Hello,
I am fitting a coxph model with factors. I am running into problems when
using 'survfit'. I am unsure how R is treating the factors when I fit, say:
>DATA<-data.frame(time.sec,done,f.pom=factor(f.pom),po,vo)
>final<-coxph(Surv(time.sec,done)~f.pom*vo+po,data=DATA)
>
On Thursday 14 July 2005 00:51, Peter Dalgaard wrote:
> Adrian Dusa <[EMAIL PROTECTED]> writes:
> > <...snip...>
>
> This comes up every now and then, and while it seems that everyone
> thinks fill patterns would be nice to have, I suspect that every
> attempt to actually implement it have gotten k
Uwe Ligges wrote:
> Dongseok Choi wrote:
>
>> Thank you very much for your help!!
>> Now, it runs without any problem.
>>
>> Is it going to be fixed in the next release?
>
>
> Of course, Brian
[hmmm, looks like some wrong shortcut has been used - and it must have
been me who forgot to drink c
Dongseok Choi wrote:
> Thank you very much for your help!!
> Now, it runs without any problem.
>
> Is it going to be fixed in the next release?
Of course, Brian
> Thanks again,
> Dongseok
>
>
>
>
> Dongseok Choi, Ph.D.
> Assistant Professor
> Division of Biostatistics
> Department of Public
Hi,
I had a quick question regarding the pred_garch function (as part of the
tseries library, garch.c), and specifically: the function calculates the 'h'
vector of conditional variances, h[n+1] if the data is genuine. I am very
much a beginner, but from what I understand I would think that it is
An easy to check what you have is to use month.day.year:
> eh <- data.frame(t = c("06/05/2005 01:15:25", "06/07/2005 01:15:25"))
>
> # substring converts factor to character and extracts substring
> chron(dates = substring(eh$t, 1, 10), times = substring(eh$t, 12))
[1] (06/05/05 01:15:25) (06/07/
are those dates in m/d/y or d/m/y ?
?chron and watch out for
format = c(dates = "d/m/y", times = "h:m:s")
On 13/07/05, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> [I had some emails problems so I am sending this again. Sorry
> if you get it twice.]
>
> On 7/13/05, Gabor Grothendieck <[EMAI
R-colleagues
I have adapted the anova.lmlist function to use the model object name as
the first column in the output instead of the string "Model n".
If there is general agreement can the change be implemented into the
stats package?
Regards
Ross Darnell
--
University of Queensland, Brisbane
What kinds of matrices? There are facilities in the Matrix and
SparseM packages that might help for sparse matrices. If they are N x k
where N is large and k is not, can you compute something like the QR
decomposition and get away with keeping only the R part for most of your
matric
My preference is to test see if the smallest eigenvalue is less than
something like sqrt(.Machine$double.eps) times the largest. This may be
too conservative, but if the ratio of the smallest to the largest is
less than some small number like that, the inverse of such a real
symmetri
At 09:35 AM 14/07/2005, Emilio A. Laca wrote:
>I need to specify a model similar to this
>
>lme.formula(fixed = sqrt(lbPerAc) ~ y + season + y:season, data = cy,
> random = ~y | observer/set, correlation = corARMA(q = 6))
>
>except that observer and set are actually crossed instead of nested.
r=1:10
u=c("a","a","a","b","b","b","c","d","e","e")
uf = factor(u)
rm = tapply(r, uf, mean)
stripchart(r~u,vertical=TRUE,pch=21)
stripchart(rm~levels(uf),vertical=TRUE,pch=3,add=TRUE)
--
the above code creates a scatter plot of nominal data
are there alternatives to generate the same or
See ?which Hint: arr.ind=TRUE
Simon.
At 09:28 AM 14/07/2005, wu sz wrote:
>Hello,
>
>I have a data set matrix of 1200 * 15. How can I get the position of a
>specific value in the matrix?
