Re: [R] tcltk error compiling R 2.13.0 under Windows 7 x64
On 14.12.2010 02:26, Peter Langfelder wrote: On Mon, Dec 13, 2010 at 4:33 PM, Joshua Wileyjwiley.ps...@gmail.com wrote: Hi All, I am trying to compile R 2.13.0 r53834 on 64-bit Windows 7 (home premium) using Rtools212.exe from Duncan Murdoch's site. I (think) I followed the installation guide for Windows. When installing Rtools, made sure I was installing 64-bit tools (including tcltk), etc. I created a copy of MkRules.local where I selected 64 bit instead of 32 architecture. When I run: C:\R\R-devel\src\gnuwin32 make all recommended I get the following error at the very end: building package 'tcltk' making init.d from init.c making tcltk.d from tcltk.c making tcltk_win.d from tcltk_win.c x86_64-w64-mingw32-gcc -I../../../../include -I ../../../../Tcl/include -DWi Note also that the quotes above are suspicious in ../../../../Tcl/include ... Uwe Ligges n32 -O2 -Wall -std=gnu99 -c init.c -o init.o In file included from init.c:22:0: tcltk.h:23:17: fatal error: tcl.h: No such file or directory compilation terminated. make[4]: *** [init.o] Error 1 make[3]: *** [mksrc-win2] Error 1 make[2]: *** [all] Error 2 make[1]: *** [R] Error 1 make: *** [all] Error 2 Just in case no one knowledgeable answers, here's my shot in the dark (I'm mostly a linux user)... the compiler is looking for .h files in the directories listed in the --I options to gcc, which seems to be Tcl, not Tcl64. The compiler also defines a macro Win32 which to my untrained eyes seems suspicious since you are trying to produce a 64-bit executable, right (although perhaps Win32 also applies to 64-bit version, I'm not sure)? Maybe your selection of 64 bit architecture wasn't honored - is the compiler reading the modified copy or the original of MkRules? Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rpart.object help
On Mon, 2010-12-13 at 01:55 -0800, jagdeesh_mn wrote: Prof Brian Ripley wrote: snip / Thanks Mr. Brian. That kind of answers my query. On the same note I would like to ask few other questions. Sorry if you find them naive, I am a novice in this subject and am trying to get a grip on things. 1. I am using R package using my code and the fitted object looks like this : The Model representation : n= 60 node), split, n, deviance, yval * denotes terminal node 1) root 60 983551500 12615.670 2) dataFrame[, 6]='Small' 13 21804710 7682.385 * 3) dataFrame[, 6]='Compact','Large','Medium','Sporty','Van' 47 557851600 13980.190 6) dataFrame[, 3]='Japan/USA','Korea','USA' 29 13105 12673.030 12) dataFrame[, 6]='Compact','Sporty' 14 11426050 11055.570 * 13) dataFrame[, 6]='Large','Medium','Van' 15 48812470 14182.670 * 7) dataFrame[, 3]='France','Germany','Japan','Sweden' 18 297418200 16086.170 * What does the term deviance here stand for? At this point, go an read up on the theory of classification and regression trees. Depending on how you fitted your tree (what options used, what type of response modelled) the deviance could be computed in different ways. In short it is a measure of how impure each node of the tree is. See the References section of ?rpart HTH G 2. Could you also suggest me some readings on the topic of CnR trees specific to R with case studies? Regards, Jagdeesh -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re : Re: descriptive statistics
On 12/14/2010 01:14 AM, effeesse wrote:
I am sorry, but I cannot understand how to use the summary function. Maybe,
if I describe my needs, you could sketch a line that could work.
In the data set variable V can take values 1 to 14. For the subgroup of
individuals where V takes value =1 I want the mean and variance of a
certain set of other variables (V1, V2, V3, V4, V5). And this for all the
other subgroups for values 2 to 14.
What do you suggest?
Step 1 - In a reproducible example one makes up some data and does
something to it to show how it is or isn't working. Clearly, you don't
know how to do that yet, so here's how.
mydataframe-data.frame(V=sample(1:14,100,TRUE),
V1=rnorm(100),V2=runif(100),V3=sample(-3:3,100,TRUE),
V4=sample(0:1,100,TRUE),V5=rpois(100,3))
If you run this code, you will then have a data frame that may not look
like what you want, but it will serve as an example. In my initial post,
I assumed that you wanted some summary statistic for each of the
variables V1 to V5, broken down by V. That's easy:
by(mydataframe[c(V1,V2,V3,V4,V5)],
mydataframe$V,mean)
If you run that code, you will get a big array of all of the means of
all of the V1-V5 columns broken down by the V column as you asked. Now
maybe you want both the mean and variance in one shot:
by_many-function(x,by,stats) {
nfun=length(stats)
myoutputlist-vector(list,nfun)
for(fun in 1:nfun)
myoutputlist[[fun]]-by(x,by,get(stats[fun]))
names(myoutputlist)-stats
return(myoutputlist)
}
by_many(mydataframe[c(V1,V2,V3,V4,V5)],
mydataframe$V,stats=c(mean,var))
The first part defines a function that will call by for each statistic
that you pass in stats, which now has to be the name of the function
rather than the function. You will have to pick your variances out of
the diagonal of the matrices due to the way var works.
So have a look at these and work out if they come close to doing what
you want.
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] Saving iterative components
Thanks a lot Uwe Ligges, I've abandoned R for a few time but now I'm working with it, so I've a question about the last problem that you solved: instead to write cv_1994- idw.cv(X01_1994) cv_1995- idw.cv(X01_1995) cv_1996- idw.cv(X01_1996) cv_1997- idw.cv(X01_1997) . cv_2006- idw.cv(X01_2006) cv_2007- idw.cv(X01_2007) cv_2008- idw.cv(X01_2008) can I use a more compact expression Thanks -- View this message in context: http://r.789695.n4.nabble.com/Saving-iterative-components-tp2553676p3086697.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] Drawing Maps of detailed Australian regions
Hi all, I'm trying to draw a bubble plot over a map -- where the map shows local government areas (LGA) in Australia (or other structures under the Australian Standard Geographical Classification), but am not sure where I can find the data to plot such maps (e.g. draw the state of Northern Territory but also showing LGA divisions). Any help will be greatly appreciated, apologies if this is a simple question -- this is the first time I'm trying to draw a map in R :). Cheers Kev Kevin Wang Senior Advisor, Health and Human Services Practice Government Advisory Services KPMG 10 Shelley Street Sydney NSW 2000 Australia Tel +61 2 9335 8282 Fax +61 2 9335 7001 kevinw...@kpmg.com.au ** The information in this e-mail is confidential and may be legally privileged. It is intended solely for the addressee. Access to this e-mail by anyone else is unauthorised. If you have received this communication in error, please notify us immediately by return e-mail with the subject heading Received in error or telephone +61 2 93357000, then delete the email and destroy any copies of it. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. Any opinions or advice contained in this e-mail are subject to the terms and conditions expressed in the governing KPMG client engagement letter. Opinions, conclusions and other information in this e-mail and any attachments that do not relate to the official business of the firm are neither given nor endorsed by it. KPMG cannot guarantee that e-mail communications are secure or error-free, as information could be intercepted, corrupted, amended, lost, destroyed, arrive late or incomplete, or contain viruses. KPMG, an Australian partnership and a member firm of the KPMG network of independent member firms affiliated with KPMG International Cooperative (“KPMG International”), a Swiss entity. KPMG International provides no services to clients. Liability limited by a scheme approved under Professional Standards Legislation. ** __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Saving iterative components
On Tue, Dec 14, 2010 at 10:34 AM, Annalaura annalaura.ru...@imaa.cnr.it wrote: Thanks a lot Uwe Ligges, I've abandoned R for a few time but now I'm working with it, so I've a question about the last problem that you solved: instead to write cv_1994- idw.cv(X01_1994) cv_1995- idw.cv(X01_1995) cv_1996- idw.cv(X01_1996) cv_1997- idw.cv(X01_1997) . cv_2006- idw.cv(X01_2006) cv_2007- idw.cv(X01_2007) cv_2008- idw.cv(X01_2008) can I use a more compact expression Yes. There are nicer ways to formulate the problem (and the solution), but without access to the object all I can is a shot in the dark. ##obtain a vector of variable names vars - names(Tmese[, -(1:2)]) ##apply the function 'idw.cv' iteratively, each time ##using a 'vars' element as the functions first argument; ##store the results in a list res - lapply(vars, idw.cv) res Regards Liviu PS For more on loops and the apply family check [1]. [1] http://promberger.info/files/rnews-vectorvsloops2008.pdf Thanks -- View this message in context: http://r.789695.n4.nabble.com/Saving-iterative-components-tp2553676p3086697.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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Saving iterative components
Hi, I'm trying to understand when and how to use do.call(). In this case, would it work with do.call() instead of lapply(), like this? vars - list(names(Tmese[, -(1:2)])) res - do.call(idw.cv, vars) Thanks, Ivan Le 12/14/2010 11:48, Liviu Andronic a écrit : On Tue, Dec 14, 2010 at 10:34 AM, Annalauraannalaura.ru...@imaa.cnr.it wrote: Thanks a lot Uwe Ligges, I've abandoned R for a few time but now I'm working with it, so I've a question about the last problem that you solved: instead to write cv_1994- idw.cv(X01_1994) cv_1995- idw.cv(X01_1995) cv_1996- idw.cv(X01_1996) cv_1997- idw.cv(X01_1997) . cv_2006- idw.cv(X01_2006) cv_2007- idw.cv(X01_2007) cv_2008- idw.cv(X01_2008) can I use a more compact expression Yes. There are nicer ways to formulate the problem (and the solution), but without access to the object all I can is a shot in the dark. ##obtain a vector of variable names vars- names(Tmese[, -(1:2)]) ##apply the function 'idw.cv' iteratively, each time ##using a 'vars' element as the functions first argument; ##store the results in a list res- lapply(vars, idw.cv) res Regards Liviu PS For more on loops and the apply family check [1]. [1] http://promberger.info/files/rnews-vectorvsloops2008.pdf Thanks -- View this message in context: http://r.789695.n4.nabble.com/Saving-iterative-components-tp2553676p3086697.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. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] spatial clusters
SKATER is included in the spdep package, and was contributed by its original authors. Roger Georg Ruß wrote: On 10/12/10 23:26:28, dorina.lazar wrote: I am looking for a clustering method usefull to classify the countries in some clusters taking account of: a) the geographical distance (in km) between countries and b) of some macroeconomic indicators (gdp, life expectancy...). Hi Dorina, ... Efficient regionalization techniques for socio-economic geographical units using minimum spanning trees (SKATER) I haven't seen too many R implementations yet, though. You may also try the R-sig-geo mailing list, because your data look geo :-) https://stat.ethz.ch/mailman/listinfo/r-sig-geo Regards, Georg. -- Research Assistant Otto-von-Guericke-Universität Magdeburg resea...@georgruss.de http://research.georgruss.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Roger Bivand Economic Geography Section Department of Economics Norwegian School of Economics and Business Administration Helleveien 30 N-5045 Bergen, Norway -- View this message in context: http://r.789695.n4.nabble.com/spatial-clusters-tp3082728p3086903.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] Drawing Maps of detailed Australian regions
On Tue, Dec 14, 2010 at 10:45 AM, Wang, Kevin (SYD) kevinw...@kpmg.com.au wrote: Hi all, I'm trying to draw a bubble plot over a map -- where the map shows local government areas (LGA) in Australia (or other structures under the Australian Standard Geographical Classification), but am not sure where I can find the data to plot such maps (e.g. draw the state of Northern Territory but also showing LGA divisions). Any help will be greatly appreciated, apologies if this is a simple question -- this is the first time I'm trying to draw a map in R :). A second's googling leads me to an aussie gov site with downloadable shapefiles: http://www.abs.gov.au/AUSSTATS/a...@.nsf/DetailsPage/1259.0.30.001July%202010?OpenDocument shapefiles can be read in R using the rgdal package. Get that, get 'sp', get 'maptools' - get mapping! Any other problems, head over to the r-sig-geo mailing list! Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Drawing Maps of detailed Australian regions
Please consider posting to R-sig-geo at https://stat.ethz.ch/mailman/listinfo/r-sig-geo https://stat.ethz.ch/mailman/listinfo/r-sig-geo , where maybe someone will be able to point you to a data source. Roger Wang, Kevin (SYD) wrote: Hi all, I'm trying to draw a bubble plot over a map -- where the map shows local government areas (LGA) in Australia (or other structures under the Australian Standard Geographical Classification), but am not sure where I can find the data to plot such maps (e.g. draw the state of Northern Territory but also showing LGA divisions). Any help will be greatly appreciated, apologies if this is a simple question -- this is the first time I'm trying to draw a map in R :). Cheers Kev Kevin Wang Senior Advisor, Health and Human Services Practice Government Advisory Services KPMG 10 Shelley Street Sydney NSW 2000 Australia Tel +61 2 9335 8282 Fax +61 2 9335 7001 kevinw...@kpmg.com.au ** The information in this e-mail is confidential and may be legally privileged. It is intended solely for the addressee. Access to this e-mail by anyone else is unauthorised. If you have received this communication in error, please notify us immediately by return e-mail with the subject heading Received in error or telephone +61 2 93357000, then delete the email and destroy any copies of it. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. Any opinions or advice contained in this e-mail are subject to the terms and conditions expressed in the governing KPMG client engagement letter. Opinions, conclusions and other information in this e-mail and any attachments that do not relate to the official business of the firm are neither given nor endorsed by it. KPMG cannot guarantee that e-mail communications are secure or error-free, as information could be intercepted, corrupted, amended, lost, destroyed, arrive late or incomplete, or contain viruses. KPMG, an Australian partnership and a member firm of the KPMG network of independent member firms affiliated with KPMG International Cooperative (“KPMG International”), a Swiss entity. KPMG International provides no services to clients. Liability limited by a scheme approved under Professional Standards Legislation. ** __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Roger Bivand Economic Geography Section Department of Economics Norwegian School of Economics and Business Administration Helleveien 30 N-5045 Bergen, Norway -- View this message in context: http://r.789695.n4.nabble.com/Drawing-Maps-of-detailed-Australian-regions-tp3086816p3086909.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] multivariate multi regression
On Dec 14, 2010, at 03:21 , Bastiaan Bergman wrote: That doesn't work, one would get two different answers depending on the order of execution. The physics is: Overlay error on a Silicon wafer. One wafer has many flash fields, each flash field has multiple locations where the overlay error is measured (as: dX,dY offset). If one contemplates that the error is caused by a rotation of the flash field then we can say (dX,dY)=(-Y,X)*RotAngle. If in addition we have a scaling error: (dX,dY)=(X*XScale,Y*YScale) than the total model is: dX~X*XScale-Y*RotAngle dY~Y*YScale+X*RotAngle Now I want to find the values for XScale, YScale and RotAngle Length(dX)==length(dY)==length(X)==length(Y)==number of measured sites on a wafer Hope this clarifies... Not completely, but I can see the general picture. I think the main thing is to rewrite the formula as dX = XScale * X + YScale * 0 + RotAngle * (-Y) + eX dY = XScale * 0 + YScale * Y + RotAngle * X+ eY (where eX and eY are noise terms) Then string together all your data pairs as dX1 dY1 dX2 dY2 ... and the corresponding design matrix X1 0 -Y1 0 Y1 X1 X2 0 -Y2 0 Y2 X2 ... This sets up the joint linear model that you want. (Technically, it could be easier to put all the dX's first, then all the dY's, though.) The remaining question is what you assume about the eX and eY terms. If they can be assumed to be independent and have the same variance, you're done. If they are correlated and/or have different variance, then I think you need to look in the direction of lme() with a suitable variance function. -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Monday, December 13, 2010 6:06 PM To: Bastiaan Bergman Cc: r-help@r-project.org Subject: Re: [R] multivariate multi regression On Dec 13, 2010, at 8:46 PM, Bastiaan Bergman wrote: Hello, I want to model my data with the following model: Y1=X1*coef1+X2*coef2 Y2=X1*coef2+X2*coef3 Note: coef2 appears in both lines Xi, Yi is input versus output data respectively How can I do this in R? I got this far: lm(Y1~X1+X2,mydata) now how do I add the second line of the model including the cross dependency? The usual way would be to extract coef2 from the object returned from the first invocation of lm(...) and use it to calculate an offset term in a second model. It would not have any variance calculated since you are forcing it to be what was returned in the first model. Now, what is it that you are really trying to do with this procedure? -- David. David Winsemius, MD 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. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multivariate multi regression / aligning objects anything in bio or mapping?
