[R] Extracting only multiple occurrences
Hoping someone here can help me with this small problem. set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) This gives a vector of 36 letters with some muliples (in this case, g,m,s,t,u,v,x,y). Now what I need is to get rid of the ones that only occur once and keep the multiples. I need the opposite of the unique() function. I expect this should be pretty easy but I can't see it. Anyone know a solution? Thanks in advance! _ Kevin Parent, Ph.D Korea Maritime University [[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] Extracting only multiple occurrences
On Aug 7, 2013, at 10:37 PM, Kevin Parent wrote: Hoping someone here can help me with this small problem. set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) This gives a vector of 36 letters with some muliples (in this case, g,m,s,t,u,v,x,y). Now what I need is to get rid of the ones that only occur once and keep the multiples. I need the opposite of the unique() function. I expect this should be pretty easy but I can't see it. Anyone know a solution? Thanks in advance! ?duplicated x[ duplicated(x) ] David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting only multiple occurrences
On Aug 7, 2013, at 11:03 PM, David Winsemius wrote: On Aug 7, 2013, at 10:37 PM, Kevin Parent wrote: Hoping someone here can help me with this small problem. set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) This gives a vector of 36 letters with some muliples (in this case, g,m,s,t,u,v,x,y). Now what I need is to get rid of the ones that only occur once and keep the multiples. I need the opposite of the unique() function. I expect this should be pretty easy but I can't see it. Anyone know a solution? Thanks in advance! ?duplicated x[ duplicated(x) ] Also this may be of interest: cran.r-project.org/doc/contrib/Short-refcard.pdf ... but I was suprised to find that both duplicated and rle are absent from Short's refcards. -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting only multiple occurrences
Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] Playing around with it, I got this but can't helping thinking there must be a less awkward way: set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) x-x[duplicated(x)] x-sort(c(x,unique(x))) _ Kevin Parent, Ph.D Korea Maritime University From: David Winsemius dwinsem...@comcast.net Cc: r-help@r-project.org r-help@r-project.org Sent: Thursday, August 8, 2013 3:03 PM Subject: Re: [R] Extracting only multiple occurrences On Aug 7, 2013, at 10:37 PM, Kevin Parent wrote: Hoping someone here can help me with this small problem. set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) This gives a vector of 36 letters with some muliples (in this case, g,m,s,t,u,v,x,y). Now what I need is to get rid of the ones that only occur once and keep the multiples. I need the opposite of the unique() function. I expect this should be pretty easy but I can't see it. Anyone know a solution? Thanks in advance! ?duplicated x[ duplicated(x) ] David Winsemius Alameda, CA, USA [[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] Extracting only multiple occurrences
On Aug 7, 2013, at 11:23 PM, Kevin Parent wrote: Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] x[ duplicated(x) | duplicated(x, fromLast=TRUE) ] -- David. Playing around with it, I got this but can't helping thinking there must be a less awkward way: set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) x-x[duplicated(x)] x-sort(c(x,unique(x))) _ Kevin Parent, Ph.D Korea Maritime University From: David Winsemius dwinsem...@comcast.net To: Kevin Parent kspar...@yahoo.com Cc: r-help@r-project.org r-help@r-project.org Sent: Thursday, August 8, 2013 3:03 PM Subject: Re: [R] Extracting only multiple occurrences On Aug 7, 2013, at 10:37 PM, Kevin Parent wrote: Hoping someone here can help me with this small problem. set.seed(2013) x-sort(c(letters,letters[sample(26,10,1)])) This gives a vector of 36 letters with some muliples (in this case, g,m,s,t,u,v,x,y). Now what I need is to get rid of the ones that only occur once and keep the multiples. I need the opposite of the unique() function. I expect this should be pretty easy but I can't see it. Anyone know a solution? Thanks in advance! ?duplicated x[ duplicated(x) ] David Winsemius Alameda, CA, USA David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How is a file descriptor stored ?
Hi, The file handling code sometimes throws this exception. Error in UseMethod(close) : no applicable method for 'close' applied to an object of class c ('integer', 'numeric') Is there a sample based on my code that I can test ? I want to extract the file descriptors from the hashmap and close them. I think that is causing the exception. Sometimes just closing - close(fd) - is causing this too. Thanks, Mohan RE: [R] How is a file descriptor stored ? William Dunlap to: Berend Hasselman, mohan.radhakrish...@polarisft.com 07-08-2013 08:01 PM Cc: r-help@r-project.org Use assign(key, file( key, w ), envir=cpufile) In your assign expression you are assigning cpufile to the third formal argument which is pos. You meant the envir argument, I presume. Or use the syntax cpufile[[key]] - file(key, w) instead of assign(key, file( key, w ), envir=cpufile) The former works for lists and environments and corresponds to your later usage of listoffiles[[key]] to retrieve the data. From what I've seen of your example, a list might be a better way to store your data, because of its copy-on-write semantics and because it doesn't keep a parent environment in memory. By using '[[' instead of 'get' and 'assign' you minimize the number of changes required to switch between a list and an environment. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Berend Hasselman Sent: Wednesday, August 07, 2013 4:10 AM To: mohan.radhakrish...@polarisft.com Cc: r-help@r-project.org Subject: Re: [R] How is a file descriptor stored ? On 07-08-2013, at 12:13, mohan.radhakrish...@polarisft.com wrote: Hi, I thought that 'R' like java will allow me to store file names (keys) and file descriptors(values) in a hashmap. filelist.array - function(n){ sink(nmon.log) cpufile - new.env(hash=T, parent=emptyenv()) for (i in 1:n) { key - paste(output, i, .txt, sep = ) assign(key, file( key, w ), cpufile) } sink() return (cpufile) } But when I try to test it like this there is an exception [1] Exception is Error in UseMethod(\close\): no applicable method for 'close' applied to an object of class \c('integer', 'numeric')\\n test.simple.filelist.array - function() { execution - tryCatch({ sink(nmon.log) listoffiles - filelist.array(3) for (v in ls(listoffiles)) { print(paste(Map value is [, listoffiles[[v]], ])) fd - listoffiles[[v]] close(fd) } sink() }, error = function(err){ print(paste(Exception is ,err)) }) } I think I am missing some fundamentals. Read the help page for assign more carefully. Use assign(key, file( key, w ), envir=cpufile) In your assign expression you are assigning cpufile to the third formal argument which is pos. You meant the envir argument, I presume. Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-Mail may contain proprietary and
Re: [R] Running time complexity of Seasonal ARIMA model (forecast package)
On 08/08/2013 05:08, Mohit Dhingra wrote: *Dear All,* I am using Seasonal ARIMA model for predicting cloud workloads. I want to know the running time complexity of building model by the algorithm implemented in R (I am not sure, is it Yule-Walker?). I want to know if it It is not Yule-Walker (which is for AR models only). is polynomial O(n^2) etc. or exponential or linear (O(n)). Can someone please help. What is 'n' here? Please read the references for yourself: they will tell you enough to deduce the answer -- or you could experiment. PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] diallel analysis question
plz solve this question and send me commands.. this is a question for diallel analysis.. thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] laf_open_fwf
Dear Jan Many thanks for your help. In fact, all lines are shorter than my column width... my.column.widths: 238 range(nchar(lines)):235 237 So, it seems I have an inconsistent file structure... I guess there is no way to handle this in an automated way? Best Regard Christian Kamenik Project Manager Federal Department of the Environment, Transport, Energy and Communications DETEC Federal Roads Office FEDRO Division Road Traffic Road Accident Statistics Mailing Address: 3003 Bern Location: Weltpoststrasse 5, 3015 Bern Tel +41 31 323 14 89 Fax +41 31 323 43 21 christian.kame...@astra.admin.ch www.astra.admin.ch -Ursprüngliche Nachricht- Von: Jan van der Laan [mailto:rh...@eoos.dds.nl] Gesendet: Mittwoch, 7. August 2013 20:57 An: r-help@r-project.org Cc: Kamenik Christian ASTRA Betreff: Re: [R] laf_open_fwf Dear Christian, Well... it shouldn't normally do that. The only way I can currently think of that might cause this problem is that the file has \r\n\r\n, which would mean that every line is followed by an empty line. Another cause might be (although I would not really expect the results you see) that the sum of your column widths is larger than the actual with of the line. You can check your line lengths using: lines - readLines(my.filename) nchar(lines) Each line should have the same length and be equal to (or at least larger than) sum(my.column.widths) If this is not the problem: would it be possible that you send me a small part of your file so that I could try to reproduce the problem? Or if you cannot share your data: replace the actual values with nonsense values. Regards, Jan PS I read your mail by chance as I am not a regular r-help reader. When you have specific LaF problems it is better to also cc me directly. On 08/06/2013 12:35 PM, christian.kame...@astra.admin.ch wrote: Dear all I was trying the (fairly new) LaF package, and came across the following problem: I opened a connection to a fixed width ASCII file using laf_open_fwf(my.filename, my.column_types, my.column_widths, my.column_names) When looking at the data, it turned out that \n (newline) and \r (carriage return) were considered as characters, thus destroying the structure in my data (the second column does not include any numbers): my.data[1565:1575,1:3] MF_FARZ1 Fahrzeugarttext MF_MARKE 1 \n043 Landwirt. Traktor2140 2 \n043 Landwirt. Traktor6206 3 \n001 Personenwagen2026 4 \n001 Personenwagen2026 5\r\n00 1Personenwagen404 6\r\n02 0Gesellschaftswagen 710 7\r\n00 1Personenwagen505 8\r\n00 1Personenwagen505 9\r\n00 1Personenwagen301 10 \r\n00 1Personenwagen553 11 \r\n04 3Landwirt. Traktor257 I am working on Windows 7 32-bit. Any help would be highly appreciated. Best Regard Christian Kamenik Project Manager Federal Department of the Environment, Transport, Energy and Communications DETEC Federal Roads Office FEDRO Division Road Traffic Road Accident Statistics Mailing Address: 3003 Bern Location: Weltpoststrasse 5, 3015 Bern Tel +41 31 323 14 89 Fax +41 31 323 43 21 christian.kame...@astra.admin.chmailto:christian.kamenik@astra.admin. ch www.astra.admin.chhttp://www.astra.admin.ch/ [[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] Upgrading to R and keeping packages
Hi! I just installed the latest R 3.01. I then wanted to update my packages. I believe the advice given is to take the library folder from the old R version and copy it on top of (overwrite) the library folder of the new R version, in my case the library of R 2.15.2 to library of R 3.01. When I did this, the next time I started R 3.01, I had an error message Error in .Call(R_isMethodsDispatchOn, onOff, PACKAGE = base) : R_isMethodsDispatchOn not available for .Call() for package base also the update.packages() function was 'not available'. Where did I go wrong? I think I will just update the packages I want rather than the whole folder and see if this works, Pancho Mulongeni Research Assistant PharmAccess Foundation 1 Fouché Street Windhoek West Windhoek Namibia Tel: +264 61 419 000 Fax: +264 61 419 001/2 Mob: +264 81 4456 286 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 na.action within own function
Dear R Community, I am trying to build a very simple function which uses lm and coeftest to return a coefficient matrix with heteroskedasticity robust standard errors. The function is the following: reg=function(formula,data,na.action){ res=lm(formula=formula,data=data,na.action=na.action) hc3=coeftest(res, vcov = vcovHC(res, type = HC3)) residuals=resid(res) return(list(coef.hc3=hc3,R2=summary(res)$r.squared, R2.adj=summary(res)$adj.r.squared, residuals=residuals)) } The function works perfect as long as the data contains no missing values. I.e. test1=seq(1,30,1) test2=seq(1,30,1) testdata=data.frame(test1,test2) reg(formula=test1~test2, data=testdata, na.action=na.exclude) However, as soon as I have a missing value, it does not work any more (the error message is: Error in estfun(x)/X : non-conformable arrays): test1=seq(1,30,1) test2=seq(1,30,1) test2[5]=NA testdata=data.frame(test1,test2) reg(formula=test1~test2, data=testdata, na.action=na.exclude) My feeling is that it has something to do with na.exclude being a function itself and hence R having a problem with inputing a function into another function. Does anyone have an idea? Thanks a lot!!! 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] Extracting only multiple occurrences
On 08/08/2013 04:23 PM, Kevin Parent wrote: Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] Hi Kevin, How about: x[x %in% duplicated(x)] 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] Upgrading to R and keeping packages