>
>I use "seq(along = x)[x > value]" to look for the position of the
>value in the matrix, but "seq" can just
I need to specify a model similar to this
lme.formula(fixed = sqrt(lbPerAc) ~ y + season + y:season, data = cy,
random = ~y | observer/set, correlation = corARMA(q = 6))
except that observer and set are actually crossed instead of nested.
observer and set are factors
y and lbPerAc are numer
Use which(..., arr.ind=TRUE); e.g.,
> m <- matrix(runif(12), 3, 4)
> which(m > .8, arr.ind=TRUE)
row col
[1,] 1 3
[2,] 2 3
[3,] 3 3
[4,] 3 4
> m
[,1] [,2] [,3] [,4]
[1,] 0.2148183 0.08251853 0.9444718 0.4487148
[2,] 0.5386863 0.49673282 0.8054240 0.51
Hello,
I have a data set matrix of 1200 * 15. How can I get the position of a
specific value in the matrix?
I use "seq(along = x)[x > value]" to look for the position of the
value in the matrix, but "seq" can just find the sequence position row
by row in the matrix, not a real position (like "row
Hi,
I wish to analyze with R the results of a perception experiment in which
subjects had to recognize each stimulus among three choices (this was a
forced-choice design). The experiment runs under two different
conditions and the data is like the following:
N1 : count of trials in condition
Luis Tercero <[EMAIL PROTECTED]> writes:
> Dear R-help community,
>
> would any of you have a (preferably simple) example of a
> presentation-quality .png plot, i.e. one that looks like the .eps plots
> generated by R? I am working with R 2.0.1 in WindowsXP and am having
> similar problems as
Sorry for last post.
I don't know why i got the error message last time.
but if i did in the following way:
t<-scan('train1.dat', sep='|', na.string='.')
t2<-matrix(t, nrow=195, ncol=273529)
t3<-t(t2)
t4<-as.data.frame(t3)
now I got what i needed.
Thanks a lot for Gabor's prompt help.
weiwei
Adrian Dusa <[EMAIL PROTECTED]> writes:
> On Wednesday 13 July 2005 17:36, Knut Krueger wrote:
> > Adrian Dusa schrieb:
> > >Is it possible to draw barplots using a texture instead of colors, for a
> > > black and white printer?
> >
> > barplot(height,.,density=c(4,6,8,10) ...)
> >
> > for
i think what you meant is
> trn<-matrix(scan('train1.dat', sep='|', na.string='.'), nrow=195,
> ncol=273529)
and then transpose it. However:
Error: cannot allocate vector of size 512000 Kb
the answer is no :(
I think i am going to write my own function to split the result from
scan but not
On 7/13/05, Ruben Roa <[EMAIL PROTECTED]> wrote:
> Hi:
> Where is the iris data set actually
> located in the R 2.1.0 folder (under W XP)?
> Is it a text file or it is a binary file?
> Ruben
Uwe has already explained how to get it in text
form; however, if you are curious about its original
format
Try reading it into and transposing the matrix afterwards. Don't know if
that would work but its worth a try. Actually if you
are having problems read it into a vector, check that its of the required
size, just in case, and then turn it into a matrix and transpose it.
On 7/13/05, Weiwei Shi <[E
there is another problem since last time i forgot "byrow" :(
> trn<-matrix(scan('train1.dat', sep='|', na.string='.'), nrow=273529,
> ncol=195, byrow=T)
Read 53338155 items
Error: cannot allocate vector of size 416704 Kb
please help with this 'simple' reading task.
weiwei
On 7/13/05, Weiwei Sh
Maybe you don't really need a data frame in the first place?
You were concerned with speed and matrices tend to
have higher performance than data frames.
On 7/13/05, Weiwei Shi <[EMAIL PROTECTED]> wrote:
> that sort of works for my purpose.
>
> btw, is there a bettter way to get data.frame by pa
that sort of works for my purpose.
btw, is there a bettter way to get data.frame by passing around
matrix(). Since I could not find data.frame() with nrow or ncol
arguments. so i have to use matrix first and then as.data.frame to
convert it.
is there any other (better) way?
weiwei
On 7/13/05, G
[I had some email problems so I am sending this again.
Sorry if you get this twice.]
On 7/13/05, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>
>
> On 7/13/05, Luis Tercero
> <[EMAIL PROTECTED]> wrote:
> > Dear R-help community,
> >
> > would any of you have a (preferably simple) example of a
[I had some emails problems so I am sending this again. Sorry
if you get it twice.]