From: pda...@gmail.com Date: Tue, 14 Dec 2010 13:50:27 +0100 To: bastiaan.berg...@wdc.com CC: r-help@r-project.org Subject: Re: [R] multivariate multi regression On Dec 14, 2010, at 03:21 , Bastiaan Bergman wrote: That doesn't work, one would get two different answers depending on the order of execution. The physics is: Overlay error on a Silicon wafer. One wafer has many flash fields, each flash field has multiple locations where the overlay error is measured (as: dX,dY offset). If one contemplates that the error is caused by a rotation of the flash field then we can say (dX,dY)=(-Y,X)*RotAngle. If in addition we have a scaling error: (dX,dY)=(X*XScale,Y*YScale) than the total model is: dX~X*XScale-Y*RotAngle dY~Y*YScale+X*RotAngle First of course if you have a model it would help to get the equations right for tanslation, scale, and rotation around an arbitrary origin. If you can live with 3D, there may be something in the bio packages for aligning molecules with no assumptions about their relative initial orientations. Anyone know off hand if objection alignment exists in any bio or mapping packages? I would imagine related tasks come up in building a coherent map out of overlapping pieces, something which may have significant support too. Now I want to find the values for XScale, YScale and RotAngle Length(dX)==length(dY)==length(X)==length(Y)==number of measured sites on a wafer Hope this clarifies... Not completely, but I can see the general picture. I think the main thing is to rewrite the formula as dX = XScale * X + YScale * 0 + RotAngle * (-Y) + eX dY = XScale * 0 + YScale * Y + RotAngle * X + eY (where eX and eY are noise terms) Then string together all your data pairs as dX1 dY1 dX2 dY2 ... and the corresponding design matrix X1 0 -Y1 0 Y1 X1 X2 0 -Y2 0 Y2 X2 ... This sets up the joint linear model that you want. (Technically, it could be easier to put all the dX's first, then all the dY's, though.) The remaining question is what you assume about the eX and eY terms. If they can be assumed to be independent and have the same variance, you're done. If they are correlated and/or have different variance, then I think you need to look in the direction of lme() with a suitable variance function. -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Monday, December 13, 2010 6:06 PM To: Bastiaan Bergman Cc: r-help@r-project.org Subject: Re: [R] multivariate multi regression On Dec 13, 2010, at 8:46 PM, Bastiaan Bergman wrote: Hello, I want to model my data with the following model: Y1=X1*coef1+X2*coef2 Y2=X1*coef2+X2*coef3 Note: coef2 appears in both lines Xi, Yi is input versus output data respectively How can I do this in R? I got this far: lm(Y1~X1+X2,mydata) now how do I add the second line of the model including the cross dependency? The usual way would be to extract coef2 from the object returned from the first invocation of lm(...) and use it to calculate an offset term in a second model. It would not have any variance calculated since you are forcing it to be what was returned in the first model. Now, what is it that you are really trying to do with this procedure? -- David. David Winsemius, MD 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. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problems with lattice barchart
Hello everybody!
I´m trying to create figures summarizing the abundance of a given species
(y), by its length classes (x), year (conditional factor) and a grouping
named as fraction. I am using the lattice´s barchart but some problems
arises:
When I run the function, the sizes become characters, and this is the
starting problem for me when trying to adjust the x-axis in order to achieve
a good-looking. In particular, I have found that when range of
length-classes is high, the output yield a matted looking in the x-axis,
which in some cases is unreadable. In order to achieve a wider space between
x-axis annotations, I was trying to reduce the number of sizes shown in the
figure, but so far I got a few improvements in the final results. Here you
will find a piece of the data frame,
x[1:25,]
Length.class year fraction y
1 6 2006 Fraction.a 0
2 7 2006 Fraction.a 0
3 8 2006 Fraction.a 191
4 9 2006 Fraction.a 784
510 2006 Fraction.a38
611 2006 Fraction.a 1135
712 2006 Fraction.a 1810
813 2006 Fraction.a 2836
914 2006 Fraction.a 6625
10 15 2006 Fraction.a 6414
11 16 2006 Fraction.a 10554
12 17 2006 Fraction.a 12423
13 18 2006 Fraction.a 15260
14 19 2006 Fraction.a 11126
15 20 2006 Fraction.a 9830
16 21 2006 Fraction.a 8460
17 22 2006 Fraction.a 6062
18 23 2006 Fraction.a 4382
19 24 2006 Fraction.a 3555
20 25 2006 Fraction.a 1792
21 26 2006 Fraction.a 2220
22 27 2006 Fraction.a 1235
23 28 2006 Fraction.a 465
24 29 2006 Fraction.a 0
25 30 2006 Fraction.a59
and this is the barchart specifications,
plot.i-barchart(x[,4]~x[,1]|x[,2],horizontal=F,stack=TRUE,layout=paneis,
par.settings = simpleTheme(col = c(red3,olivedrab)),groups=x[,3],
strip = strip.custom (bg = c('transparent'),par.strip.text = list(cex =
0.8)),auto.key=list(columns=2),
panel=function(...,box.width,border){
panel.grid(h=-1,v=0)
panel.abline(v=TML,col=red,lty=2)
panel.barchart(...,box.width = .99,border=black)},
,xlab=Length Size,ylab=Y)
Can someone guide me to solve the problem?
--
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 plot Ellipsoid like function
Dear R-Users,
I am trying to plot an ellipsoid like function that represents some
physical threshold in its eigenvalue space. I am facing a few problems
generating a figure I need for my thesis. A small example looks as
follwos where the two contour3d plots do NOT overlay as desired so you
may try plotting the surfaces one by one to see what I mean.
# begin example
require(rgl)
require(misc3d)
require(MASS);
f - function(x, y, z){
chi0=-0.6603368
eps0=0.006590395
xi0=0.01117194
(x^2 + y^2 + z^2 - chi0*(x*y + x*z + y*z))/eps0^2 + (x + y + z)/xi0
}
ff - function(x, y, z)x + y + z
open3d()
clear3d(all)
bg3d(color=#88)
light3d()
x - seq(-.02,.02,len=20)
# plot ellipsoid
contour3d(f,1,x,x,x,color=#FF,alpha=0.5)
# plot plane
contour3d(ff,1,x,x,x,color=#FF,alpha=0.5)
# plot data points
spheres3d(c(-0.009379952, 0.007899338), c(-0.00879318, 0.00700924),
c(-0.009009740, 0.007656409),radius=0.0005,color=#FF)
# plot hydrostatic pressure line
lines3d(c(-0.012, 0.012), c(-0.012, 0.012), c(-0.012, 0.012),
col=#A8A8A8, lwd=4)
# end example
I have three questions regarding this problem and I hope you could help
me.
1. How can I overlay the plan plotted using contours3d(ff, ...) and
the ellipsoid plotted with contours3d(f, ...)
2. Instead of using spheres3d(...) I would love to use plot3d to
obtain proper x, y and z coordinate axes. Is there a possibility
to overly the contour3d() and line3d() commands with pot3d?
Otherwise is there a possibility to plot proper coordinate axes
with tics and such as usual R plots?
3. How can I save the scene to an image? pdf(...) ... dev.off()
seems not to work on my machine. I am using Ubuntu on a 32 Bit
Laptop.
Thanks a million for your help!
Uwe
Am Montag, den 13.12.2010, 10:20 -0500 schrieb Duncan Murdoch:
On 13/12/2010 10:13 AM, Uwe Wolfram wrote:
I am currently trying to fit a tensorial function in its principal
coorinate system. The function is given by:
1~(x1^2 + x2^2 + x3^2 - chi0*(x1*x2 + x1*x3 + x2*x3))/eps0^2 + (x1 + x2
+ x3)/xi0
Where eps0 = 0.0066, chi0 = -0.66 and xi0 = 0.011 are obtained from
experimental data using nls().I am able to plot the experimental points
that delivered the parameters of the function. For my thesis, however, I
need to overlay the fitted surface. So far I am using the following code
which wonderfully plots the experimental points in 3D:
===
# from demo(bivar)
require(rgl)
require(misc3d)
require(MASS);
# New window
open3d()
# clear scene:
clear3d(all)
# setup env. That is, background, light and so on:
bg3d(color=#88)
light3d()
# spheres at points in principal strain space
#spheres3d(e1,e2,e3,radius=0.00025,color=#FF)
# draws points alternatively
plot3d(e1,e2,e3, col=#FF)
===
According to the examples on http://rgl.neoscientists.org/gallery.shtml
I tried to overlay the point plot using surface3d. However, these were
only functions of type y ~f(x1, x2). I think that the surface could be
plotted if I could provide the gridpoints correctly. Using
xyz.coords(1~(x1^2 + x2^2 + x3^2 - chi0*(x1*x2 + x1*x3 + x2*x3))/eps0^2
+ (x1 + x2 + x3)/xi0, y = NULL, z = NULL) did unfortunately not solve
the problem.
Is there any function that can generate the surface for the given
function such as ContourPlot3D in Mathematica.
See ?misc3d::contour3d
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 plot Ellipsoid like function
On 14/12/2010 8:45 AM, Uwe Wolfram wrote:
Dear R-Users,
I am trying to plot an ellipsoid like function that represents some
physical threshold in its eigenvalue space. I am facing a few problems
generating a figure I need for my thesis. A small example looks as
follwos where the two contour3d plots do NOT overlay as desired so you
may try plotting the surfaces one by one to see what I mean.
# begin example
require(rgl)
require(misc3d)
require(MASS);
f- function(x, y, z){
chi0=-0.6603368
eps0=0.006590395
xi0=0.01117194
(x^2 + y^2 + z^2 - chi0*(x*y + x*z + y*z))/eps0^2 + (x + y + z)/xi0
}
ff- function(x, y, z)x + y + z
open3d()
clear3d(all)
bg3d(color=#88)
light3d()
x- seq(-.02,.02,len=20)
# plot ellipsoid
contour3d(f,1,x,x,x,color=#FF,alpha=0.5)
# plot plane
contour3d(ff,1,x,x,x,color=#FF,alpha=0.5)
# plot data points
spheres3d(c(-0.009379952, 0.007899338), c(-0.00879318, 0.00700924),
c(-0.009009740, 0.007656409),radius=0.0005,color=#FF)
# plot hydrostatic pressure line
lines3d(c(-0.012, 0.012), c(-0.012, 0.012), c(-0.012, 0.012),
col=#A8A8A8, lwd=4)
# end example
I have three questions regarding this problem and I hope you could help
me.
1. How can I overlay the plan plotted using contours3d(ff, ...) and
the ellipsoid plotted with contours3d(f, ...)
Use add=TRUE in the second call to contour3d. However, the level you
chose (i.e. 1) never occurs in the range of your x,y,z data, so you
won't see anything unless you change it.
2. Instead of using spheres3d(...) I would love to use plot3d to
obtain proper x, y and z coordinate axes. Is there a possibility
to overly the contour3d() and line3d() commands with pot3d?
Otherwise is there a possibility to plot proper coordinate axes
with tics and such as usual R plots?
Use decorate3d() to add the axes etc, or axes3d() for just axes.
3. How can I save the scene to an image? pdf(...) ... dev.off()
seems not to work on my machine. I am using Ubuntu on a 32 Bit
Laptop.
That's harder, but you can try rgl.snapshot() (for a bitmap copy in PNG
format) or rgl.postscript() (for vector graphics in
ps, eps, tex, pdf, svg, or pgf, but with incomplete rendering of some
scenes).
Duncan Murdoch
Thanks a million for your help!
Uwe
Am Montag, den 13.12.2010, 10:20 -0500 schrieb Duncan Murdoch:
On 13/12/2010 10:13 AM, Uwe Wolfram wrote:
I am currently trying to fit a tensorial function in its principal
coorinate system. The function is given by:
1~(x1^2 + x2^2 + x3^2 - chi0*(x1*x2 + x1*x3 + x2*x3))/eps0^2 + (x1 + x2
+ x3)/xi0
Where eps0 = 0.0066, chi0 = -0.66 and xi0 = 0.011 are obtained from
experimental data using nls().I am able to plot the experimental points
that delivered the parameters of the function. For my thesis, however, I
need to overlay the fitted surface. So far I am using the following code
which wonderfully plots the experimental points in 3D:
===
# from demo(bivar)
require(rgl)
require(misc3d)
require(MASS);
# New window
open3d()
# clear scene:
clear3d(all)
# setup env. That is, background, light and so on:
bg3d(color=#88)
light3d()
# spheres at points in principal strain space
#spheres3d(e1,e2,e3,radius=0.00025,color=#FF)
# draws points alternatively
plot3d(e1,e2,e3, col=#FF)
===
According to the examples on http://rgl.neoscientists.org/gallery.shtml
I tried to overlay the point plot using surface3d. However, these were
only functions of type y ~f(x1, x2). I think that the surface could be
plotted if I could provide the gridpoints correctly. Using
xyz.coords(1~(x1^2 + x2^2 + x3^2 - chi0*(x1*x2 + x1*x3 + x2*x3))/eps0^2
+ (x1 + x2 + x3)/xi0, y = NULL, z = NULL) did unfortunately not solve
the problem.
Is there any function that can generate the surface for the given
function such as ContourPlot3D in Mathematica.
See ?misc3d::contour3d
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] rpart.object help
On Sun, 12 Dec 2010, jagdeesh_mn wrote: Hi, Suppose i have generated an object using the following : fit - rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) And when i print fit, i get the following : n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 17 absent (0.7901235 0.2098765) 2) Start=8.5 62 6 absent (0.9032258 0.0967742) 4) Start=14.5 29 0 absent (1.000 0.000) * 5) Start 14.5 33 6 absent (0.8181818 0.1818182) 10) Age 55 12 0 absent (1.000 0.000) * 11) Age=55 21 6 absent (0.7142857 0.2857143) 22) Age=111 14 2 absent (0.8571429 0.1428571) * 23) Age 111 7 3 present (0.4285714 0.5714286) * 3) Start 8.5 19 8 present (0.4210526 0.5789474) * Is it possible to extract the splits alone as a matrix using rpart.object? If so, how? Regards, Jagdeesh The best description of the rpart object is obtained with help(rpart.object). Each row of $frame describes one primary split. More detailed descriptions of the (1 + ncompete + nprimary) split variables for the node are found in the $splits and $csplits component. You would need to look at summary.rpart to see how that is all indexed. I would suggest grabbing a copy of the source code, since that contains comments, which are stripped out when you print the R internal version. Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] rpart - how to estimate the “meaningful” predictors for an outcome (in classification t rees)
Hi dear R-help memebers, When building a CART model (specifically classification tree) using rpart, it is sometimes obvious that there are variables (X's) that are meaningful for predicting some of the outcome (y) variables - while other predictors are relevant for other outcome variables (y's only). *How can it be estimated, which explanatory variable is used for which of the predicted value in the outcome variable?* Here is an example code in which x2 is the only important variable for predicting b (one of the y outcomes). There is no predicting variable for c, and x1 is a predictor for a, assuming that x2 permits it. How can this situation be shown using the an rpart fitted model? N - 200 set.seed(5123) x1 - runif(N) x2 - runif(N) x3 - runif(N) y - sample(letters[1:3], N, T) y[x1 .5] - a y[x2 .1] - b fit - rpart(y ~ x1+x2) fit2 - prune(fit, cp= 0.07) plot(fit2) text(fit2, use.n=TRUE) Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[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] curve
Val, Here's the complete console output. The graph produced is at: http://www.functionaldiversity.org/temp/curve.png R version 2.12.0 (2010-10-15) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-redhat-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. ls() character(0) test- rnorm(5000,1000,100) test1 - subset(test, subset=(test 1100)) d - density(test) png(curve.png) plot(d, main=Density of production, xlab=) xarea - c(1100, d$x[d$x 1100], c(max(test))) yarea - c(0, d$y[d$x 1100], 0) polygon(xarea, yarea, col=blue) curveheight - d$y[abs((d$x - mean(test1))) == min(abs((d$x - mean(test1] segments(x0=mean(test1), y0=0, x1=mean(test1), y1=curveheight, col=red, lwd=2) dev.off() null device 1 Sarah On Tue, Dec 14, 2010 at 8:59 AM, Val valkr...@gmail.com wrote: Hi Sara, Could you please send me your output as an attached file if it is possible? Thanks 1. to shade or color (blue) the curve using the criterion that any values greater than 11,000 I think I was not clear in the above point. I want shade not the line but the area under the curve, Here's an example of how to do that using polygon: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=7 and Your last line of code, segments(x0=mean(test1), y0=0, y1=curveheight) gave me the following error message Error in segments(x0 = mean(test1), y0 = 0, y1 = curveheight) : element 3 is empty; the part of the args list of '.Internal' being evaluated was: (x0, y0, x1, y1, col = col, lty = lty, lwd = lwd, ...) could you check it please I checked it before I sent it to you. The code I provided works correctly on my computer. (R 2.12.0, Linux). You could try this statement instead: segments(x0 = mean(test1), y0 = 0, x1=mean(test1), y1 = curveheight) Sarah -- -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Browsing through a dataframe page by page (like with shell command more)
Thank you all for your input. View is quite nice, despite the fact that you need to be in a X session and it seems(?) difficult to copy paste data out of the view. page(my_var) works great and is most similar to what I had in mind. TkListView seems interesting but quite slow for large list or frames. Alexandre On Mon, Dec 13, 2010 at 7:48 PM, Greg Snow greg.s...@imail.org wrote: For data frames the best is probably the View function (note capitol V) which opens the data frame in a spreadsheet like window that you can scroll through. For more complicated, list or list-like objects, look at TkListView in the TeachingDemos package. For more general investigation of data objects look at ?page and ?options specifically the pager section. -- 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 Alexandre CESARI Sent: Monday, December 13, 2010 3:49 AM To: r-help@r-project.org Subject: [R] Browsing through a dataframe page by page (like with shell command more) Hello, I'm looking for an easy way to display a data.frame (or other variables) page by page, similarly to what is possible on a file using the more command in a standard UNIX shell. Any help would be greatly appreciated. Thanks Alexandre [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question about cut()
Dear all, I would like to use cut() to make numerics to factors, the sample codes are as follows. However, the result is not what I want, since r[3] = 9 should be in the interval of 8-10% rather than 2-4%. Maybe cut() is not the right function to use for my situation. Please help. r - c(1,1,9,1,1,1) col_no - as.factor(as.numeric(cut(r,c(0,2,4,6,8,10,100 levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%) col_no [1] 2% 2% 2-4% 2% 2% 2% Levels: 2% 2-4% 4-6% 6-8% 8-10% 10%Thanks, Tianchan__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lines and points without margin
hi, i am sure that this is a trivial question but i have not been able to find an answer by searching the mailing lists. i want to plot points on a graph, joined by lines. the command that i am using is points(x, y, type = b, pch = 21) this plots nice open circles at the data points and draws lines between them. however, the lines do not come all the way up to the edge of the circles but stop some small distance away so that there is an empty margin around the circles. is there a way to get rid of this margin? my first guess was that there would be an option to par() but i did not find anything there. any suggestions would be appreciated. thanks! best regards, andrew. -- Andrew B. Collier Physicist Waves and Space Plasmas Group Hermanus Magnetic Observatory Honorary Senior Lecturer tel: +27 31 2601157 Space Physics Research Institute fax: +27 31 2607795 University of KwaZulu-Natal, Durban, South Africagsm: +27 83 3813655 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] labelling of axis in plot.ca
hi all, how can i get correspondence analysis plots (package ca, plot.ca()) with the inertia on the two dimensions of the map? thanks, kat [[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] lines and points without margin
Hi, Try the following, plot(1:10,rnorm(10),t=o) ## fill the points in white plot(1:10,rnorm(10),t=o,pch=21,bg=white) You could also try this with Grid graphics, library(gridExtra) # like type=o grid.barbed(space=0) # like type=b grid.barbed(space=1) # like the example above, but without filling the dots grid.barbed(space=0.5) (I may be wrong but don't recall that this parameter is tunable with base graphics) HTH, baptiste On 14 December 2010 16:33, Andrew Collier collie...@gmail.com wrote: hi, i am sure that this is a trivial question but i have not been able to find an answer by searching the mailing lists. i want to plot points on a graph, joined by lines. the command that i am using is points(x, y, type = b, pch = 21) this plots nice open circles at the data points and draws lines between them. however, the lines do not come all the way up to the edge of the circles but stop some small distance away so that there is an empty margin around the circles. is there a way to get rid of this margin? my first guess was that there would be an option to par() but i did not find anything there. any suggestions would be appreciated. thanks! best regards, andrew. -- Andrew B. Collier Physicist Waves and Space Plasmas Group Hermanus Magnetic Observatory Honorary Senior Lecturer tel: +27 31 2601157 Space Physics Research Institute fax: +27 31 2607795 University of KwaZulu-Natal, Durban, South Africa gsm: +27 83 3813655 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Re : Re: descriptive statistics
Thanks! I got the results! Now I would like to put them in a nice and readable plot to see if there are outlier means or variances. Using plot(by_many$mean) I got Error in by_many$mean : object of type 'closure' is not subsettable. Another question: I have done the principal components analysis on the same group of variables (V1--V5) I used before. I'd like to get a similar descriptive analysis of these principal components by variable V. Correct me if I am wrong; should I run it on the scores of the 2 or 3 principal components I obtained in the PCA? -- View this message in context: http://r.789695.n4.nabble.com/descriptive-statistics-tp3085197p3086849.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question
How do I determine if my data deviate from the normal distribution? The sample size is 1000 (weights of people). __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Selecting non-empty elements after strsplit of string
Hello, I am attempting to manipulate strings in which there are differing amounts of whitespace before and after each element taht I want to keep (any word, letter, or number). However, after strsplit and unlist, I know how to select specific elements with [ ], but I want to select instead all elements that are not missing. That is, do not select any elements that are simply . output - c(a b 6) gsub( ,,unlist(strsplit(output, ,fixed=TRUE))) [1] a b 6 Thanks! Scott Chamberlain [[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] Selecting non-empty elements after strsplit of string
Hello, I am attempting to manipulate strings in which there are differing amounts of whitespace before and after each element taht I want to keep (any word, letter, or number). However, after strsplit and unlist, I know how to select specific elements with [ ], but I want to select instead all elements that are not missing. That is, do not select any elements that are simply . output - c(a b 6) gsub( ,,unlist(strsplit(output, ,fixed=TRUE))) [1] a b 6 Thanks! Scott Chamberlain [[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] lines and points without margin
Andrew Collier collierab at gmail.com writes: i am sure that this is a trivial question but i have not been able to find an answer by searching the mailing lists. i want to plot points on a graph, joined by lines. the command that i am using is points(x, y, type = b, pch = 21) this plots nice open circles at the data points and draws lines between them. however, the lines do not come all the way up to the edge of the circles but stop some small distance away so that there is an empty margin around the circles. is there a way to get rid of this margin? my first guess was that there would be an option to par() but i did not find anything there. any suggestions would be appreciated. Does type=o do what you want? (See ?plot ) However, this actually overplots (which would be OK with filled point types, or if you don't care if you have lines running across your open points) -- it doesn't draw the line up to the edge of the point. It looks like the size of the spacing is hard-coded at a fairly deep level: in plot.c, within the do_plot_xy function, there is a spacing factor which seems to be encoded as half the width of standard character in the current plotting coordinates. Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing sample values in to a file
chandu wrote:
I am relatively new to R. I would like to know how can we write the
realizations (for example generated through rnorm or runif) in to a data
file. It would be very inefficient to first generate values and then write
them in to file using write function. Instead, is there a way to
generate 1 value at a time and append them in to the file.
