Hi,
my question is the following one:
predict(VAR(y,p=1,type="none"),n.ahead=500)
How can I extract the forecasts of this prediction? y is a dataset of
four parameters with 500 values each.One parameter for instance is
lambda the head looks like this:
$lambda
fcst lower
10,20,30).
Spot is the sort of interest rate I want to determine.
The error request I receive now is: Error in hasTsp(x) : invalid time
series parameters specified. How fix this concrete case? I have not
found an example to that.
Best Regards Jonas
Hello everybody, I have to confess that I am relatively new to R.
My problem is the following one:
I have a data set of 500 values. For each value I wanna make prediction
of four steps ahead via autogregression. Because I have to repeat this
procedure on three other data sets I wanna make a lo
around?
Thanks,
Jonas
Jonas Josefsson (PhD student)
Swedish University of Agricultural Sciences (SLU)
Department of Ecology
Box 7044
750 07 Uppsala
Sweden
jonas.josefs...@slu.se<mailto:jonas.josefs...@slu.se>
0046 (0)18
ying your procedure.
Thanks again!
Best,
Jonas
Zitat von PIKAL Petr :
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Jonas Walter
Sent: Monday, February 25, 2013 2:38 PM
To: r-help@r-project.org
Subject: [R] creatin
nes.
min(mydat$match)
[1] 1
> max(mydat$match)
[1] 1
# example: row 1699: no match Stimulus - Prediction
mydat$Stimulus[1699] == mydat$Prediction[1699]
# [1] FALSE
# but:
mydat$match[1699]
# [1] 1
How can I get the right coding? Where is the mistake?
Thanks!
Best,
Jonas Walter
e a binary font file?
Can someone give me a link to an example?
Kind regards and thanks a lot,
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Hi,
how can i plot two different x axis in a ggplot2 qplot?
I want to plot Farenheit and Celsius in one diagram.
x1:Farenheit x2:Celsius
kind regards,
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he classic "plot' command i could add
the curves with something like
for... {
data<-read.csv
points(data$x, data$y, col=myrainbow, type="l")
}
How would you solve it in ggplot?
Kind regards,
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ion like sum and a copy mysum:
> mysum <- sum
> is.object(sum)
[1] FALSE
> is.object(mysum)
[1] FALSE
If we change the class of mysum, the original function sum becomes an
object, too!
> class(mysum)<-class(mysum)
> is.object(mysum)
[1] TRUE
> is.object(sum)
[1] TRUE
Is th
Dear all,
I have some network data - about 300 vertices and several thousand edges. I
am looking for a way to turn multiple edges into weights of the edges. I
looked around and - surprisingly? - haven't found anything. Is there an
easy way to do this?
Best, Jonas
[[alternative
R vers. 2.15.0 and ggplot 0.9.1, win xp
Best wishes
Jonas Hal
_
BRFkredit sender e-mails og vedhaeftede dokumenter i ikke-krypteret form. Hvis
du ikke ?nsker at
I am using win xp, R 2.15.0 and texi2dvi from the tools package.
Regards
Jonas
-Oprindelig meddelelse-
Fra: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sendt: 8. august 2012 16:43
Til: Jonas Hal
Cc: r-help@r-project.org
Emne: Re: [R] sweave problem with special danish characters
On
c}
\begin{document}
This doesn't æ ø å and pdf document only contains NA
\end{document}
Any suggestions?
Best wishes
Jonas Hal
_
BRFkredit sender e-mails o
On 07/12/2012 01:39 AM, Duncan Murdoch wrote:
On 12-07-11 2:34 PM, Jonas Stein wrote:
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as though overflow is
occurring. I'd recommend expanding tanh() in terms of expone
low.
