Re: [R] documentation of wts object returned by nnet() function
See the summary() method. It labels the weights for you. On Tue, 10 Jun 2008, wmg wrote: In MASS there is a nnet() function which returns, among other things, a wts[] vector giving the weights in the neural network. However, in neither the book nor in the online documentation for the routine can I find documentation of which element of wts[] goes with which link in the network---input unit to hidden layer unit, hidden layer unit to output unit, skip layer links, etc. Anybody know the answer to that, presumably in form of an algorithm for pairing them up? Thanks in advance. Regards, Will Grove U. of Minnesota Psychology Dept. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] computing and showing mean
Hello to everyone, I am trying to compute a mean and show it to the screen and then save to later be used as a boxplot.I have used the following code: dat-read.table(file=C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt,header=T,row.names=1) file.show(file=C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt) x-dat[2,23:46] y=mean(x,trim=0,na.rm=T) save(y,file=ydata) Am I computing the mean correctly and how to I show the value of the mean to the screen? [[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] Polynomial Goal Programming
Hello R,
Is there a package to perform Polynomial goal programming in R?
BR, Shubha
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] samr result
Yes, here it is:
samr.obj-samr(data,resp.type=Two class unpaired, nperms=100,
center.arrays=T)
where *data *is a matrix of microarray gene expressions with genes as rows
and tissues as columns. With putting *center.arrays=T *the *data* matrix is
normalized such as each column has median=0. I would like to retrieve the
new normalized matrix, but it seems that it is not returned by *samr.* If
you have any idea on how I can find this transformed matrix I would be glad
to hear that!
Thanks again,
E.
On Tue, Jun 10, 2008 at 11:30 PM, Richardson, Patrick
[EMAIL PROTECTED] wrote:
Could you post your code so we can see what you are trying to do?
Thanks,
Patrick
From: [EMAIL PROTECTED] [EMAIL PROTECTED] On
Behalf Of Eleni Christodoulou [EMAIL PROTECTED]
Sent: Tuesday, June 10, 2008 11:20 AM
To: r-help@r-project.org
Subject: [R] samr result
Hello list!
I have a proble trying to perform a SAM analysis using the function samr
from the samr package. I have put the option *center.arrays=TRUE *in order
to scale all the experiments to median=0. I would like to retrieved the
scaled data but it seems that samr does not return it...Does anyone have
any
idea on this?
Thanks a lot!!!
E.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
This email message, including any attachments, is for ...{{dropped: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] Comparing two groups of proportions
Hi Ivan, It was not clear from your original post that QA was a repeated factor. But your problem may be reframed very much like you would do with a McNemar chi-square: Just count the number of times both procedure give the same result, of each kind, and different results, again of both kinds, to get four counts by condition. These events are probably independent within your setting. You should then be able to test various binomial or Poisson models with the proper equality constraints. HTH, Yvonnick Noel, PhD University of Brittany at Rennes France Re: [R] Comparing two groups of proportions To: r-help@r-project.org Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=iso-8859-1 Hi Rolf, On Monday 09 June 2008 11:16:57 pm Rolf Turner wrote: Your approach tacitly assumes --- as did the poster's question --- that the probability of passing an item by one method is *independent* of whether it is passed by the other method. Which makes the methods effectively independent of the nature of the item being assessed! So it seems I can't just block my primary factor (QA procedure) by nuisance one (production line) and run Cochran test to see if effects of primary factor are identical for both its levels. Not much actual quality being assured there! In fact, I am not interested in quality of QA procedures as much as in how different the results are (error component). Thanks, Ivan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: newbie nls question
Hi [EMAIL PROTECTED] napsal dne 11.06.2008 00:56:19: I'm tyring to fit a relatively simple nls model to some data, but keep coming up against the same error (code follows): Oto=nls(Otolith ~ Linf*(1-exp(-k(AGE-to))), data = ages, start = list(Linf=1000, k=0.1, to=0.1), trace = TRUE) The error message I keep getting is Error in eval(expr, envir, enclos) : could not find function k. I've used this line of code for other nls models (with different data, parameter estimates, etc.), but have never gotten this error. The data is Does Oto=nls(Otolith ~ Linf*(1-exp(-k*(AGE-to))), data = ages, start = list(Linf=1000, k=0.1, to=0.1), trace = TRUE) help? Regards Petr AGE,Otolith,Scale 1,207.1052632,207.1052632 2,329.962963,332.7586207 3,401.9473684,406 4,422,413.111 5,452.6785714,458.34375 6,510.75,533 7,477,674 8,643,704 9,,615 10,695.5, 12,615, Are the missing values (e.g., no value for Otolith, AGE 9) having an effect? I am using Tinn-R as an editor I'm new to R, trying to get away from Rcmdr (though it has been helpful), and still trying to learn the language with the aid of several books on the R. A search of the R Archive did not prove fruitful. Thanks in advance, SR Steven H. Ranney Graduate Research Assistant (Ph.D) USGS Montana Cooperative Fishery Research Unit Montana State University PO Box 173460 Bozeman, MT 59717-3460 phone: (406) 994-6643 fax: (406) 994-7479 [[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] Matrix transformation problem
ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),ncol=3,byrow=T) x [,1] [,2] [,3] [1,]100 [2,]001 [3,]010 [4,]001 [5,]010 [6,]100 In the matrix above, on the first row, the 1 is in column 1, hence '1' on the first row in the matrix below. On the second row in the matrix above, the 1 is in column 3, hence the '3' on the second row in the matrix below. And so on... y [,1] [1,]1 [2,]3 [3,]2 [4,]3 [5,]2 [6,]1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Bayesian Analysis using the BUGS Language
DrakeGis wrote: WinBugs doesn't run in LINUX. Neither OpenBugs. Ah, sorry, haven't read carefully enough. The only possible thing is to run WinBUGS under wine and use R2WinBUGS, which supports the wine/WinBUGS combination. Best wishes, Uwe Ligges DrakeGis wrote: Hi all, I observed that the 'link' between R and BUGS (winBugs/linBugs) has totally disappeared. I'm wondering what package can be use to run bayesian models specified using the BUGS language in R (specifically under Linux). Is there any other option besides JAGS ? Thanks Ther are several package, if you want to use WinBUGS (not recommended), there is the CRAN package R2WinBUGS, and if you want to use OpenBUGS (the new version of WinBUGS), use BRugs, which is available from the CRAN extras repository hosted by Brian Ripley (which is a default repository for R for Windows, hence install.packages(BRugs) should work). Best wishes, Uwe Ligges - Stay ahead of the information curve. Receive EDA news and jobs on your desktop daily. Subscribe today to the EDA CafeNews newsletter. [ http://www10.EDACafe.com/nl/newsletter_subscribe.php ] It's informative and essential. This message was sent to you from a machine at 141.211.153.44 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Stay ahead of the information curve. Receive EDA news and jobs on your desktop daily. Subscribe today to the EDA CafeNews newsletter. [ http://www10.EDACafe.com/nl/newsletter_subscribe.php ] It's informative and essential. This message was sent to you from a machine at 141.211.153.44 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] difference between nlm and nlminb
I believe nlminb() performs *constrained* optimization, where as nlm() is for *unconstrained* opimization So I guess nlm() is for solving min(f[a,b]), and nlminb() min(f[a,b]) given a+b = c FYI I think optim() also does constrained optimization, well I've used for min(f[a,b]) given a = a* and b = b*. David ae2356 wrote: Hi, I was wondering if someone could give a brief, big picture overview of the difference between the two optimization functions nlm and nlminb. I'm not familiar with PORT routines, so I was hoping someone could give an explanation. Thanks, Angelo _ Instantly invite friends from Facebook and other social networks to join yo https://www.invite2messenger.net/im/?source=TXT_EML_WLH_InviteFriends [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/difference-between-nlm-and-nlminb-tp17769859p17772440.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] Matrix transformation problem
try this: x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0), ncol = 3, byrow = TRUE) which(x == 1, arr.ind = TRUE)[, col, drop = FALSE] I hope it helps. Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: [EMAIL PROTECTED] To: r-help@r-project.org Sent: Wednesday, June 11, 2008 10:10 AM Subject: [R] Matrix transformation problem ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),ncol=3,byrow=T) x [,1] [,2] [,3] [1,]100 [2,]001 [3,]010 [4,]001 [5,]010 [6,]100 In the matrix above, on the first row, the 1 is in column 1, hence '1' on the first row in the matrix below. On the second row in the matrix above, the 1 is in column 3, hence the '3' on the second row in the matrix below. And so on... y [,1] [1,]1 [2,]3 [3,]2 [4,]3 [5,]2 [6,]1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Matrix transformation problem
sorry, my previous answer was not correct; you need: x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0), ncol = 3, byrow = TRUE) which(t(x == 1), arr.ind = TRUE)[, row, drop = FALSE] Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: [EMAIL PROTECTED] To: r-help@r-project.org Sent: Wednesday, June 11, 2008 10:10 AM Subject: [R] Matrix transformation problem ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),ncol=3,byrow=T) x [,1] [,2] [,3] [1,]100 [2,]001 [3,]010 [4,]001 [5,]010 [6,]100 In the matrix above, on the first row, the 1 is in column 1, hence '1' on the first row in the matrix below. On the second row in the matrix above, the 1 is in column 3, hence the '3' on the second row in the matrix below. And so on... y [,1] [1,]1 [2,]3 [3,]2 [4,]3 [5,]2 [6,]1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Matrix transformation problem
you may try a matrix multiplication, which has a very high performance in R x%*%1:ncol(x) hth. [EMAIL PROTECTED] schrieb: ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),ncol=3,byrow=T) x [,1] [,2] [,3] [1,]100 [2,]001 [3,]010 [4,]001 [5,]010 [6,]100 In the matrix above, on the first row, the 1 is in column 1, hence '1' on the first row in the matrix below. On the second row in the matrix above, the 1 is in column 3, hence the '3' on the second row in the matrix below. And so on... y [,1] [1,]1 [2,]3 [3,]2 [4,]3 [5,]2 [6,]1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/42803-8243 F ++49/40/42803-7790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Bayesian Analysis using the BUGS Language
On Wed, 11 Jun 2008, Uwe Ligges wrote: DrakeGis wrote: WinBugs doesn't run in LINUX. Neither OpenBugs. Ah, sorry, haven't read carefully enough. The only possible thing is to run WinBUGS under wine and use R2WinBUGS, which supports the wine/WinBUGS combination. 'Linux' is too vague. AFAIK none of this runs on Linux on Sparc or PPC for example. Tobias Verbeke referred to a thread which is AFAICS about building an ix86 Linux bugs executable from the distributed brugs.so in OpenBUGS. If you can get that to run (and like Uwe, I get a trap), you can then communicate with that bugs executable via files (in the same way that R2WinBUGS communicates with WinBUGS). It is not clear to me that the Windows classicbugs.exe or winbugs.exe in OpenBUGS could not be run under Wine with communication via files, but R2WinBUGS is not set up to do so. Best wishes, Uwe Ligges DrakeGis wrote: Hi all, I observed that the 'link' between R and BUGS (winBugs/linBugs) has totally disappeared. I'm wondering what package can be use to run bayesian models specified using the BUGS language in R (specifically under Linux). Is there any other option besides JAGS ? Thanks Ther are several package, if you want to use WinBUGS (not recommended), there is the CRAN package R2WinBUGS, and if you want to use OpenBUGS (the new version of WinBUGS), use BRugs, which is available from the CRAN extras repository hosted by Brian Ripley (which is a default repository for R for Windows, hence install.packages(BRugs) should work). Best wishes, Uwe Ligges - Stay ahead of the information curve. Receive EDA news and jobs on your desktop daily. Subscribe today to the EDA CafeNews newsletter. [ http://www10.EDACafe.com/nl/newsletter_subscribe.php ] It's informative and essential. This message was sent to you from a machine at 141.211.153.44 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Stay ahead of the information curve. Receive EDA news and jobs on your desktop daily. Subscribe today to the EDA CafeNews newsletter. [ http://www10.EDACafe.com/nl/newsletter_subscribe.php ] It's informative and essential. This message was sent to you from a machine at 141.211.153.44 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: computing and showing mean
Hi [EMAIL PROTECTED] napsal dne 11.06.2008 08:41:27: Hello to everyone, I am trying to compute a mean and show it to the screen and then save to later be used as a boxplot.I have used the following code: dat-read.table(file=C:\\Documents and Settings\\Owner\\My Documents\\Yeast\ here you get data frame \Yeast.txt,header=T,row.names=1) file.show(file=C:\\Documents and Settings\\Owner\\My Documents\\Yeast\\Yeast.txt) x-dat[2,23:46] x is data frame consisting from row 2 and columns 23:46 of data frame dat y=mean(x,trim=0,na.rm=T) y is named vector numerically the same as x (you compute mean of one value) if you want to see how your objects look like just print them on your console by simply typing x or y. If you want to see internal structure do str(x) or str(y) save(y,file=ydata) Am I computing the mean correctly and how to I show the value of the mean to the screen? Maybe its time to have a small glimpse into some basic documentation like R-intro. I may guess you want mean of columns 25:46. If this is true, then colMeans(dat[, 23:46]) prints means for defined columns. And boxplot(dat[, 23:46]) gives you boxplots for each column. But enough of speculation what do you really want. Regards Petr [[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] Fwd: Hello
