Re: [R] Variable Wildcard Value
Francis Smart fsmart at gmail.com writes: Is there a wildcard value for vector values in r? For instance: M - *wildcard (M==1) TRUE (M==peanut butter) TRUE is.na(M) FALSE If grep on a vector does not help, maybe the following comes closer? Finding all variables rm(list=ls(all=TRUE)) a = 33 aa = 44 dd = 33 c = 3 names = ^a.* Vars - ls() r - Vars[grep(names,Vars)] r # not really correct for names.* and Vars.* Dieter __ 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 can I use R through command line?
you can use source() to Read R Code from a File. 2009/4/1 minben minb...@gmail.com: Suppose I have written a R program and saved it in test.R . How can I call the program in the command line? __ 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. -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ 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] problem with 'loading file size'
Hi all, I am working in a project which needed to load *.csv files of size more than 200MB is it posible to load 200MB size file to r-project and do subsetting as per requirement i am able to load maximum of 90 mb is there any way to increase memory limits and how much maximum memory we can exten please some one help me to get it work thanks in advance. [[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] Variable Wildcard Value
I would be truly amazed if the answer were yes. I find this the most fascinating question on R-help for a long time, maybe ever. Can you tell us what you have in mind and what your ultimate purpose is? Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) Francis Smart wrote: Is there a wildcard value for vector values in r? For instance: M - *wildcard (M==1) TRUE (M==peanut butter) TRUE is.na(M) FALSE thanks, Francis __ 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] scope of variables in R
Stavros Macrakis wrote:
On Tue, Mar 31, 2009 at 3:41 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
Stavros Macrakis wrote:
...All that being said, programming with global variables makes certain
classes of bug much more likely
... in a language like r, that heavily relies on setting global variables
(par(), options(), ...).
Yes, par, options, etc. are problematic, too, especially since there's no
easy way I know of to make them local (like dynamic binding in old Lisps).
indeed, i too was thinking about something like perl's local:
print $$;
# 123456
{ local $$;
print $$; }
# (empty string)
or fluid-let in scheme.
i don't know of any construct in r that would localize global variables
in this manner, but for options, par, etc., you do have an option, the
(arguably ugly) idiom with storing the previous value and restoring it
on exit:
print(getOption('digits'))
# 7
local({
options = options(digits=1)
on.exit(options(options))
print(getOption('digits')) })
# 1
print(getOption('digits'))
# 7
it's like manually localizing a global variable.
vQ
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[R] Plotting lines(...type=b, pch=1) vs. legend(pch=1)
I'd like to do a lineplot with type=b (both lines and points), opposed to type=o (points overplotted on lines). I'd like the legend to match the line, with the point character plotted distinctly from the line, but I can't figure out from the manual ( http://stat.ethz.ch/R-manual/R-patched/library/graphics/html/legend.html ) which option to set on the legend() call. Is there a way to get the legend to plot lines and point characters without overplotting the character that I'm missing? -- Michael R. Head bur...@suppressingfire.org http://www.cs.binghamton.edu/~mike/ __ 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 generate natural cubic spline in R?
SK == Stephan Kolassa stephan.kola...@gmx.de on Tue, 31 Mar 2009 22:20:50 +0200 writes: SK Hi David, SK David Winsemius schrieb: The splinefun documentation indicates that natural is one of the types of cubic spline options available. SK That sounds good, didn't know that... rcs() has the advantage of coming SK with a book (Harrell's Regression Modeling Strategies). well, and splinefun() comes with R !!! Does rcs actually do fitting? Such would not be my expectation on reading the documentation and I do not see any examples of such functionality in the help pages. SK Nope, but you can include rcs() within fitting functions, SK lm(foo~rcs(bar,3)), which makes more sense to me than having a spline SK function fit... Looks like better encapsulation to me. well, but ns() has been part of the S language for ages, and part of R since its very early days, exactly for the purpose to be used in something like lm(y ~ ns(.) + ..) So while rcs() may have extra merits, (I assume it does, as I assume Frank Harrell will have known about ns()), in order to solve the OP's problem, I still believe that standard R [and good old S version 3 for that matter] contains all functionality needed. Martin Maechler, ETH Zurich SK Best, SK Stephan SK __ SK R-help@r-project.org mailing list SK https://stat.ethz.ch/mailman/listinfo/r-help SK PLEASE do read the posting guide http://www.R-project.org/posting-guide.html SK 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] How can I use R through command line?
minben minbenh at gmail.com writes: Suppose I have written a R program and saved it in test.R . How can I call the program in the command line? Assuming Windows (might work under linux) rterm --vanilla --no-save myrfile.r Dieter __ 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 can I use R through command line?
r == ronggui ronggui.hu...@gmail.com on Wed, 1 Apr 2009 16:00:16 +0800 writes: r you can use source() to Read R Code from a File. yes, but that's not in the command line. Slightly more closely to the OP's question: R CMD BATCH test.R or Rscript test.R (orr test.Rif you have installed 'littler' ) Martin Maechler, ETH Zurich r 2009/4/1 minben minb...@gmail.com: Suppose I have written a R program and saved it in test.R . How can I call the program in the command line? __ 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] Generating EPS figures automatically (like Sweave)
eariasca eariasca at math.ucsd.edu writes:
Is there a way to generate EPS figures automatically out of a chunk
of code? Basically, I would like to do something like Sweave does (I
just find it a little cumbersome to create a .Rnw file and then keep
track of the figure numbering).
I normally use this type of code when not using Sweave
ToPostscript = TRUE
if (ToPostscript){
trellis.device(postscript,file=myfile.ps,width=6,height=4,color=FALSE)
} else
{
windows( height=4,width=6)
}
# Do the plot-works
if (ToPostscript){
dev.off()
sink()
}
## And use GhostFriend from
http://www.noliturbare.com/ReadGhostFriend.htm
to create TIFF if your publisher asks for it.
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[R] FW: How to specify axis interval unit...
Dear Duncan Murdoch: Thanks for your tips. By the way, I think I didn't mentioned my question very clear. What I mean to say is that, in Excel one could decide upon the axis interval unit. Please see the attached file. Likewise, is it possible in R...? Regards, Prasanth. -Original Message- From: Duncan Murdoch [mailto:murd...@stats.uwo.ca] Sent: Tuesday, March 31, 2009 6:25 PM To: V Prasanth Cc: r-help@r-project.org Subject: Re: [R] How to specify axis interval unit... On 3/31/2009 5:08 AM, V Prasanth wrote: Dear R Users: Greetings! Is there any way to specify the axis interval unit for barplots in R.? Any help is deeply appreciated. You can set axes=FALSE, names= and then use whatever axes you like. Modifying the first example in examples(barplot): tN - table(Ni - stats::rpois(100, lambda=5)) r - barplot(tN, col=rainbow(20), axes=F, names=) axis(2, at=2^(0:5)) axis(1, at=r, labels=letters[1:length(tN)]) Duncan Murdoch Thanks in advance! Prasanth VP, Global Manager - Biometrics, Delta Technology Management Services Pvt Ltd, Plot No: 13/2, Sector - I, Third Floor, HUDA Techno Enclave, Madhapur, Hyderabad - 500 033. Office: +91-40-3028 0659 Mobile : +91-9848 290025 http://www.deltaintech.com/ www.deltaintech.com http://www.deltaintech.com/ ** ** ** http://www.deltaintech.com/ 'The information contained in this email is confidential and may contain proprietary information. It is meant solely for the intended recipient. Access to this email by anyone else is unauthorized. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted in reliance on this, is prohibited and may be unlawful. No liability or responsibility is accepted if information or data is, for whatever reason corrupted or does not reach its intended recipient. No warranty is given that this email is free of viruses. The views expressed in this email are, unless otherwise stated, those of the author and not those of DELTA Technology and Management Services pvt ltd or its management. DELTA Technology and Management Services pvt ltd reserves the right to monitor intercept and block emails addressed to its users or take any other action in accordance with its email use policy' http://www.deltaintech.com/ Thank you in advance for your cooperation. http://www.deltaintech.com/ ** ** ** http://www.deltaintech.com/ P Please don't print this e-mail unless you really need to. [[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. Document1.pdf Description: Adobe PDF document __ 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] Variable Wildcard Value
Hi,
Since you insist, here is something that I think matches the
specifications :
wildcard - function( ) structure( NULL, class = wildcard )
Ops.wildcard - function (e1, e2){
if (nargs() == 1L)
return( e1 )
result - switch(.Generic,
`` = , `` = , `==` = ,
`=` = , `=` = TRUE, `!=` = FALSE)
result
}
is.na.wildcard - function( x ) FALSE
w == 1
[1] TRUE
w == peanut butter
[1] TRUE
is.na( w )
[1] FALSE
peanut butter == w
[1] TRUE
w == w
[1] TRUE
w != w
[1] FALSE
# Is negation of a wildcard also a wildcard, or should it be a bizarro
wildcard ?
(!w) == 2
[1] TRUE
Not really sure how this could be useful though, and would also be
interested in Francis end game.
Romain
Patrick Burns wrote:
I would be truly amazed if the answer were yes.
I find this the most fascinating question on R-help
for a long time, maybe ever. Can you tell us what
you have in mind and what your ultimate purpose is?
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
Francis Smart wrote:
Is there a wildcard value for vector values in r?
For instance:
M - *wildcard
(M==1)
TRUE
(M==peanut butter)
TRUE
is.na(M)
FALSE
thanks,
Francis
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--
Romain Francois
Independent R Consultant
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
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Re: [R] Variable Wildcard Value
Patrick Burns wrote: I would be truly amazed if the answer were yes. I find this the most fascinating question on R-help for a long time, maybe ever. Can you tell us what you have in mind and what your ultimate purpose is? this seems a request for a 'match all' value, somewhat the inverse of NA (~'match nothing'). i could think of this being useful when you want to check for duplicates, and want to mark that some entries (e.g., the unknown ones) are irrelevant for the comparison (i.e., they should match everything): identical(c(1, NA), c(1, 2)) # FALSE identical(c(1, NA), c(1, NA)) # TRUE # confusing! # hypothetical identical(c(1, *), c(1, 2)) # TRUE identical(c(1, *), c(1, NA)) # TRUE not sure about the last one, though, since the original post demanded that is.na(*) == FALSE i don't claim this would be useful in practice, just speculating. vQ Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) Francis Smart wrote: Is there a wildcard value for vector values in r? For instance: M - *wildcard (M==1) TRUE (M==peanut butter) TRUE is.na(M) FALSE thanks, Francis __ 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] How to specify axis interval unit...
V Prasanth wrote: Dear R Users: Greetings! Is there any way to specify the axis interval unit for barplots in R.? Any help is deeply appreciated. Hi Prasanth, If I understand your question, you might want to look at the barp function in the plotrix package that automatically centers bars and groups of bars on integer values. Jim __ 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] Variable Wildcard Value
Sure thing. I realize that it is an unusual request and not the type of thing that I have seen used in any other language that I know of. So right now I am using some of the statistical functions of R to get some summary statistics and visual output from this historic data set. I have a lot functions that look something like this: summary(lm(SLV_DIE_PER[tontype==TONNAGE_TYPE]~SLV_PER_TON[tontype==TONNAGE_TYPE])) SLV_DIE_PER - being the percent of slaves that died between purchase and delivery SLV_PER_TON - being the number of slaves per standardized ton (ship capacity) tontype - being the type of ton that the ship capacity was recorded as. There are various factors mostly in the form of 1,2,3,4,5... representing Spanish Ton, British Ton, etc. Now I want to run a simple linear model and graphs and other things by specifying TONNAGE_TYPE=1 or 2 etc. Giving me a regression that is only using looking at a particular type of tonnage over that of another. All of that works fine. But, it gets a little ugly when I want to generalize the linear model to include all values irrespective of tontype. Of course I could duplicate and trim the statement as such: summary(lm(SLV_DIE_PER~SLV_PER_TON)) I am sure you can see how a wildcard could be more important given a series of similar expressions. Perhaps something that looks like this: summary(SLV_DIE_PER[(SLV_DIE_PER=SLV_Min_Value)(Nationality==Select_Nationality)((SLV_PER_TON=SLV_Max_Value))(tontype==TONNAGE_TYPE)])) Well thanks for your interest. Any suggestions that can help clean up my code could be extremely helpful right now. Btw thanks Dieter for that hint. Not exactly what I was looking for but I am sure to use it in the future. Romain. Thanks for your code, though I don't see readily how to fit w into your functions. Perhaps you could add an additional line between: is.na.wildcard - function( x ) FALSE and w == 1 Thanks! Francis On Wed, Apr 1, 2009 at 2:02 AM, Patrick Burns pbu...@pburns.seanet.com wrote: I would be truly amazed if the answer were yes. I find this the most fascinating question on R-help for a long time, maybe ever. Can you tell us what you have in mind and what your ultimate purpose is? Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) Francis Smart wrote: Is there a wildcard value for vector values in r? For instance: M - *wildcard (M==1) TRUE (M==peanut butter) TRUE is.na(M) FALSE thanks, Francis -- Francis Smart (406) 223-8108 cell -- Francis Smart (406) 223-8108 cell __ 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 R Group SFBA April meeting reminder; video of Feb k
Rolf Turner wrote: I get to the video screen OK --- there's a large greenish sideways triangle waiting to be clicked on. I do so; there's a message that says it's downloading, with a little progress bar. That seems to complete quite rapidly. Then nothing for a while. Then an error message on the video screen saying ``Fatal error --- video source not ready.'' Then that error message goes away. Long wait. Then I get audio, but never any video. Give up. I'm using Firefox on an Imac; the ``About Mozilla Firefox'' button on the Firefox dropdown menu says I've got Mozilla 5.0, Firefox 2.0.0.2 --- whatever that means. Bottom line --- I can't watch the video. Firefox 3.0.8 on OS X doesn't work either. But you can go directly to http://www.lecturemaker.com/wp-content/uploads/2009/02/lmpremovie.swf and it will work. -- Gad Abraham MEng Student, Dept. CSSE and NICTA The University of Melbourne Parkville 3010, Victoria, Australia email: gabra...@csse.unimelb.edu.au web: http://www.csse.unimelb.edu.au/~gabraham __ 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] Does R support double-exponential smoothing?
