
##Ticks only on left
histogram(~height|voice.part, data=singer,
scales=list(y=list(relation=free)))
Providing tck as a vector of length 2 doesn't seem to work, neither does
setting alternating=3 (since the argument is ignored, and I don't really
want the labels on both sides - just the ticks).
I'm sure I'm missing something obvious.
Regards,
Richie.
Mathematical Sciences Unit
HSL
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Re: [R] Fitted probabilities in conditional logit regression
On Sun, 2008-08-31 at 09:49 -0400, Min Chen wrote: Dear R-help, I'm doing conditional logit regression for a discrete choice model. I want to know whether there's a way to get the fitted probabilities. In Stata, predict works for clogit, but it seems that in R predict does not. Thank you very much! Best wishes. Can you explain/demonstrate how/why predict doesn't work for clogit? I can get predictions from a clogit model: ## from ?clogit mod - clogit(case~spontaneous+induced+strata(stratum),data=infert) predict(mod) predict(mod, type = expected) all work for that example to give the fitted values. See ?predict.coxph for details of the predict function. I'm not familiar with either the code or the methods, but 'type = expected' seems to give what you want. HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ 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] how to write a escape sequences in to a file
Try this: cat(Str1, Str2, file = out.txt, sep = \n) On Fri, Sep 5, 2008 at 2:21 AM, Kurapati, Ravichandra (Ravichandra) [EMAIL PROTECTED] wrote: Hi Str1-hai Str2-hru I want to write these 2 strings in a file separated by newline. How can I get this. Thanks K.Ravichandra [[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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] Plotting using 'if' statements
Sorry! Hopefully just one more question! I'm now trying to plot columns 3 and 6. I've tried: plot(January[3:6]) ...but this plots columns 4 and 5 too! I've also tried: plot(January[3],January[6]) plot(January[3,6]) plot(January[3]:January[6]) But these don't produce the right results (instead, errors!) Again, I'm sure it's a just a small (but important) detail that I'm missing, and I'd be very grateful if anyone could help me out. Many thanks again, Steve Date: Thu, 4 Sep 2008 13:52:03 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Plotting using 'if' statements CC: r-help@r-project.org Use '': plot(January[January[,3] 0 January[,3] 2, 3:4]) On Thu, Sep 4, 2008 at 1:47 PM, Steve Murray wrote: Ah that's great, thank you very much. As a follow-on, in the same format, how would I plot where column 3 is greater than 0 *but also less than 2*? Once again, any help is much appreciated. Thanks, Steve Date: Thu, 4 Sep 2008 13:39:12 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Plotting using 'if' statements CC: r-help@r-project.org Try this: plot(January[January[,3] 0, 3:4]) On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray wrote: Dear all, I have a dataset of four columns, and I wish to plot (as a scatter graph) the values of the third column where the values are greater than zero, and the fourth column. I tried doing this via the plot command itself, but got into a bit of a mess (resulting in errors!). My dataframe is called 'January': plot(January[3(0):4]) Error: unexpected '' in plot(January[3( After a few variations on this, I thought I'd try making a new object which includes all values from the third column of January0 (to plot in a separate step) as follows: JanFilter - January[3]0 No error here. However, when I display the 'values' of JanFilter, it shows that instead of keeping the numerical values, the above operation simply displays the results of the logical test: head(JanFilter) Value [1,] FALSE [2,]TRUE [3,]TRUE [4,]TRUE [5,]TRUE [6,]TRUE [[elided Hotmail spam]] So my question is, how do I perform 'if' statements in order to filter out various parts of a dataset, for plotting on a graph. Many thanks, Steve __ R-help@r-project.org mailing list 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 _ -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O _ Win New York holidays with Kellogg’s Live Search __ 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] Plotting using 'if' statements
Use this: plot(January[c(3, 6)]) On Fri, Sep 5, 2008 at 8:52 AM, Steve Murray [EMAIL PROTECTED] wrote: Sorry! Hopefully just one more question! I'm now trying to plot columns 3 and 6. I've tried: plot(January[3:6]) ...but this plots columns 4 and 5 too! I've also tried: plot(January[3],January[6]) plot(January[3,6]) plot(January[3]:January[6]) But these don't produce the right results (instead, errors!) Again, I'm sure it's a just a small (but important) detail that I'm missing, and I'd be very grateful if anyone could help me out. Many thanks again, Steve Date: Thu, 4 Sep 2008 13:52:03 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Plotting using 'if' statements CC: r-help@r-project.org Use '': plot(January[January[,3] 0 January[,3] 2, 3:4]) On Thu, Sep 4, 2008 at 1:47 PM, Steve Murray wrote: Ah that's great, thank you very much. As a follow-on, in the same format, how would I plot where column 3 is greater than 0 *but also less than 2*? Once again, any help is much appreciated. Thanks, Steve Date: Thu, 4 Sep 2008 13:39:12 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Plotting using 'if' statements CC: r-help@r-project.org Try this: plot(January[January[,3] 0, 3:4]) On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray wrote: Dear all, I have a dataset of four columns, and I wish to plot (as a scatter graph) the values of the third column where the values are greater than zero, and the fourth column. I tried doing this via the plot command itself, but got into a bit of a mess (resulting in errors!). My dataframe is called 'January': plot(January[3(0):4]) Error: unexpected '' in plot(January[3( After a few variations on this, I thought I'd try making a new object which includes all values from the third column of January0 (to plot in a separate step) as follows: JanFilter - January[3]0 No error here. However, when I display the 'values' of JanFilter, it shows that instead of keeping the numerical values, the above operation simply displays the results of the logical test: head(JanFilter) Value [1,] FALSE [2,]TRUE [3,]TRUE [4,]TRUE [5,]TRUE [6,]TRUE This is obviously no good for plotting the numerical values on axes! So my question is, how do I perform 'if' statements in order to filter out various parts of a dataset, for plotting on a graph. Many thanks, Steve __ 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 _ Make a mini you and download it into Windows Live Messenger http://clk.atdmt.com/UKM/go/111354029/direct/01/ -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O _ Win New York holidays with Kellogg's Live Search http://clk.atdmt.com/UKM/go/111354033/direct/01/ -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] how to draw the legend about color from 3d picture
I have drawed a picture with persp, it's 3d map with different color, indicate different altitude. In gnuplot, the corresponding command 'splot' will generate a picture beside to indicate the relationship between color and altitude. But in R, how to draw it? I have read the manual of legend, but they are all about how to draw a legend with colored text, not a continuous varing color with corresponding number. __ 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] Orthogonalization algorithms
Hi, I have eight vectors that I would like to orthogonalize preferably using R. The vectors are of considerable length, however due to their nature I know they satisfy the conditions needed to apply the Gram-Schmidt algorithm. Before I embark on some R coding, I wanted to check that there is no facility / function already around that computes the orthogonalized set of vectors? I have performed an RSiteSearch etc with no luck. Anything that could help me along the way would be much appreciated. Thanks in advance, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[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] Orthogonalization algorithms
Please ignore this, it appears I have found what I'm looking for in the far package. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Williams, Robin Sent: Friday, September 05, 2008 1:18 PM To: r-help@r-project.org Subject: [R] Orthogonalization algorithms Hi, I have eight vectors that I would like to orthogonalize preferably using R. The vectors are of considerable length, however due to their nature I know they satisfy the conditions needed to apply the Gram-Schmidt algorithm. Before I embark on some R coding, I wanted to check that there is no facility / function already around that computes the orthogonalized set of vectors? I have performed an RSiteSearch etc with no luck. Anything that could help me along the way would be much appreciated. Thanks in advance, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[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] Error: can not allocate vector of 117.3 Mb
Hi R users, I am doing multiscale bootstrapping for clustering through pvclust package. I have large data set (observations 182 and variables 5546). When i tried to make bootstrapping, then i got message as Bootstrap (r = 0.5)Error: cannot allocate vector of size 117.3 Mb. I am new R user and could not understand what is the problem and also don't know the easiet solution. If someone knows, pls help me. Thanks in advance. Ram Kumar Basnet Graduate student. [[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] Error: can not allocate vector of 117.3 Mb
ram basnet wrote: Hi R users, I am doing multiscale bootstrapping for clustering through pvclust package. I have large data set (observations 182 and variables 5546). When i tried to make bootstrapping, then i got message as Bootstrap (r = 0.5)Error: cannot allocate vector of size 117.3 Mb. I am new R user and could not understand what is the problem and also don't know the easiet solution. If someone knows, pls help me. Please read ?Memory Uwe Ligges Thanks in advance. Ram Kumar Basnet Graduate student. [[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] Plotting using 'if' statements
Thanks again for a very quick reply! It worked really well too! _ Discover Bird's Eye View now with Multimap from Live Search __ 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] how to draw the legend about color from 3d picture
On 9/5/2008 8:06 AM, yk wrote:
I have drawed a picture with persp, it's 3d map with different color,
indicate different altitude. In gnuplot, the corresponding command
'splot' will generate a picture beside to indicate the relationship
between color and altitude. But in R, how to draw it? I have read the
manual of legend, but they are all about how to draw a legend with
colored text, not a continuous varing color with corresponding number.