It would be very inefficient. Sounds like you tried it.
n = 3
system.time(for (i in 1:n){
cat(rnorm(1),file=a.txt,append=TRUE)
}
)
system.time(write.table(rnorm(n),file=b.txt))
)
user system elapsed
3.988.66 16.11
system.time(write.table(rnorm(n),file=b.txt))
user system elapsed
0.590.010.61
So the inefficient method is about 25 times faster.
Dieter
--
View this message in context:
http://r.789695.n4.nabble.com/writing-sample-values-in-to-a-file-tp3086286p3087354.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] Question about cut()
On Tue, 2010-12-14 at 15:40 +, Tianchan Niu wrote: Dear all, I would like to use cut() to make numerics to factors, the sample codes are as follows. However, the result is not what I want, since r[3] = 9 should be in the interval of 8-10% rather than 2-4%. Maybe cut() is not the right function to use for my situation. Please help. r - c(1,1,9,1,1,1) col_no - as.factor(as.numeric(cut(r,c(0,2,4,6,8,10,100 levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%) col_no [1] 2% 2% 2-4% 2% 2% 2% Levels: 2% 2-4% 4-6% 6-8% 8-10% 10%Thanks, Tianchan It would really help if you looked at the intermediary steps of your calculations to see what went wrong. The problem is here: as.factor(as.numeric(cut(r,c(0,2,4,6,8,10,100 [1] 1 1 5 1 1 1 Levels: 1 5 There are 2 levels in the data passed to as.factor, 1 and 5. levels(...)[1] is 1, and levels(...)[2] is 5. You then assign the values 2% 2-4% to these two levels. Hence the result One option is to use factor and specify the levels as 1:6: factor(as.numeric(cut(r,c(0,2,4,6,8,10,100))), levels = 1:6) [1] 1 1 5 1 1 1 Levels: 1 2 3 4 5 6 Using this we have: col_no - factor(as.numeric(cut(r,c(0,2,4,6,8,10,100))), levels = 1:6) levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%) col_no [1] 2% 2% 8-10% 2% 2% 2% Levels: 2% 2-4% 4-6% 6-8% 8-10% 10% I'm sure there are other, possibly better, ways of doing this, but that is one. HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing sample values in to a file
Hi:
On Mon, Dec 13, 2010 at 4:28 PM, chandu chandrasekhar.kar...@gmail.comwrote:
Dear all,
I am relatively new to R. I would like to know how can we write the
realizations (for example generated through rnorm or runif) in to a data
file. It would be very inefficient to first generate values and then write
them in to file using write function. Instead, is there a way to generate
1 value at a time and append them in to the file.
On the contrary, it is very inefficient in R to generate one value at a time
and then append it to a file. R can do vectorized calculations, so for
generating random data, it takes one line of code; e.g.,
rnorm(100, 0, 5)
generates a vector of 100 random numbers from a normal distribution with
mean 0 and standard deviation 5. To generate the code and write it to a file
can also take one line:
write.csv(rnorm(100, 0, 5), file = 'myRandomNumbers.csv'), row.names =
FALSE, quote = FALSE)
On my system, it took 4.24 seconds to write 100 random numbers to a
file (its size is 18.6 Mb).
Now, let's try your for loop approach, without writing to a file:
# Pre-allocate space for the vector:
u - vector('numeric', 100)
system.time(for(i in seq_along(u)) u[i] - rnorm(1, 0, 5))
user system elapsed
6.860.006.88
# Initialize an empty object and populate it one element at a time:
u - NULL
system.time(for(i in 1:100) u - c(u, rnorm(1, 0, 5)))
The reason the second one is so inefficient is because of two important
features in R that generally don't arise in most programming languages:
fixed memory for workspaces and lazy evaluation. Because you are repeatedly
appending to a object that grows and grows (this is where the lazy
evaluation come into play), R has to work harder to find new memory after a
while and so it slows down precipitously as it expends more and more effort
finding memory. I got impatient with waiting, so
Timing stopped at: 571.83 9.08 585.53
system.time(for(i in 1:1000) u - c(u, rnorm(1, 0, 5)))
user system elapsed
2.640.082.74
system.time(for(i in 1:1) u - c(u, rnorm(1, 0, 5)))
user system elapsed
27.470.50 28.08
Multiply the last one (total time is on the far right) by 100 to get a
probable lower bound for how long this takes. There are more efficient ways
to do this (use of the function force(), for example), but the point is that
one thing you definitely do NOT want to do in R is to append to an object
one value at a time. It wouldn't be much different if you were writing to an
external file.
The question might be trivial to many experts. I appreciate your help.
The question is far from trivial, and many people have put in great amounts
of effort to make R efficient. If possible, vectorized operations are a
good way to go because they are generally fast. Having said that, there are
occasions where it is more efficient to perform loops, but for novices who
are used to Fortran/C/Java looping constructs, it is usually the case that
there are very fast ways to do the same thing in R without a loop using
vectorized operations.
HTH,
Dennis
Thank you
--
View this message in context:
http://r.789695.n4.nabble.com/writing-sample-values-in-to-a-file-tp3086286p3086286.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] Question about cut()
Tianchan Niu niu at isis.georgetown.edu writes: Dear all, I would like to use cut() to make numerics to factors, the sample codes are as follows. However, the result is not what I want, since r[3] = 9 should be in the interval of 8-10% rather than 2-4%. Maybe cut() is not the right function to use for my situation. Please help. r - c(1,1,9,1,1,1) col_no - as.factor(as.numeric(cut(r,c(0,2,4,6,8,10,100 levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%) col_no [1] 2% 2% 2-4% 2% 2% 2% Levels: 2% 2-4% 4-6% 6-8% 8-10% 10%Thanks, Tianchan I think you're making the problem a little harder than it needs to be ... r - c(1,1,9,1,1,1) col_no - cut(r,c(0,2,4,6,8,10,100), labels = c(2%,2-4%,4-6%,6-8%,8-10%,10%)) [1] 2% 2% 8-10% 2% 2% 2% Levels: 2% 2-4% 4-6% 6-8% 8-10% 10% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Selecting non-empty elements after strsplit of string
Hi, I have just recently started to use regular expression, so I'm not sure that my solution would be valid in every case. But it works in yours: unlist(strsplit(output, split= +)) [1] ab 6 It would split them whatever the number of spaces between them is (because of the + in the pattern of splitting). HTH, Ivan Le 12/14/2010 14:32, Scott Chamberlain a écrit : Hello, I am attempting to manipulate strings in which there are differing amounts of whitespace before and after each element taht I want to keep (any word, letter, or number). However, after strsplit and unlist, I know how to select specific elements with [ ], but I want to select instead all elements that are not missing. That is, do not select any elements that are simply . output- c(a b 6) gsub( ,,unlist(strsplit(output, ,fixed=TRUE))) [1] a b 6 Thanks! Scott Chamberlain [[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. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Selecting non-empty elements after strsplit of string
unlist(strsplit(output, +)) On Tue, Dec 14, 2010 at 2:09 PM, Scott Chamberlain myrmecocys...@gmail.com wrote: Hello, I am attempting to manipulate strings in which there are differing amounts of whitespace before and after each element taht I want to keep (any word, letter, or number). However, after strsplit and unlist, I know how to select specific elements with [ ], but I want to select instead all elements that are not missing. That is, do not select any elements that are simply . output - c(a b 6) gsub( ,,unlist(strsplit(output, ,fixed=TRUE))) [1] a b 6 Thanks! Scott Chamberlain [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about cut()
Tianchan, why aren't you just using col_no - cut(r,c(0,2,4,6,8,10,100)) levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%) ? Your use of as.numeric() is nonsensical; check step by step what is happening with that. Hth, Gerrit On Tue, 14 Dec 2010, Tianchan Niu wrote: Dear all, I would like to use cut() to make numerics to factors, the sample codes are as follows. However, the result is not what I want, since r[3] = 9 should be in the interval of 8-10% rather than 2-4%. Maybe cut() is not the right function to use for my situation. Please help. r - c(1,1,9,1,1,1) col_no - as.factor(as.numeric(cut(r,c(0,2,4,6,8,10,100 levels(col_no) - c(2%,2-4%,4-6%,6-8%,8-10%,10%) col_no [1] 2% 2% 2-4% 2% 2% 2% Levels: 2% 2-4% 4-6% 6-8% 8-10% 10%Thanks, Tianchan Best regards -- Gerrit Best regards -- Gerrit Eichner Viele Grüße -- Gerrit Viele Grüße -- Gerrit Eichner Viele Grüße -- GE Freundliche Grüße -- Gerrit Eichner Freundliche Grüße -- GE Grüße -- Gerrit Grüße -- Gerrit Eichner Grüße -- GE Gruß -- G - AOR Dr. Gerrit Eichner Mathematical Institute, Room 212 gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany Fax: +49-(0)641-99-32109http://www.uni-giessen.de/cms/eichner -__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question
On Tue, 14 Dec 2010, Matthew Rosett wrote: How do I determine if my data deviate from the normal distribution? The sample size is 1000 (weights of people). See ?qqnorm and/or ?shapiro.test and/or a text book on applied statistics and/or google for testing normality. Hth, Gerrit PS: With n = 1000 your data will almost inevitably deviate from the normal distribution. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] factor predictor using random forest
Hello everyone, I have been doing a binary classification using random forest from the library randomForest. One of the predictors is a factor variable, which is known to be highly related to the binary response I am trying to predict. Other 80 predictors are numeric. Totally I have 44 subjects. However, the random forest returns the factor variable as the least important one based on the decreased accuracy measurement. I specified the classification in the way below: myrf - randomForest (disease ~ ., data = mydata, ntree = 500, importance = T, do.trace = F, keep.forest = T) disease is factor of cases and controls. Are there any specific things I need to do differently when a factor predictor is included? Thank you very much, Jing [[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] Date variable error
Kurt_Helf at nps.gov writes: Greetings In attempting to create a date variable based on month (e.g., February, April, etc.) and year (e.g., 2006) data, wherein I converted Month to a factor with Jan=1...Dec=12, I used the following command: data$Date-mdy.date(month=data$Month,day=15,year=data$Year) however, I get a message Error: trunc not meaningful for factors Have you tried data$Date-mdy.date(month=as.numeric(data$Month),day=15,year=data$Year) (I like with() or transform() for making these cleaner: data - transform(data,Date=mdy.date(month=as.numeric(Month),day=15,year=Year)) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] curve
You are most likely to get reasonable help if you provide at a minimum your OS, the version of R you're using, and the error message you get. And also if you send your replies to the R-help list, not just me. Sarah On Tue, Dec 14, 2010 at 10:58 AM, Val valkr...@gmail.com wrote: Thank you Sarah, It worked in my Linux machine as well but not in Windows. On Tue, Dec 14, 2010 at 9:37 AM, Sarah Goslee sarah.gos...@gmail.com wrote: Val, Here's the complete console output. The graph produced is at: http://www.functionaldiversity.org/temp/curve.png R version 2.12.0 (2010-10-15) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-redhat-linux-gnu (64-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. ls() character(0) test- rnorm(5000,1000,100) test1 - subset(test, subset=(test 1100)) d - density(test) png(curve.png) plot(d, main=Density of production, xlab=) xarea - c(1100, d$x[d$x 1100], c(max(test))) yarea - c(0, d$y[d$x 1100], 0) polygon(xarea, yarea, col=blue) curveheight - d$y[abs((d$x - mean(test1))) == min(abs((d$x - mean(test1] segments(x0=mean(test1), y0=0, x1=mean(test1), y1=curveheight, col=red, lwd=2) dev.off() null device 1 Sarah On Tue, Dec 14, 2010 at 8:59 AM, Val valkr...@gmail.com wrote: Hi Sara, Could you please send me your output as an attached file if it is possible? Thanks 1. to shade or color (blue) the curve using the criterion that any values greater than 11,000 I think I was not clear in the above point. I want shade not the line but the area under the curve, Here's an example of how to do that using polygon: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=7 and Your last line of code, segments(x0=mean(test1), y0=0, y1=curveheight) gave me the following error message Error in segments(x0 = mean(test1), y0 = 0, y1 = curveheight) : element 3 is empty; the part of the args list of '.Internal' being evaluated was: (x0, y0, x1, y1, col = col, lty = lty, lwd = lwd, ...) could you check it please I checked it before I sent it to you. The code I provided works correctly on my computer. (R 2.12.0, Linux). You could try this statement instead: segments(x0 = mean(test1), y0 = 0, x1=mean(test1), y1 = curveheight) Sarah -- -- Sarah Goslee http://www.functionaldiversity.org -- Sarah Goslee http://www.stringpage.com http://www.astronomicum.com http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stepAIC: plot predicted versus observed
I figured out how to call coefficients and R² from the summary. - fit-lm(...) #multiple regression function stepComp - stepAIC(fit, direction=both) summary(stepComp)$coef[1,1] #call first coefficient summary(stepComp)$coef[1,2] #call Std. Error of first coefficient summary(stepComp)$coef[2,1] #call second coefficient #... so on... #And the R² can be called by summary(stepComp)$adj.r.squared #call adjusted R² summary(stepComp)$r.squared #call multiple R² -- View this message in context: http://r.789695.n4.nabble.com/stepAIC-plot-predicted-versus-observed-tp3085991p3087345.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] Installing R-packages in Windows
Hi there,
I have the following problem and I hope somebody might help me.
First of all: I am using WinXP SP3 (english and/or german) with R in
Version 2.10.0.
Now I am trying to install some packages but unfortunately I am getting
a weird error. No matter which package I am trying to install - I nearly
get the same error.