Fit an ARIMA(0,2,2) model - it's the same thing and you'll get the MLE of the smoothing parameter for free. Use logs if you want a multiplicative model. Gerard Stephan Kolassa Stephan.Kolassa@ gmx.deTo Sent by: minben minb...@gmail.com r-help-boun...@r- cc project.org r-help@r-project.org Subject Re: [R] Does R support 31/03/2009 11:11 double-exponential smoothing? Hi, ets() in Hyndman's forecast package allows you to specify which one of the many smoothing variants (additive/multiplicative season, damped trend, additive/multiplicative errors) you want. HTH, Stephan minben schrieb: I want to use double-exponential smoothing to forecast time series datas,but I couldn't find it in the document,does R support this method? __ 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. ** The information transmitted is intended only for the person or entity to which it is addressed and may contain confidential and/or privileged material. Any review, retransmission, dissemination or other use of, or taking of any action in reliance upon, this information by persons or entities other than the intended recipient is prohibited. If you received this in error, please contact the sender and delete the material from any computer. It is the policy of the Department of Justice, Equality and Law Reform and the Agencies and Offices using its IT services to disallow the sending of offensive material. Should you consider that the material contained in this message is offensive you should contact the sender immediately and also mailminder[at]justice.ie. Is le haghaidh an duine nó an eintitis ar a bhfuil sí dírithe, agus le haghaidh an duine nó an eintitis sin amháin, a bheartaítear an fhaisnéis a tarchuireadh agus féadfaidh sé go bhfuil ábhar faoi rún agus/nó faoi phribhléid inti. Toirmisctear aon athbhreithniú, atarchur nó leathadh a dhéanamh ar an bhfaisnéis seo, aon úsáid eile a bhaint aisti nó aon ghníomh a dhéanamh ar a hiontaoibh, ag daoine nó ag eintitis seachas an faighteoir beartaithe. Má fuair tú é seo trí dhearmad, téigh i dteagmháil leis an seoltóir, le do thoil, agus scrios an t-ábhar as aon ríomhaire. Is é beartas na Roinne Dlí agus Cirt, Comhionannais agus Athchóirithe Dlí, agus na nOifígí agus na nGníomhaireachtaí a úsáideann seirbhísí TF na Roinne, seoladh ábhair cholúil a dhícheadú. Más rud é go measann tú gur ábhar colúil atá san ábhar atá sa teachtaireacht seo is ceart duit dul i dteagmháil leis an seoltóir láithreach agus le mailminder[ag]justice.ie chomh maith. *** __ 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] Discussion on hosting of local user group web portals under r-project
Hi all, Driven by the success of R, a number of local R users groups have been formed in several places (London, NY, etc.). We are also currently considering setting up our own local group in Spain. We have found that setting up a web page in order to centralise the information relevant to the group members, keep a calendar of events, and other functionalities is very important as a tool to keep the community together and active. Other local groups have developed their own web portals. We are discussing the implementations details of our own. We believe that nurturing these local communities could help to build stronger user base. We are aware that r-forge and r-wiki projects, currently under the r-project umbrella, started from discussions on this list. Therefore, we would like to propose that a new discussion were open in the list in order to consider the convenience of setting up a new environment under r-project (and parallel to r-forge and r-wiki) where local user groups could build their web pages and provide their members the required services under a common architecture and sharing a number of tools. We believe that this would be a great addition to the community and we would be delighted to hear the community express their views on the topic. Best regards. [[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] use R Group SFBA April meeting reminder; video of Feb k
On 01-Apr-09 09:37:49, Gad Abraham wrote: Rolf Turner wrote: I get to the video screen OK --- there's a large greenish sideways triangle waiting to be clicked on. I do so; there's a message that says it's downloading, with a little progress bar. That seems to complete quite rapidly. Then nothing for a while. Then an error message on the video screen saying ``Fatal error --- video source not ready.'' Then that error message goes away. Long wait. Then I get audio, but never any video. Give up. I'm using Firefox on an Imac; the ``About Mozilla Firefox'' button on the Firefox dropdown menu says I've got Mozilla 5.0, Firefox 2.0.0.2 --- whatever that means. Bottom line --- I can't watch the video. Firefox 3.0.8 on OS X doesn't work either. But you can go directly to http://www.lecturemaker.com/wp-content/uploads/2009/02/lmpremovie.swf and it will work. -- Gad Abraham AND it works beautfully using FlashPlayer v.9 on Linux too! This was the version I already had installed when I first failed to locate the video -- instead seeing the little panel for installing FlashPlayer, which I ignored, since I already had it, not realising that it was telling me I should install FlashPhayer 10. It is now clear that this advice was erroneous, since version 9 in fact works fine! Furthermore, when (as I previously reported) I did install version 10, that didn't work either. I later found that version 10 was looking for versions of Linux libraries which I didn't have; and also that FlashPlayer no longer worked (e.g.) on the BBC website. So I re-installed version 9, and everything again worked on the BBC and for videos in Press reports, etc. But of course still the same failure for the SFBA video. Now, using the direct URL for the same video as kindly revealed by Gad, it works and could have worked all along! It is, therefore, clear that the HTML in the original URL is doing the wrong thing, by somehow detecting a version 10 of FlashPlayer and refusing to cooperate, when this was erroneous. I (and Jim Porzak and Mike Driscoll) received a private email from Ron Fredericks Video director and Multimedia expressionist http://www.lecturemaker.com Ron @ Lecturemaker r...@lecturemaker.com (whom I'm adding to the address-list for this mail) which stated: The video may be hard find, admittedly, if you do not have version 10 of flash player installed. If you do not have a recent version of Flash in your web browser, I present a smallish Adobe image with the words you do not have version 10 of flash installed, click to install. Attached is a screenshot of what you should be seeing at this link http://www.lecturemaker.com/2009/02/r-kickoff-video/ I'll take your feedback from the R User Group list server as a suggestion to make the need to update your flash player a lot bigger on the screen. Ron Fredericks So, message to Ron Fredericks: The video works with version 9. If (for some reason) it is essential for the original URL to detect when version 10 is not installed, then there should be a pointer on the folloowing lines: You appear not to have version 10 of Flach Player installed. If you are using an earlier version, please click on the following link . which would take people to the URL provided by Gad, which does work on earlier versions (well, version 9 at least). Then it would no longer be the case that The video may be hard find, admittedly, if you do not have version 10 of flash player installed.! My thanks to everyone who hsd helped to delve into this tangle, and especially to Gad who found where the solution was lurking! Best wishes to all, Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 01-Apr-09 Time: 11:17:02 -- XFMail -- __ 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] SAS Institute to invest upto $20 m with R Project
A SAS spokesperson has confirmed to this blog that they have invested in the R –Core project to help build next generation algorithms . Details are sketchy but indications of some shift on cloud hosted SAS ,called SaaS are emerging.Also includes some details on Jim Davis ,SVP SAS marketing's statement on BI and Anne Milley having a new assignment within SAS Institute. Read more here - http://www.decisionstats.com/2009/04/sas-institute-invests-in-r-project/ __ 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] Variable Wildcard Value
Francis Smart wrote: Sure thing. I realize that it is an unusual request and not the type of thing that I have seen used in any other language that I know of. So right now I am using some of the statistical functions of R to get some summary statistics and visual output from this historic data set. I have a lot functions that look something like this: summary(lm(SLV_DIE_PER[tontype==TONNAGE_TYPE]~SLV_PER_TON[tontype==TONNAGE_TYPE])) SLV_DIE_PER - being the percent of slaves that died between purchase and delivery SLV_PER_TON - being the number of slaves per standardized ton (ship capacity) tontype - being the type of ton that the ship capacity was recorded as. There are various factors mostly in the form of 1,2,3,4,5... representing Spanish Ton, British Ton, etc. Now I want to run a simple linear model and graphs and other things by specifying TONNAGE_TYPE=1 or 2 etc. Giving me a regression that is only using looking at a particular type of tonnage over that of another. All of that works fine. But, it gets a little ugly when I want to generalize the linear model to include all values irrespective of tontype. Of course I could duplicate and trim the statement as such: summary(lm(SLV_DIE_PER~SLV_PER_TON)) I am sure you can see how a wildcard could be more important given a series of similar expressions. Perhaps something that looks like this: summary(SLV_DIE_PER[(SLV_DIE_PER=SLV_Min_Value)(Nationality==Select_Nationality)((SLV_PER_TON=SLV_Max_Value))(tontype==TONNAGE_TYPE)])) Well thanks for your interest. Any suggestions that can help clean up my code could be extremely helpful right now. Btw thanks Dieter for that hint. Not exactly what I was looking for but I am sure to use it in the future. Romain. Thanks for your code, though I don't see readily how to fit w into your functions. Perhaps you could add an additional line between: is.na.wildcard - function( x ) FALSE w - wildcard() Note that it answers your previous question, but not this one. This is untested because we do not have your data, but you could go with something like that, assuming the variables tontype, SLV_DIE_PER, SLV_PER_TON are in the data frame data. That way, you seperate your model to the data it is applied to: with( subset( data, tontype == TONNAGE_TYPE ), lm( SLV_DIE_PER ~ SLV_PER_TON ) ) Not sure how the wildcard you asked for in your first email fits into this. Romain and w == 1 Thanks! Francis On Wed, Apr 1, 2009 at 2:02 AM, Patrick Burns pbu...@pburns.seanet.com wrote: I would be truly amazed if the answer were yes. I find this the most fascinating question on R-help for a long time, maybe ever. Can you tell us what you have in mind and what your ultimate purpose is? Patrick Burns patr...@burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of The R Inferno and A Guide for the Unwilling S User) Francis Smart wrote: Is there a wildcard value for vector values in r? For instance: M - *wildcard (M==1) TRUE (M==peanut butter) TRUE is.na(M) FALSE thanks, Francis -- Francis Smart (406) 223-8108 cell -- Romain Francois Independent R Consultant +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fedora 10 KDE plasma font rendering issue
On Tuesday 31 March 2009 21:02:23 dfermin wrote: Hello. I've got a new workstation running Fedora 10 linux and I use the KDE 4.2 desktop which uses some kind of new desktop environment called 'plasma'. If I start up R and generate a plot (for example: hist(rnorm(1, mean=0, sd=1), breaks=100) ). The plot appears but all text (the x/y axes, title, etc..) is replaced by a square box. No font is rendered at all. Has anyone else got this problem? If so do you have a work around or a solution? What graphics card do you have? I saw recently reports about this happening on grace when using intel graphic cards. I am starting to suspect that this could be the same problem. I'm using R version 2.8.1 installed from the Fedora 10 repositories if that helps. Thanks in advance. -- José Abílio __ 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 message obtained when plotting survival curves (error not previously obtained)
Hello Thomas, I had been using both survival design packages. I have just tested both packages and can report that the error only appears after loading Design. If I just run survival there is no error. regards Bob At 06:08 PM 1/04/2009, Thomas Lumley wrote: On Wed, 1 Apr 2009, Bob Green wrote: Hello, I now receive an error message when obtaining a survival plot, which was not previously received with the same code. I recently updated all my packages. It does not seem to be a peculiarity of my data as I receive the error using data available in R. A plot is produced but I am uncertain regarding the error message. library(surv2sample) data(gastric) fit - survfit(Surv(time, event) ~ treatment, data = gastric) Error in get(x, envir = ns, inherits = FALSE) : variable survfit.km was not found #Default plot: plot(fit) You don't show what packages you have loaded. My guess would be that you are calling Design::survfit rather than survival::survfit, and the problem is due to the incompatibility between the most recent versions of Design and survival. Frank Harrell and the folks at Vanderbilt are working on a new version of Design. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ 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] Fedora 10 KDE plasma font rendering issue
On Tue, 2009-03-31 at 18:36 -0700, dfermin wrote:
Nope. I checked this. Both those fonts are installed.
Well it is some kind of font rendering problem. The default device is
the Cairo X11 device, which uses the Pango layout engine for font
rendering.
If you set the environment variable FC_DEBUG to 1 before launching your
R session, you will get some debugging information. It is very verbose,
but we only need to see this bit:
First font Pattern has 15 elts (size 15)
family: Nimbus Sans L(w)
style: Regular(w)
slant: 0(i)(w)
weight: 80(i)(w)
width: 100(i)(w)
foundry: urw(w)
file: /usr/share/fonts/default/Type1/n019003l.pfb(w)
index: 0(i)(w)
outline: FcTrue(w)
scalable: FcTrue(w)
Martyn Plummer-2 wrote:
Quoting dfermin dfer...@umich.edu:
Hello.
I've got a new workstation running Fedora 10 linux and I use the KDE 4.2
desktop which uses some kind of new desktop environment called 'plasma'.
If I start up R and generate a plot (for example: hist(rnorm(1,
mean=0,
sd=1), breaks=100) ). The plot appears but all text (the x/y axes, title,
etc..) is replaced by a square box. No font is rendered at all.
Has anyone else got this problem? If so do you have a work around or a
solution?
I'm using R version 2.8.1 installed from the Fedora 10 repositories if
that
helps.
Thanks in advance.
It sounds like you are missing some fonts. Check that the urw-fonts and
liberation-fonts RPMs are installed.
---
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Re: [R] re commended computing server for R (March 2009)?