Hii The contour plot is somewhat similar to what I want. I am attaching the data here. What I want to do is divide all the three variable( Recency, frequency,and monetory ) in 5 intervals based on their values.That will make 25 rectangle(5(recency)X 5(frequency) boxes on heat plot. and then plot a heat map with recency on X axis and frquency on Y axis, and also I want to show the no. of elements in each box( total 25) by heat colours i.e increasing no. of elements with the darker colour box. And one more plot with the same variable on XY as mentioned above but this time box colour depicting the average monetary value of that box. Hope this clears the problem I am facing. Thank you. Regards, Sumit On 6/11/08, Petr PIKAL [EMAIL PROTECTED] wrote: sumit gupta [EMAIL PROTECTED] napsal dne 11.06.2008 07:53:54: Hello thanx for the info.. I still have tthe doubt regarding the heat map. I am attching a plot of a 70X3 data matrix.In which 3rd variable has been shown as colour of different boxes.Could you help me to draw this type of graph using R Well, as you did not provided some data mat-cbind(sample(1:5, 70, replace=T), sample(1:5, 70, replace=T), rnorm(70)) dim(mat) [1] 70 3 library(akima) filled.contour(interp(mat[,1], mat[,2], mat[,3], duplicate=mean)) image(interp(mat[,1], mat[,2], mat[,3], duplicate=mean)) is this what you want? Regards Petr Regards, Sumit On 6/10/08, Petr PIKAL [EMAIL PROTECTED] wrote: Hi [EMAIL PROTECTED] napsal dne 10.06.2008 14:49:55: Hello, I am facing a problem in drawing heat map using R. I have a 70X3 matrix and I want to draw a heat map with 1 coloumn on X axis another on Y axis and want to show the value of 3rd coloumn using heat colours. Could you please help me with this . You can use function interp from akima package Petr Thank you . Regards, Sumit [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sumit Kumar Gupta 4thYear ,Biotechnology IIT Madras [příloha rfm.bmp odstraněna uživatelem Petr PIKAL/CTCAP] -- Sumit Kumar Gupta 4thYear ,Biotechnology IIT Madras -- Sumit Kumar Gupta 4thYear ,Biotechnology IIT Madras Recency Frequency Monetary 562 2 4795.5 570 2 1200 578 3 1719 585 2 728 575 3 2326. 583 1 1822 612 2 9508.9637 578 2 2127.5 581 4 2781.5 562 5 1504.4 563 3 1447 575 1 3163 577 2 995 564 6 1198. 569 2 1740.5 562 6 1294.5 566 8 881.25 587 1 1913 578 2 1553.5 585 13 2472.0769 567 1 3049 560 1 964 568 2 1452 569 2 3370 575 6 1386. 585 3 1643 560 3 4554.2631 559 2 1536.7438 559 1 1319.6811 559 5 3169.5263 560 3 875.5572 561 2 5093.6068 559 8 2394.0242 561 1 4076.4088 559 1 1625.0949 562 5 1775.1878 560 1 1500.597 562 14 1876.4794 563 3 1582.5177 564 3 4268.6134 564 2 3799.6765 567 4 4590.8274 563 1 2140 564 2 1475 569 1 6970 568 1 1500 567 7 2070.2857 568 3 1681. 568 3 697.5219 571 1 2000 571 5 1438 570 4 1393 570 4 2355.25 571 1 756.3875 576 1 1720.8184 571 5 1613 574 3 1903. 577 6 1828.2312 571 2 2087.5 581 3 3456. 582 13 2135.3076 582 1 3876 581 1 2424 576 1 3061 583 1 1534 583 3 2810 577 13 1042.7692 576 4 756.3875 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Hello
sumit gupta [EMAIL PROTECTED] napsal dne 11.06.2008 07:53:54: Hello thanx for the info.. I still have tthe doubt regarding the heat map. I am attching a plot of a 70X3 data matrix.In which 3rd variable has been shown as colour of different boxes.Could you help me to draw this type of graph using R Well, as you did not provided some data mat-cbind(sample(1:5, 70, replace=T), sample(1:5, 70, replace=T), rnorm(70)) dim(mat) [1] 70 3 library(akima) filled.contour(interp(mat[,1], mat[,2], mat[,3], duplicate=mean)) image(interp(mat[,1], mat[,2], mat[,3], duplicate=mean)) is this what you want? Regards Petr Regards, Sumit On 6/10/08, Petr PIKAL [EMAIL PROTECTED] wrote: Hi [EMAIL PROTECTED] napsal dne 10.06.2008 14:49:55: Hello, I am facing a problem in drawing heat map using R. I have a 70X3 matrix and I want to draw a heat map with 1 coloumn on X axis another on Y axis and want to show the value of 3rd coloumn using heat colours. Could you please help me with this . You can use function interp from akima package Petr Thank you . Regards, Sumit [[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. -- Sumit Kumar Gupta 4thYear ,Biotechnology IIT Madras [příloha rfm.bmp odstraněna uživatelem Petr PIKAL/CTCAP] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Matrix transformation problem
For precisely this particular type of question, the following seems to be the simplest, most direct, and most transparent solution: rowSums(x%*%(1:ncol(x))) # [1] 1 3 2 3 2 1 Ted. On 11-Jun-08 09:21:35, Dimitris Rizopoulos wrote: sorry, my previous answer was not correct; you need: x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0), ncol = 3, byrow = TRUE) which(t(x == 1), arr.ind = TRUE)[, row, drop = FALSE] Best, Dimitris Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: [EMAIL PROTECTED] To: r-help@r-project.org Sent: Wednesday, June 11, 2008 10:10 AM Subject: [R] Matrix transformation problem ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),ncol=3,byrow=T) x [,1] [,2] [,3] [1,]100 [2,]001 [3,]010 [4,]001 [5,]010 [6,]100 In the matrix above, on the first row, the 1 is in column 1, hence '1' on the first row in the matrix below. On the second row in the matrix above, the 1 is in column 3, hence the '3' on the second row in the matrix below. And so on... y [,1] [1,]1 [2,]3 [3,]2 [4,]3 [5,]2 [6,]1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 11-Jun-08 Time: 10:49:13 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Finding Coordinate of Max/Min Value in a Data Frame
Hi, Suppose I have the following data frame. __BEGIN__ library(MASS) data(crabs) crab.pca - prcomp(crabs[,4:8],retx=TRUE) crab.pca$rotation PC1PC2PC3PC4PC5 FL 0.2889810 0.3232500 -0.5071698 0.7342907 0.1248816 RW 0.1972824 0.8647159 0.4141356 -0.1483092 -0.1408623 CL 0.5993986 -0.1982263 -0.1753299 -0.1435941 -0.7416656 CW 0.6616550 -0.2879790 0.4913755 0.1256282 0.4712202 BD 0.2837317 0.1598447 -0.5468821 -0.6343657 0.4386868 __END__ Is there a way to identify the coordinate of a max/min value of all the points above. For example the coord of maximum value is (RW,PC2) = 0.865, and coord of min value is (CW,PC2) = - 0.288. -- Gundala Viswanath __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] MCA in R
Dear Kimmo, MCA is a rather old name (introduced, I think, in the 1960s by Songuist and Morgan in the OSIRIS package) for a linear model consisting entirely of factors and with only additive effects -- i.e., an ANOVA model will no interactions. You can fit such a model with lm() -- e.g., lm(y ~ f1 + f2 + etc.). Typically, the results of an MCA are reported using adjusted means. You could compute these manually, or via the effects package. I hope this helps, John -- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario, Canada web: socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of K. Elo Sent: June-11-08 1:07 AM To: r-help@r-project.org Subject: [R] MCA in R Hi! Is there any possibilities to do multiple classification analysis (MCA) in R? (MCA examines the relationships between several categorical independent variables and a single dependent variable, and determines the effects of each predictor before and after adjustment for its inter-correlations with other predictors in the analysis). Kind regrads, Kimmo Elo --- University of Turku, Finland Dep. of political 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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] piper diagram
Jim Lemon wrote: Dylan Beaudette wrote: Hi, Is anyone on the list familiar with an R implementation of Piper Diagrams? Example: http://faculty.uml.edu/nelson_eby/89.315/IMAGES/Figure%209-78.jpg I am thinking that two calls to triax.plot (plotrix) along with some kind of affine-transformed standard plot would do the trick. Not so sure about the final layout, or a nice generalized version for something like lattice. Okay, I've had a look at the Piper diagram and with better labeled ones and some explanations, I think I've got the picture. No, triax.plot won't do the job, but it's a similar sort of illustration. Give me a few days to think about it and I may be able to put one together. 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] model simplification using Crawley as a guide
ChCh wrote: Hello, I have consciously avoided using step() for model simplification in favour of manually updating the model by removing non-significant terms one at a time. I'm using The R Book by M.J. Crawley as a guide. It comes as no surprise that my analysis does proceed as smoothly as does Crawley's and being a beginner, I'm struggling with what to do next. I have a model: lm(y~A * B * C) where A is a categorical variable with three levels and B and C are continuous covariates. Following Crawley, I execute the model, then use summary.aov() to identify non-significant terms. I begin deleting non-significant interaction terms one at a time (using update). After each update() statement, I use anova(modelOld,modelNew) to contrast the previous model with the updated one. After removing all the interaction terms, I'm left with: lm(y~ A + B + C) again, using summary.aov() I identify A to be non-significant, so I remove it, leaving: lm(y~B + C) both of which are continuous variables Does it still make sense to use summary.aov() or should I use summary.lm() instead? Has the analysis switched from an ANCOVA to a regression? Both give different results so I'm uncertain which summary to accept. Any help would be appreciated! Does he really recommend using summary.aov() on an lm object??? I wouldn't. It _might_ give sensible results, but in general, aov() and its methods rely on balancedness and orthogonality properties of the design, to the extent that I'm inclined to say that if you do not know exactly what is going on, it is probably the wrong thing. I'd use drop1 throughout. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] model simplification using Crawley as a guide
ChCh wrote: Hello, I have consciously avoided using step() for model simplification in favour of manually updating the model by removing non-significant terms one at a time. I'm using The R Book by M.J. Crawley as a guide. It comes as no surprise that my analysis does proceed as smoothly as does Crawley's and being a beginner, I'm struggling with what to do next. I have a model: lm(y~A * B * C) where A is a categorical variable with three levels and B and C are continuous covariates. Following Crawley, I execute the model, then use summary.aov() to identify non-significant terms. I begin deleting non-significant interaction terms one at a time (using update). After each update() statement, I use anova(modelOld,modelNew) to contrast the previous model with the updated one. After removing all the interaction terms, I'm left with: lm(y~ A + B + C) again, using summary.aov() I identify A to be non-significant, so I remove it, leaving: lm(y~B + C) both of which are continuous variables Does it still make sense to use summary.aov() or should I use summary.lm() instead? Has the analysis switched from an ANCOVA to a regression? Both give different results so I'm uncertain which summary to accept. Any help would be appreciated! What is the theoretical basis for removing insignificant terms? How will you compensate for this in the final analysis (e.g., how do you unbias your estimate of sigma squared)? -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Finding Coordinate of Max/Min Value in a Data Frame
Try this: which(pca == min(pca), arr.ind = TRUE) which(pca == max(pca), arr.ind = TRUE) On Wed, Jun 11, 2008 at 7:34 AM, Gundala Viswanath [EMAIL PROTECTED] wrote: Hi, Suppose I have the following data frame. __BEGIN__ library(MASS) data(crabs) crab.pca - prcomp(crabs[,4:8],retx=TRUE) crab.pca$rotation PC1PC2PC3PC4PC5 FL 0.2889810 0.3232500 -0.5071698 0.7342907 0.1248816 RW 0.1972824 0.8647159 0.4141356 -0.1483092 -0.1408623 CL 0.5993986 -0.1982263 -0.1753299 -0.1435941 -0.7416656 CW 0.6616550 -0.2879790 0.4913755 0.1256282 0.4712202 BD 0.2837317 0.1598447 -0.5468821 -0.6343657 0.4386868 __END__ Is there a way to identify the coordinate of a max/min value of all the points above. For example the coord of maximum value is (RW,PC2) = 0.865, and coord of min value is (CW,PC2) = - 0.288. -- Gundala Viswanath __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] searching for specific row in matrix
Hi,
I have matrix of bits and a target vector. Is there an
efficient way to search the rows of the matrix for the target?
I am interested in the first row index where target is found.
Example:
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]10110
[2,]11010
[3,]00100
[4,]10011
[5,]10111
[6,]11001
[7,]10011
[8,]00111
[9,]01101
[10,]00010
target: 1 1 0 1 1
Should return -1 (or some other indicator) since the
target was not found in any of the rows.
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]00110
[2,]10000
[3,]10000
[4,]11000
[5,]11100
[6,]00110
[7,]01110
[8,]00110
[9,]11011
[10,]10100
target: 1 1 0 1 1
Should return 9 since the target was found in row 9
If the target is found, it is no longer necessary to keep
searching the rest of the matrix (which may be quite large)
The data/size etc may change of course, but target will
always have the same number of columns as the matrix.
I tried variations of which, and a for loop
comparing pop[i,] to target without much success, nor
did google yield any results. I am hoping someone here
can provide a suggestion.
Thanks,
EB
-
# Here is the code that generates the above data
create_bin_string - function(len)
{
sample(0:1, len, replace=T)
}
ROWS = 10
COLS = 5
pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
target=c(1, 1, 0, 1, 1)
# my population
print(pop)