>
> Duncan Murdoch
Hi Duncan,
Thank you for your suggestion. I wrote a function "mytanh" and
nls terminates a bit later with another error message:
Error in nls(data = dd, y ~ 1/2 * (1 - mytanh((x - ttt)/1e-04) * exp(-x/tau2)),
:
number of iterations exceeded maximum of
E)
# get error:
# Error in nls(data = dd, y ~ 1/2 * (1 - tanh((x - ttt)/smallc) *
exp(-x/tau2)), :
# singular gradient
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Dear list,
I am currently scraping some text data from several PDFs using the
readPDF() function in the tm package. This all works very well and in most
cases the encoding seems to be "latin1" - in some, however, it is not. Is
there a good way in R to check character encodings? I found the functio
cairo_pdf() vs. pdf() what is the difference between them and
which one would you suggest to use in combination with pdflatex?
kind regards,
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PLEASE do
regards,
--
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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.
important.
Thanks,
Jonas Fransson
Ph.D.stud.
IVA / Det Informationsvidenskabelige Akademi
Royal School of Library and Information Science
Birketinget 6
DK-2300 Copenhagen S
T +45 32 58 60 66
D +45 32 34 15 10
www.iva.dk/jf
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ta object
nlmod <- nls(X ~ A+B*sin(C* Y), data=mydata, start=list(A=2, B=4, C=1),
trace=TRUE)
# this nls fit fails.
Who can help me to fit this type of data?
Kind regards,
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https://stat.eth
and we can see the code used.
sure
http://www.r-bloggers.com/ggheat-a-ggplot2-style-heatmap-function/
kind regards,
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PLEASE do read the posting guide htt
gend like this in a standard plot like
plot(1:10) too?
http://www.r-bloggers.com/wp-content/uploads/2011/03/heatmap.png
kind regards,
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PLEASE do read the
s for the text need adjustment
(linespacing is quite large and so on)
Is there an alternative to legend()?
Is it possible to place the legend() outside of the plot area?
Kind regards,
--
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sequence at 0.5.
> l <- 0.6
> u <- seq(0.5, 0.7, 0.1)
> u
[1] 0.5 0.6 0.7
> mygrid <- expand.grid("l" = l, "u" = u)
> mygrid
l u
1 0.6 0.5
2 0.6 0.6
3 0.6 0.7
> mygridcollapsed <
> Jonas, I've just seen your function 'sistring' code and it's different from
> the code in
Thanks a lot for reporting this bug. It is fixed now in the git
repository.
I added some examples, but they do not work:
R CMD check sitools
= snip
>
On 2012-01-18, William Dunlap wrote:
> Try adding
> LazyData: yes
> to the DESCRIPTION file.
>> [3] https://github.com/jonasstein/sitools
Thank you. Now it works and I could add all SI prefixes.
--
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]
kind regatds,
[1] https://github.com/jonasstein/sitools/blob/master/init.R
[2] https://github.com/jonasstein/sitools/downloads
[3] https://github.com/jonasstein/sitools
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> I don't believe you can. However, you need not install it into a system-wide
> library directory... your personal library (e.g.
> /home/jonas/R/x86_64-pc-linux-gnu-library/2.14) should be sufficient.
Finally i created a new testuser to install the library locally as you w
to directory
‘/usr/local/lib/R/site-library’
R tries to install something in my system. That may confuse my
debian packagemanagement.
kind regards,
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P
3 * kilo
[1] 3000
[*] https://github.com/jonasstein/sitools
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and prov
,ylim=c(45,184),ylab='FACIT-PAL',xlab='Time
> to death',color=FALSE,lwd=1)
i can not use your example, as i have no enable2r.csv, but perhaps
you have luck if you change
xlab='Time to death'
to
xlab="Time to death
ite set of points you can at least deal with it
> exhaustively.
my real data is not limited to integer. Do you know a ready to use code
example for this?
Would it be a good idea to create a function and make it public to the
community? And if yes as single .R file, or as a library?
kind rega
Paradis (R for Beginners)
I have uploaded the source to
https://github.com/jonasstein/R-Reference-Card
It is my favourite R Reference Card and I am looking forward to recieve
your suggestion for improvement.
Is there a prominent place where i should push the compiled .PDF to?
Kind regards,
--
Hi,
i'd like to update my R-Reference card and commit some edits,
but i could not get the source from rpad.org
Did it move?
kind regards,
--
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PLEA
On 2012-01-05, Jonas Stein wrote:
> i want to plot values with frequency on a logarithmic x axis.