A SAS spokesperson has confirmed to this blog that they have invested in the R âCore project to help build next generation algorithms . Details are sketchy but indications of some shift on cloud hosted SAS ,called SaaS are emerging.Also includes some details on Jim Davis ,SVP SAS marketing's statement on BI and Anne Milley having a new assignment within SAS Institute. Read more here - http://www.decisionstats.com/2009/04/sas-institute-invests-in-r-project/ ivowel wrote: dear r-experts: I need to speed up my monte-carlo simulations. my code is written in R (and it was also the cause of my many questions here over the last few days). my code is almost all matrix/vector algebra on panel data sets---long-difference, fixed-effects, blundell-bond, etc.. the data set is about 10MB, so 1GB per CPU core should be plenty for my operations, and with $10/GB of DRAM, this is no longer a bottleneck. For my application, parallelism is a given, since most of it is monte-carlo simulations. (I guess the diametrically opposite need would be when one cannot parallelize, in which case the recommendations would be quite different.) My operating system will probably be ubuntu. (I also run a little of it on an OSX Mac Pro I own.) I want to use an Intel/AMD system with a prebuilt R executable. I do not want to fiddle (too much) with building R myself, unless it is real easy and makes a real speed difference. I wish I could ask R to load something exotic like CUDA, but I presume that this is not yet ready for prime-time. PS3 is probably silly, too. in fact, if I am not mistaken (and I may well be), R pre-built does not even take advantage of SSE3 out-of-the-box. software-wise, is there anything unusual that I should heed, or should I just pick of R 2.8.1 from the CRAN archives and be done with it? now, I also have to make some simple hardware decisions. Right now, a dual-socket quad-core AMD opteron shanghai 2.3GHz system seems cheap. $174/CPU + $70/motherboard. is there a system that dominates this in terms of $/MFlops? (I presume the fact that core i7 has threads is irrelevant to R.) I am not trying to ignite a flame-war---in fact, I don't care about any other features that AMD or Intel or anything might have for this particular computer. Other needs may warrant different choices. Any other thoughts would be appreciated. Although you can just email them to me, I presume that this question has enough interest to others that posting it is ok. regards, /ivo welch [[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/recommended-computing-server-for-R-%28March-2009%29--tp22756946p22824049.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SAS Institute to invest upto $20 m with R Project
And a happy April Fools day to you to... On Wed, Apr 1, 2009 at 5:44 AM, Ajay ohri ohri2...@gmail.com wrote: A SAS spokesperson has confirmed to this blog that they have invested in the R –Core project to help build next generation algorithms . Details are sketchy but indications of some shift on cloud hosted SAS ,called SaaS are emerging.Also includes some details on Jim Davis ,SVP SAS marketing's statement on BI and Anne Milley having a new assignment within SAS Institute. Read more here - http://www.decisionstats.com/2009/04/sas-institute-invests-in-r-project/ __ 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] repeated measures ANOVA - among group differences
If Month is nested within Quadrat I think you want: aov(ProportioninTreatment ~ Treatment*Month +Error(Quadrat/Month), RM) If Treatment is also nested within Quadrat, you want: aov(ProportioninTreatment ~ Treatment*Month +Error(Quadrat/(Treatment*Month)), RM) On Wed, Apr 1, 2009 at 12:42 AM, Jessica L Hite/hitejl/O/VCU hit...@vcu.edu wrote: I have data on the proportion of clutches experiencing different fates (e.g., 4 different sources of mortality) for 5 months . I need to test 1) if the overall proportion of these different fates is different over the entire study and 2) to see if there are monthly differences within (and among) fate types. Thus, I am pretty sure this is an RM analysis -( I measure the same quadrats each month). I am fine running the analysis in R - with the code below, however, there is no output for the among group variation...this is an important component - any ideas on how to solve this problem? I have included code and sample data below. Many thanks in advance for help and suggestions. J both.aov - aov(ProportioninTreatment ~ factor(Treatment)*factor(Month) + Error(factor(Quadrat)), RM) Error: factor(id) Df Sum Sq Mean Sq F value Pr(F) Residuals 3 0.51619 0.17206 #why only partial output here? ### Error: Within Df Sum Sq Mean Sq F value Pr(F) factor(Fate1) 3 1.2453 0.4151 3.5899 0.017907 * time 1 0.9324 0.9324 8.0637 0.005929 ** factor(Fate1):time 3 0.9978 0.3326 2.8763 0.042272 * Residuals 69 7.9783 0.1156 Fate1 Proportion in Fate ASIN Month Quadrat 1 0.117647059 0.350105778 1 1 1 0 0 2 1 1 0.1 0.339836909 3 1 1 0 0 4 1 1 0 0 5 1 1 0 0 1 2 1 0 0 2 2 1 0.2 0.463647609 3 2 1 0.25 0.523598776 4 2 1 0.1 0.339836909 5 2 1 0 0 1 3 1 0 0 2 3 1 0 0 3 3 1 0.384615385 0.668964075 4 3 1 0 0 5 3 1 0 0 1 4 1 0 0 2 4 1 0 0 3 4 1 0.16667 0.420534336 4 4 1 0 0 5 4 2 0.352941176 0.636132062 1 1 2 0.2 0.463647609 2 1 2 0.3 0.615479708 3 1 2 1 1.570796327 4 1 2 0 0 5 1 2 0.5 0.785398163 1 2 2 0 0 2 2 2 0.6 0.886077124 3 2 2 0.41667 0.701674124 4 2 2 0.2 0.490882678 5 2 2 0 0 1 3 2 0.2 0.463647609 2 3 2 0 0 3 3 2 0.461538462 0.746898594 4 3 2 0 0 5 3 2 0 0 1 4 2 0 0 2 4 2 0.307692308 0.588002604 3 4 2 0.7 0.955316618 4 4 2 0 0 5 4 3 0 0 1 1 3 0 0 2 1 3 0.4 0.729727656 3 1 3 0 0 4 1 3 1 1.570796327 5 1 3 0.5 0.785398163 1 2 3 0 0 2 2 3 0 0 3 2 3 0.25 0.523598776 4 2 3 0.6 0.841068671 5 2 3 0 0 1 3 3 0 0 2 3 3 0 0 3 3 3 0.153846154 0.403057075 4 3 3 0.7 0.955316618 5 3 3 0 0 1 4 3 0 0 2 4 3 0 0 3 4 3 0 0 4 4 3 0.875 1.209429203 5 4 4 0.294117647 0.573203309 1 1 4 0.2 0.463647609 2 1 4 0 0 3 1 4 0 0 4 1 4 0 0 5 1 4 0 0 1 2 4 0 0 2 2 4 0 0 3 2 4 0.08333 0.292842771 4 2 4 0.1 0.339836909 5 2 4 0 0 1 3 4 0 0 2 3 4 0 0 3 3 4 0 0 4 3 4 0.16667 0.420534336 5 3 4 0 0 1 4 4 0 0 2 4 4 0.461538462 0.746898594 3 4 4 0 0 4 4 4 0.125 0.361367124 5 4 [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tinyurl.com/mikes-public-calendar ~ Certainty is folly... I think. ~ __ 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] DCT function?
Not in signal, indeed, but there are functions dct() and dst() in package dtt. Uwe Ligges Red Roo wrote: Looking for the DCT function, but don't see it in the signal pkg. http://rss.acs.unt.edu/Rdoc/library/signal/html/signal.package.html http://rss.acs.unt.edu/Rdoc/library/signal/html/00Index.html As I understand it, the 'signal' functions are ports of the corresponding matlab/octave code, where the DCT exists. Did I miss it (different name?) or is someone working on a port for 'signal'? --RR __ 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] problem with 'loading file size'
It is certainly possible since I loaded a 825 MB csv file yesterday that had 1,5 milion lines, but the answer about how might very system specific. Please re-read the Posting Guide and offer the requested information and any error reports. You could try to do a bit os self-education: ?Memory On Apr 1, 2009, at 4:03 AM, venkata kirankumar wrote: Hi all, I am working in a project which needed to load *.csv files of size more than 200MB is it posible to load 200MB size file to r-project and do subsetting as per requirement i am able to load maximum of 90 mb is there any way to increase memory limits and how much maximum memory we can exten please some one help me to get it work thanks in advance. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] repeated measures ANOVA - among group differences
Hm, it seems I possibly used the technical term nested inappropriately in my response. I meant: If Month is a repeated measure within each Quadrat... and If Treatment is also a repeated measure within each Quadrat... On Wed, Apr 1, 2009 at 8:21 AM, Mike Lawrence mike.lawre...@dal.ca wrote: If Month is nested within Quadrat I think you want: aov(ProportioninTreatment ~ Treatment*Month +Error(Quadrat/Month), RM) If Treatment is also nested within Quadrat, you want: aov(ProportioninTreatment ~ Treatment*Month +Error(Quadrat/(Treatment*Month)), RM) On Wed, Apr 1, 2009 at 12:42 AM, Jessica L Hite/hitejl/O/VCU hit...@vcu.edu wrote: I have data on the proportion of clutches experiencing different fates (e.g., 4 different sources of mortality) for 5 months . I need to test 1) if the overall proportion of these different fates is different over the entire study and 2) to see if there are monthly differences within (and among) fate types. Thus, I am pretty sure this is an RM analysis -( I measure the same quadrats each month). I am fine running the analysis in R - with the code below, however, there is no output for the among group variation...this is an important component - any ideas on how to solve this problem? I have included code and sample data below. Many thanks in advance for help and suggestions. J both.aov - aov(ProportioninTreatment ~ factor(Treatment)*factor(Month) + Error(factor(Quadrat)), RM) Error: factor(id) Df Sum Sq Mean Sq F value Pr(F) Residuals 3 0.51619 0.17206 #why only partial output here? ### Error: Within Df Sum Sq Mean Sq F value Pr(F) factor(Fate1) 3 1.2453 0.4151 3.5899 0.017907 * time 1 0.9324 0.9324 8.0637 0.005929 ** factor(Fate1):time 3 0.9978 0.3326 2.8763 0.042272 * Residuals 69 7.9783 0.1156 Fate1 Proportion in Fate ASIN Month Quadrat 1 0.117647059 0.350105778 1 1 1 0 0 2 1 1 0.1 0.339836909 3 1 1 0 0 4 1 1 0 0 5 1 1 0 0 1 2 1 0 0 2 2 1 0.2 0.463647609 3 2 1 0.25 0.523598776 4 2 1 0.1 0.339836909 5 2 1 0 0 1 3 1 0 0 2 3 1 0 0 3 3 1 0.384615385 0.668964075 4 3 1 0 0 5 3 1 0 0 1 4 1 0 0 2 4 1 0 0 3 4 1 0.16667 0.420534336 4 4 1 0 0 5 4 2 0.352941176 0.636132062 1 1 2 0.2 0.463647609 2 1 2 0.3 0.615479708 3 1 2 1 1.570796327 4 1 2 0 0 5 1 2 0.5 0.785398163 1 2 2 0 0 2 2 2 0.6 0.886077124 3 2 2 0.41667 0.701674124 4 2 2 0.2 0.490882678 5 2 2 0 0 1 3 2 0.2 0.463647609 2 3 2 0 0 3 3 2 0.461538462 0.746898594 4 3 2 0 0 5 3 2 0 0 1 4 2 0 0 2 4 2 0.307692308 0.588002604 3 4 2 0.7 0.955316618 4 4 2 0 0 5 4 3 0 0 1 1 3 0 0 2 1 3 0.4 0.729727656 3 1 3 0 0 4 1 3 1 1.570796327 5 1 3 0.5 0.785398163 1 2 3 0 0 2 2 3 0 0 3 2 3 0.25 0.523598776 4 2 3 0.6 0.841068671 5 2 3 0 0 1 3 3 0 0 2 3 3 0 0 3 3 3 0.153846154 0.403057075 4 3 3 0.7 0.955316618 5 3 3 0 0 1 4 3 0 0 2 4 3 0 0 3 4 3 0 0 4 4 3 0.875 1.209429203 5 4 4 0.294117647 0.573203309 1 1 4 0.2 0.463647609 2 1 4 0 0 3 1 4 0 0 4 1 4 0 0 5 1 4 0 0 1 2 4 0 0 2 2 4 0 0 3 2 4 0.08333 0.292842771 4 2 4 0.1 0.339836909 5 2 4 0 0 1 3 4 0 0 2 3 4 0 0 3 3 4 0 0 4 3 4 0.16667 0.420534336 5 3 4 0 0 1 4 4 0 0 2 4 4 0.461538462 0.746898594 3 4 4 0 0 4 4 4 0.125 0.361367124 5 4 [[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. -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tinyurl.com/mikes-public-calendar ~ Certainty is folly... I think. ~ -- Mike Lawrence
[R] How to prevent inclusion of intercept in lme with interaction
Dear friends of lme, After so many year with lme, I feel ashamed that I cannot get this to work. Maybe it's a syntax problem, but possibly a lack of understanding. We have growth curves of new dental bone that can well be modeled by a linear growth curve, for two different treatments and several subjects as random parameter. By definition, newbone is zero at t=0, so I tried to force the curve through 0. In Pinheiro/Bates, this is done by including the -1 term, and it works well when treatment is not included (newbone~t-1), but seems to have no effect in (newone ~ t*treat-1). What's wrong? Some bracket missing? I tried a few variants. Dieter Menne # library(nlme) library(lattice) # Generated data set.seed(4711) subject = as.factor(letters[1:5]) varslope = rnorm(length(subject),0,0.02) cslope = c (0.1,0.15) grd = expand.grid(t=seq(5,15,by=5), subject=subject,treat=c(contr,test)) grd$slope = varslope[grd$subject] + cslope[grd$treat] grd$newbone = grd$slope*grd$t+rnorm(nrow(grd),0,0.2) xyplot(newbone~t|treat,groups=subject,data=grd, type=l,xlim=c(0,20),ylim=c(0,3)) # With intercept grd.lme1 = lme(newbone~t*treat,data=grd,random=~1|subject) grd$pred1 = predict(grd.lme1,level=0) summary(grd.lme1) # How go force intercept = 0 ??? grd.lme0 = lme(newbone~t*treat-1,data=grd,random=~1|subject) grd$pred0 = predict(grd.lme0,level=0) summary(grd.lme0) # Gives true, all.equal(grd$pred1,grd$pred0) # Everything as expected without treat grd.lme2 = lme(newbone~t,data=grd,random=~1|subject) grd$pred2 = predict(grd.lme2,level=0) summary(grd.lme2) # Forced intercept = 0 grd.lme3 = lme(newbone~t-1,data=grd,random=~1|subject) grd$pred3 = predict(grd.lme3,level=0) summary(grd.lme3) # As expected: not equal all.equal(grd$pred2,grd$pred3) #--- R version 2.9.0 Under development (unstable) (2009-03-13 r48127) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252; LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices datasets utils methods base other attached packages: [1] lattice_0.17-20 nlme_3.1-90 loaded via a namespace (and not attached): [1] grid_2.9.0 tools_2.9.0 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R.matlab_1.2.4.tgz best library for loading *.mat files?