I don't think there's an automatic way to do this (though probably some
package provides one).
If you want to write your own, take a look at example(filled.contour).
The legend it draws is done by this code:
plot.new()
plot.window(xlim = c(0, 1), ylim = range(levels), xaxs = i,
yaxs = i)
rect(0, levels[-length(levels)], 1, levels[-1], col = col)
if (missing(key.axes)) {
if (axes)
axis(4)
}
else key.axes
box()
(but it has done a lot of setup using layout() and par() before this).
Duncan Murdoch
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[R] Use of colour in plots
Dear all, I have 3 datasets all of which share the same longitude and latitude values, which I'm looking to plot onto a scattergraph. The third dataset has values which can only be either '1' or '2'. So to incorporate all three datasets onto two axes, I'm wondering if I can plot dataset1 and dataset2 as normal, but then use colour to determine whether these points are either values '1' or '2' according to the third dataset. If so, how would I go about doing this in R, and what format would the command take? Thanks for any help offered, Steve __ 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] cluster a distance(analogue)-object using agnes(cluster)
On Thu, 2008-09-04 at 11:28 +0200, Martin Maechler wrote:
B == Birgitle [EMAIL PROTECTED]
on Tue, 2 Sep 2008 03:02:31 -0700 (PDT) writes:
B I try to perform a clustering using an existing dissimilarity matrix
that I
B calculated using distance (analogue)
B I tried two different things. One of them worked and one not and I
don`t
B understand why.
B Here the code:
B not working example
B library(cluster)
B library(analogue)
B iris2 - as.data.frame(iris)
why that? After the above, iris2 is identical() to iris !
B str(iris2)
B 'data.frame': 150 obs. of 5 variables:
B $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
B $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
B $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
B $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
B $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1
1 1
B 1 1 1 ...
B Test.Gower - distance(iris2, method =mixed)
why not just
daisy(iris2, metric = gower)
daisy() is in cluster which has been a recommended R package
forever.
So the solution (here, not in general!)
is to stay with package 'cluster' and use
daisy() before agnes().
And as the author of distance(), I agree *completely*, and say so in the
Environmetrics Task View.
distance() was written for a very specific task and that it does what
you want it to, Birgit, is side effect of the way I wrote distance().
Anyway, Birgit, you'll be glad to know that the example you included
works in the R-forge version of analogue (0.5-4 to be). Get it from:
https://r-forge.r-project.org/R/?group_id=69
or directly from within R
install.packages(analogue, repos=http://R-Forge.R-project.org;)
if you want to use distance(), but I'd be using the cluster package
tools myself for the problem you emailed about.
You also probably want to use as.dist() on the output from distance() to
store the matrix in a more compact form. Because of how I wanted to
calculate distances (between two data sets), the matrix output was
essential, but this output is storing (and computing) redundant data in
the case you cite (a single matrix).
All the best,
G
Regards,
Martin Maechler, ETH Zurich
{same city! feel free to phone me..}
B Test.Gower.agnes-agnes(Test.Gower, diss=T)
B Fehler in agnes(Test.Gower, diss = T) :
B (list) Objekt kann nicht nach 'logical' umgewandelt werden
B Error in agnes(Test.Gower, diss=T).
B (list) object can`t be transformed to logical
B working example only numerics used:
B library(cluster)
B library(analogue)
B irisPart-subset(iris, select= Sepal.Length:Petal.Width)
B Dist.Gower - distance(irisPart, method =mixed)
B AgnesA - agnes(Dist.Gower, method=average, diss=TRUE)
B Would be great if somebody could help me.
B The dataset that I would like to use for the clustering also contains
B factors.
B and gives me the same Error message as in the not working example.
B Thanks in advance
B B.
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[R] Confidence Intervals on Hazard Plots
Hello, Is it possible to create confidence intervals for hazard rates? I'm creating two muhaz objects: haz1 - muhaz(NumDaysCustomer[cRV==true],status[cRV==true]) haz2 - muhaz(NumDaysCustomer[cRV==false],status[cRV==false]) and plotting them. There are many, many more observations in the cohort cRV==false than ==true. And, there are many more observations with lifetimes in the middle of the range than at the ends. I suspect that this is a common phenomenon. When I plot the two hazard rate curves, haz1 looks very different than haz2, but I'd like to see if it's different enough. How can I go about creating confidence intervals and plotting them? Thanks, Alan -- Alan Cox Director, User Experience iContact, Corp. p 919.459.1038 f 919.287.2475 __ 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] text processing for plots
Hi R people, I want to write some functions to automate the plotting of some expressions that use some of the plotmath capabilities. An example is a string supplied to a plot call such as: plot(1, 1, ylab = pm10 (ug/m3)) This should actually appear like: plot(1, 1, ylab = expression(PM[10] * ( * mu * g m ^-3 * ))) i.e. pollutant name, units What I would like to do is write a function that will automatically detect certain strings, reformat them, and supply it as an argument to plot. I could just search for the whole string, but I would like something more flexible because there will be many combinations of pollutant (pm10, nox, no2...) and units (ug/m3, mg/m3, ng/m3...). My question is this - is there a way to separately process the pollutant and units and supply the correct overall expression to plot? I'm not sure of the best strategy for this having read through previous posts etc., and would appreciate your help! Many thanks. David Carslaw - Institute for Transport Studies University of Leeds -- View this message in context: http://www.nabble.com/text-processing-for-plots-tp19331219p19331219.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.
Re: [R] Maximum likelihood estimation
Hi,
You should re-write your function `fr' as follows:
fr - function(par){
a - par[1]
b - par[2]
p - par[3]
lambda - par[4]
l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 lambda*b*p
delta - sqrt(abs(l^2 - b*p*lambda))
mt - a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}
However, I don't understand what the following fragment is doing in `fr':
l^2 lambda*b*p
Can you clarfy that?
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of toh
Sent: Thursday, September 04, 2008 9:15 PM
To: r-help@r-project.org
Subject: Re: [R] Maximum likelihood estimation
Yes I'm trying to optimize the parameters a, b, p and lambda where a 0, b
0 and 0 p 1. I attached the error message that I got when I run mle.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,2
17,226,230,
+
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,367,375,
381,
+
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,463,463,
464,
+
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
473,
+ 473,473,473,475,475,475,475)
library(stats4)
fr - function(a, b, p, lambda){
+ l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
+ l^2 lambda*b*p
+ delta - sqrt(abs(l^2 - b*p*lambda))
+ mt -
+ a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
+ logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
+ return(-logl)
+ }
mle(start=list(a=12,b=0.01,p=0.5,lambda=0.01),fr, method=L-BFGS-B,
+ lower = c(0.002, 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
Error in optim(start, f, method = method, hessian = TRUE, ...) :
non-finite finite-difference value [3]
Prof Brian Ripley wrote:
From ?optim
fn: A function to be minimized (or maximized), with first
argument the vector of parameters over which minimization is
to take place. It should return a scalar result.
I think you intended to optimize over c(a,b,p, lambda), so you need to
specify them as a single vector.
You may be making life unnecessarily hard for yourself: see function
mle() in package stats4.
Showing your code without a verbal description of what you are doing
nor the error message you got is less helpful than we need.
On Wed, 3 Sep 2008, toh wrote:
Hi R-experts,
I'm new to R in mle. I tried to do the following but just couldn't
get it right. Hope someone can point out the mistakes. thanks a lot.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,
217,226,230,
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,3
67,375,381,
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,4
63,463,464,
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
473,
473,473,473,475,475,475,475)
fr - function(a, b, p, lambda){
l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 lambda*b*p
delta - sqrt(abs(l^2 - b*p*lambda))
mt -
a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}
optim(c(15,0.01,0.5,0.01),fr, method=L-BFGS-B, lower = c(0.002,
0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
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49.html Sent from the R help mailing list archive at Nabble.com.
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University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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View this message in
Re: [R] Error: can not allocate vector of 117.3 Mb
To see some useful discussions of this problem in various settings search R-Help for the phrase cannot allocate vector of size. Tom ram basnet wrote: Hi R users, I am doing multiscale bootstrapping for clustering through pvclust package. I have large data set (observations 182 and variables 5546). When i tried to make bootstrapping, then i got message as Bootstrap (r = 0.5)Error: cannot allocate vector of size 117.3 Mb. I am new R user and could not understand what is the problem and also don't know the easiet solution. If someone knows, pls help me. Thanks in advance. Ram Kumar Basnet Graduate student. [[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. -- View this message in context: http://www.nabble.com/Error%3A-can-not-allocate-vector-of-117.3-Mb-tp19330258p19331629.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] Question about col.names in write.csv
Dear support, I don't ignore col.names in write.csv. I want to write names for the firts row. How can I do? Thanks in advance __ 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] (with subject)
Oh, YES, thank you! This weekend I'll try and figure out how to plot these events on a 24 hour scale, i.e I'll aggregate the SQL query on the time But not on the date) to see how many of those fall outside of normal working hours :-)-O greetings, el On 05 Sep 2008, at 00:59 , Gabor Grothendieck wrote: I assume the problem is that you want the axis to have all 12 months but your data is much shorter. Try this: mos - seq(as.Date(2008-01-01), length = 12, by = month) plot(range(mos), range(rawData$y), type = n, xaxt = n) lines(rawData$Date, rawData$y) axis(1, mos, month.abb) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Use of colour in plots
Here is an example doing the same type of thing. It should be easy enough to adapt. Good luck === x - runif(100, 0, 1) y - runif(100, 0, 1) z - data.frame(x,y) plot(subset(z, z$y =.5), col=red, ylim=c(min(z$y), max(z$y)), pch=16) points(subset(z, z$y =.49), col=blue, pch=16) === --- On Fri, 9/5/08, Steve Murray [EMAIL PROTECTED] wrote: From: Steve Murray [EMAIL PROTECTED] Subject: [R] Use of colour in plots To: r-help@r-project.org Received: Friday, September 5, 2008, 9:10 AM Dear all, I have 3 datasets all of which share the same longitude and latitude values, which I'm looking to plot onto a scattergraph. The third dataset has values which can only be either '1' or '2'. So to incorporate all three datasets onto two axes, I'm wondering if I can plot dataset1 and dataset2 as normal, but then use colour to determine whether these points are either values '1' or '2' according to the third dataset. If so, how would I go about doing this in R, and what format would the command take? Thanks for any help offered, Steve __ 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] Example on Maximum likelihood estimation using R
Hi all,
I am very new to R too, but I read that R is powerful.