> similar to this example that i found in the web:
> http://www.usspeaker.com/jensen%20p15n-graph.gif
>
> I would like to convert long numbers to si prefix notation
> like in the e
thing like mailman does here
https://stat.ethz.ch/mailman/listinfo/r-help
If he can reply he will be permanently accepted as member.
If you like this idea, could the responible person check
the optionflag in gmane?
Kind regards,
--
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__
R
create labels by hand, but
i have many files so i need some automatic function.
Has anyone done that in R?
Kind regards and thank you for your help,
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i have a list of values like this
x y
1 3
2 2
3 3
4 4
5 5
6 4
7 3
8 2
9 3
and need the inflexion points (and all max and min).
Is there a nice way to get the local max, min and inflexion points?
kind regards,
--
Jonas Stein
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system.time(int2bin(x))
user system elapsed
0.310.000.32
On Thu, Dec 1, 2011 at 7:14 AM, Jonas Jägermeyr wrote:
Dear R-help members,
I'm processing a large amount of MODIS data where quality assessment
information is stored as an integer value for each pixel. I have to
converte
ndigits-q+1] <- (x %% 2)
x <- (x %/% 2)
}
bin<- apply(base,1,paste,collapse="")
return(bin)
}
Since it is still slow, I have to find a way to express this more
elegantly. I'd really appreciate any help.
Thanking you, with best regards
Jon
Thank you very much for your help.
I'm using the second suggestion in my program and it works very well.
Jonas
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Thu 11/10/2011 5:58 AM
To: Richter-Dumke, Jonas
Cc: r-help@r-project.org
Subject: R
irck Kleve",
"Bernardus Engelb Franciscus Linde")?
This expression would very reliably seperate the person names from the
additional information in my historic church register transcription.
Thank you very much for your effort,
Jonas
--
This mai
umn or a vector where all the DEPTHs are listed: 2, 3, 3, 2, 2, 2,
5 (NUMBER=1 in all cases).
Is there any simple solution to the problem?
Thanks,
Jonas
Jonas Fransson
Ph.D.stud.
IVA / Det Informationsvidenskabelige Akademi
Royal School of Library and Information Science
Birketinget 6
DK-2300 C
m(25),rnorm(25), xaxs ="i", yaxs="i", xlim=c(-2,2),
> ylim=c(-2,2))
That is exactly what i have been looking for. Thank you.
>> [1]
>> http://commons.wikimedia.org/wiki/File:Atmospheric_radiocarbon_1954-1993.svg
--
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regards,
--
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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.
ta<- data.frame(y,x,a,b,c)
mod1<- lmer(y ~ x + (1|a:b) + (1|b:c), data=mydata)
mod2.lme<- lme(y ~ x, random=list(a=~1, b=~1, c=~1), data=mydata)
mod2.lmer<- lmer(y ~ x + (1|a) + (1|a:b) + (1|a:b:c), data=mydata)
My objective is to specify mod1 using function lme.
Anyone knows how
nslate mod1 into lme language?
Any help on this would be much appreciated.
Jonas
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Hi,
I'm using heatmap.2 to cluster my data, using the centroid method for
clustering and the maximum method for calculating the distance matrix:
library("gplots")
library("RColorBrewer")
test <- matrix(c(0.96, 0.07, 0.97, 0.98, 0.50, 0.28, 0.29, 0.77,
0.08, 0.96, 0.51, 0.51, 0.
entries and merging them...but there must be a
better solution, or?)
thx. bye
Jonas
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and provide
Dear all,
First, I would like to thank you for your immense work. My question is
about a frequent topic which I am not able to solve - even after hours
of search in the mailing lisy.
I would like to analyse random-effects (and fixed-effects)models of
longitudinal / panel data with sampling wei
to try it again now.
kind regards,
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and provide commented, minimal, self
Many thanks for your help!