Just curious - Is R.matlab_1.2.4.tgz still the best library for loading *.mat files? Want to start using R to process some *.mat files. Thanks again for any feedback and insight provided. Jason __ 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 prevent inclusion of intercept in lme with interaction
Dear Dieter, With t*treat the model allows for a different slope AND a different intercept for each treatment. If you only want different slopes and all intercepts equal to 0, then t:treat - 1 or t + t:treat - 1 is the model you are looking for. 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 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Dieter Menne Verzonden: woensdag 1 april 2009 14:11 Aan: r-h...@stat.math.ethz.ch Onderwerp: [R] How to prevent inclusion of intercept in lme with interaction Dear friends of lme, After so many year with lme, I feel ashamed that I cannot get this to work. Maybe it's a syntax problem, but possibly a lack of understanding. We have growth curves of new dental bone that can well be modeled by a linear growth curve, for two different treatments and several subjects as random parameter. By definition, newbone is zero at t=0, so I tried to force the curve through 0. In Pinheiro/Bates, this is done by including the -1 term, and it works well when treatment is not included (newbone~t-1), but seems to have no effect in (newone ~ t*treat-1). What's wrong? Some bracket missing? I tried a few variants. Dieter Menne # library(nlme) library(lattice) # Generated data set.seed(4711) subject = as.factor(letters[1:5]) varslope = rnorm(length(subject),0,0.02) cslope = c (0.1,0.15) grd = expand.grid(t=seq(5,15,by=5), subject=subject,treat=c(contr,test)) grd$slope = varslope[grd$subject] + cslope[grd$treat] grd$newbone = grd$slope*grd$t+rnorm(nrow(grd),0,0.2) xyplot(newbone~t|treat,groups=subject,data=grd, type=l,xlim=c(0,20),ylim=c(0,3)) # With intercept grd.lme1 = lme(newbone~t*treat,data=grd,random=~1|subject) grd$pred1 = predict(grd.lme1,level=0) summary(grd.lme1) # How go force intercept = 0 ??? grd.lme0 = lme(newbone~t*treat-1,data=grd,random=~1|subject) grd$pred0 = predict(grd.lme0,level=0) summary(grd.lme0) # Gives true, all.equal(grd$pred1,grd$pred0) # Everything as expected without treat grd.lme2 = lme(newbone~t,data=grd,random=~1|subject) grd$pred2 = predict(grd.lme2,level=0) summary(grd.lme2) # Forced intercept = 0 grd.lme3 = lme(newbone~t-1,data=grd,random=~1|subject) grd$pred3 = predict(grd.lme3,level=0) summary(grd.lme3) # As expected: not equal all.equal(grd$pred2,grd$pred3) #--- R version 2.9.0 Under development (unstable) (2009-03-13 r48127) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252; LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices datasets utils methods base other attached packages: [1] lattice_0.17-20 nlme_3.1-90 loaded via a namespace (and not attached): [1] grid_2.9.0 tools_2.9.0 __ 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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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 prevent inclusion of intercept in lme with interaction
Hi Dieter, the following model assumes a linear relationship between the response newbone and the independent variable t with a common intercept equal to 0 and treatment-dependent slopes: grd.lme0 - lme(newbone~t:treat-1, data=grd, random=~1|subject) summary(grd.lme0) Christian __ 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 prevent inclusion of intercept in lme with interaction
ONKELINX, Thierry Thierry.ONKELINX at inbo.be writes: With t*treat the model allows for a different slope AND a different intercept for each treatment. If you only want different slopes and all intercepts equal to 0, then t:treat - 1 or t + t:treat - 1 is the model you are looking for. Thanks Thierry and Christian, I knew it was something stupid. Dieter __ 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] Obtaining average ranking from matrix of frequencies
I have a small matrix where the columns represents a ranking and the values are the number of times each ranking was obtained eg 1 2 3 x 1 2 0 y 0 1 2 z 2 0 1 I'd like to be able to return an average of the ranking obtained average x 1.67 y 2.67 z 1.67 Whats the nicest way to do this? I'm new to the language and looking for an elegant solution :) Thanks Ben -- View this message in context: http://www.nabble.com/Obtaining-average-ranking-from-matrix-of-frequencies-tp22825754p22825754.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] Obtaining average ranking from matrix of frequencies
say 'm' is your matrix, then try rowMeans(m * col(m)) I hope it helps. Best, Dimitris bwgoudey wrote: I have a small matrix where the columns represents a ranking and the values are the number of times each ranking was obtained eg 1 2 3 x 1 2 0 y 0 1 2 z 2 0 1 I'd like to be able to return an average of the ranking obtained average x 1.67 y 2.67 z 1.67 Whats the nicest way to do this? I'm new to the language and looking for an elegant solution :) Thanks Ben -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ 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] Obtaining average ranking from matrix of frequencies
Thanks heaps. Is there a neat way to do the same for the standard deviation? Cheers Ben On Wed, Apr 1, 2009 at 11:59 PM, Dimitris Rizopoulos d.rizopou...@erasmusmc.nl wrote: say 'm' is your matrix, then try rowMeans(m * col(m)) I hope it helps. Best, Dimitris bwgoudey wrote: I have a small matrix where the columns represents a ranking and the values are the number of times each ranking was obtained eg 1 2 3 x 1 2 0 y 0 1 2 z 2 0 1 I'd like to be able to return an average of the ranking obtained average x 1.67 y 2.67 z 1.67 Whats the nicest way to do this? I'm new to the language and looking for an elegant solution :) Thanks Ben -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ 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] Need Advice on Matrix Not Positive Semi-Definite with cholesky decomposition
Dear fellow R Users: I am doing a Cholesky decomposition on a correlation matrix and get error message the matrix is not semi-definite. Does anyone know: 1- a work around to this issue? 2- Is there any approach to try and figure out what vector might be co-linear with another in thr Matrix? 3- any way to perturb the data to work around this? Thanks for any suggestions. __ 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] Principal components vs. raw variables
Hello to everyone, I am starting to work on classification procedures. I usualy do a principal component analysis (PCA) as a previous step in order to reduce variables and after I apply a cluster procedure. My question is if it will be better to use raw variables instead of use principal components obtained from these variables since the original variables keep all the variability. Now i am thinking to use a variable group analysis (VGA) and a correlation analysis together in order to identify which of my original variables could explain differences on my data better, and after apply a cluster analysis on selected variables. What do you think about it? What would be better: work with PCA or with raw variables. Thanks in advance. Manuel - Manuel Ramón Fernández Group of Reproductive Biology (GBR) University of Castilla-La Mancha (Spain) mra...@jccm.es -- View this message in context: http://www.nabble.com/Principal-components-vs.-raw-variables-tp22824280p22824280.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] Fwd: 'for Loop'
Hello, A nice guy call Jun Shen was helping me out with this, but I require a bit more help. Below is my data set or list called 'test'. I'm trying to calculate the %RSD for each pair of index and keep it in cronological order if you can imagine a 3rd column with 'date' beside index. Result Index 1 0.290117 2 0.292817 3 0.291318 4 0.289318 5 0.3526 9 6 0.3316 9 7 0.349210 8 0.346710 9 0.338511 10 0.334111 11 0.344912 12 0.340612 13 0.369013 14 0.332713 15 0.332414 16 0.324514 17 0.324515 18 0.322915 19 0.333116 20 0.320816 now this bit of programming works for it how ever it puts the mean of my index back into numeric order. aggregate(test[1],test[2],mean)-inter names(inter)[2]='mean' merge(test,inter,all=T)-inter2 inter2$RSV=inter2$Result/inter2$mean Would anyone know away around this so I could keep the data in cronological order? Kind Regards, Al [[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] R hangs when trying to invoke through java in linux environment
Hello, I have a .R file which I am trying to invoke via java with command line parameters. Basically java invokes the command as follows: Path to R binary --slave --args Path to CSV File Direcotry path Date Script path I am using Rutime class of Java to execute this command. I have written a test program to do this. This program works well in windows, generating the .png file. While in linux this leaves the java console hung i.e. waiting for something. I had tried to change the --slave arguements to --vanilla in both the windows and linux version of the program and compared the output. The output in windows version confirmed that it had completed successfully. While for the linux version of the program, the last line that was printed was Type 'q()' to quit R.. After this it remained hung. My assumption is that after this R tries to read the graph file and outputs that to the console and this task is failing in linux. However if I manually type the same command in the command prompt for linux, it executes properly, generating the .png file. Has anyone had similar problems in invoking R via java in linux? Any information regarding this issue will be appreciated. -- View this message in context: http://www.nabble.com/R-hangs-when-trying-to-invoke-through-java-in-linux-environment-tp22825081p22825081.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] list substring
hi I ve a list of item x = ( x1 x2 x3) I need to extract a new vector y = ( , , ) I mean, for each item I need to extract the first 4 digit How is that possible? Thank you -- View this message in context: http://www.nabble.com/list-substring-tp22826435p22826435.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] Request: Optimum value of cost complexity parameter k in tree package
Dear R community I have a question regarding the value of cost complexity parameter k used in tree package for pruning purpose. Any help in finding the optimum value of k is requested. Please give some suggestion in this regard. In the example below i used k=0 but i don't know why? But if i use k=NULL, then it will not plot the resultant tree. library(tree) ds=iris; iris=transform(iris, Species = factor(Species, labels = letters[1:3])) miris - tree(Species ~ ., data = iris, control=tree.control(nobs = 150, minsize = 5, mincut = 2)); iris.prun=prune.tree(miris, method=c(misclass), best = NULL, k=0); iris.prun; summary(iris.prun); plot(iris.prun) best regards Muhammad Azam [[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] list substring
calpeda wrote:
hi
I ve a list of item x = ( x1
x2
x3)
I need to extract a new vector y = ( ,
,
)
I mean, for each item I need to extract the first 4 digit
How is that possible?
sub('^(\\d{4}).*', '\\1', x, perl=TRUE)
vQ
__
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] list substring
try this: x - c(x1, x2, x3) substr(x, 1, 4) Best, Dimitris calpeda wrote: hi I ve a list of item x = ( x1 x2 x3) I need to extract a new vector y = ( , , ) I mean, for each item I need to extract the first 4 digit How is that possible? Thank you -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ 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] list substring
Dear celpeda, Try this: x = c( x1, x2, x3) substr(x,1,4) [1] See ?substr for more details. HTH, Jorge On Wed, Apr 1, 2009 at 9:30 AM, calpeda mauro.bias...@calpeda.it wrote: hi I ve a list of item x = ( x1 x2 x3) I need to extract a new vector y = ( , , ) I mean, for each item I need to extract the first 4 digit How is that possible? Thank you -- View this message in context: http://www.nabble.com/list-substring-tp22826435p22826435.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] Constrined dependent optimization.
Thank you I had not considered using gradient in this fashion. Now as an add
on question. You (an others) have suggested using SANN. Does your answer change
if instead of 100 variables or bins there are 20,000? From the documentation
L-BFGS-B is designed for a large number of variables. But maybe SANN can handle
this as well.
Kevin
Paul Smith phh...@gmail.com wrote:
Apparently, the convergence is faster if one uses this new swap function:
swapfun - function(x,N=100) {
loc - c(sample(1:(N/2),size=1,replace=FALSE),sample((N/2):100,1))
tmp - x[loc[1]]
x[loc[1]] - x[loc[2]]
x[loc[2]] - tmp
x
}
It seems that within 20 millions of iterations, one gets the exact
optimal solution, which does not take too long.
Paul
On Mon, Mar 30, 2009 at 5:11 PM, Paul Smith phh...@gmail.com wrote:
Optim with SANN also solves your example:
---
f - function(x) sum(c(1:50,50:1)*x)
swapfun - function(x,N=100) {
loc - sample(N,size=2,replace=FALSE)
tmp - x[loc[1]]
x[loc[1]] - x[loc[2]]
x[loc[2]] - tmp
x
}
N - 100
opt1 -
optim(fn=f,par=sample(1:N,N),gr=swapfun,method=SANN,control=list(maxit=5,fnscale=-1,trace=10))
opt1$par
opt1$value
---
We need to specify a large number of iterations to get the optimal
solution. The objective function at the optimum is 170425, and one
gets a close value with optim and SANN.
Paul
On Mon, Mar 30, 2009 at 2:22 PM, Hans W. Borchers
hwborch...@googlemail.com wrote:
Image you want to minimize the following linear function
f - function(x) sum( c(1:50, 50:1) * x / (50*51) )
on the set of all permutations of the numbers 1,..., 100.
I wonder how will you do that with lpSolve? I would simply order
the coefficients and then sort the numbers 1,...,100 accordingly.
I am also wondering how optim with SANN could be applied here.
As this is a problem in the area of discrete optimization resp.
constraint programming, I propose to use an appropriate program
here such as the free software Bprolog. I would be interested to
learn what others propose.
Of course, if we don't know anything about the function f then
it amounts to an exhaustive search on the 100! permutations --
probably not a feasible job.
Regards, Hans Werner
Paul Smith wrote:
On Sun, Mar 29, 2009 at 9:45 PM, rkevinbur...@charter.net wrote:
I have an optimization question that I was hoping to get some suggestions
on how best to go about sovling it. I would think there is probably a
package that addresses this problem.
This is an ordering optimzation problem. Best to describe it with a
simple example. Say I have 100 bins each with a ball in it numbered
from 1 to 100. Each bin can only hold one ball. This optimization is that
I have a function 'f' that this array of bins and returns a number. The
number returned from f(1,2,3,4) would return a different number from
that of f(2,1,3,4). The optimization is finding the optimum order of
these balls so as to produce a minimum value from 'f'.I cannot use the
regular 'optim' algorithms because a) the values are discrete, and b) the
values are dependent ie. when the variable representing the bin
location is changed (in this example a new ball is put there) the
existing ball will need to be moved to another bin (probably swapping
positions), and c) each variable is constrained, in the example above
the only allowable values are integers from 1-100. So the problem becomes
finding the optimum order of the balls.
Any suggestions?
If your function f is linear, then you can use lpSolve.
Paul
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Re: [R] Need Advice on Matrix Not Positive Semi-Definite with cholesky decomposition
Look at the nearPD() function in the package Matrix. require(Matrix) ?nearPD In particular, pay attention to the arguments eig.tol and posd.tol, which you can tweak to define how much positiveness you would like to have. 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: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Gottlieb, Neil Sent: Wednesday, April 01, 2009 9:57 AM To: 'r-help@r-project.org' Subject: [R] Need Advice on Matrix Not Positive Semi-Definite with cholesky decomposition Dear fellow R Users: I am doing a Cholesky decomposition on a correlation matrix and get error message the matrix is not semi-definite. Does anyone know: 1- a work around to this issue? 2- Is there any approach to try and figure out what vector might be co-linear with another in thr Matrix? 3- any way to perturb the data to work around this? Thanks for any suggestions. __ 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] Definition of = vs. -
NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
(e.g. f(a=1), which is equivalent to
eval(structure(quote(f(1)),names=c('','a'))).
As far as I can tell from the documentation, assignment with = is precisely
equivalent to assignment with -. Yet they call different primitives:
`=`
.Primitive(=)
`-`
.Primitive(-)
(Perhaps these are different names for the same internal function?)
Also, the difference is preserved by the parser:
quote({a=b})
{
a = b
}
quote({a-b})
{
a - b
}
even though in other cases the parser canonicalizes variant syntax, e.g. -
to -:
quote({a-b})
{
b - a
}
`-`
Error: object - not found
Is there in fact some semantic difference between = and - ?
If not, why do they use a different operator internally, each calling a
different primitive?
Or is this all just accidental inconsistency resulting from the '=', '-',
and '-' features being added at different times by different people working
off different stylistic conventions?
-s
[[alternative HTML version deleted]]
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[R] Learning development concepts in R for newbie users
Hi R users, I apologize for a seemingly trivial question, but I felt this forum would be the best place to seek advice. I have been an R user for a year now, but I am limited to using R and its various contributed packages. I strongly feel that users of a free and open source software tool must eventually provide development expertise. I am aware of the extensive documentation available on the R project page regarding Internals, Language definitions and the like, but I would like to find tutorials on R development, much the same way as there are tutorials for the use of R in statistics and data analysis. Information on Book titles, online sources, etc that you maybe aware of which could allow a user of R to also develop R (packages, improvise on functions, etc) would be much appreciated. For example, I have used the adf.test in package 'tseries' but the test assumes a certain form the the time series process. There are 3 forms in literature, and the other 2 are not used. I would like to modify the adf.test to incorporate the other two forms as well. Thanks Harsh Singhal [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need Advice on Matrix Not Positive Semi-Definite with cholesky decomposition
Neil, pls tell why do you need the correlation matrix? if you are trying to simulate correlated variables then you can go around the cholesky by using svd. if you really need the correlation ( I think it is always possible to avoid it ) then Rmetrics have a function to turn your matrix into positive semi-definite. you can also consider a factor model which will reduce the dimensionality tremendously. Stephen C. Bond On Apr 1, 2009, ngottl...@marinercapital.com wrote: Dear fellow R Users: I am doing a Cholesky decomposition on a correlation matrix and get error message the matrix is not semi-definite. Does anyone know: 1- a work around to this issue? 2- Is there any approach to try and figure out what vector might be co-linear with another in thr Matrix? 3- any way to perturb the data to work around this? Thanks for any suggestions. __ 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] Need Advice on Matrix Not Positive Semi-Definite with cholesky decomposition
Hi Stephen: It's the inputs given to me by a end-user. Ultimately trying to fit a student-t copula to a bunch of simulated price returns while maintaining the structure of the estimated correlation matrix. The other challenge is I use R to test and work a solution but then have also done in matlab as I had to write a C# wrapper around the code for the end-user. In a way I am stuck with Matalb but was hoping to try and solve with R which is great. I appreciate your responding to my email help request. Neil From: sten...@go.com [mailto:sten...@go.com] Sent: Wednesday, April 01, 2009 10:47 AM To: Gottlieb, Neil; r-help@r-project.org Subject: Re: [R] Need Advice on Matrix Not Positive Semi-Definite with cholesky decomposition Neil, pls tell why do you need the correlation matrix? if you are trying to simulate correlated variables then you can go around the cholesky by using svd. if you really need the correlation ( I think it is always possible to avoid it ) then Rmetrics have a function to turn your matrix into positive semi-definite. you can also consider a factor model which will reduce the dimensionality tremendously. Stephen C. Bond On Apr 1, 2009, ngottl...@marinercapital.com wrote: Dear fellow R Users: I am doing a Cholesky decomposition on a correlation matrix and get error message the matrix is not semi-definite. Does anyone know: 1- a work around to this issue? 2- Is there any approach to try and figure out what vector might be co-linear with another in thr Matrix? 3- any way to perturb the data to work around this? Thanks for any suggestions. __ 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] Definition of = vs. -
On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
(e.g. f(a=1), which is equivalent to
eval(structure(quote(f(1)),names=c('','a'))).
