Re: [R] Minimum cutsets
I don't know the specifics of this package, but generally the C code is called internally by R: it, however, requires compilation before R can talk to it. You wont need to learn C though. Look at the link David suggested for a precompiled version, but it may be older than the development-version functionality discussed up thread. If you don't have access to a pre-compiled binary (the author may be willing to send one to you), it is possible to compile one yourself with not too much trouble. Theoretically, you won't have to touch the C code other than to pass it through the package building process, which is rather well documented. If the source is too development-y, this might be a little tricky, so I'd ask the author for a binary first. Michael On Thu, Oct 13, 2011 at 10:24 PM, David Winsemius dwinsem...@comcast.net wrote: On Oct 13, 2011, at 7:47 PM, malhomidi wrote: Hi again, I've looked at the links above No links visible to the rest of us non-Nabble users. When will nabble-users ever learn to include context? (or to post with new Subject: lines when the topic changes?) and I see the development version of the igraph library. I see the src folder implemented in C. Are these source codes available in R or I just would have to use the C code? I'm guessing from context that you mean to ask: are their binary versions of the package? (There are always source versions of packages, especially so for development versions.) Here is one CRAN mirror's igraph (non-development) page. Look for something like: Package source: igraph_0.5.5-2.tar.gz http://lib.stat.cmu.edu/R/CRAN/web/packages/igraph/index.html The reason is that I just started learning R and I really want to stay away from C and C++. Regards, Mohammed Alhomidi -- View this message in context: http://r.789695.n4.nabble.com/Minimum-cutsets-tp885346p3903287.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Length of data.frame column
I have a related question...I have a data frame similar with 74 rows that I created with header=TRUE, but when I try to coerce one of the data frame columns into a vector, it shows up as having length 1, even though when I print it, it shows 74 elements: VAL - c(DailyDiary[1]) VAL [1] 3 3 3 2 3 3 4 4 3 3 2 1 1 4 3 3 2 3 3 2 2 3 3 3 1 2 2 [28] 1 1 3 3 3 2 4 3 2 3 4 3 3 3 3 4 3 2 3 2 3 1 2 1 2 3 1 [55] 0 3 2 4 3 1 2 3 1 1 1 4 3 2 1 2 3 3 3 2 length(VAL) [1] 1 On the other hand, I can easily coerce the row names to a vector of length 74 partf - row.names(DailyDiary) length(partf) [1] 74 What I would like to do is make VAL into a vector with length 74 instead of length 1 so I can sort it using use partf as factors. I tried as.vector, but it doesn't let you specify the length. Any ideas? Thanks, and sorry if I'm being unclear or stupid, I'm a newbie :) Logan -- View this message in context: http://r.789695.n4.nabble.com/Length-of-data-frame-column-tp864585p3903892.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about GAMs
pre-predict(ozonea,groupA,type=terms,terms=NULL,newdata.guaranteed=FALSE,na.action=na.pass) yeah! but I don't know how to only show the value of s(ratio,bs=cr) -- View this message in context: http://r.789695.n4.nabble.com/Question-about-GAMs-tp3900848p3903753.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] Drop ALL Levels of a Data Frame Object
I'm using the package 'gdata' and 'drop.levels' in the following code as a
simplified example of what I want to do, pinpointing the issue of isolating
variable in a data frame:
/dfNamesAndHeight - data.frame(Name = c('Bill','Bob','Bo'), Height =
c(73,68,83))
name_Bob - drop.levels(dfNamesAndHeight[2,1])
cat(name_Bob, is a good guy.)
/
The output is 1 is a good guy. Print instead of cat returns Error in
print.default(xx, quote = quote, ...) : invalid 'quote' argument. When you
just enter 'name_Bob' the output is:
[1] Bob
Levels: Bob
Is there any way to isolate 'Bob' without the 'Levels: Bob'?
--
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Re: [R] Length of data.frame column
You haven't provided a reproducible example. You haven't even provided a subset of your data, or the commands you used to read it in. I might guess that VAL - DailyDiary[[1]] might be what you wanted, and the str function might help you understand why. Also, the c function does not coerce object types... it embeds the arguments into a new vector. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. jawbonemurphy jdurlands...@gmail.com wrote: I have a related question...I have a data frame similar with 74 rows that I created with header=TRUE, but when I try to coerce one of the data frame columns into a vector, it shows up as having length 1, even though when I print it, it shows 74 elements: VAL - c(DailyDiary[1]) VAL [1] 3 3 3 2 3 3 4 4 3 3 2 1 1 4 3 3 2 3 3 2 2 3 3 3 1 2 2 [28] 1 1 3 3 3 2 4 3 2 3 4 3 3 3 3 4 3 2 3 2 3 1 2 1 2 3 1 [55] 0 3 2 4 3 1 2 3 1 1 1 4 3 2 1 2 3 3 3 2 length(VAL) [1] 1 On the other hand, I can easily coerce the row names to a vector of length 74 partf - row.names(DailyDiary) length(partf) [1] 74 What I would like to do is make VAL into a vector with length 74 instead of length 1 so I can sort it using use partf as factors. I tried as.vector, but it doesn't let you specify the length. Any ideas? Thanks, and sorry if I'm being unclear or stupid, I'm a newbie :) Logan -- View this message in context: http://r.789695.n4.nabble.com/Length-of-data-frame-column-tp864585p3903892.html Sent from the R help mailing list archive at Nabble.com. _ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about GAMs
predict.gam is returning a matrix with a named column for each term. Select the appropriate column. Example below library(mgcv) n-200;sig - 2 dat - gamSim(1,n=n,scale=sig) b - gam(y~s(x0)+s(I(x1^2))+s(x2)+offset(x3),data=dat) newd - data.frame(x0=(0:30)/30,x1=(0:30)/30,x2=(0:30)/30,x3=(0:30)/30) pred - predict.gam(b,newd,type=terms) pred[,3] ## extracts s(x2) pred[,s(x2)] ## same On 14/10/11 04:54, pigpigmeow wrote: pre-predict(ozonea,groupA,type=terms,terms=NULL,newdata.guaranteed=FALSE,na.action=na.pass) yeah! but I don't know how to only show the value of s(ratio,bs=cr) -- View this message in context: http://r.789695.n4.nabble.com/Question-about-GAMs-tp3900848p3903753.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. -- Simon Wood, Mathematical Science, University of Bath BA2 7AY UK +44 (0)1225 386603 http://people.bath.ac.uk/sw283 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] binding all elements of list (character vectors) to a matrix as rows
hello again, i used the command rbind(Mymatrix, t(as.data.frame(Z))) and it works! thanks very much for your replys! marion 2011/10/11 David Winsemius dwinsem...@comcast.net On Oct 11, 2011, at 2:47 AM, Marion Wenty wrote: dear r-users, i have got a problem which i am trying to solve: i have got the following commands: Mymatrix - matrix(1:9,ncol=3) Z - list (V1 =c(a,,),V2=c(b,,**),V3=c(c,,),V4=c(**d,,)) rbind(Mymatrix, t(as.data.frame(Z))) The next method could be used if you had more lists: do.call(rbind, list(Mymatrix, t(as.data.frame(Z Mymatrix - rbind(Mymatrix,Z[[1]],Z[[2]],**Z[[3]],Z[[4]]) now this is working, but i would like to substitute Z[[1]],Z[[2]],Z[[3]],Z[[4]] for a command with which i could also use another list with a different number of elements, e.g. 5 or 6 elements. -- David Winsemius [[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] Drop ALL Levels of a Data Frame Object
Why aren't you using as.character instead of drop.levels?
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
swonder03 ramey.ste...@gmail.com wrote:
I'm using the package 'gdata' and 'drop.levels' in the following code as a
simplified example of what I want to do, pinpointing the issue of isolating
variable in a data frame:
/dfNamesAndHeight - data.frame(Name = c('Bill','Bob','Bo'), Height =
c(73,68,83))
name_Bob - drop.levels(dfNamesAndHeight[2,1])
cat(name_Bob, is a good guy.)
/
The output is 1 is a good guy. Print instead of cat returns Error in
print.default(xx, quote = quote, ...) : invalid 'quote' argument. When you
just enter 'name_Bob' the output is:
[1] Bob
Levels: Bob
Is there any way to isolate 'Bob' without the 'Levels: Bob'?
--
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http://r.789695.n4.nabble.com/Drop-ALL-Levels-of-a-Data-Frame-Object-tp3903788p3903788.html
Sent from the R help mailing list archive at Nabble.com.
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[R] Analysis
i'm actually new at R, but to me its going to be big time, i want to do an analysis for the attached data, like define variables and then analyse get means, variances, histogram, and other important attributes including cross tabs Regards, Peter Countycode,Provincecode,Interviewercode,Locationoftowncentre:Latitude,Locationoftowncentre:Longitude,Locationoftowncentre:Altitude,Locationoftowncentre:Accuracy,Dateofstudy,Durationofbusinessoperation,Levelofeducation,Equitybankaccount,Anotherbankaccount,Visittobranch,Monthlyaveragedeposit,Distancetobranch,Preferredbranchcode,Meanoftransport,Durationtobranch,DistancetoChiefcamp,Distancetoapcamporpolicestation,Walkingdurationtochiefaporpolicestation,Drivingdurationchiefsaporpolicestation,Outletpowersource,Timesforpoweruse,Lackedelectricity,Powersourcefunding,Whentotakethefunding,Cellphonenetworks,Mostreliablenetwork,Monthlyincome,Agencyadverteffectiveness,Oralorverbalinformation,CTOVCview,viewsondistribution,Viewtowardpatneringwithbank,Equityagentoroutlet,Districtname,Towncentre 1,4,7,-1.5187021000,37.2623512000,0E-10,2804.00,Thu Oct 06 00:00:00 UTC 2011,5,2,1,,3,7,7,3,1,45,6,5,4,73,1,30,3,3,4,1 2,1 2,1,1,1,They deserve,New business,Banking and savings,1,Machakos,Kathiani 29,8,7,null,null,null,null,Fri Oct 07 00:00:00 UTC 2011,3,5,1,,9,5,17,49,3,30,2,5,10,15,1,30,3,1,2,1 2 3 4,1,2,2,2,Discrimination, lack information, lack exposure,Good idea to transfer through equity agents,Would love to do the transfers,1,Nandi east,Himaki 29,8,7,null,null,null,null,Mon Oct 10 00:00:00 UTC 2011,6,3,1,,25,35000,10,49,3,30,3,300,30,5,1,30,2,1,2,1 3 4,1 4,3,2,1,Necessary because the people are poor,Will promote business attract more customers,Excellent idea take service closer to people,1,Central nandi,Baraton 27,8,7,null,null,null,null,Mon Oct 10 00:00:00 UTC 2011,7,5,1,,5,1,1,30,1,5,1,300,20,5,1,30,2,1,2,1 2,1,3,2,1,They need financial literacy,Equity will do well in that,Would love that since more income,1,Wareng,Eldoret town 41,5,8,-0.0963076000,34.2857898000,0E-10,3837.00,Thu Oct 06 00:00:00 UTC 2011,20,3,1,,30,10,1,75,1,3,1,1,30,1,1,30,4,2,null,1 2 3 4,1,5,3,null,It is good,There is no problem,It is okay,1,Bondo,Bondo 41,5,8,-0.0724487000,34.2750895000,0E-10,5000.00,Thu Oct 06 00:00:00 UTC 2011,23,3,1,,5,2,20,75,2 3,30,3,3,60,60,5,1,30,1,1,1 2,1,5,3,1,Na,It will give us more customers and save time for them,It will give us more revenue,1,41,Got' Agulu mkt 41,5,8,-0.2684695000,34.348734,0E-10,4019.00,Thu Oct 06 00:00:00 UTC 2011,16,2,1,2,15,15,40,75,2 3,60,1,15,15,30,1,30,30,1,1,1 2 3 4 5,2,5,3,null,It helps them,It will increase our customer base,It is good for marketing,1,Rarieda,Ragengni 42,5,8,-0.0890969000,34.6036899000,0E-10,4064.00,Fri Oct 07 00:00:00 UTC 2011,5,3,1,2,8,20,23,29,3 4,25,1,3,5,5,1,1,4,2,null,1 2 3 4,1 2,5,2,1,N/a,It will bring services closer to the people,It is a source of income for agents,1,Kisumu west,Holo 42,5,8,-0.0715916000,34.6584807000,0E-10,2811.00,Fri Oct 07 00:00:00 UTC 2011,7,3,1,,15,5,12,29,3,30,12,30,120,30,1,1,4,1,1,1 2 3 4,1 2 3,4,2,1,No comment,It will promote local business.,It will make agents have more customers,1,Kisumu west,Kisian 1,4,1,-0.3277709800,36.9052373600,1953.1999511719,7.00,Thu Oct 06 00:00:00 UTC 2011,null,4,1,,30,50,15,11,4,15,5,5,2,2,1,30,null,2,4,1 2 3 4,1,null,3,1,None,None,Okay,1,Nyeri,Mweiga null,2,1,-0.3527577000,36.9052095000,0E-10,4109.00,Fri Oct 07 00:00:00 UTC 2011,null,3,1,1,30,50,5,41,1,10,3,2,2,30,1 2,30,4,2,4,1 2 4,1,null,3,1,null,null,Okay,1,Isiolo,Isiolo 1,2,1,-0.3527577000,36.9052095000,0E-10,4109.00,Fri Oct 07 00:00:00 UTC 2011,null,3,1,1,24,2,3,41,1,3,2,2,10,15,1,30,4,1,1,1 2 3 4,1,null,2,1,null,null,Okay,1,Isiolo,Isiolo 3,6,2,-3.953335,39.7436864000,0E-10,1873.00,Fri Oct 07 00:00:00 UTC 2011,4,3,1,2,40,120,2,119,1 2,10,2,1,15,4,1,24,2,1,1,1 2 3 4,1,2,3,1,A great idea,Will ensure trust is build and work diligently,A continuity since they already support children,1,Kilifi,Mtwapa 3,6,2,-3.6339741000,39.8550807000,0E-10,1625.00,Thu Oct 06 00:00:00 UTC 2011,4,3,1,6,3,900,70,106,3,90,5,5,45,10,1,30,3,1,1,1 2,1,5,1,1,Vulnerable and no dependants,It is okay,Nice idea,1,Bomba,Bomba 3,6,2,-3.9485545000,39.745171,0E-10,75.00,Fri Oct 07 00:00:00 UTC 2011,3,4,1,1,90,180,1,119,1,10,5,2,20,5,1,30,3,1,3,1 2 3 4,1,3,3,1,No view,A good idea reach to people,Give thanks,2,Mtwapa,Mtwapa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ctest package
Please send me the r package : ctest to the following e-mail mak.sta...@gmail.com Thank you. Best regards MAK [[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] Length of data.frame column
Yup, DailyDiary[[1]] did it, thanks! On Thu, Oct 13, 2011 at 11:43 PM, Jeff Newmiller [via R] ml-node+s789695n3903955...@n4.nabble.com wrote: You haven't provided a reproducible example. You haven't even provided a subset of your data, or the commands you used to read it in. I might guess that VAL - DailyDiary[[1]] might be what you wanted, and the str function might help you understand why. Also, the c function does not coerce object types... it embeds the arguments into a new vector. --- Jeff Newmiller The . . Go Live... DCN:[hidden email] http://user/SendEmail.jtp?type=nodenode=3903955i=0 Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. jawbonemurphy [hidden email]http://user/SendEmail.jtp?type=nodenode=3903955i=1 wrote: I have a related question...I have a data frame similar with 74 rows that I created with header=TRUE, but when I try to coerce one of the data frame columns into a vector, it shows up as having length 1, even though when I print it, it shows 74 elements: VAL - c(DailyDiary[1]) VAL [1] 3 3 3 2 3 3 4 4 3 3 2 1 1 4 3 3 2 3 3 2 2 3 3 3 1 2 2 [28] 1 1 3 3 3 2 4 3 2 3 4 3 3 3 3 4 3 2 3 2 3 1 2 1 2 3 1 [55] 0 3 2 4 3 1 2 3 1 1 1 4 3 2 1 2 3 3 3 2 length(VAL) [1] 1 On the other hand, I can easily coerce the row names to a vector of length 74 partf - row.names(DailyDiary) length(partf) [1] 74 What I would like to do is make VAL into a vector with length 74 instead of length 1 so I can sort it using use partf as factors. I tried as.vector, but it doesn't let you specify the length. Any ideas? Thanks, and sorry if I'm being unclear or stupid, I'm a newbie :) Logan -- View this message in context: http://r.789695.n4.nabble.com/Length-of-data-frame-column-tp864585p3903892.html Sent from the R help mailing list archive at Nabble.com. _ [hidden email] http://user/SendEmail.jtp?type=nodenode=3903955i=2mailing list https://stat.ethz.ch/mailman/listinfo/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]] __ [hidden email] http://user/SendEmail.jtp?type=nodenode=3903955i=3mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Length-of-data-frame-column-tp864585p3903955.html To unsubscribe from Length of data.frame column, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=864585code=amR1cmxhbmRzdGVyQGdtYWlsLmNvbXw4NjQ1ODV8MTg1MjIxNzE3OQ==. -- View this message in context: http://r.789695.n4.nabble.com/Length-of-data-frame-column-tp864585p3903963.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ctest package
On Fri, 14 Oct 2011, Khalek wrote: Please send me the r package : ctest to the following e-mail mak.sta...@gmail.com The package is in the CRAN archives, check out http://CRAN.R-project.org/package=ctest Do note however the time stamps. You may also want to read FAQ 5.1.1 http://CRAN.R-project.org/doc/FAQ/R-FAQ.html#Add_002don-packages-in-R Thank you. Best regards MAK [[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] Analysis