Thanks to you guys I manage to solve my problem in an efficient way
All the best
J
On Tue, Oct 26, 2010 at 11:04 PM, Remko Duursma wrote:
>
> I don't know why I forgot that you can do this as well :
>
> area.poly(intersect(p1,p2))
>
> ... a bit more straightforward.
>
there any other package
that does this?
Thanks
On Tue, Oct 26, 2010 at 3:38 AM, Remko Duursma wrote:
>
> Dear Jonas,
>
> if you can write the difference in y-values between your polygons as a
> function, you can use
> integrate() to get the area between the polygons.
>
> It sou
orward algorithm to calculate the intersection area
between polygons poly1 and poly2?
Thank you so much in advance
Jonas
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n bxp help page.
Regards
Petr
http://img51.imageshack.us/img51/9011/plott.jpg
Jonas
2010-10-20 17:14, David Winsemius skrev:
On Oct 20, 2010, at 10:45 AM, fugelpitch wrote:
I have a factor which is species and the variables are a few sites
where this
species live.
When I p
are clearly clustered (in this example there are 3 clusters).
Is there any R algorithm that
allows me to identify the three groups of points and allocate to each pair
the respective cluster?
Many thanks in advance
Jonas
[[alternative HTML version dele
img51/9011/plott.jpg
Jonas
2010-10-20 17:14, David Winsemius skrev:
On Oct 20, 2010, at 10:45 AM, fugelpitch wrote:
I have a factor which is species and the variables are a few sites
where this
species live.
When I plot:
plot(species.factor, minlatitudemaxlatitude)
the plot produced is autom
the two values...
Thanks!
Jonas Josefsson
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and provide commented, minimal, self-contained
I can recommend RKWard which is more than an editor but is great. There
is also an option to patch gedit to send commands, but it was to much of
a hassle to set up to be of interest to me at least.
Jonas
2010-10-05 15:52, Mehdi Zarrei skrev:
Hello R-Users!
I am looking for an editor to be
I have a two-column table as follows where age is in the 1st column and
the number of individuals is in the 2nd.
age;no
1;21
2;31
3;9
4;12
5;6
Can I use mean() and sd() to calculate the mean and standard deviation
from this or do I have to manually multiplicate 21*1+31*2 etc. / N?
_
te(PrH[i]) == F & tempHER > tempSPR) {PrH[i] <- 1 }
if (is.finite(PrH[i]) == F & tempHER < tempSPR) {PrH[i] <- 0 }}
PrH, tempSPR and tempHER are equally long vectors.
Thanks in advance!Jonas
[[alternative HTML version deleted]]
___
say that I have a script that uses read.csv to read a textfile in a
certain directory, then transforming the data a bit and then spit out a
new file in an output directory.
Is it possible to make this universal so the same procedure is made for
all files in a certain directory and also make it s
I am trying to find a convenient way to control line colors when
printing from a for loop using the lines command.
Right now I have solved this by creating a colors vector that is refered
to in the loop with index. However, the colors choosen here are just
1,2,3,4,5...
I would like to get colo
Hello,
try this way : any(as.integer(c(1, 3))==3)
cheers,
Jonas
Christofer Bogaso a écrit :
> Hi I have following line of code:
>
>> any(as.integer(c(1, 3))) == 3
> [1] FALSE
>
> Shouldn't I expect it is true?
>
> Thanks,
>
> ___
Yes, thanks! :)
2010-09-23 11:55, Ivan Calandra skrev:
>Hi,
> xlim and ylim should be given the extremes only:
>
> plot(x,y, xlim=c(pheno.dt$year[1],pheno.dt$year[nrow(pheno.dt)]),
> ylim=c(50,150), xlab="Year", ylab="Julian Day")
>^^
Hello,
Check the function diff() it can do it for you, no need for a loop.