May I know given a set of data, are there any simple examples on using mle()
to estimate parameters of a lognormal and weibull distribution ?
Hope to hear from you soon.
Thank you
Vincent
Ravi Varadhan wrote:
Hi,
You should re-write your function `fr' as follows:
fr - function(par){
a - par[1]
b - par[2]
p - par[3]
lambda - par[4]
l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 lambda*b*p
delta - sqrt(abs(l^2 - b*p*lambda))
mt - a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}
However, I don't understand what the following fragment is doing in `fr':
l^2 lambda*b*p
Can you clarfy that?
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: [EMAIL PROTECTED]
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On
Behalf Of toh
Sent: Thursday, September 04, 2008 9:15 PM
To: r-help@r-project.org
Subject: Re: [R] Maximum likelihood estimation
Yes I'm trying to optimize the parameters a, b, p and lambda where a 0,
b
0 and 0 p 1. I attached the error message that I got when I run mle.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,2
17,226,230,
+
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,367,375,
381,
+
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,463,463,
464,
+
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
473,
+ 473,473,473,475,475,475,475)
library(stats4)
fr - function(a, b, p, lambda){
+ l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
+ l^2 lambda*b*p
+ delta - sqrt(abs(l^2 - b*p*lambda))
+ mt -
+ a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
+ logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
+ return(-logl)
+ }
mle(start=list(a=12,b=0.01,p=0.5,lambda=0.01),fr, method=L-BFGS-B,
+ lower = c(0.002, 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
Error in optim(start, f, method = method, hessian = TRUE, ...) :
non-finite finite-difference value [3]
Prof Brian Ripley wrote:
From ?optim
fn: A function to be minimized (or maximized), with first
argument the vector of parameters over which minimization is
to take place. It should return a scalar result.
I think you intended to optimize over c(a,b,p, lambda), so you need to
specify them as a single vector.
You may be making life unnecessarily hard for yourself: see function
mle() in package stats4.
Showing your code without a verbal description of what you are doing
nor the error message you got is less helpful than we need.
On Wed, 3 Sep 2008, toh wrote:
Hi R-experts,
I'm new to R in mle. I tried to do the following but just couldn't
get it right. Hope someone can point out the mistakes. thanks a lot.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,
217,226,230,
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,3
67,375,381,
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,4
63,463,464,
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
473,
473,473,473,475,475,475,475)
fr - function(a, b, p, lambda){
l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 lambda*b*p
delta - sqrt(abs(l^2 - b*p*lambda))
mt -
a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}
optim(c(15,0.01,0.5,0.01),fr, method=L-BFGS-B, lower = c(0.002,
0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
--
View this message in context:
http://www.nabble.com/Maximum-likelihood-estimation-tp19304249p193042
49.html Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44
[R] using nls to fit a curve to data
Hi, I am trying to fit a curve to data. My command line is: model10=nls(offspring~((A*c^k)/gamma(k))*((degdays-alpha)^(k-1))*exp(-c*(degdays-alpha)), start=list(A=30,k=2,c=.018,alpha=131)) I get the error message: Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an infinity produced when evaluating the model In addition: Warning message: full precision was not achieved in 'gammafn' My starting values yield a curve very similar t to other models which I fitted successfully. So any suggestions? I've tried different starting values. -- View this message in context: http://www.nabble.com/using-nls-to-fit-a-curve-to-data-tp19332210p19332210.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.
Re: [R] Use of colour in plots
Hi [EMAIL PROTECTED] napsal dne 05.09.2008 16:24:35: Here is an example doing the same type of thing. It should be easy enough to adapt. Good luck === x - runif(100, 0, 1) y - runif(100, 0, 1) z - data.frame(x,y) plot(subset(z, z$y =.5), col=red, ylim=c(min(z$y), max(z$y)), pch=16) points(subset(z, z$y =.49), col=blue, pch=16) === Or third - (z$y =.5)+1 plot(z, col=third, pch=16) Just tell to col a vector of colors with appropriate use of logical. Or you can use col = as.numeric(some factor), which is quite convenient use of factors feature which is not desired in other cases. See warning section of factor help page. Regards --- On Fri, 9/5/08, Steve Murray [EMAIL PROTECTED] wrote: From: Steve Murray [EMAIL PROTECTED] Subject: [R] Use of colour in plots To: r-help@r-project.org Received: Friday, September 5, 2008, 9:10 AM Dear all, I have 3 datasets all of which share the same longitude and latitude values, which I'm looking to plot onto a scattergraph. The third dataset has values which can only be either '1' or '2'. So to incorporate all three datasets onto two axes, I'm wondering if I can plot dataset1 and dataset2 as normal, but then use colour to determine whether these points are either values '1' or '2' according to the third dataset. If so, how would I go about doing this in R, and what format would the command take? Thanks for any help offered, Steve __ 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-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] Density estimates in modelling framework
Thanks for all your replies. The suggestions I got were: * gss * logspline * locfit I ended up using locfit because the interface was just right for my needs, and it was faster than the alternatives. Hadley On Fri, Aug 29, 2008 at 2:03 PM, hadley wickham [EMAIL PROTECTED] wrote: Hi all, Do any packages implement density estimation in a modelling framework? I want to be able to do something like: dmodel - density(~ a + b, data = mydata) predict(dmodel, newdata) This isn't how sm or KernSmooth or base density estimation works. Are there other packages that do density estimation? Or is there some reason that this is a bad idea. Hadley -- http://had.co.nz/ -- http://had.co.nz/ __ 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] ggplot2: Changes to grobs not saved to file output
Unfortunately it's not particularly easy in the current version. In the next version, you can do: p - qplot(wt, mpg, data=mtcars, colour=cyl) # Get the plot grob grob - ggplotGrob(p) # Modify it place grob - geditGrob(grob, gPath(strip,label), gp=gpar(fontface=bold)) # Draw it pdf(...) grid.newpage() grid.draw(grob) dev.off() I think in the current version you can do qplot(wt, mpg, data=mtcars, colour=cyl) grob - grid.grab() and then follow the remaining steps. Regards, Hadley On Fri, Aug 29, 2008 at 8:58 AM, btcruiser [EMAIL PROTECTED] wrote: Hello, Maybe I missed something - most likely .:-( I create a gplot and then makes some changes to the plot using grid graphics functions. These changes show up on the display OK, but when I save using ggsave() the grid changes do not show up. How do I save the plot with these changes? Thanks in advance. -- View this message in context: http://www.nabble.com/ggplot2%3A-Changes-to-grobs-not-saved-to-file-output-tp19220492p19220492.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. -- http://had.co.nz/ __ 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] binary order combinations
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2 names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
__
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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] binary order combinations
Does ?combos in the quantreg package do what you want?
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,
13] [,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
V13 V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2
names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
__
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] controlling lattice plot ticks with relation=free
How do you persuade lattice to draw tick marks on both the left and
right
side of the y-axis, when relation=free in the scales component?
#Ticks appear on both sides
histogram(~height|voice.part, data=singer)
##Ticks only on left
histogram(~height|voice.part, data=singer,
scales=list(y=list(relation=free)))
I had an off list reply that was very helpful, suggesting the use of
panel.axis. For posterity, here's my own answer to my problem, using
yscale.component. It's not a particularly natural solution, and it
requires manually increasing the horizontal spacing between plots (with
the between agrument), but it works.
myyscale.component - function(...)
{
ans - yscale.components.default(...)
ans$right - ans$left
ans$right$labels$labels - NULL
ans
}
histogram(~height|voice.part, data=singer,
scales=list(y=list(relation=free)),
yscale.component=myyscale.component,
between=list(x=.5))
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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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.