On 10/14/2011 06:29 AM, Peter Kaiga wrote: i'm actually new at R, but to me its going to be big time, i want to do an analysis for the attached data, like define variables and then analyse get means, variances, histogram, and other important attributes including cross tabs Hi, Thanks for sharing your enthousiasm for R, this is definitely the right mailing list for that ;). But do you have a question? Before sending one, please do check out the posting guide: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. regards, Paul Regards, Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Paul Hiemstra, Ph.D. Global Climate Division Royal Netherlands Meteorological Institute (KNMI) Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39 P.O. Box 201 | 3730 AE | De Bilt tel: +31 30 2206 494 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 [[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] is there an option to turn off scientific notation in write.csv
On 10/14/2011 05:25 AM, Chris Conner wrote: Dear Help-Rs, I'm working with a file that contains large numbers and I need to export them as is. for example take: x - c(27104010002005,27104020001805,27104090001810,90050013000140,90050013000120) y - c(1:5) df - data.frame(cbind(x,y)) When I then try a simple: write.csv(df,file=df.csv) I get: x y 1 2.7104E+13 1 2 2.7104E+13 2 3 2.71041E+13 3 4 9.005E+13 4 5 9.005E+13 5 Then I tried: options(scipen=999) df$x - as.character(df$x) Hi, You can use format in this case: df$x -format(df$x, scientific = FALSE) write.csv(df,file=df.csv) regards, Paul write.csv(df,file=df.csv) and I still get: x y 1 2.7104E+13 1 2 2.7104E+13 2 3 2.71041E+13 3 4 9.005E+13 4 5 9.005E+13 5 How can I have R write the file so it looks like this: x y 1 27104010002005 2 27104020001805 3 27104090001810 4 90050013000140 5 90050013000120 [[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. -- Paul Hiemstra, Ph.D. Global Climate Division Royal Netherlands Meteorological Institute (KNMI) Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39 P.O. Box 201 | 3730 AE | De Bilt tel: +31 30 2206 494 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 [[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] Wilcoxon and the use of simulation
Dear forum users, It's 3:35am and I am swamped with statistics homework lol I'm terrible with R and this time I have no idea what the prof wants. Here is the question: Consider the (two-‐sample) Wilcoxon rank statistic T = Σrank(Xi). For n1=106 and n2=192, determine by simulation the α=.05 critical point for testing H0: θ=0, H1:θ0. We can do this as follows: For m=1 (no wimpy m=200 or 500 as in the book), draw m=1 subsets of size 106 from the integers 1:298 using repeatedly the command xdat = sample(1:298, size=106). For each such subset, the value of the Wilcoxon is sum(xdat). Be sure to answer the following question: why is it unnecessary to calculate ranks? Any help would be greatly appreciately at this time. -- View this message in context: http://r.789695.n4.nabble.com/Wilcoxon-and-the-use-of-simulation-tp3904036p3904036.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] Length of data.frame column
Hi I have a related question...I have a data frame similar with 74 rows that I created with header=TRUE, but when I try to coerce one of the data frame columns into a vector, it shows up as having length 1, even though when I print it, it shows 74 elements: VAL - c(DailyDiary[1]) VAL [1] 3 3 3 2 3 3 4 4 3 3 2 1 1 4 3 3 2 3 3 2 2 3 3 3 1 2 2 [28] 1 1 3 3 3 2 4 3 2 3 4 3 3 3 3 4 3 2 3 2 3 1 2 1 2 3 1 [55] 0 3 2 4 3 1 2 3 1 1 1 4 3 2 1 2 3 3 3 2 length(VAL) [1] 1 your VAL is still data frame. See ?[ and link to data.frame method. Use str(object) to see the structure of your object not just printing it to console as print is a function which has also different methods for different objects and does not inform you about nature of your object. You shall profit from reading R-intro which I believe you have overlooked. It shall be installed in doc/manual directory of your R installation. Regards Petr On the other hand, I can easily coerce the row names to a vector of length 74 partf - row.names(DailyDiary) length(partf) [1] 74 What I would like to do is make VAL into a vector with length 74 instead of length 1 so I can sort it using use partf as factors. I tried as.vector, but it doesn't let you specify the length. Any ideas? Thanks, and sorry if I'm being unclear or stupid, I'm a newbie :) Logan -- View this message in context: http://r.789695.n4.nabble.com/Length-of- data-frame-column-tp864585p3903892.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] verbose source command?
On 11-10-13 10:16 PM, R. Michael Weylandt wrote: source(FILE, print.eval = TRUE) or source(FILE, echo = TRUE) which will also echo the inputs - the students will find this helpful. Duncan Murdoch Hope this helps good work on getting the next round of R enthusiasts up and going! Michael On Thu, Oct 13, 2011 at 8:59 PM, Stephen Daviesstep...@umw.edu wrote: (Apologies for the n00b question.) Hello, I'm teaching R in an introductory programming course and am walking the students through the baby steps. One thing I'd like to be able to do is have them copy the commands they type at the R console into a text file, and then execute the text file to see the results. For instance, if their session looks like this: 3+4 [1] 7 factorial(10) [1] 3628800 I'd like for them to create a text file called myCommands.R with contents: 3+4 factorial(10) and then run it at the R console using source(myCommands.R), and see this output: [1] 7 [1] 3628800 This would help with many things, including my grading their lab work. Unfortunately, if I/they do this using the source() command as it stands, the result is of course no output at all, because nothing is being explicitly printed (with print() or cat() command, for example). My question is: is there a command to do what I'm trying to do here? Is there some kind of verbose source command (or mode) that will run a .R script/program/file and print all the results from it exactly as if those commands had been entered at the console? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] binomial GLM quasi separation
On 13.10.2011 21:46, Ben Bolker wrote: lincolnmiseno77at hotmail.com writes: Hi all, I have run a (glm) analysis where the dependent variable is the gender (family=binomial) and the predictors are percentages. I get a warning saying fitted probabilities numerically 0 or 1 occurred that is indicating that quasi-separation or separation is occurring. This makes sense given that one of these predictors have a very influential effect that is depending on a specific threshold separating these effects, in other words in my analysis one of these variables predicts males about the 80% of times when its values are less or equal to zero and females about the 80% when its values are greater than zero. I have been looking at other posts about this but I haven’t understood how I should act when the separation (or quasi separation) is not a statistical artifact but it is something real. As suggested in http://r.789695.n4.nabble.com/ OT-quasi-separation-in-a-logistic-GLM-td875726.html#a3850331 http://r.789695.n4.nabble.com/ OT-quasi-separation-in-a-logistic-GLM-td875726.html#a3850331 [warning, broke URLs to make gmane happy] (the last post is mine) I tried to use brglm procedure that uses a penalized maximum likelihood but it made no difference. I'm not sure what's going on here, and I don't know why brglm() shouldn't work ... from a squint at your Nabble post (I can't really see the figure very well), I agree that the hcp profile is funky, but I wouldn't immediately conclude that the profile is bad -- in particular, it seems that the x-axis range is -45 to -15, rather than something like (-600,-300) as I would expect from the estimated parameter (ca. -400) and standard error (ca. 60). I would start by setting which=3 (to confine your attention to the hcp parameter) and messing around with the gridsize, stepsize, stdn parameters in profileModel to see what's going on. If that doesn't work you might have to post data, or a subset of data, in order to get any more help ... Or if just the separating hyperplane is to be found (and no tests have to be considered), I'd use an lda rather than logistic regression in such a case. Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Wilcoxon and the use of simulation
On Oct 14, 2011, at 09:39 , shl2a wrote: Dear forum users, It's 3:35am and I am swamped with statistics homework lol I'm terrible with R and this time I have no idea what the prof wants. Here is the question: Consider the (two-‐sample) Wilcoxon rank statistic T = Σrank(Xi). For n1=106 and n2=192, determine by simulation the α=.05 critical point for testing H0: θ=0, H1:θ0. We can do this as follows: For m=1 (no wimpy m=200 or 500 as in the book), draw m=1 subsets of size 106 from the integers 1:298 using repeatedly the command xdat = sample(1:298, size=106). For each such subset, the value of the Wilcoxon is sum(xdat). Be sure to answer the following question: why is it unnecessary to calculate ranks? Any help would be greatly appreciately at this time. Well, he wants you to use your brain, not ours... This is not a list for helping people with their homework. Look through your teaching materials and notes and see if your friendly prof hasn't already provided an example of doing replicated simulations, drawing a histogram of the results, etc. -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ctest package
On 14.10.2011 09:26, Achim Zeileis wrote: On Fri, 14 Oct 2011, Khalek wrote: Please send me the r package : ctest to the following e-mail mak.sta...@gmail.com The package is in the CRAN archives, check out http://CRAN.R-project.org/package=ctest Do note however the time stamps. You may also want to read FAQ 5.1.1 http://CRAN.R-project.org/doc/FAQ/R-FAQ.html#Add_002don-packages-in-R ... and note that there was certainly a reason to remove the package from CRAN. Best, Uwe Ligges Thank you. Best regards MAK [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to plot two surfaces with lattice::wireframe
Hi Carl,
I have no idea what z or f(z) are, but maybe outer will help you:
wireframe(outer(seq(0,5,length.out=50),seq(2,4,length.out=40),function(x,y)sin(x*y)))
cheers.
Am 13.10.2011 23:37, schrieb Carl Witthoft:
Hi all,
I'd like to plot the Real and Imaginary parts of some f(z) as two
different surfaces in wireframe (the row/column axes are the real and
imag axes). I know I can do it by, roughly speaking, something like
plotz - expand.grid(x={range of Re(z)}, y={range of Im(z), groups=1:2)
plotz$func-c(Re(f(z),Im(f(z))
wireframe(func~x*y,data=plotz,groups=groups)
But that seems like a clunky way to go, especially if I happen to have
started out with a nice matrix of the f(z) values.
So, is there some simpler way to write the formula in wireframe? I
envision, for a matrix of complex values zmat, pseudocode:
wireframe(c(Re(zmat),Im(zmat), groups=1:2)
-- and yes, I'm fully aware that without a connection between the
'groups' variable and zmat, this won't work as written.
All suggestions (including read the help file for {some lattice func I
didn't know about} ) greatfully accepted.
Carl
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Re: [R] binomial GLM quasi separation
As you suggested I had a further look at the profile by changing default values of stepsize (I tried to modify the others but apparently there was any change). Here they go the scripts I have used: dati-read.table(simone.txt,header=T,sep=\t,as.is=T) glm.sat-glm(sex~twp+hwp+hcp+hnp,binomial,data=dati) Mensajes de aviso perdidos glm.fit: fitted probabilities numerically 0 or 1 occurred pp-profileModel(glm.sat,quantile=qchisq(0.95,1),objective=ordinaryDeviance,which=4) Preliminary iteration . Done Profiling for parameter hcp ... Done pp1-profileModel(glm.sat,quantile=qchisq(0.95,1),objective=ordinaryDeviance,which=4,stepsize=20) Preliminary iteration . Done Profiling for parameter hcp ... Done pp2-profileModel(glm.sat,quantile=qchisq(0.95,1),objective=ordinaryDeviance,which=4,stepsize=100) Preliminary iteration . Done Profiling for parameter hcp ... Done plot(pp) mtext(Default stepsize,adj=0,cex=2,line=1) dev.new() plot(pp) mtext(Stepsize=20,adj=0,cex=2,line=1) dev.new() plot(pp) mtext(Stepsize=100,adj=0,cex=2,line=1) And these are the plots as they look like: http://r.789695.n4.nabble.com/file/n3904261/plot1.png http://r.789695.n4.nabble.com/file/n3904261/plot2.png http://r.789695.n4.nabble.com/file/n3904261/plot3.png I have tried to understand what is going on but I don't know how to interpret this. It's quite a long time that I am trying to solve this but I have not been able to. Here ( http://r.789695.n4.nabble.com/file/n3904261/simone.txt simone.txt ) I attach a subset of the data I am working with that comprises the variables specified in the above glm model and by the way the funky variable called hcp. Thank you for take your time to help me in this. -- View this message in context: http://r.789695.n4.nabble.com/binomial-GLM-quasi-separation-tp3901687p3904261.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] Generating Data
Dear All I need to generate multivariate NON-NORMAL data in R, which follows a given mean vector and covariance matrix, say multivariate exponential data. How can I do that? Best regards mra __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [Related Topic] need help on read.spss
On 13.10.2011 18:28, Matt Shotwell wrote: Would it be worthwhile to update the read.spss implementation using the more recent discoveries from the PSPP group? If there are important features, I think so. Getting code into foreign is not too trivial. Smaller changes are more likely to be accepted than a re-implementation which would need really careful and extensive tests. Note also this won't resolve the underlying problem I described in my last message. I don't mean to copy their code; Iff the code is reasonable, why not talk to the PSPP people, take the code and adapt it, given it is GPL'ed. but to use the ideas in their code. Is anyone working on this? I wouldn't want the effort to be duplicated. Not that I know, at least. Best, Uwe Ligges On Thu, 2011-10-13 at 16:22 +0200, Uwe Ligges wrote: On 11.10.2011 12:07, Smart Guy wrote: Hi, I have one doubt about one of the parameter of 'read.spss()' from 'foreign' package. Here is the syntax :- read.spss ( file, use.value.labels = TRUE, to.data.frame = FALSE, max.value.labels = Inf, trim.factor.names = FALSE, trim_values = TRUE, reencode = NA, use.missings = to.data.frame ) In above syntax when I pass *'to.data.frame= FALSE*' it gives me missing values from SPSS file (that I try to read using read.spss() ). But when I pass '*to.data.frame = TRUE*' then its not giving me missing values. And need to get missing values. According to read.spss() documentation *to.data.frame : return a data frame?* I am curious to know, if we pass *'to.data.frame = TRUE*' , is it going to cause some issue or effect something? I didn't understand the read.spss() documentation correctly. Please explain. Thanks in Advance An R data.frame cannot represent different kinds of missing values, since R just has NA. Therefore, there are two way to import data: to.data.frame=FALSE will read all the information, but into a format you will likely have to postprocess to make it conveniently usable. to.data.frame=TRUE will import into a data.frame, but that cannot represent all the nuances known from the SPSS representation. Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Fitting a Harmonic Function to Time Series Data
On 14/10/2011 1:00 a.m., ashz wrote: Dear All, I have some time series data where X=month and Y=nutrient concentration (I can have several concentration data for one month). Is there a way to fit for it an Harmonic Function. Is there a package, script,etc which I can use? Thx Possibly there is this functionality in Rob Hyndman's forecast package. See this post by him: http://robjhyndman.com/researchtips/longseasonality/ which has some code which should also be of use. David Scott _ David Scott Department of Statistics The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Legend symbols (line, points) in one column.
