This thread strikes me as pretty far off-topic for a forum dedicated to
software support on R.
https://www.r-project.org/mail.html#instructions
"The ‘main’ R mailing list, for discussion about problems and solutions
using R, announcements (not covered by ‘R-announce’ or ‘R-packages’,
see above), a
To check whether the data are being read in appropriately, what happens
when you plot the distribution of each of the independent variables on
the respective systems?
-A
On Wed, 5 Jun 2019 12:32:28 +0200
Olivier Crouzet wrote:
> Hi,
>
> 32bit vs. 64bit systems?
>
> Another thing I would look
Hi all,
Following some updates to R that I received via Synaptic Package
Manager on Ubuntu 16.04 (looks like I now have R 3.4.4-1xenial0), I
have been unable to reinstall rgdal, and I need help.
Initially I was getting error messages about dependencies on GDAL
1.11.4, but after following instruct
I'm trying to develop a linear model for crop productivity based on
variables published as part of the SSURGO database released by the
USDA. My default is to just run lm() with continuous predictor
variables as numeric, and discrete predictor variables as factors, but
some of the discrete variable
Hello.
I am running into difficulties running some older scripts I produced in
2014 to handle raster data using the R package "raster". I have the
feeling that some behavior has changed with an upgrade but don't
remember what version I was using when I wrote the scripts. Can
somebody help me tro
. - a:b), which
means I have to translate a1:b1 somehow to a:b.
Overall, I want to remove any continuous/quantitative predictor if any
associated coefficient cannot be estimated.
Is that clearer now?
KR,
-Thorn
-Original Message-
From: Eik Vettorazzi [mailto:e.vettora...@uke.de]
Sen
cient of a LM back to
the term, something like:
termFromCoef("a1") ## a1
termFromCoef("a1:b1") ## a:b
With this I could simply translate the rownames from alias into the terms
needed for the model update.
Thanks for your help.
Kind Regards,
Thorn Thaler
NRC Lau
tructure for spatial access methods [at least I learnt something
new ;)])
Any pointer to the right function is highly appreciated.
Cheers,
Thorn Thaler
NRC Lausanne
Applied Mathematics
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t do that at
all and I would be curious to hear these things as well. For now I am yet
interested to know how I can remove layers of a plot conveniently.
Thanks for your help!
Kind Regards,
Thorn Thaler
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https:/
\\U\\2",
gsub("\\.", " ", gsub("\\.f", "", txt)),
perl = TRUE)
Thanks for your input!
Kind Regards,
Thorn Thaler
NRC Lausanne
Applied Mathematics
+41 21 785 8220
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d <- d[-c(2, 21)]
Thanks for your input, I am looking forward to your suggestions.
Kind Regards,
Thorn Thaler
Mathematician
Applied Mathematics
Nestec Ltd,
Nestlé Research Center
PO Box 44
CH-1000 Lausanne 26
Phone: +41 21 785 8220
Fax: +41 21 785 9486
s the root
cause?
Thanks for your help!
KR,
-Thorn
> -Original Message-
> From: arun [mailto:smartpink...@yahoo.com]
> Sent: Mittwoch, 31. Oktober 2012 13:15
> To: Thaler,Thorn,LAUSANNE,Applied Mathematics
> Cc: R help
> Subject: Re: [R] aggregate.formula: formula fro
expected, and where my mistake lies? (that means in
particular I am not asking for a solution of how to get the thing done - there
are plenty of alternatives - but instead to understand why this very approach
does not work)
Thanks for your help!
Kind Regards,
Thorn Thaler
Mathematician
Applied
(that is all the three items in one legend)? For that I tried to assign the
same names to the legends but this did not work either.
So any help would be highly appreciated.
Kind Regards,
Thorn Thaler
Mathematician
Applied Mathematics
Nestec Ltd,
Nestlé Research Center
PO Box 44
CH-1000 La
Well, that is exactly what I wanted to have. And you were right, it had
something to do with package versions. So I updated R and the plot looks
exactly the way I wanted it. Thanks a lot for your help and time.
KR,
-Thorn
> -Original Message-
> From: John Kane [mailto:
igned meaning that all
green plots seem to have a different position on the x-axis, while all the
green points for x == "A" should align exactly with "A". Am I clearer now?
KR,
-Thorn
> -Original Message-
> From: John Kane [mailto:jrkrid...@inbox.com]
> Sen
tions on the x-axis for geom_boxplot determined? Any
ideas?
Thanks for the help, anyways.