On 7/13/05, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>
>
> On 7/13/05, Young Cho <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > I have a column of a dataframe which has time stamps
> > like:
> >
> > > eh$t[1
[I had some email problems and am sending this again. Sorry
if you get it twice.]
You could use the nlines= argument to scan to read in a
portion at a time.
>
>
> On 7/13/05, Weiwei Shi <[EMAIL PROTECTED]> wrote:
> > add:
> > I used
> > trn<-matrix(scan('train1.dat', sep='|', na.string=
On 7/13/05, Luis Tercero <[EMAIL PROTECTED]> wrote:
>
> Dear R-help community,
>
> would any of you have a (preferably simple) example of a
> presentation-quality .png plot, i.e. one that looks like the .eps plots
> generated by R? I am working with R 2.0.1 in WindowsXP and am having
> similar p
You could use the nlines= argument to scan to read in a
portion at a time.
On 7/13/05, Weiwei Shi <[EMAIL PROTECTED]> wrote:
>
> add:
> I used
> trn<-matrix(scan('train1.dat', sep='|', na.string='.'), nrow=273529,
> ncol=195)
>
> it is done.
> so it seems that I just have no patience to wait
On 7/13/05, Young Cho <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I have a column of a dataframe which has time stamps
> like:
>
> > eh$t[1]
> [1] 06/05/2005 01:15:25
>
> and was wondering how to convert it to chron variable.
> Thanks a lot.
Try this:
# test data frame eh containing a factor v
add:
I used
trn<-matrix(scan('train1.dat', sep='|', na.string='.'), nrow=273529, ncol=195)
it is done.
so it seems that I just have no patience to wait for half an hour :)
but i still have that question:
is there a way to track the process if it takes too long. Could we
stop in the middle to se
Hi,
I have a question on read.table.
I have a dataset with 273,000 lines and 195 columns. I used the
read.table to load the data into R:
trn<-read.table('train1.dat', header=F, sep='|', na.strings='.')
I found it takes forever.
then I run 1/10 of the data (test) using read.table again. And this
t
Dear R-users,
Is there a preferred method for testing whether a real symmetric matrix is
positive definite? [modulo machine rounding errors.]
The obvious way of computing eigenvalues via "E <- eigen(A, symmetric=T,
only.values=T)$values" and returning the result of "!any(E <= 0)" seems
less effic
You can do the following (don't know it this is the most efficient way but
it works)
temp<-read.table("your file to read the data", header=T)
temp1<-table(temp)
plot(temp$x, temp$y, cex=0)
text(as.numeric(rownames(temp1)), as.numeric(colnames(temp1)), temp1)
HTH
On Wed, 13 Jul 2005, Kerry Bush
Hi,
I have a column of a dataframe which has time stamps
like:
> eh$t[1]
[1] 06/05/2005 01:15:25
and was wondering how to convert it to chron variable.
Thanks a lot.
Young.
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Dear R-helper,
I want to plot the following-like data:
x y
1 1
1 1
1 2
1 3
1 3
1 4
..
In the plot that produced, I don't want to show the
usual circles or points. Instead, I want to show the
number of replicates at that point. e.g. at the
position of (1,1), there are 2 obsevations, so a
num
Thank you very much for your help!!
Now, it runs without any problem.
Is it going to be fixed in the next release?
Thanks again,
Dongseok
Dongseok Choi, Ph.D.