As far as I can tell from the documentation, assignment with = is precisely
equivalent to assignment with -. Yet they call different primitives:
The parser does treat them differently:
if (x - 2) cat(assigned\n)
assigned
if (x = 2) cat(assigned\n)
Error: unexpected '=' in if (x =
The ?= man page explains this:
The operator '-' can be used anywhere,
whereas the operator '=' is only allowed at the top level (e.g.,
in the complete expression typed at the command prompt) or as one
of the subexpressions in a braced list of expressions.
though the restriction on '=' seems to be described incorrectly:
if ((x = 2)) cat(assigned\n)
assigned
in which the assignment is in parentheses, not a braced list.
As to the difference between the operations of the two primitives: see
do_set in src/main/eval.c. The facility is there to distinguish between
them, but it is not used.
Duncan Murdoch
`=`
.Primitive(=)
`-`
.Primitive(-)
(Perhaps these are different names for the same internal function?)
Also, the difference is preserved by the parser:
quote({a=b})
{
a = b
}
quote({a-b})
{
a - b
}
even though in other cases the parser canonicalizes variant syntax, e.g. -
to -:
quote({a-b})
{
b - a
}
`-`
Error: object - not found
Is there in fact some semantic difference between = and - ?
If not, why do they use a different operator internally, each calling a
different primitive?
Or is this all just accidental inconsistency resulting from the '=', '-',
and '-' features being added at different times by different people working
off different stylistic conventions?
-s
[[alternative HTML version deleted]]
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[R] Discriminant analyse
Hi everyone, I intend to do a discriminant analyse for 2 measures(eye diameter and body length) and for different areas to show differences between those areas if there are any. The raw data (eye diameter, body length) make one cloud of points so it seems there aren't any differences between those areas. If I take the mean values (eye diameter) for length classes, I nearly get linear functions that differ in slope and intercept. The question is: Can I do something like a discriminant analyse for those lines to statistically separate them from each other or is there a totally different method for this case. If you know any method, please, could you let me know. Thanks,Ben -- __ 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] CORRECTION: Re: Multicollinearity with brglm?
I'm running brglm to do binomial loguistic regression. The perhaps multicollinearity-related feature(s) are: (1) the k IVs are all binary categorical, coded as 0 or 1; (2) each row of the IVs contains exactly C ( k) 1's; (I think this is the source of the problem) (3) there are n * k unique rows, where n is as much as 10; (4) when brglm is run, at least 1 IV is reported as involving a singularity and this occurs for nearly every choice of k, n. How should I go about computing estimates for the offending IVs? I'm interested primarily in the reliability of the parameter estimates. -- View this message in context: http://www.nabble.com/Multicollinearity-with-brglm--tp22814696p22827727.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] smv() in e1071 and the BreastCancer data from mlbench
R-help, I am trying to perform a basic anlaysis of the BreastCancer data from mlbench using the svm() function in e1071. I use the following code library(e1071) library(mlbench) data(BreastCancer) BC - subset(BreastCancer, select=-Id) pairs(BC) model - svm(Class ~ ., data=BC, cross=10) ## plot(model, BC, ) tobj - tune.svm(Class ~ ., data=BC, gamma=seq(0.0001,0.2,length=3), cost=seq(10,100,length=3)) ...And here are the results from my session (with contextual information) system(uname -a) Linux hmhuw015 2.6.9-34.ELsmp #1 SMP Fri Feb 24 16:56:28 EST 2006 x86_64 x86_64 x86_64 GNU/Linux version _ platform x86_64-unknown-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 8.1 year 2008 month 12 day22 svn rev47281 language R version.string R version 2.8.1 (2008-12-22) tobj - tune.svm(Class ~ ., data=BC, gamma=seq(0.0001,0.2,length=3), + cost=seq(1,100,length=3)) Error in names(ret2) - rowns : 'names' attribute [70] must be the same length as the vector [68] Sorry, but could someone tell me what the error is trying to get at? thanks, Brandon --- This e-mail was sent by GlaxoSmithKline Services Unlimited (registered in England and Wales No. 1047315), which is a member of the GlaxoSmithKline group of companies. The registered address of GlaxoSmithKline Services Unlimited is 980 Great West Road, Brentford, Middlesex TW8 9GS. --- [[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] list substring
thank you,
but I m importing data from a txt file and I have a matrix of n*1
The function str seems to work only from 1*n
Wacek Kusnierczyk wrote:
calpeda wrote:
hi
I ve a list of item x = ( x1
x2
x3)
I need to extract a new vector y = ( ,
,
)
I mean, for each item I need to extract the first 4 digit
How is that possible?
sub('^(\\d{4}).*', '\\1', x, perl=TRUE)
vQ
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[R] Package tcltk
Hi, When I type library(tcltk) under R 2.8.1 I get the error message: Loading Tcl/Tk interface ...Error in inDL(x, as.logical(local), as.logical(now), ...) : unable to load shared library 'C:/PROGRA~1/R/R-28~1.1/library/tcltk/libs/tcltk.dll': LoadLibrary failure: The specified module could not be found. Error : .onLoad failed in 'loadNamespace' for 'tcltk' Error: package/namespace load failed for 'tcltk' I have the same problem with 2.7.0 but not in R 2.6.2. This problem appeared from the R 2.8.1 installation. Can you help me? Many Thanks, --- Rita Sousa Departamento de Metodologia e Sistemas de Informação INE - DP: Instituto Nacional de Estatística - Delegação do Porto Tel.: 22 6072016 (Extensão: 4116) --- [[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] Learning development concepts in R for newbie users
Hi: I have had a similar issue so below are ways that I deal with that.  i don't think there are manuals/ documentation for becoming more of a developer ( someone can correct me if I'm wrong and I'd be happy to be wrong ) but there are other ways: 1) Staying on this list as much as possible and watching what people send as solutions and trying to solve them yourself 2) printing out package source code. For adf.test, look at the source and then try to figure out from that what you want to do. Most of the code from packages is written beautifully and therefore it's often not that difficult to understand. 3) Reading the R Language Defintion manual. I found that about 6 months ago and it was really interesting and helpful. 4) Asking the R guRus questions without overwhelming/annoying them. If you stay on this list, you will fquickly figure out who they are.   There are many and some know everything and others are experts in specific areas. They are quite generous with   their expertise and time. But, you should keep in mind that they help for free and it's not their job so try to minimize your   bothers.                                                                                                                                  Good luck. On Apr 1, 2009, Harsh singhal...@gmail.com wrote: Hi R users, I apologize for a seemingly trivial question, but I felt this forum would be the best place to seek advice. I have been an R user for a year now, but I am limited to using R and its various contributed packages. I strongly feel that users of a free and open source software tool must eventually provide development expertise. I am aware of the extensive documentation available on the R project page regarding Internals, Language definitions and the like, but I would like to find tutorials on R development, much the same way as there are tutorials for the use of R in statistics and data analysis. Information on Book titles, online sources, etc that you maybe aware of which could allow a user of R to also develop R (packages, improvise on functions, etc) would be much appreciated. For example, I have used the adf.test in package 'tseries' but the test assumes a certain form the the time series process. There are 3 forms in literature, and the other 2 are not used. I would like to modify the adf.test to incorporate the other two forms as well. Thanks Harsh Singhal [[alternative HTML version deleted]] __ [1]r-h...@r-project.org mailing list [2]https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide [3]http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. References 1. mailto:R-help@r-project.org 2. https://stat.ethz.ch/mailman/listinfo/r-help 3. http://www.R-project.org/posting-guide.html __ 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] Definition of = vs. -
Duncan Murdoch murd...@stats.uwo.ca writes:
On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
(e.g. f(a=1), which is equivalent to
eval(structure(quote(f(1)),names=c('','a'))).
As far as I can tell from the documentation, assignment with = is
precisely
equivalent to assignment with -. Yet they call different primitives:
The parser does treat them differently:
if (x - 2) cat(assigned\n)
assigned
if (x = 2) cat(assigned\n)
Error: unexpected '=' in if (x =
The ?= man page explains this:
The operator '-' can be used anywhere,
whereas the operator '=' is only allowed at the top level (e.g.,
in the complete expression typed at the command prompt) or as one
of the subexpressions in a braced list of expressions.
though the restriction on '=' seems to be described incorrectly:
if ((x = 2)) cat(assigned\n)
assigned
in which the assignment is in parentheses, not a braced list.
As to the difference between the operations of the two primitives:
see do_set in src/main/eval.c. The facility is there to distinguish
between them, but it is not used.
10.4.2 of R-lang shows they differ in precedence
x = y - 1
rm(x,y)
x - y = 1
Error in (x - y) = 1 : object 'x' not found
as reflected in main/names.c (PREC_LEFT vs. PREC_EQ)
{-, do_set, 1, 100,-1, {PP_ASSIGN, PREC_LEFT,
1}},
{=, do_set, 3, 100,-1, {PP_ASSIGN, PREC_EQ,
1}},
and include/Defn.h
PREC_LEFT= 1,
PREC_EQ = 2,
Martin
Duncan Murdoch
`=`
.Primitive(=)
`-`
.Primitive(-)
(Perhaps these are different names for the same internal function?)
Also, the difference is preserved by the parser:
quote({a=b})
{
a = b
}
quote({a-b})
{
a - b
}
even though in other cases the parser canonicalizes variant syntax,
e.g. -
to -:
quote({a-b})
{
b - a
}
`-`
Error: object - not found
Is there in fact some semantic difference between = and - ?
If not, why do they use a different operator internally, each
calling a
different primitive?
Or is this all just accidental inconsistency resulting from the '=',
-',
and '-' features being added at different times by different people working
off different stylistic conventions?
-s
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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--
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N.
PO Box 19024 Seattle, WA 98109
Location: Arnold Building M1 B861
Phone: (206) 667-2793
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Re: [R] Definition of = vs. -
On Wed, Apr 1, 2009 at 10:55 AM, Duncan Murdoch murd...@stats.uwo.cawrote: On 4/1/2009 10:38 AM, Stavros Macrakis wrote: As far as I can tell from the documentation, assignment with = is precisely equivalent to assignment with -. Yet they call different primitives: The parser does treat them differently: if (x - 2) cat(assigned\n) assigned if (x = 2) cat(assigned\n) Error: unexpected '=' in if (x = Interesting way of handling the classic C glitch (some of us would say design flaw in C, but...) The ?= man page explains this: The operator '-' can be used anywhere, whereas the operator '=' is only allowed at the top level (e.g., in the complete expression typed at the command prompt) or as one of the subexpressions in a braced list of expressions. though the restriction on '=' seems to be described incorrectly: if ((x = 2)) cat(assigned\n) assigned The restriction is incorrect in many other cases as well, e.g. the following are all assignments: function()a=3; if(...)a=3; while(...)a=3; a=b=3 (two assignments), and even a*b=3 (parses as assignment, but `*-` happens not to be defined in the default environment). In fact, the only cases I have found where = does *not* mean assignment is in functional or array argument position (f(a=2) and f[a=2]), the following contexts with function-like syntax: function(XXX)..., if(XXX)..., and while(XXX)...; and for (i in XXX) Are there any others? Perhaps the documentation could be updated? As to the difference between the operations of the two primitives: see do_set in src/main/eval.c. The facility is there to distinguish between them, but it is not used. So are you saying that it is planned to make = and - non-synonymous, unlike a-b and b-a, which parse the same and are therefore guaranteed to be synonymous? -s [[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] Odp: Fwd: 'for Loop'
Hi r-help-boun...@r-project.org napsal dne 01.04.2009 11:16:26: Hello, A nice guy call Jun Shen was helping me out with this, but I require a bit more help. Below is my data set or list called 'test'. I'm trying to calculate the %RSD for each pair of index and keep it in cronological order if you can imagine a 3rd column with 'date' beside index. Result Index 1 0.290117 2 0.292817 3 0.291318 4 0.289318 5 0.3526 9 6 0.3316 9 7 0.349210 8 0.346710 9 0.338511 10 0.334111 11 0.344912 12 0.340612 13 0.369013 14 0.332713 15 0.332414 16 0.324514 17 0.324515 18 0.322915 19 0.333116 20 0.320816 now this bit of programming works for it how ever it puts the mean of my index back into numeric order. aggregate(test[1],test[2],mean)-inter names(inter)[2]='mean' merge(test,inter,all=T)-inter2 inter2$RSV=inter2$Result/inter2$mean ?ave was also recommended and for your case it is strightforward test$mean-ave(test$Result, test$Index) test$RSV-test$Result/test$mean Regards Petr Would anyone know away around this so I could keep the data in cronological order? Kind Regards, Al [[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] Still confused about R method of data exchanging between caller and called function
First of all I'd like to thank all those who answered me back teaching me
different ways to get the calledfunction modify global data rather than its
own. I fixed that.
Now I have a similar pproblem. Whenever the caller passes a matrix to the
called function I thought the called function would make a copy of the passed
array in its own memory frame, modify its own copy, but be able to pass back
the results to the caller as follows ... but I am mistaken (I find all NA on
the second call ... how caome ?
ford - function(X,N,Nfour,Kfour,LevMin,LevMax) {
Y - X
Nord - Kord
if(Nord == 2){
cat(\n call HAAR \n)
Y - Haar(Y,Nfour,1)
}else{
cat(\n call pwtset \n)
pwtset(Nord)
cat(\n call wt1 \n)
Y - wt1(Y,Nfour,1)
cat(\n function 'ford' Y = ,Y,\n)
}
#
wt1 - function (a,n,isign){
cat(\n BEGIN 'wt1' \n)
cat(\n 'wt1': n = ,n,\n)
cat(\n 'wt1': a = ,a,\n)
if (n 4){
cat(\n function 'wt1': WRONG INPUT SIGNAL LENGTH \n)
return()
}
if(isign = 0){
nn - n
while(nn = 4){
cat(\n call pwt \n)
a - pwt(a,nn,isign)
nn - nn/2
}
...
cat(\n END 'wt1' \n)
a #RETURN WAVELET COEFFICIENTS
}# end function wt1
tutti i telefonini TIM!
[[alternative HTML version deleted]]
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[R] Fit unequal variance model in R
I'am trying to develop some code if R, which would correspond to what I did in SAS. The data look like: TreatmentReplicategroup1 GSI Control A 1 0.81301 Control B 1 1.06061 Control C 1 1.26350 Control D 1 0.93284 Low A 2 0.79359 Low B 2 0.89111 Low C 2 1.03485 Low D 2 1.29166 Mid A 3 1.26900 Mid B 3 . Mid C 3 1.58666 Mid D 3 1.35759 HighA 4 2.02680 HighB 4 1.52372 HighC 4 2.19167 HighD 4 1.29949 The SAS code is: proc mixed data=data_name order=data method=ml; *scoring=10; classes group1; model GSI=group1/residual influence solution; repeated /group=group1; run; Basically, I need different variance for each treatment group. I want to do the similar thing in R. Here is what I get so far: lm1-lme(response~treatment,data=o,random=~1|as.factor(dummy),weights=varIdent(form=~1|treatment),method=ML) There should no random term in the model. However If I don't specify one, lme won't work, so I made a dummy variable, which equals to 1 for every observation. If anyone could help, it will be greatly appreciated. Thanks, Jingyu [[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] A query about na.omit
Dear all, Say I have the following dataset: DF x y z [1] 1 1 1 [2] 2 2 2 [3] 3 3NA [4] 4 NA 4 [5] NA 5 5 And I want to omit all the rows which have NA, but only in columns X and Y, so that I get: x y z 1 1 1 2 2 2 3 3 NA If I use na.omit(DF), I would delete the row for which z=NA, obtaining thus x y z 1 1 1 2 2 2 But this is not what I want, of course. If I use na.omit(DF[,1:2]), then I obtain x y 1 1 2 2 3 3 which is OK for x and y columns, but I wouldn't get the corresponding values for z (ie 1 2 NA) Any suggestions about how to obtain the desired results efficiently (the actual dataset has millions of records and almost 50 columns, and I would apply the procedure on 12 of these columns)? Sincerely, Jose Luis Jose Luis Iparraguirre Senior Research Economist Economic Research Institute of Northern Ireland [[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 with mixed-effects model with temporal pseudoreplication!