KR,
-Thorn
> -Original Message-
> From: John Kane [mailto:jrkrid...@inbox.com]
> Sent: Montag, 2. Juli 2012 15:04
> To: Thaler,Thorn,LAUSANNE,Applied Mathematics; r-help@r-project.o
Yes indeed. Sorry for the typo, I just added the library(ggplot) thing
afterwards and I did not check for the spelling. So it should read as
library(ggplot2) and there the issue is still unsettled.
Thx for pointing that out.
KR,
-Thorn
> -Original Message-
> From: Rui Ba
eas?
Kind Regards,
Thorn Thaler
Mathematician
Applied Mathematics
Nestec Ltd,
Nestlé Research Center
PO Box 44
CH-1000 Lausanne 26
Phone: +41 21 785 8220
Fax: +41 21 785 9486
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valid column selectors
->8-
So my question is, whether there are smarter/better/easier/more R-like
ways of doing that?
Any input appreciated.
KR,
-Thorn
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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.
unlist(tapply(grp, grp, seq_along))
xyplot(y ~ index | factor(grp), xlab = "Index")
This should work, but it seems to be a rather elaborate solution,
especially since an index plot is nothing too fancy.
So maybe I'm not seeing the wood for trees, but does anybody know an
easier way?
f these components [i.e. xlim, ylim, xat, yat, dx and dy]; any
missing component will be replaced by the corresponding default."
I'd understand that if I do not specify ylim it is calculated
automatically? Not a big thing though, but it seems to me t
the name of the data
frame is changed earlier?
Finally, your suggestion with
> update(models[[1]], . ~ ., data = model.frame(models[[1]]))
solved all the issues (and I was wondering why I did not try it out
myself, so obviously I was not seeing the wood for the trees). So,
thanks a lot for
LM object within a function, which could be
processed outside this particular function in the usual way? Or is it
simply a bug in update?
Any help highly appreciated.
Thanks,
-Thorn
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rfect for making
comparisons!
Thanks again to everybody who's tried to help me out on this!
Alexandra
On Thu, 2011-06-23 at 21:29 +0200, Jan van der Laan wrote:
> Alexandra,
>
> Have a look at add1 and drop1.
>
> Regards,
> Jan
>
>
> On 06/23/2011 07:32 PM, Ale
Here's a more general question following up on the specific question I
asked earlier:
Can anybody recommend an R command other than mle.aic() (from the wle
package) that will give back a ranked list of submodels? It seems like
a pretty basic piece of functionality, but the closest I've been able
On Thu, 2011-06-23 at 09:29 -0400, Alexandra Thorn wrote:
> Ok, here's some example code showing how I get different output for
AIC
> vs. mle.aic(). Now that I've taken another look at the independent
> variables, I'm wondering whether missing values in one of the
vari
01 1 1 1 -57.34
[20,] 01 1 1 0 -56.35
Printed the first 20 best models
R> AIC(lm(y1~xA)) # Model 1 above
[1] -120.3801
R> AIC(lm(y1~xA+x15)) # Model 2 above
[1] -110.8642
R> AIC(lm(y1~xA+x5)) # Model 3 above
[1] -118.9906
On Thu, 2011-06-23 at 09:0
>
>
> Dr. Rubén Roa-Ureta
> AZTI - Tecnalia / Marine Research Unit
> Txatxarramendi Ugartea z/g
> 48395 Sukarrieta (Bizkaia)
> SPAIN
>
>
>
> > -Mensaje original-
> > De: r-help-boun...@r-project.org
> > [mailto:r-h
I know this a newbie question, but I've only just started using AIC for
model comparison and after a bunch of different keyword searches I've
failed to find a page laying out what the differences are between the
AIC scores assigned by AIC() and mle.aic() using default settings.
I started by usin
ng like:
d <- data.frame(A=1:3, B=2:4)
Reduce(cbind, mapply(function (col, nam) {
td <- as.data.frame(do.call(cbind, rep(list(col), 3)))
names(td) <- paste(nam, 1:3, sep = "_")
td}, d, names(d), SIMPLIFY=FALSE))
Any suggestions? Or is it alre
perm <- sample(values)
}
tmp <- perm[as.numeric(as.factor(x))]
dim(tmp) <- dim(x)
list(x = x, perm.x = tmp, perm = perm)
}
Thanks.
> -Original Message-
> From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
> Sent: mardi 8 mars 2011 16:21
> To: Thaler,Thor
we would use
perm <- c(2,1,3)
all.equal(m.perm, permutateMatrix(m, perm)$mat.perm) # TRUE
What do you think of this solution? Are there more elegant ways of doing
that? Any comments appreciated.