Assistant Professor
Division of Biostatistics
Department of Public Health & Preventive Medicine
Oregon Health & Science University
318
Hi Joan,
You need to do:
res1 <- mnp(PREVOTE3 ~ 1, choiceX = list("1"=UCLC, "2"=UDLC, "3"=UPLC),
cXnames = "ut", data = small, verbose = TRUE)
Hello,
a proposed solution of Bill Venables is archieved on the S-News mailing
list:
http://www.biostat.wustl.edu/archives/html/s-news/2001-07/msg00035.html
and if I remember it correctly (and if the variance matrix is estimated
from the data), another similar way is simply to use the Euclidean
>>> Ted Harding <[EMAIL PROTECTED]> 07/13/05 02:12AM >>>
[snip]
>> I'm not sufficiently acquainted with the internals of "plot"
>> and friends to anticipate the answer to this question; but,
>> anyway, the question is:
>>
>>Is it feasible to include, as a parameter to "plot", "lines"
>>
On Wed, 13 Jul 2005, EJ Nikelski wrote:
> Hello,
>
> Thanks for your help Brian. You are correct in assuming that I am
> trying to use write.foreign to export a data frame for use in SPSS,
> using the usual format:
>
> >write.foreign(df, dataFile, codeFile, package="SPSS")
>
> Your suggestion
On Wed, 13 Jul 2005, EJ Nikelski wrote:
>
> Your suggestion that the unprintable characters represent UTF-8 encoded
> Unicode left and right double quotes also appears correct. Now, although
> the suggested work-around may well help, the foreign package does seem
> to be creating a corrupted file.
help.search("asdf") only works if you have "asdf" in something that
is installed. RSiteSearch("asdf"), on the other hand, works for
anything in the R archives. This does NOT include, however, the
contents of R News, which you can search via http://www.r-project.org/
-> Newsletter ->
Dear Fernando
Please read the posting guide. If you want to get an answer to your question
you need to be specific about your analysis, and provide examples of the
data structure and code that you tried and didn't work.
Francisco
-
Folks
I have modified an existing function to calculate 'ec/ld/lc' 50 values
and their associated Fieller's confidence limits. It is based on
EC50.calc (writtien by John Bailer) - but also borrows from the dose.p
(MASS) function. My goal was to make the original EC50.calc function
flexible
Try RSiteSearch("map") or help.search("map")
Cheers
Francisco
>From: m p <[EMAIL PROTECTED]>
>To: r-help@stat.math.ethz.ch
>Subject: [R] maps drawing
>Date: Wed, 13 Jul 2005 09:15:33 -0700 (PDT)
>
>Hello,
>is there a package in R that would allow map drawing:
>coastlines, country/state boundari
One way I do this is to use Luke Tierney's "active bindings". I make an
active binding of a name to a function which either loads or saves the
object. Then the name behaves like the R object it's replacing. This works
nicely as long as I don't need lots of random accesses to the matrix.
I'd be hap
m p wrote:
> Hello,
> is there a package in R that would allow map drawing:
> coastlines, country/state boundaries, maybe
> topography,
> rivers etc?
What about package maps?
Moreover, what about reading the posting guide and trying to search
yourself at first. I think it is almost impossible n
Hello,
is there a package in R that would allow map drawing:
coastlines, country/state boundaries, maybe
topography,
rivers etc?
Thanks for any guidance,
Mark
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PLEASE do
so my chi-square approximation was not very good:
> pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE)
[1] -19775.52
> (pchisq(39540, 1, lower.tail=FALSE, log.p=TRUE)
+ /log(10))
[1] -8588.398
... roughly 1e-8588. With a few hours with Abramowitz and Stegun, I
suspect I could
Ruben Roa wrote:
> Hi:
> Where is the iris data set actually
> located in the R 2.1.0 folder (under W XP)?
> Is it a text file or it is a binary file?
It is a special binary file in package datasets in the binary
distribution. Just dump() or write.table() on the data to get a text
representatio
help.search("iris") tells you.
You should always try R's built-in help resources **before** posting.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
"The business of the statistician is to catalyze the scientific learning
process." - George E. P. Box
> -Origi
Knut Krueger schrieb:
>Adrian Dusa schrieb:
>
>
>
>>Is it possible to draw barplots using a texture instead of colors, for a
>>black
>>and white printer?
>>
>>
>>
>>
>>
> barplot(height,.,density=c(4,6,8,10) ...)
>
>for each bar one number - this example is for a barplot with 4 ba
Hi:
Where is the iris data set actually
located in the R 2.1.0 folder (under W XP)?
Is it a text file or it is a binary file?