Hi David:
Thank you for your answer.
El vie, 14-10-2011 a las 00:32 -0400, David Winsemius escribió:
Legends are built in columns. You need to find a graphics symbol to
put in the points column or you need to find something that the
lines paramater will turn into a dot (and I'm not sure what that might
be.) My suggestion would be to change the line type to dashed and use
- - - for the pch argument.
No, because I want a continous line for the model.
But the data as points, and in the legend to identify each one
as the example shows but in only in one column in the legend!.
This is what I came up with using the panel.lmline rather than
building it outside the plotting function:
If the model is a linear one, it is fine, but the model could
be a complex one that it is not programmed as panel.xxmodel, or else.
Thank you very much for your interest in finding the solution.
x-1:5
y-1*x+rnorm(10)
data1-data.frame(x,y,type=rep(data,length(x)))
require(lattice)
xyplot(y~x,group=type,
type=c(p,l), lty=3, lwd=2, cex=4,
key=list(space=right,text=list(c(Data,Model)),
points=list(pch = c(., -),cex=c(4,1.5)) ),
par.settings=confMisc1,
panel=function(x,y,...){panel.xyplot(x,y,...)
panel.lmline(x,y,...)},
distribute.type=TRUE,
data=data1)
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Re: [R] Minimum cutsets
Hi, Thank you David and Michael for your suggestions and comments. I'll try to get the binary version of the package. I'm just waiting for Gabor to let me know about the development package which works on Windows. Regards, Mohammed -- View this message in context: http://r.789695.n4.nabble.com/Minimum-cutsets-tp885346p3904424.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] binomial GLM quasi separation
On Fri, 2011-10-14 at 02:32 -0700, lincoln wrote: As you suggested I had a further look at the profile by changing default values of stepsize (I tried to modify the others but apparently there was any change). Have you read ?glm, specifically this bit: Details: For the background to warning messages about ‘fitted probabilities numerically 0 or 1 occurred’ for binomial GLMs, see Venables Ripley (2002, pp. 197-8). There, VR say (me paraphrasing) that if there are some large fitted \beta_i the curvature of the log-likelihood at the fitted \beta can be much less than at \beta_i = 0. The Wald approximation underestimates the change in the LL on setting \beta_i = 0. As the absolute value of the fitted \beta_i becomes large (tends to infinity) the t statistic tends to 0. This is the Hauck Donner effect. Whilst I am (so very) far from being an expert here - this does seem to fit the results you showed. Furthermore, did you follow the steps Ioannis Kosmidis took me through with my data in that email thread? I have with your data and everything seems to follow the explanation/situation given by Ioannis. Namely, if you increase the number of iterations and tolerance in the glm() call you get the same fit as with a standard glm() call: ## normal glm() summary(mod) Call: glm(formula = sex ~ twp + hwp + hcp + hnp, family = binomial, data = dat) Deviance Residuals: Min 1Q Median 3Q Max -2.9703 -0.1760 0.3181 0.6061 3.5235 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)1.4362 0.2687 5.345 9.02e-08 *** twp5.5673 1.3602 4.093 4.26e-05 *** hwp -4.2793 2.3781 -1.799 0.0719 . hcp -450.191856.6823 -7.942 1.98e-15 *** hnp -4.5302 3.2825 -1.380 0.1676 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 581.51 on 425 degrees of freedom Residual deviance: 294.00 on 421 degrees of freedom (41 observations deleted due to missingness) AIC: 304 Number of Fisher Scoring iterations: 8 ## now with control = glm.control(epsilon = 1e-16, maxit = 1000) ## in the call summary(mod3) Call: glm(formula = sex ~ twp + hwp + hcp + hnp, family = binomial, data = dat, control = glm.control(epsilon = 1e-16, maxit = 1000)) Deviance Residuals: Min 1Q Median 3Q Max -2.9703 -0.1760 0.3181 0.6061 3.5235 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)1.4362 0.2687 5.345 9.02e-08 *** twp5.5673 1.3602 4.093 4.26e-05 *** hwp -4.2793 2.3781 -1.799 0.0719 . hcp -450.191856.6823 -7.942 1.98e-15 *** hnp -4.5302 3.2825 -1.380 0.1676 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 581.51 on 425 degrees of freedom Residual deviance: 294.00 on 421 degrees of freedom (41 observations deleted due to missingness) AIC: 304 Number of Fisher Scoring iterations: 9 From the thread you link to, we would expect the effects to diverge. Profiling the model shows that the LL starts to increase again at low values, but does so slowly. The LL is very flat around the estimates and is far from 0, which seems to correspond with the description of the Hauck Donner effect given By Venables and Ripley in their book. In your case however, the statistic is still sufficiently large for it to be identified as significant via the Wald test. If we fit the model via brglm() we get essentially the same result as fitted by glm(): summary(mod2) Call: brglm(formula = sex ~ twp + hwp + hcp + hnp, family = binomial, data = dat) Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept)1.4262 0.2662 5.357 8.47e-08 *** twp5.3696 1.3323 4.030 5.57e-05 *** hwp -4.2813 2.3504 -1.821 0.0685 . hcp -435.921255.0566 -7.918 2.42e-15 *** hnp -4.6295 3.2459 -1.426 0.1538 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 560.48 on 425 degrees of freedom Residual deviance: 294.08 on 421 degrees of freedom Penalized deviance: 302.8212 (41 observations deleted due to missingness) AIC: 304.08 Ioannis points out that if there were separation, the brglm results would differ markedly from the glm() ones, and they don't. As Ioannis mentioned in that thread, you can get the warning about the fitted probabilities without a separation problem - In my case it was because there were some very small fitted probabilities, which R was just warning about. HTH G Here they go the scripts I have used: dati-read.table(simone.txt,header=T,sep=\t,as.is=T)
Re: [R] getting data associated with coordinates in a spatial data frame
On 13-Oct-11 20:33, Sarah Goslee wrote: Hi, On Thu, Oct 13, 2011 at 2:05 PM, Bailey, Danielbai...@spu.edu wrote: Thank you Sarah. I tried your suggestion, and if I coerce it into a normal data.frame, that method works. But if you've already made the data into a SpatialPixelsDataFrame and run coordinates (both from the package sp) so that the columns x and y become a single column coordinates with the format (0, 17) for x and y, how do you then call or manipulate data at a specific location? The following: e[e$coordinates==(0,17),] Doesn't work. They don't become a single column but rather a single matrix with two columns, and (0, 17) isn't the correct way to specify a vector. You can identify particular coordinates using the form I offered earlier, and then use that to subset the data slot of your SGPF. Using built-in data: library(sp) data(meuse.grid) m = SpatialPixelsDataFrame(points = meuse.grid[c(x, y)], data = meuse.grid) m@data[coordinates(m)[,x] == 181100 coordinates(m)[,y] == 333660,] There ought to be a more elegant way to match coordinates (other than the do.call() and paste() approach), but I'm not sure what it is. Not sure if it is nicer, but another possibility is: data(meuse.grid) coordinates(meuse.grid) = ~x+y meuse.grid@data[which(duplicated(rbind(c(181100, 333660), coordinates(meuse.grid-1,] = factor(c(1,2,3,4,1)) To avoid numerical problems, you can also find the data of the location closest to the point you are interested in: meuse.grid@data[which.min(spDistsN1(meuse.grid, c(181100, 333660))),] = factor(c(1,1,1,1,1)) For questions about spatial data you can also use the mailing-list r-sig-geo. Cheers, Jon __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Compiling R extension: undefined reference to `__mingw_vsprintf'
I'm trying to link a hydrological model in FORTRAN with R. I have a subroutine inside wetall.f90 which calls two contained functions. When I try rcmd SHLIB -o wetall.dll wetall.f90 I get a bunch of errors stating undefined reference to `__mingw_vsprintf' from dos (see below). When the same is run from cygwin the dll is compiled but hangs R when dyn.load(ed) The code does not do any I/O so the errors seem unusual. Any suggestions? Microsoft Windows [Version 6.0.6002] Copyright (c) 2006 Microsoft Corporation. All rights reserved. C: \wetland_testrcmd SHLIB -o wetall2.dll wetall.f90 gfortran -shared -s -static-libgcc -o wetall2.dll tmp.def wetall.o -LC:/PROGRA~1/R/R-212~1.1/bin/i386 -lR c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(unix.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(main.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(format.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(transfer.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$nml_read_obj.clone.1+0x199): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$nml_read_obj.clone.1+0x345): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$nml_get_obj_data.clone.0+0x4ab): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$nml_get_obj_data.clone.0+0x56b): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$nml_get_obj_data.clone.0+0x602): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(list_read.o):(.text$nml_get_obj_data.clone.0+0x69c): more undefined references to `__mingw_snprintf' follow c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(write.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(write.o):(.text$output_float.clone.5+0x536): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(write.o):(.text$write_float+0xf9): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(write.o):(.text$write_float+0x18d): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(write.o):(.text$write_float+0x36e): undefined reference to `__mingw_snprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(open.o):(.text$sprintf+0x1b): undefined reference to `__mingw_vsprintf' c:/rtools/mingw/bin/../lib/gcc/mingw32/4.5.0/libgfortran.a(open.o):(.text$__gfortrani_new_unit+0x560): undefined reference to `__mingw_snprintf' collect2: ld returned 1 exit status [[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] Length of data.frame column
On Oct 14, 2011, at 5:11 AM, Petr PIKAL wrote: Hi I have a related question...I have a data frame similar with 74 rows that I created with header=TRUE, but when I try to coerce one of the data frame columns into a vector, it shows up as having length 1, even though when I print it, it shows 74 elements: VAL - c(DailyDiary[1]) VAL [1] 3 3 3 2 3 3 4 4 3 3 2 1 1 4 3 3 2 3 3 2 2 3 3 3 1 2 2 [28] 1 1 3 3 3 2 4 3 2 3 4 3 3 3 3 4 3 2 3 2 3 1 2 1 2 3 1 [55] 0 3 2 4 3 1 2 3 1 1 1 4 3 2 1 2 3 3 3 2 length(VAL) [1] 1 your VAL is still data frame. See ?[ and link to data.frame method. Use str(object) Or at least a list. The c function stripped its attributes: str( c( data.frame(a=1:10) ) ) List of 1 $ a: int [1:10] 1 2 3 4 5 6 7 8 9 10 Which also is an explanation for why its length is still reported as 1. to see the structure of your object not just printing it to console as print is a function which has also different methods for different objects and does not inform you about nature of your object. You shall profit from reading R-intro which I believe you have overlooked. It shall be installed in doc/manual directory of your R installation. Regards Petr On the other hand, I can easily coerce the row names to a vector of length 74 partf - row.names(DailyDiary) length(partf) [1] 74 What I would like to do is make VAL into a vector with length 74 instead of length 1 so I can sort it using use partf as factors. I tried as.vector, but it doesn't let you specify the length. Any ideas? One might think that as.vector might be able to extract a vector from a list if length 1 but one would be wrong and should then got to : ?unlist Thanks, and sorry if I'm being unclear or stupid, I'm a newbie :) Logan David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] quantile function nonlinear in parameters.
Dear all,
I need to run a quantile regression to estimate the coefficients of the
following model: Q_{Y}(Ï|X)=exp(βâ(Ï)+Xâ²Î²â(Ï)).
Since the model is nonlinear, I need to use nlrq(.). However, if I try
nlrq(Y~exp(X), tau=Ï), the software does not accept and also does not
understand that my coefficients are inside the exponential. The software would
allow me to run rq(Y~exp(X), tau=Ï) since it would understand that beta is
outside exp(X).
I could not find one way to write my model in a proper way in the software. Can
someone help me? Sorry if this is too stupid and I haven't see it yet.
Thanks a lot,
Julia
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Re: [R] how to plot two surfaces with lattice::wireframe
No, the point of my question is how to plot the Re and Im parts as two
separate surfaces in one chart.
On 10/14/11 5:46 AM, Eik Vettorazzi wrote:
Hi Carl,
I have no idea what z or f(z) are, but maybe outer will help you:
wireframe(outer(seq(0,5,length.out=50),seq(2,4,length.out=40),function(x,y)sin(x*y)))
cheers.
Am 13.10.2011 23:37, schrieb Carl Witthoft:
Hi all,
I'd like to plot the Real and Imaginary parts of some f(z) as two
different surfaces in wireframe (the row/column axes are the real and
imag axes). I know I can do it by, roughly speaking, something like
plotz- expand.grid(x={range of Re(z)}, y={range of Im(z), groups=1:2)
plotz$func-c(Re(f(z),Im(f(z))
wireframe(func~x*y,data=plotz,groups=groups)
But that seems like a clunky way to go, especially if I happen to have
started out with a nice matrix of the f(z) values.
So, is there some simpler way to write the formula in wireframe? I
envision, for a matrix of complex values zmat, pseudocode:
wireframe(c(Re(zmat),Im(zmat), groups=1:2)
-- and yes, I'm fully aware that without a connection between the
'groups' variable and zmat, this won't work as written.
All suggestions (including read the help file for {some lattice func I
didn't know about} ) greatfully accepted.
Carl
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Sent from my Cray XK6
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[R] Split a list
I have a list of dataframes i.e. each list element is a dataframe with three columns and differing number of rows. The third column takes on only two values. I wish to split the list into two sublists based on the value of the third column of the list element. Second issue with lists as well. I would like to reduce each of the sublist based on the range of the second column, i.e. if the range of the second column is greater than twenty for example keep the list element. Could someone help me with a code to implement these two issues. Thanks in advance for your help, JN [[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] Split a list
It would be nice if you could provide a sample. However, if the data
in the list have the same colnames, you can combine them by
df-do.call('rbind',your_list_data_frame)
Then you can do what you want on the dataframe instead of a list
HTH
Weidong Gu
On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum jpnts...@yahoo.com wrote:
I have a list of dataframes i.e. each list element is a dataframe with three
columns and differing number of rows. The third column takes on only two
values. I wish to split the list into two sublists based on the value of the
third column of the list element.
Second issue with lists as well. I would like to reduce each of the sublist
based on the range of the second column, i.e. if the range of the second
column is greater than twenty for example keep the list element.
Could someone help me with a code to implement these two issues. Thanks in
advance for your help,
JN
[[alternative HTML version deleted]]
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Re: [R] Split a list
Comments inline: On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum jpnts...@yahoo.com wrote: I have a list of dataframes i.e. each list element is a dataframe with three columns and differing number of rows. The third column takes on only two values. I wish to split the list into two sublists based on the value of the third column of the list element. Assuming the third column always takes a constant value for each data.frame, something this will work: lst - list(a = data.frame(1:3, 4:6, 7), b = data.frame(1:3, 4:6, 8), c = data.frame(1:3, 4:6, 7)) idx - sapply(lst, function(x) x[1,3] == 7) lst1 = lst[idx] lst2 = lst[!idx] Second issue with lists as well. I would like to reduce each of the sublist based on the range of the second column, i.e. if the range of the second column is greater than twenty for example keep the list element. Similar line of argument: idx - sapply(lst, function(x) range(x[,2]) 20) lst1 = lst[idx] Could someone help me with a code to implement these two issues. Thanks in advance for your help, JN [[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] Split a list
On Oct 14, 2011, at 9:26 AM, Weidong Gu wrote:
It would be nice if you could provide a sample.