Sorry if this is the wrong ml for this question, I am new to R. I am trying to use R to analyze the data from my thesis experiment and I am having troubles accounting for the pseudoreplication properly from having each participant repeat each treatment combination (combination of fixed factors) 5 times. The design of the experiment is as follows... Responses: CompletionTIme VisitedTargets Fixed-factors: Targets (4-levels): 4, 9, 14, 19 Entropy (3-levels): Low, Medium, High Random-factors: Participants: 31 total participants Replicates: 5 (this could also be viewed as a time factor I think) BlockOrder: 1 though 60 (the order of the trials was random for each participant, but I am not so concerned about this right now) The fixed part of the model seems pretty intuitive: fixed=log(CompletionTime)~(Targets*Entropy) The random part of the model is where I get stuck on, I've tried many combinations and all give me the wrong degrees of freedom. I really don't know what to use. Any help would be greatly appreciated Here is the code I am using in R: library(nlme) datafile=http://people.rit.edu/rmh3093/mot.csv; master1 = read.table(datafile,header=T) Block=factor(master1$Block) BlockOrder=factor(master1$Block_Order) Replicate=factor(master1$Replicate) Participant=factor(master1$Participant_ID) Targets=factor(master1$Targets) Entropy=factor(master1$Entropy) CompletionTime=master1$Completion_Time summary(lme(log(CompletionTime)~(Entropy*Targets),random=~1|Participant,method=ML)) Thanks in advance! -Ryan __ 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] Calculating First Occurance by a factor
I have another question regarding ddply. In my actual data.frame, I have many
other column variables corresponding to the type of the trial. I'm wondering
if I can have ddply include those in firstfixtime as well.
I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it
to df[which.min(df$FixInx)] or adding new lines with the additional columns
that I want to include, but nothing seemed to work. I'll admit I only have a
mild understanding of what is going on with the function .fun. :-)
Mike Lawrence wrote:
I discovered Hadley Wickham's plyr package last week and have found
it very useful in circumstances like this:
library(plyr)
firstfixtime = ddply(
.data = data
, .variables = c('Sub','Tr','IA')
, .fun - function(df){
df$FixTime[which.min(df$FixInx)]
}
)
On Mon, Mar 30, 2009 at 3:40 PM, jwg20 jason.gulli...@gmail.com wrote:
I'm having difficulty finding a solution to my problem that without
using a
for loop. For the amount of data I (will) have, the for loop will
probably
be too slow. I tried searching around before posting and couldn't find
anything, hopefully it's not embarrassingly easy.
Consider the data.frame, Data, below
Data
Sub Tr IA FixInx FixTime
p1 t1 1 1 200
p1 t1 2 2 350
p1 t1 2 3 500
p1 t1 3 4 600
p1 t1 3 5 700
p1 t1 4 6 850
p1 t1 3 7 1200
p1 t1 5 8 1350
p1 t1 5 9 1500
What I'm trying to do is for each unique IA get the first occurring
FixTime.
This will eventually need to be done by each Trial (Tr) and each Subject
Number (Sub). FixInx is essentially the number of rows in a trial. The
resulting data.frame is below.
Sub Tr IA FirstFixTime
p1 t1 1 200
p1 t1 2 350
p1 t1 3 600
p1 t1 4 850
p1 t1 5 1350
Here is the solution I have now.
agg = aggregate(data$FixInx, list(data$Sub, data$Tr, data$IA), min) #get
the
minimum fix index by Sub, Tr, and IA... I can use this min fix index to
pull
out the desired fixtime
agg$firstfixtime = 0 # new column for results
for (rown in 1:length(rownames(agg))){ #cycle through rows and get each
data$firstfixtime from FixTime in matching rows
agg$firstfixtime[rown] = as.character(data[data$Tr == agg$Group.2[rown]
data$Sub == agg$Group.1[rown] data$IA == agg$Group.3[rown]
data$FixInx
== agg$x[rown], ]$FixTime)
}
--
View this message in context:
http://www.nabble.com/Calculating-First-Occurance-by-a-factor-tp22789964p22789964.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student
Department of Psychology
Dalhousie University
Looking to arrange a meeting? Check my public calendar:
http://tinyurl.com/mikes-public-calendar
~ Certainty is folly... I think. ~
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating First Occurance by a factor
I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it to df[which.min(df$FixInx)] or adding new lines with the additional columns that I want to include, but nothing seemed to work. I'll admit I only have a mild understanding of what is going on with the function .fun. :-) You probably want: df[which.min(df$FixInx), ] Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating First Occurance by a factor
On Wed, Apr 1, 2009 at 11:00 AM, hadley wickham h.wick...@gmail.com wrote: I tried messing with the line df$FixTime[which.min(df$FixInx)] changing it to df[which.min(df$FixInx)] or adding new lines with the additional columns that I want to include, but nothing seemed to work. I'll admit I only have a mild understanding of what is going on with the function .fun. :-) You probably want: df[which.min(df$FixInx), ] Or alternatively: ddply(data, .(Sub, Tr, IA), subset, FixInx == min(FixInx)) which might be a bit easier to understand. Hadley -- 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] Latex symbols in R (\perp and \parallel)
Dear All,
I am sure this is a one-liner, but I cannot find the R command to
generate the LaTex symbols \perp and \parallel. Consider for instance
the figure
(one can use any kind of data for the plot)
pdf(friction_linear_chain_perpendicular.pdf)
par( mar = c(4.5,5, 2, 1) + 0.1)
plot(data[ ,1], data[
,4],col=blue,lwd=1.5,lty=1,pch=5,ylab=expression(*beta[n]^{perpendicular}/beta[1]*),
xlab=expression(*n*)
,cex.axis=1.4,cex.lab=1.6)
lines(x,beta_perp(as.list(coef(nls.out)), x), col=black, lwd=2)
dev.off()
Greek letters are understood on the spot, but this is not the case for
the perpendicular symbol in LaTex.
Does anyone know where I can find a list of LaTex symbols translated for R?
Many thanks
Lorenzo
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Re: [R] Deriving Samples from specific, not implemented PDF for a QQ-Plot
David, thank you very much for the quick response: The sample example helped and works fine for me. I'm sorry for not providing an example. In order to explain my problem see following example: test-rt(1000,df=5) Repeat only the following code to see how the second and third plot changes par(mfrow=c(1,3)) qqnorm(test) sample.from.normal-sample(seq(-3,3,by=0.0001), size=1000, prob=dnorm(seq(-3,3,by=0.0001)), replace=TRUE) qqplot(sample.from.normal,test, main = Q-Q Plot with sample from dnorm) quanto-quantile(test,seq(0,1,by=0.0001)) quanta2-quantile(sample.from.normal,seq(0,1,by=0.0001)) plot(quanta2,quanto, main = Plot of Quantiles with sample from dnorm) The first example shows the QQ-Plot against the normal (qqnorm), the second shows the QQ-Plot against a sample of the normal, the third one is a self-made QQ-Plot (not using the qqplot function). If you fix the data sample (test) and repeat only the plotting, the second and third plot change a little bit every time. This is clearly because of the sample drawing from the theoretical distribution (sample.from.normal). Besides, this method seems to be circuitous and imprecisely. Why does the qqnorm plot not change? I guess it is due to the precisely calculated quantiles of the normal distribution. How can I derive the quantiles directly from the pdf (i.e. from dnorm(x) without sampling or how to get qnorm from dnorm without using qnorm)? Remember, I've got only the pdf for use. Best regards Arndt Zimmermann == On Mar 31, 2009, at 11:53 AM, Arndt Zimmermann wrote: trimmed superfluous white-space this is my first post to the R-help, so please don't be too strict. Reproducible examples should not be too much to expect, since that is requested in the posting guide and multiple other locations. My problem concerns a QQ-Plot: I want to show how well empirical samples match with a theoretical distribution. The theoretical distribution has got several parameters, but I made it to fit via ML. Anyway, the theoretical function gives me the density for a given data point x. As far as I'm aware, the qqplot (generic) function in R does only take data samples as Input. How can I derive samples from the pdf? Am I right in the way I see the problem? Rather hard to tell, since you haven't described what I understand to be a problem, nor described how you see it. Or How can I modify the qqplot function, (or rebuild for myself) to get it going? It seems to be going fine for us. Did you break it? I'm sure, this is very likely just a misunderstanding problem from myself, but anybody who can shed further light on this is very welcome. Since you offer no examples, it is rather difficult to offer specific advice. You should be able to use the sample function on a pdf by assigning a vector of probabilities to the prob argument. mean(sample(seq(-2,2,by=0.2), size=100, prob=dnorm(seq(-2,2,by=0.2)), replace=TRUE)) [1] -0.08 sd(sample(seq(-2,2,by=0.2), siez=100, prob=dnorm(seq(-2,2,by=0.2)), replace=TRUE)) [1] 0.911101 David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fit unequal variance model in R
Feng, Jingyu wrote: I'am trying to develop some code if R, which would correspond to what I did in SAS. The data look like: TreatmentReplicategroup1 GSI .. The SAS code is: proc mixed data=data_name order=data method=ml; *scoring=10; classes group1; model GSI=group1/residual influence solution; repeated /group=group1; run; Basically, I need different variance for each treatment group. I want to do the similar thing in R. Here is what I get so far: lm1-lme(response~treatment,data=o,random=~1|as.factor(dummy),weights=varIdent(form=~1|treatment),method=ML) There should no random term in the model. However If I don't specify one, lme won't work, so I made a dummy variable, which equals to 1 for every observation. The much underused (quote Frank Harrell) gls in package nlme should do that. Quote PB (p250): It can be viewed as an lme function without the random argument. Dieter -- View this message in context: http://www.nabble.com/Fit-unequal-variance-model-in-R-tp22829549p22830566.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] Definition of = vs. -
On 4/1/2009 11:39 AM, Stavros Macrakis wrote: On Wed, Apr 1, 2009 at 10:55 AM, Duncan Murdoch murd...@stats.uwo.cawrote: On 4/1/2009 10:38 AM, Stavros Macrakis wrote: As far as I can tell from the documentation, assignment with = is precisely equivalent to assignment with -. Yet they call different primitives: The parser does treat them differently: if (x - 2) cat(assigned\n) assigned if (x = 2) cat(assigned\n) Error: unexpected '=' in if (x = Interesting way of handling the classic C glitch (some of us would say design flaw in C, but...) The ?= man page explains this: The operator '-' can be used anywhere, whereas the operator '=' is only allowed at the top level (e.g., in the complete expression typed at the command prompt) or as one of the subexpressions in a braced list of expressions. though the restriction on '=' seems to be described incorrectly: if ((x = 2)) cat(assigned\n) assigned The restriction is incorrect in many other cases as well, e.g. the following are all assignments: function()a=3; if(...)a=3; while(...)a=3; a=b=3 (two assignments), and even a*b=3 (parses as assignment, but `*-` happens not to be defined in the default environment). In fact, the only cases I have found where = does *not* mean assignment is in functional or array argument position (f(a=2) and f[a=2]), the following contexts with function-like syntax: function(XXX)..., if(XXX)..., and while(XXX)...; and for (i in XXX) Are there any others? Perhaps the documentation could be updated? As to the difference between the operations of the two primitives: see do_set in src/main/eval.c. The facility is there to distinguish between them, but it is not used. So are you saying that it is planned to make = and - non-synonymous, unlike a-b and b-a, which parse the same and are therefore guaranteed to be synonymous? No, I don't know of any plans like that. That doesn't mean there aren't any, nor does the current implementation of - and - guarantee no changes there, but I wouldn't expect either to change. 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.
Re: [R] Help with mixed-effects model with temporal pseudoreplication!