Ruben
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PLEASE do read the posting guide! ht
Hello,
Thanks for your help Brian. You are correct in assuming that I am
trying to use write.foreign to export a data frame for use in SPSS,
using the usual format:
>write.foreign(df, dataFile, codeFile, package="SPSS")
Your suggestion that the unprintable characters represent UTF-8 enco
Dear R-help community,
would any of you have a (preferably simple) example of a
presentation-quality .png plot, i.e. one that looks like the .eps plots
generated by R? I am working with R 2.0.1 in WindowsXP and am having
similar problems as Knut Krueger in printing high-quality plots. I have
Dear Ghislain
I do not know a general elegant solution, but for some
applications the following example may be helpful:
## Artificial data for demonstration: group is fixed, species is random
dat <- data.frame(group = c(rep("A",20),rep("B",17),rep("C",24)),
species = c(rep("sp1
Knut Krueger schrieb:
>Adrian Dusa schrieb:
>
>
>
>>Is it possible to draw barplots using a texture instead of colors, for a
>>black
>>and white printer?
>>
>>
>>
>>
>>
> barplot(height,.,density=c(4,6,8,10) ...)
>
>for each bar one number - this example is for a barplot with 4 ba
Apologies for cross-posting
-- Final call. There are still 3 places available on each course --
We would like to announce three statistics courses in and around Aberdeen, UK
Course 1: Regression, GLM, GAM, mixed modelling and tree models
Course 2: Multivariate analysis and multivariate time se
Welcome to R. The learning curve is well worth the benefits.
If you are used to Eisen clustering and other fancy softwares to do
clustering, then you might be a little disappointed with R's clustering
ability of thousands of genes. But then again clustering is an
exploratory tool and I see no reas
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Beale, Colin
> Sent: 13 July 2005 10:15
> To: Prof Brian Ripley
> Cc: r-help@stat.math.ethz.ch
> Subject: Re: [R] nlme, MASS and geoRglm for spatial autocorrelation?
>
>
> My data are indeed bernoulli and
On Wednesday 13 July 2005 17:36, Knut Krueger wrote:
> Adrian Dusa schrieb:
> >Is it possible to draw barplots using a texture instead of colors, for a
> > black and white printer?
>
> barplot(height,.,density=c(4,6,8,10) ...)
>
> for each bar one number - this example is for a barplot with
Dear All,
Just start to use the long expected R, my focus will be
doing clustering on microarray data, just wonder, anyone can
show me any references to conquer the steep learning curve?
Thanks!
Best regards,
Baoqiang Cao
__
R-help@stat.math.ethz.ch
Adrian Dusa schrieb:
>Is it possible to draw barplots using a texture instead of colors, for a black
>and white printer?
>
>
>
barplot(height,.,density=c(4,6,8,10) ...)
for each bar one number - this example is for a barplot with 4 bars.
with regards
Knut Krueger
http://www.biostatis
Does one of the packages of R include functions to generate random
fractals as for instance outlined in http://classes.yale.edu/fractals ?
Heinz Schild
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PLEASE do read
Dear R list,
For some reason I am unable to access neither search.r-project.org, nor
http://finzi.psych.upenn.edu/ so I cannot search the archives for a possible
answer (I Googled for this but didn't find anything).
Is it possible to draw barplots using a texture instead of colors, for a blac
Hi
I'm using Sweave with tex4th to generate xhtml documents
However there seems to be a problem with \Link defined in
Rd.sty, since this is also defined in tex4ht.sty
My workaround was to replace the lines with
\newcommand{\Link} by \providecommand{\Link}
in Rd.sty
I'm therefore wondering whether
hi everybody,
I noticed the following: in one of my scripts 'layout' is used to
generate a (approx. square) grid of variable dimensions (depending on
no. of input files). if the no. of subplots (grid cells) becomes
moderately large (say > 9) I use a construct like
###layout grid computatio
I do not know if current XML package suppose to work for "windows R Version
2.0.1"; however, current version of XML works fine for current version of R
(2.1.1). Also, version of XML available when R Version 2.0.1 was current,
worked just fine as well. So the answer might be to update your R version
On 7/13/05, Soren Wilkening <[EMAIL PROTECTED]> wrote:
> Dear all
>
> the regular XML package does not work correctly with the R 2.0.1 windows
> version.
> Can anybody indicate a suitable alternative ?
> I need to dynamically read, parse and process a HTML table in R that is
> available at a certa
Dear all
the regular XML package does not work correctly with the R 2.0.1 windows
version.