It looks like this:
-
C:\Program Files\R\R-2.10.0\binR CMD INSTALL
%SystemDrive%\rPAcsTemp\car_2.0-2.
zip
Warning in rawToChar(block[seq_len(ns)]) :
truncating string with embedded nul:
'PK\003\004\n\0\0\0\0\01áG=\0\0\0\0\0\0\0
\0\0\0\0\0\004\0\0\0car/PK\003\004\n\0\0\0\0\0-áG=\0\0\0\0\0\0\0\0\0\0\0\0\t\0\0
\0car/data/PK\003\004\024\0\002\0\b\0-áG=+2ô[\t\003\0¥-\003\0\022'
Warning in rawToChar(block[101:107]) :
truncating string with embedded nul: '\0\0\0car/'
Error in switch(rawToChar(z), ` ` = { : invalid octal digit
-
I am trying to install with a batch-script that looks like this:
-
SET rVersion=2.10.0
SET RPfad=%ProgramFiles%\R\R-%rVersion%\bin\
set RPacPfad=software\R_packages
REM #
REM copyying to a folder without spaces
REM #
cd %windir%
cd ..
mkdir rPAcsTemp
cd %OrdnerDIR%\%RPACPFAD%
copy *.zip %SystemDrive%\rPAcsTemp
cd %RPFAD%
R CMD INSTALL %SystemDrive%\rPAcsTemp\car*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\dbi*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\zipfR*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\lme4*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\coda*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\languageR*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\lattice_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\mvtnorm_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\multcomp_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\R.oo_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\R.methodsS3_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\R.matlab_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\RPostgreSQL_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\tree_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\xtable_*.zip
R CMD INSTALL %SystemDrive%\rPAcsTemp\emu*.tar.gz
cd %Back-to-Normal%
-
Does anybody have an idea why I am recieving these error-messages? I
thought about buggy files but after downloading the files (from
http://cran.r-project.org/bin/windows/contrib/2.10/ ) again I could
reproduce the messages.
By the way: Is there a better method installing packages? I tried within R
install.packages from online and local repository but that only works
for the user I am installing with.
Thanks in advance for the replies
Erik__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] peak detection
Thank you. I will look at that also. On Mon, Dec 13, 2010 at 9:49 PM, Michael Bedward michael.bedw...@gmail.com wrote: Hi Joe, Just for info, I've done this in the past by applying lowess followed by diff to a vector, then identifying points with change of sign in the diffs. Michael On 14 December 2010 14:22, Joe Stuart joe.stu...@gmail.com wrote: Never mind. I did find this package, which seems to do the trick. Thanks http://rgm2.lab.nig.ac.jp/RGM2/R_man-2.9.0/library/msProcess/man/msExtrema.html On Mon, Dec 13, 2010 at 9:05 PM, Joe Stuart joe.stu...@gmail.com wrote: Hi, I was wondering if anyone knows of a package that contains the ability for peak/valley detection. Here is an example of what I'm looking for, only problem is that it's written in Matlab. http://www.billauer.co.il/peakdet.html Thanks for any help in advance. -Joe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] survreg vs. aftreg (eha) - the relationship between fitted coefficients?
On Fri, Dec 10, 2010 at 5:21 PM, Eleni Rapsomaniki
er...@medschl.cam.ac.uk wrote:
Dear R-users,
I need to use the aftreg function in package 'eha' to estimate failure times
for left truncated survival data.
Be careful! This is only possible under strong assumptions about the
(unobserved) covariate vector. This is not (yet) reflected on the help
page for aftreg, but has been discussed here earlier.
Apparently, survreg still cannot fit such models. Both functions
should be fitting the accelerated failure time (Weibull) model.
However, as Göran Broström points out in the help file for aftreg, the
parameterisation is different giving rise to different coefficients.
The betas for adjusted covariates are opposite in sign but otherwise
identical, whereas the intercept is quite different in a non-obvious
way. The log-likelihoods are similar also, but not identical. I would
like to find out how I can convert one set of coefficients to the
other so as to obtain the same linear predictors using either model.
Any ideas???
Yes. Since you work with Weibull data, try 'phreg' instead. You will
have to change sign of estimated coefficients, and divide them by
estimated shape, see the help page for 'weibreg', to compare to
parameters from survreg, but they are directly comparable with those
from coxph and coxreg.
Göran
#the example below uses right-censored data for simplicity (the principle
should be the same with left truncation I hope)
library(survival)
library(eha)
# COMPARE coefs between survreg ('survival' pkg) and aftreg ('eha' pkg)
#Fitting NULL models (no covariates) results in (approximately) the same
coefs (which is good!)
m1_NULL=survreg(Surv(futime/365, status==1) ~ 1, data=pbcseq)
m2_NULL=aftreg(Surv(futime/365, status==1) ~ 1, data=pbcseq)
c(m1_NULL$coef, 1/m1_NULL$scale) #-- intercept= 3.878656 , shape = 1.478177
c(m2_NULL$coef[1], exp(m2_NULL$coef[2])) #-- intercept= 3.878859 ,
shape=1.478150
# NOW I adjust for covariates
m1=survreg(Surv(futime/365, status==1) ~ chol+stage, data=pbcseq)
m2= aftreg(Surv(futime/365, status==1) ~ chol+stage, data=pbcseq)
### m2 ###
#Coefficients:
# (Intercept) chol stage
# 5.944641913 -0.001692574 -0.470861324
#Scale= 0.6416744
#Loglik(model)= -483.9 Loglik(intercept only)= -506.8
# Chisq= 45.91 on 2 degrees of freedom, p= 1.1e-10
#n=1124 (821 observations deleted due to missingness)
### m2 ###
#Covariate W.mean Coef Exp(Coef) se(Coef) Wald p
#chol 303.777 0.002 1.002 0.000 0.000
#stage 3.298 0.460 1.584 0.119 0.000
#
#log(scale) 5.029 152.807 0.477 0.000
#log(shape) 0.467 1.595 0.095 0.000
#
#Events 92
#Total time at risk 9017
#Max. log. likelihood -484.31
#LR test statistic 45.0
#Degrees of freedom 2
#Overall p-value 1.64669e-10
Many thanks for any help you may be able to provide.
Eleni Rapsomaniki
Research Associate
University of Cambridge
Institute of Primary and Public Health
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Göran Broström
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question
... (in addition to the very useful suggestion to plot your data): (Sounds like a homework question... ?). Sigh. [mount soapbox] 1. Data never deviate from normality. They only provide provide evidence to challenge (test is the formal term) the assumption that the population from which the data were sampled (how? -- see below) can be modeled as normal (e.g. whether the data provide strong evidence against this assumption). This is a philosophical brain twister, I know; but understanding what it means is actually very important for how one uses evidence (data) to inform science. It took me about 20 years after grad school to (partially, anyway) figure it out. Bear of little brain and all that.. 2. Define: Deviate from normality. With a sample of 1000, normality tests at conventional significance levels will typically come out statistically significant/contradict normality (which is why a whole school of statistics, the gang of Bayesians, do not think that statistical significance and evidence in the data have much to do with one another). But that's not the real question, is it? 3. The real question is: Does whatever I do to analyze the data and draw scientific conclusions depend crucially on the assumption of normality of the underlying population from which the data are sampled? Of course, it depends on exactly what you do, but, by and large, basic statistical texts continue to teach that the answer is yes. Unfortunately, that is mostly (not always -- and it depends on what's at issue) a lie, as we have known for about 50 years. The crucial matter in practice is not normality but how the sampled data were obtained: the study design and, especially, the issue of independence. Unfortunately, that is rather complicated to deal with, so the Intro Stats texts prefer to ignore it and teach hogwash. [dismount soapbox] Thoughtful nasty rejoinders welcome. Please send your thought-less nasty ones to me privately to spare our colleagues. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Installing R-packages in Windows
( hotmail just randomly decides not to prefix original text, my comments below) Date: Tue, 14 Dec 2010 17:36:11 +0100 From: er...@phonetik.uni-muenchen.de To: r-help@r-project.org Hi there, I have the following problem and I hope somebody might help me. First of all: I am using WinXP SP3 (english and/or german) with R in Version 2.10.0. Now I am trying to install some packages but unfortunately I am getting a weird error. No matter which package I am trying to install - I nearly get the same error. It looks like this: - C:\Program Files\R\R-2.10.0\binR CMD INSTALL %SystemDrive%\rPAcsTemp\car_2.0-2. zip Warning in rawToChar(block[seq_len(ns)]) : truncating string with embedded nul: M:: I used to build all from source then that stopped working and I have seen others here complain about installation failures. I have had good luck with what others have suggested, and use install.packages() It goes into a script ok, for example, ( note that myR is my script that invokes R but you can change etc). This points to ucla mirror, writes a script into $tf to install $pkf, installs $pkf by running the gerated script, and then shows you the results and deletes junk ( normally you want to check error codes before deling junk however) tf=$$_temp cat - DUMMTY $tf options(repos=c(http://cran.stat.ucla.edu;)) install.packages(c($pkf),dep=TRUE) DUMMTY echo executing cat $tf myR -run $tf echo removing $tf rm $tf echo removing R output cat $tf.Rout rm $tf.Rout __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Installing R-packages in Windows
Or in short, type install.packages(car) within R. Note that you won't get new versions of car for the outdated version of R. Note also that your zip file might not fit to the R version you are using. Hence recommendation is to upgrade to R-2.12.0 patched (or 2.12.1 which will be released this Thursday) and run install.packages() after the upgrade. Uwe Ligges On 14.12.2010 18:17, Mike Marchywka wrote: ( hotmail just randomly decides not to prefix original text, my comments below) Date: Tue, 14 Dec 2010 17:36:11 +0100 From: er...@phonetik.uni-muenchen.de To: r-help@r-project.org Hi there, I have the following problem and I hope somebody might help me. First of all: I am using WinXP SP3 (english and/or german) with R in Version 2.10.0. Now I am trying to install some packages but unfortunately I am getting a weird error. No matter which package I am trying to install - I nearly get the same error. It looks like this: - C:\Program Files\R\R-2.10.0\binR CMD INSTALL %SystemDrive%\rPAcsTemp\car_2.0-2. zip Warning in rawToChar(block[seq_len(ns)]) : truncating string with embedded nul: M:: I used to build all from source then that stopped working and I have seen others here complain about installation failures. I have had good luck with what others have suggested, and use install.packages() It goes into a script ok, for example, ( note that myR is my script that invokes R but you can change etc). This points to ucla mirror, writes a script into $tf to install $pkf, installs $pkf by running the gerated script, and then shows you the results and deletes junk ( normally you want to check error codes before deling junk however) tf=$$_temp cat - DUMMTY $tf options(repos=c(http://cran.stat.ucla.edu;)) install.packages(c($pkf),dep=TRUE) DUMMTY echo executing cat $tf myR -run $tf echo removing $tf rm $tf echo removing R output cat $tf.Rout rm $tf.Rout __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Question
What about thoughtless un-nasty comments? I would suggest the original poster (and others thinking that they want to do normality testing) read the help page for SnowsPenultimateNormalityTest in the TeachingDemos package, which I think agrees mostly with what Bert wrote below (though it goes more with the idea that data always deviates from normality, but that may be differences in interpreting the language rather than the theory). -- 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 Bert Gunter Sent: Tuesday, December 14, 2010 10:12 AM To: Gerrit Eichner Cc: r-help@r-project.org; Matthew Rosett Subject: Re: [R] Question ... (in addition to the very useful suggestion to plot your data): (Sounds like a homework question... ?). Sigh. [mount soapbox] 1. Data never deviate from normality. They only provide provide evidence to challenge (test is the formal term) the assumption that the population from which the data were sampled (how? -- see below) can be modeled as normal (e.g. whether the data provide strong evidence against this assumption). This is a philosophical brain twister, I know; but understanding what it means is actually very important for how one uses evidence (data) to inform science. It took me about 20 years after grad school to (partially, anyway) figure it out. Bear of little brain and all that.. 2. Define: Deviate from normality. With a sample of 1000, normality tests at conventional significance levels will typically come out statistically significant/contradict normality (which is why a whole school of statistics, the gang of Bayesians, do not think that statistical significance and evidence in the data have much to do with one another). But that's not the real question, is it? 3. The real question is: Does whatever I do to analyze the data and draw scientific conclusions depend crucially on the assumption of normality of the underlying population from which the data are sampled? Of course, it depends on exactly what you do, but, by and large, basic statistical texts continue to teach that the answer is yes. Unfortunately, that is mostly (not always -- and it depends on what's at issue) a lie, as we have known for about 50 years. The crucial matter in practice is not normality but how the sampled data were obtained: the study design and, especially, the issue of independence. Unfortunately, that is rather complicated to deal with, so the Intro Stats texts prefer to ignore it and teach hogwash. [dismount soapbox] Thoughtful nasty rejoinders welcome. Please send your thought-less nasty ones to me privately to spare our colleagues. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] colour-plot of point intensities?
Hi, I've spent a while scrabbling around at this problem, to no avail. I'm sure there /should/ be a simple answer... I have a large (14 million rows) data set, with two columns. Each row contains the number of times an individual moved, and the distance that individual moved in total. I would like a figure where the colour indicates the density of points, rather like this one my colleague produced in Matlab (with an older version of the data): http://www.warwick.ac.uk/~lssgah/warwick1.pdf Sadly, my colleague is away at the moment, so I can't find out what they did in Matlab. How should I be approaching this problem? [ideally, I'd like a colour-scale like that, too, but] My current approach, based on a bit in the back of Crawley's R book, was to create a new data file where I rounded distance off, and counted the number of similar rows to give a third column (a count). I then did: library(akima) obs - read.table(numvsdistscount.dat) zz - with(obs,interp(V1,V2,V3)) filled.contour(zz,col=heat.colors(12)) But: i) I think doing a rounding step to create the third column (separate from the plot generation) is unwise ii) This results in a shaded area less than the full area of the plot iii) The colour comes out almost uniform I'd be grateful for any pointers. Thanks, Matthew -- Matthew Vernon, Research Fellow Ecology and Epidemiology Group, University of Warwick http://blogs.warwick.ac.uk/mcvernon __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] curve
On 2010-12-14 08:43, Sarah Goslee wrote: You are most likely to get reasonable help if you provide at a minimum your OS, the version of R you're using, and the error message you get. And also if you send your replies to the R-help list, not just me. Sarah Sarah's code works perfectly well on Windows Vista. Is Val using Windows 95? Peter Ehlers On Tue, Dec 14, 2010 at 10:58 AM, Valvalkr...@gmail.com wrote: Thank you Sarah, It worked in my Linux machine as well but not in Windows. [...snip...] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] colour-plot of point intensities?
Look at the hexbin package (bioconductor). -- 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 Matthew Vernon Sent: Tuesday, December 14, 2010 9:48 AM To: r-h...@stat.math.ethz.ch Subject: [R] colour-plot of point intensities? Hi, I've spent a while scrabbling around at this problem, to no avail. I'm sure there /should/ be a simple answer... I have a large (14 million rows) data set, with two columns. Each row contains the number of times an individual moved, and the distance that individual moved in total. I would like a figure where the colour indicates the density of points, rather like this one my colleague produced in Matlab (with an older version of the data): http://www.warwick.ac.uk/~lssgah/warwick1.pdf Sadly, my colleague is away at the moment, so I can't find out what they did in Matlab. How should I be approaching this problem? [ideally, I'd like a colour-scale like that, too, but] My current approach, based on a bit in the back of Crawley's R book, was to create a new data file where I rounded distance off, and counted the number of similar rows to give a third column (a count). I then did: library(akima) obs - read.table(numvsdistscount.dat) zz - with(obs,interp(V1,V2,V3)) filled.contour(zz,col=heat.colors(12)) But: i) I think doing a rounding step to create the third column (separate from the plot generation) is unwise ii) This results in a shaded area less than the full area of the plot iii) The colour comes out almost uniform I'd be grateful for any pointers. Thanks, Matthew -- Matthew Vernon, Research Fellow Ecology and Epidemiology Group, University of Warwick http://blogs.warwick.ac.uk/mcvernon __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 300 dpi and eps:
Hi,
I have a run of 5 graphs that I want to place them under the same page.
Everything works fine to place them in a pdf file , or eps file, but
when it comes to have a high quality of
300 dpi these graphs are not good. For example I open the eps file with
Adobe Illustrator (AI) and it shows that it is a 72dpi graph. If I start
with a 72dpi graph AI cannot improve this to 300 dpi. Q: HOW CAN A GRAPH
IN R DIRECTLY SAVED AS 300dpi? What options do I need to add to the
postscript function to have a 1 page graph that has these 5 plots and is
a 300 dpi graph? Thank you in advance, Aldi
Here is what I am using right now:
postscript(file='./plotST.eps',paper='special',width=10,height=10,horizontal=FALSE)
par(mfrow=c(5 ,1))
plot(sortord , X1 , cex=0.5 ,pch=21 , ylim=c(1.02 , 8.08771 ),
xlab='ST: 1-5 position (unit)',ylab='ylabel',font.lab = 1, cex.lab =
1,cex.main=0.5 +0.4, col=chrom )
text(midpointsbyc+minadj+5,0-5,chnum ,cex=0.5 +0.2 ,col=chnum+1 )
abline(h = -1*log10(9.7e-08 ), lty = 2,col = 'gray' )
# Draw the legend
legend('topright',legend=c('X1'))
plot(sortord , X2
# Put a box around the plot
box(lwd = 1)
dev.off()
--
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How to left or right truncate a character string?
Hi R-helpers, I have a character string, for example: lm(y ~ X2 + X3 + X4) from which I would like to strip off the leading and trailing quotation marks resulting in this: lm(y ~ X2 + X3 + X4) I have tried using gsub() but I can't figure out how to specify the quotation mark using a regular expression. Alternatively, I would like a function that lets me delete the leading (or trailing) X characters, and in this case X=1 (but it could be used more flexibly to delete several leading or trailing characters). I would appreciate help with either of these potential solutions (gsub and regex, or delete leading/trailing characters). Many thanks! Mark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to left or right truncate a character string?