# I am looking for the index of this in pop
# if present (else -1?)
cat(\ntarget: , target, \n)
##
## this is NOT working
## plus it would continue to search
## after it found the target
##
for(i in ROWS)
if (pop[i,] == target)
cat(\nfound in row: , i, \n\n)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Finding Coordinate of Max/Min Value in a Data Frame
This suggests that one nice to have item would be to support arr.ind= on which.min and which.max On Wed, Jun 11, 2008 at 7:57 AM, Henrique Dallazuanna [EMAIL PROTECTED] wrote: Try this: which(pca == min(pca), arr.ind = TRUE) which(pca == max(pca), arr.ind = TRUE) On Wed, Jun 11, 2008 at 7:34 AM, Gundala Viswanath [EMAIL PROTECTED] wrote: Hi, Suppose I have the following data frame. __BEGIN__ library(MASS) data(crabs) crab.pca - prcomp(crabs[,4:8],retx=TRUE) crab.pca$rotation PC1PC2PC3PC4PC5 FL 0.2889810 0.3232500 -0.5071698 0.7342907 0.1248816 RW 0.1972824 0.8647159 0.4141356 -0.1483092 -0.1408623 CL 0.5993986 -0.1982263 -0.1753299 -0.1435941 -0.7416656 CW 0.6616550 -0.2879790 0.4913755 0.1256282 0.4712202 BD 0.2837317 0.1598447 -0.5468821 -0.6343657 0.4386868 __END__ Is there a way to identify the coordinate of a max/min value of all the points above. For example the coord of maximum value is (RW,PC2) = 0.865, and coord of min value is (CW,PC2) = - 0.288. -- Gundala Viswanath __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 idiom list? [was: Matrix transformation problem
Dimitris Rizopoulos schrieb: sorry, my previous answer was not correct; you need: x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0), ncol = 3, byrow = TRUE) which(t(x == 1), arr.ind = TRUE)[, row, drop = FALSE] Dear helpeRs, I would like to use this thread for a proposal which lingered in my head for quite a while. Several years ago I was an APL aficionado and I very much liked the idea of an APL idiom list (see e.g http://www.pyr.fi/apl/texts/Idiot.htm). Looking at the solution of Dimitris Rizopoulos wouldn't it be nice not to have it disappear in the archives? To be explicit I would like to propose an R idiom list. If there does exist one please forgive my ignorance. Best regards Dietrich -- Dietrich Trenkler c/o Universitaet Osnabrueck Rolandstr. 8; D-49069 Osnabrueck, Germany email: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Finding Coordinate of Max/Min Value in a Data Frame
Firstly surely (CL,PC5) = -0.7416656 is the minima? I think something like this should work, though I'm not sure library(MASS) data(crabs) crab.pca - prcomp(crabs[,4:8],retx=TRUE) x- crab.pca$rotation c.len = length(x[,1]) r.len = length(x[1,]) maxy = which(x == min(x)) x.co - ceiling(maxy/c.len) y.co - maxy-(x.co-1)*c.len Though I'm sure their is a more simple solution to your problem, Best, David U Bristol Edward Wijaya-2 wrote: Hi, Suppose I have the following data frame. __BEGIN__ library(MASS) data(crabs) crab.pca - prcomp(crabs[,4:8],retx=TRUE) crab.pca$rotation PC1PC2PC3PC4PC5 FL 0.2889810 0.3232500 -0.5071698 0.7342907 0.1248816 RW 0.1972824 0.8647159 0.4141356 -0.1483092 -0.1408623 CL 0.5993986 -0.1982263 -0.1753299 -0.1435941 -0.7416656 CW 0.6616550 -0.2879790 0.4913755 0.1256282 0.4712202 BD 0.2837317 0.1598447 -0.5468821 -0.6343657 0.4386868 __END__ Is there a way to identify the coordinate of a max/min value of all the points above. For example the coord of maximum value is (RW,PC2) = 0.865, and coord of min value is (CW,PC2) = - 0.288. -- Gundala Viswanath __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Finding-Coordinate-of-Max-Min-Value-in-a-Data-Frame-tp17775656p17776020.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] Finding Coordinate of Max/Min Value in a Data Frame
Thanks so much both. - Gundala On 6/11/08, Henrique Dallazuanna [EMAIL PROTECTED] wrote: Try this: which(pca == min(pca), arr.ind = TRUE) which(pca == max(pca), arr.ind = TRUE) On Wed, Jun 11, 2008 at 7:34 AM, Gundala Viswanath [EMAIL PROTECTED] wrote: Hi, Suppose I have the following data frame. __BEGIN__ library(MASS) data(crabs) crab.pca - prcomp(crabs[,4:8],retx=TRUE) crab.pca$rotation PC1PC2PC3PC4PC5 FL 0.2889810 0.3232500 -0.5071698 0.7342907 0.1248816 RW 0.1972824 0.8647159 0.4141356 -0.1483092 -0.1408623 CL 0.5993986 -0.1982263 -0.1753299 -0.1435941 -0.7416656 CW 0.6616550 -0.2879790 0.4913755 0.1256282 0.4712202 BD 0.2837317 0.1598447 -0.5468821 -0.6343657 0.4386868 __END__ Is there a way to identify the coordinate of a max/min value of all the points above. For example the coord of maximum value is (RW,PC2) = 0.865, and coord of min value is (CW,PC2) = - 0.288. -- Gundala Viswanath __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 -- Gundala Viswanath __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R idiom list? [was: Matrix transformation problem
Google for Rtips. Also Rwiki On Wed, Jun 11, 2008 at 6:35 AM, Dietrich Trenkler [EMAIL PROTECTED] wrote: Dimitris Rizopoulos schrieb: sorry, my previous answer was not correct; you need: x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0), ncol = 3, byrow = TRUE) which(t(x == 1), arr.ind = TRUE)[, row, drop = FALSE] Dear helpeRs, I would like to use this thread for a proposal which lingered in my head for quite a while. Several years ago I was an APL aficionado and I very much liked the idea of an APL idiom list (see e.g http://www.pyr.fi/apl/texts/Idiot.htm). Looking at the solution of Dimitris Rizopoulos wouldn't it be nice not to have it disappear in the archives? To be explicit I would like to propose an R idiom list. If there does exist one please forgive my ignorance. Best regards Dietrich -- Dietrich Trenkler c/o Universitaet Osnabrueck Rolandstr. 8; D-49069 Osnabrueck, Germanyemail: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] searching for specific row in matrix
This should work for you:
create_bin_string - function(len)
+ {
+ sample(0:1, len, replace=T)
+ }
ROWS = 10
COLS = 5
set.seed(2)
pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
target=c(1, 1, 0, 1, 1)
# my population
print(pop)
[,1] [,2] [,3] [,4] [,5]
[1,]01101
[2,]10101
[3,]10100
[4,]11000
[5,]10100
[6,]00010
[7,]00111
[8,]11010
[9,]10001
[10,]11011
# determine which data matches
matches - t(pop) == target # 't' due to matching in column order
# colSums equal to COLS will indicate matches
which(colSums(matches) == COLS)
[1] 10
On Wed, Jun 11, 2008 at 7:58 AM, Esmail Bonakdarian [EMAIL PROTECTED] wrote:
Hi,
I have matrix of bits and a target vector. Is there an
efficient way to search the rows of the matrix for the target?
I am interested in the first row index where target is found.
Example:
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]10110
[2,]11010
[3,]00100
[4,]10011
[5,]10111
[6,]11001
[7,]10011
[8,]00111
[9,]01101
[10,]00010
target: 1 1 0 1 1
Should return -1 (or some other indicator) since the
target was not found in any of the rows.
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]00110
[2,]10000
[3,]10000
[4,]11000
[5,]11100
[6,]00110
[7,]01110
[8,]00110
[9,]11011
[10,]10100
target: 1 1 0 1 1
Should return 9 since the target was found in row 9
If the target is found, it is no longer necessary to keep
searching the rest of the matrix (which may be quite large)
The data/size etc may change of course, but target will
always have the same number of columns as the matrix.
I tried variations of which, and a for loop
comparing pop[i,] to target without much success, nor
did google yield any results. I am hoping someone here
can provide a suggestion.
Thanks,
EB
-
# Here is the code that generates the above data
create_bin_string - function(len)
{
sample(0:1, len, replace=T)
}
ROWS = 10
COLS = 5
pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
target=c(1, 1, 0, 1, 1)
# my population
print(pop)
# I am looking for the index of this in pop
# if present (else -1?)
cat(\ntarget: , target, \n)
##
## this is NOT working
## plus it would continue to search
## after it found the target
##
for(i in ROWS)
if (pop[i,] == target)
cat(\nfound in row: , i, \n\n)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] searching for specific row in matrix
try this:
match.pat - function (mat, target, nomatch = -1) {
f1 - do.call(paste, c(as.data.frame(mat), sep = \r))
f2 - paste(target, collapse = \r)
ind - f1 %in% f2
if (any(ind)) which(ind)[1] else nomatch
}
##
set.seed(1234)
mat - matrix(sample(0:1, 50, TRUE), 10, 5)
targ1 - mat[2, ]
match.pat(mat, targ1)
targ2 - rep(0, 5)
match.pat(mat, targ2)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Esmail Bonakdarian [EMAIL PROTECTED]
To: r-help@r-project.org
Sent: Wednesday, June 11, 2008 1:58 PM
Subject: [R] searching for specific row in matrix
Hi,
I have matrix of bits and a target vector. Is there an
efficient way to search the rows of the matrix for the target?
I am interested in the first row index where target is found.
Example:
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]10110
[2,]11010
[3,]00100
[4,]10011
[5,]10111
[6,]11001
[7,]10011
[8,]00111
[9,]01101
[10,]00010
target: 1 1 0 1 1
Should return -1 (or some other indicator) since the
target was not found in any of the rows.
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]00110
[2,]10000
[3,]10000
[4,]11000
[5,]11100
[6,]00110
[7,]01110
[8,]00110
[9,]11011
[10,]10100
target: 1 1 0 1 1
Should return 9 since the target was found in row 9
If the target is found, it is no longer necessary to keep
searching the rest of the matrix (which may be quite large)
The data/size etc may change of course, but target will
always have the same number of columns as the matrix.
I tried variations of which, and a for loop
comparing pop[i,] to target without much success, nor
did google yield any results. I am hoping someone here
can provide a suggestion.
Thanks,
EB
-
# Here is the code that generates the above data
create_bin_string - function(len)
{
sample(0:1, len, replace=T)
}
ROWS = 10
COLS = 5
pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
target=c(1, 1, 0, 1, 1)
# my population
print(pop)
# I am looking for the index of this in pop
# if present (else -1?)
cat(\ntarget: , target, \n)
##
## this is NOT working
## plus it would continue to search
## after it found the target
##
for(i in ROWS)
if (pop[i,] == target)
cat(\nfound in row: , i, \n\n)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Default Argument Passing in Script
Hi all, Currently I run R script with arguments the following ways $ R --vanilla myscript.R ARGUMENT1 And in my script it is encoded as: __BEGIN__ args-commandArgs() do_sth(args[3]) My question is that is there a way to set a default argument inside the R script? In Perl analogically would be: my $param = $ARGV[0] || default_argument; I am wondering how can this be done in R. -- Gundala Viswanath Jakarta - Indonesia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] searching for specific row in matrix
Try this:
which(apply(t(m) == target, 2, all))
On Wed, Jun 11, 2008 at 8:58 AM, Esmail Bonakdarian [EMAIL PROTECTED]
wrote:
Hi,
I have matrix of bits and a target vector. Is there an
efficient way to search the rows of the matrix for the target?
I am interested in the first row index where target is found.
Example:
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]10110
[2,]11010
[3,]00100
[4,]10011
[5,]10111
[6,]11001
[7,]10011
[8,]00111
[9,]01101
[10,]00010
target: 1 1 0 1 1
Should return -1 (or some other indicator) since the
target was not found in any of the rows.
source(lookup.R)
[,1] [,2] [,3] [,4] [,5]
[1,]00110
[2,]10000
[3,]10000
[4,]11000
[5,]11100
[6,]00110
[7,]01110
[8,]00110
[9,]11011
[10,]10100
target: 1 1 0 1 1
Should return 9 since the target was found in row 9
If the target is found, it is no longer necessary to keep
searching the rest of the matrix (which may be quite large)
The data/size etc may change of course, but target will
always have the same number of columns as the matrix.
I tried variations of which, and a for loop
comparing pop[i,] to target without much success, nor
did google yield any results. I am hoping someone here
can provide a suggestion.
Thanks,
EB
-
# Here is the code that generates the above data
create_bin_string - function(len)
{
sample(0:1, len, replace=T)
}
ROWS = 10
COLS = 5
pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
target=c(1, 1, 0, 1, 1)
# my population
print(pop)
# I am looking for the index of this in pop
# if present (else -1?)
cat(\ntarget: , target, \n)
##
## this is NOT working
## plus it would continue to search
## after it found the target
##
for(i in ROWS)
if (pop[i,] == target)
cat(\nfound in row: , i, \n\n)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] searching for specific row in matrix
Henrique Dallazuanna wrote:
Try this:
which(apply(t(m) == target, 2, all))
Wow! .. talk about concise! Neat! Thanks.
This will return all matches correct? So if I only wanted
the first I'd simply subscript [1] into it.
Do you think the fact that it searches the whole matrix instead
of stopping when it finds a match may slow it down?
This is my own solution I came up with in the meantime,
looks rather pedestrian compared to your one line, but it will
drop out immediately once if finds the target. Yours looks (based
simply on appearance :-) faster. Having no feel for the language
I may just have to time them.
I would assume that your solution would be faster simple since it's
using built-in language constructs which are optimized (and implemented
in C?) instead of my own interpreted way.
# return index of target in pop, else -1
searchPop - function(pop, target)
{
rows = length(pop[1,])
for(i in 1:rows)
{
result = (pop[i,] == target)
if (sum(which(result==FALSE)) == 0)
return(i)
}
return (-1)
}
idx=searchPop(pop, target)
if (idx 0)
{
cat(NOT found\n)
} else
cat(Found at position , idx, \n)
Esmail
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] searching for specific row in matrix
Dimitris Rizopoulos wrote:
try this:
match.pat - function (mat, target, nomatch = -1) {
f1 - do.call(paste, c(as.data.frame(mat), sep = \r))
f2 - paste(target, collapse = \r)
ind - f1 %in% f2
if (any(ind)) which(ind)[1] else nomatch
}
Thanks! More R for me to sink my teeth in :-) My own solution
doesn't seem to work quite correctly as I found out from some
further testing .. so the solutions posted here are much appreciated!