That is certainly true.
However, if the data
in the list have the same colnames, you can combine them by
df-do.call('rbind',your_list_data_frame)
Then you can do what you want on the dataframe instead of a list
HTH
Weidong Gu
On Fri, Oct 14, 2011 at 9:06 AM, Juliet Ndukum jpnts...@yahoo.com
wrote:
I have a list of dataframes i.e. each list element is a dataframe
with three columns and differing number of rows. The third column
takes on only two values. I wish to split the list into two
sublists based on the value of the third column of the list element.
Perhaps something like:
list_of_firsts - lapply(dflist, function(x) X[ , X[,3]==first] )
list_of_seconds - lapply( dflist, function(x) X[ , X[,3]==second])
Ow with subset (but without that missing example it is more difficult
to show the true value of subset:
subset(X, select= X[,3]==first)
Second issue with lists as well. I would like to reduce each of the
sublist based on the range of the second column, i.e. if the range
of the second column is greater than twenty for example keep the
list element.
Same as above with an inequality sign.
Could someone help me with a code to implement these two issues.
Thanks in advance for your help,
JN
[[alternative HTML version deleted]]
David Winsemius, MD
West Hartford, CT
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Re: [R] Generating Data
See Duncan Murdoch's post here for some pointers as well as an explanation of why this isn't a totally well-formed question: http://r.789695.n4.nabble.com/generate-two-sets-of-random-numbers-that-are-correlated-td3736161.html (Specifically, the post at 1:33, but it may be worthwhile to read the whole thread) Michael On Fri, Oct 14, 2011 at 5:51 AM, Muhammed Rauf Ahmad mu...@mai.liu.se wrote: Dear All I need to generate multivariate NON-NORMAL data in R, which follows a given mean vector and covariance matrix, say multivariate exponential data. How can I do that? Best regards mra __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] pgfSweave-example not compiling
Yes, it is a little bit amazing to me too, so let's cc to the package
maintainer to see if he can do anything.
Regards,
Yihui
--
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Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Fri, Oct 14, 2011 at 12:06 AM, Peter Meilstrup
peter.meilst...@gmail.com wrote:
Thanks, that lets me move to the next step in figuring out what's
wrong with my document :) Funny enough that's not the first problem
I've had with Sweave and non-ascii characters.
Peter
On Thu, Oct 13, 2011 at 7:30 PM, Yihui Xie x...@yihui.name wrote:
OK, that is really helpful for diagnosis. In line 436 of your tex
file, there is an NA, but it should really be this:
}% this brace ends the effect of ‘include external’
So the reason for LaTeX failure was this missing bracket }. And the
reason for R to output NA here is most likely to be that in the LaTeX
comment here, there are curly quotes, which are not ASCII characters.
To solve the problem, you can either delete these non-ASCII
characters, or tell Sweave that your encoding is not latin1, but UTF8.
The latter way may not be directly possible, and this is why I brought
up the issue to pgfSweave developers:
https://github.com/cameronbracken/pgfSweave/issues/34
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Thu, Oct 13, 2011 at 8:34 PM, Peter Meilstrup
peter.meilst...@gmail.com wrote:
Sure, here's the generated .tex file:
http://pastebin.com/b9fTsr0h
Peter
On Thu, Oct 13, 2011 at 6:14 PM, Yihui Xie x...@yihui.name wrote:
Seems to be weird. I can run the example smoothly under Ubuntu with R
2.13.2. Could you also post your .tex file?
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Thu, Oct 13, 2011 at 7:01 PM, Peter Meilstrup
peter.meilst...@gmail.com wrote:
I'm trying to get pgfSweave up and running. Hopefully I can get it
working from within LyX, but first I'm just trying to get the simplest
possible thing (compiling one of the example files in the pgfSweave
package) to work. I'm using the example that comes in the pgfSweave
package unmodified, but for reference I copied it to:
http://pastebin.com/tW4RL6fs
Configuration:
R version 2.13.1 on OS X (intel) 10.5.8
pgfSweave package version 1.2.1
tikzDevice version 0.6.1
MacTeX 2011 installed
tikz/pgf package version 2.10
Here is what I tried; I copied the file pgfSweave-example.Rnw into my
working directory, created a cache dir and ran:
require(pgfSweave)
pgfSweave(pgfSweave-example.Rnw)
This did some computation and then complained about a missing bracket:
! Missing } inserted.
inserted text
}
l.439 \end{figure}
After I dismissed this, it seemed to fail with various errors.
make: *** [pgfSweave-example-boxplot.pdf] Error 1
Error in tools::texi2dvi(paste(fn, tex, sep = .), pdf = pdf, ...) :
Running 'texi2dvi' on 'pgfSweave-example.tex' failed.
LaTeX errors:
! Package tikz Warning: Some images are not up-to-date and need to be
generated
. [1
Rerunning pgfSweave led to the same errors.
The process created a .pdf file, but in place of a figure it has the
notice:
[[ Image Discarded Due To ‘/tikz/external/mode=list and make’ ]] NA
Upon inspecting the log file I found this notice:
= mode=`list and make': Use 'make -f pgfSweave-example.makefile' to
generat
e all images. Then, re-run (pdf)latex pgfSweave-example. =
I tried following this instruction, but it also produced an error:
bayanus-2:writing peter$ make -f pgfSweave-example.makefile
pdflatex -halt-on-error -interaction=batchmode -jobname
pgfSweave-example-boxplot
\def\tikzexternalrealjob{pgfSweave-example}\input{pgfSweave-example}
This is pdfTeX, Version 3.1415926-2.3-1.40.12 (TeX Live 2011)
restricted \write18 enabled.
entering extended mode
make: *** [pgfSweave-example-boxplot.pdf] Error 1
I put the full R transcript at http://pastebin.com/aZECxKyP in case
that contains additional useful information.
I also put the contents of pgfSweave-example.log at
http://pastebin.com/LzZW1z3R .
Also pgfSweave-example-boxplot.log at http://pastebin.com/AF3TCCEK .
What's going wrong?
Cheers,
Peter
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[R] qcc package
Hi All, I installed qcc package and the dependency packages. For the first time I can use the function : process.capability.sixpack(). But later when ran the code again I always got the following error: Error: could not find function process.capability.sixpack. I tried reinstalling the qcc package but didn't help. Does anyone have this kind of experience? Thank you! Regards, Yan [[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] qcc package
Did you reload the package for each new session? Sounds like you are probably missing a line like library(qcc) or require(qcc) Michael On Fri, Oct 14, 2011 at 11:10 AM, Li, Yan yan...@ibi.com wrote: Hi All, I installed qcc package and the dependency packages. For the first time I can use the function : process.capability.sixpack(). But later when ran the code again I always got the following error: Error: could not find function process.capability.sixpack. I tried reinstalling the qcc package but didn't help. Does anyone have this kind of experience? Thank you! Regards, Yan [[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] qcc package
Apologies for my brief reply -- I didn't take the time to examine the package closely. On a fresh install, I don't see process.capability.sixpack() as an available function and process.capability() isn't a generic so it's not that sort of thing. Looking at the package index, there's no indication such a thing exists, though perhaps it did in an older version of the package. Do you have a citation for this function? I don't see one in the official CRAN description... Michael On Fri, Oct 14, 2011 at 11:25 AM, Li, Yan yan...@ibi.com wrote: Yes, I did everytime. And the other functions: qcc(), process.capability() works. Thank you for your reply! -Original Message- From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] Sent: Friday, October 14, 2011 11:22 AM To: Li, Yan Cc: r-help@r-project.org Subject: Re: [R] qcc package Did you reload the package for each new session? Sounds like you are probably missing a line like library(qcc) or require(qcc) Michael On Fri, Oct 14, 2011 at 11:10 AM, Li, Yan yan...@ibi.com wrote: Hi All, I installed qcc package and the dependency packages. For the first time I can use the function : process.capability.sixpack(). But later when ran the code again I always got the following error: Error: could not find function process.capability.sixpack. I tried reinstalling the qcc package but didn't help. Does anyone have this kind of experience? Thank you! Regards, Yan [[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] non-parametric permutation and signed paired-difference distributions
Hi all Consider the classic data below from Darwin on the heights of 15 pairs of zea mays (corn) plants either cross-fertilized or self-fertilized, where the goal is to see if it makes a difference. head(ZeaMays) pair pot cross self diff 11 1 23.500 17.375 6.125 22 1 12.000 20.375 -8.375 33 1 21.000 20.000 1.000 44 2 22.000 20.000 2.000 55 2 19.125 18.375 0.750 66 2 21.500 18.625 2.875 ... I'd like to illustrate two types of non-parametric tests of whether the mean(diff) = 0. (a) Permutation test, where the values of, say self are permuted and diff=cross - self is calculated for each permutation. There are 15! permutations, but a reasonably large number of random permutations would suffice. (b) Test based on assigning each abs(diff) a + or - sign, and calculating the mean(diff). There are 2^15 such possible values, but again, a reasonably large number of random samples would do. This is obviously a case for apply and friends, but I can't quite see how to set it up. The complete data: dput(ZeaMays) structure(list(pair = 1:15, pot = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c(1, 2, 3, 4), class = factor), cross = c(23.5, 12, 21, 22, 19.125, 21.5, 22.125, 20.375, 18.25, 21.625, 23.25, 21, 22.125, 23, 12), self = c(17.375, 20.375, 20, 20, 18.375, 18.625, 18.625, 15.25, 16.5, 18, 16.25, 18, 12.75, 15.5, 18), diff = c(6.125, -8.375, 1, 2, 0.75, 2.875, 3.5, 5.125, 1.75, 3.625, 7, 3, 9.375, 7.5, -6)), row.names = c(NA, -15L), .Names = c(pair, pot, cross, self, diff), class = data.frame) -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street Web: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] boxplot
A few things in play here: 1) I'm guessing you are new to R, so I'd advise you to take some time to read some introductory materials at this point. If you type help.start() into your R session, a good introductory manual will be available. 2) You don't need any of this textConnection business, that's just one way of reading in data from an email. If you are reading in your data via read.table() you don't need anything else. 3) Your code should perhaps look something like this, but I can't be certain without an actual sample of your data: lista - read.table(lista.txt,header=T,dec=,) vit-subset(lista, (lista$familia == Vittariaceae) , select=c(ingreso, familia, genero,categoria,ira5,ira6,ira7,ira8,ira9,ira10,irah5,irah6,irah7,irah8,irah9,irah10,ed5)) boxplot( t(vit[grepl(ira,rownames(vit)),]), col = ifelse(vit[,NCOL(vit)] == C, 2,3)) but at this point, I'd spend some a little more time learning how to manipulate and subset in R before trying to handle this finer sort of detail. I'll quickly parse that last line for you, though, just in case this is a time sensitive project: working from the inside out: grepl(ira, rownames(vit)) # Identify those rows that have ira in the name vit[grepl(ira, rownames(vit)),] # Select only those rows from vit t( ) # transpose since boxplot takes in columns (see my above note) vit[,NCOL(vit)] # isolate the last column ** == C # Test for being of type C ifelse( **, 2, 3) ## assign either a two or a three depending whethere it was a C or not col = # color the boxplots 4) On another note, are you sure you want to do your boxplots rowwise? It's pretty standard in R that columns represent similar data while rows are independent measurements/observations. 5) And just for the record, the error you got is because you passed boxplot something that wasn't a data object, but rather an empty textConnection(). Hope this helps, Michael PS -- It's good form (and usually more helpful to you) to cc the whole list on each step of the correspondance. It lets other voices chime in if I start leading you astray as well as ensuring that there are replies coming even at all hours of the day since R-help has a world-wide readership. Also, it makes sure everything gets archived nicely so someone can read them in the future. On Fri, Oct 14, 2011 at 9:19 AM, Ruth Arias rueu...@yahoo.es wrote: thanks for answer me michel I made some modifications to your suggestion. I have tried this: lista - read.table(lista.txt,header=T,dec=,) vit-subset(lista, (lista$familia == Vittariaceae) , select=c(ingreso, familia, genero,categoria,ira5,ira6,ira7,ira8,ira9,ira10,irah5,irah6,irah7,irah8,irah9,irah10,ed5)) vitbp-textConnection(vit) closeAllConnections() boxplot(vitbp[,1:5], col = ifelse(V[,6] == C, 2,3)) and I have this answer: Error en plot.window(xlim = xlim, ylim = ylim, log = log, yaxs = pars$yaxs) : se necesitan valores finitos de 'ylim' = needs finite values Además: Mensajes de aviso perdidos = also: message 1: In min(x) : ningún argumento finito para min; retornando Inf = no finite argument for minimun 2: In max(x) : ningun argumento finito para max; retornando -Inf = no finite argument for maximun and what I want is to make a box plot to let me see the differences in different iras (ira5, ira6 ...) between the category C and E I'm not sure what the column to identify other the other columns maps to graphically, but perhaps something like this will get you started V - read.table(textConnection( 4 5 6 7 8 site 23 56 41 45 63 C 21 89 42 10 63 E 32 45 14 17 96 E 45 74 13 63 41 C 68 32 10 20 03 E 95 10 84 45 96 C ),header=TRUE) closeAllConnections() boxplot(V[,1:5], col = ifelse(V[,6] == C, 2,3)) Michael On Thu, Oct 13, 2011 at 5:54 PM, Ruth Arias rueu...@yahoo.es wrote: hello I want to make a boxplot with diferents rows and also include a column to sort into two groups to each of the other columns my date set looks like this: 4 5 6 7 8 site 23 56 41 45 63 C 21 89 42 10 63 E 32 45 14 17 96 E 45 74 13 63 41 C 68 32 10 20 03 E 95 10 84 45 96 C THANKS [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
[R] Party package: varimp(..., conditional=TRUE) error: term 1 would require 9e+12 columns
I would like to build a forest of regression trees to see how well some covariates predict a response variable and to examine the importance of the covariates. I have a small number of covariates (8) and large number of records (27368). The response and all of the covariates are continuous variables. A cursory examination of the covariates does not suggest they are correlated in a simple fashion (e.g. the variance inflation factors are all fairly low) but common sense suggests there should be some relationship: one of them is the day of the year and some of the others are environmental parameters such as water temperature. For this reason I would like to follow the advice of Strobl et al. (2008) and try the authors' conditional variable importance measure. This is implemented in the party package by calling varimp(..., conditional=TRUE). Unfortunately, when I call that on my forest I receive the error: varimp(myforest, conditional=TRUE) Error in model.matrix.default(as.formula(f), data = blocks) : term 1 would require 9e+12 columns Does anyone know what is wrong? I noticed a post in June 2011 where a user reported this message and the ultimate problem was that the importance measure was being conditioned on too many variables (47). I have only a small number of variables here so I guessed that was not the problem. Another suggestion was that there could be a factor with too many levels. In my case, all of the variables are continuous. Term 1 (x1 below) is the day of the year, which does happen to be integers 1 ... 366. But the variable is class numeric, not integer, so I don't believe cforest would treat it as a factor, although I do not know how to tell whether cforest is treating something as continuous or as a factor. Thank you for any help you can provide. I am running R 2.13.1 with party 0.9-4. You can download the data from http://www.duke.edu/~jjr8/data.rdata (512 KB). Here is the complete code: load(\\Temp\\data.rdata) nrow(df) [1] 27368 summary(df) y x1 x2 x3 x4 x5 x6 x7 x8 Min. : 0.000 Min. : 1.0 Min. :0. Min. : 1.00 Min. : 52 Min. : 0.008184 Min. :16.71 Min. :0.000 Min. : 0.02727 1st Qu.: 0.000 1st Qu.:105.0 1st Qu.:0. 1st Qu.: 30.00 1st Qu.:1290 1st Qu.: 6.747035 1st Qu.:23.92 1st Qu.:0.000 1st Qu.: 0.11850 Median : 1.282 Median :169.0 Median :0.2353 Median : 38.00 Median :1857 Median :11.310277 Median :26.35 Median :0.0001569 Median : 0.14625 Mean : 5.651 Mean :178.7 Mean :0.2555 Mean : 55.03 Mean :1907 Mean :12.889021 Mean :26.31 Mean :0.0162043 Mean : 0.20684 3rd Qu.: 5.353 3rd Qu.:262.0 3rd Qu.:0.4315 3rd Qu.: 47.00 3rd Qu.:2594 3rd Qu.:18.427410 3rd Qu.:28.95 3rd Qu.:0.0144660 3rd Qu.: 0.20095 Max. :195.238 Max. :366.0 Max. :1. Max. :400.00 Max. :3832 Max. :29.492380 Max. :31.73 Max. :0.3157486 Max. :11.76877 library(HH) output deleted vif(y ~ ., data=df) x1 x2 x3 x4 x5 x6 x7 x8 1.374583 1.252250 1.021672 1.218801 1.015124 1.439868 1.075546 1.060580 library(party) output deleted mycontrols - cforest_unbiased(ntree=50, mtry=3) # Small forest but requires a few minutes myforest - cforest(y ~ ., data=df, controls=mycontrols) varimp(myforest) x1 x2 x3 x4 x5 x6 x7 x8 11.924498 103.180195 16.228864 30.658946 5.053500 12.820551 2.113394 6.911377 varimp(myforest, conditional=TRUE) Error in model.matrix.default(as.formula(f), data = blocks) : term 1 would require 9e+12 columns __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] change uppercase variables