Try this: noquote(lm(y ~ X2 + X3 + X4)) To remove X characters: gsub(^.|.$, , lm(y ~ X2 + X3 + X4)) On Tue, Dec 14, 2010 at 6:27 PM, Mark Na mtb...@gmail.com wrote: Hi R-helpers, I have a character string, for example: lm(y ~ X2 + X3 + X4) from which I would like to strip off the leading and trailing quotation marks resulting in this: lm(y ~ X2 + X3 + X4) I have tried using gsub() but I can't figure out how to specify the quotation mark using a regular expression. Alternatively, I would like a function that lets me delete the leading (or trailing) X characters, and in this case X=1 (but it could be used more flexibly to delete several leading or trailing characters). I would appreciate help with either of these potential solutions (gsub and regex, or delete leading/trailing characters). Many thanks! Mark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Discriminant Correspondence Analysis
Hello everyone, I am totally new to the R program. I have had a look at some pdf documents that I downloaded and that explain how to do many things in R; however, I still cannot figure out how to do what I want to do, which is to perform Discriminant Correspondence Analysis on a rectangular matrix of data that I have in an Excel file. I know R users frown upon Excel and recommend converting Excel files to .csv format, which I have done, no problem. That is not an issue. There are several parts to my problem. 1) When I try the read.table command, even if I include the directory name in the filename, R still cannot read the file, even if it is in .csv format 2) I was able to copy my file and then read the clipboard contents into R but then I do not know to assign a name to the data frame in order to conduct any operations on it 3) I need the ADE4 program in order to perform Discriminant Correspondence Analysis, so I used the install.packages command to install it. It installed no problem but I do not know how to access the ADE4 program in R. I am unable to open it directly, either. 4) I thought that using the ADE4 GUI (called ade4TkGUI) would be easier because I do not know many of the R commands; but, again, I downloaded it but cannot open or access it. The following is the suggested coding that I found through the R website, but when I try to use this code, I don't know how to assign a name for the df, or what to put for fac, and what is worse, I get an error message saying that the program cannot find the discrimin.coa command. Usage discrimin.coa(df, fac, scannf = TRUE, nf = 2) Arguments df a data frame containing positive or null values fac a factor defining the classes of discriminant analysis scannf a logical value indicating whether the eigenvalues bar plot should be displayed nf if scannf FALSE, an integer indicating the number of kept axes Examples data(perthi02) plot(discrimin.coa(perthi02$tab, perthi02$cla, scan = FALSE)) For clarification, my data consists of measurements of morphological characters of an assemblage of biological specimens. I have already performed Principal Co-ordinates Analysis, Principal Compionents Analysis and Cluster Analysis in another program (PAST) in order to see if the data fall into distinct groupings that might represent different morphological species. I now want to test the groupings that I found on my test data set using Discriminant Correspondence Analysis.There are both continuous and categorical characters, which is the reason why I need to perform Discriminant Correspondence Analysis, instead of Linear Discriminant Analysis, which is only valid for continuous measurements. R seems to be the only program in which I can perform Discriminant Correspondence Analysis. Thanks for any help offered on any of these points. Wayne [[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] Discriminant Correspondence Analysis
Read files, if you're on windows remember to include the path like this: C:\\Documents and Settings\\USER\\My Documents\\MyFile.csv -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Wayne Sawtell Sent: Tuesday, December 14, 2010 12:36 PM To: r-help@r-project.org Subject: [R] Discriminant Correspondence Analysis Hello everyone, I am totally new to the R program. I have had a look at some pdf documents that I downloaded and that explain how to do many things in R; however, I still cannot figure out how to do what I want to do, which is to perform Discriminant Correspondence Analysis on a rectangular matrix of data that I have in an Excel file. I know R users frown upon Excel and recommend converting Excel files to .csv format, which I have done, no problem. That is not an issue. There are several parts to my problem. 1) When I try the read.table command, even if I include the directory name in the filename, R still cannot read the file, even if it is in .csv format 2) I was able to copy my file and then read the clipboard contents into R but then I do not know to assign a name to the data frame in order to conduct any operations on it 3) I need the ADE4 program in order to perform Discriminant Correspondence Analysis, so I used the install.packages command to install it. It installed no problem but I do not know how to access the ADE4 program in R. I am unable to open it directly, either. 4) I thought that using the ADE4 GUI (called ade4TkGUI) would be easier because I do not know many of the R commands; but, again, I downloaded it but cannot open or access it. The following is the suggested coding that I found through the R website, but when I try to use this code, I don't know how to assign a name for the df, or what to put for fac, and what is worse, I get an error message saying that the program cannot find the discrimin.coa command. Usage discrimin.coa(df, fac, scannf = TRUE, nf = 2) Arguments df a data frame containing positive or null values fac a factor defining the classes of discriminant analysis scannf a logical value indicating whether the eigenvalues bar plot should be displayed nf if scannf FALSE, an integer indicating the number of kept axes Examples data(perthi02) plot(discrimin.coa(perthi02$tab, perthi02$cla, scan = FALSE)) For clarification, my data consists of measurements of morphological characters of an assemblage of biological specimens. I have already performed Principal Co-ordinates Analysis, Principal Compionents Analysis and Cluster Analysis in another program (PAST) in order to see if the data fall into distinct groupings that might represent different morphological species. I now want to test the groupings that I found on my test data set using Discriminant Correspondence Analysis.There are both continuous and categorical characters, which is the reason why I need to perform Discriminant Correspondence Analysis, instead of Linear Discriminant Analysis, which is only valid for continuous measurements. R seems to be the only program in which I can perform Discriminant Correspondence Analysis. Thanks for any help offered on any of these points. Wayne [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to left or right truncate a character string?
Mark -
Since regular expressions in R are just character
strings, it's pretty easy to assemble a regular expression
to delete leading or trailing characters. For example:
delchars = function(str,n,lead=TRUE){
+dots = paste(rep('.',n),collapse='')
+pat = if(lead)paste('^',dots,sep='') else paste(dots,'$',sep='')
+sub(pat,'',str)
+ }
str = this is a test
delchars(str,4)
[1] is a test
delchars(str,4,lead=FALSE)
[1] this is a
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Tue, 14 Dec 2010, Mark Na wrote:
Hi R-helpers,
I have a character string, for example:
lm(y ~ X2 + X3 + X4)
from which I would like to strip off the leading and trailing
quotation marks resulting in this:
lm(y ~ X2 + X3 + X4)
I have tried using gsub() but I can't figure out how to specify the
quotation mark using a regular expression.
Alternatively, I would like a function that lets me delete the leading
(or trailing) X characters, and in this case X=1 (but it could be used
more flexibly to delete several leading or trailing characters).
I would appreciate help with either of these potential solutions (gsub
and regex, or delete leading/trailing characters).
Many thanks!
Mark
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How to bind models into a list of models?
Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: dd [1] lm(y ~ 1,data=Cement) lm(y ~ X,data=Cement) lm(y ~ X1,data=Cement) [4] lm(y ~ X2,data=Cement)lm(y ~ X3,data=Cement)lm(y ~ X4,data=Cement) [7] lm(y ~ X + X1,data=Cement)lm(y ~ X + X2,data=Cement)lm(y ~ X + X3,data=Cement) [10] lm(y ~ X + X4,data=Cement)lm(y ~ X1 + X2,data=Cement) lm(y ~ X1 + X3,data=Cement) [13] lm(y ~ X1 + X4,data=Cement) lm(y ~ X2 + X3,data=Cement) lm(y ~ X2 + X4,data=Cement) [16] lm(y ~ X3 + X4,data=Cement) lm(y ~ X + X1 + X2,data=Cement) lm(y ~ X + X1 + X3,data=Cement) [19] lm(y ~ X + X1 + X4,data=Cement) lm(y ~ X + X2 + X3,data=Cement) lm(y ~ X + X2 + X4,data=Cement) [22] lm(y ~ X + X3 + X4,data=Cement) lm(y ~ X1 + X2 + X3,data=Cement) lm(y ~ X1 + X2 + X4,data=Cement) [25] lm(y ~ X1 + X3 + X4,data=Cement) lm(y ~ X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3,data=Cement) [28] lm(y ~ X + X1 + X2 + X4,data=Cement) lm(y ~ X + X1 + X3 + X4,data=Cement) lm(y ~ X + X2 + X3 + X4,data=Cement) [31] lm(y ~ X1 + X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3 + X4,data=Cement) I would like to convert this object into a list called Cand.models with 32 list elements each of which would contain one of the above model formulae. When I print the list, the models should run, so the first few elements of the list would look like this (see below output from a list I created by hand). Many thanks for any help you can provide! Mark Cand.models [[1]] Call: lm(formula = y ~ 1, data = Cement) Coefficients: (Intercept) 95.42 [[2]] Call: lm(formula = y ~ X, data = Cement) Coefficients: (Intercept)X 82.3081.874 [[3]] Call: lm(formula = y ~ X1, data = Cement) Coefficients: (Intercept) X1 81.4791.869 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 bind models into a list of models?
Try step() ? On Tue, Dec 14, 2010 at 10:05 PM, Mark Na mtb...@gmail.com wrote: Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: dd [1] lm(y ~ 1,data=Cement) lm(y ~ X,data=Cement) lm(y ~ X1,data=Cement) [4] lm(y ~ X2,data=Cement) lm(y ~ X3,data=Cement) lm(y ~ X4,data=Cement) [7] lm(y ~ X + X1,data=Cement) lm(y ~ X + X2,data=Cement) lm(y ~ X + X3,data=Cement) [10] lm(y ~ X + X4,data=Cement) lm(y ~ X1 + X2,data=Cement) lm(y ~ X1 + X3,data=Cement) [13] lm(y ~ X1 + X4,data=Cement) lm(y ~ X2 + X3,data=Cement) lm(y ~ X2 + X4,data=Cement) [16] lm(y ~ X3 + X4,data=Cement) lm(y ~ X + X1 + X2,data=Cement) lm(y ~ X + X1 + X3,data=Cement) [19] lm(y ~ X + X1 + X4,data=Cement) lm(y ~ X + X2 + X3,data=Cement) lm(y ~ X + X2 + X4,data=Cement) [22] lm(y ~ X + X3 + X4,data=Cement) lm(y ~ X1 + X2 + X3,data=Cement) lm(y ~ X1 + X2 + X4,data=Cement) [25] lm(y ~ X1 + X3 + X4,data=Cement) lm(y ~ X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3,data=Cement) [28] lm(y ~ X + X1 + X2 + X4,data=Cement) lm(y ~ X + X1 + X3 + X4,data=Cement) lm(y ~ X + X2 + X3 + X4,data=Cement) [31] lm(y ~ X1 + X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3 + X4,data=Cement) I would like to convert this object into a list called Cand.models with 32 list elements each of which would contain one of the above model formulae. When I print the list, the models should run, so the first few elements of the list would look like this (see below output from a list I created by hand). Many thanks for any help you can provide! Mark Cand.models [[1]] Call: lm(formula = y ~ 1, data = Cement) Coefficients: (Intercept) 95.42 [[2]] Call: lm(formula = y ~ X, data = Cement) Coefficients: (Intercept) X 82.308 1.874 [[3]] Call: lm(formula = y ~ X1, data = Cement) Coefficients: (Intercept) X1 81.479 1.869 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 300 dpi and eps:
Hi,
I have a run of 5 graphs that I want to place them under the same page.
Everything works fine to place them in a pdf file , or eps file, but
when it comes to have a high quality of
300 dpi these graphs are not good. For example I open the eps file with
Adobe Illustrator (AI) and it shows that it is a 72dpi graph. If I start
with a 72dpi graph AI cannot improve this to 300 dpi. Q: HOW CAN A GRAPH
IN R DIRECTLY SAVED AS 300dpi? What options do I need to add to the
postscript function to have a 1 page graph that has these 5 plots and is
a 300 dpi graph? Thank you in advance, Aldi
Here is what I am using right now:
postscript(file='./plotST.eps',paper='special',width=10,height=10,horizontal=FALSE)
par(mfrow=c(5 ,1))
plot(sortord , X1 , cex=0.5 ,pch=21 , ylim=c(1.02 , 8.08771 ),
xlab='ST: 1-5 position (unit)',ylab='ylabel',font.lab = 1, cex.lab =
1,cex.main=0.5 +0.4, col=chrom )
text(midpointsbyc+minadj+5,0-5,chnum ,cex=0.5 +0.2 ,col=chnum+1 )
abline(h = -1*log10(9.7e-08 ), lty = 2,col = 'gray' )
# Draw the legend
legend('topright',legend=c('X1'))
plot(sortord , X2
# Put a box around the plot
box(lwd = 1)
dev.off()
--
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] from table to matrix
I have a table like this: Date TIME Q A a1 A b2 A c3 B a4 B b5 B c6 C a7 C b8 C c9 I want use R language to turn it to a matrix like : a b c A 1 2 3 B 4 5 6 C 7 8 9 I am new to R , anyone can help? Thanks in advance! have a table like this: Date TIME Q A a1 A b2 A c3 B a4 B b5 B c6 -- View this message in context: http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3087972.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] How to bind models into a list of models?
Mark - I believe lapply(dd,function(m)eval(parse(text=m))) will do what you want. - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Tue, 14 Dec 2010, Mark Na wrote: Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: dd [1] lm(y ~ 1,data=Cement) lm(y ~ X,data=Cement) lm(y ~ X1,data=Cement) [4] lm(y ~ X2,data=Cement)lm(y ~ X3,data=Cement)lm(y ~ X4,data=Cement) [7] lm(y ~ X + X1,data=Cement)lm(y ~ X + X2,data=Cement)lm(y ~ X + X3,data=Cement) [10] lm(y ~ X + X4,data=Cement)lm(y ~ X1 + X2,data=Cement) lm(y ~ X1 + X3,data=Cement) [13] lm(y ~ X1 + X4,data=Cement) lm(y ~ X2 + X3,data=Cement) lm(y ~ X2 + X4,data=Cement) [16] lm(y ~ X3 + X4,data=Cement) lm(y ~ X + X1 + X2,data=Cement) lm(y ~ X + X1 + X3,data=Cement) [19] lm(y ~ X + X1 + X4,data=Cement) lm(y ~ X + X2 + X3,data=Cement) lm(y ~ X + X2 + X4,data=Cement) [22] lm(y ~ X + X3 + X4,data=Cement) lm(y ~ X1 + X2 + X3,data=Cement) lm(y ~ X1 + X2 + X4,data=Cement) [25] lm(y ~ X1 + X3 + X4,data=Cement) lm(y ~ X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3,data=Cement) [28] lm(y ~ X + X1 + X2 + X4,data=Cement) lm(y ~ X + X1 + X3 + X4,data=Cement) lm(y ~ X + X2 + X3 + X4,data=Cement) [31] lm(y ~ X1 + X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3 + X4,data=Cement) I would like to convert this object into a list called Cand.models with 32 list elements each of which would contain one of the above model formulae. When I print the list, the models should run, so the first few elements of the list would look like this (see below output from a list I created by hand). Many thanks for any help you can provide! Mark Cand.models [[1]] Call: lm(formula = y ~ 1, data = Cement) Coefficients: (Intercept) 95.42 [[2]] Call: lm(formula = y ~ X, data = Cement) Coefficients: (Intercept)X 82.3081.874 [[3]] Call: lm(formula = y ~ X1, data = Cement) Coefficients: (Intercept) X1 81.4791.869 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] from table to matrix
Here's one way: df = data.frame(Date=rep(LETTERS[1:3],each=3),TIME=rep(letters[1:3],3),Q=1:9) df Date TIME Q 1Aa 1 2Ab 2 3Ac 3 4Ba 4 5Bb 5 6Bc 6 7Ca 7 8Cb 8 9Cc 9 mat = matrix(0,3,3,dimnames=list(LETTERS[1:3],letters[1:3])) mat[as.matrix(df[,1:2])] = df[,3] mat a b c A 1 2 3 B 4 5 6 C 7 8 9 - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Tue, 14 Dec 2010, JESSICA [via R] wrote: I have a table like this: Date TIME Q A a1 A b2 A c3 B a4 B b5 B c6 C a7 C b8 C c9 I want use R language to turn it to a matrix like : a b c A 1 2 3 B 4 5 6 C 7 8 9 I am new to R , anyone can help? Thanks in advance! have a table like this: Date TIME Q A a1 A b2 A c3 B a4 B b5 B c6 __ This email was sent by JESSICA (via Nabble) Your replies will appear at http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3087972.html To receive all replies by email, subscribe to this discussion: http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=subscribe_by_codenode=3087972code=ci1oZWxwQHItcHJvamVjdC5vcmd8MzA4Nzk3MnwtNzg0MjM1NTA4 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to bind models into a list of models?
