hi Cameron
It so happens that the particular example P you chose has no
partners under the symmetry you describe.
Can you explain what this is to be used for?
Could you also give an example of what the operation you desire
might look like if it was successful, and
Hello!
I'm looking forward for a package which has included the deviance
information criterion.
Is there any implemenation in R? I could only find an implementation of
the bayesian information criterion (BIC).
Regrads
Alex
--
Alexander Geisler * Moserhofgasse 36/1 * A-8010 Graz
StV
Thanks for the helping links. Now, I worked out that I have to use the lme4
package (with the lmer function) for my analysis. But now I do not
understand the input to the lmer function.
In the lme function (of the nlme package) the correct input would in my case
be:
Dear all,
Has anyone solved the problem of Automatic adjustment of axis. I
checked it in R Site Search where I was delighted to find the
statement of the problem but now solution.
Thanks in Advance
Samir
Samir K.C.
IIASA
International Institute for Applied Systems Analysis
A-2361 Laxenburg,
Hi
On 19 Oct 2006 at 13:19, Ethan Johnsons wrote:
Date sent: Thu, 19 Oct 2006 13:19:40 -0400
From: Ethan Johnsons [EMAIL PROTECTED]
To: Chuck Cleland [EMAIL PROTECTED]
Copies to: r-help@stat.math.ethz.ch
Subject:Re:
Hi,
I want to know where I can get the sweave package
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
- Message d'origine
De : Marc Schwartz [EMAIL PROTECTED]
À : Gabor Grothendieck [EMAIL PROTECTED]
Cc : R-Help r-help@stat.math.ethz.ch
Envoyé
justin bem wrote:
Hi,
I want to know where I can get the sweave package
Sweave() is a function in the utils package which ships with R.
Uwe Ligges
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
- Message d'origine
De : Marc
Hello everyone,
I have successfully made an error bar graph using the points() command
with the arrows() command to maually add on the standard errors.
However, one slightly annoying feature of using this method is that the
points dont line up exactly with the arrows (if you look carefully the
Simon Pickett [EMAIL PROTECTED] writes:
Hello everyone,
I have successfully made an error bar graph using the points() command
with the arrows() command to maually add on the standard errors.
However, one slightly annoying feature of using this method is that the
points dont line up
This question comes up periodically, probably enough to give it a proper
thread and maybe point to this thread for reference (similar to the
'conservative anova' thread not too long ago).
Moving from lme syntax, which is the function found in the nlme package,
to lmer syntax (found in lme4) is
Has anyone solved the problem of Automatic adjustment of axis. I
checked it in R Site Search where I was delighted to find the
statement of the problem but now solution.
If you're interested in learning a new way of making plots, ggplot
(http://had.co.nz/ggplot) does this automatically.
It appears that dotchart is not restoring some of the 'par'
parameters. It only saves some specific one:
oldpar - par(no.readonly=TRUE)
dotchart(1:10, cex = 0.7) # first call
newpar - par(no.readonly=TRUE)
identical(oldpar,newpar)
[1] FALSE
lapply(names(oldpar), function(x)if
It would be very helpful to have a reproducible example, including the OS
and graphics device used.
For example, this may only happen with certain values of 'pch' -- e.g.
some graphics devices (pdf is one) plot circles using completely different
code from squares. And we frequently see
Dear all,
I have looked for an answer for a couple of days, but can't come with any
solution.
I have a set of functions, say:
t0 - function(x) {1}
t1 - function(x) {x}
t2 - function(x) {x^2}
t3 - function(x) {x^3}
I would like to find a way to add up the previous 4 functions and obtain a
In a mdb table, I have a text field with values of 1, 2, When I
use rodbc to read it into R, it becomes numeric. Is it a bug or
something?
Thanks.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Try this:
L - list(function(x) 1, function(x) x, sin, cos)
sumL - function(x) sum(sapply(L, function(f) f(x)))
sumL(pi) # pi
On 10/20/06, James Foadi [EMAIL PROTECTED] wrote:
Dear all,
I have looked for an answer for a couple of days, but can't come with any
solution.
I have a set of
James Foadi [EMAIL PROTECTED] writes:
Dear all,
I have looked for an answer for a couple of days, but can't come with any
solution.