Esmail
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] searching for specific row in matrix
# determine which data matches matches - t(pop) == target # 't' due to matching in column order # colSums equal to COLS will indicate matches which(colSums(matches) == COLS) Neat! .. somewhat similar to the solution I came up with in the meantime, only yours works :-) Thanks Jim. Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with densityplot from the lattice package
Dear list, I just experienced some strange behavior of the densityplots provided by the lattice package. Depending on the values for from and to that are passed on to the function for kernel density estimates, the densityplot excludes the jittered stripplot that is routinely included in lattice densityplot. Example: library(lattice) test-runif(100) densityplot(test) #the standard density plot without problems densityplot(test, from=0, to=1) #stripplot is excluded densityplot(test, from=-1, to=2) #stripplot is again included Any hints how I could get the stripplot back in the second example? Tanks in advance Malte sessionInfo() R version 2.7.0 (2008-04-22) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252;LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] lattice_0.17-8 loaded via a namespace (and not attached): [1] grid_2.7.0 sent via xelos.net | the communication and groupware solution __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] model simplification using Crawley as a guide
On Wed, Jun 11, 2008 at 6:42 AM, Frank E Harrell Jr [EMAIL PROTECTED] wrote: ChCh wrote: Hello, I have consciously avoided using step() for model simplification in favour of manually updating the model by removing non-significant terms one at a time. I'm using The R Book by M.J. Crawley as a guide. It comes as no surprise that my analysis does proceed as smoothly as does Crawley's and being a beginner, I'm struggling with what to do next. I have a model: lm(y~A * B * C) where A is a categorical variable with three levels and B and C are continuous covariates. Following Crawley, I execute the model, then use summary.aov() to identify non-significant terms. I begin deleting non-significant interaction terms one at a time (using update). After each update() statement, I use anova(modelOld,modelNew) to contrast the previous model with the updated one. After removing all the interaction terms, I'm left with: lm(y~ A + B + C) again, using summary.aov() I identify A to be non-significant, so I remove it, leaving: lm(y~B + C) both of which are continuous variables Does it still make sense to use summary.aov() or should I use summary.lm() instead? Has the analysis switched from an ANCOVA to a regression? Both give different results so I'm uncertain which summary to accept. Any help would be appreciated! What is the theoretical basis for removing insignificant terms? How will you compensate for this in the final analysis (e.g., how do you unbias your estimate of sigma squared)? And in a similar vein, where are your exploratory graphics? How do you know that there is a linear relationship between your response and your predictors? Are the distributional assumptions you are making appropriate? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Goodness of fit tests for copula
How can i calculate the discrete L2 distance between the empirical copula and the estimated one. -- View this message in context: http://www.nabble.com/Goodness-of-fit-tests-for-copula-tp17779318p17779318.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] Matrix transformation problem
stefan.petersson wrote: ng, I have a matrix (x) with binary content. Each row of the matrix holds exactly one 1, and the rest of the row is zeros. The thing is that I need to 'collapse' the matrix to one column where each row holds the original column index of the 1's (y). Sometimes, the matrix is quite large, so I have a perfomance problem. x - matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),ncol=3,byrow=T) x [,1] [,2] [,3] [1,]100 [2,]001 [3,]010 [4,]001 [5,]010 [6,]100 In the matrix above, on the first row, the 1 is in column 1, hence '1' on the first row in the matrix below. On the second row in the matrix above, the 1 is in column 3, hence the '3' on the second row in the matrix below. And so on... y [,1] [1,]1 [2,]3 [3,]2 [4,]3 [5,]2 [6,]1 max.col(x) -- View this message in context: http://www.nabble.com/Matrix-transformation-problem-tp17773270p17779431.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] Problem with densityplot from the lattice package
Look at the y axis ... so try something like densityplot(test, from=0, to=1, ylim=c(-0.1, 1.5)) The jittered plot is plotted centred on y=0. On Wed, 11 Jun 2008, Malte Brockmann wrote: Dear list, I just experienced some strange behavior of the densityplots provided by the lattice package. Depending on the values for from and to that are passed on to the function for kernel density estimates, the densityplot excludes the jittered stripplot that is routinely included in lattice densityplot. Example: library(lattice) test-runif(100) densityplot(test) #the standard density plot without problems densityplot(test, from=0, to=1) #stripplot is excluded densityplot(test, from=-1, to=2) #stripplot is again included Any hints how I could get the stripplot back in the second example? Tanks in advance Malte sessionInfo() R version 2.7.0 (2008-04-22) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252;LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] lattice_0.17-8 loaded via a namespace (and not attached): [1] grid_2.7.0 sent via xelos.net | the communication and groupware solution __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Some problem with the OLS
Dear all!! I estimated the following OLS model with R 2.5.6: Output R 2.5.6 Call: lm(formula = UN ~ log(x) + time2, data = dati) Residuals: Min 1Q Median 3QMax -5.649 -2.753 -1.015 1.225 16.199 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 7.3036294 0.6871025 10.630 2e-16 *** log(x) -0.0028542 0.0270730 -0.1050.916 time2 -0.0002670 0.0003823 -0.6980.487 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 4.323 on 87 degrees of freedom Multiple R-Squared: 0.02933,Adjusted R-squared: 0.007017 F-statistic: 1.314 on 2 and 87 DF, p-value: 0.2739 is significant only the intercept After I estimated the same model with R 2.7 and this is the new output: Output R 2.7 regressione14-lm(UN ~ log(x)+ time2,data=dati) summary(regressione14) Call: lm(formula = UN ~ log(x) + time2, data = dati) Residuals: Min 1Q Median 3QMax -8.825 -2.094 -0.861 2.233 12.664 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 22.724208 3.304670 6.876 8.89e-10 *** log(x) -7.117981 1.499482 -4.747 8.05e-06 *** time20.002051 0.000523 3.922 0.000175 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 3.853 on 87 degrees of freedom Multiple R-Squared: 0.2289, Adjusted R-squared: 0.2112 F-statistic: 12.91 on 2 and 87 DF, p-value: 1.227e-05 Well, how you can see the two outputs are very differents, so i try to estimated the same model with another package and i show the output: *Variabile* *Coefficiente* *Errore Std.* *Statistica t* *p-value* const 22,7242 3,30467 6,8764 0,1 *** logx -7,11798 1,49948 -4,7470 0,1 *** time2 0,00205144 0,000523012 3,9224 0,00017 *** Media della variabile dipendente = 6,47833 Scarto quadratico medio della variabile dipendente = 4,33827 Somma dei quadrati dei residui = 1291,58 Errore standard dei residui = 3,85302 R2 = 0,228922 R2 corretto = 0,211196 Statistica F (2, 87) = 12,9145 (p-value = 1,23e-005) Log-verosimiglianza = -247,576 Criterio di informazione di Akaike = 501,152 Criterio bayesiano di Schwarz = 508,652 Criterio di Hannan-Quinn = 504,176 How you can see this package show the same output of the R 2.7 My question is where is the bug? Kind regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] newbie nls question
Worked liked a charm. Thanks for your help. SR Steven H. Ranney Graduate Research Assistant (Ph.D) USGS Montana Cooperative Fishery Research Unit Montana State University PO Box 173460 Bozeman, MT 59717-3460 phone: (406) 994-6643 fax: (406) 994-7479 -Original Message- From: Michael Manning [mailto:[EMAIL PROTECTED] Sent: Tue 6/10/2008 5:08 PM To: Ranney, Steven; r-help@r-project.org Subject: Re: [R] newbie nls question Hi On 11/06/2008 at 10:56 a.m., Ranney, Steven [EMAIL PROTECTED] wrote: I'm tyring to fit a relatively simple nls model to some data, but keep coming up against the same error (code follows): Oto=nls(Otolith ~ Linf*(1-exp(-k(AGE-to))), data = ages, start = list(Linf=1000, k=0.1, to=0.1), trace = TRUE) The error message I keep getting is Error in eval(expr, envir, enclos) : could not find function k. I've used this line of code for other nls models (with different data, parameter estimates, etc.), but have never gotten this error. The data is AGE,Otolith,Scale 1,207.1052632,207.1052632 2,329.962963,332.7586207 3,401.9473684,406 4,422,413.111 5,452.6785714,458.34375 6,510.75,533 7,477,674 8,643,704 9,,615 10,695.5, 12,615, Are the missing values (e.g., no value for Otolith, AGE 9) having an effect? I am using Tinn-R as an editor I'm new to R, trying to get away from Rcmdr (though it has been helpful), and still trying to learn the language with the aid of several books on the R. A search of the R Archive did not prove fruitful. Thanks in advance, SR Steven H. Ranney You've made a typo. Check out the formula you passed to nls -- you forget to include a * after k. What I think you want is something like Oto - nls(Otolith ~ Linf * (1 - exp(-k * (AGE - t0) ) ), ...) Cheers, Michael [[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] model simplification using Crawley as a guide
And to follow FH and HW What level of significance are you using? .05 is excessively liberal. Are you adjusting your p-values for the number of possible models? Do you realize the p-values for dropping a term, being selected as the maximum of a set of p-values, do not follow their usual distributions? How are you compensating for sample size, as a p-value's being significant is a function of sample size? How are you compensating for the fact that the current model choice is dependent on the previous model choices? How do you know your tree of model choices is the optimal one? Have you considered cross-validation? Are you looking for a model that true describes a phenomenon or a predictive model that can be used for practical purposes? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of hadley wickham Sent: Wednesday, June 11, 2008 9:34 AM To: Frank E Harrell Jr Cc: r-help@r-project.org; ChCh Subject: Re: [R] model simplification using Crawley as a guide On Wed, Jun 11, 2008 at 6:42 AM, Frank E Harrell Jr [EMAIL PROTECTED] wrote: ChCh wrote: Hello, I have consciously avoided using step() for model simplification in favour of manually updating the model by removing non-significant terms one at a time. I'm using The R Book by M.J. Crawley as a guide. It comes as no surprise that my analysis does proceed as smoothly as does Crawley's and being a beginner, I'm struggling with what to do next. I have a model: lm(y~A * B * C) where A is a categorical variable with three levels and B and C are continuous covariates. Following Crawley, I execute the model, then use summary.aov() to identify non-significant terms. I begin deleting non-significant interaction terms one at a time (using update). After each update() statement, I use anova(modelOld,modelNew) to contrast the previous model with the updated one. After removing all the interaction terms, I'm left with: lm(y~ A + B + C) again, using summary.aov() I identify A to be non-significant, so I remove it, leaving: lm(y~B + C) both of which are continuous variables Does it still make sense to use summary.aov() or should I use summary.lm() instead? Has the analysis switched from an ANCOVA to a regression? Both give different results so I'm uncertain which summary to accept. Any help would be appreciated! What is the theoretical basis for removing insignificant terms? How will you compensate for this in the final analysis (e.g., how do you unbias your estimate of sigma squared)? And in a similar vein, where are your exploratory graphics? How do you know that there is a linear relationship between your response and your predictors? Are the distributional assumptions you are making appropriate? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Word wrapping for character objects (WINDOWS R ONLY)
Can anybody help me with this problem? ** ONLY WINDOWS R - PROBLEM DOESN'T OCCUR ON LINUX ** I want to print a long character to screen: getOption(width) [1] 60 z=(1:20)/10#z is a vector of length between 20 and 30 (depending on user options) containing lengths in mm (i.e. each element is 1-5 characters long) str1=paste(The depths chosen are (,toString(z),) mm, and more text ...\n) cat(str1) The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2, 1.3, 1.$ The problem is that on R for Windows the string is cropped by the window size (hence the $). On R for Linux, this doesn't happen and the text is word wrapped (the default for the shell, I guess): cat(str1) The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2 ) mm, and more text ... I can't find any option for word wrapping in the cat command (fill=TRUE has no effect). I also checked the menu Edit - GUI preferences..., but there doesn't seem to be a Word Wrap option there either. How do I get word wrapping like this in Windows? Perhaps the attached screenshots clarify this question. THANKS FOR ANY HELP! Toby Marthews Previous relevant posts: - The post from 2006 about Screen Wrapping (http://tolstoy.newcastle.edu.au/R/help/06/05/26673.html) which Brian Ripley answered was about controlling how long vectors are cropped to the screen. Unfortunately, the width option in options() does not affect character objects, so I can't use that control here. - I sent the same question to [EMAIL PROTECTED] in Oct 2007, but noone there could help me with it. - Try the following command on Windows R with a small window (getOption(width)117) and a large window (getOption(width)117) and you'll see you get extra nonexistent options in the menu: a=c(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec);menu(a) I guess this is a related 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.
[R] applying a function recursively
Hi, I have a question about applying a function recursively through a list. Suppose I have a list where the different elements have different levels of recursion: test.list-list(I=list(A=c(a, b, c), B=c(d, e, f), C=c(g, h, i)), + II=list(A=list(a=c(a, b, c), b=c(d, e, f), + c=c(g, h, i)), + B=c(d, e, f), C=c(g, h, i))) test.list $I $I$A [1] a b c $I$B [1] d e f $I$C [1] g h i $II $II$A $II$A$a [1] a b c $II$A$b [1] d e f $II$A$c [1] g h i $II$B [1] d e f $II$C [1] g h i I would like to apply a function recursively to that list, in a way that the function does someting with each vector (eg. rev()) and returns a list of modified vectors that has the same structure as the input list, in my example: $I $I$A [1] c b a $I$B [1] f e d $I$C [1] i h g $II $II$A $II$A$a [1] c b a $II$A$b [1] f e d $II$A$c [1] i h g $II$B [1] f e d $II$C [1] i h g I understand that with a fixed number of recursion levels one can use lapply() in a nested way, but what if the numbers of recursion levels is not fixed or is different between the list elements as it is in my example? Any hint will be appreciated. Best, Georg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Some problem with the OLS
On Wed, 11 Jun 2008, roberto laforgia wrote: Dear all!! I estimated the following OLS model with R 2.5.6: Output R 2.5.6 You need to read and follow the posting guide AND provide a minimal, reproducible example. What is R 2.5.6 ?? What do sessionInfo() or version report? Where can I get a copy? ;-) (Hint: Correct answer is NOT on CRAN. There is not such version.) Call: lm(formula = UN ~ log(x) + time2, data = dati) Residuals: Min 1Q Median 3QMax -5.649 -2.753 -1.015 1.225 16.199 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 7.3036294 0.6871025 10.630 2e-16 *** log(x) -0.0028542 0.0270730 -0.1050.916 time2 -0.0002670 0.0003823 -0.6980.487 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 4.323 on 87 degrees of freedom Multiple R-Squared: 0.02933,Adjusted R-squared: 0.007017 F-statistic: 1.314 on 2 and 87 DF, p-value: 0.2739 is significant only the intercept After I estimated the same model with R 2.7 and this is the new output: Output R 2.7 regressione14-lm(UN ~ log(x)+ time2,data=dati) summary(regressione14) Call: lm(formula = UN ~ log(x) + time2, data = dati) Residuals: Min 1Q Median 3QMax -8.825 -2.094 -0.861 2.233 12.664 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 22.724208 3.304670 6.876 8.89e-10 *** log(x) -7.117981 1.499482 -4.747 8.05e-06 *** time20.002051 0.000523 3.922 0.000175 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 3.853 on 87 degrees of freedom Multiple R-Squared: 0.2289, Adjusted R-squared: 0.2112 F-statistic: 12.91 on 2 and 87 DF, p-value: 1.227e-05 Well, how you can see the two outputs are very differents, so i try to estimated the same model with another package and i show the output: *Variabile* *Coefficiente* *Errore Std.* *Statistica t* *p-value* const 22,7242 3,30467 6,8764 0,1 *** logx -7,11798 1,49948 -4,7470 0,1 *** time2 0,00205144 0,000523012 3,9224 0,00017 *** Media della variabile dipendente = 6,47833 Scarto quadratico medio della variabile dipendente = 4,33827 Somma dei quadrati dei residui = 1291,58 Errore standard dei residui = 3,85302 R2 = 0,228922 R2 corretto = 0,211196 Statistica F (2, 87) = 12,9145 (p-value = 1,23e-005) Log-verosimiglianza = -247,576 Criterio di informazione di Akaike = 501,152 Criterio bayesiano di Schwarz = 508,652 Criterio di Hannan-Quinn = 504,176 How you can see this package show the same output of the R 2.7 My question is where is the bug? Kind regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Word wrapping for character objects (WINDOWS R ONLY)