Many thanks Phil. This is perfect. I usually forget about lapply and try something more complicated. Your solution works really well. Best, Mark On Tue, Dec 14, 2010 at 3:45 PM, Phil Spector spec...@stat.berkeley.edu wrote: Mark - I believe lapply(dd,function(m)eval(parse(text=m))) will do what you want. - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Tue, 14 Dec 2010, Mark Na wrote: Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: dd [1] lm(y ~ 1,data=Cement) lm(y ~ X,data=Cement) lm(y ~ X1,data=Cement) [4] lm(y ~ X2,data=Cement) lm(y ~ X3,data=Cement) lm(y ~ X4,data=Cement) [7] lm(y ~ X + X1,data=Cement) lm(y ~ X + X2,data=Cement) lm(y ~ X + X3,data=Cement) [10] lm(y ~ X + X4,data=Cement) lm(y ~ X1 + X2,data=Cement) lm(y ~ X1 + X3,data=Cement) [13] lm(y ~ X1 + X4,data=Cement) lm(y ~ X2 + X3,data=Cement) lm(y ~ X2 + X4,data=Cement) [16] lm(y ~ X3 + X4,data=Cement) lm(y ~ X + X1 + X2,data=Cement) lm(y ~ X + X1 + X3,data=Cement) [19] lm(y ~ X + X1 + X4,data=Cement) lm(y ~ X + X2 + X3,data=Cement) lm(y ~ X + X2 + X4,data=Cement) [22] lm(y ~ X + X3 + X4,data=Cement) lm(y ~ X1 + X2 + X3,data=Cement) lm(y ~ X1 + X2 + X4,data=Cement) [25] lm(y ~ X1 + X3 + X4,data=Cement) lm(y ~ X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3,data=Cement) [28] lm(y ~ X + X1 + X2 + X4,data=Cement) lm(y ~ X + X1 + X3 + X4,data=Cement) lm(y ~ X + X2 + X3 + X4,data=Cement) [31] lm(y ~ X1 + X2 + X3 + X4,data=Cement) lm(y ~ X + X1 + X2 + X3 + X4,data=Cement) I would like to convert this object into a list called Cand.models with 32 list elements each of which would contain one of the above model formulae. When I print the list, the models should run, so the first few elements of the list would look like this (see below output from a list I created by hand). Many thanks for any help you can provide! Mark Cand.models [[1]] Call: lm(formula = y ~ 1, data = Cement) Coefficients: (Intercept) 95.42 [[2]] Call: lm(formula = y ~ X, data = Cement) Coefficients: (Intercept) X 82.308 1.874 [[3]] Call: lm(formula = y ~ X1, data = Cement) Coefficients: (Intercept) X1 81.479 1.869 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] defining contrasts in lm() - is there a simpler way?
Dear R users, I am looking for a simple way to define a contrast in a linear model. I have a data frame with two factors (f1, f2) and a dependent y. x - data.frame(y =rnorm(30), f1=gl(2, 15), f2=gl(3, 5, 30)) Now I want to specify the following contrast: f1= 1 or 2 and f2=1 vs. f1= 1 or 2 and f2=3 The best I can come up with is the following: x$new[x$f1==1 x$f2==1] - 1 x$new[x$f1==1 x$f2==2] - 2 x$new[x$f1==1 x$f2==3] - 3 x$new[x$f1==2 x$f2==1] - 4 x$new[x$f1==2 x$f2==2] - 5 x$new[x$f1==2 x$f2==3] - 6 x$new - as.factor(x$new) contrasts(x$new) - cbind( my constrast = c( .5, -.5, 0, .5, 0, -.5)) summary(lm(y ~ new, x)) I have two questions concerning this: 1) if I take a look at the constrast matrix derived from the one contrast I specified, I assume that R automatically adds the missing constrasts so they are orthogonal. round(cor(attributes(x$new)$contrast), 1) Is that always the case? 2) Can I get this in a simpler way? I find it a bit tedious to define the constrast like above. Is there something simpler, like: lm(y ~ f1:f2, + user defined contrast definition inside lm that is equivalent to the above) Thanks in advance, Mark ––– Mark Heckmann Blog: www.markheckmann.de R-Blog: http://ryouready.wordpress.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] Discriminant Correspondence Analysis
Wayne, I don't know how to assign a name for the df, or what to put for fac, and what is worse, I get an error message saying that the program cannot find the discrimin.coa command. Before you can use a package you have downloaded you need to activate it. There are different ways of doing this. Simplest is to type library(ade4). ## library(ade4) ?discrimin.coa Follow Bastiaan and read in your file as follows (single forward slashes also work): ## See ?read.csv as you may need to change some switches MyFile - read.csv(C:\\Documents and Settings\\USER\\My Documents\\MyFile.csv) str(MyFile) Without data it is difficult to help you further, but your general call to discrimin.coa is ## This may or may not work; depends what's in MyFile T.discrimin - discrimin.coa(MyFile, fac = someFacInMyFile, scann=F, nf=4) T.discrimin plot(T.discrimin) Regards, Mark. -- View this message in context: http://r.789695.n4.nabble.com/Discriminant-Correspondence-Analysis-tp3087929p3088091.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] from table to matrix
That's so weird, I just signed up on here to ask exactly the same question! However, I think my issue is like Jessica's who says that her data is like that, not actually that... So the issue is not in generating that data on-the-fly but in transforming it from a data frame to a matrix. As a more concrete example, I have read the following data in from a file (around 300,000 rows): xy z 00687 0164 0271 0355 0452 0551 0638 0738 0854 0949 . . . 3049792829771 35137715479801 38383517405411 41813360247101 42154910285721 47131417518361 57960218173931 71351555243851 So what I want to do is transform this into a matrix where at position (x,y) in the matrix I have value z. I am doing this so that I can then do a filled.contour plot on the data. I think this is the same as what Jessica is asking... Regards and many thanks, Jonathan UCL Computer Science -- View this message in context: http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088096.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] survival: ridge log-likelihood workaround
I would like to clarify statistics that ridge coxph returns. Here is my understanding and please correct me where I am wrong. 1) In his paper Gray [JASA 1992] suggests a Wald-type statistics with the formula for degree of freedom. The summary function for ridge coxph returns the degree of freedom and Wald test which are equivalent to what Gray wrote. 2) Summary for ridge coxph prints a likelihood ratio test which is not, if I may say, a proper likelihood ratio test. First it is based on unpenalized log-likelihoods and the above defined degree of freedom is used. I accept that there is nothing which suggests that one should use penalized log-likelihoods. However, there is also nothing published which suggests that unpenalized log-likelihoods should be used with the above defined degree of freedom. I have found that TherneauGrambsch in their excellent book discuss this in a paragraph and mention that the p-value thus returned is somewhat conservative (p too large). Therefore, the likelihood ratio test that ridge coxph returns is not a true one and the statistics returned (i.e. 2*(loglik(beta)-loglik(0))) has the distribution which is somewhat more compact than the chi-square. I like conservative p-values and like to be on a safe side. However, in my work Wald test p-values for ridge regression are much higher than the l! ikelihood ratio test's p-value and I don't get impression that they are conservative. 3) There is no efficient score test for ridge regression, as there is no penalized efficient score test. That is OK. 4) Rsquare and max possible are returned and I have so far failed to find exact references for them. I would like to add a note here that the coxph algorithm works really fast with high-dimensional covariates and all compliments to people who developed it. There was evidently a lot of effort put to make it all work fast and correctly and, in my opinion, it is a shame that the last bit - summary for ridge coxph - is a bit, if I may say, shaky. In my opinion, only Wald test and degree of freedom as defined by Gray deserve to be part of summary for ridge coxph and I look forward to be corrected. On my behalf I am prepared in my spare time to write the code so that the summary for ridge coxph does not return NULL and that the statistics printed in summary for ridge coxph are based on published papers. Damjan Krstajic Subject: Re: [R] survival: ridge log-likelihood workaround From: thern...@mayo.edu To: r-help@r-project.org; dkrsta...@hotmail.com Date: Fri, 10 Dec 2010 09:07:42 -0600 -- begin inclusion - Dear all, I need to calculate likelihood ratio test for ridge regression. In February I have reported a bug where coxph returns unpenalized log-likelihood for final beta estimates for ridge coxph regression. In high-dimensional settings ridge regression models usually fail for lower values of lambda. As the result of it, in such settings the ridge regressions have higher values of lambda (e.g. over 100) which means that the difference between unpenalized log-likelihood and penalized log-likelihood is not insignificant. I would be grateful if someone can confirm that the below code is correct workaround. --- end included message First, the bug you report is not a bug. The log partial likelihood from a Cox model LPL(beta) is well defined for any vector of coefficients beta, whether they are result of a maximization or taken from your daily horoscope. The loglik component of coxph is the LPL for the reported coefficients. For a ridge regression the coxph function maximizes LPL(beta) - penalty(beta) = penalized partial likelihood = PPL(beta). You have correctly recreated the PPL. Second: how do you do formal tests on such a model? This is hard. The difference LPL1- LPL2 is a chi-square when each is the result of maximizing the Cox LPL over a set of coefficients; when using a PPL we are maximizing over something else. The distribution of the difference of constrained LPL values can be argued to be a weighed sum of squared normals where the weights are in (0,1), which is something more complex than a chisq distribution. In a world with infinite free time I'd have pursued this, worked it all out, and added appropriate code to coxph. What about the difference in PPL values, which is the test you propose? I'm not aware of any theory showing that these have any relation to a chi-square distribution. (Said theory may well exist, and I'd be happy for pointers.) Terry Therneau [[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] GLM with different factors
Hi R-Helpers, I am stuck with my analysis. I wonder if anybody could have with this. My data set looks like: Squares Hours SppRichness Feeding_guild natprop 1 10 1 aq inverts 0.118697 2 25 2 aq inverts 0.605874 3 35 2 fish 0.61255 4 20 4 fish 0.764418 5 40 3 insects 0.810646 6 20 2 insects 0.832063 7 20 2 Omnivores 0.838509 n 20 3 Omnivores 0.839228 My interest is to run a GLM model between the response variable y = SppRichness (species richness) and y = natprop )Proportion of Natural Area considering each Feeding Guild (factor) separately. In other words I want to run with one single code the richness within guild as a function of natural proportion area. Is there a way to write a code to run all the correlations for each guild at once? And then to keep the coefficients? And plot the correlations in separately files? Thanks in advance. Rafael -- ** RAFAEL CAMARGO Postgraduate Student Biology Department of University of Ottawa 30 Marie Currie, room # 251 Ottawa, ON, CANADA Tel: +1 (613) 562-5800 ext. 6366 Cel: +1 (613) 869-3772 e-mail: rcama...@uottawa.ca rafael.x.cama...@gmail.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] from table to matrix
If the goal is to produce a filled contour plot, then look at the levelplot function in the trellis package, it takes the data in the form that you already have it, no need to transform. -- 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 jonathan Sent: Tuesday, December 14, 2010 3:14 PM To: r-help@r-project.org Subject: Re: [R] from table to matrix That's so weird, I just signed up on here to ask exactly the same question! However, I think my issue is like Jessica's who says that her data is like that, not actually that... So the issue is not in generating that data on-the-fly but in transforming it from a data frame to a matrix. As a more concrete example, I have read the following data in from a file (around 300,000 rows): xy z 00687 0164 0271 0355 0452 0551 0638 0738 0854 0949 . . . 3049792829771 35137715479801 38383517405411 41813360247101 42154910285721 47131417518361 57960218173931 71351555243851 So what I want to do is transform this into a matrix where at position (x,y) in the matrix I have value z. I am doing this so that I can then do a filled.contour plot on the data. I think this is the same as what Jessica is asking... Regards and many thanks, Jonathan UCL Computer Science -- View this message in context: http://r.789695.n4.nabble.com/from-table- to-matrix-tp3087972p3088096.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] from table to matrix
There are many ways to do this in R. For very simple problems, this one is convenient: library(ecodist) newdata - crosstab(mydata$x, mydata$y, mydata$z) For more complicated problems, reshape is very powerful. Sarah On Tue, Dec 14, 2010 at 5:13 PM, jonathan j...@than.biz wrote: That's so weird, I just signed up on here to ask exactly the same question! However, I think my issue is like Jessica's who says that her data is like that, not actually that... So the issue is not in generating that data on-the-fly but in transforming it from a data frame to a matrix. As a more concrete example, I have read the following data in from a file (around 300,000 rows): x y z 0 0 687 0 1 64 0 2 71 0 3 55 0 4 52 0 5 51 0 6 38 0 7 38 0 8 54 0 9 49 . . . 304979 282977 1 351377 1547980 1 383835 1740541 1 418133 6024710 1 421549 1028572 1 471314 1751836 1 579602 1817393 1 713515 5524385 1 So what I want to do is transform this into a matrix where at position (x,y) in the matrix I have value z. I am doing this so that I can then do a filled.contour plot on the data. I think this is the same as what Jessica is asking... Regards and many thanks, Jonathan UCL Computer Science -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] from table to matrix
Jonathan - Same problem, same solution: Suppose your data frame is called df: thematrix = matrix(NA,max(df$x),max(df$y)) thematrix[as.matrix(df[,1:2])] = df[,3] - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Tue, 14 Dec 2010, jonathan wrote: That's so weird, I just signed up on here to ask exactly the same question! However, I think my issue is like Jessica's who says that her data is like that, not actually that... So the issue is not in generating that data on-the-fly but in transforming it from a data frame to a matrix. As a more concrete example, I have read the following data in from a file (around 300,000 rows): xy z 00687 0164 0271 0355 0452 0551 0638 0738 0854 0949 . . . 3049792829771 35137715479801 38383517405411 41813360247101 42154910285721 47131417518361 57960218173931 71351555243851 So what I want to do is transform this into a matrix where at position (x,y) in the matrix I have value z. I am doing this so that I can then do a filled.contour plot on the data. I think this is the same as what Jessica is asking... Regards and many thanks, Jonathan UCL Computer Science -- View this message in context: http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088096.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] peak detection
THere's the peaks package which will do some peak detection, or 'peaks' in the simecol package. Or msc.peaks.find in caMassClass, I'm madly trying to remember what spectral package I found which has a tool similar to msextrema but returns a couple more useful lists describing the data and the results... Carl *** From: Michael Bedward michael.bedward_at_gmail.com Date: Tue, 14 Dec 2010 14:49:04 +1100 Hi Joe, Just for info, I've done this in the past by applying lowess followed by diff to a vector, then identifying points with change of sign in the diffs. Michael On 14 December 2010 14:22, Joe Stuart joe.stuart_at_gmail.com wrote: Never mind. I did find this package, which seems to do the trick. Thanks http://rgm2.lab.nig.ac.jp/RGM2/R_man-2.9.0/library/msProcess/man/msExtrema.html On Mon, Dec 13, 2010 at 9:05 PM, Joe Stuart joe.stuart_at_gmail.com wrote: Hi, I was wondering if anyone knows of a package that contains the ability for peak/valley detection. Here is an example of what I'm looking for, only problem is that it's written in Matlab. http://www.billauer.co.il/peakdet.html Thanks for any help in advance. -Joe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Use generalised additive model to plot curve
Readers, I have been reading 'the r book' by Crawley and think that the generalised additive model is appropriate for this problem. The package 'gam' was installed using the command (as root) install.package(gam) ... library(gam) library(gam) Loading required package: splines Loading required package: akima library(mgcv) This is mgcv 1.3-25 Attaching package: 'mgcv' The following object(s) are masked from package:gam : gam, gam.control, gam.fit, plot.gam, predict.gam, s, summary.gam x-c(30,50,80,90,100) y-c(160,180,250,450,300) model-gam(y~s(x)) Error in smooth.construct.tp.smooth.spec(object, data, knots) : A term has fewer unique covariate combinations than specified maximum degrees of freedom The objective is to plot y against x, finally to produce a graph with a smooth curve (and then remove the data points). What is my mistake please? yours, r251 gnu/linux mandriva2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] from table to matrix
For contour plots, some of the contour functions in graphics packages (lattice for one IIRC) are pretty good at understanding that columns in a matrix correspond to x,y, and z values already. There are many ways to do this in R. For very simple problems, this one is convenient: library(ecodist) newdata - crosstab(mydata$x, mydata$y, mydata$z) For more complicated problems, reshape is very powerful. Sarah On Tue, Dec 14, 2010 at 5:13 PM, jonathan jon_at_than.biz wrote: That's so weird, I just signed up on here to ask exactly the same question! However, I think my issue is like Jessica's who says that her data is like that, not actually that... So the issue is not in generating that data on-the-fly but in transforming it from a data frame to a matrix. As a more concrete example, I have read the following data in from a file (around 300,000 rows): xy z 00687 0164 0271 0355 0452 0551 0638 0738 0854 0949 . . . 3049792829771 35137715479801 38383517405411 41813360247101 42154910285721 47131417518361 57960218173931 71351555243851 So what I want to do is transform this into a matrix where at position (x,y) in the matrix I have value z. I am doing this so that I can then do a filled.contour plot on the data. I think this is the same as what Jessica is asking... Regards and many thanks, Jonathan UCL Computer Science __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 generalised additive model to plot curve
Dear Lurker, If all you art trying to do is to plot something, isn't all you need something like the following? x - c( 30, 50, 80, 90, 100) y - c(160, 180, 250, 450, 300) sp - spline(x, y, n = 500) plot(sp, type = l, xlab = x, ylab = y, las = 1, main = A Spline Interpolation) points(x, y, pch = 3, col = red, lwd = 2) Bill Venables. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of e-letter Sent: Wednesday, 15 December 2010 8:37 AM To: r-help@r-project.org Subject: [R] Use generalised additive model to plot curve Readers, I have been reading 'the r book' by Crawley and think that the generalised additive model is appropriate for this problem. The package 'gam' was installed using the command (as root) install.package(gam) ... library(gam) library(gam) Loading required package: splines Loading required package: akima library(mgcv) This is mgcv 1.3-25 Attaching package: 'mgcv' The following object(s) are masked from package:gam : gam, gam.control, gam.fit, plot.gam, predict.gam, s, summary.gam x-c(30,50,80,90,100) y-c(160,180,250,450,300) model-gam(y~s(x)) Error in smooth.construct.tp.smooth.spec(object, data, knots) : A term has fewer unique covariate combinations than specified maximum degrees of freedom The objective is to plot y against x, finally to produce a graph with a smooth curve (and then remove the data points). What is my mistake please? yours, r251 gnu/linux mandriva2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Use generalised additive model to plot curve
On Tue, 2010-12-14 at 22:36 +, e-letter wrote: Readers, I have been reading 'the r book' by Crawley and think that the generalised additive model is appropriate for this problem. The package 'gam' was installed using the command (as root) install.package(gam) ... library(gam) library(gam) Loading required package: splines Loading required package: akima library(mgcv) This is mgcv 1.3-25 Why do you want to use both gam:::gam and mgcv:::gam at the same time? Use one or the other package; not both. Attaching package: 'mgcv' The following object(s) are masked from package:gam : gam, gam.control, gam.fit, plot.gam, predict.gam, s, summary.gam These warnings/messages are because you loaded two packages that both provide a gam() and other similarly named functions. x-c(30,50,80,90,100) y-c(160,180,250,450,300) model-gam(y~s(x)) Error in smooth.construct.tp.smooth.spec(object, data, knots) : A term has fewer unique covariate combinations than specified maximum degrees of freedom You are trying to fit a thin-plate spline that is of greater complexity than the available data. You need to reduce the complexity of the fitted spline whilst fitting the model. For these data the maximum complexity is 5: mod - gam(y ~ s(x, k = 5)) HTH G The objective is to plot y against x, finally to produce a graph with a smooth curve (and then remove the data points). What is my mistake please? yours, r251 gnu/linux mandriva2008 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] from table to matrix
OK well I don't mean to hijack Jessica's thread with a tangent on graphics and plots, but I'm still having some trouble. I'm a total newbie here so if the correct etiquette would be for me to start a new thread at this point then please do advise me!! The code I'm now trying to run is: library(lattice) df - read.table(data, sep=\t, header=TRUE) levelplot(df$z ~ df$x * df$y) It seems to run OK but I don't seem to be getting any output... I'm running the latest R on OS X. All the normal plotting functions, e.g. standard plot() seem to work fine and display in a Quartz window, but when I run this I'm not getting anything =/ Any tips? Jonathan -- View this message in context: http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088161.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] rpart - how to estimate the “meaningful” predictors for an outcome (in classification t rees)
Hi, Tal, Here is a quick way of getting around. First create two responses via dummy variables y1 - ifelse(y==a, 1, 0) y2 - ifelse(y==b, 1, 0) and then built two separate tree models for y1 and y2 separately. Hope it helps. Xiaogang On Tue, Dec 14, 2010 at 8:33 AM, Tal Galili tal.gal...@gmail.com wrote: Hi dear R-help memebers, When building a CART model (specifically classification tree) using rpart, it is sometimes obvious that there are variables (X's) that are meaningful for predicting some of the outcome (y) variables - while other predictors are relevant for other outcome variables (y's only). *How can it be estimated, which explanatory variable is used for which of the predicted value in the outcome variable?* Here is an example code in which x2 is the only important variable for predicting b (one of the y outcomes). There is no predicting variable for c, and x1 is a predictor for a, assuming that x2 permits it. How can this situation be shown using the an rpart fitted model? N - 200 set.seed(5123) x1 - runif(N) x2 - runif(N) x3 - runif(N) y - sample(letters[1:3], N, T) y[x1 .5] - a y[x2 .1] - b fit - rpart(y ~ x1+x2) fit2 - prune(fit, cp= 0.07) plot(fit2) text(fit2, use.n=TRUE) Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[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. -- == Xiaogang Su, Ph.D. Associate Professor, Statistician School of Nursing, University of Alabama Birmingham, AL 35294-1210 (205) 934-2355 [Office] x...@uab.edu xiaogan...@gmail.com http://homepage.uab.edu/xgsu/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] from table to matrix
Maybe I can help here: In OS X at least you can open the R Package Installer from the Packages Data menu, select a mirror, etc, then type ecodist into the search bar and then Install Selected... this will install the ecodist package on your machine. Not exactly sure how it would be done under other operating systems but that's the basic idea. Jonathan -- View this message in context: http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088187.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] Modifying values outside a function using apply.