Cheers,
Jonas Mandel
ecvet...@uwaterloo.ca a écrit :
> I have a data frame with 2 columns, one for day and one for average. The
> day starts at 97 all the way to 279. I want to subtract day 98 average-
> day 9
Hi,
I have a number of multilevel models and I would like some Latex outputs of
them. I usually use the "apsrtable" package, but it does not accept "lme"
outputs. Neither does the "mtable" function in the "memisc" package. Is
there any good altern
Hi,
I have a number of multilevel models and I would like some Latex outputs
of them. I usually use the "apsrtable" package, but it does not accept
"lme" outputs. Neither does the "mtable" function in the "memisc"
package. Is there any good alterna
s not at all balanced :
> table(data$Série, data$Traitemen)
1 2 3
7 0 28 31
8 24 17 0
In batch 7 there is no mouse treated with treatment 1, and in batch 8
there is no mouse treated with treatment 3. Maybe the error come from
here, even if lm() is fine with this ?
Cheers,
Jonas
, components], 1, prod) : index out of range
In the example given in the help page there are factor variables and
numeric variables so I don't think the problem comes from here. Any help
is welcome here.
Cheers,
Jonas
Joris Meys a écrit :
> Please, read the posting guide:
> - provide a minim
t this error:
> eff.lm1 <- allEffect(mod=lm1,
Error in apply(mod.matrix[, components], 1, prod) : index out of range
I read the help and the example but I don't understand the reason why it
doesn't work. Can you help me ?
Thanks
--
Jonas Mandel
_
gt;> On 04/03/2010 10:32 PM, David Winsemius wrote:
>>>
>>>> On Mar 4, 2010, at 9:47 PM, jonas garcia wrote:
>>>>
>>>>> When I opened the file with a hex-editor, the problematic character
>>>>> turned out to be 1a
>>&
e in a binary/raw, finding the
> offending character and replacing it with a blank (or whatever and
> then writing the file back out). You can then probably process it
> using read.table.;
>
> On Thu, Mar 4, 2010 at 12:50 PM, jonas garcia
> wrote:
> > Thank you so much for your rep
go for text editors
separately.
Help on this would be much appreciated.
Thanks again
J
On 3/4/10, David Winsemius wrote:
>
>
> On Mar 3, 2010, at 2:22 PM, jonas garcia wrote:
>
> Dear R users,
>>
>> I am trying to read a huge file in R. For some reason, only a
= header, sep = sep, quote = quote, :
incomplete final line found by readTableHeader on 'new3.dat'
I am working with R 2.10.1 in windows XP.
Thanks in advance
Jonas
[[alternative HTML version deleted]]
__
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10) to a 8-digit zero fill-up (x
= 00010110) in order to address digit indices?
Or other way round, how can I make the 'substr' function to count from the
right hand side?
Thank you.
With kind regards,
Jonas Jägermeyr
Department of Geography
Humboldt-University of Berlin
jonas.jaegerm
e but with opposing
> orientation.
>
> Try fooling about with the mgp argument in axis():
>
> par(mfrow=c(1,1), cex.axis = 0.5, cex.lab = 0.5)
> plot(1,1, axes = F)
> axis(1, mgp=c(3,0.7,0))
> axis(2, mgp=c(3,1,0))
>
> >>> jonas garcia 10/13/09 5:58 PM >>>
&
, cex.lab = 0.5)
plot(1,1, axes = F)
axis(1)
axis(2)
Thanks in advance
Jonas
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PLEASE do read the posting guide http://www.R
get, env=env))
> }
>
> testA()
>
> But more generally, I doubt your construct using assign and get is the most
> natural way to reach your goal in R.
>
> HTH,
>
> baptiste
>
>
> 2009/9/4 jonas garcia
>
>> Hi all,
>>
>>
>>
>> I ha
they both work...
Why is this?
Thanks in advance
Jonas
testA<-function(input)
{
dat<- data.frame(A=seq(input,5), B=seq(6,10))
vec.names<- c("a", "b")
for(i in 1:ncol(dat))
{
tab<- dat[,i]-1
assign(vec.name
Thanks Erik and Henrique,
That's what I was after.