Bugzilla from rmh3...@gmail.com wrote: Responses: CompletionTIme VisitedTargets Fixed-factors: Targets (4-levels): 4, 9, 14, 19 Entropy (3-levels): Low, Medium, High Random-factors: Participants: 31 total participants Replicates: 5 (this could also be viewed as a time factor I think) BlockOrder: 1 though 60 (the order of the trials was random for each participant, but I am not so concerned about this right now) The fixed part of the model seems pretty intuitive: fixed=log(CompletionTime)~(Targets*Entropy) The random part of the model is where I get stuck on, I've tried many combinations and all give me the wrong degrees of freedom. I really don't know what to use. Any help would be greatly appreciated Here is the code I am using in R: library(nlme) datafile=http://people.rit.edu/rmh3093/mot.csv; master1 = read.table(datafile,header=T) Block=factor(master1$Block) BlockOrder=factor(master1$Block_Order) Replicate=factor(master1$Replicate) Participant=factor(master1$Participant_ID) Targets=factor(master1$Targets) Entropy=factor(master1$Entropy) CompletionTime=master1$Completion_Time summary(lme(log(CompletionTime)~(Entropy*Targets),random=~1|Participant,method=ML)) Looks perfectly fine to me. To paraphrase Kubrick: Stop worrying about degrees of freedoms, and love your results. You might find a few flames and insights when searching the list for degrees freedom bates lme lmer. Using log(CompletionTime) is fine if that's what is usually done in your field, but think of using the weights argument in lme to get results directly in linear units. Most readers prefer time to log(time) in the final result. Dieter -- View this message in context: http://www.nabble.com/Help-with-mixed-effects-model-with-temporal-pseudoreplication%21-tp22829729p22830759.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] Vector of Vectors
I have a matrix of data. I need to scan the matrix and find every sequence from maxima to maxima across a row. I can write a loop to do this easily. Problem is, I can't figure out how to store the results. Each result is a vector of widely varying lengths. Ideally I'd like a vector of these, i.e. a vector of vectors, so I can quickly iterate through them and compute correlation coefficients. Here's a transcript of my fuddling to date: x - c(1,2,3) y - c(4,5) v - list(1=x, 2=y) unlist(v[1]) 11 12 13 1 2 3 unlist(v[1])[1] 11 1 unlist(v[1])[2] 12 2 unlist(v[1])[3] 13 3 unlist(v[2])[3] NA NA unlist(v[2])[2] 22 5 v - c(x,y) v [1] 1 2 3 4 5 v - vector() v - merge(v, x) v [,1] [,2] attr(,row.names) integer(0) v[1] [1] NA As you can see, vectors aren't very cooperative and lists are downright baffling to me. Shawn Garbett shawn.p.garb...@vanderbilt.edu Vanderbilt Cancer Biology 220 Pierce Ave, PRB 715AA Nashville, TN 37232 Office: 615.936.1975 Cell: 615.397.8737 __ 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] A query about na.omit
On 01-Apr-09 15:49:40, Jose Iparraguirre D'Elia wrote: Dear all, Say I have the following dataset: DF x y z [1] 1 1 1 [2] 2 2 2 [3] 3 3NA [4] 4 NA 4 [5] NA 5 5 And I want to omit all the rows which have NA, but only in columns X and Y, so that I get: x y z 1 1 1 2 2 2 3 3 NA Roll up your sleeves, and spell out in detail the condition you need: DF-data.frame(x=c(1,2,3,4,NA),y=c(1,2,3,NA,5),z=c(1,2,NA,4,5)) DF #x y z # 1 1 1 1 # 2 2 2 2 # 3 3 3 NA # 4 4 NA 4 # 5 NA 5 5 DF[!(is.na(rowSums(DF[,(1:2)]))),] # x y z # 1 1 1 1 # 2 2 2 2 # 3 3 3 NA Hoping this helps, Ted. If I use na.omit(DF), I would delete the row for which z=NA, obtaining thus x y z 1 1 1 2 2 2 But this is not what I want, of course. If I use na.omit(DF[,1:2]), then I obtain x y 1 1 2 2 3 3 which is OK for x and y columns, but I wouldn't get the corresponding values for z (ie 1 2 NA) Any suggestions about how to obtain the desired results efficiently (the actual dataset has millions of records and almost 50 columns, and I would apply the procedure on 12 of these columns)? Sincerely, Jose Luis Jose Luis Iparraguirre Senior Research Economist Economic Research Institute of Northern Ireland [[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. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 01-Apr-09 Time: 18:00:53 -- XFMail -- __ 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] Latex symbols in R (\perp and \parallel)
Lorenzo Isella wrote: I am sure this is a one-liner, but I cannot find the R command to generate the LaTex symbols \perp and \parallel. As often, the most helpful how-to resource is by Prof. Brian Ripley http://markmail.org/thread/kauzftprydrhqq5m if you manage to get around the many lines telling me that I am an idiot. What works depends on your system. Mine is Windows. plot(1:5, type=n) #http://www.fileformat.info/info/unicode/char/27c2/index.htm text(1,1, \u27C2) # does not work for me #http://www.stat.auckland.ac.nz/~paul/R/CM/AdobeSym.html text(2,2, \u22a5) # not for me text(3,3, \x5e,font=5) # works for me Please simplify your examples and make them self-running next time. Dieter -- View this message in context: http://www.nabble.com/Latex-symbols-in-R-%28%5Cperp-and-%5Cparallel%29-tp22829919p22831262.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] A query about na.omit
First input the data frame: Lines - x y z +1 1 1 +2 2 2 +3 3NA +4 NA 4 + NA 5 5 DF - read.table(textConnection(Lines), header = TRUE) # Now uses complete.cases to get required rows: DF[complete.cases(DF[1:2]),] x y z 1 1 1 1 2 2 2 2 3 3 3 NA On Wed, Apr 1, 2009 at 11:49 AM, Jose Iparraguirre D'Elia j...@erini.ac.uk wrote: Dear all, Say I have the following dataset: DF x y z [1] 1 1 1 [2] 2 2 2 [3] 3 3 NA [4] 4 NA 4 [5] NA 5 5 And I want to omit all the rows which have NA, but only in columns X and Y, so that I get: x y z 1 1 1 2 2 2 3 3 NA If I use na.omit(DF), I would delete the row for which z=NA, obtaining thus x y z 1 1 1 2 2 2 But this is not what I want, of course. If I use na.omit(DF[,1:2]), then I obtain x y 1 1 2 2 3 3 which is OK for x and y columns, but I wouldn't get the corresponding values for z (ie 1 2 NA) Any suggestions about how to obtain the desired results efficiently (the actual dataset has millions of records and almost 50 columns, and I would apply the procedure on 12 of these columns)? Sincerely, Jose Luis Jose Luis Iparraguirre Senior Research Economist Economic Research Institute of Northern Ireland [[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] Recommended packages for a statistician
The company I work for require users to request what packages they want from the IT department (user cannot download themselves). I intend to request installation of the latest version of R plus the 23 Cran task views. As a statistician what are the recommended packages or packages that statisticians using R recommend to install. I have started a new position and want to (greedily) get everything that I may or may not use as I want to avoid multiple requests to our IT dept. Thanks [[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] Noobie ANOVA intercept question
Dear R list, I've been attempting to interpret the results from a three-way ANOVA. I think I understand contrasts and the R defaults for these (treatment contrasts). My question is: what is the intercept in this test? As far as I can tell, its NOT the expected value of a point that belongs to the first level of all three explanatory factors (because there is only one point that satisfies these requirements and their values differ). Its not the grand mean, or any of the treatment means. What is this thing? (Note: this dataset is from an example I'm working through in Grafen Hails 2002 text) Q2: Just noticed that in pasting I lose mono-spaced formatting. Is it possible to post to the list such that format is maintained? Thanks in advance! Relevant output: anova(mod1) Analysis of Variance Table Response: SQBLOOMS Df Sum Sq Mean Sq F valuePr(F) BED2 4.1323 2.0661 9.4570 0.0007277 *** WATER 2 3.7153 1.8577 8.5029 0.0013016 ** SHADE 3 1.6465 0.5488 2.5120 0.0789451 . Residuals 28 6.1173 0.2185 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 summary(mod1) Call: lm(formula = SQBLOOMS ~ BED + WATER + SHADE) Residuals: Min 1Q Median 3Q Max -1.23992 -0.18979 -0.01840 0.17471 0.74686 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 3.7765 0.2203 17.139 2.23e-16 *** BED2 0.3185 0.1908 1.669 0.106242 BED3 -0.5044 0.1908 -2.643 0.013293 * WATER20.7842 0.1908 4.109 0.000313 *** WATER30.4489 0.1908 2.353 0.025905 * SHADE20.1969 0.2203 0.894 0.379172 SHADE3 -0.2157 0.2203 -0.979 0.336068 SHADE4 -0.3673 0.2203 -1.667 0.106641 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.4674 on 28 degrees of freedom (51 observations deleted due to missingness) Multiple R-squared: 0.6081, Adjusted R-squared: 0.5102 F-statistic: 6.208 on 7 and 28 DF, p-value: 0.0001911 model.frame(mod1) SQBLOOMS BED WATER SHADE 1 4.359 1 1 1 2 3.317 1 1 2 3 3.606 1 1 3 4 4.123 1 1 4 5 4.472 1 2 1 6 4.583 1 2 2 7 4.359 1 2 3 8 4.690 1 2 4 9 4.123 1 3 1 104.123 1 3 2 113.464 1 3 3 123.873 1 3 4 133.606 2 1 1 144.000 2 1 2 153.464 2 1 3 163.873 2 1 4 174.690 2 2 1 185.000 2 2 2 195.385 2 2 3 204.583 2 2 4 214.690 2 3 1 224.690 2 3 2 234.690 2 3 3 244.243 2 3 4 253.317 3 1 1 263.606 3 1 2 273.317 3 1 3 282.828 3 1 4 293.873 3 2 1 305.000 3 2 2 313.742 3 2 3 322.449 3 2 4 334.000 3 3 1 344.583 3 3 2 353.162 3 3 3 363.162 3 3 4 model.tables(mod1,means,se=TRUE) Tables of means Grand mean 4.029028 BED BED 1 2 3 4.091 4.409 3.587 WATER WATER 1 2 3 3.618 4.402 4.067 SHADE SHADE 1 2 3 4 4.126 4.322 3.910 3.758 Standard errors for differences of means BED WATER SHADE 0.1908 0.1908 0.2203 replic. 12 12 9 Design matrix: model.matrix(mod1) (Intercept) BED2 BED3 WATER2 WATER3 SHADE2 SHADE3 SHADE4 1100 0 0 0 0 0 2100 0 0 1 0 0 3100 0 0 0 1 0 4100 0 0 0 0 1 5100 1 0 0 0 0 6100 1 0 1 0 0 7100 1 0 0 1 0 8100 1 0 0 0 1 9100 0 1 0 0 0 10 100 0 1 1 0 0 11 100 0 1 0 1 0 12 100 0 1 0 0 1 13 110 0 0 0 0 0 14 110 0 0 1 0 0 15 110 0 0 0 1 0 16 110 0 0 0 0 1 17 110 1 0 0 0 0 18 110 1 0 1 0 0 19 110 1 0 0 1 0 20 110 1 0 0 0 1 21 110 0 1 0 0 0 22 110 0 1 1 0 0 23 110 0 1
[R] Lattice plot with an extra strip showing group weights
Dear r-help, How can I add a strip to show group weights using lattice package? For example, in the following code, I'd like to using wt variable in a trip to demonstrate the relative size of groups. (Following is just the simplest form to demonstrate the question. A conditional variable will be added to the bwplot in real problems). # library(lattice) # generate data x1 - cbind(rnorm(50,1,1),rnorm(50,10,1),1) x2 - cbind(rnorm(50,2,1),rnorm(50,20,1),2) x - as.data.frame(rbind(x1,x2)) names(x) - c(x,wt,group) # plot bwplot(x ~ group, data=x,horizontal=FALSE) ## Thanks in advance for this great R group! Regards, Marvel1812 [[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] SNOW: Error in socketSelect(socklist) : not a socket connection
I'm trying to use snow in my dual-core (hopefully later this is going to
run in a cluster). So, at this moment I create a cluster using SOCK
connection (MPI in the future). However when I try to use
clusterApplyLB I got Error in socketSelect(socklist) : not a socket
connection. Any ideas ? Do you know if that is going to be an isuue too
when I swith from SOCK to MPI ?
Sample code is attached.
Thanks.
# TODO: Add comment
#
# Author: diego
###
rm(list=ls())
library(snow)
t1-Sys.time()
rates-c(0.5,0.5,0.7)
initialState-c(0,0,0,0,0)
AllEvents-c(1,1,1,1,2,2,2,3,1,2)
AllLocations-c(1,2,3,4,1,4,2,1,5,4)
AllTimes-1:10
AllConfigurations-rbind(
c(0,0,0,0,0),
c(1,0,0,0,0),
c(1,1,0,0,0),
c(1,1,1,0,0),
c(1,1,1,1,0),
c(2,1,1,1,0),
c(2,1,1,2,0),
c(2,2,1,2,0),
c(0,2,1,2,0),
c(0,2,1,2,1),
c(0,2,1,0,1)
)
nevents-length(AllEvents)
clusterSize-2
cl-makeCluster(clusterSize,type=SOCK)
arguments-vector(list,nevents)
completedTimes-c(0,AllTimes)
for(i in 1:nevents){
element-c(AllConfigurations[i,],
rates,
completedTimes[i],
AllEvents[i],
AllLocations[i],
AllTimes[i])
arguments[[i]]-element
}
source(GetLogLikelihood.R)
parallelOutputs-clusterApplyLB(cl,arguments,GetLogLikelihood) # This give me
error
#parallelOutputs-clusterApply(cl,arguments,GetLogLikelihood) # This Work OK
print(sum(unlist(parallelOutputs)))
stopCluster(cl)
t2-Sys.time()
print(t2-t1)
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[R] Plotting multiple ablines
I really want to do this:
abline(
a=tan(-kT*pi/180),
b=kY-tan(-kT*pi/180)*kX
)
where kX,kY and kT are vectors of equal length. But I can't do that
with abline unless I use a loop, and I haven't figured out the least
unelegant way of writing the loop yet. So is there a way to do this
without a loop?
Or if I am to resort to the loop, what's the best way of doing it
considering that I have some missing data? Here's the mess that I
wrote.
converge - na.omit(data.frame(kX,kY,kT))
for (z in (length(converge$kT)))
{abline(
a=tan(converge$kT[z]*pi/180),
b=converge$kY[z]-tan(-converge$kT[z]*converge$kX[z]*pi/180)
)}
I think the missing data are causing the problem; this happens when I run:
Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) :
'a' and 'b' must be finite
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[R] Bootstrap Confidence Intervals
How can I performing Bootstrap Confidence Intervals for the estimates of nonparametric regression y=f(x) such as loess and spline smoothing Thanks in advance [[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] feature selection problem,urgent help need
Hello,
I have a problem in feature selection I would be thankful if you can help
me.
I have a dataset with limited samples (for example 100) and a lot of
features (for example 3000) and i have to do feature selection.
if i use cross validation (for example *10 fold*) i rank the features based
on 90 samples (using svmrfe method) i achieve ranked feature for example
{f2,f4,f1,f3,...} (it means f2 is ranked first with svmrfe) now, I want to
know how many features i should use? so i should compute the performance for
n feature selected from first of the ranked list and compute the performance
of it. for example train learner with f2, another time with f2,f4, another
time with f2,f4,f1 ... and see which is better , but my problem is:
1) first of all,for comparison should i use the performance of 9 fold that
has been used for ranking or the performance of learner on the fold which
has been left out?I mean in the *feature selection* *step (not in the final
evaluation)*,for example to see I should select only f2 or select {f2 , f4}
how should I compare?
2) in each stage of cross validation different feature subset will be
created . i can compute for each feature the number of times it has repeated
in each folding result, but after that how can i conclude the final feature
set?
can you please help me? I need your urgent help.
thanks in advance
Azadeh
[[alternative HTML version deleted]]
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[R] Plotting a time series
I have data that I read in using: data-read.table(RAVK2.obs.data,sep=\t) 'data' looks like this: V1V2 1 2009-03-25 06:00:00 12.86 2 2009-03-25 12:00:00 12.80 3 2009-03-25 18:00:00 12.76 4 2009-03-26 00:00:00 12.68 5 2009-03-26 06:00:00 12.66 6 2009-03-26 12:00:00 12.64 7 2009-03-26 18:00:00 12.83 8 2009-03-27 00:00:00 13.33 9 2009-03-27 06:00:00 13.84 10 2009-03-27 12:00:00 14.13 11 2009-03-27 18:00:00 14.29 12 2009-03-28 00:00:00 14.41 13 2009-03-28 06:00:00 14.48 14 2009-03-28 12:00:00 14.58 15 2009-03-28 18:00:00 14.75 16 2009-03-29 00:00:00 15.02 17 2009-03-29 06:00:00 15.40 18 2009-03-29 12:00:00 15.88 19 2009-03-29 18:00:00 16.50 20 2009-03-30 00:00:00 16.77 21 2009-03-30 06:00:00 16.73 22 2009-03-30 12:00:00 16.55 23 2009-03-30 18:00:00 16.31 24 2009-03-31 00:30:00 16.04 25 2009-03-31 06:00:00 15.80 26 2009-03-31 12:00:00 15.65 27 2009-03-31 18:00:00 15.53 28 2009-04-01 00:00:00 15.39 29 2009-04-01 06:00:00 15.20 30 2009-04-01 12:00:00 15.00 when I use plot(data) the plot looks fine except that the plot symbols are broken horizontal lines. How do I change the plot symbol? I have tried: plot(data, type=p) and I get the same result. Thanks, Tom -- Thomas E Adams National Weather Service Ohio River Forecast Center 1901 South State Route 134 Wilmington, OH 45177 EMAIL: thomas.ad...@noaa.gov VOICE: 937-383-0528 FAX:937-383-0033 __ 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] Public R servers?
Hello, Earlier I posted a question about memory usage, and the community's input was very helpful. However, I'm now extending my dataset (which I use when running a regression using lm). As a result, I am continuing to run into problems with memory usage, and I believe I need to shift to implementing the analysis on a different system.. I know that R supports R servers through Rserve. Are there any public servers where I could upload my datasets (either as a text file, or through a connection to a SQL server), execute the analysis, then download the results? I identifed Wessa.net (http://www.wessa.net/mrc.wasp?outtype=Browser%20Blue%20-%20Charts%20White), but it's not clear it will meet my needs. Can anyone suggest any other resources? Thanks in advance, Aaron Barzilai [[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] list substring
calpeda wrote: thank you, but I m importing data from a txt file and I have a matrix of n*1 The function str seems to work only from 1*n you see, it would help if you provided more details from the start. you may still need to do it; it seems that both solutions you were given (mine and the substr one) work for both n*1 and 1*n matrices: # 5*1 (strings = matrix(as.character(sample(10^6, 5)), ncol=1)) substr(strings, 1, 4) substr(t(strings), 1, 4) with a fixed pattern of four digits from the beginning of the string, the substr solution is better -- simpler and a few times faster. vQ __ 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] Variable Wildcard Value
This whole thing is an April Fool's joke. Isn't it?
