search for: lnscntmcs01

Displaying 9 results from an estimated 9 matches for "lnscntmcs01".

2005 Nov 14
2
change some levels of a factor column in data frame according to a condition
...0.24368402 3 1 c -0.24958322 4 1 small -1.37577955 5 1 e -0.01713288 6 2 f -1.25203573 7 2 small -1.94348533 8 2 h -0.16041719 9 2 small -1.91572616 10 2 j -0.20256478 Thank you for advice! Markus Gesmann ************LNSCNTMCS01*************************************************** The information in this E-Mail and in any attachments is CON...{{dropped}}
2005 Apr 14
2
Legend in xyplot two columns
...John","Paul","George","Ringo", "The Beatles")), columns=3, ) xyplot(1~1, key=schluessel2) So I think each list has to have 6 items, but some with "no" content. How do I do this? Thank you very much! Markus ************LNSCNTMCS01*************************************************** The information in this E-Mail and in any attachments is CON...{{dropped}}
2005 Mar 14
1
'pch' plot symbol with more than one character
R-help, Argument 'pch' in 'plot' can only represent a single character. Is it possible to represent, let's say, two instead? Thanks in advance. I'm running on Windows Xp > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 0.1
2005 Feb 24
1
Place more than one key with xyplot
...key1) xyplot(1~1, key = key2) Unfortunately I don't really understand which "fun" (for legend) I have to use. The following line does not work. xyplot(1~1, legend = list(right=list(key1, fun="grob"),top=list(key2, fun="grob"))) Kind Regards Markus ************LNSCNTMCS01*************************************************** The information in this E-Mail and in any attachments is CONFIDENTIAL and may be privileged. If you are NOT the intended recipient, please destroy this message and notify the sender immediately. You should NOT retain, copy or use this E-mail for...
2005 May 31
1
GLM question
I am unfamiliar with R and I’m trying to do few statistical things like GLM and GAM with it. I hope my following questions will be clear enough: My datas ( y(i,j ))are run off triangles for example : J=1 J=2 J=3 I=1 1 2 3 I=2 4 5 I=3 6 My model is : E[y(i,j)] =m(i,j) Var[y(i,j)] =constant *m(i,j) Log(m(i,j)) = eta (i,j) eta (i,j) = c + alpha(i)
2006 Jan 26
2
footnote in postscript lattice
I would like to add a footnote to this graph but do not see a "footnote" command in the package:lattice documentation. I would like to note the "span=.8" as the footnote. postscript(file= ?C:/Documents and Settings/dsonneborn/My Documents/Slovak/output/pcb_tables/smooth_PCB_lines_four.ps?, bg=?transparent?, onefile=FALSE, pointsize=20,paper=?letter?, horizontal=TRUE,
2005 Mar 08
1
To convert an adjacency list model into a nested set model
...& rgt > 7) emp lft rgt 1 Albert 1 12 3 Chuck 4 11 In the site mentioned above John provides also some code to transform a adjacency list model into a nested set model. Does somebody know if there is already a package for this in R? Kind Regards Markus Gesmann ************LNSCNTMCS01*************************************************** The information in this E-Mail and in any attachments is CONFIDENTIAL and may be privileged. If you are NOT the intended recipient, please destroy this message and notify the sender immediately. You should NOT retain, copy or use this E-mail for...
2005 Mar 22
5
r under linux: creating high quality bmp's for win users
Hi I produce graphics with R under linux, but my collaborators often use windows and cannot import eps pics e.g. in msword what is the standard way to get e.g. bmp's with the same quality as eps. going the way: creating eps, convert eps2bmp using 'convert' doesn't yield good enough bmp's thanks for a short hint cheers christoph
2005 May 18
2
R -SQL
Hello, I've got a problem in a sql query! for(j in 1:length(criteria$Title)) { graphe_par<-sqlQuery(channel,"select q.type,crit.Title, r.Value from criteria crit, reply r,question_reply qr, question q, question_criteria qc, form_question fq where qr.reply=r.ID and qr.question=q.ID and qc.question=q.ID and crit.ID=qc.criteria and fq.question=q.ID and fq.form=4 and