Hello everyone, I'm trying to change the name variables of a big dataset. Here's more than 300 variables. The point is that I have to match it with another dataset that have same variables, but in lowercase, the I can use rbind and do my work. Is there any function for changing uppercase or lowercase variables? Thank you in advance! José [[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] qcc package
You'll note that the package version for which you are reading the manual is not the version that you downloaded from CRAN. As I suspected, this functionality seems to have existed in an older version of the package, but seems to have since been deprecated (cf. the current CRAN files). You can install an older version from the CRAN archives (2.0.1) is still available if you wish, but I'd suggest that you also look around the new package and see if there's an even better version available. I don't know the package so I can't direct you to whatever updated/improved versions there may be. I don't know how widely used the package is, but if no one on this list can help, it may be worth contacting the maintainer directly to see if he can help. If you do choose to reinstall the older version, there doesn't seem to be any C/FORTRAN code in it, so something like install.packages(qcc.tar.gz, type=source, repos=NULL) *should* work directly (once you point R to the file), no promises though. Hope this helps, Michael PS -- It's good form (and usually more helpful to you) to cc the whole list on each step of the correspondance. It lets other voices chime in if I start leading you astray as well as ensuring that there are replies coming even at all hours of the day since R-help has a world-wide readership. Also, it makes sure everything gets archived nicely so someone can read them in the future. On Fri, Oct 14, 2011 at 11:36 AM, Li, Yan yan...@ibi.com wrote: I attached the .pdf for qcc package. For process.capability.sixpack() the reference is : Minitab (2000) Users's Guide 2: Data Analysis and Quality Tools, rel. 13, Chapter 14. (I cannot find it in the web...) Thank you! -Original Message- From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] Sent: Friday, October 14, 2011 11:28 AM To: Li, Yan; r-help Subject: Re: [R] qcc package Apologies for my brief reply -- I didn't take the time to examine the package closely. On a fresh install, I don't see process.capability.sixpack() as an available function and process.capability() isn't a generic so it's not that sort of thing. Looking at the package index, there's no indication such a thing exists, though perhaps it did in an older version of the package. Do you have a citation for this function? I don't see one in the official CRAN description... Michael On Fri, Oct 14, 2011 at 11:25 AM, Li, Yan yan...@ibi.com wrote: Yes, I did everytime. And the other functions: qcc(), process.capability() works. Thank you for your reply! -Original Message- From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com] Sent: Friday, October 14, 2011 11:22 AM To: Li, Yan Cc: r-help@r-project.org Subject: Re: [R] qcc package Did you reload the package for each new session? Sounds like you are probably missing a line like library(qcc) or require(qcc) Michael On Fri, Oct 14, 2011 at 11:10 AM, Li, Yan yan...@ibi.com wrote: Hi All, I installed qcc package and the dependency packages. For the first time I can use the function : process.capability.sixpack(). But later when ran the code again I always got the following error: Error: could not find function process.capability.sixpack. I tried reinstalling the qcc package but didn't help. Does anyone have this kind of experience? Thank you! Regards, Yan [[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] heritability estimation
Hello, I'm looking for a method to estimate narrow sense heritability of traits in a RIL population. Papers I've checked either use either SAS or SPSS or do not give any details at all. I've found some reference to using variance components in ANOVA, using the kinship or wgaim packages, but I don't have a clue as to how to do any of this. Is there any way fro a very R illiterate user to do it? Thanks -- View this message in context: http://r.789695.n4.nabble.com/heritability-estimation-tp3904908p3904908.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] Selecting multiple vectors from a list of lists of matrices
Hi all,
I was unable to find a solution to my problem in the archives, but this might
be due to a lack knowledge on the correct terminology on my part. Please
forgive me if this has been explained before and please forgive me my probably
clumsy way of explaining things.
This is what I want to do:
I have a list made up of 6 lists containing 7 4x4 matrices each.
My goal is to select a large number of rows from all those matrices (thus
giving vectors of length 4), and stack them into a matrix or data frame.
Creating a single such vector is easy using the normal extract functionality,
say I want the 2nd row of the 5th matrix in the 3rd list, I use:
my.list[[3]] [[5]] [2, ]
Problem is, I need to get a matrix or data frame of about 8000 of these rows...
My idea was to create three vectors for each of the three indices, and put
those vector names where the 3, 5 and 2 are in my example. I did something
similar before with a 3-dimensional array, where you can use a matrix with
dimensions [n,3] as an index to create a vector of length n. This, however,
does not work with lists. I created a workaround using the lapply function
(shown below) which works but is incredibly slow.
extract - function (selector) {
matrix.list[[selector[1]]][[selector[2]]][selector[3],]
}
output - function(var1,var2,var3) {
selector - as.data.frame(rbind(var1,var2,var3))
stack - do.call('rbind',lapply(selector,FUN=extract))
stack
}
The object matix.list is the said object consisting of 6 lists containing 7 4x4
matrices each. Var 1 2 and 3 are three vectors of length 8000.
My problem that, although this is working, it is really slow. The output is
used in a monte carlo simulation with many iterations, so the function is
called over and over again. What parts of this code are slowing it down the
most, and how can I speed things up?
Thanks for all the help!
Cheers,
Gimon
De inhoud van dit bericht is vertrouwelijk en alleen bes...{{dropped:15}}
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[R] Multi t tests
hi all I have R object look like this: spl $SB012XSB044 DPW Cross 1 66.6 SB012XSB044 2 96.5 SB012XSB044 3 78.8 SB012XSB044 4 68.6 SB012XSB044 5 62.0 SB012XSB044 6 72.1 SB044XSB012 7 72.2 SB044XSB012 8 69.6 SB044XSB012 9 87.9 SB044XSB012 10 84.4 SB044XSB012 11 51.9 SB044XSB012 12 65.5 SB044XSB012 $SB012XSB099 DPW Cross 13 100.9 SB012XSB099 14 44.4 SB012XSB099 15 83.5 SB012XSB099 16 89.9 SB012XSB099 17 78.0 SB012XSB099 18 83.0 SB012XSB099 19 114.3 SB099XSB012 20 173.1 SB099XSB012 21 114.3 SB099XSB012 22 58.8 SB099XSB012 23 98.1 SB099XSB012 24 12.8 SB099XSB012 I want to do a t test for DPW values by the 2 different Cross types for each of spl components. Any suggestions? I just got errors till now Thanks, Imri. -- View this message in context: http://r.789695.n4.nabble.com/Multi-t-tests-tp3905075p3905075.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] change uppercase variables
What sort of variables are these? generally toupper() or tolower() would do the trick, but it can be little trickier if things are stored as factors. Michael On Fri, Oct 14, 2011 at 12:12 PM, Jose Bustos Melo jbustosm...@yahoo.es wrote: Hello everyone, I'm trying to change the name variables of a big dataset. Here's more than 300 variables. The point is that I have to match it with another dataset that have same variables, but in lowercase, the I can use rbind and do my work. Is there any function for changing uppercase or lowercase variables? Thank you in advance! José [[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] Minimization/Optimization under functional constraints
Thanks for your help...actually there is monotonicity in beta so minimizing
the square of the functional constraint works. I verified it with a brute
force search (while loop).
For the sake of knowledge this is what someone else suggested (but didn't
work in my case)
Since x is fixed (given the data), you are really just trying to find
inf{beta0 | g(beta) = 0}
where g() is defined in the obvious way.
If you can be sure that the infimum is not 0, then you can get rid of
the constraint beta0 to transforming the problem to
inf{gamma | h(gamma) = 0}
where, e.g., h(gamma) = g(exp(gamma)).
Now, if your original f is continuous and isn't constant over any
interval, say, then you could try to solve for the zeros of h, and the
smallest one should be what you're looking for. Finding all the zeros
of h could still be hard of course, and I suspect your problem isn't
that nice anyway or you wouldn't be asking me.
--
View this message in context:
http://r.789695.n4.nabble.com/Minimization-Optimization-under-functional-constraints-tp3899020p3905137.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] help to ... import the data from Excel
I am tried z - read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls, colNames=FALSE, rowNames=FALSE) and the following massege is appeared: Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Incorrect number of arguments (11), expecting 10 for 'ReadXls' My dear , Jean thanks a lot ( Flower) . */My best wishes/* - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3905334.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] date and time
Dear R users, I got date and time as two separate characters [1] 2008-04-11 [1] 22:00:00 which correspond to my starting point in time domain. Now I need to produce series over, 5 sec epochs up to end point, say: [1] 2008-04-12 [1] 23:00:00 So something like 2008-04-11 22:00:00 2008-04-11 22:00:05 2008-04-11 22:00:10 . . . 2008-04-12 23:00:00 Is there any strightforward way to do it? Any suggestions? Best, robert -- View this message in context: http://r.789695.n4.nabble.com/date-and-time-tp3905358p3905358.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] aov(variable~group*(speed*person)) but How to get 95% confidence intervals ?
Greetings and gratitude, I have 19 persons in each group, and each person walks at 3 different speeds. I can do a nice p value and f value with summary command Please help me learn how to report the 95% confidence interval for this anova? This is easier for my fourth, separate condition: preferred walking speed, where there is no speed factor in the model: aov(variable~group) although I compute the 95%CI in SPSS for this model :( Any advice on how to do this for the single speed anova in R, because it is really nice! Thank you Jebb Jebb Remelius PhDc j...@kin.umass.edu 413-545-4959 office 224-645-3490 mobile __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help to ... import the data from Excel
I can't remember the specifics right now, but when you load the package used to provide read.xls() you probably get a long warning message saying you need to run xls.getshlib() before using the package. If I remember right, this solves your problem. Michael On Fri, Oct 14, 2011 at 11:54 AM, Sarah_R_edu sarah_r_...@hotmail.com wrote: I am tried z - read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls, colNames=FALSE, rowNames=FALSE) and the following massege is appeared: Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Incorrect number of arguments (11), expecting 10 for 'ReadXls' My dear , Jean thanks a lot ( Flower) . */My best wishes/* - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3905334.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] change uppercase variables
On 14.10.2011 18:12, Jose Bustos Melo wrote: Hello everyone, I'm trying to change the name variables of a big dataset. Here's more than 300 variables. The point is that I have to match it with another dataset that have same variables, but in lowercase, the I can use rbind and do my work. Is there any function for changing uppercase or lowercase variables? See ?tolower Uwe Ligges Thank you in advance! José [[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] change uppercase variables
?tolower Weidong Gu On Fri, Oct 14, 2011 at 12:12 PM, Jose Bustos Melo jbustosm...@yahoo.es wrote: Hello everyone, I'm trying to change the name variables of a big dataset. Here's more than 300 variables. The point is that I have to match it with another dataset that have same variables, but in lowercase, the I can use rbind and do my work. Is there any function for changing uppercase or lowercase variables? Thank you in advance! José [[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] heritability estimation
On Oct 14, 2011, at 9:49 AM, Moohbear wrote: Hello, I'm looking for a method to estimate narrow sense heritability of traits in a RIL population. I admit to not knowing that TLA. Papers I've checked either use either SAS or SPSS or do not give any details at all. I've found some reference to using variance components in ANOVA, using the kinship or wgaim packages, but I don't have a clue as to how to do any of this. Is there any way fro a very R illiterate user to do it? (I will also admit to at first reading that as illegitimate. Apologies.) To become less illiterate (and also less illegitimate in the r-help context) you should consider first using one of the many search facilities. For instance, at an r-console session, try this: install.packages(sos) require(sos) findFn(heritability) # found 60 matches; retrieving 3 pages 2 3 I don't have the knowledge or experience to pick through all those candidates ... but you presumably do. Thanks -- View this message in context: http://r.789695.n4.nabble.com/heritability-estimation-tp3904908p3904908.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multi t tests
It's usually standard to provide an example of what code you've tried
and also to put your data in a form that can be more easily
cut-and-pasted into R.
That said, would something like this work if you know you only have
two sorts of cross in each level?
lapply(spl, function(x) {x - split(x[,1],x[,2]); t.test(x[[1]], x[[2]])})
Michael
On Fri, Oct 14, 2011 at 10:38 AM, Imri bisr...@agri.huji.ac.il wrote:
hi all
I have R object look like this:
spl
$SB012XSB044
DPW Cross
1 66.6 SB012XSB044
2 96.5 SB012XSB044
3 78.8 SB012XSB044
4 68.6 SB012XSB044
5 62.0 SB012XSB044
6 72.1 SB044XSB012
7 72.2 SB044XSB012
8 69.6 SB044XSB012
9 87.9 SB044XSB012
10 84.4 SB044XSB012
11 51.9 SB044XSB012
12 65.5 SB044XSB012
$SB012XSB099
DPW Cross
13 100.9 SB012XSB099
14 44.4 SB012XSB099
15 83.5 SB012XSB099
16 89.9 SB012XSB099
17 78.0 SB012XSB099
18 83.0 SB012XSB099
19 114.3 SB099XSB012
20 173.1 SB099XSB012
21 114.3 SB099XSB012
22 58.8 SB099XSB012
23 98.1 SB099XSB012
24 12.8 SB099XSB012
I want to do a t test for DPW values by the 2 different Cross types for each
of spl components.
Any suggestions? I just got errors till now
Thanks,
Imri.