You could try passing your character string to the strwrap function first, then use cat on the result. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Toby Marthews Sent: Wednesday, June 11, 2008 6:52 AM To: r-help@r-project.org Subject: [R] Word wrapping for character objects (WINDOWS R ONLY) Can anybody help me with this problem? ** ONLY WINDOWS R - PROBLEM DOESN'T OCCUR ON LINUX ** I want to print a long character to screen: getOption(width) [1] 60 z=(1:20)/10#z is a vector of length between 20 and 30 (depending on user options) containing lengths in mm (i.e. each element is 1-5 characters long) str1=paste(The depths chosen are (,toString(z),) mm, and more text ...\n) cat(str1) The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2, 1.3, 1.$ The problem is that on R for Windows the string is cropped by the window size (hence the $). On R for Linux, this doesn't happen and the text is word wrapped (the default for the shell, I guess): cat(str1) The depths chosen are ( 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2 ) mm, and more text ... I can't find any option for word wrapping in the cat command (fill=TRUE has no effect). I also checked the menu Edit - GUI preferences..., but there doesn't seem to be a Word Wrap option there either. How do I get word wrapping like this in Windows? Perhaps the attached screenshots clarify this question. THANKS FOR ANY HELP! Toby Marthews Previous relevant posts: - The post from 2006 about Screen Wrapping (http://tolstoy.newcastle.edu.au/R/help/06/05/26673.html) which Brian Ripley answered was about controlling how long vectors are cropped to the screen. Unfortunately, the width option in options() does not affect character objects, so I can't use that control here. - I sent the same question to [EMAIL PROTECTED] in Oct 2007, but noone there could help me with it. - Try the following command on Windows R with a small window (getOption(width)117) and a large window (getOption(width)117) and you'll see you get extra nonexistent options in the menu: a=c(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct ,Nov,Dec);menu(a) I guess this is a related 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] applying a function recursively
See ?rapply On Wed, 11 Jun 2008, Georg Otto wrote: Hi, I have a question about applying a function recursively through a list. Suppose I have a list where the different elements have different levels of recursion: test.list-list(I=list(A=c(a, b, c), B=c(d, e, f), C=c(g, h, i)), + II=list(A=list(a=c(a, b, c), b=c(d, e, f), + c=c(g, h, i)), + B=c(d, e, f), C=c(g, h, i))) test.list $I $I$A [1] a b c $I$B [1] d e f $I$C [1] g h i $II $II$A $II$A$a [1] a b c $II$A$b [1] d e f $II$A$c [1] g h i $II$B [1] d e f $II$C [1] g h i I would like to apply a function recursively to that list, in a way that the function does someting with each vector (eg. rev()) and returns a list of modified vectors that has the same structure as the input list, in my example: $I $I$A [1] c b a $I$B [1] f e d $I$C [1] i h g $II $II$A $II$A$a [1] c b a $II$A$b [1] f e d $II$A$c [1] i h g $II$B [1] f e d $II$C [1] i h g I understand that with a fixed number of recursion levels one can use lapply() in a nested way, but what if the numbers of recursion levels is not fixed or is different between the list elements as it is in my example? Any hint will be appreciated. Best, Georg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Increase Number of Decimals
R Users, I'm new to R and was wondering how I can show more decimal places when I run commands. If I'm simply running a correlation(ES,YM) how would I increase the number of decimal places R shows? When I run this it shows me .9734044. How can I extend this further? In addition I was running histograms on high frequency data to check a Spread. The spread moves at an extremely granular level. It may move .01 frequently but again at a low level. How can I increase the graularity of the histogram plots to detect smaller moves? I appreciate any and all help. Neil [[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] applying a function recursively
See ?rapply for your example rapply( test.list, rev, how='replace' ) HTH, Chuck On Wed, 11 Jun 2008, Georg Otto wrote: Hi, I have a question about applying a function recursively through a list. Suppose I have a list where the different elements have different levels of recursion: test.list-list(I=list(A=c(a, b, c), B=c(d, e, f), C=c(g, h, i)), + II=list(A=list(a=c(a, b, c), b=c(d, e, f), + c=c(g, h, i)), + B=c(d, e, f), C=c(g, h, i))) test.list $I $I$A [1] a b c $I$B [1] d e f $I$C [1] g h i $II $II$A $II$A$a [1] a b c $II$A$b [1] d e f $II$A$c [1] g h i $II$B [1] d e f $II$C [1] g h i I would like to apply a function recursively to that list, in a way that the function does someting with each vector (eg. rev()) and returns a list of modified vectors that has the same structure as the input list, in my example: $I $I$A [1] c b a $I$B [1] f e d $I$C [1] i h g $II $II$A $II$A$a [1] c b a $II$A$b [1] f e d $II$A$c [1] i h g $II$B [1] f e d $II$C [1] i h g I understand that with a fixed number of recursion levels one can use lapply() in a nested way, but what if the numbers of recursion levels is not fixed or is different between the list elements as it is in my example? Any hint will be appreciated. Best, Georg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] applying a function recursively
on 06/11/2008 10:51 AM Georg Otto wrote: Hi, I have a question about applying a function recursively through a list. Suppose I have a list where the different elements have different levels of recursion: test.list-list(I=list(A=c(a, b, c), B=c(d, e, f), C=c(g, h, i)), + II=list(A=list(a=c(a, b, c), b=c(d, e, f), + c=c(g, h, i)), + B=c(d, e, f), C=c(g, h, i))) test.list $I $I$A [1] a b c $I$B [1] d e f $I$C [1] g h i $II $II$A $II$A$a [1] a b c $II$A$b [1] d e f $II$A$c [1] g h i $II$B [1] d e f $II$C [1] g h i I would like to apply a function recursively to that list, in a way that the function does someting with each vector (eg. rev()) and returns a list of modified vectors that has the same structure as the input list, in my example: $I $I$A [1] c b a $I$B [1] f e d $I$C [1] i h g $II $II$A $II$A$a [1] c b a $II$A$b [1] f e d $II$A$c [1] i h g $II$B [1] f e d $II$C [1] i h g I understand that with a fixed number of recursion levels one can use lapply() in a nested way, but what if the numbers of recursion levels is not fixed or is different between the list elements as it is in my example? Any hint will be appreciated. Best, Georg See ?rapply, which is a recursive version of lapply(): rapply(test.list, rev, how = list) $I $I$A [1] c b a $I$B [1] f e d $I$C [1] i h g $II $II$A $II$A$a [1] c b a $II$A$b [1] f e d $II$A$c [1] i h g $II$B [1] f e d $II$C [1] i h g HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to save an updated dataset
I wrote a package which includes a number of genome sequencing project statistics on the web like http://www.ncbi.nlm.nih.gov/genomes/lproks.cgi. I included some generic functions to summarize, plot, and update the tables with the most recent version data(lproks) update(lproks) [1] lproks successfully updated, 7 new genomes added I usually save the dataset back to my package data directory... save(lproks, file=/path/to/genomes/data/lproks.rda) but I may one day put the package on CRAN (or BioConductor), so now I need to know where the package/data directory is located, if the user has permission to save to that directory, and probably some other complications I'm missing. Any suggestions? Thanks, Chris Stubben -- View this message in context: http://www.nabble.com/how-to-save-an-updated-dataset-tp17782584p17782584.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] Increase Number of Decimals
options(digits=...) On Wed, Jun 11, 2008 at 12:17 PM, Neil Gupta [EMAIL PROTECTED] wrote: R Users, I'm new to R and was wondering how I can show more decimal places when I run commands. If I'm simply running a correlation(ES,YM) how would I increase the number of decimal places R shows? When I run this it shows me .9734044. How can I extend this further? In addition I was running histograms on high frequency data to check a Spread. The spread moves at an extremely granular level. It may move .01 frequently but again at a low level. How can I increase the graularity of the histogram plots to detect smaller moves? I appreciate any and all help. Neil [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Increase Number of Decimals
On Jun 11, 2008, at 12:17 PM, Neil Gupta wrote: R Users, I'm new to R and was wondering how I can show more decimal places when I run commands. If I'm simply running a correlation(ES,YM) how would I increase the number of decimal places R shows? When I run this it shows me . 9734044. How can I extend this further? I am not sure what you can gain by looking at more decimal points in a correlation, but there are two ways, as far as I can tell: 1) Wrap the command in print, and use the digits optional argument in print. Have a look at ?print for details 2) Change the default more permanently for the session, by something like options(digits=10) Look at ?options for details In addition I was running histograms on high frequency data to check a Spread. The spread moves at an extremely granular level. It may move .01 frequently but again at a low level. How can I increase the graularity of the histogram plots to detect smaller moves? We would need a reproducible example for this, I think. I appreciate any and all help. Neil Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Parsing a data file - Help
Hi All, I have the data in the following format: idktsaaplahto pidg 5266199119943078A 526619921005199210302968A 19930208199302093074A 2002032920020402F322 2002040220020409F322 6866198107131981091729800 6866198111091981112029550 6866198202031982021929550 I would like to parse the data and reformat into a single row for each unique idkt, something like: 5266 1991 1994 3078A 19921005 19921030 2968A I have tried with f - read.table(file.txt, sep=\t, header=TRUE); attach(f); fac - factor(f[,1]); id - matrix(length(fac), 4); for(i in fac) id[i] - f[idkt %in% fac[i], ]; I am not able make the list id into a single row. Could you please help how I can do this? Thanks in advance. Kind regards, Ezhil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Barplot help
#I am having trouble figuring out this one. I have read the help and I am at a loss. what am I missing x - structure(list(X = structure(c(6L, 5L, 9L, 2L, 10L, 8L, 7L, 3L, 13L, 12L, 11L, 4L, 1L, 1L, 1L), .Label = c(, April, August, December, February, January, July, June, March, May, November, October, September), class = factor), X2006 = c(5987.387, 6478.592, 8386, 4651.273, 4339.167, 4631.978, 5217.306, 5846.903, 3867.825, 3886.434, 3959.668, 3848.853, NA, NA, NA), X2007 = c(4354.516, 5924.315, 5559.468, 3967.5, 5053.56, 4808.694, 4017.632, 3969.883, 3910.236, 3782.094, 3961.286, 3711.262, NA, NA, NA), X2008 = c(3685.789, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c(X, X2006, X2007, X2008), class = data.frame, row.names = c(NA, -15L)) barplot(x) #I would like the months to be on the x-axis and then have the bars be the values for the year -- 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 [[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] Barplot help
on 06/11/2008 12:44 PM stephen sefick wrote: #I am having trouble figuring out this one. I have read the help and I am at a loss. what am I missing x - structure(list(X = structure(c(6L, 5L, 9L, 2L, 10L, 8L, 7L, 3L, 13L, 12L, 11L, 4L, 1L, 1L, 1L), .Label = c(, April, August, December, February, January, July, June, March, May, November, October, September), class = factor), X2006 = c(5987.387, 6478.592, 8386, 4651.273, 4339.167, 4631.978, 5217.306, 5846.903, 3867.825, 3886.434, 3959.668, 3848.853, NA, NA, NA), X2007 = c(4354.516, 5924.315, 5559.468, 3967.5, 5053.56, 4808.694, 4017.632, 3969.883, 3910.236, 3782.094, 3961.286, 3711.262, NA, NA, NA), X2008 = c(3685.789, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c(X, X2006, X2007, X2008), class = data.frame, row.names = c(NA, -15L)) barplot(x) #I would like the months to be on the x-axis and then have the bars be the values for the year Is this what you want? barplot(t(x[-1]), names.arg = x$X, las = 2) Note that the 'height' argument for barplot() needs to be either a vector or a matrix. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Barplot help
I had to to get the data as a matrix and fix a couple of other things- If you would like for me to post my solution then I can. I appoligize my fingers went faster than my brain. Stephen On Wed, Jun 11, 2008 at 1:55 PM, Marc Schwartz [EMAIL PROTECTED] wrote: on 06/11/2008 12:44 PM stephen sefick wrote: #I am having trouble figuring out this one. I have read the help and I am at a loss. what am I missing x - structure(list(X = structure(c(6L, 5L, 9L, 2L, 10L, 8L, 7L, 3L, 13L, 12L, 11L, 4L, 1L, 1L, 1L), .Label = c(, April, August, December, February, January, July, June, March, May, November, October, September), class = factor), X2006 = c(5987.387, 6478.592, 8386, 4651.273, 4339.167, 4631.978, 5217.306, 5846.903, 3867.825, 3886.434, 3959.668, 3848.853, NA, NA, NA), X2007 = c(4354.516, 5924.315, 5559.468, 3967.5, 5053.56, 4808.694, 4017.632, 3969.883, 3910.236, 3782.094, 3961.286, 3711.262, NA, NA, NA), X2008 = c(3685.789, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c(X, X2006, X2007, X2008), class = data.frame, row.names = c(NA, -15L)) barplot(x) #I would like the months to be on the x-axis and then have the bars be the values for the year Is this what you want? barplot(t(x[-1]), names.arg = x$X, las = 2) Note that the 'height' argument for barplot() needs to be either a vector or a matrix. HTH, Marc Schwartz -- 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 [[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] MLE Estimation of Gamma Distribution Parameters for data with 'zeros'
Greetings, all
I am having difficulty getting the fitdistr() function to return without
an error on my data. Specifically, what I'm trying to do is get a
parameter estimation for fracture intensity data in a well / borehole.
Lower bound is 0 (no fractures in the selected data interval), and upper
bound is ~ 10 - 50, depending on what scale you are conducting the
analysis on.
I read in the data from a text file, convert it to numerics, and then
calculate initial estimates of the shape and scale parameters for the
gamma distribution from moments. I then feed this back into the
fitdistr() function.
R code (to this point):
data.raw=c(readLines(FSM_C_9m_ENE.inp))
data.num - as.numeric(data.raw)
data.num
library(MASS)
shape.mom = ((mean(data.num))/ (sd(data.num))^2
shape.mom
med.data = mean(data.num)
sd.data = sd(data.num)
med.data
sd.data
shape.mom = (med.data/sd.data)^2
shape.mom
scale.mom = (sd.data^2)/med.data
scale.mom
fitdistr(data.num,gamma,list(shape=shape.mom,
scale=scale.mom),lower=0)
fitdistr() returns the following error:
Error in optim(x = c(0.402707037, 0.40348, 0.404383704,
2.432626667, :
L-BFGS-B needs finite values of 'fn'
Next thing I tried was to manually specify the negative log-likelihood
function and pass it straight to mle() (the method specified in Ricci's
tutorial on fitting distributions with R). Basically, I got the same
result as using fitdistr().
Finally I tried using some R code I found from someone with a similar
problem back in 2003 from the archives of this mailing list:
R code
gamma.param1 - shape.mom
gamma.param2 - scale.mom
log.gamma.param1 - log(gamma.param1)
log.gamma.param2 - log(gamma.param2)
gammaLoglik -
function(params,
negative=TRUE){
lglk - sum(dgamma(data, shape=exp(params[1]), scale=exp(params[2]),
log=TRUE))
if(negative)
return(-lglk)
else
return(lglk)
}
optim.list - optim(c(log.gamma.param1, log.gamma.param2), gammaLoglik)
gamma.param1 - exp(optim.list$par[1])
gamma.param2 - exp(optim.list$par[2])
#
If I test this function using my sample data and the estimates of shape
and scale derived from the method of moments, gammaLogLike returns as
INF. I suspect the problem is that the zeros in the data are causing the
optim solver problems when it attempts to minimize the negative
log-likelihood function.
Can anyone suggest some advice on a work-around? I have seen
suggestions online that a 'censoring' algorithm can allow one to use MLE
methods to estimate the gamma distribution for data with zero values
(Wilkes, 1990, Journal of Climate). I have not, however, found R code to
implement this, and, frankly, am not smart enough to do it myself... :-)
Any suggestions? Has anyone else run up against this and written code to
solve the problem?
Thanks in advance!
Aaron Fox
Senior Project Geologist, Golder Associates
+1 425 882 5484 || +1 425 736 3958 (mobile)
[EMAIL PROTECTED] || www.fracturedreservoirs.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] Parsing a data file - Help
On 6/11/2008 1:29 PM, A Ezhil wrote:
Hi All,
I have the data in the following format:
idktsaaplahto pidg
5266199119943078A
526619921005199210302968A
19930208199302093074A
2002032920020402F322
2002040220020409F322
6866198107131981091729800
6866198111091981112029550
6866198202031982021929550
I would like to parse the data and reformat into a single row for each unique
idkt, something like:
5266 1991 1994 3078A 19921005 19921030 2968A
I have tried with
f - read.table(file.txt, sep=\t, header=TRUE);
attach(f);
fac - factor(f[,1]);
id - matrix(length(fac), 4);
for(i in fac) id[i] - f[idkt %in% fac[i], ];
I am not able make the list id into a single row. Could you please help how I
can do this?
If you can create a variable that differentiates multiple records
from the same idkt, you can use reshape() like this:
f - idkt saap lahto pidg
5266 1991 1994 3078A
5266 19921005 19921030 2968A
19930208 19930209 3074A
20020329 20020402 F322
20020402 20020409 F322
6866 19810713 19810917 29800
6866 19811109 19811120 29550
6866 19820203 19820219 29550
fdata - read.table(textConnection(f), sep= , header=TRUE)
fdata$time - unlist(lapply(table(fdata$idkt), function(x){1:x}))
reshape(fdata, idvar = idkt, timevar = time, direction=wide)
idkt saap.1 lahto.1 pidg.1 saap.2 lahto.2 pidg.2 saap.3
lahto.3 pidg.3
1 5266 1991 1994 3078A 19921005 19921030 2968A NA
NA NA
3 19930208 19930209 3074A 20020329 20020402 F322 20020402
20020409 F322
6 6866 19810713 19810917 29800 19811109 19811120 29550 19820203
19820219 29550
Thanks in advance.