On 08/08/2013 09:07, Pancho Mulongeni wrote: Hi! I just installed the latest R 3.01. I then wanted to update my packages. I believe the advice given is to take the library folder from the old R version and copy it on top of (overwrite) the library folder of the new R version, in my case the library of R 2.15.2 to library of R 3.01. When I did this, the next time I started R 3.01, I had an error message Error in .Call(R_isMethodsDispatchOn, onOff, PACKAGE = base) : R_isMethodsDispatchOn not available for .Call() for package base also the update.packages() function was 'not available'. Where did I go wrong? Copy over the base packages. This was never the advice. See e.g. http://cran.r-project.org/bin/windows/base/rw-FAQ.html#What_0027s-the-best-way-to-upgrade_003f I think I will just update the packages I want rather than the whole folder and see if this works, Pancho Mulongeni Research Assistant PharmAccess Foundation 1 Fouché Street Windhoek West Windhoek Namibia Tel: +264 61 419 000 Fax: +264 61 419 001/2 Mob: +264 81 4456 286 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting only multiple occurrences
On 08-08-2013, at 10:27, Jim Lemon j...@bitwrit.com.au wrote: On 08/08/2013 04:23 PM, Kevin Parent wrote: Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] Hi Kevin, How about: x[x %in% duplicated(x)] Don't you mean this x[x %in% x[duplicated(x)]] Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Numercial evaluation of intgral with different bounds
Hello R helpers, I'm struggling how to apply the integrate function to a data frame. Here is an example of what I'm trying to do: # Create data frame x - 0:4 tx - 10:14 T - 12:16 data - data.frame(x=x, tx=tx, T=T) # Parameter alpha - 10 beta - 11 # Integral integrand - function(y){ (y+alpha)^-(r+data$x)*(y+beta^-(s+1)) } Now I want to apply the integrate function to evaluate the integral for each line of the data frame with tx as the lower and T as the upper bound. The respektive values (and the values only) should be returned in a vector. I want to avoid the use of a loop since the integral is part of a function I want to optimize with optim and so speed is crucial. I tried to do this by something like: integral - lapply(data$tx, integrate, f=integrand, upper=data$T) integral2 - sapply(integral, function(x){x[1]}) integral3 - unlist(integral2, use.names=FALSE) But this doesn't work properly. I'd glad if you have any hints how to get this done. Many thanks and best regards, Carlos -- - Carlos Nasher Buchenstr. 12 22299 Hamburg tel:+49 (0)40 67952962 mobil:+49 (0)175 9386725 mail: carlos.nas...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting only multiple occurrences
On 08/08/2013 06:52 PM, Berend Hasselman wrote: On 08-08-2013, at 10:27, Jim Lemonj...@bitwrit.com.au wrote: On 08/08/2013 04:23 PM, Kevin Parent wrote: Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] Hi Kevin, How about: x[x %in% duplicated(x)] Don't you mean this x[x %in% x[duplicated(x)]] Berend Ah, yes, thanks. 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] Extracting only multiple occurrences
On 08/08/13 20:27, Jim Lemon wrote: On 08/08/2013 04:23 PM, Kevin Parent wrote: Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] Hi Kevin, How about: x[x %in% duplicated(x)] Uh, I think you mean x[x %in% x[duplicated(x)]] Another idear: tx - table(x) tx - tx[tx1] rep(names(tx),tx) Well, that's three lines as opposed to one, so not as good. But it perhaps demonstrates a useful tool to add to one's kit. cheers, Rolf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] gnls() and L-BFGS-B in library(nlme)
Dear R Users, I attempt to estimate a generalized nonlinear least squares model using gnls() from the nlme library. I wish to restrict some of my parameters using the L-BFGS-B method for the optim optimizer. However, in contrast to nls() in the same package, gnls() does not accept any `lower' or `upper' argument, nor does the control argument allow for such elements. Unfortunately, the help file does not provide a description of how to include lower or upper bounds, nor did I find any hint in the internet. Example library(nlme) #works but no restriction included fm1 - gnls(weight ~ SSlogis(Time, Asym, xmid, scal), Soybean, start=c(Asym=17,xmid=50,scal=7),weights = varPower(),control=gnlsControl(opt=optim,optimMethod=L-BFGS-B)) summary(fm1) #does not work fm2 - gnls(weight ~ SSlogis(Time, Asym, xmid, scal), Soybean,start=c(Asym=17,xmid=50,scal=7), weights = varPower(),control=gnlsControl(opt=optim,optimMethod=L-BFGS-B),lower=c(0,0,0),upper=c(100,100,100)) summary(fm2) fm3 - gnls(weight ~ SSlogis(Time, Asym, xmid, scal), Soybean,start=c(Asym=17,xmid=50,scal=7),lower=c(0,0,0),upper=c(100,100,100), weights = varPower(),control=gnlsControl(opt=optim,optimMethod=L-BFGS-B)) summary(fm3) Do you know a way to include lower and upper bounds? And: How do I restrict only selected parameters while I wish others to take any value. Any hints pointing to a solution are highly appreciated. Regards, Philipp Grueber - EBS Universitaet fuer Wirtschaft und Recht FARE Department Wiesbaden/ Germany http://www.ebs.edu/index.php?id=finaccL=0 -- View this message in context: http://r.789695.n4.nabble.com/gnls-and-L-BFGS-B-in-library-nlme-tp4673341.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] Running time complexity of Seasonal ARIMA model (forecast package)
*Dear Sir,* Thanks for your response. Here, I was using 'n' to denote the input size (no. of points in time series using which I am building a Seasonal ARIMA model). I can check the running time myself and I have done that as well (it takes some 1-2 minutes for 50 iterations for my input size), but I want to know more about the asysmptotic complexity of the algorithm R uses. I can see three methods CSS, CSS-ML and ML that it uses to optimize the parameters. Like bubble sort takes O(n^2) time where n is the no. of elements. Can I define something like this to build my ARIMA model which has n points? * Thanks Regards Mohit Dhingra +919611190435* On 8 August 2013 12:45, Prof Brian Ripley rip...@stats.ox.ac.uk wrote: On 08/08/2013 05:08, Mohit Dhingra wrote: *Dear All,* I am using Seasonal ARIMA model for predicting cloud workloads. I want to know the running time complexity of building model by the algorithm implemented in R (I am not sure, is it Yule-Walker?). I want to know if it It is not Yule-Walker (which is for AR models only). is polynomial O(n^2) etc. or exponential or linear (O(n)). Can someone please help. What is 'n' here? Please read the references for yourself: they will tell you enough to deduce the answer -- or you could experiment. PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~**ripley/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 [[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] laf_open_fwf
Without example data it is difficult to give suggestions on how you might read this file. Are you sure your file is fixed width? Sometimes columns are neatly aligned using whitespace (tabs/spaces). In that case you could use read.table with the default settings. Another possibility might be that the file is encoded in utf8. I expect that reading it in assuming another encoding (such as latin1) would lead to varying line sizes. Although I would expect the lengths to be larger than the sum of your column widths (as one symbol can be larger than one byte). Jan christian.kame...@astra.admin.ch schreef: Dear Jan Many thanks for your help. In fact, all lines are shorter than my column width... my.column.widths: 238 range(nchar(lines)):235 237 So, it seems I have an inconsistent file structure... I guess there is no way to handle this in an automated way? Best Regard Christian Kamenik Project Manager Federal Department of the Environment, Transport, Energy and Communications DETEC Federal Roads Office FEDRO Division Road Traffic Road Accident Statistics Mailing Address: 3003 Bern Location: Weltpoststrasse 5, 3015 Bern Tel +41 31 323 14 89 Fax +41 31 323 43 21 christian.kame...@astra.admin.ch www.astra.admin.ch -Ursprüngliche Nachricht- Von: Jan van der Laan [mailto:rh...@eoos.dds.nl] Gesendet: Mittwoch, 7. August 2013 20:57 An: r-help@r-project.org Cc: Kamenik Christian ASTRA Betreff: Re: [R] laf_open_fwf Dear Christian, Well... it shouldn't normally do that. The only way I can currently think of that might cause this problem is that the file has \r\n\r\n, which would mean that every line is followed by an empty line. Another cause might be (although I would not really expect the results you see) that the sum of your column widths is larger than the actual with of the line. You can check your line lengths using: lines - readLines(my.filename) nchar(lines) Each line should have the same length and be equal to (or at least larger than) sum(my.column.widths) If this is not the problem: would it be possible that you send me a small part of your file so that I could try to reproduce the problem? Or if you cannot share your data: replace the actual values with nonsense values. Regards, Jan PS I read your mail by chance as I am not a regular r-help reader. When you have specific LaF problems it is better to also cc me directly. On 08/06/2013 12:35 PM, christian.kame...@astra.admin.ch wrote: Dear all I was trying the (fairly new) LaF package, and came across the following problem: I opened a connection to a fixed width ASCII file using laf_open_fwf(my.filename, my.column_types, my.column_widths, my.column_names) When looking at the data, it turned out that \n (newline) and \r (carriage return) were considered as characters, thus destroying the structure in my data (the second column does not include any numbers): my.data[1565:1575,1:3] MF_FARZ1 Fahrzeugarttext MF_MARKE 1 \n043 Landwirt. Traktor2140 2 \n043 Landwirt. Traktor6206 3 \n001 Personenwagen2026 4 \n001 Personenwagen2026 5\r\n00 1Personenwagen404 6\r\n02 0Gesellschaftswagen 710 7\r\n00 1Personenwagen505 8\r\n00 1Personenwagen505 9\r\n00 1Personenwagen301 10 \r\n00 1Personenwagen553 11 \r\n04 3Landwirt. Traktor257 I am working on Windows 7 32-bit. Any help would be highly appreciated. Best Regard Christian Kamenik Project Manager Federal Department of the Environment, Transport, Energy and Communications DETEC Federal Roads Office FEDRO Division Road Traffic Road Accident Statistics Mailing Address: 3003 Bern Location: Weltpoststrasse 5, 3015 Bern Tel +41 31 323 14 89 Fax +41 31 323 43 21 christian.kame...@astra.admin.chmailto:christian.kamenik@astra.admin. ch www.astra.admin.chhttp://www.astra.admin.ch/ [[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] Lack of catalog number referencing for statistical objects
Hi, In Skew-t fits to mortality data - can a Gaussian-related distribution replace the Gompertz-Makeham as the basis for mortality studies? (J Gerontol A Biol Sci Med Sci 2013 August ;68(8):903-913; doi:10.1093/Gerona/gls239) Clark et al compares the fit of several distributions to mortality data (using the bbmle package, among others). This is not the place to comment on the paper, of course. Just wanted to share with you the following quote (p. 906) -almost a throwaway remark given the context of the paper- which may be of interest to the statisticians in the forum: [T]here was a lack of catalogue number referencing for statistical objects. Attempts at cataloguing are under way (eg the Digital Library of Mathematical Functions), but the present lack of number referencing could be contrasted with, for example, that for genes, genomes, single-nucleotide polymorphisms or proteins. This has, seemingly, led in some cases to poor characterisation: for example... the AIC is calculated differently in packages bbmle and gamlss. And then they provide one suggestion for cataloguing. An obvious question to the forum is how come the AIC is calculated differently depending on the package, but I wanted to highlight this need for cataloguing statistical objects akin to genes, proteins, etc. Kind regards, José Prof. José Iparraguirre Chief Economist Age UK The Wireless from Age UK | Radio for grown-ups. www.ageuk.org.uk/thewireless If youre looking for a radio station that offers real variety, tune in to The Wireless from Age UK. Whether you choose to listen through the website at www.ageuk.org.uk/thewireless, on digital radio (currently available in London and Yorkshire) or through our TuneIn Radio app, you can look forward to an inspiring mix of music, conversation and useful information 24 hours a day. --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confide...{{dropped:25}} __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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 na.action within own function
I found the solution. coeftest does not work with na.exclude but with na.omit only, i.e. one needs to omit missing values from the residual matrix. Cheers! On Thu, Aug 8, 2013 at 10:19 AM, ivan i.pet...@gmail.com wrote: Dear R Community, I am trying to build a very simple function which uses lm and coeftest to return a coefficient matrix with heteroskedasticity robust standard errors. The function is the following: reg=function(formula,data,na.action){ res=lm(formula=formula,data=data,na.action=na.action) hc3=coeftest(res, vcov = vcovHC(res, type = HC3)) residuals=resid(res) return(list(coef.hc3=hc3,R2=summary(res)$r.squared, R2.adj=summary(res)$adj.r.squared, residuals=residuals)) } The function works perfect as long as the data contains no missing values. I.e. test1=seq(1,30,1) test2=seq(1,30,1) testdata=data.frame(test1,test2) reg(formula=test1~test2, data=testdata, na.action=na.exclude) However, as soon as I have a missing value, it does not work any more (the error message is: Error in estfun(x)/X : non-conformable arrays): test1=seq(1,30,1) test2=seq(1,30,1) test2[5]=NA testdata=data.frame(test1,test2) reg(formula=test1~test2, data=testdata, na.action=na.exclude) My feeling is that it has something to do with na.exclude being a function itself and hence R having a problem with inputing a function into another function. Does anyone have an idea? Thanks a lot!!! 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] p values for partial correlations