[R] Time Series (attribute)
Hello, what kind of advantages does R's time series offer? Would it be possible to use data with units? For example data from a real time series with certain sampling rates - could they be used so, that the sample-times are measured in millisceonds or microseconds, instead of an index of the positions? If so, this would be much easier to handle than just working with indices of a vector. If the above mentioned way would not be possible, for what is this timeseries-attribute (?) useful? Ciao, Oliver __ 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 order combinations
I am not sure it can do it. Besides, I ran a test of combos from quantreg:
library(quantreg)
H-1:3
test.combos-lapply(1:3,function(x)
{combn(H,x)
})
Every time I tried it crashed my R...
:(
Dimitri
On 9/5/08, roger koenker [EMAIL PROTECTED] wrote:
Does ?combos in the quantreg package do what you want?
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2 names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
__
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[R] Lowest k values of list
Hi @ all, how do I get the largest or lowest k values of list? It must similar to the min() / max() function, but I just don't get it. Best wishes, Markus __ 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] Plot by column
Dear list,
I have the following matrix. How can I make the following plot?
1. The x-axis has index 1:7, and the first column is plotted against index 1,
second against 2, and so on.
2. I want the points from the left upper conner including the antidiagonal to
be plotted with col=2, and the rest with col=3
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 0.589 0.857 0.923 0.944 0.954 0.963 0.980
[2,] 0.470 0.763 0.877 0.900 0.911 0.957 0.974
[3,] 0.486 0.863 0.905 0.960 0.968 0.998 1.015
[4,] 0.653 0.888 0.943 0.952 0.962 0.992 1.009
[5,] 0.664 0.774 0.958 0.986 0.996 1.027 1.045
[6,] 0.546 0.910 1.011 1.041 1.051 1.083 1.102
[7,] 0.407 0.600 0.667 0.686 0.693 0.715 0.727
The following the my code to do the above with the matrix called r. How can I
simply the code?
plot(r[,1],type=n,ylim=c(0.4,1.2))
for (i in 1:7)
{points(rep(i,8-i),r[,i][1:(8-i)],col=2) }
for (i in 2:7)
{points(rep(i,(i-1)),r[,i][(9-i):7],col=3)}
Thanks
Sincerely,
Yanwei Zhang
Department of Actuarial Research and Modeling
Munich Re America
mailto:[EMAIL PROTECTED]
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Re: [R] Upgrade to R 2.7.2 from Debian Repository
Tom, (That's a suitable question for r-sig-debian) On 5 September 2008 at 03:08, Tom La Bone wrote: | I am running R 2.7.1 on an eeepc with the standard Xandros Linux. I | install/upgrade R binaries using synaptic from the | http://cran.r-project.org/bin/linux/debian/etch-cran/ repository. R 2.7.2 | does not appear to be available from this repository. Will it be available | soon or is there another repository I should be using? The volunteer who is providing these backports for CRAN indicated a few days ago that he is being slowed down due to network issues. So sorry, right now you either have to wait, fine someone else to build the .deb files on etch for you, or do it yourself. We do have the Ubuntu backports in place so if you were contemplating switching your Xandros install with the Ubuntu/EEE setup you'd be covered... Dirk -- Three out of two people have difficulties with fractions. __ 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] Confidence Intervals on Hazard Plots
On Sep 5, 2008, at 9:20 AM, Alan Cox wrote: Hello, Is it possible to create confidence intervals for hazard rates? I'm creating two muhaz objects: haz1 - muhaz(NumDaysCustomer[cRV==true],status[cRV==true]) haz2 - muhaz(NumDaysCustomer[cRV==false],status[cRV==false]) and plotting them. There are many, many more observations in the cohort cRV==false than ==true. And, there are many more observations with lifetimes in the middle of the range than at the ends. I suspect that this is a common phenomenon.When I plot the two hazard rate curves, haz1 looks very different than haz2, but I'd like to see if it's different enough. How can I go about creating confidence intervals and plotting them? ?muhaz.object says that var.min and bias.min could be extracted from haz1 and haz2, since you have not changed the default for bw.method. The default haz.est vector is twice as long as the default var.min vector, so you will need to take care to only plot the haz.est values only for the the event times, or perhaps you could specify that n.est.grid be the same as n.min.grid at the time of creating your fits. Not sure if the second method might have any undesirable side- effects, however. -- David Winsemius Heritage Laboratories __ 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] dealing with NAs in time series
Certain timeseries I have had outliers, which I removed by assigning NA to their positions. Now acf() refuses to go to work. What's the right way to remove outliers from ts objects, and what are teh standard ways to interpolate NAs in them? Cheers, Alexy __ 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 Series (attribute)
The zoo package can handle series with times of any class that is totally ordered and provides certain methods. That includes all common time classes and allows for you to define new time classes too. See library(zoo) ?zoo and the three vignettes (i.e. pdf documents) that come with it. On Fri, Sep 5, 2008 at 11:22 AM, Oliver Bandel [EMAIL PROTECTED] wrote: Hello, what kind of advantages does R's time series offer? Would it be possible to use data with units? For example data from a real time series with certain sampling rates - could they be used so, that the sample-times are measured in millisceonds or microseconds, instead of an index of the positions? If so, this would be much easier to handle than just working with indices of a vector. If the above mentioned way would not be possible, for what is this timeseries-attribute (?) useful? Ciao, Oliver __ 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] Lowest k values of list
Dear Markus,
Try this:
# Data set
set.seed(123)
mylist=list(a=rnorm(10),b=rpois(15,10),f=rnorm(30,12,3))
mylist
# min and max
temp=lapply(mylist,function(x){
res=c(min(x),max(x))
names(res)=c('Min.','Max.')
res
}
)
do.call(rbind,temp)
Min. Max.
a -1.265061 1.715065
b 3.00 15.00
f 7.353742 18.506868
HTH,
Jorge
On Fri, Sep 5, 2008 at 11:26 AM, Markus Mühlbacher [EMAIL PROTECTED]wrote:
Hi @ all,
how do I get the largest or lowest k values of list? It must similar to the
min() / max() function, but I just don't get it.
Best wishes,
Markus
__
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.
Re: [R] Lowest k values of list
On Sep 5, 2008, at 11:26 AM, Markus Mühlbacher wrote: Hi @ all, how do I get the largest or lowest k values of list? It must similar to the min() / max() function, but I just don't get it. Might depend on how you define high and low. Consider these experiments: z - list(1,4,5,6,,2,3) head(sort(unlist(z)),n=2) [1] 1 2 tail(sort(unlist(z)),n=2) [1] 5 6 z - list(1,2,3,4,5,6,a) tail(sort(unlist(z)),n=2) [1] 6 a head(sort(unlist(z)),n=2) [1] 1 2 -- David Winsemius __ 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 order combinations
names - sprintf(V%d, 1:4);
n - length(names);
stopifnot(n = 32); # Theoretical upper limit
x - matrix(intToBits(1:(2^n-1)), ncol=2^n-1);
x - x[1:n,,drop=FALSE];
keys - apply(x, MARGIN=2, FUN=function(z) paste(names[as.logical(z)],
collapse=:));
print(keys);
[1] V1 V2 V1:V2 V3
[5] V1:V3 V2:V3 V1:V2:V3V4
[9] V1:V4 V2:V4 V1:V2:V4V3:V4
[13] V1:V3:V4V2:V3:V4V1:V2:V3:V4
/H
On Fri, Sep 5, 2008 at 8:23 AM, Dimitri Liakhovitski [EMAIL PROTECTED] wrote:
I am not sure it can do it. Besides, I ran a test of combos from quantreg:
library(quantreg)
H-1:3
test.combos-lapply(1:3,function(x)
{combn(H,x)
})
Every time I tried it crashed my R...
:(
Dimitri
On 9/5/08, roger koenker [EMAIL PROTECTED] wrote:
Does ?combos in the quantreg package do what you want?
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2 names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
__
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-help@r-project.org mailing list
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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] Lowest k values of list
Do you mean a vector? If so, head/tail will work for you: x - runif(10) x [1] 0.26550866 0.37212390 0.57285336 0.90820779 0.20168193 0.89838968 0.94467527 [8] 0.66079779 0.62911404 0.06178627 # lowest 5 head(sort(x), 5) [1] 0.06178627 0.20168193 0.26550866 0.37212390 0.57285336 # highest 5 tail(sort(x), 5) [1] 0.6291140 0.6607978 0.8983897 0.9082078 0.9446753 On Fri, Sep 5, 2008 at 11:26 AM, Markus Mühlbacher [EMAIL PROTECTED] wrote: Hi @ all, how do I get the largest or lowest k values of list? It must similar to the min() / max() function, but I just don't get it. Best wishes, Markus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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 order combinations
Sorry,
I misread it first. I tried:
library(quantreg)
test.combos-lapply(1:15,function(x)
{combos(15,x)
})
It gives me a list with an order that is somewhat different from
before, but I am not sure it helps me much:
[[1]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]1234567891011121314
[,15]
[1,]15
[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]111111111 1 1 1 1 1
[2,]2 15 14 13 12 11 1098 7 6 5 4 3
[,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
[1,] 2 2 2 2 2 2 2 2 2 2 2 2
[2,] 3151413121110 9 8 7 6 5
Dimitri
On 9/5/08, roger koenker [EMAIL PROTECTED] wrote:
You need to read the help file ?combos, and then use it to do indexing
of your objects, it only knows how to construct the integer combinations
given a pair (n,p).