I am sure this si a simple problem but the solution is evading me.
I have a list of matrices all with the same number of columns but different
number of rows. The first two columns label the row. The labels are
allways the same for the same row numbers, just some matricies have more
rows.
For example using 3 column matrices...
q.1 - function(r){return(cbind(seq(0, 10, by=1)[1:r], seq(10, 30,
by=2)[1:r], runif(r)))}
sapply(q, q.1)
[[1]]
[,1] [,2] [,3]
[1,]0 10 0.5399220
[2,]1 12 0.1551015
[3,]2 14 0.9664470
[[2]]
[,1] [,2] [,3]
[1,]0 10 0.09678172
[2,]1 12 0.75177116
[3,]2 14 0.59927159
[4,]3 16 0.18472215
[[3]]
[,1] [,2] [,3]
[1,]0 10 0.6343689
[2,]1 12 0.8121039
Given such a list I would like to create a matrix:
0 10 mean(ThisCol for this row)
1 12 mean(ThisCol for this row)
2 14 mean(ThisCol for this row)
3 16 mean(ThisCol for this row)
I can loop using a for loop but I would like to use apply but I have no idea
how to get it to work. If I could pass arguments by reference to a function
it would be easy but as far as I can tell there is only pass by value.
cheers
Worik
[[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] Installing R-packages in Windows
Date: Tue, 14 Dec 2010 23:48:20 +0100 From: er...@phonetik.uni-muenchen.de To: marchy...@hotmail.com CC: r-help@r-project.org; lig...@statistik.tu-dortmund.de Hi Mike, Hi Uwe, thanks for your reply. The main thing is that we have mixed OSs. On the one hand we have Linux, on the other WinXP. I need the same version of packages and R on both systems. Otherwise I cant explain my colleagues why error messages are reproduceable on only one system. Installing by compiling the .tar.gz-files does work on Linux, but not on Win because make is not avaivable :( I already tried. If you want to do anything on 'dohs where you want to use the computer to automate data processing, get something like cygwin- that was probably where my make came from. You can put that on all the relevant computers and disto a script or something, that should be allowable. Machine specific failures I would think would be quite common in most settings in any case. The disadvantage of install.packages is (if I am not right tell me please), that the package is only installed for the local user (and its home-Directory ) who is using R at the moment. In my case I need to install for all local users. Is there a way to do that automated with install.packages ? Well, without actually knowing myself I would again just suggest giving people a script or if it is just one machine that you use install multiple times/copies even if it is a bit of a waste and could create inconsistencies. In any case, however, you should be able to pick a library directory for R. I've never tried to point mine but presumably all the installs could point to the same one. Permissions may need to be modified but usually these things are read-execute world so it may not be a factor. I might think about Uwe's proposal to install the latest R with the latest packages. But there's still that problem with installing for ALL users :S I am unable to do that - even if I can use a local repo. Any ideas? Thanks in advance Erik On 12/14/2010 06:17 PM, Mike Marchywka wrote: ( hotmail just randomly decides not to prefix original text, my comments below) Date: Tue, 14 Dec 2010 17:36:11 +0100 From: er...@phonetik.uni-muenchen.de To: r-help@r-project.org Hi there, I have the following problem and I hope somebody might help me. First of all: I am using WinXP SP3 (english and/or german) with R in Version 2.10.0. Now I am trying to install some packages but unfortunately I am getting a weird error. No matter which package I am trying to install - I nearly get the same error. It looks like this: - C:\Program Files\R\R-2.10.0\binR CMD INSTALL %SystemDrive%\rPAcsTemp\car_2.0-2. zip Warning in rawToChar(block[seq_len(ns)]) : truncating string with embedded nul: M:: I used to build all from source then that stopped working and I have seen others here complain about installation failures. I have had good luck with what others have suggested, and use install.packages() It goes into a script ok, for example, ( note that myR is my script that invokes R but you can change etc). This points to ucla mirror, writes a script into $tf to install $pkf, installs $pkf by running the gerated script, and then shows you the results and deletes junk ( normally you want to check error codes before deling junk however) tf=$$_temp cat - DUMMTY $tf options(repos=c(http://cran.stat.ucla.edu;)) install.packages(c($pkf),dep=TRUE) DUMMTY echo executing cat $tf myR -run $tf echo removing $tf rm $tf echo removing R output cat $tf.Rout rm $tf.Rout __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Discriminant Correspondence Analysis
Wayne, So far, no one has said the obvious: Please do work your way through (or at least skim) An Introduction to R which you'll find right there on your computer under Help/Manuals. Your questions indicate that you have not yet done so. Do it, it really will pay off. Peter Ehlers On 2010-12-14 12:36, Wayne Sawtell wrote: Hello everyone, I am totally new to the R program. I have had a look at some pdf documents that I downloaded and that explain how to do many things in R; however, I still cannot figure out how to do what I want to do, which is to perform Discriminant Correspondence Analysis on a rectangular matrix of data that I have in an Excel file. I know R users frown upon Excel and recommend converting Excel files to .csv format, which I have done, no problem. That is not an issue. There are several parts to my problem. 1) When I try the read.table command, even if I include the directory name in the filename, R still cannot read the file, even if it is in .csv format 2) I was able to copy my file and then read the clipboard contents into R but then I do not know to assign a name to the data frame in order to conduct any operations on it 3) I need the ADE4 program in order to perform Discriminant Correspondence Analysis, so I used the install.packages command to install it. It installed no problem but I do not know how to access the ADE4 program in R. I am unable to open it directly, either. 4) I thought that using the ADE4 GUI (called ade4TkGUI) would be easier because I do not know many of the R commands; but, again, I downloaded it but cannot open or access it. The following is the suggested coding that I found through the R website, but when I try to use this code, I don't know how to assign a name for the df, or what to put for fac, and what is worse, I get an error message saying that the program cannot find the discrimin.coa command. Usage discrimin.coa(df, fac, scannf = TRUE, nf = 2) Arguments df a data frame containing positive or null values fac a factor defining the classes of discriminant analysis scannf a logical value indicating whether the eigenvalues bar plot should be displayed nf if scannf FALSE, an integer indicating the number of kept axes Examples data(perthi02) plot(discrimin.coa(perthi02$tab, perthi02$cla, scan = FALSE)) For clarification, my data consists of measurements of morphological characters of an assemblage of biological specimens. I have already performed Principal Co-ordinates Analysis, Principal Compionents Analysis and Cluster Analysis in another program (PAST) in order to see if the data fall into distinct groupings that might represent different morphological species. I now want to test the groupings that I found on my test data set using Discriminant Correspondence Analysis.There are both continuous and categorical characters, which is the reason why I need to perform Discriminant Correspondence Analysis, instead of Linear Discriminant Analysis, which is only valid for continuous measurements. R seems to be the only program in which I can perform Discriminant Correspondence Analysis. Thanks for any help offered on any of these points. Wayne [[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] lattice fun: multiple themes in panels with spplot()
I have some geospatial data where two layers are thematic and the third is a percentage, so the maps need to have different themes. thumbDf - as( stack( thumb), SpatialGridDataFrame) names(thum...@data) - c(pri, sec, pct) thum...@data$pri - factor(thum...@data$pri, levels=c(0:8), labels= names( peelClasses)) thum...@data$sec - factor(thum...@data$sec, levels=c(0:8), labels= names( peelClasses)) stack() comes from library(raster), which does not do lattice natively and doesn't plot thematic maps particularly well, AFAIK. Coercing the data to something that spplot() can handle and changing the thematic data to factors results in a very satisfactory legend for that theme. I would like the third map to have the same size so that they more closely resemble a set of panels even though they have different themes. I have experimented with three approaches: thumbDfPlot - spplot(thumbDf, c(pri, sec), names.attr=c(Primary cover, Secondary cover), between= list( x=0.5)) trellis.par.set( peelTheme) plot( thumbDfPlot, split= c(1,1,2,1), more=TRUE) trellis.par.set(sp.theme()) plot( spplot(thumbDf, pct), split= c(2,1,2,1)) The second plot (showing the third panel) is taller because it's legend is narrower and together they have half of the page to work with. Also, it does not have a panel title showing it's name (pct unless I give it something else, which I cannot figure out how to do for a plot of one panel. trellis.par.set( peelTheme) plot( spplot( thumbDf, pri), split= c(1,1,3,1), more=TRUE) plot( spplot( thumbDf, sec), split= c(2,1,3,1), more=TRUE) trellis.par.set(sp.theme()) plot( spplot(thumbDf, pct), split= c(3,1,3,1), position= c(0,0.25,1,0.75)) Doing it this way, each of the first two gets its own legend, which is not what I want either, but at least none of them has a panel title, so I could give them each a plot title. The size is close, but not quite right. I could probably get it pretty close changing the position vector itertively, but that is not very satisfying. trellis.par.set( peelTheme) plot( thumbDfPlot, position= c(0,0,0.625,1), more=TRUE) trellis.par.set(sp.theme()) plot( spplot(thumbDf, pct, names.attr=Confidence), position=c( 0.625,0,1,1)) Sizing is closer than the first, but the same basic issues are still in play, plus names.attr has no effect. There must be a way to link the panel sizes from one plot to the next in case something else changes. Is there a way to turn off the legend and plot it separately, extracting it from the original spplot() object? Is it possible to specify the theme panel by panel? I could post some images somewhere if it's helpful. Any suggestions on a free, public tool for doing that? Thanks for taking a look at this. Climbing the lattice learning curve, Neil -- View this message in context: http://r.789695.n4.nabble.com/lattice-fun-multiple-themes-in-panels-with-spplot-tp3088225p3088225.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] survfit
Hello R helpers: *My first message didn't pass trough filter so here it's again* I would like to obtain probability of an event for one single patient as a function of time (from survfit.coxph) object, as I want to find what is the probability of an event say at 1 month and what is the probability of an event at 80 months and compare. So I tried the following but it fails miserably. I looked at some old posts but could not figure out the solution. Here's what I did where for one single patient, the answer is NULL. Could anyone help me with the problem? R.app GUI 1.34 (5589 Leopard build 32-bit) library(glmpath) * dataall - list(x=lung.data$x[1:130,], time=lung.data$time[1:130], status=lung.data$status[1:130])* * fit.a - coxpath(dataall) testall - list(x=lung.data$x[131:137,], time=lung.data$time[131:137], status=lung.data$status[131:137]) testpred - predict(fit.a, testall, s=0.99, type='coxph', mode='lambda.fraction') testpred Call: coxph(formula = Surv(time, status) ~ x, method = object$method) coef exp(coef) se(coef) zp karno -0.00756 0.992 0.0364 -0.208 0.84 Likelihood ratio test=0.89 on 1 df, p=0.344 n= 7 newd1 - list(testall$x[1,]) survtest - survfit(testpred,newdata=newd1,individual=T) survtest Call: survfit(formula = testpred, newdata = newd1, individual = T) records n.max n.start events median 0.95LCL 0.95UCL 7 0 0 0 0 0 0 summary(survtest) Call: survfit(formula = testpred, newdata = newd1, individual = T) time n.risk n.event survival std.err lower 95% CI upper 95% CI survtest$survival NULL *** [[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] from table to matrix
I actually got it to work now using levelplot, and I also worked out how to get the graph I wanted to display, I had to assign the graph to a reference and then print it -- yes, that simple! Now I just need to work out why my levelplot doesn't look quite right, even though the data is in there... and I should be good to go! Hope that you got your stuff to work Jessica, and thanks to everyone for your help :-) I'm sure I will be posting here again very soon! Jonathan -- View this message in context: http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088230.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] survival: ridge log-likelihood workaround
You ask some good questions. I would like to clarify statistics that ridge coxph returns. Here is my understanding and please correct me where I am wrong. 1) In his paper Gray [JASA 1992] suggests a Wald-type statistics with the formula for degree of freedom. The summary function for ridge coxph returns the degree of freedom and Wald test which are equivalent to what Gray wrote. Not quite. The printout contains se and se2. The first is the suggestion of Verwiej and Van Houwlingen and the second is the suggestion of Gray. The first error estimate is always larger although usually very close to the second. However, I have seen several data sets where the second collapses and produces a tiny variance estimate. Thus the Wald statistic is based on the Verweij estimate. 2) Summary for ridge coxph prints a likelihood ratio test which is not, if I may say, a proper likelihood ratio test. First it is based on unpenalized log-likelihoods and the above defined degree of freedom is used. I accept that there is nothing which suggests that one should use penalized log-likelihoods. However, there is also nothing published which suggests that unpenalized log-likelihoods should be used with the above defined degree of freedom. I have found that TherneauGrambsch in their excellent book discuss this in a paragraph and mention that the p-value thus returned is somewhat conservative (p too large). Therefore, the likelihood ratio test that ridge coxph returns is not a true one and the statistics returned (i.e. 2*(loglik(beta)-loglik(0))) has the distribution which is somewhat more compact than the chi-square. I like conservative p-values and like to be on a safe side. However, in my work Wald test p-values for ridge regression are much higher than the likelihood ratio test's p-value and I don't get impression that they are conservative. There is literature, but it's been 10 years and I don't remember the references. One key issue, not documented I admit, is that for the difference in LR between two models to be valid at all, they have to have the same penalty parameter. I have almost never compared two penalized LR to each other, likely because I rarely use ridge(). More often I am comparing the LR of a penalized model to one without the penalty term, for which the issue does not arise. The same type of issue will arise, I believe, if one were to fit two mixed effects models that differed only in the fixed effects, and then used an LR test. Comparisons are only valid if two have the same variance values, or else the models are not nested. 3) There is no efficient score test for ridge regression, as there is no penalized efficient score test. That is OK. 4) Rsquare and max possible are returned and I have so far failed to find exact references for them. I was sure that was somewhere, but I can't find it either. The manual page summary.coxph.object fell off my list somewhere. The R^2 value is that of Nagelkirke. At the time I wrote that code it was a simple estimate of R^2. There have since been several other proposals, and a recent review by Schemper suggests that the Nagelkirke method may be nearly the worst of the bunch. Replacing this is another item on my to-do list. I would like to add a note here that the coxph algorithm works really fast with high-dimensional covariates and all compliments to people who developed it. There was evidently a lot of effort put to make it all work fast and correctly and, in my opinion, it is a shame that the last bit - summary for ridge coxph - is a bit, if I may say, shaky. In my opinion, only Wald test and degree of freedom as defined by Gray deserve to be part of summary for ridge coxph and I look forward to be corrected. On my behalf I am prepared in my spare time to write the code so that the summary for ridge coxph does not return NULL and that the statistics printed in summary for ridge coxph are based on published papers. This is the hard issue. The penalized methods addition to coxph was designed so that a new method could be added without any changes at all to the coxph (or survreg) code. The primary reason was to allow users to add arbitrary methods. It turned out that no one ever has done so, at least none that I have ever heard of, so that particular design goal now looks silly. A consequnce of the design is that the basic printouts for the model consist of a constant piece + an addition bit supplied by the inserted method. There are others: the user has to know whether a particular penalty plug in is appropriate for survreg, coxph, or both. This is true for pspline and ridge, but most of the parameter combinations for frailty only work with coxph. It is in the documentation for fraily, but has nevertheless caused many a user to stumble. I think that it should be rewritten, but this is a formidable task. Note: the people who developed the survival package is --- me, and
Re: [R] lattice fun: multiple themes in panels with spplot()
I figured out how to toggle the legend for the second scenario, but now I have three different map sizes! trellis.par.set( peelTheme) plot( spplot( thumbDf, pri, colorkey= list( space= left)), split= c( 1, 1, 3, 1), more= TRUE) plot( spplot( thumbDf, sec, colorkey= FALSE), split= c( 2, 1, 3, 1), more= TRUE) trellis.par.set( seqTheme) plot( spplot( thumbDf, pct, at= seq( 0, 100, by= 5), split= c( 3, 1, 3, 1), position= c( 0, 0.25, 1, 0.75)) -- View this message in context: http://r.789695.n4.nabble.com/lattice-fun-multiple-themes-in-panels-with-spplot-tp3088225p3088244.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] Urgent help requested using survfit(individual=T):