Dear Tal: Thank you for your help. Thats what I run: install.packages(corpcor) require(corpcor) correlations=cor(mydata) pcorrrel = cor2pcor(correlations); pcorrrel 2013/8/7 Tal Galili tal.gal...@gmail.com A short self contained code would help us help you. You can try using str on the output of the command you are using, and try to understand where the p.value is located. Tal Contact Details:--- Contact me: tal.gal...@gmail.com | Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Wed, Aug 7, 2013 at 10:01 PM, Demetrio Luis Guadagnin dlguadag...@gmail.com wrote: Dear: I needed to calculate partial correlations and used the package corpcor for that purpose. The output doesnot provide p values and I was unable to find information or posts on how to get them. Does someone can help me? Thanks. -- Dr. Demetrio Luis Guadagnin Conservação e Manejo de Vida Silvestre Universidade Federal do Rio Grande do Sul Departamento de Ecologia Av. Bento Gonçalves 9500 Setor 4, Prédio 43422, Sala 105 Caixa Postal 15007 - 91501-970 Porto Alegre RS Fone: (51) 3308 6774 Fax: (51) 3308 7626 dlguadag...@gmail.com Skype: demetriolguadagnin [[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. -- Dr. Demetrio Luis Guadagnin Conservação e Manejo de Vida Silvestre Universidade Federal do Rio Grande do Sul Departamento de Ecologia Av. Bento Gonçalves 9500 Setor 4, Prédio 43422, Sala 105 Caixa Postal 15007 - 91501-970 Porto Alegre RS Fone: (51) 3308 6774 Fax: (51) 3308 7626 dlguadag...@gmail.com Skype: demetriolguadagnin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] New package about migration indices
Dear useRs, me and my colleague (cc) have recently released a new package on CRAN about computing various migration indices like the Crude Migration Rate, the Effectiveness and Connectivity Index, different Gini indices or the Coefficient of Variation. I hope that some of you dealing with migration matrices in the R console would find this small package useful: http://cran.r-project.org/web/packages/migration.indices/ Best, Gergely [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Add column to dataframe based on code in other column
Hi all, I have a dataframe of users which contain US-state codes. Now I want to add a column named REGION based on the state code. I have already done a mapping: NorthEast - c(07, 20, 22, 30, 31, 33, 39, 41, 47) MidWest - c(14, 15, 16, 17, 23, 24, 26, 28, 35, 36, 43, 52) South - c(01, 04, 08, 09, 10, 11, 18, 19, 21, 25, 34, 37, 42, 44, 45, 49, 51) West - c(02, 03, 05, 06, 12, 13, 27, 29, 32, 38, 46, 50, 53) Other - c(40, 48, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 94, 98, 99) So for example: NameState_Code Tom 20 Harry 56 Ben 05 Sally 04 Should become like: So for example: NameState_Code REGION Tom 20 NorthEast Harry 56 Other Ben 05 West Sally 04 South Could anyone help me with a clever statement? -- View this message in context: http://r.789695.n4.nabble.com/Add-column-to-dataframe-based-on-code-in-other-column-tp4673335.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] Explaining variance in a univariate time series
Dear List, I am looking to reveal the combination of environmental factors that bets explain the observed variance in a uni-variate time series of a population. I have approached this using two methods, and have different results, therefore i was hoping somebody may have done something similar, or have knowledge of the area, such that they could advise me of the best approach. My first approach was to use canonical correlations i.e. search for significant correlations between explanatory variables x1, x2 and x3 and the time series, this made sense to me and produced perfectly plausible ( in line with a priori hypothesise) results. However, this approach doesn't take into account the other variables present. To address this i then used a dynamic factor analysis - explaining temporal variation in a set of n observed time series using linear combinations of a set of m hidden random walks, where m n. I then used a AIC framework to arrive at the most likely model. http://cran.r-project.org/web/packages/MARSS/vignettes/UserGuide.pdf However, the results differed, the variables present in the best AIC model were not necessarily the ones with the strongest canonical correlation. Why my this be the case? Is there a better way to go about this? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Varying statistical significance in estimates of linear model
Hi everyone, I have a response variable 'y' and several predictor variables 'x_i'. I start with a linear model: m1 - lm(y ~ x1); summary(m1) and I get a statistically significant estimate for 'x1'. Then, I modify my model as: m2 - lm(y ~ x1 + x2); summary(m2) At this moment, the estimate for x1 might become non-significant while the estimate of x2 significant. As I add more predictor variables (or interaction terms), the estimates for which I get a statistically significant result vary. So sometimes x1, x2, x6 are significant, while others, x2, x4, x5 are. It seems to me that I could tweak my model in such a way (by adding/removing predictor variables or suitable interaction terms) that I could prove whatever I'd like to prove. What is the proper methodology involved here ? What do you people do in such cases ? I can provide the data if anyone cares and would like to have a look at them. Best regards, Stathis Kamperis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Varying statistical significance in estimates of linear model
Dear Stathis, I recommend that you try to get some advice from a local statistician or read an introductory book on statistics. This kind of question is beyond the scope of a mailing list. Best regards, ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium + 32 2 525 02 51 + 32 54 43 61 85 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Stathis Kamperis Verzonden: donderdag 8 augustus 2013 12:43 Aan: r-help@r-project.org Onderwerp: [R] Varying statistical significance in estimates of linear model Hi everyone, I have a response variable 'y' and several predictor variables 'x_i'. I start with a linear model: m1 - lm(y ~ x1); summary(m1) and I get a statistically significant estimate for 'x1'. Then, I modify my model as: m2 - lm(y ~ x1 + x2); summary(m2) At this moment, the estimate for x1 might become non-significant while the estimate of x2 significant. As I add more predictor variables (or interaction terms), the estimates for which I get a statistically significant result vary. So sometimes x1, x2, x6 are significant, while others, x2, x4, x5 are. It seems to me that I could tweak my model in such a way (by adding/removing predictor variables or suitable interaction terms) that I could prove whatever I'd like to prove. What is the proper methodology involved here ? What do you people do in such cases ? I can provide the data if anyone cares and would like to have a look at them. Best regards, Stathis Kamperis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Varying statistical significance in estimates of linear model
Stathis: 1. This has nothing to do with R. Post on a statistics list, like stats.stackexchange.com 2. Read a basic regression/linear models text. You need to educate yourself. -- Bert On Thu, Aug 8, 2013 at 3:43 AM, Stathis Kamperis ekamp...@gmail.com wrote: Hi everyone, I have a response variable 'y' and several predictor variables 'x_i'. I start with a linear model: m1 - lm(y ~ x1); summary(m1) and I get a statistically significant estimate for 'x1'. Then, I modify my model as: m2 - lm(y ~ x1 + x2); summary(m2) At this moment, the estimate for x1 might become non-significant while the estimate of x2 significant. As I add more predictor variables (or interaction terms), the estimates for which I get a statistically significant result vary. So sometimes x1, x2, x6 are significant, while others, x2, x4, x5 are. It seems to me that I could tweak my model in such a way (by adding/removing predictor variables or suitable interaction terms) that I could prove whatever I'd like to prove. What is the proper methodology involved here ? What do you people do in such cases ? I can provide the data if anyone cares and would like to have a look at them. Best regards, Stathis Kamperis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.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] Add column to dataframe based on code in other column
On 08-08-2013, at 11:33, Dark i...@software-solutions.nl wrote: Hi all, I have a dataframe of users which contain US-state codes. Now I want to add a column named REGION based on the state code. I have already done a mapping: NorthEast - c(07, 20, 22, 30, 31, 33, 39, 41, 47) MidWest - c(14, 15, 16, 17, 23, 24, 26, 28, 35, 36, 43, 52) South - c(01, 04, 08, 09, 10, 11, 18, 19, 21, 25, 34, 37, 42, 44, 45, 49, 51) West - c(02, 03, 05, 06, 12, 13, 27, 29, 32, 38, 46, 50, 53) Other - c(40, 48, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 94, 98, 99) So for example: NameState_Code Tom 20 Harry 56 Ben 05 Sally 04 Should become like: So for example: NameState_Code REGION Tom 20 NorthEast Harry 56 Other Ben 05 West Sally 04 South dd - read.table(text=NameState_Code Tom 20 Harry 56 Ben 05 Sally 04, header=TRUE, stringsAsFactors=FALSE) # Create table for regions indexed by state_code region.table - rep(UNKNOWN,99) region.table[NorthEast] - NorthEast region.table[MidWest] - MidWest region.table[South] - South region.table[West] - West region.table[Other] - Other region.table # then this is easy dd[,REGION] - region.table[dd$State_Code] Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Revolutions blog roundup: July 2013
Revolution Analytics staff write about R every weekday at the Revolutions blog: http://blog.revolutionanalytics.com and every month I post a summary of articles from the previous month of particular interest to readers of r-help. In case you missed them, here are some articles related to R from the month of July: A new 90-second, creative commons video helps R enthusiasts share the history, community and applications of R: http://bit.ly/13mf0HX Analyst group Butler Analytics reviews 10 predictive analytics platforms, and says that real analysts use R: http://bit.ly/13mf0HY An excellent example of Simpsons Paradox: US median wages rose overall, but within every educational subgroup, they declined: http://bit.ly/13meXMk Some tips on identifying R functions that will benefit most from byte compilation, and how to enable automatic package compilation: http://bit.ly/13mf0HZ Joe Rickert's poster at useR! 2013 lists the ways that Revolution Analytics supports the R community: http://bit.ly/13meXMl Andrie deVries describes the applications of survival analysis techniques to solve the problem of marketing attribution: http://bit.ly/13mf0I0 A Shiny app by Ramnath Vaidyanathan displays the real-time status of bike-sharing programs in more than 100 cities: http://bit.ly/13mf0I1 The new (and free) O'Reilly mini-book on real-time analytics includes a section on a big-data architecture with R: http://bit.ly/13mf0I2 Thomas Levine's R spells includes some useful R tricks you might not know about: http://bit.ly/13mf0I3 A review of the Rcpp tutorial at useR! 2013, with some benchmarked examples combining R and C++ code: http://bit.ly/13meXMo Slides from the Hadoop Summit talk High Performance Predictive Analytics in R and Hadoop by Revolution Analytics' US Chief Scientist Mario Inchosa: http://bit.ly/13mf0I4 My two-part review of some talks from the useR! 2013 conference: part 1 http://bit.ly/13mf0I5 and part 2 http://bit.ly/13mf0I6 Joe Rickert looks at the new big-data tree algorithm behind the rxDTree function in the RevoScaleR package: http://bit.ly/13mf3n0 Digital marketing company X+1 uses Revolution R Enterprise for real-time marketing optimization based on statistical models in R: http://bit.ly/13mf3mZ Some highlights from the June 2013 issue of the R Journal: http://bit.ly/13mf0I7 Meet the members of the Revolution Analytics team in London: http://bit.ly/13mf3n1 A map of R user groups worldwide: http://bit.ly/13mf0Yk Highlights and photos from some recent R user group meetings around the world: http://bit.ly/13mf3n2 Some non-R stories in the past month included: what not to do when analyzing data (http://bit.ly/13mf0Ym), results of a survey of data scientists (http://bit.ly/13mf0Yn), a red-hot ball of nickel meets a block of ice (http://bit.ly/13mf0Yl), a lion reunites with his caretakers (http://bit.ly/13mf0Yo) and one picture that looks like four (http://bit.ly/13mf3n3). Meeting times for local R user groups (http://bit.ly/eC5YQe) can be found on the updated R Community Calendar at: http://bit.ly/bb3naW If you're looking for more articles about R, you can find summaries from previous months at http://blog.revolutionanalytics.com/roundups/. Join the Revolution mailing list at http://revolutionanalytics.com/newsletter to be alerted to new articles on a monthly basis. As always, thanks for the comments and please keep sending suggestions to me at da...@revolutionanalytics.com . Don't forget you can also follow the blog using an RSS reader, or by following me on Twitter (I'm @revodavid). Cheers, # David -- David M Smith da...@revolutionanalytics.com VP of Marketing, Revolution Analytics http://blog.revolutionanalytics.com Tel: +1 (650) 646-9523 (Seattle WA, USA) Twitter: @revodavid We're hiring! www.revolutionanalytics.com/careers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Reason for difference in singular value decomposition produced by function La.svd (via prcomp)?