Can anybody indicate a suitable alternative ?
I need to dynamically read, parse and process a HTML table in R that is
available at a certain url.
Regards
Soren Wilkening
--
CENSIX Consulting
[EMAIL PROT
Hi all,
Does anybody have a hint on what may be going wrong in this R code? I
mimic the sample code from the MNP developers but I seem unable to get the
choice specific variables right.
Thanks,
Joan Serra
> rm(list=ls())
> library(foreign)
> small<-read.spss("small.sav")
Warning message:
small
On 7/12/05, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hi, I would like to know whether it is possible to obtain a value of
> significance for random effects when aplying the lme or related
> functions. The default output in R is just a variance and standard
> deviation measurement.
>
> I feel
On Mon, Jul 11, 2005 at 08:27:40AM -0700, Rob J Goedman wrote:
> Ajay,
>
> After installing both setRNG (2004.4-1, source or binary) and dse
> (2005.6-1, source only), it works fine.
Thanks! :-) Now dse1 works, but I get:
> library(dse2)
Warning message:
replacing previous import: acf in: name
Hello!
How to set function for the whole R-class to be executed when the object
is no more referenced from R and garbage collection takes place? I need
the function to be applied for the whole class (let it be "someRClass")
like this someRClass.on.finalize<-function(...){...}. I have found
reg.
(Ted Harding) <[EMAIL PROTECTED]> writes:
>
> This is definitely a case where "dynamic rescaling" could save
> hassle! Brian Ripley's suggestion involves first building a
> matrix whose columns are the replications and rows the time-points,
> and Robin Hankin's could be easily adapted to do the s
My data are indeed bernoulli and not binomial, as I indicated. The
dataset consists of points (grid refs) that are either locations of
events (animals) or random points (with no animal present). For each
point I have a suite of environmental covariates describing the habitat
at this point. I was an
Hi R users and developers:
I want to know how can I save memory in R
for example:
- saving on disk a matrix.
- using again the matrix (changing their values)
- saving again the matrix on disk in a different file.
The idea is that I have a process that generate several
matrices, but if I kee
On 13 Jul 2005, at 11:01, (Ted Harding) wrote:
> On 13-Jul-05 Prof Brian Ripley wrote:
>
>> For most purposes it is easiest to use matplot() to plot superimposed
>> plots like this. E.g.
>>
>> x <- 0.1*(0:20)
>> matplot(x, cbind(sin(x), cos(x)), "pl", pch=1)
>>
>
> This, and Robin's suggestion,
If you want to append to the first/last column or row, then use cbind or
rbind. It is a little tricky if you want to insert a row in the middle
somewhere. See insertRow in micEcon package.
Regards, Adai
On Wed, 2005-07-13 at 12:08 +0200, ecoinfo wrote:
> Hi,
> May I ask a related question, how t
Hello
Thanks for the replies. Merge was what I needed! But Christoph, I will
keep your email. What you described is something else I have been
wondering how to do in R...
Thanks again
Karen
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You seem to want to model spatially correlated bernoulli variables.
That's a difficult task, especially as these are bernoulli and not
binomial(n>1). With a much fuller description of the problem we may be
able to help, but I at least have no idea of the aims of the analysis.
glmmPQL is designe
Ivy_Li wrote:
> Dear all,
> I really appreciate your help. I think I have a little advancement. ^_^
> Now I use the package.skeleton() function to create a template. I type:
> f <- function(x,y) x+y
> g <- function(x,y) x-y
> d <- data.frame(a=1, b=2)
> e <- rnorm(1000)
>
Hi , R Users
I'm trying to fit a Weibull ditribution on observed percentiles using nls but
it doesn't work. Here is the code I use: is there something wrong ?
# p corresponds to percentiles
# and q to the observed values
# the datas are from the livebirth in france in 1998 distribution
ined1998
On 13-Jul-05 Prof Brian Ripley wrote:
> For most purposes it is easiest to use matplot() to plot superimposed
> plots like this. E.g.
>
> x <- 0.1*(0:20)
> matplot(x, cbind(sin(x), cos(x)), "pl", pch=1)
This, and Robin's suggestion, are good practical solutions especially
when only a few graphs
Hi,
May I ask a related question, how to insert a row/column to a matrix?
