Re: [R] Fitting models in a loop
Murray,
How about creating an empty list and filling it during your loop:
mod - list()
for (i in 1:6) {
mod[[i]] - lm(y ~ poly(x,i))
print(summary(mod[[i]]))
}
All your models are than stored in one object and you can use lapply to
do something on them, like:
lapply(mod, summary) or lapply(mod, coef)
Kind Regards
Markus Gesmann
FPMA
Lloyd's Market Analysis
Lloyd's * One Lime Street * London * EC3M 7HA
Telephone +44 (0)20 7327 6472
Facsimile +44 (0)20 7327 5718
http://www.lloyds.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: 01 August 2006 06:16
To: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch
Subject: Re: [R] Fitting models in a loop
Murray,
Here is a general paradigm I tend to use for such problems. It extends
to fairly general model sequences, including different responses, c
First a couple of tiny, tricky but useful functions:
subst - function(Command, ...) do.call(substitute, list(Command,
list(...)))
abut - function(...) ## jam things tightly together
do.call(paste, c(lapply(list(...), as.character), sep = ))
Name - function(...) as.name(do.call(abut, list(...)))
Now the gist.
fitCommand - quote({
MODELi - lm(y ~ poly(x, degree = i), theData)
print(summary(MODELi))
})
for(i in 1:6) {
thisCommand - subst(fitCommand, MODELi = Name(model_, i), i =
i)
print(thisCommand) ## only as a check
eval(thisCommand)
}
At this point you should have the results and
objects(pat = ^model_)
should list the fitted model objects, all of which can be updated,
summarised, plotted, c, because the information on their construction
is all embedded in the call.
Bill.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Murray Jorgensen
Sent: Tuesday, 1 August 2006 2:09 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Fitting models in a loop
If I want to display a few polynomial regression fits I can do something
like
for (i in 1:6) {
mod - lm(y ~ poly(x,i))
print(summary(mod))
}
Suppose that I don't want to over-write the fitted model objects,
though. How do I create a list of blank fitted model objects for later
use in a loop?
Murray Jorgensen
--
Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html
Department of Statistics, University of Waikato, Hamilton, New Zealand
Email: [EMAIL PROTECTED]Fax 7 838 4155
Phone +64 7 838 4773 wkHome +64 7 825 0441Mobile 021 1395 862
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Re: [R] trim function in R
See ?substr and ?nchar
Try:
substr(Hello World, 0, nchar(Hello World)-2)
Regards
Markus Gesmann
FPMA
Lloyd's Market Analysis
Lloyd's * One Lime Street * London * EC3M 7HA
Telephone +44 (0)20 7327 6472
Facsimile +44 (0)20 7327 5718
http://www.lloyds.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Wade Wall
Sent: 24 July 2006 14:42
To: r-help@stat.math.ethz.ch
Subject: [R] trim function in R
Hi all,
I am looking for a function in R to trim the last two characters of an
8 character string in a vector. For example, I have the codes
37-079-2, 370079-3,37-079-8 and want to trim them to 37-079 by
removing the last two characters. Is sub the correct function to use,
and if so how can I specify trimming the last 2 characters? I have
read the help file, but can't quite figure out how to do it.
Thanks,
Wade
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Re: [R] about overdispersed poisson model
Dear Chi Kai,
Three years ago there was a similar thread.
At that time David Firth offered a solution for the quasipoission
problem with negative observations, see:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/16143.html
I remember that his code gave you slightly different answers than the
example in England and Verrall's paper.
Kind Regards
Markus Gesmann
FPMA
Lloyd's Market Analysis
Lloyd's * One Lime Street * London * EC3M 7HA
Telephone +44 (0)20 7327 6472
Facsimile +44 (0)20 7327 5718
http://www.lloyds.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of g0354502
Sent: 10 July 2006 05:26
To: r-help@stat.math.ethz.ch
Subject: [R] about overdispersed poisson model
Dear R users
I have been looking for functions that can deal with overdispersed
poisson
models. According to actuarial literature (England Verall, Stochastic
Claims
Reserving in General Insurance , Institute of Actiuaries 2002) this can
be handled through the
use of quasi likelihoods instead of normal likelihoods. However, we see
them frequently
in this type of data, and we would like to be able to fit the model
anyway.
If it is possible, would you please show me how to find the
corresponding package and utilize them?