I have a set of functions, say:
t0 - function(x) {1}
t1 - function(x) {x}
t2 - function(x) {x^2}
t3 - function(x) {x^3}
I would like to find a
will this work for you?
t0 - function(x) {1}
t1 - function(x) {x}
t2 - function(x) {x^2}
t3 - function(x) {x^3}
t.l - list(t0,t1,t2,t3)
t.l
[[1]]
function(x) {1}
[[2]]
function(x) {x}
[[3]]
function(x) {x^2}
[[4]]
function(x) {x^3}
arg.val - 4 # evaluate for 4
Hi,
I have a dataframe Wash2005, which has daily weather data. I am doing a
regression by month as follows:
# FM10 Regression by Month
# Plot 12 Month of OFM10, FFM10
for(i in 1:12) {
Temp - subset (Wash2005, MM == i)
assign( paste('Wash2005FM10', strtrim(month.name[i],3),
Here is one way. To have a vectorized version you need to redefine
't0', though
t0 - function(x) {1}
t1 - function(x) {x}
t2 - function(x) {x^2}
t3 - function(x) {x^3}
ttt - list(t0,t1,t2,t3)
rrr - function(x) sum(sapply(seq(along=ttt), function(i) ttt[[i]](x)))
## vectorized version
ttt[[1]]
Has anyone developed a function to generate a forest plot, the one used
a lot in meta-analysis?
Joseph F. Lucke, PhD
Biostatistician
Center for Clinical Research and Evidence-based Medicine
Department of Pediatrics
School of Medicine
University of Texas Health Science Center at Houston
Voice:
Oh yes, you are right. It seems that the default distribution used for
sampling is uniform. And whether the resampling generates a random
distribution or not depends on the distribution being sampled.
Thanks everyone for your help! I appreciate your support very much.
Tom
On 10/19/06, Ted
On Fri, 2006-10-20 at 10:25 -0400, Wensui Liu wrote:
In a mdb table, I have a text field with values of 1, 2, When I
use rodbc to read it into R, it becomes numeric. Is it a bug or
something?
I don't think it's a bug. This behavior is documented.
See ?sqlQuery. Try the as.is=T argument.
Lucke, Joseph F wrote:
Has anyone developed a function to generate a forest plot, the one used
a lot in meta-analysis?
RSiteSearch(forest plot meta-analysis) points you to metaplot in the
rmeta package.
Joseph F. Lucke, PhD
Biostatistician
Center for Clinical Research and Evidence-based
Cameron,
In your example, I think P is the only matrix with 0-1 entries that
has the given row and column sums.
In general, I would solve the problem simulating a Markov Chain. Start
from a given incidence matrix A and iterate the following steps:
(1) select two rows, a, b, and two columns,
Many thanks to those who have answered my question.
Could I ask Gabor and Peter the meaning of:
sum(sapply(ttt,function(f) f(x)))
I gather that a mysterious function f(x) is applied to all components of
ttt, and sum can act on this modified object. But what is exactly f? And how
does the list
Create a function f which, given Species s, calculates the residual sum
of squares and then sapply it to all Species. In your case Species is
the month and the levels of the Species correspond to the months 1:12
# calculate rss for Species s
f - function(s) {
iris.lm - lm(Sepal.Width ~
ttt is a list of functions so each function in ttt is passed in turn to
the anonymous function as argument f.
On 10/20/06, James Foadi [EMAIL PROTECTED] wrote:
Many thanks to those who have answered my question.
Could I ask Gabor and Peter the meaning of:
sum(sapply(ttt,function(f) f(x)))
hi,
i am wondering if i could use lda$scaling (i.e. coeff) to evaluate
variables' importance if all the x's are normalized before put into
model?
thanks.
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
On Fri, 20 Oct 2006, Chuck Cleland wrote:
Lucke, Joseph F wrote:
Has anyone developed a function to generate a forest plot, the one used
a lot in meta-analysis?
RSiteSearch(forest plot meta-analysis) points you to metaplot in the
rmeta package.