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Re: [R] help to ... import the data from Excel
eyildiz ... if i installed the package (getshlib) appeared to me to selecting a CRAN mirror ... Any one to be selected ? Note: i am beginner to used the languge R /thanks eyildiz ( my prayers to you)/ http://r.789695.n4.nabble.com/file/n3905447/47756_429620173219_788668219_5025394_743078_n.jpg - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3905447.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] help to ... import the data from Excel
On Oct 14, 2011, at 11:54 AM, Sarah_R_edu wrote: I am tried z - read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls, colNames=FALSE, rowNames=FALSE) The read.xls that I have (from pkg:gdata) does not have arguments named colNames or rowNames, and those do not look correct to be passed to read.table, so maybe you are using a different package or version? and the following massege is appeared: Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Incorrect number of arguments (11), expecting 10 for 'ReadXls' -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] non-parametric permutation and signed paired-difference distributions
On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendly frien...@yorku.ca wrote: Hi all Consider the classic data below from Darwin on the heights of 15 pairs of zea mays (corn) plants either cross-fertilized or self-fertilized, where the goal is to see if it makes a difference. head(ZeaMays) pair pot cross self diff 1 1 1 23.500 17.375 6.125 2 2 1 12.000 20.375 -8.375 3 3 1 21.000 20.000 1.000 4 4 2 22.000 20.000 2.000 5 5 2 19.125 18.375 0.750 6 6 2 21.500 18.625 2.875 ... I'd like to illustrate two types of non-parametric tests of whether the mean(diff) = 0. (a) Permutation test, where the values of, say self are permuted and diff=cross - self is calculated for each permutation. There are 15! permutations, but a reasonably large number of random permutations would suffice. (b) Test based on assigning each abs(diff) a + or - sign, and calculating the mean(diff). There are 2^15 such possible values, but again, a reasonably large number of random samples would do. What do you mean by 'assigning each abs(diff) a + or - sign, and calculating the mean(diff)'? abs(diff) should be all positive, right? This is obviously a case for apply and friends, but I can't quite see how to set it up. The complete data: dput(ZeaMays) structure(list(pair = 1:15, pot = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c(1, 2, 3, 4), class = factor), cross = c(23.5, 12, 21, 22, 19.125, 21.5, 22.125, 20.375, 18.25, 21.625, 23.25, 21, 22.125, 23, 12), self = c(17.375, 20.375, 20, 20, 18.375, 18.625, 18.625, 15.25, 16.5, 18, 16.25, 18, 12.75, 15.5, 18), diff = c(6.125, -8.375, 1, 2, 0.75, 2.875, 3.5, 5.125, 1.75, 3.625, 7, 3, 9.375, 7.5, -6)), row.names = c(NA, -15L), .Names = c(pair, pot, cross, self, diff), class = data.frame) -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street Web: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] is there an option to turn off scientific notation in write.csv
Paul, Many thanks for your help! I tried the suggestions below and, unfortunately, they didn't work for me. I actually tried a tweak even: Here is what I tried (might I be missing something?): df$x -format(df$x, scientific = FALSE) write.csv(df ,file=df.csv) df$x -format(df$x, scientific = FALSE) write.csv(format(df, scientific = FALSE) ,file=df.csv) For some reason write.csv still wants to coerce the long numeric into scientific notation prior to writing the .csv file... I'm running 2.13.1 on a Win 7 machine... I wonder if this is a bug? One work around I've figured out is to paste a character to df$x before exporting, then trimming the character after export... but there has to be a more elegant solution? From: Paul Hiemstra paul.hiems...@knmi.nl Cc: r-help@r-project.org r-help@r-project.org Sent: Friday, October 14, 2011 12:39 AM Subject: Re: [R] is there an option to turn off scientific notation in write.csv On 10/14/2011 05:25 AM, Chris Conner wrote: Dear Help-Rs, I'm working with a file that contains large numbers and I need to export them as is. for example take: x - c(27104010002005,27104020001805,27104090001810,90050013000140,90050013000120) y - c(1:5) df - data.frame(cbind(x,y)) When I then try a simple: write.csv(df,file=df.csv) I get: x y 1 2.7104E+13 1 2 2.7104E+13 2 3 2.71041E+13 3 4 9.005E+13 4 5 9.005E+13 5 Then I tried: options(scipen=999) df$x - as.character(df$x) Hi, You can use format in this case: df$x -format(df$x, scientific = FALSE) write.csv(df,file=df.csv) regards, Paul write.csv(df,file=df.csv) and I still get: x y 1 2.7104E+13 1 2 2.7104E+13 2 3 2.71041E+13 3 4 9.005E+13 4 5 9.005E+13 5 How can I have R write the file so it looks like this: x y 1 27104010002005 2 27104020001805 3 27104090001810 4 90050013000140 5 90050013000120 [[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. -- Paul Hiemstra, Ph.D. Global Climate Division Royal Netherlands Meteorological Institute (KNMI) Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39 P.O. Box 201 | 3730 AE | De Bilt tel: +31 30 2206 494 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 [[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] heritability estimation
Thanks for the sos function, I didn't know about it. Unfortunately, almost
all the entries listed are not relevant to my problem.
qgen seems to be doing what I want, but gives error message when I type
library(qgen): Error: package 'qgen' was built before R 2.10.0: please
re-install it. I've tried and succeeded (I think) to install it from the
source, but I get the same error.
rrBLUP should in theory do, but the I don't really understand the sample
code and it doesn't work either anyway (the sample code for mixed.solve).
G - matrix(rep(0,200*1000),200,1000)
for (i in 1:200) {
+ G[i,] - ifelse(runif(1000)0.5,-1,1)
+ }
u - rnorm(1000)
g - as.vector(crossprod(t(G),u))
h2 - 0.5
y - g + rnorm(200,mean=0,sd=sqrt((1-h2)/h2*var(g)))
ans - mixed.solve(y,Z=G)
Error in dim(x) : 'x' is missing
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Re: [R] Selecting multiple vectors from a list of lists of matrices
This is somewhat faster on my machine:
t(sapply(seq(var1), function(i) my.list[[var1[i]]] [[var2[i]]] [var3[i],
]))
Jean
Graaf, G de wrote on 10/14/2011 09:23:24 AM:
Hi all,
I was unable to find a solution to my problem in the archives, but
this might be due to a lack knowledge on the correct terminology on
my part. Please forgive me if this has been explained before and
please forgive me my probably clumsy way of explaining things.
This is what I want to do:
I have a list made up of 6 lists containing 7 4x4 matrices each.
My goal is to select a large number of rows from all those matrices
(thus giving vectors of length 4), and stack them into a matrix or data
frame.
Creating a single such vector is easy using the normal extract
functionality, say I want the 2nd row of the 5th matrix in the 3rd
list, I use:
my.list[[3]] [[5]] [2, ]
Problem is, I need to get a matrix or data frame of about 8000 of
these rows...
My idea was to create three vectors for each of the three indices,
and put those vector names where the 3, 5 and 2 are in my example. I
did something similar before with a 3-dimensional array, where you
can use a matrix with dimensions [n,3] as an index to create a
vector of length n. This, however, does not work with lists. I
created a workaround using the lapply function (shown below) which
works but is incredibly slow.
extract - function (selector) {
matrix.list[[selector[1]]][[selector[2]]][selector[3],]
}
output - function(var1,var2,var3) {
selector - as.data.frame(rbind(var1,var2,var3))
stack - do.call('rbind',lapply(selector,FUN=extract))
stack
}
The object matix.list is the said object consisting of 6 lists
containing 7 4x4 matrices each. Var 1 2 and 3 are three vectors of
length 8000.
My problem that, although this is working, it is really slow. The
output is used in a monte carlo simulation with many iterations, so
the function is called over and over again. What parts of this code
are slowing it down the most, and how can I speed things up?
Thanks for all the help!
Cheers,
Gimon
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Re: [R] is there an option to turn off scientific notation in write.csv
Have you tried setting options(scipen=500) # big number of digits ? E.g., df - data.frame(x=pi*10^seq(-30,30,by=10), d=seq(-30,30,by=10), s=state.name[31:37]) getOption(scipen) [1] 0 write.csv(df, stdout()) ,x,d,s 1,3.14159265358979e-30,-30,New Mexico 2,3.14159265358979e-20,-20,New York 3,3.14159265358979e-10,-10,North Carolina 4,3.14159265358979,0,North Dakota 5,31415926535.8979,10,Ohio 6,3.14159265358979e+20,20,Oklahoma 7,3.14159265358979e+30,30,Oregon oldScipen - options(scipen=500) write.csv(df, stdout()) ,x,d,s 1,0.0314159265358979,-30,New Mexico 2,0.000314159265358979,-20,New York 3,0.0314159265358979,-10,North Carolina 4,3.14159265358979,0,North Dakota 5,31415926535.8979,10,Ohio 6,314159265358979334144,20,Oklahoma 7,3141592653589793216422042866402,30,Oregon options(oldScipen) # reset to the previous value 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 Chris Conner Sent: Friday, October 14, 2011 10:31 AM To: Paul Hiemstra Cc: r-help@r-project.org Subject: Re: [R] is there an option to turn off scientific notation in write.csv Paul, Many thanks for your help! I tried the suggestions below and, unfortunately, they didn't work for me. I actually tried a tweak even: Here is what I tried (might I be missing something?): df$x -format(df$x, scientific = FALSE) write.csv(df ,file=df.csv) df$x -format(df$x, scientific = FALSE) write.csv(format(df, scientific = FALSE) ,file=df.csv) For some reason write.csv still wants to coerce the long numeric into scientific notation prior to writing the .csv file... I'm running 2.13.1 on a Win 7 machine... I wonder if this is a bug? One work around I've figured out is to paste a character to df$x before exporting, then trimming the character after export... but there has to be a more elegant solution? From: Paul Hiemstra paul.hiems...@knmi.nl Cc: r-help@r-project.org r-help@r-project.org Sent: Friday, October 14, 2011 12:39 AM Subject: Re: [R] is there an option to turn off scientific notation in write.csv On 10/14/2011 05:25 AM, Chris Conner wrote: Dear Help-Rs, I'm working with a file that contains large numbers and I need to export them as is. for example take: x - c(27104010002005,27104020001805,27104090001810,90050013000140,90050013000120) y - c(1:5) df - data.frame(cbind(x,y)) When I then try a simple: write.csv(df,file=df.csv) I get: x y 1 2.7104E+13 1 2 2.7104E+13 2 3 2.71041E+13 3 4 9.005E+13 4 5 9.005E+13 5 Then I tried: options(scipen=999) df$x - as.character(df$x) Hi, You can use format in this case: df$x -format(df$x, scientific = FALSE) write.csv(df,file=df.csv) regards, Paul write.csv(df,file=df.csv) and I still get: x y 1 2.7104E+13 1 2 2.7104E+13 2 3 2.71041E+13 3 4 9.005E+13 4 5 9.005E+13 5 How can I have R write the file so it looks like this: x y 1 27104010002005 2 27104020001805 3 27104090001810 4 90050013000140 5 90050013000120 [[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. -- Paul Hiemstra, Ph.D. Global Climate Division Royal Netherlands Meteorological Institute (KNMI) Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39 P.O. Box 201 | 3730 AE | De Bilt tel: +31 30 2206 494 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 [[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] Analysis
Here are some tutorials to R: http://www.cyclismo.org/tutorial/R/ http://www.statmethods.net/ Or just search on the web. Good luck, A. On Fri, 14 Oct 2011 09:29:52 +0300 Peter Kaiga kaigape...@gmail.com wrote: i'm actually new at R, but to me its going to be big time, i want to do an analysis for the attached data, like define variables and then analyse get means, variances, histogram, and other important attributes including cross tabs Regards, Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] heritability estimation
On Oct 14, 2011, at 1:44 PM, Moohbear wrote:
Thanks for the sos function, I didn't know about it. Unfortunately,
almost
all the entries listed are not relevant to my problem.
qgen seems to be doing what I want, but gives error message when I
type
library(qgen): Error: package 'qgen' was built before R 2.10.0:
please
re-install it. I've tried and succeeded (I think) to install it
from the
source, but I get the same error.
rrBLUP should in theory do, but the I don't really understand the
sample
code and it doesn't work either anyway (the sample code for
mixed.solve).
G - matrix(rep(0,200*1000),200,1000)
for (i in 1:200) {
+ G[i,] - ifelse(runif(1000)0.5,-1,1)
+ }
u - rnorm(1000)
g - as.vector(crossprod(t(G),u))
h2 - 0.5
y - g + rnorm(200,mean=0,sd=sqrt((1-h2)/h2*var(g)))
ans - mixed.solve(y,Z=G)
Error in dim(x) : 'x' is missing
I'm not getting the same error with that code. I installed the binary
rrBLUP_1.8 for Mac from a CRAN mirror with only a warning that it was
compiled on R 2.13.2. The copy-pasted code runs without error on a
slightly older version of R 2.13.1 in an even older version of my OS
(Mac OS 10.5.8) on a three year-old machine. The next lines also run
without error:
accuracy - cor(u,ans$u)
accuracy
[1] 0.3117429
--
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Re: [R] non-parametric permutation and signed paired-difference distributions
Michael Friendly wrote on 10/14/2011 10:38:44 AM: Hi all Consider the classic data below from Darwin on the heights of 15 pairs of zea mays (corn) plants either cross-fertilized or self-fertilized, where the goal is to see if it makes a difference. head(ZeaMays) pair pot cross self diff 11 1 23.500 17.375 6.125 22 1 12.000 20.375 -8.375 33 1 21.000 20.000 1.000 44 2 22.000 20.000 2.000 55 2 19.125 18.375 0.750 66 2 21.500 18.625 2.875 ... I'd like to illustrate two types of non-parametric tests of whether the mean(diff) = 0. (a) Permutation test, where the values of, say self are permuted and diff=cross - self is calculated for each permutation. There are 15! permutations, but a reasonably large number of random permutations would suffice. You have paired data. To conduct a permutation test you would randomly assign one member of each pair to being either cross or self-fertilized. The other member of the pair would be assigned to the opposite treatment. That would lead to 2^15 = 32,768 permutations. See the perm.test() function in the R package exactRankTests. (b) Test based on assigning each abs(diff) a + or - sign, and calculating the mean(diff). There are 2^15 such possible values, but again, a reasonably large number of random samples would do. Sounds like you are attempting to combine a sign test and a permutation test. You could do this by using the logical crossself rather than the difference cross-self. Jean This is obviously a case for apply and friends, but I can't quite see how to set it up. The complete data: dput(ZeaMays) structure(list(pair = 1:15, pot = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L), .Label = c(1, 2, 3, 4), class = factor), cross = c(23.5, 12, 21, 22, 19.125, 21.5, 22.125, 20.375, 18.25, 21.625, 23.25, 21, 22.125, 23, 12), self = c(17.375, 20.375, 20, 20, 18.375, 18.625, 18.625, 15.25, 16.5, 18, 16.25, 18, 12.75, 15.5, 18), diff = c(6.125, -8.375, 1, 2, 0.75, 2.875, 3.5, 5.125, 1.75, 3.625, 7, 3, 9.375, 7.5, -6)), row.names = c(NA, -15L), .Names = c(pair, pot, cross, self, diff), class = data.frame) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] date and time
Good afternoon Robert, Suppose you have your date and time in characters like this: d.start = 2008-04-11 t.start = 22:00:00 d.end = 2008-04-12 t.end = 15:00:00 then use POSIXct to convert them to a unified time object: start - as.POSIXct(paste(d.start, t.start)) end - as.POSIXct(paste(d.end, t.end)) Then simply use seq() on your objects, here: OUT - seq(start, end, by = 5) # 5 second steps. Hope this helps, (Robert) Michael Weylandt On Fri, Oct 14, 2011 at 12:02 PM, threshold r.kozar...@gmail.com wrote: Dear R users, I got date and time as two separate characters [1] 2008-04-11 [1] 22:00:00 which correspond to my starting point in time domain. Now I need to produce series over, 5 sec epochs up to end point, say: [1] 2008-04-12 [1] 23:00:00 So something like 2008-04-11 22:00:00 2008-04-11 22:00:05 2008-04-11 22:00:10 . . . 2008-04-12 23:00:00 Is there any strightforward way to do it? Any suggestions? Best, robert -- View this message in context: http://r.789695.n4.nabble.com/date-and-time-tp3905358p3905358.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] aov(variable~group*(speed*person)) but How to get 95% confidence intervals ?