Kind regards,
Ezhil
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
fax: (917) 438-0894
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Search entire table for a value
Hi, I have a table of numbers without any repeats. Is there a function that will search the entire table for a specified value and return the indices if it is found? Also, some rows may have columns without values. Thank you so much, -Nina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] specifying ranges in scatter plot
Hi, there: Does anyone know how to specify the ranges in the axises when I make scatter plots using pairs()? In the general plot function, I can use xlim and ylim option. But how can I do this if I use pairs()? Thanks. Yulei [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Search entire table for a value
See ?which On Wed, Jun 11, 2008 at 4:19 PM, [EMAIL PROTECTED] wrote: Hi, I have a table of numbers without any repeats. Is there a function that will search the entire table for a specified value and return the indices if it is found? Also, some rows may have columns without values. Thank you so much, -Nina __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Help!!! Agnes dendogram (Clustering)
The data one is a vector of 553 observations agglone-agnes(one, metric = manhattan, stand = TRUE) plot(agglone,which.plots=2, nmax=150) My problem is in the dendogram, I can not see the nodes because it is too crowded. I have attached the diagram. Any help is more than welcome. Thank you a lot!!! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 and Fortran
Dear Fellow R-Users, I am having some difficulties loading a Fortran subroutine into R and wondering if anyone could lend me some insight to this problem. When I load some simpler Fortran codes into R, it loads fine, but when I load more complex codes I get the following error: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /nfs/home/grad/d.lee/Testing/chol.o: ld.so.1: R: fatal: relocation error: file /nfs/home/grad/d.lee/Testing/chol.o: symbol s_wsle: referenced symbol not found From what I have gathered, it appears to be a problem when with linking the library to the Fortran system library, but I am unsure how to deal with this. The current 32-bit machine that runs R is: platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major2 minor0.1 year 2004 month11 day 15 language R Any help to this query would be very much appreciated. Cheers. Sincerely - Derrick Derrick Lee, MSc Candidate Department of Statistics The University of British Columbia LSK-314A | 604 - 822 - 1299 x532 d.lee at stat dot ubc dot ca | dgylee at mun dot ca www.stat.ubc.ca/~d.lee/ | www.math.mun.ca/~derrick0/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] specifying ranges in scatter plot
You can use also xlim and ylim, from the help page: ...: arguments to be passed to or from methods. pairs(iris[1:4], main = Anderson's Iris Data -- 3 species, pch = 21, bg = c(red, green3, blue)[unclass(iris$Species)]) pairs(iris[1:4], main = Anderson's Iris Data -- 3 species, pch = 21, bg = c(red, green3, blue)[unclass(iris$Species)], xlim = c(0,10), ylim = c(0,10)) On Wed, Jun 11, 2008 at 2:05 PM, He, Yulei [EMAIL PROTECTED] wrote: Hi, there: Does anyone know how to specify the ranges in the axises when I make scatter plots using pairs()? In the general plot function, I can use xlim and ylim option. But how can I do this if I use pairs()? Thanks. Yulei [[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 and Fortran
Dear Fellow R-Users, I am having some difficulties loading a Fortran subroutine into R and wondering if anyone could lend me some insight to this problem. When I load some simpler Fortran codes into R, it loads fine, but when I load more complex codes I get the following error: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /nfs/home/grad/d.lee/Testing/chol.o: ld.so.1: R: fatal: relocation error: file /nfs/home/grad/d.lee/Testing/chol.o: symbol s_wsle: referenced symbol not found From what I have gathered, it appears to be a problem when with linking the library to the Fortran system library, but I am unsure how to deal with this. As well, I was wondering why I am unsuccessful loading those same successful codes when I go from a 32-bit machine to a 64-bit machine, why does that cause such a big issue? The current 32-bit machine that runs R is: platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major2 minor0.1 year 2004 month11 day 15 language R Any help to these queries would be very much appreciated. Cheers. Sincerely - Derrick Derrick Lee, MSc Candidate Department of Statistics The University of British Columbia LSK-314A | 604 - 822 - 1299 x532 d.lee at stat dot ubc dot ca | dgylee at mun dot ca www.stat.ubc.ca/~d.lee/ | www.math.mun.ca/~derrick0/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] piper diagram
On Wednesday 11 June 2008, Jim Lemon wrote: Jim Lemon wrote: Dylan Beaudette wrote: Hi, Is anyone on the list familiar with an R implementation of Piper Diagrams? Example: http://faculty.uml.edu/nelson_eby/89.315/IMAGES/Figure%209-78.jpg I am thinking that two calls to triax.plot (plotrix) along with some kind of affine-transformed standard plot would do the trick. Not so sure about the final layout, or a nice generalized version for something like lattice. Okay, I've had a look at the Piper diagram and with better labeled ones and some explanations, I think I've got the picture. No, triax.plot won't do the job, but it's a similar sort of illustration. Give me a few days to think about it and I may be able to put one together. Jim Any ideas would be appreciated! Cheers, Dylan -- Dylan Beaudette Soil Resource Laboratory http://casoilresource.lawr.ucdavis.edu/ University of California at Davis 530.754.7341 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and Fortran
Hi Erin, Unfortunately no, its not named chol, I know about the cholesky decomposition in R, I was just testing out a more complex subroutine as the subroutine I intend to use gets similar errors. Thanks for the insight, though. Cheers. - Derrick Erin Hodgess wrote: Hi Derrick! Is your subroutine named chol? There is a function in R by the name of chol. You might be having a problem with that. thanks, Erin On Wed, Jun 11, 2008 at 2:39 PM, Derrick Lee [EMAIL PROTECTED] wrote: Dear Fellow R-Users, I am having some difficulties loading a Fortran subroutine into R and wondering if anyone could lend me some insight to this problem. When I load some simpler Fortran codes into R, it loads fine, but when I load more complex codes I get the following error: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /nfs/home/grad/d.lee/Testing/chol.o: ld.so.1: R: fatal: relocation error: file /nfs/home/grad/d.lee/Testing/chol.o: symbol s_wsle: referenced symbol not found From what I have gathered, it appears to be a problem when with linking the library to the Fortran system library, but I am unsure how to deal with this. The current 32-bit machine that runs R is: platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major2 minor0.1 year 2004 month11 day 15 language R Any help to this query would be very much appreciated. Cheers. Sincerely - Derrick Derrick Lee, MSc Candidate Department of Statistics The University of British Columbia LSK-314A | 604 - 822 - 1299 x532 d.lee at stat dot ubc dot ca | dgylee at mun dot ca www.stat.ubc.ca/~d.lee/ | www.math.mun.ca/~derrick0/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sincerely - Derrick Derrick Lee, MSc Candidate Department of Statistics The University of British Columbia LSK-314A | 604 - 822 - 1299 x532 d.lee at stat dot ubc dot ca | dgylee at mun dot ca www.stat.ubc.ca/~d.lee/ | www.math.mun.ca/~derrick0/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] MLE Estimation of Gamma Distribution Parameters for data with 'zeros'
Fox, Aaron wrote:
Greetings, all
I am having difficulty getting the fitdistr() function to return without
an error on my data. Specifically, what I'm trying to do is get a
parameter estimation for fracture intensity data in a well / borehole.
Lower bound is 0 (no fractures in the selected data interval), and upper
bound is ~ 10 - 50, depending on what scale you are conducting the
analysis on.
I read in the data from a text file, convert it to numerics, and then
calculate initial estimates of the shape and scale parameters for the
gamma distribution from moments. I then feed this back into the
fitdistr() function.
R code (to this point):
data.raw=c(readLines(FSM_C_9m_ENE.inp))
data.num - as.numeric(data.raw)
data.num
library(MASS)
shape.mom = ((mean(data.num))/ (sd(data.num))^2
shape.mom
med.data = mean(data.num)
sd.data = sd(data.num)
med.data
sd.data
shape.mom = (med.data/sd.data)^2
shape.mom
scale.mom = (sd.data^2)/med.data
scale.mom
fitdistr(data.num,gamma,list(shape=shape.mom,
scale=scale.mom),lower=0)
fitdistr() returns the following error:
Error in optim(x = c(0.402707037, 0.40348, 0.404383704,
2.432626667, :
L-BFGS-B needs finite values of 'fn'
Next thing I tried was to manually specify the negative log-likelihood
function and pass it straight to mle() (the method specified in Ricci's
tutorial on fitting distributions with R). Basically, I got the same
result as using fitdistr().
Finally I tried using some R code I found from someone with a similar
problem back in 2003 from the archives of this mailing list:
R code
gamma.param1 - shape.mom
gamma.param2 - scale.mom
log.gamma.param1 - log(gamma.param1)
log.gamma.param2 - log(gamma.param2)
gammaLoglik -
function(params,
negative=TRUE){
lglk - sum(dgamma(data, shape=exp(params[1]), scale=exp(params[2]),
log=TRUE))
if(negative)
return(-lglk)
else
return(lglk)
}
optim.list - optim(c(log.gamma.param1, log.gamma.param2), gammaLoglik)
gamma.param1 - exp(optim.list$par[1])
gamma.param2 - exp(optim.list$par[2])
#
If I test this function using my sample data and the estimates of shape
and scale derived from the method of moments, gammaLogLike returns as
INF. I suspect the problem is that the zeros in the data are causing the
optim solver problems when it attempts to minimize the negative
log-likelihood function.
Can anyone suggest some advice on a work-around? I have seen
suggestions online that a 'censoring' algorithm can allow one to use MLE
methods to estimate the gamma distribution for data with zero values
(Wilkes, 1990, Journal of Climate). I have not, however, found R code to
implement this, and, frankly, am not smart enough to do it myself... :-)
Any suggestions? Has anyone else run up against this and written code to
solve the problem?
It's fairly easy. You decide that the zeros really represent values less
than delta (e.g. 0.5 if your data are integers), then replace
dgamma(0,) with pgamma(delta,...) in the likelihood. (And, BTW, the
problem is not that the optimizers get in trouble, but rather that the
log-likelihood *is* +/- Inf if there are zeros in data unless the shape
parameter is exactly 1 -- the x^(a-1) factor in the gamma density causes
this).
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 and Fortran
Dear Fellow R-Users, I am having some difficulties loading a Fortran subroutine into R and wondering if anyone could lend me some insight to this problem. When I load some simpler Fortran codes into R, it loads fine, but when I load more complex codes I get the following error: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /nfs/home/grad/d.lee/Testing/chol.o: ld.so.1: R: fatal: relocation error: file /nfs/home/grad/d.lee/Testing/chol.o: symbol s_wsle: referenced symbol not found From what I have gathered, it appears to be a problem when with linking the library to the Fortran system library, but I am unsure how to deal with this. As well, I was wondering why I am unsuccessful loading those same successful codes when I go from a 32-bit machine to a 64-bit machine, why does that cause such a big issue? The current 32-bit machine that runs R is: platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major2 minor0.1 year 2004 month11 day 15 language R Any help to these queries would be very much appreciated. Cheers. Sincerely - Derrick Derrick Lee, MSc Candidate Department of Statistics The University of British Columbia LSK-314A | 604 - 822 - 1299 x532 d.lee at stat dot ubc dot ca | dgylee at mun dot ca www.stat.ubc.ca/~d.lee/ | www.math.mun.ca/~derrick0/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Copula fitting
first, verify the class of your data by the following code: class(data) the class must be numeric if ok the problem can be caused from the starting value. Try choose other value. -- View this message in context: http://www.nabble.com/-R--Copula-fitting-tp6947714p17786789.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] Close Window Button Problems
i have created a wrapper C++ class that popen()s R. When i create a x11 window and plot, i cannot close it using the window button. Also, when i minimize or maximize the window, the plot does not redraw. Can anybody tell me if there is a way i can get back this window functionality or if it is not possible? thank you kindly. -damon __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and Fortran
Are you serious, reporting on a 2004 version of R? Please do as the posting guide requested, and update your R. (Only posting one copy is also appreciated around here.) Beyond that, normally foo.o is not a shared library but a compiled object. You have not told us what you did, and probably the issue is that you don't know how to make a shared library from Fortran. Please ask your IT advisers about that (and hint, R CMD SHLIB may help you). On Wed, 11 Jun 2008, Derrick Lee wrote: Dear Fellow R-Users, I am having some difficulties loading a Fortran subroutine into R and wondering if anyone could lend me some insight to this problem. When I load some simpler Fortran codes into R, it loads fine, but when I load more complex codes I get the following error: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library /nfs/home/grad/d.lee/Testing/chol.o: ld.so.1: R: fatal: relocation error: file /nfs/home/grad/d.lee/Testing/chol.o: symbol s_wsle: referenced symbol not found From what I have gathered, it appears to be a problem when with linking the library to the Fortran system library, but I am unsure how to deal with this. The current 32-bit machine that runs R is: platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major2 minor0.1 year 2004 month11 day 15 language R Any help to this query would be very much appreciated. Cheers. Sincerely - Derrick Derrick Lee, MSc Candidate Department of Statistics The University of British Columbia LSK-314A | 604 - 822 - 1299 x532 d.lee at stat dot ubc dot ca | dgylee at mun dot ca www.stat.ubc.ca/~d.lee/ | www.math.mun.ca/~derrick0/ -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] the title is too long for a graph
Thank you all! The answers are very helpful! Best regards, Hua --- On Wed, 6/11/08, Prof Brian Ripley [EMAIL PROTECTED] wrote: Hua Li wrote: Hi All, I have a problem of putting long titles on a graph: for example, x= seq(1:100) y=seq(1:100) plot(x,y,main=p=0.05:A-B=3,C-D=10,D-E=100,A-F=2,AFR-E=3,ACE-D=1,ADEF-M=0,AED-E=10,DE-F=3,AB-J=4,AC-J=10,ED-F=1,ED-B=4,AF-B=10,CD-S=10,AM-C=4) R seems not able to print the whole title. The __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ETH Internship - Dynamic Portfolio Asset Allocation