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 10:23 AM, Dimitri Liakhovitski wrote:
I am not sure it can do it. Besides, I ran a test of combos from quantreg:
library(quantreg)
H-1:3
test.combos-lapply(1:3,function(x)
{combn(H,x)
})
Every time I tried it crashed my R...
:(
Dimitri
On 9/5/08, roger koenker [EMAIL PROTECTED] wrote:
Does ?combos in the quantreg package do what you want?
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
V13
V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2
names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
__
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-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] Lowest k values of list
Dear Markus,
I think that I missed something in my previous reply. Here is another
approach using the first k values of each list:
# Data set
set.seed(123)
mylist=list(a=rnorm(10),b=rpois(15,10),f=rnorm(30,12,3))
# Function to report the first k values
k.values=function(x,k){
res= x[order(x)][1:k]
names(res)=paste('value_',1:k,sep=)
res
}
# Example: first 3 values
res=lapply(mylist,k.values,k=3)
do.call(rbind,res)
value_1value_2value_3
a -1.265061 -0.6868529 -0.5604756
b 3.00 4.000 6.000
f 7.353742 8.2038109 8.5855892
HTH,
Jorge
On Fri, Sep 5, 2008 at 11:26 AM, Markus Mühlbacher [EMAIL PROTECTED]wrote:
Hi @ all,
how do I get the largest or lowest k values of list? It must similar to the
min() / max() function, but I just don't get it.
Best wishes,
Markus
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
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 order combinations
You need to read the help file ?combos, and then use it to do indexing
of your objects, it only knows how to construct the integer combinations
given a pair (n,p).
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 10:23 AM, Dimitri Liakhovitski wrote:
I am not sure it can do it. Besides, I ran a test of combos from
quantreg:
library(quantreg)
H-1:3
test.combos-lapply(1:3,function(x)
{combn(H,x)
})
Every time I tried it crashed my R...
:(
Dimitri
On 9/5/08, roger koenker [EMAIL PROTECTED] wrote:
Does ?combos in the quantreg package do what you want?
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,
13]
[,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11
V12 V13
V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2
names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
V1 V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12
V13 V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of
the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
__
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-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] how to draw the legend about color from 3d picture
Look at the color.legend function in the plotrix package to see if it does what you want. You may want to use the layout function to split the graphics device into 2 sections, one for the persp plot and the other to hold the legend. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of yk Sent: Friday, September 05, 2008 6:07 AM To: r-help@r-project.org Subject: [R] how to draw the legend about color from 3d picture I have drawed a picture with persp, it's 3d map with different color, indicate different altitude. In gnuplot, the corresponding command 'splot' will generate a picture beside to indicate the relationship between color and altitude. But in R, how to draw it? I have read the manual of legend, but they are all about how to draw a legend with colored text, not a continuous varing color with corresponding number. __ 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] lm and time series.
I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. Thank you. Kevin __ 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 order combinations
Henrik,
this is amazing! wow!
Thank you so much!
Dimitri
On 9/5/08, Henrik Bengtsson [EMAIL PROTECTED] wrote:
names - sprintf(V%d, 1:4);
n - length(names);
stopifnot(n = 32); # Theoretical upper limit
x - matrix(intToBits(1:(2^n-1)), ncol=2^n-1);
x - x[1:n,,drop=FALSE];
keys - apply(x, MARGIN=2, FUN=function(z) paste(names[as.logical(z)],
collapse=:));
print(keys);
[1] V1 V2 V1:V2 V3
[5] V1:V3 V2:V3 V1:V2:V3V4
[9] V1:V4 V2:V4 V1:V2:V4V3:V4
[13] V1:V3:V4V2:V3:V4V1:V2:V3:V4
/H
On Fri, Sep 5, 2008 at 8:23 AM, Dimitri Liakhovitski [EMAIL PROTECTED]
wrote:
I am not sure it can do it. Besides, I ran a test of combos from quantreg:
library(quantreg)
H-1:3
test.combos-lapply(1:3,function(x)
{combn(H,x)
})
Every time I tried it crashed my R...
:(
Dimitri
On 9/5/08, roger koenker [EMAIL PROTECTED] wrote:
Does ?combos in the quantreg package do what you want?
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
Dear all!
I have a vector of names
names-(V1, V2, V3,., V15)
I could create all possible combinations of these names (of all
lengths) using R:
combos-lapply(1:15,function(x)
{combn(names,x)
})
I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):
[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,14]
[1,] V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
V14
[,15]
[1,] V15
[[2]] - the second element contains all possible combinations of 2 names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[1,] V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1 V1
[2,] V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13
V14
.
.
.
etc.
My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc
So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.
Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?
Thank you very much!
Dimitri Liakhovitski
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Re: [R] lm and time series.
The Time Series section in ?lm should be self explanatory. If you are using diff's and lag's then look at the dyn package. On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. Thank you. Kevin __ 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] Lowest k values of list
In addition to the other responses, you may want to look at the 'partial' argument to sort. For large vectors it may speed things up to not sort everything if all you care about is the top and bottom few. Try: tmp - rnorm(100) plot( sort(tmp, partial=c(1,2,99,100) ) ) Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Markus Mühlbacher Sent: Friday, September 05, 2008 9:26 AM To: r-help@r-project.org Subject: [R] Lowest k values of list Hi @ all, how do I get the largest or lowest k values of list? It must similar to the min() / max() function, but I just don't get it. Best wishes, Markus __ 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] Plot by column
You can try something like about this:
r[upper.tri(r)] - NA
matplot(t(r), lty = 1, type = 'p', pch = 1, ylim = c(0.4, 1.2), col =
'black')
On Fri, Sep 5, 2008 at 12:27 PM, Zhang Yanwei - Princeton-MRAm
[EMAIL PROTECTED] wrote:
Dear list,
I have the following matrix. How can I make the following plot?
1. The x-axis has index 1:7, and the first column is plotted against index
1, second against 2, and so on.
2. I want the points from the left upper conner including the antidiagonal
to be plotted with col=2, and the rest with col=3
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 0.589 0.857 0.923 0.944 0.954 0.963 0.980
[2,] 0.470 0.763 0.877 0.900 0.911 0.957 0.974
[3,] 0.486 0.863 0.905 0.960 0.968 0.998 1.015
[4,] 0.653 0.888 0.943 0.952 0.962 0.992 1.009
[5,] 0.664 0.774 0.958 0.986 0.996 1.027 1.045
[6,] 0.546 0.910 1.011 1.041 1.051 1.083 1.102
[7,] 0.407 0.600 0.667 0.686 0.693 0.715 0.727
The following the my code to do the above with the matrix called r. How
can I simply the code?
plot(r[,1],type=n,ylim=c(0.4,1.2))
for (i in 1:7)
{points(rep(i,8-i),r[,i][1:(8-i)],col=2) }
for (i in 2:7)
{points(rep(i,(i-1)),r[,i][(9-i):7],col=3)}
Thanks
Sincerely,
Yanwei Zhang
Department of Actuarial Research and Modeling
Munich Re America
mailto:[EMAIL PROTECTED]
[[alternative HTML version deleted]]
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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[R] Adding a legend to R graph device with several plots (no to individual plots!)
Hi Nelson,
I don't know if you got your answer or not but here it is my low-tech solution
for what is worth. I like to do a layout as I want my graphs to be and add some
space where I plot an invisible graph with dummy data and add the legend
there. In this case I decided to put the legend on middle top of the layout.
Because you used lapply to do your graphs it seems that I have to change the
numbers in the p.names parameter because somehow it establishes the order of
your graphs as well. I usually use the simple for loop because you never have
loads of graphs in one page, so it is safe enough.
I hope this helps a little.
Monica
# code starts
l - layout(matrix(c(rep(1, 2), seq(2,5)), 3,2, byrow = T), c(3,3,3),
c(1,3,3,3))
layout.show(l)
par(mar = c(1,1,1,1))
plot(seq(1,10), seq(1,10), type = n, axes = F)
smartlegend(x= center, y=top, c(1995,2006), lty=2:1)
par(mar = c(4,4,3,2))
p.names - c(2,3,4,5)### change p.names definition
lapply(p.names, function(x) {
d1 - density(rnorm(100,0,1))
d2 - density(rnorm(100,.3,1.2))
plot(range(d1$x, d2$x), range(d1$y, d2$y), type = n, xlab = , ylab = )
## change xlab, ylab in plot
lines(d1, col = black, lty = 2)
lines(d2, col = black, lty = 1)
})
mtext(Learning R the hard way, side = 1, line=-1.5, outer=TRUE)
code ends
--
Message: 16
Date: Thu, 4 Sep 2008 09:35:42 -0400
From: Nelson Villoria
Subject: [R] Adding a legend to R graph device with several plots (no
to individual plots!)
To: r-help@r-project.org
Message-ID:
Content-Type: text/plain; charset=ISO-8859-1
Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
plots in my device. each one has a density for 1995 and one for 2006.