There is also a more up-to-date and
Hi there,
I did what Peter Dalgaard very kindly suggested and saved it as a pdf and
viewed it using adobe, which as if by magic resolves the problem. It must
have been a pixel issue with the r graphics device.
Sorry to have wasted your time, thanks for your help, always appreciated.
Simon.
It
An example:
n - 3
f - function(x) x^n
f(2)
# [1] 8
n - 2
f(2)
# [1] 4
f
# function(x) x^n
Ok, I know this is trivial, because function f is foverer bound
to the variable n. But how can I _fix_ n when I define _f_, so
that changing _n_ won't change the function f?
Alberto Monteiro
It sounds like you want to use 'local' to create private variables:
# your example
n - 3
f - function(x) x^n
f(2)
[1] 8
n-1
f(2)
[1] 2
# now using 'local' to make 'n' private
f - local({
+ n - 3
+ function(x) x^n
+ })
f(2)
[1] 8
n - 1
f(2)
[1] 8
Or why not just pass 'n' to the
On Fri, 20 Oct 2006, Alberto Monteiro wrote:
An example:
n - 3
f - function(x) x^n
f(2)
# [1] 8
n - 2
f(2)
# [1] 4
f
# function(x) x^n
Ok, I know this is trivial, because function f is foverer bound
to the variable n. But how can I _fix_ n when I define _f_, so
that changing _n_
Is this what you want?
f - function(x,n=3) x^n
f(2)
[1] 8
n - 2
f(2)
[1] 8
f(2,2)
[1] 4
Alberto Monteiro wrote:
An example:
n - 3
f - function(x) x^n
f(2)
# [1] 8
n - 2
f(2)
# [1] 4
f
# function(x) x^n
Ok, I know this is trivial, because function f is foverer bound
to the
Jim Holtman wrote:
It sounds like you want to use 'local' to create private variables:
Ok, so, in a real world example, I should do something
like this:
n - 3 # but here there is some very complex computations
# that give me a value for n
f - local({
f.n - n
function(x) x^f.n
})
f(2)
# 8
n
May I am missing something, but it seems to me that the easiest way to
solve your problem, if you don't want to change 'n', is to define
f - function(x) x^3
If you want to allow the possibility for 'n' to change, you can
include it as an argument of 'f'
f - function(x,n=3) x^n
Best,
Giovanni
Hello,
I am a chemical student and I make use of 'lme/lmer function'
to handle experiments in split-plot structures.
I know about the mcmcsamp and I think that it's very promissory.
I would like knowing the concept behind of the mcmcsamp function.
I do not want the C code of the MCMCSAMP
Enio Jelihovschi [EMAIL PROTECTED]
Date: Fri, 20 Oct 2006 11:28:12 -0200
Subject: CORRESPONCE ANALYSIS
Dear All
I am new R user, trying to do correspondence analysis using the function mca
of the package MASS. My question is: In the following example
farms.mca - mca(farms, abbrev = T) # Use
What's the best / simplest way to create 95% confidence bands for a
model created with lm() that can be plotted around teh regression
line? I've looked everywhere for this - I guess I must be missing
something.
- Jason
__
On 10/20/06, Doran, Harold [EMAIL PROTECTED] wrote:
This question comes up periodically, probably enough to give it a proper
thread and maybe point to this thread for reference (similar to the
'conservative anova' thread not too long ago).
Moving from lme syntax, which is the function found
jim holtman ([EMAIL PROTECTED]) wrote:
It sounds like you want to use 'local' to create private variables:
Hello,
I hope I do not disturb your discussion very much writing in this
thread but the topic sounded extremely familiar for me :)
My only question is:
Is there a good book or more
On Oct 20, 2006, at 2:46 PM, Jason Horn wrote:
What's the best / simplest way to create 95% confidence bands for a
model created with lm() that can be plotted around teh regression
line? I've looked everywhere for this - I guess I must be missing
something.
- Jason
library(effects)
On Fri, 2006-10-20 at 14:46 -0400, Jason Horn wrote:
What's the best / simplest way to create 95% confidence bands for a
model created with lm() that can be plotted around teh regression
line? I've looked everywhere for this - I guess I must be missing
something.
- Jason
See the
Thank you so much.