Dear expeRts, I have run some simulations under R 2.15.1 on a Mac, and I have rerun a sample of them under R 3.0.1 on Windows (and also for comparison under R2.14.1 on Windows). For most cases, I get exactly the same results in all three runs. However, for those cases that depend on principal components computed with prcomp, where the particular choice of the orthogonalization is arbitrary because of several identical singular values, I get different results between the two Windows versions on the one hand and the Mac version on the other hand. I did not find anything documented about the difference; maybe I didn't know where to look. Can someone help me understand the reason? Best, Ulrike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Mac Os X 10.6.8, R and edit/vi
I tried using various versions of the 'edit' command. Here is an account of how this failed. I hope I have included all relevant information. I haven't used R for a couple of years. Before restarting with R, I downloaded the latest version I could find in its binary version, and installed it without any problems. Mac Os X Finder command About R responds with R 3.0.1 GUI 1.61 Snow Leopard build (6492) From inside R version platform x86_64-apple-darwin10.8.0 arch x86_64 os darwin10.8.0#(However, my os is in fact 10.6.8) system x86_64, darwin10.8.0 status major 3 minor 0.1 year 2013 month 05 day16 svn rev62743 language R version.string R version 3.0.1 (2013-05-16) nickname Good Sport edit(file='2.9.R') Error in file(con, r) : cannot open the connection In addition: Warning message: In file(con, r) : cannot open file '2.9.R': No such file or directory getOption('editor') [1] vi edit(file='2.9.R',editor='/opt/local/bin/vim') Error in file(con, r) : cannot open the connection In addition: Warning message: In file(con, r) : cannot open file '2.9.R': No such file or directory vi(file='try') Error in file(con, r) : cannot open the connection In addition: Warning message: In file(con, r) : cannot open file 'try': No such file or directory And here is my interaction with tcsh (my default shell) H2:~% echo $VISUAL /opt/local/bin/vim H2:~% echo $EDITOR /opt/local/bin/vim H2:~% which vi vi: aliased to /opt/local/bin/vim H2:~/4Chap2% ls -ld drwxr-xr-x 11 dbae dbae 374 8 Aug 10:54 ./ What am I doing wrong? Thanks for any help. David __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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 dea.boot under R 3.0.1
Dear Vera, I had a similar problem once and as far as I can remember the reason were some negative inputs or outputs. # Check for negative values x1a.neg - apply(x1a, 1, function(x) any(x0)) y1.neg - apply(y1, 1, function(x) any(x0)) exclude - x1a.neg | y1.neg # Exclude negative rows x1a.neg - x1a.neg[!exclude,] y1.neg - y1.neg[!exclude,] # Try again! Daniel -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-dea-boot-under-R-3-0-1-tp4669964p4673363.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] Add column to dataframe based on code in other column
Dark: 1. In future, please use dput() to post data to enable us to more easily read them from your email. 2. As Berend demonstrates, using a more appropriate data structure is what's required. Here is a slightly shorter, but perhaps trickier alternative to his solution: df ## Your example data frame Name State_Code 1 Tom 20 2 Harry 56 3 Ben 5 4 Sally 4 l -list(MidWest=MidWest,South=South,NorthEast=NorthEast,Other=Other,West=West) df - within(df,regions - rep(names(l),sapply(l,length))[match(State_Code,unlist(l))]) df Name State_Code regions 1 Tom 20 NorthEast 2 Harry 56 Other 3 Ben 5 West 4 Sally 4 South 3. Need I say that there may be other alternatives that might be better. Cheers, Bert On Thu, Aug 8, 2013 at 7:14 AM, Berend Hasselman b...@xs4all.nl wrote: On 08-08-2013, at 11:33, Dark i...@software-solutions.nl wrote: Hi all, I have a dataframe of users which contain US-state codes. Now I want to add a column named REGION based on the state code. I have already done a mapping: NorthEast - c(07, 20, 22, 30, 31, 33, 39, 41, 47) MidWest - c(14, 15, 16, 17, 23, 24, 26, 28, 35, 36, 43, 52) South - c(01, 04, 08, 09, 10, 11, 18, 19, 21, 25, 34, 37, 42, 44, 45, 49, 51) West - c(02, 03, 05, 06, 12, 13, 27, 29, 32, 38, 46, 50, 53) Other - c(40, 48, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 94, 98, 99) So for example: NameState_Code Tom 20 Harry 56 Ben 05 Sally 04 Should become like: So for example: NameState_Code REGION Tom 20 NorthEast Harry 56 Other Ben 05 West Sally 04 South dd - read.table(text=NameState_Code Tom 20 Harry 56 Ben 05 Sally 04, header=TRUE, stringsAsFactors=FALSE) # Create table for regions indexed by state_code region.table - rep(UNKNOWN,99) region.table[NorthEast] - NorthEast region.table[MidWest] - MidWest region.table[South] - South region.table[West] - West region.table[Other] - Other region.table # then this is easy dd[,REGION] - region.table[dd$State_Code] Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.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] How is a file descriptor stored ?
I cannot reproduce your problem. You will have to give more details. (I assume you have already made the suggested changes to your code - either label the 3rd argument to your assign call 'envir=' or use the syntax 'cpufile[[key]] - value' instead of assign.) To start debugging this, have your function print the class of the object your object before you try to close it. You can do this by adding a cat() statement or by using options(error=recover) so you can inspect things after an error. Also, get rid of the tryCatch business and the calls to sink(). The first hides where an error might be coming from and the latter may be hiding some printed messages that might help you track down the problem. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: mohan.radhakrish...@polarisft.com [mailto:mohan.radhakrish...@polarisft.com] Sent: Wednesday, August 07, 2013 11:56 PM To: r-help@r-project.org Cc: Berend Hasselman; William Dunlap Subject: RE: [R] How is a file descriptor stored ? Hi, The file handling code sometimes throws this exception. Error in UseMethod(close) : no applicable method for 'close' applied to an object of class c ('integer', 'numeric') Is there a sample based on my code that I can test ? I want to extract the file descriptors from the hashmap and close them. I think that is causing the exception. Sometimes just closing - close(fd) - is causing this too. Thanks, Mohan RE: [R] How is a file descriptor stored ? William Dunlap to: Berend Hasselman, mohan.radhakrish...@polarisft.com 07-08-2013 08:01 PM Cc: r-help@r-project.org Use assign(key, file( key, w ), envir=cpufile) In your assign expression you are assigning cpufile to the third formal argument which is pos. You meant the envir argument, I presume. Or use the syntax cpufile[[key]] - file(key, w) instead of assign(key, file( key, w ), envir=cpufile) The former works for lists and environments and corresponds to your later usage of listoffiles[[key]] to retrieve the data. From what I've seen of your example, a list might be a better way to store your data, because of its copy-on-write semantics and because it doesn't keep a parent environment in memory. By using '[[' instead of 'get' and 'assign' you minimize the number of changes required to switch between a list and an environment. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Berend Hasselman Sent: Wednesday, August 07, 2013 4:10 AM To: mohan.radhakrish...@polarisft.com Cc: r-help@r-project.org Subject: Re: [R] How is a file descriptor stored ? On 07-08-2013, at 12:13, mohan.radhakrish...@polarisft.com wrote: Hi, I thought that 'R' like java will allow me to store file names (keys) and file descriptors(values) in a hashmap. filelist.array - function(n){ sink(nmon.log) cpufile - new.env(hash=T, parent=emptyenv()) for (i in 1:n) { key - paste(output, i, .txt, sep = ) assign(key, file( key, w ), cpufile) } sink() return (cpufile) } But when I try to test it like this there is an exception [1] Exception is Error in UseMethod(\close\): no applicable method for 'close' applied to an object of class \c('integer', 'numeric')\\n test.simple.filelist.array - function() { execution - tryCatch({ sink(nmon.log) listoffiles - filelist.array(3) for (v in ls(listoffiles)) { print(paste(Map value is [, listoffiles[[v]], ])) fd - listoffiles[[v]] close(fd) } sink() }, error = function(err){ print(paste(Exception is ,err)) }) } I think I am missing some fundamentals. Read the help page for assign more carefully. Use assign(key, file( key, w ), envir=cpufile) In your assign expression you are assigning cpufile to the third formal argument which is pos. You meant the envir argument, I presume. Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing
Re: [R] Add column to dataframe based on code in other column
dat1- read.table(text= Name State_Code Tom 20 Harry 56 Ben 05 Sally 04 ,sep=,header=TRUE,stringsAsFactors=FALSE) dat2- do.call(cbind,list(NorthEast,MidWest,South,West,Other)) colnames(dat2)- c(NorthEast,MidWest,South,West,Other) dat2- as.data.frame(dat2) library(reshape2) datM-melt(dat2) colnames(datM)- c(REGION,State_Code) library(plyr) join(dat1,datM,type=left,match=first,by=State_Code)[,c(2,1,3)] # Name State_Code REGION #1 Tom 20 NorthEast #2 Harry 56 Other #3 Ben 5 West #4 Sally 4 South A.K. - Original Message - From: Dark i...@software-solutions.nl To: r-help@r-project.org Cc: Sent: Thursday, August 8, 2013 5:33 AM Subject: [R] Add column to dataframe based on code in other column Hi all, I have a dataframe of users which contain US-state codes. Now I want to add a column named REGION based on the state code. I have already done a mapping: NorthEast - c(07, 20, 22, 30, 31, 33, 39, 41, 47) MidWest - c(14, 15, 16, 17, 23, 24, 26, 28, 35, 36, 43, 52) South - c(01, 04, 08, 09, 10, 11, 18, 19, 21, 25, 34, 37, 42, 44, 45, 49, 51) West - c(02, 03, 05, 06, 12, 13, 27, 29, 32, 38, 46, 50, 53) Other - c(40, 48, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 94, 98, 99) So for example: Name State_Code Tom 20 Harry 56 Ben 05 Sally 04 Should become like: So for example: Name State_Code REGION Tom 20 NorthEast Harry 56 Other Ben 05 West Sally 04 South Could anyone help me with a clever statement? -- View this message in context: http://r.789695.n4.nabble.com/Add-column-to-dataframe-based-on-code-in-other-column-tp4673335.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] cbind with headers
Hi, You can save it in file. I copy and paste: Subtype,Gender,Expression A,m,-0.54 A,f,-0.8 B,f,-1.03 C,m,-0.41 on the gedit and save it as data1.csv. You might be able to do the same with notepad. x - read.csv(data1.csv,header=T,sep=,) x2 - read.csv(data2N.csv,header=T,sep=,) x3 - cbind(x,x2) x3 # Subtype Gender Expression Age City #1 A m -0.54 32 New York #2 A f -0.80 21 Houston #3 B f -1.03 34 Seattle #4 C m -0.41 67 Houston #or if the dataset is small as in the example x- read.table(text= Subtype,Gender,Expression A,m,-0.54 A,f,-0.8 B,f,-1.03 C,m,-0.41 ,sep=,,header=TRUE,stringsAsFactors=FALSE) x2- read.table(text= Age,City 32,New York 21,Houston 34,Seattle 67,Houston ,sep=,,header=TRUE,stringsAsFactors=FALSE) cbind(x,x2) # Subtype Gender Expression Age City #1 A m -0.54 32 New York #2 A f -0.80 21 Houston #3 B f -1.03 34 Seattle #4 C m -0.41 67 Houston A.K. Hi, I can't seem to get this to work: http://www.endmemo.com/program/R/cbind.php Do I save the data as data1.csv in note pad and pull in the file? Do I type data1.csv-Subtype,Gender,Expression,A,m,-0.54,A,f,-0.8,B,f,-1.03,C,m,-0.41?? I can do a simple matrix. But, I want to have headers and data to combine. Simple Matrix I combined. m-as.data.frame(matrix(c(1:6),ncol=2)) n-as.data.frame(matrix(c(7:12),ncol=2)) m V1 V2 1 1 4 2 2 5 3 3 6 n V1 V2 1 7 10 2 8 11 3 9 12 Mary-cbind(m,n) Mary V1 V2 V1 V2 1 1 4 7 10 2 2 5 8 11 3 3 6 9 12 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Why is mclappy slower than apply in this case?