***Please***!!! (Let it be an April Fool's joke.)
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] A query about na.omit
On Wed, 2009-04-01 at 16:49 +0100, Jose Iparraguirre D'Elia wrote: Dear all, Say I have the following dataset: DF x y z [1] 1 1 1 [2] 2 2 2 [3] 3 3NA [4] 4 NA 4 [5] NA 5 5 And I want to omit all the rows which have NA, but only in columns X and Y, so that I get: x y z 1 1 1 2 2 2 3 3 NA If I use na.omit(DF), I would delete the row for which z=NA, obtaining thus x y z 1 1 1 2 2 2 But this is not what I want, of course. If I use na.omit(DF[,1:2]), then I obtain x y 1 1 2 2 3 3 which is OK for x and y columns, but I wouldn't get the corresponding values for z (ie 1 2 NA) Any suggestions about how to obtain the desired results efficiently (the actual dataset has millions of records and almost 50 columns, and I would apply the procedure on 12 of these columns)? Sincerely, Jose Luis Jose Luis Iparraguirre Senior Research Economist Economic Research Institute of Northern Ireland Hi Jose Luis, I think this script is sufficient for your problem: tab-matrix(c(1,1,1,2,2,2,3,3,NA,4,NA,4,NA,5,5),ncol=3,byrow=T) tab[!is.na(tab[,1])!is.na(tab[,2]),] -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil __ 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] Noobie ANOVA intercept question
Dear Allen, On Wed, 1 Apr 2009 10:44:33 -0700 (PDT) AllenL allen.laroc...@gmail.com wrote: Dear R list, I've been attempting to interpret the results from a three-way ANOVA. I think I understand contrasts and the R defaults for these (treatment contrasts). My question is: what is the intercept in this test? As far as I can tell, its NOT the expected value of a point that belongs to the first level of all three explanatory factors (because there is only one point that satisfies these requirements and their values differ). Its not the grand mean, or any of the treatment means. What is this thing? Actually, the intercept IS the estimated expected value of an observation that belongs to the first level of all three factors, but conditional on the model that you've fit, which is an additive model. Had you fit a model with all interactions, SQBLOOMS ~ BED*WATER*SHADE, then the intercept would have been the mean (i.e., single observation) in this cell. BTW, more than half the data were deleted due to NAs. (Note: this dataset is from an example I'm working through in Grafen Hails 2002 text) Q2: Just noticed that in pasting I lose mono-spaced formatting. Is it possible to post to the list such that format is maintained? This is a function of your mail client, not R. The r-help list sends out plain-text email, so the monospaced font was restored. I hope this helps, John Thanks in advance! Relevant output: anova(mod1) Analysis of Variance Table Response: SQBLOOMS Df Sum Sq Mean Sq F valuePr(F) BED2 4.1323 2.0661 9.4570 0.0007277 *** WATER 2 3.7153 1.8577 8.5029 0.0013016 ** SHADE 3 1.6465 0.5488 2.5120 0.0789451 . Residuals 28 6.1173 0.2185 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 summary(mod1) Call: lm(formula = SQBLOOMS ~ BED + WATER + SHADE) Residuals: Min 1Q Median 3Q Max -1.23992 -0.18979 -0.01840 0.17471 0.74686 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 3.7765 0.2203 17.139 2.23e-16 *** BED2 0.3185 0.1908 1.669 0.106242 BED3 -0.5044 0.1908 -2.643 0.013293 * WATER20.7842 0.1908 4.109 0.000313 *** WATER30.4489 0.1908 2.353 0.025905 * SHADE20.1969 0.2203 0.894 0.379172 SHADE3 -0.2157 0.2203 -0.979 0.336068 SHADE4 -0.3673 0.2203 -1.667 0.106641 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 0.4674 on 28 degrees of freedom (51 observations deleted due to missingness) Multiple R-squared: 0.6081, Adjusted R-squared: 0.5102 F-statistic: 6.208 on 7 and 28 DF, p-value: 0.0001911 model.frame(mod1) SQBLOOMS BED WATER SHADE 1 4.359 1 1 1 2 3.317 1 1 2 3 3.606 1 1 3 4 4.123 1 1 4 5 4.472 1 2 1 6 4.583 1 2 2 7 4.359 1 2 3 8 4.690 1 2 4 9 4.123 1 3 1 104.123 1 3 2 113.464 1 3 3 123.873 1 3 4 133.606 2 1 1 144.000 2 1 2 153.464 2 1 3 163.873 2 1 4 174.690 2 2 1 185.000 2 2 2 195.385 2 2 3 204.583 2 2 4 214.690 2 3 1 224.690 2 3 2 234.690 2 3 3 244.243 2 3 4 253.317 3 1 1 263.606 3 1 2 273.317 3 1 3 282.828 3 1 4 293.873 3 2 1 305.000 3 2 2 313.742 3 2 3 322.449 3 2 4 334.000 3 3 1 344.583 3 3 2 353.162 3 3 3 363.162 3 3 4 model.tables(mod1,means,se=TRUE) Tables of means Grand mean 4.029028 BED BED 1 2 3 4.091 4.409 3.587 WATER WATER 1 2 3 3.618 4.402 4.067 SHADE SHADE 1 2 3 4 4.126 4.322 3.910 3.758 Standard errors for differences of means BED WATER SHADE 0.1908 0.1908 0.2203 replic. 12 12 9 Design matrix: model.matrix(mod1) (Intercept) BED2 BED3 WATER2 WATER3 SHADE2 SHADE3 SHADE4 1100 0 0 0 0 0 2100 0 0 1 0 0 3100 0 0 0 1 0 4100 0 0 0 0 1 5100 1 0 0 0 0 6100 1 0 1 0 0 7100 1 0 0 1 0 8100 1 0 0 0 1 9100 0 1 0 0 0 10
[R] something equivalent to switch condition
Is there any syntax in R that allows a switch-type condition to be used?
switch(variable){
case CONSTANT_VALUE;
break;
default:
break;
}
Thanks,
Jason
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Re: [R] unicode only works with a second one
One of the points of my.symbols is that you can define your own symbols to use
with it (hence the my).
I downloaded a graphic of the aries symbol (your original attempt in unicode I
belive) and used the following code to trace the left half of the symbol
(starting bottom center), then used that to create a matrix for the aries
symbol:
library(EBImage)
tmp - chooseImage()
image(tmp)
library(TeachingDemos)
tmp2 - par('usr')
updateusr(tmp2[1:2], tmp2[3:4], c(-1,1), c(-1,1))
out1 - locator(type='l', col='red', lwd=3)
out1$x - out1$x - out1$x[1]
ms.aries - rbind( cbind( out1$x, out1$y), NA, cbind( -out1$x, out1$y ) )
this resulted in the matrix:
ms.aries -
structure(c(0, 0, -0.00361689814814814, -0.047019675925926,
-0.0795717592592593,
-0.1085069, -0.14105902778, -0.1953125, -0.242332175925926,
-0.285734953703704, -0.332754629629630, -0.365306712962963, -0.390625,
-0.419560185185185, -0.43041087962963, -0.437644675925926, -0.441261574074074,
-0.44487847222, -0.452112268518518, -0.44849537037037, -0.44487847222,
-0.441261574074074, -0.426793981481481, -0.408709490740741, -0.390625,
-0.3689236, -0.339988425925926, -0.318287037037037, -0.285734953703704,
-0.253182870370370, -0.220630787037037, NA, 0, 0, 0.00361689814814814,
0.047019675925926, 0.0795717592592593, 0.1085069, 0.14105902778,
0.1953125, 0.242332175925926, 0.285734953703704, 0.332754629629630,
0.365306712962963, 0.390625, 0.419560185185185, 0.43041087962963,
0.437644675925926, 0.441261574074074, 0.44487847222, 0.452112268518518,
0.44849537037037, 0.44487847222, 0.441261574074074, 0.426793981481481,
0.408709490740741, 0.390625, 0.3689236, 0.339988425925926,
0.318287037037037, 0.285734953703704, 0.253182870370370, 0.220630787037037,
-0.534428351808399, -0.453515656606522, 0.254470426409904, 0.307063678291124,
0.35561129541225, 0.387976373493001, 0.40820454729347, 0.432478355854034,
0.448660894894409, 0.436523990614127, 0.420341451573752, 0.404158912533376,
0.375839469212719, 0.339428756371875, 0.298972408770936, 0.262561695930091,
0.246379156889716, 0.214014078808965, 0.185694635488308, 0.157375192167651,
0.120964479326806, 0.0885994012460556, 0.0602799579253985, 0.0238692450845536,
-0.000404563476009408, -0.0246783720365727, -0.0449065458370419,
-0.0570434501173234, -0.0732259891576988, -0.0772716239177927,
-0.0772716239177927, NA, -0.534428351808399, -0.453515656606522,
0.254470426409904, 0.307063678291124, 0.35561129541225, 0.387976373493001,
0.40820454729347, 0.432478355854034, 0.448660894894409, 0.436523990614127,
0.420341451573752, 0.404158912533376, 0.375839469212719, 0.339428756371875,
0.298972408770936, 0.262561695930091, 0.246379156889716, 0.214014078808965,
0.185694635488308, 0.157375192167651, 0.120964479326806, 0.0885994012460556,
0.0602799579253985, 0.0238692450845536, -0.000404563476009408,
-0.0246783720365727, -0.0449065458370419, -0.0570434501173234,
-0.0732259891576988, -0.0772716239177927, -0.0772716239177927
), .Dim = c(63L, 2L))
Which can be used as:
my.symbols(1:10,1:10, ms.aries, add=FALSE, inches=0.3, lwd=2)
If aries is the only one that you need, just use the above matrix, if you need
others, you can follow the above steps to trace out the other symbols (just
remember to put in a row of NA everywhere you want a jump without connecting
line).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: Thomas Steiner [mailto:finbref.2...@gmail.com]
Sent: Tuesday, March 31, 2009 4:07 PM
To: Jim Lemon; Greg Snow
Cc: r-help@r-project.org
Subject: Re: [R] unicode only works with a second one
unfortunately in my.symbols there are no zodiac signs, so I have to
choose unicode which does not work for me yet.
Thomas
2009/3/31 Thomas Steiner finbref.2...@gmail.com:
Thanks for the feedback.
I did now try Vista (2.8.1), XP (2.9.0alpha) and Win2000 (2.8.1) and
non did work compeletely, only on Vista/2.8.1 I got some symbols if I
add the other sign. I will try the my.symbols later today, when i am
at home.
Thomas
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Re: [R] something equivalent to switch condition
How about reading ?switch ?
Best,
Gabor
On Wed, Apr 1, 2009 at 9:37 PM, Jason Rupert jasonkrup...@yahoo.com wrote:
Is there any syntax in R that allows a switch-type condition to be used?
switch(variable){
case CONSTANT_VALUE;
break;
default:
break;
}
Thanks,
Jason
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--
Gabor Csardi gabor.csa...@unil.ch UNIL DGM
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Re: [R] something equivalent to switch condition
On 2/04/2009, at 8:37 AM, Jason Rupert wrote:
Is there any syntax in R that allows a switch-type condition to
be used?
switch(variable){
case CONSTANT_VALUE;
break;
default:
break;
}
?switch
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Re: [R] Convert Character to Date
Just to correct/expand/clarify the parenthetical below (how often is there a chance to correct or clarify something posted by Bill Venables?), the ides are the 15th of March, May, July, and October, but the 13th of the other months. So if you want to use the ides as the date to use, you will need a vector rather than the scalar 15, if you use 15, then it will only be the ides of some months. I wonder if the roman calendar was created on kalends April, but no one got the joke. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of bill.venab...@csiro.au Sent: Tuesday, March 31, 2009 1:29 AM To: quagmire54...@yahoo.com; r-help@r-project.org Subject: Re: [R] Convert Character to Date If you want the vector to be a Date you need to specify a date at least down to the day. Otherwise the date is not well defined and becomes NA as you noted. Perhaps the easiest thing is to give it a particular day of the month, e.g. the first, or the 15 (the ides), or ... x - as.Date(paste(1990-January, 1, sep=-), format = %Y-%B-%d) x [1] 1990-01-01 Now if you want to display the date suppressing the dummy day, you can y - format(x, %Y-%B) y [1] 1990-January Bill Venables http://www.cmis.csiro.au/bill.venables/ -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Bob Roberts Sent: Tuesday, 31 March 2009 5:14 PM To: r-help@r-project.org Subject: [R] Convert Character to Date Hello, I have a date in the format Year-Month Name (e.g. 1990-January) and R classes it as a character. I want to convert this character into a date format, but when I try as.Date(1990-January, %Y-%B), I get back NA. The function strptime also gives me NA back. Thanks. [[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-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] Calculating First Occurance by a factor
pmatch() facilitates a very simple solution: #Data IA - factor(c(1,2,2,3,3,4,3,5,5)) FixTime - c(200,350,500,600,700,850,1200,1350,1500) #First occurrence of each level first. - pmatch(levels(IA),IA) #Use first occurrence to subscript a vector or data frame FixTime[first.] A simple way to apply pmatch across unique combinations of levels of several factors is to create a new composite factor with paste() and factor(), then proceed as above. -- View this message in context: http://www.nabble.com/Calculating-First-Occurance-by-a-factor-tp22789964p22834451.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] Definition of = vs. -
Stavros Macrakis wrote: `-` Error: object - not found that's weird! vQ __ 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 multiple ablines
On 2/04/2009, at 7:04 AM, Thomas Levine wrote:
I really want to do this:
abline(
a=tan(-kT*pi/180),
b=kY-tan(-kT*pi/180)*kX
)
where kX,kY and kT are vectors of equal length. But I can't do that
with abline unless I use a loop, and I haven't figured out the least
unelegant way of writing the loop yet. So is there a way to do this
without a loop?
Or if I am to resort to the loop, what's the best way of doing it
considering that I have some missing data? Here's the mess that I
wrote.
converge - na.omit(data.frame(kX,kY,kT))
for (z in (length(converge$kT)))
{abline(
a=tan(converge$kT[z]*pi/180),
b=converge$kY[z]-tan(-converge$kT[z]*converge$kX[z]*pi/180)
)}
I think the missing data are causing the problem; this happens when
I run:
Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) :
'a' and 'b' must be finite
The help for abline explicitly states that a and b must be ``single
values'';
so no vectorization appears to be possible, as abline is currently
written.
Hence you are stuck with a for-loop.
There appears to be nothing wrong with the for-loop that you've written,
at first blush at least.
There won't be NAs in ``converge'' since you've very cleverly used
na.omit.
So ``missing data'' are NOT the problem.
The problem is then (probably) that some of your data are yielding
infinite
values of tan().
***Look*** at the values in converge. ***Look*** at the values of a
and b
produced in your loop and see where you're getting infinite values.
cheers,
Rolf Turner
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