I see your point.
ej
On 10/20/06, Petr Pikal [EMAIL PROTECTED] wrote:
Hi
On 19 Oct 2006 at 13:19, Ethan Johnsons wrote:
Date sent: Thu, 19 Oct 2006 13:19:40 -0400
From: Ethan Johnsons [EMAIL PROTECTED]
To: Chuck Cleland
I received a private request for my complete dataset from my previous
questions. In gratitude to the developers of R, and especially to the
helpful members of r-help, I'm happy to attach it. Anyone is free to use
this dataset in any manner they wish, including published books.
Attribution is not
Hello fellow R's,
I'm sure there must be an easy way to do this. But after digging in the
documentation and thinking about it for a while I couldn't figure it
out. I need to get a decreasing recursive vector in. I mean something
like this: if starting at 2, and ending at 6, the vector
On Fri, 2006-10-20 at 12:51 -0700, Julian Burgos wrote:
Hello fellow R's,
I'm sure there must be an easy way to do this. But after digging in the
documentation and thinking about it for a while I couldn't figure it
out. I need to get a decreasing recursive vector in. I mean something
On Fri, 2006-10-20 at 15:11 -0500, Marc Schwartz wrote:
On Fri, 2006-10-20 at 12:51 -0700, Julian Burgos wrote:
Hello fellow R's,
I'm sure there must be an easy way to do this. But after digging in the
documentation and thinking about it for a while I couldn't figure it
out. I need
I have installed Microsoft Vista Release Candidate 1, and R-2.4.0, on a 4
year old DELL box with a 2.26 GHz P4 and 1 gig. It was a clean install R
is the only non-MS program running.
I cannot install packages from CRAN, nor from local zipped files. (I have
R-2.4.0 installed on a Windows XP
On Fri, 2006-10-20 at 15:11 -0500, Marc Schwartz wrote:
On Fri, 2006-10-20 at 12:51 -0700, Julian Burgos wrote:
Hello fellow R's,
I'm sure there must be an easy way to do this. But after digging in the
documentation and thinking about it for a while I couldn't figure it
out. I need
try this:
start.val - 2
end.val - 6500
system.time(res - unlist(lapply(start.val:end.val, :, to = end.val)))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven,
On 10/20/06, Cleber N. Borges [EMAIL PROTECTED] wrote:
Hello,
I am a chemical student and I make use of 'lme/lmer function'
to handle experiments in split-plot structures.
I know about the mcmcsamp and I think that it's very promising
I would like knowing the concept behind of the mcmcsamp
On 10/19/06, Paul Brewin [EMAIL PROTECTED] wrote:
Dear R-listers,
I would like to know if there is a way to programmatically generate
parameter start values for the model y~(a*exp(b*x)+c*exp(d*x)) in R.
I've scoured the help files and archives for nls() and similar searches,
and have read
Charles Annis, P.E. [EMAIL PROTECTED] writes:
I have installed Microsoft Vista Release Candidate 1, and R-2.4.0, on a 4
year old DELL box with a 2.26 GHz P4 and 1 gig. It was a clean install R
is the only non-MS program running.
I cannot install packages from CRAN, nor from local zipped
On Fri, 20 Oct 2006 15:16:12 -0500 Marc Schwartz wrote:
Range - c(2:6)
Gack
Disregard the 'c' and parens there. Left over from a first attempt
at a solution using c(2, 6)...
Like this one:
myseq4 - function(start, end)
unlist(lapply(start:end, function(x) x:end))
which is
This week's eweek has an article on Vista's security and system
administration--I'm guessing (a Linux user guess) that you are running
afoul of Vista's User Account Control feature.
Clint BowmanINTERNET: [EMAIL PROTECTED]
Air Dispersion Modeler INTERNET:
Peter Dalgaard wrote:
Charles Annis, P.E. [EMAIL PROTECTED] writes:
I have installed Microsoft Vista Release Candidate 1, and R-2.4.0, on a 4
year old DELL box with a 2.26 GHz P4 and 1 gig. It was a clean install R
is the only non-MS program running.
I cannot install packages from
A quick question, please.
46 e coli lab samples are tested, 6 of them returned positive.