Summer Internship at ETH Zurich Dynamic Portfolio Asset Allocation We offer a 3-months internship starting midth July 2008. The topic addresses Dynamic Portfolio Asset Allocation including alternative instruments and hedge funds. The goal will be to compare the robust mean-variance, the lower partial moment and the conditional value-at-risk approaches for portfolio construction and optimization using the Rmetrics package fPortfolio. Moreover, we will investigate the influence of quadratic covariance and copulae related tail risk budget constraints as an option to limit and control the risk attributed by individual assets. We offer a generous compensation for traveling, accomodation, and living costs in the beautiful city of Zurich. The candidate should have a strong quantitative background and have experience in data modelling with the R language. The application should include a letter of motivation (highlighting your interest and prior knowledge in the internship topic) and an up-to-date CV. Please send your application to [EMAIL PROTECTED] If you need further information, please contact us. PD Dr. Diethelm Wuertz Econophysics Group at the Institute of Theoretical Physics ETH Zurich www.ethz.ch www.phys.ethz.ch www.rmetrics.org NOTE: Rmetrics Workshop: http://www.rmetrics.org/meielisalp.htm June 29th - July 3rd Meielisalp, Lake Thune, Switzerland __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] MLE Estimation of Gamma Distribution Parameters for data with 'zeros'
Fox, Aaron Afox at golder.com writes: Greetings, all I am having difficulty getting the fitdistr() function to return without an error on my data. Specifically, what I'm trying to do is get a parameter estimation for fracture intensity data in a well / borehole. Lower bound is 0 (no fractures in the selected data interval), and upper bound is ~ 10 - 50, depending on what scale you are conducting the analysis on. You're right that the basic problem is with the gamma distribution. P(x,shape) dx = 0 (shape1), 1 (shape=1), or Inf (shape1). A quick cheat would be to add a small number (0.001?) to your data, try it again, and see how sensitive the estimate is to how small that number is. You could also try a negative binomial fit, which is the discrete analog of the gamma (and hence won't have any problem with zeros). People who do beta regressions with zero values in them often talk about adding a small Bayesian 'fudge factor' to deal with this problem ... (see http://psychology.anu.edu.au/people/smithson/details/betareg/Readme.pdf ) 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] Close Window Button Problems
This should be posted on R-devel, not here. -- Bert -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Wednesday, June 11, 2008 1:30 PM To: r-help@r-project.org Subject: [R] Close Window Button Problems i have created a wrapper C++ class that popen()s R. When i create a x11 window and plot, i cannot close it using the window button. Also, when i minimize or maximize the window, the plot does not redraw. Can anybody tell me if there is a way i can get back this window functionality or if it is not possible? thank you kindly. -damon __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] model simplification using Crawley as a guide
Lucke, Joseph F Joseph.F.Lucke at uth.tmc.edu writes: And to follow FH and HW What level of significance are you using? .05 is excessively liberal. Are you adjusting your p-values for the number of possible models? Do you realize the p-values for dropping a term, being selected as the maximum of a set of p-values, do not follow their usual distributions? How are you compensating for sample size, as a p-value's being significant is a function of sample size? How are you compensating for the fact that the current model choice is dependent on the previous model choices? How do you know your tree of model choices is the optimal one? Have you considered cross-validation? Are you looking for a model that true describes a phenomenon or a predictive model that can be used for practical purposes? Ouch. While Frank Harrell and Joseph Lucke are raising serious issues about model selection, maybe we could keep in mind that we don't want to scare off all the students who ever try to use R to figure out basic statistics. I would follow Peter Dalgaard's advice (about drop1) and Hadley Wickham's (about graphical diagnostics), and if possible bring up the other issues about model selection with others around you -- if you're a student, ask your prof. or someone in the stats department. It can be tough to try to do things right if those around you are still doing them wrong ... If you tell us what field you're in we may be able to point you to more subject-specific references (e.g. Whittingham, Mark J., Philip A. Stephens, Richard B. Bradbury, and Robert P. Freckleton. 2006. Why do we still use stepwise modelling in ecology and behaviour? Journal of Animal Ecology 75, no. 5: 1182-1189) 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] converting a data set to a format for time series analysis
Jim, it worked perfectly. thanks a lot On Mon, Jun 9, 2008 at 8:58 PM, jim holtman [EMAIL PROTECTED] wrote: This should do it: x - read.table(textConnection(subject hospitaldate_enrollment hospital_beds + 1 hospitalA 1/3/2002300 + 2 hospitalA 1/6/2002300 + 3 hospitalB 2/4/2002150 + 4 hospitalC 3/2/2002200), header=TRUE) closeAllConnections() y - as.Date(x$date_enrollment, %m/%d/%Y) z - cbind(x, year=format(y, %Y), month=format(y, %m)) # partition the data z.s - split(z, list(z$year, z$month, z$hospital), drop=TRUE) # now aggregate do.call(rbind, lapply(z.s, function(a) data.frame(hospital=a$hospital[1], cases=nrow(a), + year=a$year[1], month=a$month[1], beds=a$hospital[1]))) hospital cases year month beds 2002.01.hospitalA hospitalA 2 200201 hospitalA 2002.02.hospitalB hospitalB 1 200202 hospitalB 2002.03.hospitalC hospitalC 1 200203 hospitalC On Mon, Jun 9, 2008 at 1:51 PM, Ricardo Pietrobon [EMAIL PROTECTED] wrote: Jim, thanks a lot. This does the trick for dates, but what I have been struggling the most with is actually the conversion from having one subject per row to having one month per row. I didn't explain that well at all in my previous email and so let me try again. The idea is that the current data set is displayed with one subject per row. I would like to have it displayed having one hospital per month per row. For example, the new data set would look like this: month yearsite number_enrolled_subjects hospital_beds 1 2002 hospitalA 22 300 meaning that hospital A enrolled 22 subjects in 01/2002, and hospital A has 300 beds -- the beds variable is one variable in a vector that would display all the covariates for my ARIMA model your suggestion solved the problem for the dates, but the command I am looking for now is something that would count the number of subjects per site per month of a year and then displayed it in the format above. any thoughts? I really appreciate your help On Mon, Jun 9, 2008 at 1:04 PM, jim holtman [EMAIL PROTECTED] wrote: Will something like this work for you: x - read.table(textConnection(subject hospitaldate_enrollment hospital_beds + 1 hospitalA 1/3/2002300 + 2 hospitalA 1/6/2002300 + 3 hospitalB 2/4/2002150 + 4 hospitalC 3/2/2002200), header=TRUE) closeAllConnections() y - as.Date(x$date_enrollment, %m/%d/%Y) cbind(x, year=format(y, %Y), month=format(y, %m)) subject hospital date_enrollment hospital_beds year month 1 1 hospitalA1/3/2002 300 200201 2 2 hospitalA1/6/2002 300 200201 3 3 hospitalB2/4/2002 150 200202 4 4 hospitalC3/2/2002 200 200203 On Mon, Jun 9, 2008 at 12:45 PM, Ricardo Pietrobon [EMAIL PROTECTED] wrote: I currently have a data set describing human subjects enrolled into an international clinical trial, the name of the hospital enrolling this human subject, the date when the subject was enrolled, and a vector with variables representing characteristics of the site (e.g., number of beds in a hospital). my data sets looks like this: subject hospitaldate_enrollment hospital_beds 1 hospitalA 1/3/2002300 2 hospitalA 1/6/2002300 3 hospitalB 2/4/2002150 4 hospitalC 3/2/2002200 to perform a time series analysis I am now trying to get to a format that would give me the following variables: month yearsitenumber_enrolled_subjectshospital_beds the data would be displayed on one-month intervals, and number of subjects clustered around sites. any help would be greatly appreciate thanks Ricardo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] I wonder if cch function in Survival package can calculate time dependent covariate
Hi In case cohort study, we can fit proportional hazard regression model to case-cohort data. In R, the function is cch() in Survival package Now I am working on case cohort analysis with time dependent covariates using cch() of Survival R package. I wonder if cch() provide this utility or not? The cch() manual does not say if time dependent covariate is allowed I know coxph() in Survival package can estimate time dependent covariates. Thanks Jin Wang [[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] difference between nlm and nlminb
nlminb provides unconstrained optimization and optimization subject to box constraints (i.e. upper and/or lower constraints on individual elements of the parameter vector). The nlm function provides unconstrained optimization. I created the nlminb function because I was unable to get reliable convergence on some difficult optimization problems for the nlme and lme4 packages using nlm and optim. The nlme package was originally written for S from Bell Labs (the forerunner of S-PLUS) and the PORT package was the optimization code used. Even though it is very old style Fortran code I find it quite reliable as an optimizer. It allows for what is called reverse communication which is convenient in an environment like R. It is a technical issue that has to do with what code is in control when your R expression needs to be evaluated. That said, I still don't feel that I have seen good, modern Open-Source optimization code. I would welcome suggestions of where one might find such code. On Wed, Jun 11, 2008 at 3:16 AM, DavidM.UK [EMAIL PROTECTED] wrote: I believe nlminb() performs *constrained* optimization, where as nlm() is for *unconstrained* opimization So I guess nlm() is for solving min(f[a,b]), and nlminb() min(f[a,b]) given a+b = c FYI I think optim() also does constrained optimization, well I've used for min(f[a,b]) given a = a* and b = b*. David ae2356 wrote: Hi, I was wondering if someone could give a brief, big picture overview of the difference between the two optimization functions nlm and nlminb. I'm not familiar with PORT routines, so I was hoping someone could give an explanation. Thanks, Angelo _ Instantly invite friends from Facebook and other social networks to join yo https://www.invite2messenger.net/im/?source=TXT_EML_WLH_InviteFriends [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/difference-between-nlm-and-nlminb-tp17769859p17772440.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] model simplification using Crawley as a guide
Ben Bolker wrote: Lucke, Joseph F Joseph.F.Lucke at uth.tmc.edu writes: And to follow FH and HW What level of significance are you using? .05 is excessively liberal. Are you adjusting your p-values for the number of possible models? Do you realize the p-values for dropping a term, being selected as the maximum of a set of p-values, do not follow their usual distributions? How are you compensating for sample size, as a p-value's being significant is a function of sample size? How are you compensating for the fact that the current model choice is dependent on the previous model choices? How do you know your tree of model choices is the optimal one? Have you considered cross-validation? Are you looking for a model that true describes a phenomenon or a predictive model that can be used for practical purposes? Ouch. While Frank Harrell and Joseph Lucke are raising serious issues about model selection, maybe we could keep in mind that we don't want to scare off all the students who ever try to use R to figure out basic statistics. I would follow Peter Dalgaard's advice (about drop1) and Hadley Wickham's (about graphical diagnostics), and if possible bring up the other issues about model selection with others around you -- if you're a student, ask your prof. or someone in the stats department. It can be tough to try to do things right if those around you are still doing them wrong ... If you tell us what field you're in we may be able to point you to more subject-specific references (e.g. Whittingham, Mark J., Philip A. Stephens, Richard B. Bradbury, and Robert P. Freckleton. 2006. Why do we still use stepwise modelling in ecology and behaviour? Journal of Animal Ecology 75, no. 5: 1182-1189) Ben Bolker Good points Ben. For now I'd recommend simply that the allergic reaction to insignificant statistical tests be treated with an antihistimine :-) Frank __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem when combining dotplot() and textplot() using grid
Hi
Bernhard wrote:
Hi everyone.
I want to solve the following problem. I have a data.frame and I
create a dotplot using lattice.
Then I want to use the grid-package to create a combined graphic which
contains the dotplot as well as a textplot() (using package gplots) of
the data.frame next to the dotplot.
Example code:
library(lattice)
library(grid)
library(gplots)
xx - data.frame(f=factor(rep(1:5, each=5)), gr= rep(c(gr1, gr2,
gr3, gr4, gr5), 5), val=rnorm(25))
grid.newpage()
pushViewport(viewport(layout = grid.layout(1, 2)))
pushViewport(viewport(layout.pos.col=1, layout.pos.row=1))
p - dotplot(f ~ val, groups=gr, xx, cex=1.7, pch=20)
print(p, newpage=FALSE)
popViewport(1)
pushViewport(viewport(layout.pos.col=2, layout.pos.row=1))
textplot(xx, show.rownames=FALSE)
popViewport(1)
Obviously, this does not work since textplot() seems to call plot.new
no matter what options are set with par().
My question: Is it possible to archieve my goal to plot the data.frame
with textplot() next to the dotplot?
This should work (adjustments to your code are accompanied by comments):
library(lattice)
library(grid)
library(gplots)
# Also need gridBase
library(gridBase)
# Start a traditional graphics page
plot.new()
xx - data.frame(f=factor(rep(1:5, each=5)), gr= rep(c(gr1, gr2,
gr3, gr4, gr5), 5), val=rnorm(25))
grid.newpage()
pushViewport(viewport(layout = grid.layout(1, 2)))
pushViewport(viewport(layout.pos.col=1, layout.pos.row=1))
p - dotplot(f ~ val, groups=gr, xx, cex=1.7, pch=20)
print(p, newpage=FALSE)
popViewport(1)
pushViewport(viewport(layout.pos.col=2, layout.pos.row=1))
# Prevent textplot() starting new page
par(new=TRUE)
# Locate traditional plot region based on grid viewport
par(plt=gridPLT())
textplot(xx, show.rownames=FALSE)
popViewport(1)
And furthermore, if it is possible to do so, is it even possible to
adjust the textplot in a way that the each row of the data.frame when
plotted using textplot() is at the same height than the corresponding
row in the dotplot? I guess that this is quite tricky and any hint on
which package or functions I could use would be very helpful.
This is starting to push things a bit; you are getting to the point
where it might be easier to start drawing things yourself. However,
here's one approach that might give you what you want. The idea is
to draw the textplot() for each level of the factor in its own viewport
(to get the alignment with the scale on the lattice plot).
# Start a traditional graphics page
plot.new()
grid.newpage()
# Viewport in left half of page to leave space for text
pushViewport(viewport(x=0, width=.5, just=left))
p - dotplot(f ~ val, groups=gr, xx, cex=1.7, pch=20)
print(p, newpage=FALSE,
# Control name of lattice viewport so we can get back to it
prefix=lattice)
# Leave all viewports so we can navigate back
upViewport(1)
# Go back into the lattice plot viewport to add text bits
downViewport(lattice.panel.1.1.off.vp)
# For each level of 'f'
for (i in levels(xx$f)) {
# Push a viewport off to the side
pushViewport(viewport(x=unit(1, npc) + unit(1, cm),
# Here's the vertical alignment bit
y=unit(as.numeric(as.character(i)),
native) +
unit(1, lines),
height=1/(length(levels(xx$f)) + 1),
just=c(left, top)))
# Just to show where we are (can be removed)
grid.rect(gp=gpar(col=grey))
# Align the traditional plot region
par(new=TRUE)
par(fig=gridFIG())
# Fiddle with size of text
par(cex=0.7)
# Draw the relevant subset of the data frame
textplot(xx[xx$f == i, ], halign=left, valign=top,
mar=rep(0, 4))
popViewport()
}
# Final tidy up
upViewport(0)
Bits of this are very your-example-specific, but this might be enough
for a one-off.
Paul
Thank you very much.