Hello, i'm pretty confused. I want to speed up my algorithm by using mclapply: parallel, but when I compare time efficiency, apply still wins. I'm smoothing log2ratio data by rq.fit.fnb:quantreg which is called by my function quantsm and I'm wrapping my data into matrix/list for apply/lapply (mclapply) usage. I adjust my data like this: codespan class='pln'q /spanspan class='pun'=/spanspan class='pln' matrix/spanspan class='pun'(/spanspan class='pln'data/spanspan class='pun',/spanspan class='pln' ncol/spanspan class='pun'=/spanspan class='pln'N/spanspan class='pun')/spanspan class='pln'/spanspan class='com'# wrapping into matrix (using N = 2, 4, 6 or 8)/spanspan class='pln' ql /spanspan class='pun'=/spanspan class='pln' /spanspan class='kwd'as/spanspan class='pun'./spanspan class='pln'list/spanspan class='pun'(/spanspan class='kwd'as/spanspan class='pun'./spanspan class='pln'data/spanspan class='pun'./spanspan class='pln'frame/spanspan class='pun'(/spanspan class='pln'q/spanspan class='pun'))/spanspan class='pln' /spanspan class='com'# making list/span/code And time comparing: codespan class='pln'apply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'apply/spanspan class='pun'(/spanspan class='pln'q/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'1/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'lapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' mc2lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' mc4lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'4/spanspan class='pun'))/spanspan class='pln' mc6lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'6/spanspan class='pun'))/spanspan class='pln' mc8lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'8/spanspan class='pun'))/spanspan class='pln' timing/spanspan class='pun'=/spanspan class='pln'rbind/spanspan class='pun'(/spanspan class='pln'apply/spanspan class='pun',/spanspan class='pln'lapply/spanspan class='pun',/spanspan class='pln'mc2lapply/spanspan class='pun',/spanspan class='pln'mc4lapply/spanspan class='pun',/spanspan
[R] mgcv predict.bam strange results
Dear useR, I don't understand the results of the predict.bam function of mgcv package when constucting a varying-coefficient model with bam instead of gam: library(mgcv) dat - gamSim(4) b - gam(y ~ fac+s(x2,by=fac)+s(x0), data=dat) predict(b, dat[1,], type = terms) with gam everything is fine: only s(x2):fac1 is different of zero but using bam: b1 - bam(y ~ fac+s(x2,by=fac, bs = cc)+s(x0),data=dat) predict(b1, dat[1,], type = terms) all terms s(x2):fac1, s(x2):fac2 and s(x2):fac3 are differnt of zero. Thanks for your help Florian -- View this message in context: http://r.789695.n4.nabble.com/mgcv-predict-bam-strange-results-tp4673364.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] cbind with headers
Hi, I can't seem to get this to work: http://www.endmemo.com/program/R/cbind.php Do I save the data as data1.csv in note pad and pull in the file? Do I type data1.csv-Subtype,Gender,Expression,A,m,-0.54,A,f,-0.8,B,f,-1.03,C,m,-0.41?? I can do a simple matrix. But, I want to have headers and data to combine. Simple Matrix I combined. m-as.data.frame(matrix(c(1:6),ncol=2)) n-as.data.frame(matrix(c(7:12),ncol=2)) m V1 V2 1 1 4 2 2 5 3 3 6 n V1 V2 1 7 10 2 8 11 3 9 12 Mary-cbind(m,n) Mary V1 V2 V1 V2 1 1 4 7 10 2 2 5 8 11 3 3 6 9 12 -- View this message in context: http://r.789695.n4.nabble.com/cbind-with-headers-tp4673354.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] For loop output
I am having difficulty storing the output of a for loop I have generated. All I want to do is find all the files that I have, create a string with all of the names in quotes and separated by commas. This is proving more difficult than I initially anticipated. I am sure it is either very simple or the construction of the for loop is not quite right The result gets automatically printed after the loop but I can't seem to save it. I have tried to create the element in advance but the result is the same: NULL individual.proj = Sys.glob(Arabica/proj_current/individual_projections/*.img, dirmark = FALSE) individual.proj [1] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img [2] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GLM.img [3] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_MARS.img [4] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_RF.img [5] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_RUN10_GBM.img ##generate loop to create string out of the table of projected files. L.ip = length(individual.proj) for (i in 1:L.ip){ individual.proj.i - individual.proj[i] individual.proj.quote = cat(paste('', individual.proj.i, '', ',',sep=)) } Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img,Arabica/proj_current/individual_projections/proj_current ##print output string individual.proj.quote NULL #command to be applied to individual.proj.quote to removed the final comma from the string substr(individual.proj.quote, 1, nchar(individual.proj.quote)-1) Any help or pointers would be greatly appreciated, no amount of extensive google searches have been fruitful so far. ** Jenny Williams Spatial Information Scientist, GIS Unit Herbarium, Library, Art Archives Directorate Royal Botanic Gardens, Kew Richmond, TW9 3AB, UK Tel: +44 (0)208 332 5277 email: jenny.willi...@kew.orgmailto:jenny.willi...@kew.org ** Film: The Forgotten Home of Coffee - Beyond the Gardenshttp://www.youtube.com/watch?v=-uDtytKMKpAsns=tw Stories: Coffee Expedition - Ethiopiahttp://storify.com/KewGIS/coffee-expedition-ethiopia Kew in Harapan Rainforest Sumatrahttp://storify.com/KewGIS/kew-in-harapan-rainforest Articles: Seeing the wood for the treeshttp://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf How Kew's GIS team and South East Asia botanists are working to help conserve and restore a rainforest in Sumatra. Download a pdf of this article here.http://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf The Royal Botanic Gardens, Kew is a non-departmental public body with exempt charitable status, whose principal place of business is at Royal Botanic Gardens, Kew, Richmond, Surrey TW9 3AB, United Kingdom. The information contained in this email and any attachments is intended solely for the addressee(s) and may contain confidential or legally privileged information. If you have received this message in error, please return it immediately and permanently delete it. Do not use, copy or disclose the information contained in this email or in any attachment. Any views expressed in this email do not necessarily reflect the opinions of RBG Kew. Any files attached to this email have been inspected with virus detection software by RBG Kew before transmission, however you should carry out your own virus checks before opening any attachments. RBG Kew accepts no liability for any loss or damage which may be caused by software viruses. [[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] For loop output
Hi Jenny, Firstly, to my knowledge you cannot assign the output of cat to an object (i.e. it only prints it). Second, you can just add the 'collapse' option of the paste function. individual.proj.quote - paste(individual.proj, collapse = ,) if you really want the quotes individual.proj.quote - paste(individual.proj, collapse=',') but you will be stuck with some backslashes I can't recall the syntax to remove. Hope this serves your purposes Cheers, Charles On Thu, Aug 8, 2013 at 10:05 AM, Jenny Williams jenny.willi...@kew.orgwrote: I am having difficulty storing the output of a for loop I have generated. All I want to do is find all the files that I have, create a string with all of the names in quotes and separated by commas. This is proving more difficult than I initially anticipated. I am sure it is either very simple or the construction of the for loop is not quite right The result gets automatically printed after the loop but I can't seem to save it. I have tried to create the element in advance but the result is the same: NULL individual.proj = Sys.glob(Arabica/proj_current/individual_projections/*.img, dirmark = FALSE) individual.proj [1] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img [2] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GLM.img [3] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_MARS.img [4] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_RF.img [5] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_RUN10_GBM.img ##generate loop to create string out of the table of projected files. L.ip = length(individual.proj) for (i in 1:L.ip){ individual.proj.i - individual.proj[i] individual.proj.quote = cat(paste('', individual.proj.i, '', ',',sep=)) } Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img,Arabica/proj_current/individual_projections/proj_current ##print output string individual.proj.quote NULL #command to be applied to individual.proj.quote to removed the final comma from the string substr(individual.proj.quote, 1, nchar(individual.proj.quote)-1) Any help or pointers would be greatly appreciated, no amount of extensive google searches have been fruitful so far. ** Jenny Williams Spatial Information Scientist, GIS Unit Herbarium, Library, Art Archives Directorate Royal Botanic Gardens, Kew Richmond, TW9 3AB, UK Tel: +44 (0)208 332 5277 email: jenny.willi...@kew.orgmailto:jenny.willi...@kew.org ** Film: The Forgotten Home of Coffee - Beyond the Gardens http://www.youtube.com/watch?v=-uDtytKMKpAsns=tw Stories: Coffee Expedition - Ethiopia http://storify.com/KewGIS/coffee-expedition-ethiopia Kew in Harapan Rainforest Sumatra http://storify.com/KewGIS/kew-in-harapan-rainforest Articles: Seeing the wood for the trees http://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf How Kew's GIS team and South East Asia botanists are working to help conserve and restore a rainforest in Sumatra. Download a pdf of this article here. http://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf The Royal Botanic Gardens, Kew is a non-departmental public body with exempt charitable status, whose principal place of business is at Royal Botanic Gardens, Kew, Richmond, Surrey TW9 3AB, United Kingdom. The information contained in this email and any attachments is intended solely for the addressee(s) and may contain confidential or legally privileged information. If you have received this message in error, please return it immediately and permanently delete it. Do not use, copy or disclose the information contained in this email or in any attachment. Any views expressed in this email do not necessarily reflect the opinions of RBG Kew. Any files attached to this email have been inspected with virus detection software by RBG Kew before transmission, however you should carry out your own virus checks before opening any attachments. RBG Kew accepts no liability for any loss or damage which may be caused by software viruses. [[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 Determan Integrated Biosciences PhD Candidate University of Minnesota [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do
Re: [R] For loop output
On Thu, Aug 08, 2013 at 04:05:57PM +0100, Jenny Williams wrote: I am having difficulty storing the output of a for loop I have generated. All I want to do is find all the files that I have, create a string with all of the names in quotes and separated by commas. This is proving more difficult than I initially anticipated. I am sure it is either very simple or the construction of the for loop is not quite right The result gets automatically printed after the loop but I can't seem to save it. I have tried to create the element in advance but the result is the same: NULL This is a somewhat frequent confusion of the (very different!) concepts of assignment and printing. The fact that something becomes visible to the human user via printing has nothing to do really with the fact that an object is generated and assigned to a variable, thereby becoming accessible (and, in that allegorical sense, visible) to the subsequent code. The cat function prints values but it does not return them. It returns NULL, which is what you get. individual.proj = Sys.glob(Arabica/proj_current/individual_projections/*.img, dirmark = FALSE) individual.proj [1] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img [2] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GLM.img [3] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_MARS.img [4] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_RF.img [5] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_RUN10_GBM.img ##generate loop to create string out of the table of projected files. L.ip = length(individual.proj) for (i in 1:L.ip){ individual.proj.i - individual.proj[i] individual.proj.quote = cat(paste('', individual.proj.i, '', ',',sep=)) } Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img,Arabica/proj_current/individual_projections/proj_current ##print output string individual.proj.quote NULL #command to be applied to individual.proj.quote to removed the final comma from the string substr(individual.proj.quote, 1, nchar(individual.proj.quote)-1) Any help or pointers would be greatly appreciated, no amount of extensive google searches have been fruitful so far. If you don't mind me suggesting this, reading the documentation page of the function(s) you use is an approach that is more targeted and therefore often quicker. (Googling cat is probably especially bad, as there are various cat applications and functions in several languages, not to mention several species of mammals... ;-) ) As a further remark, you don't need a loop, one line composed of sprintf and paste (check the collapse parameter) should do the trick you're after. Best regards, Jan ** Jenny Williams Spatial Information Scientist, GIS Unit Herbarium, Library, Art Archives Directorate Royal Botanic Gardens, Kew Richmond, TW9 3AB, UK Tel: +44 (0)208 332 5277 email: jenny.willi...@kew.orgmailto:jenny.willi...@kew.org ** Film: The Forgotten Home of Coffee - Beyond the Gardenshttp://www.youtube.com/watch?v=-uDtytKMKpAsns=tw Stories: Coffee Expedition - Ethiopiahttp://storify.com/KewGIS/coffee-expedition-ethiopia Kew in Harapan Rainforest Sumatrahttp://storify.com/KewGIS/kew-in-harapan-rainforest Articles: Seeing the wood for the treeshttp://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf How Kew's GIS team and South East Asia botanists are working to help conserve and restore a rainforest in Sumatra. Download a pdf of this article here.http://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf The Royal Botanic Gardens, Kew is a non-departmental public body with exempt charitable status, whose principal place of business is at Royal Botanic Gardens, Kew, Richmond, Surrey TW9 3AB, United Kingdom. The information contained in this email and any attachments is intended solely for the addressee(s) and may contain confidential or legally privileged information. If you have received this message in error, please return it immediately and permanently delete it. Do not use, copy or disclose the information contained in this email or in any attachment. Any views expressed in this email do not necessarily reflect the opinions of RBG Kew. Any files attached to this email have been inspected with virus detection software by RBG Kew before transmission, however you should carry out your own virus checks before opening any attachments. RBG Kew accepts no liability for any loss or damage which may be caused by software viruses. [[alternative HTML version deleted]]
Re: [R] Why is mclappy slower than apply in this case?