I have found that using legend or smartlegend I can add a legend to
each plot, but I am looking for something in the spirit of mtext. That
is, putting the legend anywhere I want on the device. In my situation
I have:
par(mfrow=c(2,2), ann = FALSE)
p.names - c(1,2,3,4)
lapply(p.names, function(x) {
d1 - density(rnorm(100,0,1))
d2 - density(rnorm(100,.3,1.2))
plot(range(d1$x, d2$x), range(d1$y, d2$y), type = n, xlab = NULL,
ylab = NULL)
lines(d1, col = black, lty = 2)
lines(d2, col = black, lty = 1)
})
mtext(Learning R the hard way, side = 1, line=-1.5, outer=TRUE)
smartlegend(x= center, y=top, c(1995,2006), lty=2:1)
the last line puts the legend on the lower-right plot. I'd like to put
it in the middle - or top of the device. Any suggestion?
Thanks!
Nelson
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50F681DAD532637!5295.entry?ocid=TXT_TAGLM_WL_domore_092008
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[R] casting help please
I have a data.frame which I believe is melted already and am having trouble casting it to 'wide' format. It looks something like (x - data.frame(ticker=c(rep(A,5),rep(B,6)), date=c(1:5, 1:6), value=c(NA,100*exp(rnorm(10,0,.1) cast(x, date ~ ticker) # this does what I want with toy data But when I use my real data frame str(prices) 'data.frame': 308188 obs. of 3 variables: $ Ticker: chr ticker1 ticker1 ticker1 ticker1 ... $ Date:Class 'Date' num [1:308188] 12296 12297 12298 12299 12300 ... $ Price : num NA NA NA NA NA NA NA NA NA NA ... I get prices.wide - cast(prices, Date ~ Ticker, add.missing=TRUE) Error in data.frame(data[, c(variables), drop = FALSE], result = data$value) : arguments imply differing number of rows: 308188, 0 (I tried various other arguments to cast - all gave the same error message.) It is a fact that the various tickers have data for different date ranges in the data frame and there are lots of NA's, but the toy example above has different date ranges for the two tickers and an NA, so I don't know what else to look for in my data, or what args to cast might make it work. Any insights or direction would be much appreciated. David L. Reiner, PhD Head Quant Rho Trading Securities, LLC [[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] Question about col.names in write.csv
Hi Luz, No entiendo bien tu pregunta. Querés grabar una tabla con nombres en las columnas y tambien en la primera fila? Si es asi, tenés que asignarle los nombres a la tabla antes de grabar. Por ejemplo: mi.tabla=matrix(runif(30),ncol=3) colnames(mi.tabla)=c(A,B,C) rownames(mi.tabla)=c(D,rep(,9)) write.csv(mi.tabla,file=mi.archivo.csv) Cuando asignás nombres a las filas o columnas, el vector tiene que tener la misma longitud que el numero de filas o columnas. Entonces, para darle un nombre solamente a la primera fila, hice un vector con el nombre y nueve espacios en blanco (para completar las diez filas que tiene mi tabla). Al grabar la tabla usando write.csv, el comportamiento default es guardar los nombres de las columnas y filas. Saludos, Julian Luz Milena Zea Fernandez wrote: Dear support, I don't ignore col.names in write.csv. I want to write names for the firts row. How can I do? Thanks in advance __ 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] casting help please
Hi David, Does getting rid of the missings help? prices - prices[complete.cases(prices), ] Also, you shouldn't need add.missing=TRUE - what are you trying to do? Hadley On Fri, Sep 5, 2008 at 12:12 PM, [EMAIL PROTECTED] wrote: I have a data.frame which I believe is melted already and am having trouble casting it to 'wide' format. It looks something like (x - data.frame(ticker=c(rep(A,5),rep(B,6)), date=c(1:5, 1:6), value=c(NA,100*exp(rnorm(10,0,.1) cast(x, date ~ ticker) # this does what I want with toy data But when I use my real data frame str(prices) 'data.frame': 308188 obs. of 3 variables: $ Ticker: chr ticker1 ticker1 ticker1 ticker1 ... $ Date:Class 'Date' num [1:308188] 12296 12297 12298 12299 12300 ... $ Price : num NA NA NA NA NA NA NA NA NA NA ... I get prices.wide - cast(prices, Date ~ Ticker, add.missing=TRUE) Error in data.frame(data[, c(variables), drop = FALSE], result = data$value) : arguments imply differing number of rows: 308188, 0 (I tried various other arguments to cast - all gave the same error message.) It is a fact that the various tickers have data for different date ranges in the data frame and there are lots of NA's, but the toy example above has different date ranges for the two tickers and an NA, so I don't know what else to look for in my data, or what args to cast might make it work. Any insights or direction would be much appreciated. David L. Reiner, PhD Head Quant Rho Trading Securities, LLC [[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. -- http://had.co.nz/ __ 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] casting help please
I think I just remembered that in a melted data.frame the values have to be called 'value' and tried renaming my third column so. Now it works (without add.missing). Thanks for this powerful piece of software! -- David (I was trying to convert a set of data in 'long' format to 'wide' format. Basically I have a pretty long price history for many tickers basically stacked up as indicated in the toy example and want them reshaped with each ticker in a column with dates in the first column. Maybe cast was not the best tool? But it works now.) -Original Message- From: hadley wickham [mailto:[EMAIL PROTECTED] Sent: Friday, September 05, 2008 1:10 PM To: David Reiner [EMAIL PROTECTED] Cc: r-help@r-project.org Subject: [SPAM] - Re: [R] casting help please - Found word(s) list error in the Text body Hi David, Does getting rid of the missings help? prices - prices[complete.cases(prices), ] Also, you shouldn't need add.missing=TRUE - what are you trying to do? Hadley On Fri, Sep 5, 2008 at 12:12 PM, [EMAIL PROTECTED] wrote: I have a data.frame which I believe is melted already and am having trouble casting it to 'wide' format. It looks something like (x - data.frame(ticker=c(rep(A,5),rep(B,6)), date=c(1:5, 1:6), value=c(NA,100*exp(rnorm(10,0,.1) cast(x, date ~ ticker) # this does what I want with toy data But when I use my real data frame str(prices) 'data.frame': 308188 obs. of 3 variables: $ Ticker: chr ticker1 ticker1 ticker1 ticker1 ... $ Date:Class 'Date' num [1:308188] 12296 12297 12298 12299 12300 ... $ Price : num NA NA NA NA NA NA NA NA NA NA ... I get prices.wide - cast(prices, Date ~ Ticker, add.missing=TRUE) Error in data.frame(data[, c(variables), drop = FALSE], result = data$value) : arguments imply differing number of rows: 308188, 0 (I tried various other arguments to cast - all gave the same error message.) It is a fact that the various tickers have data for different date ranges in the data frame and there are lots of NA's, but the toy example above has different date ranges for the two tickers and an NA, so I don't know what else to look for in my data, or what args to cast might make it work. Any insights or direction would be much appreciated. David L. Reiner, PhD Head Quant Rho Trading Securities, LLC [[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. -- http://had.co.nz/ __ 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] upgraded maps database for italy?
Dear UseRs, I'm using the library maps. I'm drawing maps of Italy. The map available for this library were prepared around *1989*: This italy database comes from the NUTS III (Tertiary Administrative Units of the European Community) database of the United Nations Environment Programme (UNEP) GRID-Geneva data sets. These were prepared around *1989* [cited: help(italy)]. The details of this map is by province, which is not maybe optimal, but is OK. However, after 1989 many new province are born. So for example the new provincia of Rimini has born and the sardinan Island has doubled them. Some of you do know how to get upgraded maps of italy? I think it would useful for all Italian mapers. The second question is as follow: Italy is subdivided in regioni. Each regione is a set of many province; so, if I want to plot map region-based index, how can I omit the borders of prince belonging to the same regione? so, how plot maps to a higher unit level? [[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] upgraded maps database for italy?