Hey, If I understand correctly, library(gplots) plotmeans(). You might also try TukeyHSD() to see if that gets you where you are trying to go. Good luck! Ken Hutchison On Fri, Oct 14, 2011 at 12:11 PM, Jebb Remelius j...@kin.umass.edu wrote: Greetings and gratitude, I have 19 persons in each group, and each person walks at 3 different speeds. I can do a nice p value and f value with summary command Please help me learn how to report the 95% confidence interval for this anova? This is easier for my fourth, separate condition: preferred walking speed, where there is no speed factor in the model: aov(variable~group) although I compute the 95%CI in SPSS for this model :( Any advice on how to do this for the single speed anova in R, because it is really nice! Thank you Jebb Jebb Remelius PhDc j...@kin.umass.edu 413-545-4959 office 224-645-3490 mobile __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] help to ... import the data from Excel
(Sorry if I'm repeating things: working blind b/c of the nabble-listserv interface) Since you haven't actually told us what package you are using, I'm guessing that your problem seems to be the same as the one discussed here: http://r.789695.n4.nabble.com/ReadWrite-xls-problem-td3078348.html If that's true (i.e, if you are using the xlsReadWrite package) which is where this 11 vs 10 arguments thing tends to show up, then you need to run this set of commands verbatim once (and only once on each machine unless you reinstall R / the package) before trying to read an xls file library(xlsReadWrite) xls.getshlib() Note that this is different than picking a CRAN mirror (I'm not sure how shlibs came up in that process) This should fix the error message. By the way, you also haven't told us what version of R you are running: I, for one, can't get xlsReadWrite on R2.13.1 from CRAN so there may be something else at work here. As David points out, there are also other useful R/xls interface packages that seem to be a little less trouble. Michael On Fri, Oct 14, 2011 at 12:27 PM, Sarah_R_edu sarah_r_...@hotmail.com wrote: eyildiz ... if i installed the package (getshlib) appeared to me to selecting a CRAN mirror ... Any one to be selected ? Note: i am beginner to used the languge R /thanks eyildiz ( my prayers to you)/ http://r.789695.n4.nabble.com/file/n3905447/47756_429620173219_788668219_5025394_743078_n.jpg - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3905447.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] Party package: varimp(..., conditional=TRUE) error: term 1 would require 9e+12 columns
Hi, That's a tough one, I'll do my best and hope a more knowledgeable person will correct me. Since you can measure conditional importance by permuting predictors and re-evaluating importance, perhaps try the randomForest package and examine how your results change based on permutation of each predictor. I understand permutation would take a prohibitively large amount of time for certain applications. Try (clumsy) shortcuts: Some pseudocode First Option myforest=randomForest(y~.,data=df) imp=myforest$importance #Just for your info importance is here. #permute an x newx=sample(x,length(x),replace=F) #make new forest newforest=randomForest(y~newx+all else...) #predict oldx with newforest #If somewhat accurate, problems afoot. Second Option: Predicting your held out variable 1000 times with new forest(pretty quick to do) and examining the quantile of the predicted value relative to the old (non permuted) distribution of the variable, should be uniformly distributed between 0 and 1 if truly random inside the forest (and random outside since we know it has been permuted)... could measure this with Chi-square statistic. #Third option Permute the x's and plot importance for each variable when the others are held out (inferential only) Weak I know, but I hope it helps! Ken Hutchison On Fri, Oct 14, 2011 at 12:06 PM, Jason Roberts jason.robe...@duke.eduwrote: I would like to build a forest of regression trees to see how well some covariates predict a response variable and to examine the importance of the covariates. I have a small number of covariates (8) and large number of records (27368). The response and all of the covariates are continuous variables. A cursory examination of the covariates does not suggest they are correlated in a simple fashion (e.g. the variance inflation factors are all fairly low) but common sense suggests there should be some relationship: one of them is the day of the year and some of the others are environmental parameters such as water temperature. For this reason I would like to follow the advice of Strobl et al. (2008) and try the authors' conditional variable importance measure. This is implemented in the party package by calling varimp(..., conditional=TRUE). Unfortunately, when I call that on my forest I receive the error: varimp(myforest, conditional=TRUE) Error in model.matrix.default(as.formula(f), data = blocks) : term 1 would require 9e+12 columns Does anyone know what is wrong? I noticed a post in June 2011 where a user reported this message and the ultimate problem was that the importance measure was being conditioned on too many variables (47). I have only a small number of variables here so I guessed that was not the problem. Another suggestion was that there could be a factor with too many levels. In my case, all of the variables are continuous. Term 1 (x1 below) is the day of the year, which does happen to be integers 1 ... 366. But the variable is class numeric, not integer, so I don't believe cforest would treat it as a factor, although I do not know how to tell whether cforest is treating something as continuous or as a factor. Thank you for any help you can provide. I am running R 2.13.1 with party 0.9-4. You can download the data from http://www.duke.edu/~jjr8/data.rdata (512 KB). Here is the complete code: load(\\Temp\\data.rdata) nrow(df) [1] 27368 summary(df) y x1 x2 x3 x4 x5 x6 x7 x8 Min. : 0.000 Min. : 1.0 Min. :0. Min. : 1.00 Min. : 52 Min. : 0.008184 Min. :16.71 Min. :0.000 Min. : 0.02727 1st Qu.: 0.000 1st Qu.:105.0 1st Qu.:0. 1st Qu.: 30.00 1st Qu.:1290 1st Qu.: 6.747035 1st Qu.:23.92 1st Qu.:0.000 1st Qu.: 0.11850 Median : 1.282 Median :169.0 Median :0.2353 Median : 38.00 Median :1857 Median :11.310277 Median :26.35 Median :0.0001569 Median : 0.14625 Mean : 5.651 Mean :178.7 Mean :0.2555 Mean : 55.03 Mean :1907 Mean :12.889021 Mean :26.31 Mean :0.0162043 Mean : 0.20684 3rd Qu.: 5.353 3rd Qu.:262.0 3rd Qu.:0.4315 3rd Qu.: 47.00 3rd Qu.:2594 3rd Qu.:18.427410 3rd Qu.:28.95 3rd Qu.:0.0144660 3rd Qu.: 0.20095 Max. :195.238 Max. :366.0 Max. :1. Max. :400.00 Max. :3832 Max. :29.492380 Max. :31.73 Max. :0.3157486 Max. :11.76877 library(HH) output deleted vif(y ~ ., data=df) x1 x2 x3 x4 x5 x6 x7 x8 1.374583 1.252250 1.021672 1.218801 1.015124 1.439868 1.075546 1.060580 library(party) output deleted mycontrols - cforest_unbiased(ntree=50, mtry=3) # Small forest but requires a few minutes myforest - cforest(y ~ ., data=df, controls=mycontrols) varimp(myforest) x1 x2 x3
Re: [R] Split a list
Hi:
Following the lead of others, here's a reproducible example that I
believe achieves what you want.
# Q1:
L - lapply(1:3, function(n)
data.frame(x = rnorm(6), y = rnorm(6), g = rep(1:2, each = 3)))
# Using David's suggestion:
L1 - lapply(L, function(d) subset(d, g == 1L))
L2 - lapply(L, function(d) subset(d, g == 2L))
# Q2:
# Let range 2 to retain in this small example:
# Find the range of the second column of each list component:
sapply(L, function(x) diff(range(x[, 2], na.rm = TRUE)))
# The code retains the data frame if the range of the second
# column is 2, otherwise it is set to NULL:
lapply(L, function(d) if(diff(range(d[, 2], na.rm = TRUE)) 2) d else NULL)
# If you want to collapse the result into a data frame, the base R
approach would be
do.call('rbind', lapply(L, function(d) if(diff(range(d[, 2], na.rm =
TRUE)) 2) d else NULL))
# An equivalent way to do all of this in the plyr package is:
library('plyr')
L1 - llply(L, function(d) subset(d, g == 1L))
L2 - llply(L, function(d) subset(d, g == 2L))
ldply(L, function(d) if(diff(range(d[, 2], na.rm = TRUE)) 2) d else NULL)
There are advantages to naming the list components if this is what you
have in mind, since both ldply() and the rbind from do.call() will
output indicators of which component data frame each observation
belongs; ldply() uses an .id variable to designate the list component
name whereas do.call(rbind, ...) uses rownames to distinguish
observations. For this example, try
names(L) - paste('d', 1:3, sep = '')
and run the code above again to see the difference.
HTH,
Dennis
On Fri, Oct 14, 2011 at 6:06 AM, Juliet Ndukum jpnts...@yahoo.com wrote:
I have a list of dataframes i.e. each list element is a dataframe with three
columns and differing number of rows. The third column takes on only two
values. I wish to split the list into two sublists based on the value of the
third column of the list element.
Second issue with lists as well. I would like to reduce each of the sublist
based on the range of the second column, i.e. if the range of the second
column is greater than twenty for example keep the list element.
Could someone help me with a code to implement these two issues. Thanks in
advance for your help,
JN
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Re: [R] Minimum cutsets
On Thu, Oct 13, 2011 at 7:47 PM, malhomidi mal...@essex.ac.uk wrote: Hi again, I've looked at the links above and I see the development version of the igraph library. I see the src folder implemented in C. Are these source codes available in R or I just would have to use the C code? The reason is that I just started learning R and I really want to stay away from C and C++. From the googlecode page I mentioned, you can download an R source package, that you can compile and install in R. Compiling it for windows is slightly more difficult, so if you need a windows package, I can provide one. Gabor Regards, Mohammed Alhomidi -- View this message in context: http://r.789695.n4.nabble.com/Minimum-cutsets-tp885346p3903287.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. -- Gabor Csardi csa...@rmki.kfki.hu MTA KFKI RMKI __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strucchange Nyblom-Hansen Test?
Thanks a lot for your immediate help and detailed explanation! About one thing I'm not quite clear: When the default fit = glm in gefp() is used: sctest(gefp(Employed ~ Year + GNP.deflator + GNP + Armed.Forces, data = longley, fit = lm), functional = meanL2BB) is this then the original Nyblom's Parameter Stability Test (1989) or is it the joint (Nyblom-)Hansen test with theta = (beta) constant instead of theta = (beta, sigma^2) constant ? -- View this message in context: http://r.789695.n4.nabble.com/strucchange-Nyblom-Hansen-Test-tp3887208p3905669.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to mean for groups of matrix rows?
Dear All Big Thanks! R is wonderful language! -- *Thanks, Andrei* [[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] Counting the number of integers at one swoop
You guys are working too hard. Rgames y - c(0,1,1,3,3,3,5,5,6) Rgames rle(sort(y)) Run Length Encoding lengths: int [1:5] 1 2 3 2 1 values : num [1:5] 0 1 3 5 6 -- - Sent from my Cray XK6 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] sorting dataframe by arbitrary order
This has been dogging me for a while. I've started making a lot of tables
via xtable so the way I want to sort things is not always in alphabetical or
numerical order.
As an example, consider I have a dataframe as follows
set.seed(100)
a - data.frame(V1=sample(letters[1:4],100, replace=T),V2=1:100)
I know I can sort the columns first by V1 first and then by V2 by:
sorted.a - a[do.call(order,a[c('V1','V2')]),]
What I want to do is exactly that but I do not want V1 sorted
alphabetically. Rather, I would like it sorted as 'a','c','d','b'.
I know I could do it with a subset, rbind function but I thought there may
be a more elegant way?
Thanks for the help.
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] non-parametric permutation and signed paired-difference distributions
On 10/14/2011 1:20 PM, Weidong Gu wrote: On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendlyfrien...@yorku.ca wrote: Hi all Consider the classic data below from Darwin on the heights of 15 pairs of zea mays (corn) plants either cross-fertilized or self-fertilized, where the goal is to see if it makes a difference. head(ZeaMays) pair pot cross self diff 11 1 23.500 17.375 6.125 22 1 12.000 20.375 -8.375 33 1 21.000 20.000 1.000 44 2 22.000 20.000 2.000 55 2 19.125 18.375 0.750 66 2 21.500 18.625 2.875 ... I'd like to illustrate two types of non-parametric tests of whether the mean(diff) = 0. (a) Permutation test, where the values of, say self are permuted and diff=cross - self is calculated for each permutation. There are 15! permutations, but a reasonably large number of random permutations would suffice. (b) Test based on assigning each abs(diff) a + or - sign, and calculating the mean(diff). There are 2^15 such possible values, but again, a reasonably large number of random samples would do. What do you mean by 'assigning each abs(diff) a + or - sign, and calculating the mean(diff)'? abs(diff) should be all positive, right? I mean to calculate mean (sign * abs(diff)) where sign is a vector of length 15 composed of some combination of (-1, +1). This is actually what Fisher did. It corresponds to assigning one member of each pair to cross / self -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sorting dataframe by arbitrary order
Set the levels of the factor a$V1 to the order
in which you want them to be sorted. E.g.,
a - data.frame(V1=letters[rep(4:1,2)], V2=1001:1008)
a[do.call(order,a[c('V1','V2')]),]
V1 V2
4 a 1004
8 a 1008
3 b 1003
7 b 1007
2 c 1002
6 c 1006
1 d 1001
5 d 1005
a$V1 - factor(a$V1, levels=c(a,c,d,b))
a[do.call(order,a[c('V1','V2')]),]
V1 V2
4 a 1004
8 a 1008
2 c 1002
6 c 1006
1 d 1001
5 d 1005
3 b 1003
7 b 1007
This means that tables and plots will be ordered in
the way as well. E.g.,
with(a, table(V1, V2))
V2
V1 1001 1002 1003 1004 1005 1006 1007 1008
a00010001
c01000100
d10001000
b00100010
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 Trevor Davies
Sent: Friday, October 14, 2011 12:05 PM
To: r-help@r-project.org
Subject: [R] sorting dataframe by arbitrary order
This has been dogging me for a while. I've started making a lot of tables
via xtable so the way I want to sort things is not always in alphabetical or
numerical order.
As an example, consider I have a dataframe as follows
set.seed(100)
a - data.frame(V1=sample(letters[1:4],100, replace=T),V2=1:100)
I know I can sort the columns first by V1 first and then by V2 by:
sorted.a - a[do.call(order,a[c('V1','V2')]),]
What I want to do is exactly that but I do not want V1 sorted
alphabetically. Rather, I would like it sorted as 'a','c','d','b'.
I know I could do it with a subset, rbind function but I thought there may
be a more elegant way?
Thanks for the help.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] non-parametric permutation and signed paired-difference distributions
If I understand what you want to do, it's simple. 2^15 is small (only about 33000), so you can generate all the possible means (sums, actually) and and find the population quantile for your result. If avals is the vector of 15 absolute values, the complete distribution is: allsums - as.matrix(expand.grid(as.data.frame(matrix(rep(c(-1,1),15), nr =2 %*% avals This was instantaneous on my machine. Cheers, Bert On Fri, Oct 14, 2011 at 12:42 PM, Michael Friendly frien...@yorku.cawrote: On 10/14/2011 1:20 PM, Weidong Gu wrote: On Fri, Oct 14, 2011 at 11:38 AM, Michael Friendlyfrien...@yorku.ca wrote: Hi all Consider the classic data below from Darwin on the heights of 15 pairs of zea mays (corn) plants either cross-fertilized or self-fertilized, where the goal is to see if it makes a difference. head(ZeaMays) pair pot cross self diff 11 1 23.500 17.375 6.125 22 1 12.000 20.375 -8.375 33 1 21.000 20.000 1.000 44 2 22.000 20.000 2.000 55 2 19.125 18.375 0.750 66 2 21.500 18.625 2.875 ... I'd like to illustrate two types of non-parametric tests of whether the mean(diff) = 0. (a) Permutation test, where the values of, say self are permuted and diff=cross - self is calculated for each permutation. There are 15! permutations, but a reasonably large number of random permutations would suffice. (b) Test based on assigning each abs(diff) a + or - sign, and calculating the mean(diff). There are 2^15 such possible values, but again, a reasonably large number of random samples would do. What do you mean by 'assigning each abs(diff) a + or - sign, and calculating the mean(diff)'? abs(diff) should be all positive, right? I mean to calculate mean (sign * abs(diff)) where sign is a vector of length 15 composed of some combination of (-1, +1). This is actually what Fisher did. It corresponds to assigning one member of each pair to cross / self -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help 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. -- 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 [[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] sorting dataframe by arbitrary order
Works great. I did a couple changes so as to not affect the original
data.frame (and had to add levels back b/c I removed them in the original
read.csv).
a - data.frame(V1=letters[rep(4:1,2)], V2=1001:1008)
b - a
levels(b) - unique(a$V1)
b$V1 - factor(b$V1,levels=c('c','d','a','b'))
a.sorted- a[do.call(order,b[c('V1','V2')]),]
Thank you.