Tomas: Do some reading on parallelization. Parallelizing code requires the overhead of setting up, keeping track of, synching the separate threads. Whether that overhead is worth the cost depends on the problem,the size, the algorithms, the machines/hardware,... Cheers, Bert On Thu, Aug 8, 2013 at 4:00 AM, Tomas Reigl inciv...@seznam.cz wrote: Hello, i'm pretty confused. I want to speed up my algorithm by using mclapply: parallel, but when I compare time efficiency, apply still wins. I'm smoothing log2ratio data by rq.fit.fnb:quantreg which is called by my function quantsm and I'm wrapping my data into matrix/list for apply/lapply (mclapply) usage. I adjust my data like this: codespan class='pln'q /spanspan class='pun'=/spanspan class='pln' matrix/spanspan class='pun'(/spanspan class='pln'data/spanspan class='pun',/spanspan class='pln' ncol/spanspan class='pun'=/spanspan class='pln'N/spanspan class='pun')/spanspan class='pln'/spanspan class='com'# wrapping into matrix (using N = 2, 4, 6 or 8)/spanspan class='pln' ql /spanspan class='pun'=/spanspan class='pln' /spanspan class='kwd'as/spanspan class='pun'./spanspan class='pln'list/spanspan class='pun'(/spanspan class='kwd'as/spanspan class='pun'./spanspan class='pln'data/spanspan class='pun'./spanspan class='pln'frame/spanspan class='pun'(/spanspan class='pln'q/spanspan class='pun'))/spanspan class='pln' /spanspan class='com'# making list/span/code And time comparing: codespan class='pln'apply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'apply/spanspan class='pun'(/spanspan class='pln'q/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'1/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'lapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' mc2lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' mc4lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'4/spanspan class='pun'))/spanspan class='pln' mc6lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'6/spanspan class='pun'))/spanspan class='pln' mc8lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan
Re: [R] For loop output
On Thu, Aug 08, 2013 at 11:38:33AM -0500, Charles Determan Jr wrote: Hi Jenny, Firstly, to my knowledge you cannot assign the output of cat to an object (i.e. it only prints it). Second, you can just add the 'collapse' option of the paste function. individual.proj.quote - paste(individual.proj, collapse = ,) if you really want the quotes individual.proj.quote - paste(individual.proj, collapse=',') No -- the backslashes are part of the visualisation of the value, not part of the value itself. Please see FAQ 7.37 for details. And also notice that the collapse parameter specifies a separator, and the quotes are intended to enclose the individual words here, not to separate them. In this case, the left quote of the first word and the right quote of the last word will not be generated as a result of this misuse of a separator... Best regards, Jan but you will be stuck with some backslashes I can't recall the syntax to remove. Hope this serves your purposes Cheers, Charles On Thu, Aug 8, 2013 at 10:05 AM, Jenny Williams jenny.willi...@kew.orgwrote: I am having difficulty storing the output of a for loop I have generated. All I want to do is find all the files that I have, create a string with all of the names in quotes and separated by commas. This is proving more difficult than I initially anticipated. I am sure it is either very simple or the construction of the for loop is not quite right The result gets automatically printed after the loop but I can't seem to save it. I have tried to create the element in advance but the result is the same: NULL individual.proj = Sys.glob(Arabica/proj_current/individual_projections/*.img, dirmark = FALSE) individual.proj [1] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img [2] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GLM.img [3] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_MARS.img [4] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_RF.img [5] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_RUN10_GBM.img ##generate loop to create string out of the table of projected files. L.ip = length(individual.proj) for (i in 1:L.ip){ individual.proj.i - individual.proj[i] individual.proj.quote = cat(paste('', individual.proj.i, '', ',',sep=)) } Arabica/proj_current/individual_projections/proj_current_arabica_pa.data.tmp$pa.tab_Full_GBM.img,Arabica/proj_current/individual_projections/proj_current ##print output string individual.proj.quote NULL #command to be applied to individual.proj.quote to removed the final comma from the string substr(individual.proj.quote, 1, nchar(individual.proj.quote)-1) Any help or pointers would be greatly appreciated, no amount of extensive google searches have been fruitful so far. ** Jenny Williams Spatial Information Scientist, GIS Unit Herbarium, Library, Art Archives Directorate Royal Botanic Gardens, Kew Richmond, TW9 3AB, UK Tel: +44 (0)208 332 5277 email: jenny.willi...@kew.orgmailto:jenny.willi...@kew.org ** Film: The Forgotten Home of Coffee - Beyond the Gardens http://www.youtube.com/watch?v=-uDtytKMKpAsns=tw Stories: Coffee Expedition - Ethiopia http://storify.com/KewGIS/coffee-expedition-ethiopia Kew in Harapan Rainforest Sumatra http://storify.com/KewGIS/kew-in-harapan-rainforest Articles: Seeing the wood for the trees http://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf How Kew's GIS team and South East Asia botanists are working to help conserve and restore a rainforest in Sumatra. Download a pdf of this article here. http://www.kew.org/ucm/groups/public/documents/document/kppcont_060602.pdf The Royal Botanic Gardens, Kew is a non-departmental public body with exempt charitable status, whose principal place of business is at Royal Botanic Gardens, Kew, Richmond, Surrey TW9 3AB, United Kingdom. The information contained in this email and any attachments is intended solely for the addressee(s) and may contain confidential or legally privileged information. If you have received this message in error, please return it immediately and permanently delete it. Do not use, copy or disclose the information contained in this email or in any attachment. Any views expressed in this email do not necessarily reflect the opinions of RBG Kew. Any files attached to this email have been inspected with virus detection software by RBG Kew before transmission, however you should carry out your own virus checks before opening any attachments. RBG Kew accepts no liability for any loss or
Re: [R] For loop output
It's not clear how you are planning to use this within R, but you don't need a loop. individual.proj.quote - capture.output(write.table(matrix(individual.proj, 1), quote=TRUE, sep=,, row.names=FALSE, col.names=FALSE)) This produces a single character string which consists of the quoted file names separated by commas. - David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jenny Williams Sent: Thursday, August 8, 2013 10:06 AM To: 'r-help@r-project.org' Subject: [R] For loop output I am having difficulty storing the output of a for loop I have generated. All I want to do is find all the files that I have, create a string with all of the names in quotes and separated by commas. This is proving more difficult than I initially anticipated. I am sure it is either very simple or the construction of the for loop is not quite right The result gets automatically printed after the loop but I can't seem to save it. I have tried to create the element in advance but the result is the same: NULL individual.proj = Sys.glob(Arabica/proj_current/individual_projections/*.img, dirmark = FALSE) individual.proj [1] Arabica/proj_current/individual_projections/proj_current_arabic a_pa.data.tmp$pa.tab_Full_GBM.img [2] Arabica/proj_current/individual_projections/proj_current_arabic a_pa.data.tmp$pa.tab_Full_GLM.img [3] Arabica/proj_current/individual_projections/proj_current_arabic a_pa.data.tmp$pa.tab_Full_MARS.img [4] Arabica/proj_current/individual_projections/proj_current_arabic a_pa.data.tmp$pa.tab_Full_RF.img [5] Arabica/proj_current/individual_projections/proj_current_arabic a_pa.data.tmp$pa.tab_RUN10_GBM.img ##generate loop to create string out of the table of projected files. L.ip = length(individual.proj) for (i in 1:L.ip){ individual.proj.i - individual.proj[i] individual.proj.quote = cat(paste('', individual.proj.i, '', ',',sep=)) } Arabica/proj_current/individual_projections/proj_current_arabic a_pa.data.tmp$pa.tab_Full_GBM.img,Arabica/proj_current/individ ual_projections/proj_current ##print output string individual.proj.quote NULL #command to be applied to individual.proj.quote to removed the final comma from the string substr(individual.proj.quote, 1, nchar(individual.proj.quote)-1) Any help or pointers would be greatly appreciated, no amount of extensive google searches have been fruitful so far. ** Jenny Williams Spatial Information Scientist, GIS Unit Herbarium, Library, Art Archives Directorate Royal Botanic Gardens, Kew Richmond, TW9 3AB, UK Tel: +44 (0)208 332 5277 email: jenny.willi...@kew.orgmailto:jenny.willi...@kew.org ** Film: The Forgotten Home of Coffee - Beyond the Gardenshttp://www.youtube.com/watch?v=-uDtytKMKpAsns=tw Stories: Coffee Expedition - Ethiopiahttp://storify.com/KewGIS/coffee-expedition-ethiopia Kew in Harapan Rainforest Sumatrahttp://storify.com/KewGIS/kew-in-harapan-rainforest Articles: Seeing the wood for the treeshttp://www.kew.org/ucm/groups/public/documents/document/kp pcont_060602.pdf How Kew's GIS team and South East Asia botanists are working to help conserve and restore a rainforest in Sumatra. Download a pdf of this article here.http://www.kew.org/ucm/groups/public/documents/document/kp pcont_060602.pdf The Royal Botanic Gardens, Kew is a non-departmental public body with exempt charitable status, whose principal place of business is at Royal Botanic Gardens, Kew, Richmond, Surrey TW9 3AB, United Kingdom. The information contained in this email and any attachments is intended solely for the addressee(s) and may contain confidential or legally privileged information. If you have received this message in error, please return it immediately and permanently delete it. Do not use, copy or disclose the information contained in this email or in any attachment. Any views expressed in this email do not necessarily reflect the opinions of RBG Kew. Any files attached to this email have been inspected with virus detection software by RBG Kew before transmission, however you should carry out your own virus checks before opening any attachments. RBG Kew accepts no liability for any loss or damage which may be caused by software viruses. [[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
[R] [R-pkgs] WriteXLS version 3.2.1 - Bug Fix Release
Hi all, WriteXLS version 3.2.1 has been submitted to CRAN, with thanks to the CRAN maintainers. This is a bug fix release with the following fixes: 1. When row.names = TRUE, the initial comments row, which contains the comments attributes for the data frame columns and is rbind()ed to the source data frame, was not properly identified by the included Perl script code, resulting in an error in the Excel file. Bug identified yesterday by Robert Zeigler via GitHub. 2. The rownames for the original data frame were not being properly preserved, hence were in error in the Excel file when row.names = TRUE. Picked up in the course of fixing the above bug. Source tarballs of the updated package are being mirrored and binaries for Windows and OSX should appear in due course. Regards, Marc Schwartz ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] matrix with standard errors of lm model
Hi Can someone give me a hint on how to create a matrix with standard errors from lm model? I have already managed to get the matrix with coefficients: coef-as.data.frame(sapply(seq_len(ncol(es.w)),function( i) {x1- summary(lm(es.w[,i]~es.median[,i]));x1$coef[,1]})) but I can't get the one like this for standard errors. I do regression for each column. Thanks 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.
Re: [R] Why is mclappy slower than apply in this case?