Thanks + BR,
Thorn
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ssage-
> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> Sent: mardi 8 mars 2011 10:43
> To: Thaler,Thorn,LAUSANNE,Applied Mathematics
> Cc: r-help@r-project.org
> Subject: Re: [R] Read data.frame from clipboard
>
> You haven't told us your OS. But assumin
pproach uses a file connection as well, so it does not
really change things (besides that it does things in one single step),
so any comments appreciated of how I could do this Excel to R thing
quickly preferably without any file transactions.
Thanks for your help.
BR Thorn
[
,
such that predict can resolve the variable, but I've no clue how to do
this.
Help is very much appreciated.
BR + thanks,
Thorn
8<
df <- data.frame(x=factor(rep(1:2, each=10)), y=c(rnorm(10), rnorm(10,
10)), z=rep(1:10,2))
test <- function(df, resp,
l the prepanel function is called?
2.) Is the only way to adjust the panel function for different panels using
panel.number() or is there another (better) way?
Thanks again for your kind help, it improves my insight on lattice a lot,
BR, Thorn
__
col=df$df.col)
which is a common mistake I make assuming that providing the data argument to a
lattice command allows for omitting the name of the data.frame in all parts of
the argument list not just in the formula part. Is there a reason why this is
not the default b
multipanel plot with x-axis
in reversed order but where the x-axes are only drawn in the panels at
the boundary (like in xyplot(y~x|z))? Do I have to provide an own axis
function?
Any help appreciated.
BR
Thorn
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http
Perfectly, works as expected. Regarding the other questions, can anybody point
me to the right direction?
BR Thorn
From: RICHARD M. HEIBERGER [mailto:r...@temple.edu]
Sent: lundi 23 août 2010 18:36
To: Thaler,Thorn,LAUSANNE,Applied Mathematics
Cc: r-help@r-project.org
Subject: Re: [R
everything
completely. So it would be great if somebody could help me out with this
specific topic and point me to some resources where I can learn more.
Thanks for your help in advance.
BR,
Thorn
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Thanks, that does the trick. Again a new command learned. Thanks.
However, any hints regarding the rownames issue?
BR Thorn
> -Original Message-
> From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
> Sent: lundi 9 août 2010 11:07
> To: Thaler,Thorn,LAUSANNE,Applied
es, I think
ind <- pmatch(a$id, b$id, duplicates=T)
could do the job? Or do I run into troubles regarding the "partial
matching" involved in pmatch?
BTW, is there a way to prevent R of assigning [row|col]names? In the
example above I had to remove the rownames generated by rbind
explic
Works as expected, THX a lot.
BR thorn
> -Original Message-
> From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
> Sent: mardi 20 juillet 2010 11:41
> To: Thaler,Thorn,LAUSANNE,Applied Mathematics
> Cc: r-help@r-project.org
> Subject: Re: [R] apply: retur
but how can
I circumvent this mechanism? For the time being I made another *apply
call, which gives me the same result in the format I'd like: just
replace the last line in function f by
lapply(apply(bRaw, 2, function(x) list(ct(x))), function(x)
matrix(unlist(x),2,2))
How
del so in my understanding the value should be 2.
However, would it be sufficient to test
any(attr(terms(my.formula), "factors") == 2)
to see whether the given model is not valid?
Thanks for your input.
BR Thorn
Thorn Thaler
Applied Mathematics Gro
> Try this variation of my.transform that I had posted here:
> http://tolstoy.newcastle.edu.au/R/e2/help/07/09/24707.html
>
> List <- function(..., L = list()) {
>f <- function(){}
>formals(f) <- eval(substitute(as.pairlist(c(alist(...), L
>body(f) <- substitute(modifyL
ve item names. It is
still an academic question, for there are thousands of ways to solve this
issue, but I was just curious whether it is possible to find an one-liner
without the need of specifying any temp variable.
BR, Thorn
-Original Message-
From: Peter Ehlers [mailto:ehl...@uca
onal
variable? Something like
z <- list(a=1, b=a)
or even
z <- list(a=b=1)?
Both commands don't work of course, but I'm just curious whether this is
possible in principle (a rather academic question I've to admit)
Thanks + BR,
Thorn Thaler
Applied Mathematics Group
uld evaluate choices[1] not at the prompt level,
but at the moment it enters the function biplot, where choices will be known
in any case (since it has default values).
Just to make it clear once again, I'm aware of the possibilities to get this
specific task don
ike
f(z<-2,z)
but I'm just curious whether it is possible to use a fancy combination of
eval, substitute or quote ;)
BR, thorn
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