So, the best point estimate for p is 6/46 = 0.1304348.
For a 95% CI for p, I thought binom.test would give me the correct
result, but it seems it is not the right function to use. What is
the R
On 10/20/2006 4:19 PM, Charles Annis, P.E. wrote:
I have installed Microsoft Vista Release Candidate 1, and R-2.4.0, on a 4
year old DELL box with a 2.26 GHz P4 and 1 gig. It was a clean install – R
is the only non-MS program running.
I normally try to support people who are using R in
Ethan,
You need to explain why you think this is not the right function to use. R
is doing exactly what you are asking it to do. Now is up to you to choose
the methodology you feel is correct.
For a good discussion on your particular issue I recommend you the following
reference:
A.
I'm getting a weird behavior using R 2.5.0 for MacOS X -- I have a csv file
with a properly formatted date/time field, e.g. After reading in the csv
to hourly_met_data, with a date field
hourly_met_data$date - as.POSIXct(hourly_met_data$date)
works exactly as it is supposed to (e.g. Min/max of
On 2006-10-20 19:48, Enio Jelihovschi wrote:
Enio Jelihovschi [EMAIL PROTECTED]
Date: Fri, 20 Oct 2006 11:28:12 -0200
Subject: CORRESPONCE ANALYSIS
Dear All
I am new R user, trying to do correspondence analysis using the function mca
of the package MASS. My question is: In the following
Thanks! Worked like a charm!
hourly_met_data$date -
as.POSIXct(trunc(as.POSIXct(hourly_met_data$date),day))
--j
On 10/20/06 3:31 PM, jim holtman [EMAIL PROTECTED] wrote:
trunc returns a type of POSIXlt. You have to apply 'as.POSIXct' to the
result.
On 10/20/06, Jonathan Greenberg
Is there a way to calculate, say, the mean, min and max using aggregate
using one line of code? Or do I need to call them separately (e.g.
aggregate(...,mean); aggregate(...,min)) and then merge the data back
together?
--j
--
Jonathan A. Greenberg, PhD
NRC Research Associate
NASA Ames Research
Is it possible to use a string as a variable name? For example:
foo=var1
frame$foo # frame is a data frame with with a column titled var1
This does not work, unfortunately. Am I just missing the correct
syntax to make this work?
- Jason
__
On 10/20/2006 7:28 PM, Jason Horn wrote:
Is it possible to use a string as a variable name? For example:
foo=var1
frame$foo # frame is a data frame with with a column titled var1
This does not work, unfortunately. Am I just missing the correct
syntax to make this work?
Yes, you
frame[[foo]]
On 10/20/06, Jason Horn [EMAIL PROTECTED] wrote:
Is it possible to use a string as a variable name? For example:
foo=var1
frame$foo # frame is a data frame with with a column titled var1
This does not work, unfortunately. Am I just missing the correct
syntax to make this
Try summaryBy in package doBy. e.g. using the built in dataset CO2:
summaryBy(uptake ~ Plant, CO2, FUN = c(mean, min, max))
On 10/20/06, Jonathan Greenberg [EMAIL PROTECTED] wrote:
Is there a way to calculate, say, the mean, min and max using aggregate
using one line of code? Or do I need to
I run XP, not Vista, but have always installed R in
c:\Program Files\R\R...
and have never had a problem with the space in Program Files
so I doubt that that would be the problem.
On 20 Oct 2006 22:31:09 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote:
Charles Annis, P.E. [EMAIL PROTECTED]
Hello,
How do I implement a cardinality constraint with constrOptim?
I want to minimize (least square) a%*%x = 4
subject to
x12
x21
x34
count(x1, x2, x3)= 2 (cardinality constraint)
Is there a way to specify binary integer variables with constrOptim?
Here's my code so far:
a -matrix(1:3,1,3)
Try summaryBy in package doBy. e.g. using the built in dataset CO2:
summaryBy(uptake ~ Plant, CO2, FUN = c(mean, min, max))
Or with reshape with a little more work:
cm - melt(CO2, id=1:4)
cast(cm, Type ~ Treatment, c(min,mean,max))
but you get some extra flexibility:
cast(cm,
Problem solved!