Thanks
Xiaohua
On 13 Jul 2005 11:38:30 +0200, Peter Dalgaard <[EMAIL PROTECTED]>
wrote:
>
> Navarre Sabine <[EMAIL PROTECTED]> writes:
>
> > Hi,
> > I would like to know if it's possible to delete a rox from a matrix?
Fernando Espíndola wrote:
> Hi user R,
>
> I am try to calculate the spectrum function in two time series. But when plot
> a single serie, the labels in axes x is in the range 0.1 to 0.6 (frequency),
> but when calculate de spectrum with ts.union function, the labels x is in the
> range 1 to 6
Good morning,
I used in R contr.sum for the contrast in a lme model:
> options(contrasts=c("contr.sum","contr.poly"))
> Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit)
> intervals(Septo5.lme)$fixed
lower est. upper
(Intercept) 17.0644033 23.106110 29.147816
Variete1 9.5819873
I am trying to fit a GARCH model in fSeries but up to now without
success. I downloaded the the OxConsole Software together with the
[EMAIL PROTECTED] 4.0 package and saved the oxl.exe and GarchOxModelling.ox
files
correctly in the files C:\\Ox\\bin\\oxl.exe and
C:\\Ox\\lib\\GarchOxModelling.ox.
> "BT" == Berwin A Turlach <[EMAIL PROTECTED]> writes:
>> tmp <- sapply(1:length(L), function(j, mat, list)
kronecker(X[j,,drop=FALSE], L[[j]]), mat=X, list=L)
Uups, should proof read more carefully before hitting the send
button. This should be, of course:
tmp <- sapply(1:length(L),
Hi.
I'm trying to perform what should be a reasonably basic analysis of some
spatial presence/absence data but am somewhat overwhelmed by the options
available and could do with a helpful pointer. My researches so far
indicate that if my data were normal, I would simply use gls() (in nlme)
and one
Navarre Sabine <[EMAIL PROTECTED]> writes:
> Hi,
> I would like to know if it's possible to delete a rox from a matrix?
>
> > fig
> [,1] [,2] [,3] [,4]
> [1,]01 0.0 0.2
> [2,]01 0.2 0.8
> [3,]01 0.8 1.0
> [4,]01 NA NA
> [5,]01 NA NA
>
>
> "RH" == Robin Hankin <[EMAIL PROTECTED]> writes:
RH> I want to write a little function that takes a matrix X of
RH> size m-by-n, and a list L of length "m", whose elements are
RH> matrices all of which have the same number of columns but
RH> possibly a different number of row
Hi,
I would like to know if it's possible to delete a rox from a matrix?
> fig
[,1] [,2] [,3] [,4]
[1,]01 0.0 0.2
[2,]01 0.2 0.8
[3,]01 0.8 1.0
[4,]01 NA NA
[5,]01 NA NA
I would like to delete the 2 rows with NA!
Thanks
Sabine
Hi
I want to write a little function that takes a matrix X of size
m-by-n, and a list L of length "m", whose elements are matrices all
of which have
the same number of columns but possibly a different number of rows.
I then want to get a sort of dumbed-down kronecker product in which
X[i,j] i
Hi David, Since I am looking at very extreme values, it appears I will
need FMLIB. Is it an R lib? if so which version? How/where can I
download it?
Regards.
- Original Message -
From: David Duffy <[EMAIL PROTECTED]>
Date: Wednesday, July 13, 2005 9:46 am
Subject: [R] exact values for
For most purposes it is easiest to use matplot() to plot superimposed
plots like this. E.g.
x <- 0.1*(0:20)
matplot(x, cbind(sin(x), cos(x)), "pl", pch=1)
On Wed, 13 Jul 2005, Robin Hankin wrote:
> Hi
>
> Ted makes a good point... matlab can dynamically rescale a plot in
> response
> to plot(
Please consult the reference on the help page of that function: it _is_
support software for a book. It implements the Box-Cox procedure (as it
says). The original Box-Cox paper has three aims, two of which you have
mentioned (but perhaps the most inportant one is the one you
have not mention
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