If I remember correctly, the maps package only has a few dedicated maps that works with it. The maptools package has similar (but not exactly the same) functionality and can work with shapefiles that are more commonly available. Maptools also integrates nicely with the sp and related packages that give many more options for working with the maps. For the 2nd question, I would suggest plotting the map first using the province, then plot a second map over the top using the information on the regioni using a different color/line thickness for the boarders and no fill (so that the province show through). This requires having 2 different map files, one with province and one with regioni information (there may be a better way with overlaying information, but this method works without having to think about it too much). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of livio spam Sent: Friday, September 05, 2008 12:24 PM To: r-help@r-project.org Subject: [R] upgraded maps database for italy? Dear UseRs, I'm using the library maps. I'm drawing maps of Italy. The map available for this library were prepared around *1989*: This italy database comes from the NUTS III (Tertiary Administrative Units of the European Community) database of the United Nations Environment Programme (UNEP) GRID-Geneva data sets. These were prepared around *1989* [cited: help(italy)]. The details of this map is by province, which is not maybe optimal, but is OK. However, after 1989 many new province are born. So for example the new provincia of Rimini has born and the sardinan Island has doubled them. Some of you do know how to get upgraded maps of italy? I think it would useful for all Italian mapers. The second question is as follow: Italy is subdivided in regioni. Each regione is a set of many province; so, if I want to plot map region-based index, how can I omit the borders of prince belonging to the same regione? so, how plot maps to a higher unit level? [[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] asscii2netcdf]
Hi Marshall, This link provides very good directions and examples you can use. http://www.image.ucar.edu/GSP/Software/Netcdf/ A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter --- On Fri, 9/5/08, Marshall Mdoka [EMAIL PROTECTED] wrote: From: Marshall Mdoka [EMAIL PROTECTED] Subject: Re: [R] asscii2netcdf] To: r-help@r-project.org Date: Friday, September 5, 2008, 1:15 PM Dear All, A friend of mine seeks help in converting the attached ascii to netcdf. Any assistance will be sincerely appreciated. Regards, Marshall __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ 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] boxplot including null info from dataframe, not with SQLite dataframe
I have been trying to use R to gather some information from parsed log files (as part of examining some performance issues). I parsed the log files and put the data into an SQLite database, and then used RSQLite to load the data into R. The fields of interest are controller, action and total_time: controller and action have string values; total_time has a decimal value. I first did the following box plot to find the problem controllers. boxplot(total_time ~ controller, all_data) Having identified one controller of interest (let's say BadController), I then wanted to then focus on the actions associated with that controller. So I did this: boxplot(total_time ~ action, subset(all_data, controller == BadController)) This gave me a plot I was expecting: just the actions which are associated with BadController. However, I'd done this work on a FreeBSD system, and then I wanted to print it, and the easiest means seemed to re-plot using R on Windows. So I wrote the data to a file, moved it to Windows and loaded it up there. On FreeBSD: write.table(all_data, datafile.R) On Windows: all_data - read.table(datafile.R) However, on Windows, when I get to the boxplot of the subset of data, I'm seeing every action that's part of all_data, not just the ones that are associated with BadController. Eventually, I found that this wasn't a Windows vs. FreeBSD issue, because if I reload the data from the file on FreeBSD, I start to see the same behavior. Based on a few bug reports I found, it would seem that this is working as designed. So my question is: What do I do to get it to truly ignore the actions with no data? I've tried adding drop=FALSE to the subset call; that hasn't worked. I also tried specifically adding na.action=NULL to the boxplot call, with no change. I'm also curious what's different between a data frame loaded from SQLite versus a data frame loaded from a file. In the mean time, I'll either try to install RSQLite on Windows or get postscript working on FreeBSD. (My quick attempt with postscript in R on FreeBSD was not drawing the bounding box nor the axes.) Thanks for any help. -- Coey Minear __ 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] comparing two files
Hi there I have two object on is a vector T and the other is dataframe C. vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I want to find the the missing rows in dataframe C.That is those values that are not matchig in dataframe C[,2] Kindly give me suggestion on how to go about it. Ramya -- View this message in context: http://www.nabble.com/comparing-two-files-tp19337486p19337486.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.
Re: [R] library/function that estimates parameters of well known distributions from empirical data?
Ted Byers r.ted.byers at gmail.com writes: I found this a few months ago, but for the life of me I can't remember what the function or package was, and I have had no luck finding it this week. I have found, again, the functions for working with distributions like Cauchy, F, normal, c., and ks.test, but I have not found the functions for estimating the distribution parameters given a vector of values. Look at the fitdistr function in the MASS package. Consider AIC comparisons for ranking the fits to these non-nested models. good luck Ben Bolker __ 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] boxplot including null info from dataframe, not with SQLite dataframe
Coey Minear cminear at securecomputing.com writes: I have been trying to use R to gather some information from parsed log files (as part of examining some performance issues). I parsed the log files and put the data into an SQLite database, and then used RSQLite to load the data into R. The fields of interest are controller, action and total_time: controller and action have string values; total_time has a decimal value. I first did the following box plot to find the problem controllers. boxplot(total_time ~ controller, all_data) Having identified one controller of interest (let's say BadController), I then wanted to then focus on the actions associated with that controller. So I did this: boxplot(total_time ~ action, subset(all_data, controller == BadController)) This gave me a plot I was expecting: just the actions which are associated with BadController. However, I'd done this work on a FreeBSD system, and then I wanted to print it, and the easiest means seemed to re-plot using R on Windows. So I wrote the data to a file, moved it to Windows and loaded it up there. On FreeBSD: write.table(all_data, datafile.R) On Windows: all_data - read.table(datafile.R) I'm guessing that you want bad - subset(all_data,controller==BadController) bad$action - factor(bad$action) boxplot(total_time ~ action) Subsetting doesn't drop factor levels that don't occur, which is an unfortunate design decision ... Ben Bolker __ 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] lm and time series.
I am sorry but I looked at ?lm and could not see any guidance on writting a formula. If I have two arrays or a data set then I know how to do that (y ~ x) but for a time series I am not sure how to write y or x. Thank you. Kevin Gabor Grothendieck [EMAIL PROTECTED] wrote: The Time Series section in ?lm should be self explanatory. If you are using diff's and lag's then look at the dyn package. On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. Thank you. Kevin __ 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] comparing two files
?which which(C[,2]!= T) # untested --- On Fri, 9/5/08, Rajasekaramya [EMAIL PROTECTED] wrote: From: Rajasekaramya [EMAIL PROTECTED] Subject: [R] comparing two files To: r-help@r-project.org Received: Friday, September 5, 2008, 2:57 PM Hi there I have two object on is a vector T and the other is dataframe C. vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I want to find the the missing rows in dataframe C.That is those values that are not matchig in dataframe C[,2] Kindly give me suggestion on how to go about it. Ramya -- View this message in context: http://www.nabble.com/comparing-two-files-tp19337486p19337486.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. __ [[elided Yahoo spam]] __ 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] comparing two files
?match T.v - sample(1:20) C.df - data.frame(a=1:10, b= sample(1:20, 10)) T.v [1] 10 12 9 4 14 11 19 2 18 5 16 6 7 17 8 20 1 13 3 15 C.df a b 1 1 10 2 2 17 3 3 8 4 4 5 5 5 2 6 6 16 7 7 19 8 8 7 9 9 18 10 10 14 # find the missing values T.v[is.na(match(T.v, C.df[,2]))] [1] 12 9 4 11 6 20 1 13 3 15 On Fri, Sep 5, 2008 at 2:57 PM, Rajasekaramya [EMAIL PROTECTED] wrote: Hi there I have two object on is a vector T and the other is dataframe C. vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I want to find the the missing rows in dataframe C.That is those values that are not matchig in dataframe C[,2] Kindly give me suggestion on how to go about it. Ramya -- View this message in context: http://www.nabble.com/comparing-two-files-tp19337486p19337486.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] comparing two files
Hi there, See ?%in% Here is an example: x=sample(1:10) x [1] 2 1 10 5 4 7 3 6 8 9 y=sample(4:20) y [1] 17 16 18 10 8 19 4 5 11 13 12 14 20 6 7 9 15 x[x%in%y] # x values present in y [1] 10 5 4 7 6 8 9 HTH, Jorge On Fri, Sep 5, 2008 at 2:57 PM, Rajasekaramya [EMAIL PROTECTED]wrote: Hi there I have two object on is a vector T and the other is dataframe C. vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I want to find the the missing rows in dataframe C.That is those values that are not matchig in dataframe C[,2] Kindly give me suggestion on how to go about it. Ramya -- View this message in context: http://www.nabble.com/comparing-two-files-tp19337486p19337486.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. [[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] dealing with NAs in time series
look at zoo na.approx this will interpolate in a couple of ways On Fri, Sep 5, 2008 at 11:28 AM, Alexy Khrabrov [EMAIL PROTECTED] wrote: Certain timeseries I have had outliers, which I removed by assigning NA to their positions. Now acf() refuses to go to work. What's the right way to remove outliers from ts objects, and what are teh standard ways to interpolate NAs in them? Cheers, Alexy __ 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] lm and time series.
what do you want to do? On Fri, Sep 5, 2008 at 3:22 PM, [EMAIL PROTECTED] wrote: I am sorry but I looked at ?lm and could not see any guidance on writting a formula. If I have two arrays or a data set then I know how to do that (y ~ x) but for a time series I am not sure how to write y or x. Thank you. Kevin Gabor Grothendieck [EMAIL PROTECTED] wrote: The Time Series section in ?lm should be self explanatory. If you are using diff's and lag's then look at the dyn package. On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. Thank you. Kevin __ 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] Package for Tidying-up R-code
I don't know what kind of tidying you want, but ESS within emacs is a possibility. -Don At 6:32 PM +0900 9/5/08, Gundala Viswanath wrote: Dear all, Is there any such package? I am thinking of something equivalent to Perl::Tidy. For example with VI editor one can visual highlight a portion of Perlcode and then issue the command: !perltidy then the code will be automatically arranged. - Gundala Viswanath Jakarta - Indonesia __ 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. -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA 925-423-1062 __ 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] Using R to generate reports
Hi all, I would like to generate nicely formatted reports on a daily basis. The reports would be a combination of charts and tables, generated in an environment such as R. Ideally, this would be an automated process. Does anyone know if this is feasible, and how I could achieve well formatted reports? One idea would be to have R scripts process the data/generate the charts and then use Latex to create the report, referencing the charts produced by R. I've not used Latex before so I'm trying to get a feel for how to best proceed. Thanks for your input, Bill. [[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] Recall: Using R to generate reports
Bill Cunliffe would like to recall the message, Using R to generate reports. __ 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] lm and time series.