Tomas, Also: please don't send html emails (as is specified in the posting guide[1]). This is what your email looked like on our end of the table: https://stat.ethz.ch/pipermail/r-help/attachments/20130808/74a3c7c2/attachment.pl [1] Posting Guide: http://www.r-project.org/posting-guide.html -steve On Thu, Aug 8, 2013 at 9:52 AM, Bert Gunter gunter.ber...@gene.com wrote: Tomas: Do some reading on parallelization. Parallelizing code requires the overhead of setting up, keeping track of, synching the separate threads. Whether that overhead is worth the cost depends on the problem,the size, the algorithms, the machines/hardware,... Cheers, Bert On Thu, Aug 8, 2013 at 4:00 AM, Tomas Reigl inciv...@seznam.cz wrote: Hello, i'm pretty confused. I want to speed up my algorithm by using mclapply: parallel, but when I compare time efficiency, apply still wins. I'm smoothing log2ratio data by rq.fit.fnb:quantreg which is called by my function quantsm and I'm wrapping my data into matrix/list for apply/lapply (mclapply) usage. I adjust my data like this: codespan class='pln'q /spanspan class='pun'=/spanspan class='pln' matrix/spanspan class='pun'(/spanspan class='pln'data/spanspan class='pun',/spanspan class='pln' ncol/spanspan class='pun'=/spanspan class='pln'N/spanspan class='pun')/spanspan class='pln'/spanspan class='com'# wrapping into matrix (using N = 2, 4, 6 or 8)/spanspan class='pln' ql /spanspan class='pun'=/spanspan class='pln' /spanspan class='kwd'as/spanspan class='pun'./spanspan class='pln'list/spanspan class='pun'(/spanspan class='kwd'as/spanspan class='pun'./spanspan class='pln'data/spanspan class='pun'./spanspan class='pln'frame/spanspan class='pun'(/spanspan class='pln'q/spanspan class='pun'))/spanspan class='pln' /spanspan class='com'# making list/span/code And time comparing: codespan class='pln'apply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'apply/spanspan class='pun'(/spanspan class='pln'q/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'1/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'lapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' mc2lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'2/spanspan class='pun'))/spanspan class='pln' mc4lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'4/spanspan class='pun'))/spanspan class='pln' mc6lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply/spanspan class='pun'(/spanspan class='pln'ql/spanspan class='pun',/spanspan class='pln' FUN/spanspan class='pun'=/spanspan class='pln'quantsm/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'0.50/spanspan class='pun',/spanspan class='pln' /spanspan class='lit'2/spanspan class='pun',/spanspan class='pln' mc/spanspan class='pun'./spanspan class='pln'cores/spanspan class='pun'=/spanspan class='lit'6/spanspan class='pun'))/spanspan class='pln' mc8lapply/spanspan class='pun'=/spanspan class='pln'system/spanspan class='pun'./spanspan class='pln'time/spanspan class='pun'(/spanspan class='pln'mclapply
Re: [R] matrix with standard errors of lm model
Perhaps ?vcov is what you are looking for. -- Bert On Thu, Aug 8, 2013 at 10:37 AM, iza.ch1 iza@op.pl wrote: Hi Can someone give me a hint on how to create a matrix with standard errors from lm model? I have already managed to get the matrix with coefficients: coef-as.data.frame(sapply(seq_len(ncol(es.w)),function( i) {x1- summary(lm(es.w[,i]~es.median[,i]));x1$coef[,1]})) but I can't get the one like this for standard errors. I do regression for each column. Thanks 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.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] Creating new vectors from other dataFrames
I have two data frames data1 - as.data.frame(matrix(data=c(1:4,5:8,9:12,13:24), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,d,e,z data2 - as.data.frame(matrix(data=c(1:4,5:8,9:12,37:48), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,f,g,z that have some common column names. Comparing the names of the columns within each data frame to the other setdiff(names(data1), names(data2)) setdiff(names(data2), names(data1)) provides which columns are different. For each column that appears in data1 that DOES NOT appear in data2, I need to create those columns and fill them with NA values. The same is true for the reverse. So, I can create a vector of new column names that need to be filled with NA values, but here is where I'm stuck. I don't know how to get the names from inside the vector into the respective dataFrame. tmp1 - as.factor(paste(data2$, setdiff(names(data1), names(data2)), sep=)) tmp2 - as.factor(paste(data1$, setdiff(names(data2), names(data1)), sep=)) Of course, if it were as simple as only a few columns, I could do all of this by hand, but in my original data frames, I have 60 different columns that need to be created and filled with NA values for both data1 and data2. Eventually, the point of this exercise is so that I can rbind(data1, data2) and create a SQL table out of the merged dataFrames. Unfortunately, I can't rbind() everything until the column names are common across both data1 and data2. Thoughts? Thanks - SR Steven H. Ranney [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating new vectors from other dataFrames
This is exactly what I'm looking for. Each dataFrame will have those columns that are endemic to the other filled with NA. Thanks. Steven H. Ranney On Thu, Aug 8, 2013 at 12:17 PM, arun smartpink...@yahoo.com wrote: HI, Not sure about your expected result. library(plyr) data2New-join_all(lapply(setdiff(names(data1), names(data2)),function(x) {data2[,x]-NA; data2})) data1New-join_all(lapply(setdiff(names(data2), names(data1)),function(x){data1[,x]-NA;data1})) data1New # a b c d e z f g #1 1 5 9 13 17 21 NA NA #2 2 6 10 14 18 22 NA NA #3 3 7 11 15 19 23 NA NA #4 4 8 12 16 20 24 NA NA A.K. - Original Message - From: Steven Ranney steven.ran...@gmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Thursday, August 8, 2013 2:01 PM Subject: [R] Creating new vectors from other dataFrames I have two data frames data1 - as.data.frame(matrix(data=c(1:4,5:8,9:12,13:24), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,d,e,z data2 - as.data.frame(matrix(data=c(1:4,5:8,9:12,37:48), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,f,g,z that have some common column names. Comparing the names of the columns within each data frame to the other setdiff(names(data1), names(data2)) setdiff(names(data2), names(data1)) provides which columns are different. For each column that appears in data1 that DOES NOT appear in data2, I need to create those columns and fill them with NA values. The same is true for the reverse. So, I can create a vector of new column names that need to be filled with NA values, but here is where I'm stuck. I don't know how to get the names from inside the vector into the respective dataFrame. tmp1 - as.factor(paste(data2$, setdiff(names(data1), names(data2)), sep=)) tmp2 - as.factor(paste(data1$, setdiff(names(data2), names(data1)), sep=)) Of course, if it were as simple as only a few columns, I could do all of this by hand, but in my original data frames, I have 60 different columns that need to be created and filled with NA values for both data1 and data2. Eventually, the point of this exercise is so that I can rbind(data1, data2) and create a SQL table out of the merged dataFrames. Unfortunately, I can't rbind() everything until the column names are common across both data1 and data2. Thoughts? Thanks - SR Steven H. Ranney [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Method dispatch in S4
On 08/04/2013 02:13 AM, Simon Zehnder wrote: So, I found a solution: First in the initialize method of class C coerce the C object into a B object. Then call the next method in the list with the B class object. Now, in the initialize method of class B the object is a B object and the respective generateSpec method is called. Then, in the initialize method of C the returned object from callNextMethod has to be written to the C class object in .Object. See the code below. setMethod(initialize, C, function(.Object, value) {.Object@c - value; object - as(.Object, B); object - callNextMethod(object, value); as(.Object, B) - object; .Object - generateSpec(.Object); return(.Object)}) This setting works. I do not know though, if this setting is the usual way such things are done in R OOP. Maybe the whole class design is disadvantageous. If anyone detects a mistaken design, I am very thankful to learn. Hi Simon -- your 'simple' example is pretty complicated, and I didn't really follow it in detail! The code is not formatted for easy reading (e.g., lines spanning no more than 80 columns) and some of it (e.g., generateSpec) might not be necessary to describe the problem you're having. A good strategy is to ensure that 'new' called with no arguments works (there are other solutions, but following this rule has helped me to keep my classes and methods simple). This is not the case for new(A) new(C) The reason for this strategy has to do with the way inheritance is implemented, in particular the coercion from derived to super class. Usually it is better to provide default values for arguments to initialize, and to specify arguments after a '...'. This means that your initialize methods will respects the contract set out in ?initialize, in particular the handling of unnamed arguments: ...: data to include in the new object. Named arguments correspond to slots in the class definition. Unnamed arguments must be objects from classes that this class extends. I might have written initialize,A-method as setMethod(initialize, A, function(.Object, ..., value=numeric()){ .Object - callNextMethod(.Object, ..., a=value) generateSpec(.Object) }) Likely in a subsequent iteration I would have ended up with (using the convention that function names preceded by '.' are not exported) .A - setClass(A, representation(a = numeric, specA = numeric)) .generateSpecA - function(a) { 1 / a } A - function(a=numeric(), ...) { specA - .generateSpecA(a) .A(..., a=a, specA=specA) } setMethod(generateSpec, A, function(object) { .generateSpecA(object@a) }) ensuring that A() returns a valid object and avoiding the definition of an initialize method entirely. Martin Best Simon On Aug 3, 2013, at 9:43 PM, Simon Zehnder simon.zehn...@googlemail.com wrote: setMethod(initialize, C, function(.Object, value) {.Object@c - value; .Object - callNextMethod(.Object, value); .Object - generateSpec(.Object); return(.Object)}) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help 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 with standard errors of lm model
On Aug 8, 2013, at 10:54 AM, Bert Gunter wrote: Perhaps ?vcov is what you are looking for. -- Bert On Thu, Aug 8, 2013 at 10:37 AM, iza.ch1 iza@op.pl wrote: Hi Can someone give me a hint on how to create a matrix with standard errors from lm model? I have already managed to get the matrix with coefficients: coef-as.data.frame(sapply(seq_len(ncol(es.w)),function( i) {x1- summary(lm(es.w[,i]~es.median[,i]));x1$coef[,1]})) but I can't get the one like this for standard errors. I do regression for each column. It's a bit of a feature that coef(summary(lm(...))) returns a 4 column matrix of coefficients, standard errors, t-ratios and p-values while coef(lm(...)) just returns the estimated coefficients. ?lm ?summary.lm The third column of the coef(summary()) result should be == diag(vcov(lm(...)). -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] matrix with standard errors of lm model
Not quite, David. ... (see inline) On Thu, Aug 8, 2013 at 1:56 PM, David Winsemius dwinsem...@comcast.net wrote: On Aug 8, 2013, at 10:54 AM, Bert Gunter wrote: Perhaps ?vcov is what you are looking for. -- Bert On Thu, Aug 8, 2013 at 10:37 AM, iza.ch1 iza@op.pl wrote: Hi Can someone give me a hint on how to create a matrix with standard errors from lm model? I have already managed to get the matrix with coefficients: coef-as.data.frame(sapply(seq_len(ncol(es.w)),function( i) {x1- summary(lm(es.w[,i]~es.median[,i]));x1$coef[,1]})) but I can't get the one like this for standard errors. I do regression for each column. It's a bit of a feature that coef(summary(lm(...))) returns a 4 column matrix of coefficients, standard errors, t-ratios and p-values while coef(lm(...)) just returns the estimated coefficients. ?lm ?summary.lm The third column of the coef(summary()) result should be == diag(vcov(lm(...)). No, it's == sqrt(diag(vcov(lm(...)) -- Bert -- David Winsemius Alameda, CA, USA -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.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] Mac Os X 10.6.8, R and edit/vi
This is a question for R-sig-mac. However, try edit(file=file.choose()) Also, before your edit() command, try getwd() Is the file in that directory?? -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 8/8/13 7:51 AM, David Epstein david.epst...@warwick.ac.uk wrote: I tried using various versions of the 'edit' command. Here is an account of how this failed. I hope I have included all relevant information. I haven't used R for a couple of years. Before restarting with R, I downloaded the latest version I could find in its binary version, and installed it without any problems. Mac Os X Finder command About R responds with R 3.0.1 GUI 1.61 Snow Leopard build (6492) From inside R version platform x86_64-apple-darwin10.8.0 arch x86_64 os darwin10.8.0#(However, my os is in fact 10.6.8) system x86_64, darwin10.8.0 status major 3 minor 0.1 year 2013 month 05 day16 svn rev62743 language R version.string R version 3.0.1 (2013-05-16) nickname Good Sport edit(file='2.9.R') Error in file(con, r) : cannot open the connection In addition: Warning message: In file(con, r) : cannot open file '2.9.R': No such file or directory getOption('editor') [1] vi edit(file='2.9.R',editor='/opt/local/bin/vim') Error in file(con, r) : cannot open the connection In addition: Warning message: In file(con, r) : cannot open file '2.9.R': No such file or directory vi(file='try') Error in file(con, r) : cannot open the connection In addition: Warning message: In file(con, r) : cannot open file 'try': No such file or directory And here is my interaction with tcsh (my default shell) H2:~% echo $VISUAL /opt/local/bin/vim H2:~% echo $EDITOR /opt/local/bin/vim H2:~% which vi vi: aliased to /opt/local/bin/vim H2:~/4Chap2% ls -ld drwxr-xr-x 11 dbae dbae 374 8 Aug 10:54 ./ What am I doing wrong? Thanks for any help. David __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] For loop output