Bernhard
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] problem with as.Date
Data into R from Excel csv file xd-read.csv(court.dates1.txt,as.is=T, header = F) str(xd) 'data.frame': 5 obs. of 1 variable: $ V1: chr 6/6 5/27 5/16 5/2 ... xd V1 1 6/6 2 5/27 3 5/16 4 5/2 5 4/29 cdates - as.Date(xd, format = %m/ %d) Error in as.Date.default(xd, format = %m/ %d) : do not know how to convert 'xd' to class Date Suggestions appreciated, Don -- View this message in context: http://www.nabble.com/problem-with-as.Date-tp17788563p17788563.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] problem with as.Date
Using xd - structure(list(V1 = c(6/6, 5/27, 5/16, 5/2)), .Names = V1, row.names = c(NA, -4L), class = data.frame) Try: as.Date(paste(2008, xd$V1, sep = /), %Y/%m/%d) [1] 2008-06-06 2008-05-27 2008-05-16 2008-05-02 See R News 4/1 article on dates. On Wed, Jun 11, 2008 at 6:22 PM, Mr Natural [EMAIL PROTECTED] wrote: Data into R from Excel csv file xd-read.csv(court.dates1.txt,as.is=T, header = F) str(xd) 'data.frame': 5 obs. of 1 variable: $ V1: chr 6/6 5/27 5/16 5/2 ... xd V1 1 6/6 2 5/27 3 5/16 4 5/2 5 4/29 cdates - as.Date(xd, format = %m/ %d) Error in as.Date.default(xd, format = %m/ %d) : do not know how to convert 'xd' to class Date Suggestions appreciated, Don -- View this message in context: http://www.nabble.com/problem-with-as.Date-tp17788563p17788563.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] problem with as.Date
since it's a dataframe, use xd$V1. also, it doesn't seem like your data has a space between the / and the numbers, so take that out of the pattern: cdates - as.Date(xd$V1, format = %m/%d) On Wed, Jun 11, 2008 at 6:22 PM, Mr Natural [EMAIL PROTECTED] wrote: Data into R from Excel csv file xd-read.csv(court.dates1.txt,as.is=T, header = F) str(xd) 'data.frame': 5 obs. of 1 variable: $ V1: chr 6/6 5/27 5/16 5/2 ... xd V1 1 6/6 2 5/27 3 5/16 4 5/2 5 4/29 cdates - as.Date(xd, format = %m/ %d) Error in as.Date.default(xd, format = %m/ %d) : do not know how to convert 'xd' to class Date Suggestions appreciated, Don -- View this message in context: http://www.nabble.com/problem-with-as.Date-tp17788563p17788563.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] model simplification using Crawley as a guide
on 06/11/2008 05:53 PM Frank E Harrell Jr wrote: Ben Bolker wrote: Lucke, Joseph F Joseph.F.Lucke at uth.tmc.edu writes: And to follow FH and HW What level of significance are you using? .05 is excessively liberal. Are you adjusting your p-values for the number of possible models? Do you realize the p-values for dropping a term, being selected as the maximum of a set of p-values, do not follow their usual distributions? How are you compensating for sample size, as a p-value's being significant is a function of sample size? How are you compensating for the fact that the current model choice is dependent on the previous model choices? How do you know your tree of model choices is the optimal one? Have you considered cross-validation? Are you looking for a model that true describes a phenomenon or a predictive model that can be used for practical purposes? Ouch. While Frank Harrell and Joseph Lucke are raising serious issues about model selection, maybe we could keep in mind that we don't want to scare off all the students who ever try to use R to figure out basic statistics. I would follow Peter Dalgaard's advice (about drop1) and Hadley Wickham's (about graphical diagnostics), and if possible bring up the other issues about model selection with others around you -- if you're a student, ask your prof. or someone in the stats department. It can be tough to try to do things right if those around you are still doing them wrong ... If you tell us what field you're in we may be able to point you to more subject-specific references (e.g. Whittingham, Mark J., Philip A. Stephens, Richard B. Bradbury, and Robert P. Freckleton. 2006. Why do we still use stepwise modelling in ecology and behaviour? Journal of Animal Ecology 75, no. 5: 1182-1189) Ben Bolker Good points Ben. For now I'd recommend simply that the allergic reaction to insignificant statistical tests be treated with an antihistamine :-) A vote for Frank's comment to be added to the 'fortunes' package. :-) Regards, Marc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Adding new columns to (output) data - e.g., read 5 cols write 8
Hello, I have the following task I'd like to accomplish: A file contains 5 columns of data (several hundred rows), let's call them a, b, c, d and e (ie these are their column headers) I also have a set of definitions, e.g., f = a + b g = a * 3 h = c + d etc. I would like to write out a new .rda file that contains columns a b c d e f g h etc. I.e. , the original data plus new columns (with headers and data). It seems that there ought to be a simple way to do this, could someone provide some guidance on the best way to accomplish this task? (I tried a few things as this seemed rather trivial, but did not succeed. I still hope/assume this is a trivial thing to do if one knows R well) Thanks, Esmail ps: I want to thank everyone again who posted their solutions to my previous query. Seeing different solutions for the same problem is a tremendously effective way to learn something new. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] model simplification using Crawley as a guide
Good points Ben. For now I'd recommend simply that the allergic reaction to insignificant statistical tests be treated with an antihistamine :-) A vote for Frank's comment to be added to the 'fortunes' package. Seconded! :-) :-) Regards, Marc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Simon Blomberg, BSc (Hons), PhD, MAppStat. Lecturer and Consultant Statistician Faculty of Biological and Chemical Sciences The University of Queensland St. Lucia Queensland 4072 Australia Room 320 Goddard Building (8) T: +61 7 3365 2506 http://www.uq.edu.au/~uqsblomb email: S.Blomberg1_at_uq.edu.au Policies: 1. I will NOT analyse your data for you. 2. Your deadline is your problem. The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. - John Tukey. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Adding new columns to (output) data - e.g., read 5 cols write 8
yourDF - cbind(yourDF, f=yourDF$a+yourDF$b, g=yourDF$a * 3, h=yourDF$c + yourDF$d) On Wed, Jun 11, 2008 at 9:26 PM, Esmail Bonakdarian [EMAIL PROTECTED] wrote: Hello, I have the following task I'd like to accomplish: A file contains 5 columns of data (several hundred rows), let's call them a, b, c, d and e (ie these are their column headers) I also have a set of definitions, e.g., f = a + b g = a * 3 h = c + d etc. I would like to write out a new .rda file that contains columns a b c d e f g h etc. I.e. , the original data plus new columns (with headers and data). It seems that there ought to be a simple way to do this, could someone provide some guidance on the best way to accomplish this task? (I tried a few things as this seemed rather trivial, but did not succeed. I still hope/assume this is a trivial thing to do if one knows R well) Thanks, Esmail ps: I want to thank everyone again who posted their solutions to my previous query. Seeing different solutions for the same problem is a tremendously effective way to learn something new. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding new columns to (output) data - e.g., read 5 cols write 8
Esmail - Esmail Bonakdarian wrote: Hello, I have the following task I'd like to accomplish: A file contains 5 columns of data (several hundred rows), let's call them a, b, c, d and e (ie these are their column headers) Are these 5 vectors of data stored in a data.frame? I assume so. I also have a set of definitions, e.g., f = a + b g = a * 3 h = c + d etc. I would like to write out a new .rda file that contains columns a b c d e f g h etc. I.e. , the original data plus new columns (with headers and data). Example, test - data.frame(a = 1:10, b = 2:11, c = 3:12) test2 - transform(test, d = 2*a + b, e = 3*c) save(test2, file = test2.Rdata) Does this help? Best, Erik Iverson __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mgcv::gam error message for predict.gam
Sometimes, for specific models, I get this error from predict.gam in library mgcv: Error in complete.cases(object) : negative length vectors are not allowed Here's an example: model.calibrate - gam(meansalesw ~ s(tscore,bs=cs,k=4), data=toplot, weights=weight, gam.method=perf.magic) test - predict(model.calibrate,newdata) Error in complete.cases(object) : negative length vectors are not allowed The data is shown below: toplot[,c(meansalesw,tscore,weight)] meansalesw tscore weight 1 0.1275841 0.003446797 15224 2 0.1495748 0.004017158 15523 3 0.2245844 0.004375278 15520 4 0.2197668 0.004753941 15525 5 0.1317830 0.005049050 15524 6 0.2809621 0.005403199 15498 7 0.2933119 0.005764413 15529 8 0.4791150 0.006335145 15514 9 0.1833688 0.006617095 15528 10 0.3200599 0.007135850 15527 11 0.4931882 0.007781095 15529 12 0.4207684 0.008766088 15512 13 0.5928568 0.009731357 15514 14 0.8025296 0.010927579 15520 15 0.6286192 0.012004714 15513 16 0.7477922 0.014083143 15527 17 0.7251362 0.017382274 15531 18 1.1871948 0.025481173 15521 19 1.6495832 0.048264689 15524 20 5.1180227 0.131198022 15218 newdata tscore 1 0.5059341 2 0.4125522 3 1.4335818 4 0.7060673 5 0.3229316 Thanks! -- View this message in context: http://www.nabble.com/mgcv%3A%3Agam-error-message-for-predict.gam-tp17789318p17789318.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] Adding new columns to (output) data - e.g., read 5 cols write 8
Hi Erik, Erik Iverson wrote: Esmail - Are these 5 vectors of data stored in a data.frame? I assume so. Yes, I do a simple load() call first to read the .rda file ... test2 - transform(test, d = 2*a + b, e = 3*c) save(test2, file = test2.Rdata) Does this help? Yes it does .. this is just what I needed. I'll probably do it in this way: df=transform(df, X5=X1+X2) df=transform(df, X6=X1-X2) etc ... building it up the df line by line since the definitions have been provided one per line (instead of queuing them up in one line like you showed). I'll have to write a little awk or ruby script to take the original definitions and wrap them into the transform call, but then I should be all set to go. Thanks again! Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Adding new columns to (output) data - e.g., read 5 cols write 8
jim holtman wrote: yourDF - cbind(yourDF, f=yourDF$a+yourDF$b, g=yourDF$a * 3, h=yourDF$c + yourDF$d) Thanks Jim, I also learned about the transform() method from Erik which will also work beautifully. Esmail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Data.matrix fail to convert data.frame into matrix
Hi, With the following codes, I attempt to convert the data.frame into a matrix. However I notice that data.matrix function doesn't seem to work. __ BEGIN__ dat - read.table(mydata, comment.char = ! , na.strings = null); # Select n-genes by random sample # n = 1 nosamp - 1 geneid - sequence(nrow(dat)) geneid.samp - sample(geneid,nosamp) geneid.samp gexp- dat[geneid.samp,] gexp.arr - data.matrix(gexp, rownames.force = NA) print(is.matrix(gexp.arr)) print(gexp.arr) __END__ Yielding this output: __BEGIN__ print(is.matrix(gexp.arr)) [1] TRUE print(gexp.arr) V1 V2V3V4V5V6V7V8 10354 803.1 1107.8 431.6 349.8 386.7 646.3 744.2 620.9 __END__ I expect gexp.arr to be a plain vector (numeric). What's wrong with my code above? -- Gundala Viswanath Jakarta-Indonesia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Panel-specific colours in barchart, lattice package.
I would like, for possibly invalid reasons, to have the bars of the
barchart
in each panel to be the same colour, but with *different* colours
from panel
to panel. Can this be done? If so, how?
I've read the help as much as I am capable of, and done an RSiteSearch
() without
becoming any wiser.
Here's an example of what I'd like to do:
X - structure(list(y = c(2.8, 6.7, 12.8, 22.9, 18.4, 16.2, 10.1,
5.6, 4.5, 6.9, 19.8, 25.7, 23.8, 12.9, 5.9, 3, 1, 1, 12.9, 24.2,
24.2, 25, 8.1, 3.2, 1.6, 0.8, 0, 5, 12.5, 25, 17.5, 20, 12.5,
5, 0, 2.5, 25, 33.3, 25, 0, 0, 8.3, 8.3, 0, 0), x = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L,
9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L,
7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L), .Label = c(1,
2, 3, 4, 5, 6, 7, 8, 9), class = factor), f =
structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), .Label = c(a,
b, c, d, e), class = factor)), .Names = c(y, x,
f), row.names = c(NA, -45L), class = data.frame)
barchart(y ~ x | f, data=X, as.table=TRUE)
EXCEPT that I'd like panel ``a'' to have its bars being, say, solid red,
panel ``b'' to have its bars solid blue, panel ``c'' to have its bars
solid green, etc.
Can I arrange this? Thanks.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Panel-specific colours in barchart, lattice package.
Try this:
barchart(y ~ x | f, data=X, as.table=TRUE,
panel = function(...) panel.barchart(..., col = panel.number()))
On Wed, Jun 11, 2008 at 11:46 PM, Rolf Turner [EMAIL PROTECTED] wrote:
I would like, for possibly invalid reasons, to have the bars of the barchart
in each panel to be the same colour, but with *different* colours from panel
to panel. Can this be done? If so, how?
I've read the help as much as I am capable of, and done an RSiteSearch()
without
becoming any wiser.
Here's an example of what I'd like to do:
X - structure(list(y = c(2.8, 6.7, 12.8, 22.9, 18.4, 16.2, 10.1,
5.6, 4.5, 6.9, 19.8, 25.7, 23.8, 12.9, 5.9, 3, 1, 1, 12.9, 24.2,
24.2, 25, 8.1, 3.2, 1.6, 0.8, 0, 5, 12.5, 25, 17.5, 20, 12.5,
5, 0, 2.5, 25, 33.3, 25, 0, 0, 8.3, 8.3, 0, 0), x = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L,
9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L,
7L, 8L, 9L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L), .Label = c(1,
2, 3, 4, 5, 6, 7, 8, 9), class = factor), f =
structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), .Label = c(a,
b, c, d, e), class = factor)), .Names = c(y, x,
f), row.names = c(NA, -45L), class = data.frame)
barchart(y ~ x | f, data=X, as.table=TRUE)
EXCEPT that I'd like panel ``a'' to have its bars being, say, solid red,
panel ``b'' to have its bars solid blue, panel ``c'' to have its bars
solid green, etc.
Can I arrange this? Thanks.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Data.matrix fail to convert data.frame into matrix
Try gexp.arr - data.matrix(gexp, rownames.force = FALSE) You are still supposed to get a matrix with one row (not a vector - dim(gexp.arr) is c(1,8) and not NULL). --- On Thu, 12/6/08, Gundala Viswanath [EMAIL PROTECTED] wrote: From: Gundala Viswanath [EMAIL PROTECTED] Subject: [R] Data.matrix fail to convert data.frame into matrix To: [EMAIL PROTECTED] Received: Thursday, 12 June, 2008, 1:28 PM Hi, With the following codes, I attempt to convert the data.frame into a matrix. However I notice that data.matrix function doesn't seem to work. __ BEGIN__ dat - read.table(mydata, comment.char = ! , na.strings = null); # Select n-genes by random sample # n = 1 nosamp - 1 geneid - sequence(nrow(dat)) geneid.samp - sample(geneid,nosamp) geneid.samp gexp- dat[geneid.samp,] gexp.arr - data.matrix(gexp, rownames.force = NA) print(is.matrix(gexp.arr)) print(gexp.arr) __END__ Yielding this output: __BEGIN__ print(is.matrix(gexp.arr)) [1] TRUE print(gexp.arr) V1 V2V3V4V5V6V7V8 10354 803.1 1107.8 431.6 349.8 386.7 646.3 744.2 620.9 __END__ I expect gexp.arr to be a plain vector (numeric). What's wrong with my code above? -- Gundala Viswanath Jakarta-Indonesia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