Thanks to all and especially Clint Bowman and Uwe Ligges who pointed to the
fly in the ointment, viz. Vista's User Account Control, UAC.
To allow installation of R packages, you must open the Vista Control Panel
and choose Windows Security Center. Then turn the UAC off. This
Hi all,
I have a problem using rdirichlet{gtools}.
For Dir( a1, a2, ..., a_n), its mode can be found at $( a_i -1)/ (
\sum_{i}a_i - n)$;
The means are $a_i / (\sum_{i} a_i ) $;
I tried to study the above properties using rdirichlet from gtools. The code
are:
##
library(gtools)
OK, so that works. It reduces the number of tick marks and labels by
reduction of the x array to elements that are multiples of 100. That
is nice, but it's not what I really want.
I do want tick marks at all elements of x, but only a sparse number of
tick labels. I want to control what the
Hello All,
I found the TukeyHSD() function. Is there a Tukey-Kramer test for
unbalanced data?
Andrew
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
try this:
x - seq(-100,1000,25)
y - x * x
plot(x, y, xaxt = n)
# plot only the 100th values
y - x %% 100 == 0
axis(1, x[y], labels=x[y])
On 10/20/06, Darren Weber [EMAIL PROTECTED] wrote:
OK, so that works. It reduces the number of tick marks and labels by
reduction of the x array to
It looks weird to me too. Try this:
x - seq(-100,1000,25)
y - x * x
plot(x, y, xaxt = n)
axis(1, x, FALSE, tcl = -0.3)
axis(1, x[x %% 100 == 0] )
On 10/20/06, Darren Weber [EMAIL PROTECTED] wrote:
OK, so that works. It reduces the number of tick marks and labels by
reduction of the x
Thank you for the info. It helps.
After all, it would be:
0.1304348-1.96*(sqrt((0.1304348*(1-0.1304348))/46))
[1] 0.03310968
0.1304348+1.96*(sqrt((0.1304348*(1-0.1304348))/46))
[1] 0.2277599
Does R have a function for the calculation above?
ej
On 10/20/06, Francisco J. Zagmutt [EMAIL
Hi,
I am trying to get R to return just the slope of a linear regression line,
but it seems that R has to return both the slope and the name of the slope.
For example,
a=coef(lm(y~miles))
a
(Intercept) miles
360.3778 -7.2875
names(a)
[1] (Intercept) miles
a[1]
(Intercept)
Using the builtin BOD data frame:
as.vector(coef(lm(demand ~ Time, BOD)))[2]
On 10/21/06, tom soyer [EMAIL PROTECTED] wrote:
Hi,
I am trying to get R to return just the slope of a linear regression line,
but it seems that R has to return both the slope and the name of the slope.
For
Gabor,
Thanks for the code example, but it seems that BOD is not needed. I still
don't understand what is going on with the data structure returned by
coef(). The strangness is illustrated by the following example:
a=coef(lm(y~miles))
is.vector(a)
[1] TRUE
a[2]
miles
-7.2875
a=as.vector(a)
On 10/21/06, tom soyer [EMAIL PROTECTED] wrote:
Gabor,
Thanks for the code example, but it seems that BOD is not needed. I still
demand and Time are columns of BOD so if you omit it then it won't know
where they are.
don't understand what is going on with the data structure returned by
Tom,
coef returns a named vector, which is a vector with an extra attribute
called names.
To remove the extra attribute you can:
names(a) - NULL# through the accessor function
[EMAIL PROTECTED] - NULL # directly accessing the attribute names
or by creating a new vector as you did
Also try this to look inside the objects:
cc - coef(lm(demand ~ Time, BOD))
dput(cc)
structure(c(8.52142857142858, 1.72142857142857), .Names = c((Intercept),
Time))
cc2 - as.vector(cc)
dput(cc2)
c(8.52142857142858, 1.72142857142857)
from which we see that the only difference in their
Thanks Christos and Gabor. I didn't know there is such a thing called named
vector in R. Very cool.
Tom
On 10/21/06, Christos Hatzis [EMAIL PROTECTED] wrote:
Tom,
coef returns a named vector, which is a vector with an extra attribute
called names.
To remove the extra attribute you can:
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