I want to fit a function to time series. If I had: x - 1:4 y - 1:4 lm(y~x) This would fit a simple line to the four points. But if it is represented as a time series x - 1:4 t - ts(x) lm() So I have a time series in the object t. How do I write a formula for lm? What do I put in the formula for x and y when I only have t (the time series). Kevin stephen sefick [EMAIL PROTECTED] wrote: what do you want to do? On Fri, Sep 5, 2008 at 3:22 PM, [EMAIL PROTECTED] wrote: I am sorry but I looked at ?lm and could not see any guidance on writting a formula. If I have two arrays or a data set then I know how to do that (y ~ x) but for a time series I am not sure how to write y or x. Thank you. Kevin Gabor Grothendieck [EMAIL PROTECTED] wrote: The Time Series section in ?lm should be self explanatory. If you are using diff's and lag's then look at the dyn package. On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. Thank you. Kevin __ 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] Area of density
Hello! Please, anybody help me. Can I calculate area of density was created by: D - density(x) In other words I want to know area under curve 'plot(D)' and It's good to calculate area before and after 0 separately. -- View this message in context: http://www.nabble.com/Area-of-density-tp19338958p19338958.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] Articles about comparision between R and others softwares
Hi Do you know some articles, papers, something than tell about comparision between R and others softwares statisticals. Thank You Ricardo -- View this message in context: http://www.nabble.com/Articles-about-comparision-between-R-and-others-softwares-tp19338210p19338210.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.
Re: [R] upgraded maps database for italy?
The italy (and france) databases only exists in the maps package because somebody supplied me with the data in an appropriate format - latitude/longitude pairs separately for each polygon (provinces in this case). If you have the appropriate data for what you want i.e. updated provinces and regioni, then I could generate a new database (no promises as to how quickly this will happen though). You *could* do it yourself if you read all the references contain in the maps documentation, but I wouldn't recommend learning how to do this unless you intend doing it frequently. However as Greg Snow comments, maptools may well be a better way to go and uses a more generally available data format. The maps package, which traces its roots back to 'New' S (the Blue book), is rather showing its age, but there is no reason to delete it while some people still find it useful. Regards, Ray Brownrigg livio spam wrote: Dear UseRs, I'm using the library maps. I'm drawing maps of Italy. The map available for this library were prepared around *1989*: This italy database comes from the NUTS III (Tertiary Administrative Units of the European Community) database of the United Nations Environment Programme (UNEP) GRID-Geneva data sets. These were prepared around *1989* [cited: help(italy)]. The details of this map is by province, which is not maybe optimal, but is OK. However, after 1989 many new province are born. So for example the new provincia of Rimini has born and the sardinan Island has doubled them. Some of you do know how to get upgraded maps of italy? I think it would useful for all Italian mapers. The second question is as follow: Italy is subdivided in regioni. Each regione is a set of many province; so, if I want to plot map region-based index, how can I omit the borders of prince belonging to the same regione? so, how plot maps to a higher unit level? [[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] Multiple Correspondence Analysis
Is there a way to get the coordinates from a plot of an MCA object? [[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] library/function that estimates parameters of well known distributions from empirical data?
Thanks Ben That was the one I'd remembered but couldn't find. Mark Leeds also told me about DistributionFits(fBasics), which I hadn't seen. There seems to be only a little overlap between the two. Could I trouble you to expand on AIC (esp. what the function name and package is to apply it to the output from these two functions)? I just read the help provided for each and neither mentions AIC. Thanks again Ben Ted Ben Bolker wrote: Ted Byers r.ted.byers at gmail.com writes: I found this a few months ago, but for the life of me I can't remember what the function or package was, and I have had no luck finding it this week. I have found, again, the functions for working with distributions like Cauchy, F, normal, c., and ks.test, but I have not found the functions for estimating the distribution parameters given a vector of values. Look at the fitdistr function in the MASS package. Consider AIC comparisons for ranking the fits to these non-nested models. good luck Ben Bolker __ 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. -- View this message in context: http://www.nabble.com/library-function-that-estimates-parameters-of-well-known-distributions-from-empirical-data--tp19323700p19339442.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.
Re: [R] Multiple Correspondence Analysis
Dear Bill, See http://finzi.psych.upenn.edu/R/library/ade4/html/dudi.acm.html HTH, Jorge On Fri, Sep 5, 2008 at 5:00 PM, Bill Vorias [EMAIL PROTECTED]wrote: Is there a way to get the coordinates from a plot of an MCA object? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ 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] Using R to generate reports
Dear Bill, See http://www.statistik.lmu.de/~leisch/Sweave/ HTH, Jorge On Fri, Sep 5, 2008 at 5:40 PM, Bill Cunliffe [EMAIL PROTECTED] wrote: Hi all, I would like to generate nicely formatted reports on a daily basis. The reports would be a combination of charts and tables, generated in an environment such as R. Ideally, this would be an automated process. Does anyone know if this is feasible, and how I could achieve well formatted reports? One idea would be to have R scripts process the data/generate the charts and then use Latex to create the report, referencing the charts produced by R. I've not used Latex before so I'm trying to get a feel for how to best proceed. Thanks for your input, Bill. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ 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] lm and time series.
So you want time as the independent variable? Let's say that the units of y in your first example were seconds- couldn't you just use a regular lm and say that the units were seconds, minutes, or what ever? I am probably out of my league here, but I am just not understanding what it is that you want. a time series is just a series of data points indexed by time. Arima maybe, or some other cool times series modeling approach- wavelet, spectral density- for frequency domain type things... What are you trying to accomplish? On Fri, Sep 5, 2008 at 5:47 PM, [EMAIL PROTECTED] wrote: I want to fit a function to time series. If I had: x - 1:4 y - 1:4 lm(y~x) This would fit a simple line to the four points. But if it is represented as a time series x - 1:4 t - ts(x) lm() So I have a time series in the object t. How do I write a formula for lm? What do I put in the formula for x and y when I only have t (the time series). Kevin stephen sefick [EMAIL PROTECTED] wrote: what do you want to do? On Fri, Sep 5, 2008 at 3:22 PM, [EMAIL PROTECTED] wrote: I am sorry but I looked at ?lm and could not see any guidance on writting a formula. If I have two arrays or a data set then I know how to do that (y ~ x) but for a time series I am not sure how to write y or x. Thank you. Kevin Gabor Grothendieck [EMAIL PROTECTED] wrote: The Time Series section in ?lm should be self explanatory. If you are using diff's and lag's then look at the dyn package. On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. Thank you. Kevin __ 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] Area of density
On 05/09/2008 4:32 PM, pragmatic wrote: Hello! Please, anybody help me. Can I calculate area of density was created by: D - density(x) In other words I want to know area under curve 'plot(D)' That should be 1. and It's good to calculate area before and after 0 separately. That's harder, but a good approximation should be sum(x 0)/length(x) and sum(x 0)/length(x). 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] xls to csv conversion via WinXP's context menu?
Frequently I need to convert a .xls to a .csv (for import into R) and I do this by opening the file in excel and saving it as a csv. I would rather do this using WinXP's context menu (right click the xls, choose convert to csv) but I don't know of a utility that does this. Any ideas? Thanks, Mark __ 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] Area of density
On 6/09/2008, at 8:32 AM, pragmatic wrote:
Hello!
Please, anybody help me.
Can I calculate area of density was created by:
D - density(x)
In other words I want to know area under curve 'plot(D)'
and It's good to calculate area before and after 0 separately.
(1) You could use splinefun()
foo - splinefun(D$x,D$y)
integrate(foo,min(D$x),0)
integrate(foo,0,max(D$x))
or you could use simp() from
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/134918.html
with(D,simp(y=y[x0],x=x[x0]))
with(D,simp(y=y[x0],x=x[x0]))
You'll get different answers, neither of which is ``correct''.
cheers,
Rolf Turner
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Re: [R] xls to csv conversion via WinXP's context menu?
See ?xls2csv in the gdata package. On Fri, Sep 5, 2008 at 6:45 PM, Mark Na [EMAIL PROTECTED] wrote: Frequently I need to convert a .xls to a .csv (for import into R) and I do this by opening the file in excel and saving it as a csv. I would rather do this using WinXP's context menu (right click the xls, choose convert to csv) but I don't know of a utility that does this. Any ideas? Thanks, Mark __ 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] xls to csv conversion via WinXP's context menu?
Dear Mark, I think you don't need to convert your xls files into csv files to read them using R. There are several ways to read them such as the RODBC package and ?read.xls in gdata. Now, if you really want to do the conversion, try ?xls2csv in gdata as well. HTH, Jorge On Fri, Sep 5, 2008 at 6:45 PM, Mark Na [EMAIL PROTECTED] wrote: Frequently I need to convert a .xls to a .csv (for import into R) and I do this by opening the file in excel and saving it as a csv. I would rather do this using WinXP's context menu (right click the xls, choose convert to csv) but I don't know of a utility that does this. Any ideas? Thanks, Mark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ 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] annotating objects in workspace
Is there a way to associate descriptions with the objects in the workspace, and later retrieve them to know what the object was created for? Thanks, Alexy __ 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.