Hello: I would like to obtain probability of an event for one single patient as a function of time (from survfit.coxph) object, as I want to find what is the probability of an event say at 1 month and what is the probability of an event at 80 months and compare. So I tried the following but it fails miserably. I looked at some old posts but could not figure out the solution. Here's what I did where for one single patient, the answer is NULL. Could anyone help me with the problem? R.app GUI 1.34 (5589 Leopard build 32-bit) library(glmpath) * dataall - list(x=lung.data$x[1:130,], time=lung.data$time[1:130], status=lung.data$status[1:130])* * fit.a - coxpath(dataall) testall - list(x=lung.data$x[131:137,], time=lung.data$time[131:137], status=lung.data$status[131:137]) testpred - predict(fit.a, testall, s=0.99, type='coxph', mode='lambda.fraction') testpred Call: coxph(formula = Surv(time, status) ~ x, method = object$method) coef exp(coef) se(coef) zp karno -0.00756 0.992 0.0364 -0.208 0.84 Likelihood ratio test=0.89 on 1 df, p=0.344 n= 7 newd1 - list(testall$x[1,]) survtest - survfit(testpred,newdata=newd1,individual=T) survtest Call: survfit(formula = testpred, newdata = newd1, individual = T) records n.max n.start events median 0.95LCL 0.95UCL 7 0 0 0 0 0 0 summary(survtest) Call: survfit(formula = testpred, newdata = newd1, individual = T) time n.risk n.event survival std.err lower 95% CI upper 95% CI survtest$survival NULL * [[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] Multivariate binary response analysis
Dear Derek, dear list, A couple suggestions (and more) : 0) check that your problem can't be expressed in terms of what glmer() (lme4(a|b)? package(s)) can solve. I use this often, and am quite impressed with its abilities. 1) Why not modeling that directly in BUGS ? That might not be easy, will almost certainly not be fast to converge and will require lots of post- convergence analysis but offers a direct solution. Current BUGS implementations (OpenBUGS 3.1.x, JAGS 2.x) are well- interfaced to R. My current favorite is JAGS + rjags, but this preference might come from years of frustration about (Win|Open)BUG on Wine on Linux, and the current offering of OpenBUGS + BRugs, currently in beta on the Community page of the OpenBUGS website, seems quite decent. Memory use might have been a problem in the (not so) good old days, but in a era of multigigabyte netbooks, I doubt this is still relevant. Computation time is more of a concern : although Lunn al. (2009)(1) allude to a parallelized version of WinBUGS(2), I am not aware of any publicly-accessible parallelized BUGS interpreter, which is a shame in times where even said netbooks get multicore CPUs ...). 2) The nice MCMCglmm package is able to run a large subclass of DAG Bayesian models expressible as generalizations of (possibly multidependent) multivaried mixed-model GLMs, *way* faster than current BUGS implementations (the author claims an acceleration factor of about 30 on his test problem, and my (limited) experience does not disprove him). It seems (to me) that your problem *might* be expressible in terms palatable to MCMCglm(). Be aware, however, that using this package will *require* some serious understanding of the author's notation used in his (excellent) documentation. Furthermore, if you're interested in the random effects values, you're SOL : I have not (yet) been able to retrieve traces of their distributions in the resultant object. But again, my understanding of this package is limited. My personal bias goes to BUGS, as you might have guessed, but the public health warning that the Spiegelhalter gang put in front of the WinBUGS manual still apply : MCMC sampling can be dangerous. To which I add model checking, convergence assessment and validation might eat more time than modeling itself. HTH, Emmanuel Charpentier -- 1. David Lunn et al., “Rejoinder to commentaries on ‘The BUGS project: Evolution, critique and future directions’,” Statistics in Medicine 28, no. 25 (11, 2009): 3081-3082. 2. http://www.page-meeting.org/pdf_assets/6232-PARALLEL% 20Girgis_POSTER_final.pdf. Accessed on Dec, 14, 2010. On Mon, 13 Dec 2010 15:31:53 -0800, Janszen, Derek B wrote : Greetings ~ I need some assistance determining an appropriate approach to analyzing multivariate binary response data, and how to do it in R. The setting: Data from an assay, wherein 6-hours-post-fertilization zebrafish embryos (n=20, say) are exposed in a vial to a chemical (replicated 3 times, say), and 5 days later observed for the presence/absence (1/0) of defects in several organ systems (yolk sac, jaw, snout, body axis, pericardial edema, etc.) for each fish. The assay can be used as a screen (in which case a single response 'any' is first analyzed) as well as for comparing the responses of different classes of chemicals (a MANOVA-type analysis). Interest is focused on where response(s) are occurring, any associations among responses, and ultimately on possible biological mechanisms (the fish are tested for behavioral activity prior to this assay, and then ground up to provide RNA for microarray assays. A-lotta-data!). What I *wish* I could do is something like glm(response.matrix ~ treat/vial, family=binomial(logit), data=zf.dat) but I know this can't be done. I came across the baymvb (Bayesian analysis of multivariate binary data) package in the R contributed packages archives, but it is no longer supported and the author is incommunicado. In the baymvb function the model is specified as single.stacked.response ~ structure.factor + linear.model.terms. Huh? This looks suspiciously similar to analyzing repeated measures data in SAS as a univariate response with a split-plot structure (which forces the response covariance matrix to have a compound symmetric structure). If this is what's happening with this function it is definitely not appropriate. How about a GEE approach (I'm not familiar with the liter ature, or with any of the R packages that implement it)? Any other recommendations? (NB: just discovered the bayesm package. Don't know if this will work for me or not.) Any help would be greatly appreciated. Derek Janszen, PhD Sr Research Biostatistician Computational Biology Bioinformatics Pacific Northwest National Laboratory Richland, WA 99352 USA derek.jans...@pnl.gov [[alternative HTML version deleted]]
[R] Significance Help
I have a set of results where I have the eyesight power of both left and right eyes for each participant (e.g. 0.75, 0.5). Each participant then had to throw basketballs into a hoop and the number of successful throws was recorded. I would like to do two things: 1. Test whether the eyesight power of the left and right eyes are significantly different from each other. 2. Test whether the number of successful throws is significantly affected by eyesight power (i.e. is there a difference between those people who have similar powers in left and right eyesight, to those who have significantly different powers?) I know that I may not have explained this particularly well, but I'm totally stumped as to where to start! I have only had basic lessons with R and now have to do this report! Many thanks, Samaire [[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] Installing R-packages in Windows
Hi Mike, Hi Uwe, thanks for your reply. The main thing is that we have mixed OSs. On the one hand we have Linux, on the other WinXP. I need the same version of packages and R on both systems. Otherwise I cant explain my colleagues why error messages are reproduceable on only one system. Installing by compiling the .tar.gz-files does work on Linux, but not on Win because make is not avaivable :( I already tried. The disadvantage of install.packages is (if I am not right tell me please), that the package is only installed for the local user (and its home-Directory ) who is using R at the moment. In my case I need to install for all local users. Is there a way to do that automated with install.packages ? I might think about Uwe's proposal to install the latest R with the latest packages. But there's still that problem with installing for ALL users :S I am unable to do that - even if I can use a local repo. Any ideas? Thanks in advance Erik On 12/14/2010 06:17 PM, Mike Marchywka wrote: ( hotmail just randomly decides not to prefix original text, my comments below) Date: Tue, 14 Dec 2010 17:36:11 +0100 From: er...@phonetik.uni-muenchen.de To: r-help@r-project.org Subject: [R] Installing R-packages in Windows Hi there, I have the following problem and I hope somebody might help me. First of all: I am using WinXP SP3 (english and/or german) with R in Version 2.10.0. Now I am trying to install some packages but unfortunately I am getting a weird error. No matter which package I am trying to install - I nearly get the same error. It looks like this: - C:\Program Files\R\R-2.10.0\binR CMD INSTALL %SystemDrive%\rPAcsTemp\car_2.0-2. zip Warning in rawToChar(block[seq_len(ns)]) : truncating string with embedded nul: M:: I used to build all from source then that stopped working and I have seen others here complain about installation failures. I have had good luck with what others have suggested, and use install.packages() It goes into a script ok, for example, ( note that myR is my script that invokes R but you can change etc). This points to ucla mirror, writes a script into $tf to install $pkf, installs $pkf by running the gerated script, and then shows you the results and deletes junk ( normally you want to check error codes before deling junk however) tf=$$_temp cat - DUMMTY $tf options(repos=c(http://cran.stat.ucla.edu;)) install.packages(c($pkf),dep=TRUE) DUMMTY echo executing cat $tf myR -run $tf echo removing $tf rm $tf echo removing R output cat $tf.Rout rm $tf.Rout __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] from table to matrix
Thank you guys I learnt a lot . But when I tried to run the library(ecodist) function, R says there is no package called ecodist. why? __ View message @ http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088184.html To unsubscribe from from table to matrix, visit http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=3087972code=ci1oZWxwQHItcHJvamVjdC5vcmd8MzA4Nzk3MnwtNzg0MjM1NTA4 [[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] from table to matrix
Hi Jonathan,thanks a lot! Mine is windows operating system so I will try other ways . About your question, do you want to get a matrix like this: 0 1 2 3 ...1817393 5524385 0687 0 64 0 71 0 55 ... 579602 1 713515 1 Are you intending to get a matrix like above, eg. only the diagonal position are filled? __ View message @ http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088223.html To unsubscribe from from table to matrix, visit http://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=3087972code=ci1oZWxwQHItcHJvamVjdC5vcmd8MzA4Nzk3MnwtNzg0MjM1NTA4 [[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] Modifying values outside a function using apply.
Try this:
l - sapply(c(4, 3, 2), q.1)
aggregate(V3 ~ V1 + V2, as.data.frame(do.call(rbind, l)), mean)
On Tue, Dec 14, 2010 at 9:19 PM, Worik R wor...@gmail.com wrote:
I am sure this si a simple problem but the solution is evading me.
I have a list of matrices all with the same number of columns but different
number of rows. The first two columns label the row. The labels are
allways the same for the same row numbers, just some matricies have more
rows.
For example using 3 column matrices...
q.1 - function(r){return(cbind(seq(0, 10, by=1)[1:r], seq(10, 30,
by=2)[1:r], runif(r)))}
sapply(q, q.1)
[[1]]
[,1] [,2] [,3]
[1,]0 10 0.5399220
[2,]1 12 0.1551015
[3,]2 14 0.9664470
[[2]]
[,1] [,2] [,3]
[1,]0 10 0.09678172
[2,]1 12 0.75177116
[3,]2 14 0.59927159
[4,]3 16 0.18472215
[[3]]
[,1] [,2] [,3]
[1,]0 10 0.6343689
[2,]1 12 0.8121039
Given such a list I would like to create a matrix:
0 10 mean(ThisCol for this row)
1 12 mean(ThisCol for this row)
2 14 mean(ThisCol for this row)
3 16 mean(ThisCol for this row)
I can loop using a for loop but I would like to use apply but I have no
idea
how to get it to work. If I could pass arguments by reference to a
function
it would be easy but as far as I can tell there is only pass by value.
cheers
Worik
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] googleVis 0.2.2 - Using the Google Visualisation API with R
Hi all, Version 0.2.2 of the googVis package has been released on CRAN and will be available from your local CRAN mirror in due course. googleVis provides an interface between R and the Google Visualisation API. The functions of the package allow users to visualise data stored in R with the Google Visualisation API without uploading their data to Google. We presented our initial ideas on motion charts at LondonR in early October. Since then we have developed the package further to include also interfaces to Google Maps, Geo Maps, Tables, Tree Maps and Annotated Time Lines. Combining R with the Google Visualisation API enables users to create powerful analysis tools which are easily accessible also to non-data analysts. The Statistics Relating to Lloyd's site (http://www.lloyds.com/stats) is an example which used the functions of the googleVis package to create a data visualisation page. The page shows how interactive charts can be used to engage with users, allowing them to identify trends and insights over time. For more information and examples visit the project web site: http://code.google.com/p/google-motion-charts-with-r/ Please feel free to send us an email rvisualisat...@gmail.com if you would like to be kept informed of new versions, or if you have any feedback, ideas, suggestions or would like to collaborate. Markus Gesmann, Diego de Castillo And now the NEWS: Version 0.2.2 [2010-12-12] == Changes o Fixed typos in documentation NEW FEATURES o New function: - createGoogleGadget which allows users to create Google Gadgets XML output Version 0.2.1 [2010-11-30] == Changes o First version to be released on CRAN NEW FEATURES o New function: - gvisAnnotatedTimeLine to generate interactive annotated time line charts Version 0.2.0 [2010-11-14] == Changes o The package has been renamed from GoogleMotionChart to googleVis to reflect a new more flexible implementation. o More functions of the Google Visualisation API are now available. USER-VISIBLE CHANGES o New interfaces, all visualisation functions start with 'gvis'. o Output is now of class 'gvis' with generic print and plot functions. o 'gvis' objects are list of lists, which allow the user to extract the various parts of the visualisation output, e.g. the chart object. NEW FEATURES o New functions: - gvisMotionChart to generate motion charts - gvisGeoMap to generate geographical maps - gvisMap to generate maps - gvisTreeMap to generate tree maps - gvisTable to generate table output - print.gvis: generic function to print 'gvis' objects - plot.gvis: generic function to display 'gvis' objects in a browser via the R.rsp package. [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] frestimate functionality
By any chance is there an R package that may contain functionality similar to the frestimate in Matlab/Simulink. Here is the URL for a description of the frestimate functionality: http://www.mathworks.com/help/toolbox/slcontrol/ug/frestimate.html Thank you again for any feedback. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to apitalize leading letters else of personal names?
Dear R world, Do you know about a function that would capitalize in the correct manner first and family names? I found in the cwhmisc only the CapLeading function, but it just does not do the job, taking care only to capitalize the first letter of a word. I am looking for a function that would recognize |'|Mc|- and capitalize the first letter following these characters. An example: names-c(jean-francois st-john,helene o'donnel, joe mcintyre) Desired result: Jean-Francois St-John Helene O'Donnel Joe McIntyre Thanks, Rock -- View this message in context: http://r.789695.n4.nabble.com/How-to-apitalize-leading-letters-else-of-personal-names-tp3088336p3088336.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] frestimate functionality
Have you done a google search, looked on CRAN, searched the archives, or looked at the task views? I can't be of any help, but maybe one of those places could be. HTH, Stephen On Tue, Dec 14, 2010 at 7:57 PM, Jason Rupert jasonkrup...@yahoo.com wrote: By any chance is there an R package that may contain functionality similar to the frestimate in Matlab/Simulink. Here is the URL for a description of the frestimate functionality: http://www.mathworks.com/help/toolbox/slcontrol/ug/frestimate.html Thank you again for any feedback. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick | Auburn University | | Department of Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849 | |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Significance Help
Hello Samaire, I don't know much about eyesight measurements (other than that my own would probably be 0.1, 0.2) but I'll attempt some suggestions... For objective (1) you will have to define what you mean by significantly different. If you had data on left/right values from a larger population to use as a standard you could then compare each subject's left - right difference to the population. For objective (2), it would be better if you had forced them to throw the ball more than 3 times ! Anyway, something like this to start with... glm( hoops ~ lr.diff, family=binomial ) where hoops is a two column matrix with number of successful throws in col 1 and number of misses in col 2; lr.diff is absolute value of left - right value. Hope this helps. Michael On 15 December 2010 08:21, Samaire Holden samair...@hotmail.co.uk wrote: I have a set of results where I have the eyesight power of both left and right eyes for each participant (e.g. 0.75, 0.5). Each participant then had to throw basketballs into a hoop and the number of successful throws was recorded. I would like to do two things: 1. Test whether the eyesight power of the left and right eyes are significantly different from each other. 2. Test whether the number of successful throws is significantly affected by eyesight power (i.e. is there a difference between those people who have similar powers in left and right eyesight, to those who have significantly different powers?) I know that I may not have explained this particularly well, but I'm totally stumped as to where to start! I have only had basic lessons with R and now have to do this report! Many thanks, Samaire [[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] selecting certain rows from data frame
Hi, if I have a dataframe such that ID Time Earn 1 1 10 1 2 50 1 3 68 2 1 40 2 2 78 24 88 3 1 50 3 2 60 3 3 98 4 1 33 4 2 48 44 58 . . Now if I have to select the all the rows from the data frame which does not include rows with certain IDs, say for example (prime) ID == 2 3, how do I do it Thanks Rith [[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.