Best Regards,
Chi Kai
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Re: [R] The beautiful math plot
See:
?image
Markus Gesmann
FPMA
Lloyd's Market Analysis
Lloyd's * One Lime Street * London * EC3M 7HA
Telephone +44 (0)20 7327 6472
Facsimile +44 (0)20 7327 5718
http://www.lloyds.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Daniel Yang
Sent: 26 April 2006 13:39
To: R-help@stat.math.ethz.ch
Subject: [R] The beautiful math plot
Dear R-help,
How can I replicate the beautiful math plot found in the right-hand side
of http://www.r-project.org/screenshots/desktop.jpg? I tried the
following code but didn't obtain something as beautiful.
r - seq(-10, 10, len=100)
y - cos(r^2)*exp(-r/6)
par(pty=s)
plot(r,y,type=l)
Thanks in advance!
Yung-jui Yang
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Re: [R] footnote in postscript lattice
Here is one example using the page option in xyplot:
library(lattice)
library(grid)
add.footnote - function(string=Hello World, col=grey,
lineheight=0.5, cex=0.7){
grid.text(string,
x=unit(1, npc) - unit(1, mm),
y=unit(1, mm), just=c(right, bottom),
gp=gpar(col=col,lineheight=lineheight, cex=cex))
}
xyplot(1~1, page=function(n){ add.footnote()})
Markus Gesmann
FPMA
Lloyd's Market Analysis
Lloyd's * One Lime Street * London * EC3M 7HA
Telephone +44 (0)20 7327 6472
Facsimile +44 (0)20 7327 5718
http://www.lloyds.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dean Sonneborn
Sent: 26 January 2006 23:46
To: r-help@stat.math.ethz.ch
Subject: [R] 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, family=Helvetica,font=Helvetica)
xyplot(AWGT ~ lipid_adj_lpcb2_cent | malex*romanix, data=pcb_graph3,
auto.key = list(lines = TRUE, points = TRUE), ylab=Birth
Weight, xlab=Lipid Adjusted PCB,
par.settings =
list(superpose.symbol = list(col = colr, pch = plotchar),
superpose.line = list(col = colr, pch = plotchar, lty = 1)),
type=c(p, smooth), span=.8)
dev.off()
thanks
--
Dean Sonneborn, MS
Programmer Analyst
Department of Public Health Sciences
University of California, Davis
(530) 754-9516
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[R] change some levels of a factor column in data frame according to a condition
Dear R-users,
I am looking for an elegant way to change some levels of a factor column
in data frame according to a condition.
Lets look at the following data frame:
data.frame(crit1=gl(2,5), crit2=factor(letters[1:10]), x=rnorm(10))
crit1 crit2 x
1 1 a -1.06957692
2 1 b 0.24368402
3 1 c -0.24958322
4 1 d -1.37577955
5 1 e -0.01713288
6 2 f -1.25203573
7 2 g -1.94348533
8 2 h -0.16041719
9 2 i -1.91572616
10 2 j -0.20256478
Now I would like to find for each level in crit1 the two smallest values
of x and change the levels of crit2 to small, so the result would look
like this:
crit1 crit2 x
1 1 small -1.06957692
2 1 b 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
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RE: [R] GLM question
It appears that you try to do some insurance reserving analysis.
Have a look at http://finzi.psych.upenn.edu/R/Rhelp02a/archive/15315.html, this
might be helpful.
Regards
Markus
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Laetitia Mestdagh
Sent: 31 May 2005 14:44
To: r-help@stat.math.ethz.ch
Subject: [R] 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) + beta(j)
The y(i,j) are the response and they have no specified distribution.
Here is what I did and I'm not getting the right results:
y1-c(1,0,0,0,0)
y2-c(1,0,0,1,0)
y3-c(1,0,0,0,1)
y4-c(1,1,0,0,0)
y5-c(1,1,0,1,0)
y6-c(1,0,1,0,0)
C-matrix(nrow = 6, ncol = 5, byrow= TRUE)
C[1,]-y1
C[2,]-y2
C[3,]-y3
C[4,]-y4
C[5,]-y5
C[6,]-x6
m-c(1,2,3,4,5,6)
Cdata-data.frame(C[,1],C[,2],C[,3],C[,4],C[,5])
fmp-glm(m~C,family = quasipoisson(link = log),data=Cdata)
fitted.values(fmp)
123456
1.25 1.75 3.00 3.75 5.25 6.00
So my question are : - Why are the fitted wrong (except for 3 and 6)?
- Is the quasipoisson the right family for my model?
I am a little bit lost and not an expert of R, so I thank in advance for any
kind of advice
Laetitia
-
ils, photos et vidéos !