Jonas
On Tue, Sep 1, 2009 at 8:08 PM, Henrique Dallazuanna wrote:
> Try this:
>
> > sapply(vec.names, get)
>
> But for this example, you don't need for, try:
>
> > dat - 1
>
> On Tue, Sep 1, 20
1] [,2]
[1,] "a" "b"
But I was looking after the following result (using vec.names):
cbind(a,b)
a b
[1,] 0 5
[2,] 1 6
[3,] 2 7
[4,] 3 8
[5,] 4 9
Thanks in advance
Jonas
[[alternative HTML version deleted]]
ere)
I want to obtain a different vector like this:
c(ra, so, op)
Thanks in advance for your help
Jonas
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PLE
, ‘digest’, ‘cacheSweave’ are not available
| 2: In install.packages("pgfSweave", , "http://www.rforge.net/";) :
| installation of package 'pgfSweave' had non-zero exit status
|
`
what will i have to do now?
And could someone give me an example how to write a formu
akFreqHz = subset(fp, AmpNorm == max(AmpNorm))$Frequenz[1]
Is there something nicer?
And is there an easy way to do the same on "predict()"
Thank you and kind regards,
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Jonas Stein
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or a = 1.4e-05 with 98% confidence
Thank you for reading so far and thank you for any help.
--
Jonas Stein
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Could you please help me understanding which is the right model for my
question ? Thanks by advance
Jonas Mandel
U900 - Bioinformatics unit
Institut Curie
Paris
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PLEA
font that is
used in the TeX-document?
Thank you for any help.
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Jonas Stein
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and provide
Thank you! Now it's working.
Peter Dalgaard schrieb:
Jonas Weickert wrote:
Hi,
I want to do a biexponential Fit, i.e.
y ~ A1*exp(k1*x) + A2*exp(k2*x)
Is this possible? I tried nls() but it stopped with several
(different) errors. I'm using y and x as simple vectors and the
formu
Hi,
I want to do a biexponential Fit, i.e.
y ~ A1*exp(k1*x) + A2*exp(k2*x)
Is this possible? I tried nls() but it stopped with several (different)
errors. I'm using y and x as simple vectors and the formula for nls()
exactly as mentioned above.
Thanks a lot!
have been some improvement...
Does anyone know a nice and easy way to turn labels on the y axis in the
same direction as the x labels?
Thank you very much for reading and hints,
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Jonas Stein
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; Behalf Of Tony Breyal
> Sent: Wednesday, December 10, 2008 1:01 PM
> To: r-help@r-project.org
> Subject: Re: [R] missing argument
>
> ?missing
>
> never used it myself, but looks like it might help you :-)
>
> Tony Breyal.
>
> On 10 Dec, 19:09, "jonas garcia&q
Dear list,
I have a question and I'm going to give an example of my problem
f<- function(d1, d2, d3)
{
d<- d1*d2/d3
return(d)
}
v1<- 1
v2<- 2
If I try
f(v1, v2, v3)
Error in f(v1, v2, v3) : object "v3" not found
I obviously got the above error message.
I would like to add something to my functio
Thank you all for the useful responses. Problem solved!
J
On Thu, Oct 23, 2008 at 12:14 AM, Duncan Murdoch <[EMAIL PROTECTED]>wrote:
> On 22/10/2008 5:02 PM, jonas garcia wrote:
>
>> Dear list:
>>
>>
>>
>> I have the following problem: From a vector
Dear list:
I have the following problem: From a vector like this:
vec<- c("mud_1999_area_A", "gravel_2004_area_F")
I would like to get the year in a separate vector, such
y<- c("1999", "2004")
I´ve been looking to grep() but I'm not sure how to do this.
Help?
Thanks in advance
J
#here i would like to access objects not returned by NelsonSiegel
}
//Jonas
On Sun, Aug 3, 2008 at 6:05 PM, Patrick Burns <[EMAIL PROTECTED]> wrote:
> Duncan Murdoch wrote:
>>
>> On 03/08/2008 11:29 AM, Jonas wrote:
>>>
>>> hi,
>>>
>>>
unction.from.package(x,y)
plot (object.inside.package.function)
}
i'm using R 2.7.1 on linux
sincerely,
jonas
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simple and totally obvious thing, but I just do not get it. :-(
Could you please help me understand? How do I do it in MATLAB or R?
THanks in advance!
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Jonas Malmros
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PLEASE d
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