-
On Fri, Oct 14, 2011 at 12:50 PM, William Dunlap wdun...@tibco.com wrote:
Set the levels of the factor a$V1 to the order
in which you want them to be sorted. E.g.,
a - data.frame(V1=letters[rep(4:1,2)], V2=1001:1008)
a[do.call(order,a[c('V1','V2')]),]
V1 V2
4 a 1004
8 a 1008
3 b 1003
7 b 1007
2 c 1002
6 c 1006
1 d 1001
5 d 1005
a$V1 - factor(a$V1, levels=c(a,c,d,b))
a[do.call(order,a[c('V1','V2')]),]
V1 V2
4 a 1004
8 a 1008
2 c 1002
6 c 1006
1 d 1001
5 d 1005
3 b 1003
7 b 1007
This means that tables and plots will be ordered in
the way as well. E.g.,
with(a, table(V1, V2))
V2
V1 1001 1002 1003 1004 1005 1006 1007 1008
a00010001
c01000100
d10001000
b00100010
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 Trevor Davies
Sent: Friday, October 14, 2011 12:05 PM
To: r-help@r-project.org
Subject: [R] sorting dataframe by arbitrary order
This has been dogging me for a while. I've started making a lot of tables
via xtable so the way I want to sort things is not always in alphabetical
or
numerical order.
As an example, consider I have a dataframe as follows
set.seed(100)
a - data.frame(V1=sample(letters[1:4],100, replace=T),V2=1:100)
I know I can sort the columns first by V1 first and then by V2 by:
sorted.a - a[do.call(order,a[c('V1','V2')]),]
What I want to do is exactly that but I do not want V1 sorted
alphabetically. Rather, I would like it sorted as 'a','c','d','b'.
I know I could do it with a subset, rbind function but I thought there
may
be a more elegant way?
Thanks for the help.
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Re: [R] non-parametric permutation and signed paired-difference distributions
On 10/14/2011 4:10 PM, Bert Gunter wrote: If I understand what you want to do, it's simple. 2^15 is small (only about 33000), so you can generate all the possible means (sums, actually) and and find the population quantile for your result. If avals is the vector of 15 absolute values, the complete distribution is: allsums - as.matrix(expand.grid(as.data.frame(matrix(rep(c(-1,1),15), nr =2 %*% avals This was instantaneous on my machine. That's beautiful, Bert. Thanks! Here is my fleshed-out example mean(ZeaMays$diff) # complete permutation distribution of diff, for all 2^15 ways of assigning # one value to cross and the other to self allmeans - as.matrix(expand.grid(as.data.frame(matrix(rep(c(-1,1),15), nr =2 %*% abs(ZeaMays$diff) / 15 # upper-tail p-value sum(allmeans mean(ZeaMays$diff)) / 2^15 # two-tailed p-value sum(abs(allmeans) mean(ZeaMays$diff)) / 2^15 hist(allmeans, breaks=64, xlab=Mean difference, cross-self, main=Histogram of all mean differences) abline(v=mean(ZeaMays$diff), col=red, lwd=2) abline(v=-mean(ZeaMays$diff), col=red, lwd=2, lty=2) -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Chunk of code
MATLAB chunk of code
for jj = 1:numOfAA curAAPosInSeqIndex = aaPosInSeqIndex{1,jj};
for kk = 1:K p_c_a_z_given_x_contribution(1,jj,kk) =
sum(sum(p_c_a_z_given_m_x_permuted(:,:,kk) .* curAAPosInSeqIndex, 1),2); % 1 by
numOfAA by K end end
aaPosInSeqIndex is a list of size 24, with each element a
vector. p_c_a_z_given_m_x_permuted is an array of size N=1655 by J=39 by
K=151 and p_c_a_z_given_x_contribution is an array of size 1 by numOfAA = 24 by
K=151
R equivalent:app = t(sapply(1:numOfAA, function(xx) (t(sapply(1:K, function(x)
colSums(p_c_a_z_given_m_x_permuted[,,x]* aaPosInSeqIndex[[xx]])))[,3])))
p_c_a_z_given_x_contribution = app
I would be very grateful if someone could look at the R code and let me know if
there is something missing or if there is a better way of getting the exact
MATLAB output uing R code i.e. could someone write a code that would do what
the MATLAB chunk of code above is doing. The R code does not output the same
MATLAB output.
Thanks in advance for your help.JN
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[R] using do.call to call a list of functions
I am having trouble figuring out how to use do.call to call and run a list of
functions.
for example:
make.draw = function(i){i;function()runif(i)}
function.list = list()
for (i in 1:3) function.list[[i]] = make.draw(i)
will result in
function.list[[1]]()
[1] 0.2996515
function.list[[2]]()
[1] 0.7276203 0.4704813
function.list[[3]]()
[1] 0.9092999 0.7307774 0.4647443
what I want to do is create a function that calls all three functions in the
list at one go. from what I understand as.call() can be used to do this but
I am having trouble connecting the dots and getting 6 uniform random draws
from function.list. thanks, honeyoak.
--
View this message in context:
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and provide commented, minimal, self-contained, reproducible code.
[R] rpois 0
Hello what is the easiest way to generate rpois(m,lambda) but only values greater than 0 and length = m. tanks, knut -- View this message in context: http://r.789695.n4.nabble.com/rpois-0-tp3906239p3906239.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] help to ... import the data from Excel
/Michael Weylandt/ i used read.xls() and i install the packages : xlsReadWrite and getshlib before using this command but, did not work. anyway i waiting for you. thank you so much. (Flowers) - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3906354.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] using do.call to call a list of functions
function.list=c(sin, cos, function(x) tan(x))
for (f in function.list) print(f(pi))
[1] 1.224606e-16
[1] -1
[1] -1.224606e-16
On Fri, Oct 14, 2011 at 5:48 PM, honeyoak honey...@gmail.com wrote:
I am having trouble figuring out how to use do.call to call and run a list
of
functions.
for example:
make.draw = function(i){i;function()runif(i)}
function.list = list()
for (i in 1:3) function.list[[i]] = make.draw(i)
will result in
function.list[[1]]()
[1] 0.2996515
function.list[[2]]()
[1] 0.7276203 0.4704813
function.list[[3]]()
[1] 0.9092999 0.7307774 0.4647443
what I want to do is create a function that calls all three functions in
the
list at one go. from what I understand as.call() can be used to do this but
I am having trouble connecting the dots and getting 6 uniform random draws
from function.list. thanks, honeyoak.
--
View this message in context:
http://r.789695.n4.nabble.com/using-do-call-to-call-a-list-of-functions-tp3906337p3906337.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] using do.call to call a list of functions
package plyr makes it easier,
plyr::each(function.list)(pi)
HTH,
baptiste
On 15 October 2011 11:55, Richard M. Heiberger r...@temple.edu wrote:
function.list=c(sin, cos, function(x) tan(x))
for (f in function.list) print(f(pi))
[1] 1.224606e-16
[1] -1
[1] -1.224606e-16
On Fri, Oct 14, 2011 at 5:48 PM, honeyoak honey...@gmail.com wrote:
I am having trouble figuring out how to use do.call to call and run a list
of
functions.
for example:
make.draw = function(i){i;function()runif(i)}
function.list = list()
for (i in 1:3) function.list[[i]] = make.draw(i)
will result in
function.list[[1]]()
[1] 0.2996515
function.list[[2]]()
[1] 0.7276203 0.4704813
function.list[[3]]()
[1] 0.9092999 0.7307774 0.4647443
what I want to do is create a function that calls all three functions in
the
list at one go. from what I understand as.call() can be used to do this but
I am having trouble connecting the dots and getting 6 uniform random draws
from function.list. thanks, honeyoak.
--
View this message in context:
http://r.789695.n4.nabble.com/using-do-call-to-call-a-list-of-functions-tp3906337p3906337.html
Sent from the R help mailing list archive at Nabble.com.
__
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http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] help to ... import the data from Excel
This is the third time you've said you've done getshlib, but as has been pointed out to you, the command is: xls.getshlib() not just typing getshlib and it must be entered verbatim. The latter (getshlib) doesn't exist that I'm aware of. Either way, could you do xls.getshlib() again and provide the output of that command. Assuming that you're just not typing the code that you did enter, could you provide the output of sessionInfo() after the call to library(xlsReadWrite) for more diagnostics. Michael On Fri, Oct 14, 2011 at 5:57 PM, Sarah_R_edu sarah_r_...@hotmail.com wrote: /Michael Weylandt/ i used read.xls() and i install the packages : xlsReadWrite and getshlib before using this command but, did not work. anyway i waiting for you. thank you so much. (Flowers) - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3906354.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] rpois 0
On 15/10/11 10:15, knut-o wrote: Hello what is the easiest way to generate rpois(m,lambda) but only values greater than 0 and length = m. tanks, knut The rpospois() function from the VGAM package is what you are looking for. I found this by doing: RSiteSearch(truncated Poisson) and scrabbling through the results a bit. [Give a man a fish and you feed him for a day. Teach a man to fish ... and you destroy an eco-system. :-) ] cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to keep a coefficient fixed when using rq {quantreg}?
Hello all, I would like to compute a quantile regression using rq (from the quantreg package), while keeping one of the coefficients fixed. Is it possible to set an offset for rq in quantreg? (I wasn't able to make it to work) Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Irregular 3d objects with rgl
Hello, While exploring if rgl is along the lines of what I need, I checked out demo(rgl) and didn't find quite what I'm looking for, and am therefore seeking additional help/suggestions. The application is geared towards displaying a 3D rendering of a contaminant plume in the subsurface with the following highlights: Once the plume was rendered as a 3D object, a pie-like wedge could be removed (or cut away) exposing the higher concentrations within the plume as 'hotter' colors. About the closest example I could find is here: http://mclaneenv.com/graphicdownloads/plume.jpg Whereas this particular rendering shows a bullet-like object where 3/4 of the object is removed, I would like to try and show something where 3/4 of the object remains, and where the object has been cut away the colors would show concentrations within the plume, just as in the example referenced above. It would seem most software capable of this type of thing is proprietary (and perhaps for good reason if it is a difficult problem to solve). I've put together a very simple 6x6x6 cube with non-zero values internal to it representing the plume. I wondering if an isosurface where conc = 0.01 can be rendered in 3D and then if a bite or wedge can be removed from the 3d object exposing the higher concentrations inside as discussed above? x-c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6) y-c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6) z-c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6) conc-c(0,0,0,0,0,0,0,0,0.1,0.1,0,0,0,0.1,1,1,0.1,0,0,0.1,0.5,1,0.1,0,0,0,0.2,0.2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.05,0.1,0,0,0,0.05,0.8,0.8,0.05,0,0,0.05,0.4,0.8,0.05,0,0,0,0.1,0.1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.05,0,0,0,0,0.6,0.6,0.02,0,0,0,0.2,0.5,0.02,0,0,0,0.05,0.05,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.05,0.2,0,0,0,0,0,0.05,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.02,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) plume-data.frame(cbind(x=x,y=y,z=z,conc=conc)) if it helps to view the concentrations in layer by layer tabular form: Layer 1 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.10 0.10 0.00 0.00 0.00 0.10 1.00 0.50 0.20 0.00 0.00 0.10 1.00 1.00 0.20 0.00 0.00 0.00 0.10 0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 2 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.05 0.05 0.00 0.00 0.00 0.05 0.80 0.40 0.10 0.00 0.00 0.10 0.80 0.80 0.10 0.00 0.00 0.00 0.05 0.05 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.60 0.20 0.05 0.00 0.00 0.05 0.60 0.50 0.05 0.00 0.00 0.00 0.02 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 4 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.05 0.00 0.00 0.00 0.00 0.00 0.20 0.05 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 5 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Layer 6 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -- View this message in context: http://r.789695.n4.nabble.com/Irregular-3d-objects-with-rgl-tp3906573p3906573.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] help to ... import the data from Excel
*/David/* i used the following command : z - read.xls(file=C:\\Users\\user\\Desktop\\LTS.xls,colNames=FALSE,rowNames=FALSE) z - read.table(file=C:\\Users\\user\\Desktop\\LTS.xls) and i have the packages : xlsReadWrite and gdata , my R version is 2.13.2 (2011-09-30) but all these did not got me any result ! - We are all like the bright moon, we still have our darker side -- View this message in context: http://r.789695.n4.nabble.com/help-to-import-the-data-from-Excel-tp3893382p3906511.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] glmmadmb help
Hello everyone, I am using the alpha version of glmmadmb, and it works for most of the time except for one of my models. The weird thing is that it has worked before, a couple of months ago, and for some reason it won't now and nothing has changed. The code is: nbin5-glmmadmb(stainp~beetle.ev+Caged*Section/SegmentT+(1|Site)+(1|Log.code),data=dat1,family=nbinom) And it thinks for a while before giving me the error: Error in glmmadmb(stainp ~ beetle.ev + Caged * Section/SegmentT + (1 | : The function maximizer failed In addition: Warning message: running command './glmmadmb -maxfn 500' had status 1 I have narrowed it down, and the factor it has trouble with is beetle.ev. This is a column of binary data which is presence or absence of beetles. There are only 0's and 1's in this column. The really weird thing is that I have another column for a different beetle (beetle), binary also, which works fine! I have tried re-entering the data in another column, putting the beetle.ev data in the beetle column, and nothing seems to work. Any help would be much appreciated! Thanks, James -- View this message in context: http://r.789695.n4.nabble.com/glmmadmb-help-tp3906469p3906469.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] using do.call to call a list of functions
Thanks for the reference, the each function in the plyr package is exactly what I wanted. -- View this message in context: http://r.789695.n4.nabble.com/using-do-call-to-call-a-list-of-functions-tp3906337p3906574.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] help to ... import the data from Excel
On Oct 14, 2011, at 7:37 PM, Sarah_R_edu wrote: */David/* i used the following command : z - read.xls(file=C:\\Users\\user\\Desktop\ \LTS.xls,colNames=FALSE,rowNames=FALSE) As I pointed out earlier this would have produced an error on my system because the arguemtnts do not exist in eitehr read.xls or read.table and you are asked to report verbatim all error messages. z - read.table(file=C:\\Users\\user\\Desktop\\LTS.xls) and i have the packages : xlsReadWrite and gdata , my R version is 2.13.2 (2011-09-30) but all these did not got me any result ! How do we know this did not get you a result. Any errors? Warnings? What does these show after: ls() # z should be in there str(z) # information about z (It seems possible that you do not understand that objects might get created without any message about that fact.) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using do.call to call a list of functions
On Oct 14, 2011, at 8:14 PM, honeyoak wrote: Thanks for the reference, the each function in the plyr package is exactly what I wanted. I doubt it. You have not looked at the `each` code. Its just a wrapper for a for loop and on StackOverfolw you started out saying that you were rejecting for() loops and sapply() solutions because you thought they were too slow. And you did not want to put any work into doing performance analysis of your 40 or so functions that you planned on testing so, as they say these days good luck with that. -- View this message in context: http://r.789695.n4.nabble.com/using-do-call-to-call-a-list-of-functions-tp3906337p3906574.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