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RE: [R] R -SQL
Your code looks more like Visual Basic rather than R.
What you want is:
for(j in 1:length(criteria$Title)){
sqlstring - paste(select q.type,crit.Title, r.Value from criteria crit, reply
r,question_reply qr, question q, question_criteria qc, form_question fq where
qr.reply=r.ID and qr.question=q.ID and qc.question=q.ID and crit.ID=qc.criteria
and fq.question=q.ID and fq.form=4 and crit.Title=', criteria$Title[j], ';,
sep=)
graphe_par-sqlQuery(channel,sqlstring)
}
See ?paste
Regards
Markus
Your SQL statement seems to be wrong.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Navarre Sabine
Sent: 18 May 2005 13:03
To: r-help@stat.math.ethz.ch
Subject: [R] R -SQL
Hello,
I've got a problem in a sql query!
for(j in 1:length(criteria$Title))
{
graphe_par-sqlQuery(channel,select q.type,crit.Title, r.Value from
criteria crit, reply r,question_reply qr, question q, question_criteria qc,
form_question fq where qr.reply=r.ID and qr.question=q.ID and qc.question=q.ID
and crit.ID=qc.criteria and fq.question=q.ID and fq.form=4 and crit.Title=
criteria$Title[j] ;)
}
criteria$Title
[1] Content Logistic Trainer Supply
User contribution
Levels: Content Logistic Supply Trainer User contribution
The error is:
Error in 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 crit.Title= :
operations are possible only for numeric or logical types
In addition: Warning message:
not meaningful for factors in: Ops.factor(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
crit.Title=,
please help me,
Thanks a lot!
Sabine
-
ils, photos et vidéos !
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[R] Legend in xyplot two columns
Dear R-Help
I have some trouble to set the legend in a xyplot into two rows.
The code below gives me the legend in the layout I am looking for, I
just rather have it in two rows.
library(lattice)
schluessel - list(
points=list( col=red, pch=19, cex=0.5 ),
text=list(lab=John),
lines=list(col=blue),
text=list(lab=Paul),
lines=list(col=green),
text=list(lab=George),
lines=list(col=orange),
text=list(lab=Ringo),
rectangles = list(col= #CC, border=FALSE),
text=list(lab=The Beatles),
)
xyplot(1~1, key=schluessel)
The next code gives me two rows, but repeates all the points,lines, and
rectangles.
schluessel2 - list(
points=list( col=red, pch=19, cex=0.5 ),
lines=list(col=c(blue, green, orange)),
rectangles = list(col= #CC, border=FALSE),
text=list(lab=c(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
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Re: [R] Legend in xyplot two columns
Thanks Deepayan! Your solution does excatly what I want. Further experiments and thoughts on my side brought me also to a solution. If I use the option rep=FALSE, and plot the bullit with lines and split the lines argument into two groups it gives me the same result, as every item in the key list starts a new column. library(lattice) key - list( rep=FALSE, lines=list(col=c(red, blue), type=c(p,l), pch=19), text=list(lab=c(John,Paul)), lines=list(col=c(green, red), type=c(l, l)), text=list(lab=c(George,Ringo)), rectangles = list(col= #CC, border=FALSE), text=list(lab=The Beatles), ) xyplot(1~1, key=key) But your solution is much more felxible! Kind Regards Markus -Original Message- 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 any purpose, nor disclose all or any part of its contents to any other person or persons. Any views expressed in this message are those of the individual sender, EXCEPT where the sender specifically states them to be the views of Lloyd's. Lloyd's may monitor the content of E-mails sent and received via its network for viruses or unauthorised use and for other lawful business purposes. Lloyd's is authorised under the Financial Services and Markets Act 2000 From: Deepayan Sarkar [mailto:[EMAIL PROTECTED] Sent: 14 April 2005 16:01 To: r-help@stat.math.ethz.ch Cc: Gesmann, Markus Subject: Re: [R] Legend in xyplot two columns On Thursday 14 April 2005 05:30, Gesmann, Markus wrote: Dear R-Help I have some trouble to set the legend in a xyplot into two rows. The code below gives me the legend in the layout I am looking for, I just rather have it in two rows. library(lattice) schluessel - list( points=list( col=red, pch=19, cex=0.5 ), text=list(lab=John), lines=list(col=blue), text=list(lab=Paul), lines=list(col=green), text=list(lab=George), lines=list(col=orange), text=list(lab=Ringo), rectangles = list(col= #CC, border=FALSE), text=list(lab=The Beatles), ) xyplot(1~1, key=schluessel) The next code gives me two rows, but repeates all the points,lines, and rectangles. schluessel2 - list( points=list( col=red, pch=19, cex=0.5 ), lines=list(col=c(blue, green, orange)), rectangles = list(col= #CC, border=FALSE), text=list(lab=c(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? You could try using col=transparent to suppress things, but that's not a very satisfactory solution. The function to create the key is simply not designed to create unstructured legends like this. However, you can create an use an arbitrary ``grob'' (grid graphics object) for a legend, e.g.: ##- library(grid) library(lattice) fl - grid.layout(nrow = 2, ncol = 6, heights = unit(rep(1, 2), lines), widths = unit(c(2, 1, 2, 1, 2, 1), c(cm, strwidth, cm, strwidth, cm, strwidth), data = list(NULL, John, NULL, George, NULL, The Beatles))) foo - frameGrob(layout = fl) foo - placeGrob(foo, pointsGrob(.5, .5, pch=19, gp = gpar(col=red, cex=0.5)), row = 1, col = 1) foo - placeGrob(foo, linesGrob(c(0.2, 0.8), c(.5, .5), gp = gpar(col=blue)), row = 2, col = 1) foo - placeGrob(foo, linesGrob(c(0.2, 0.8), c(.5, .5), gp = gpar(col=green)), row = 1, col = 3) foo - placeGrob(foo, linesGrob(c(0.2, 0.8), c(.5, .5), gp = gpar(col=orange)), row = 2, col = 3) foo - placeGrob(foo, rectGrob(width = 0.6, gp = gpar(col=#CC, fill = #CC)), row = 1, col = 5) foo - placeGrob(foo, textGrob(lab = John), row = 1, col = 2) foo - placeGrob(foo, textGrob(lab = Paul), row = 2, col = 2) foo - placeGrob(foo, textGrob(lab = George), row = 1, col = 4) foo - placeGrob(foo
RE: [R] 'pch' plot symbol with more than one character
This might do what you want:
plot.new()
plot.window(xlim=c(1,10), ylim=c(1,10))
text(x=1:10, y=1:10, labels=Hallo)
Look at ?text
Regards
Markus
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Luis Ridao Cruz
Sent: 14 March 2005 10:37
To: r-help@stat.math.ethz.ch
Subject: [R] '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
major2
minor0.1
year 2004
month11
day 15
language R
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[R] To convert an adjacency list model into a nested set model
Dear R-help I am wondering if somebody wrote some code to convert an adjacency list model into a nested set model. In principal I want to do the same as John Celko mentioned it here with SQL: http://groups.google.co.uk/groups?hl=enlr=lang_enselm=8j0n05%24n31%241 %40nnrp1.deja.com Assume you have a tree structure like this Albert / \ / \ BertChuck /| \ / | \ /| \ / | \ DonnaEddie Fred in an adjacency list model: emp=c(Albert, Bert, Chuck, Donna, Eddie, Fred) boss=c(NA, Albert, Albert, Chuck, Chuck, Chuck) print(Personnel-data.frame(emp, boss)) emp boss 1 Albert NA 2 Bert Albert 3 Chuck Albert 4 Donna Chuck 5 Eddie Chuck 6 Fred Chuck Then it is quite hard to find the all the supervisors of one employee. John's suggestion is to convert the adjacency list model into a nested set model. The organizational chart would look like this as a directed graph: Albert (1,12) /\ /\ Bert (2,3)Chuck (4,11) /| \ / | \ /| \ / | \ Donna (5,6) Eddie (7,8) Fred (9,10) The data is than stored in the following form: lft=c(1,2,4,5,7,9) rgt=c(12,3,11,6,8,10) print(Per-data.frame(emp, lft, rgt)) emp lft rgt 1 Albert 1 12 2 Bert 2 3 3 Chuck 4 11 4 Donna 5 6 5 Eddie 7 8 6 Fred 9 10 To find now the supervisor of an employee all you have to do is to look where the employees lft figure is between lft and rgt. The supervisors of Eddie are therefore subset(Per, lft 7 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 any purpose, nor disclose all or any part of its contents to any other person or persons. Any views expressed in this message are those of the individual sender, EXCEPT where the sender specifically states them to be the views of Lloyd's. Lloyd's may monitor the content of E-mails sent and received via its network for viruses or unauthorised use and for other lawful business purposes. Lloyd's is authorised under the Financial Services and Markets Act 2000 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