If I understand the request correctly, here is an easy to follow example: (I'm using the first four letters as surrogates for the file names) (and assuming we want quotes at both the beginning and the end) tmp - letters[1:4] tmp [1] a b c d foo - paste( ', paste(tmp,collapse=','), ', sep='') cat(foo,'\n') 'a','b','c','d' In this case, cat() does a better job than print() of telling you exactly what you have. I use this sort of thing to construct arguments for the SQL in clause) -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 8/8/13 8:05 AM, Jenny Williams jenny.willi...@kew.org wrote: I am having difficulty storing the output of a for loop I have generated. All I want to do is find all the files that I have, create a string with all of the names in quotes and separated by commas. This is proving more difficult than I initially anticipated. I am sure it is either very simple or the construction of the for loop is not quite right The result gets automatically printed after the loop but I can't seem to save it. I have tried to create the element in advance but the result is the same: NULL individual.proj = Sys.glob(Arabica/proj_current/individual_projections/*.img, dirmark = FALSE) individual.proj [1] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data. tmp$pa.tab_Full_GBM.img [2] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data. tmp$pa.tab_Full_GLM.img [3] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data. tmp$pa.tab_Full_MARS.img [4] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data. tmp$pa.tab_Full_RF.img [5] Arabica/proj_current/individual_projections/proj_current_arabica_pa.data. tmp$pa.tab_RUN10_GBM.img ##generate loop to create string out of the table of projected files. L.ip = length(individual.proj) for (i in 1:L.ip){ individual.proj.i - individual.proj[i] individual.proj.quote = cat(paste('', individual.proj.i, '', ',',sep=)) } Arabica/proj_current/individual_projections/proj_current_arabica_pa.data. tmp$pa.tab_Full_GBM.img,Arabica/proj_current/individual_projections/proj _current ##print output string individual.proj.quote NULL #command to be applied to individual.proj.quote to removed the final comma from the string substr(individual.proj.quote, 1, nchar(individual.proj.quote)-1) Any help or pointers would be greatly appreciated, no amount of extensive google searches have been fruitful so far. ** Jenny Williams Spatial Information Scientist, GIS Unit Herbarium, Library, Art Archives Directorate Royal Botanic Gardens, Kew Richmond, TW9 3AB, UK Tel: +44 (0)208 332 5277 email: jenny.willi...@kew.orgmailto:jenny.willi...@kew.org ** Film: The Forgotten Home of Coffee - Beyond the Gardenshttp://www.youtube.com/watch?v=-uDtytKMKpAsns=tw Stories: Coffee Expedition - Ethiopiahttp://storify.com/KewGIS/coffee-expedition-ethiopia Kew in Harapan Rainforest Sumatrahttp://storify.com/KewGIS/kew-in-harapan-rainforest Articles: Seeing the wood for the treeshttp://www.kew.org/ucm/groups/public/documents/document/kppcont_0606 02.pdf How Kew's GIS team and South East Asia botanists are working to help conserve and restore a rainforest in Sumatra. Download a pdf of this article here.http://www.kew.org/ucm/groups/public/documents/document/kppcont_0606 02.pdf The Royal Botanic Gardens, Kew is a non-departmental public body with exempt charitable status, whose principal place of business is at Royal Botanic Gardens, Kew, Richmond, Surrey TW9 3AB, United Kingdom. The information contained in this email and any attachments is intended solely for the addressee(s) and may contain confidential or legally privileged information. If you have received this message in error, please return it immediately and permanently delete it. Do not use, copy or disclose the information contained in this email or in any attachment. Any views expressed in this email do not necessarily reflect the opinions of RBG Kew. Any files attached to this email have been inspected with virus detection software by RBG Kew before transmission, however you should carry out your own virus checks before opening any attachments. RBG Kew accepts no liability for any loss or damage which may be caused by software viruses. [[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
Re: [R] Add column to dataframe based on code in other column
Assuming your data frame of users is named 'users', and using your mapping vectors: users$regions - '' users$regions[ users$State_Code) %in% NorthEast ] - 'NorthEast' and repeat for the other regions Or, if you put your mappings in a data frame then it is as simple as merge(yourdataframe, regions) (assuming the data frame of mappings is named 'regions') The regions data frame should have two columns and 50 rows The two columns contain the state codes and their respective regions. How you get that data frame of regions could vary; here's an example using your vectors, but just two of the regions: regions - data.frame( region= c( rep('NorthEast',length(NorthEast)), rep('MidWest,length(MidWest)) ), State_Code=c(NorthEast,Midwest) ) Note that this is untested. For example, I could easily have mismatched parentheses. The whole thing could also be done using match(), without creating the dataframe of regions. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 8/8/13 2:33 AM, Dark i...@software-solutions.nl wrote: Hi all, I have a dataframe of users which contain US-state codes. Now I want to add a column named REGION based on the state code. I have already done a mapping: NorthEast - c(07, 20, 22, 30, 31, 33, 39, 41, 47) MidWest - c(14, 15, 16, 17, 23, 24, 26, 28, 35, 36, 43, 52) South - c(01, 04, 08, 09, 10, 11, 18, 19, 21, 25, 34, 37, 42, 44, 45, 49, 51) West - c(02, 03, 05, 06, 12, 13, 27, 29, 32, 38, 46, 50, 53) Other - c(40, 48, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 94, 98, 99) So for example: NameState_Code Tom 20 Harry 56 Ben 05 Sally 04 Should become like: So for example: NameState_Code REGION Tom 20 NorthEast Harry 56 Other Ben 05 West Sally 04 South Could anyone help me with a clever statement? -- View this message in context: http://r.789695.n4.nabble.com/Add-column-to-dataframe-based-on-code-in-oth er-column-tp4673335.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] p values for partial correlations
I am not an expert on shrinkage estimators of partial correlations (such as the one in corpcor), but my sense is that it is difficult to provide a good estimate of a p-value. You could try to email the authors of the package and ask them, but this may be more of a statistics rather than R question. Peter On Wed, Aug 7, 2013 at 12:01 PM, Demetrio Luis Guadagnin dlguadag...@gmail.com wrote: Dear: I needed to calculate partial correlations and used the package corpcor for that purpose. The output doesnot provide p values and I was unable to find information or posts on how to get them. Does someone can help me? Thanks. -- Dr. Demetrio Luis Guadagnin Conservação e Manejo de Vida Silvestre Universidade Federal do Rio Grande do Sul Departamento de Ecologia Av. Bento Gonçalves 9500 Setor 4, Prédio 43422, Sala 105 Caixa Postal 15007 - 91501-970 Porto Alegre RS Fone: (51) 3308 6774 Fax: (51) 3308 7626 dlguadag...@gmail.com Skype: demetriolguadagnin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting only multiple occurrences
A lot of helpful solutions that pretty much all work. Thanks, everyone! _ Kevin Parent, Ph.D Korea Maritime University From: Rolf Turner rolf.tur...@xtra.co.nz To: Jim Lemon j...@bitwrit.com.au ject.org Sent: Thursday, August 8, 2013 6:26 PM Subject: Re: [R] Extracting only multiple occurrences On 08/08/13 20:27, Jim Lemon wrote: On 08/08/2013 04:23 PM, Kevin Parent wrote: Well that almost works, and I didn't know about duplicated() so thanks for that. However, it only gives me the duplicated values. I need the original ones too. So the result I want is: [g,g,m,m,s,s,t,t,u,u,u,v,v,x,x,y,y,y]. What duplicated() gives me is [g,m,s,t,u,u,v,x,y,y] Hi Kevin, How about: x[x %in% duplicated(x)] Uh, I think you mean x[x %in% x[duplicated(x)]] Another idear: tx - table(x) tx - tx[tx1] rep(names(tx),tx) Well, that's three lines as opposed to one, so not as good. But it perhaps demonstrates a useful tool to add to one's kit. cheers, Rolf [[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] t.test error
Hi I receive a very strange error message after trying to do t-test. When I write the code t.test(x) I get an error message: error in t.test(x) : function sqr not found I don't understand this problem. Can someone help me how to do it right? Thanks 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.
Re: [R] t.test error
On Aug 8, 2013, at 6:09 PM, iza.ch1 wrote: Hi I receive a very strange error message after trying to do t-test. When I write the code t.test(x) I get an error message: error in t.test(x) : function sqr not found I don't understand this problem. Can someone help me how to do it right? Not unless you provide the code. I suspect you have written your own version of 't.test' and have overwritten the version that is from the stats package. It's possible to mask R functions. When you type: `t.test` do you see this? t.test function (x, ...) UseMethod(t.test) bytecode: 0x102c124a0 environment: namespace:stats -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating new vectors from other dataFrames
HI, Not sure about your expected result. library(plyr) data2New-join_all(lapply(setdiff(names(data1), names(data2)),function(x) {data2[,x]-NA; data2})) data1New-join_all(lapply(setdiff(names(data2), names(data1)),function(x){data1[,x]-NA;data1})) data1New # a b c d e z f g #1 1 5 9 13 17 21 NA NA #2 2 6 10 14 18 22 NA NA #3 3 7 11 15 19 23 NA NA #4 4 8 12 16 20 24 NA NA A.K. - Original Message - From: Steven Ranney steven.ran...@gmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Thursday, August 8, 2013 2:01 PM Subject: [R] Creating new vectors from other dataFrames I have two data frames data1 - as.data.frame(matrix(data=c(1:4,5:8,9:12,13:24), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,d,e,z data2 - as.data.frame(matrix(data=c(1:4,5:8,9:12,37:48), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,f,g,z that have some common column names. Comparing the names of the columns within each data frame to the other setdiff(names(data1), names(data2)) setdiff(names(data2), names(data1)) provides which columns are different. For each column that appears in data1 that DOES NOT appear in data2, I need to create those columns and fill them with NA values. The same is true for the reverse. So, I can create a vector of new column names that need to be filled with NA values, but here is where I'm stuck. I don't know how to get the names from inside the vector into the respective dataFrame. tmp1 - as.factor(paste(data2$, setdiff(names(data1), names(data2)), sep=)) tmp2 - as.factor(paste(data1$, setdiff(names(data2), names(data1)), sep=)) Of course, if it were as simple as only a few columns, I could do all of this by hand, but in my original data frames, I have 60 different columns that need to be created and filled with NA values for both data1 and data2. Eventually, the point of this exercise is so that I can rbind(data1, data2) and create a SQL table out of the merged dataFrames. Unfortunately, I can't rbind() everything until the column names are common across both data1 and data2. Thoughts? Thanks - SR Steven H. Ranney [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating new vectors from other dataFrames
Also, a more compact solution would be:library(plyr) #Creating a different dataframe as data2 columns were having almost the same as data1 set.seed(24) data3- as.data.frame(matrix(sample(1:40,6*4,replace=TRUE),ncol=6)) colnames(data3)- colnames(data2) join(data3,data1) #Joining by: a, b, c, z # a b c f g z d e #1 12 27 33 27 8 4 NA NA #2 9 37 11 27 2 23 NA NA #3 29 12 25 13 21 30 NA NA #4 21 31 15 37 6 6 NA NA join(data1,data3) #Joining by: a, b, c, z # a b c d e z f g #1 1 5 9 13 17 21 NA NA #2 2 6 10 14 18 22 NA NA #3 3 7 11 15 19 23 NA NA #4 4 8 12 16 20 24 NA NA A.K. A.K. From: Steven Ranney steven.ran...@gmail.com To: arun smartpink...@yahoo.com Cc: R help r-help@r-project.org Sent: Thursday, August 8, 2013 2:21 PM Subject: Re: [R] Creating new vectors from other dataFrames This is exactly what I'm looking for. Each dataFrame will have those columns that are endemic to the other filled with NA. Thanks. Steven H. Ranney On Thu, Aug 8, 2013 at 12:17 PM, arun smartpink...@yahoo.com wrote: HI, Not sure about your expected result. library(plyr) data2New-join_all(lapply(setdiff(names(data1), names(data2)),function(x) {data2[,x]-NA; data2})) data1New-join_all(lapply(setdiff(names(data2), names(data1)),function(x){data1[,x]-NA;data1})) data1New # a b c d e z f g #1 1 5 9 13 17 21 NA NA #2 2 6 10 14 18 22 NA NA #3 3 7 11 15 19 23 NA NA #4 4 8 12 16 20 24 NA NA A.K. - Original Message - From: Steven Ranney steven.ran...@gmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Thursday, August 8, 2013 2:01 PM Subject: [R] Creating new vectors from other dataFrames I have two data frames data1 - as.data.frame(matrix(data=c(1:4,5:8,9:12,13:24), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,d,e,z data2 - as.data.frame(matrix(data=c(1:4,5:8,9:12,37:48), nrow=4, ncol=6, byrow=F, dimnames=list(c(1:4),c(a,b,c,f,g,z that have some common column names. Comparing the names of the columns within each data frame to the other setdiff(names(data1), names(data2)) setdiff(names(data2), names(data1)) provides which columns are different. For each column that appears in data1 that DOES NOT appear in data2, I need to create those columns and fill them with NA values. The same is true for the reverse. So, I can create a vector of new column names that need to be filled with NA values, but here is where I'm stuck. I don't know how to get the names from inside the vector into the respective dataFrame. tmp1 - as.factor(paste(data2$, setdiff(names(data1), names(data2)), sep=)) tmp2 - as.factor(paste(data1$, setdiff(names(data2), names(data1)), sep=)) Of course, if it were as simple as only a few columns, I could do all of this by hand, but in my original data frames, I have 60 different columns that need to be created and filled with NA values for both data1 and data2. Eventually, the point of this exercise is so that I can rbind(data1, data2) and create a SQL table out of the merged dataFrames. Unfortunately, I can't rbind() everything until the column names are common across both data1 and data2. Thoughts? Thanks - SR Steven H. Ranney [[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.