Gabor Grothendieck wrote:
> Try:
>
> http://news.gmane.org/gmane.comp.lang.r.general
Fantastic - perfect. Saves my inbox filling up!
Thanks
Martin
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the pos
Greetings,
I have a large dendrogram that I would like to use to extract groups. But any
group with 2 or fewer members is not useful to me, so I would like to remove
all of the nodes with less than 3 members from the dendrogram.
1) Is there a good way to iterate through a dendrogram list to ch
It worked. But this student of R is left wondering why or how. In
particular the construction [.][.] was puzzling. I doubt that
matters greatly, but I am using R in both a Mac (OSX 10.2 v 2.0.1) and
WinXP (v 2.4.1.)
I have broken it into what I think are its pieces:
> table(my.mat)[-1]
my.mat
Hello to everyone!
I have a question for you..I need to predict multivariate time series, for
example sales of 2 products related one to the other, having the 2 prices
like inputs..
Is there in R a function to do it? I saw dse package but I didn't find what
a I'm looking for..
Could anyone help me
R Help
Feel sure there is a simple answer to this but answer has eluded me so far.
NB. Using R 2.6.0.
I am plotting a simple chart using plot():
plot(df, ylim=c(as.numeric(min(df)), as.numeric(max(df))), col="OliveDrab",
xlab="", ylab="")
What I would like to do is have this chart appear such
On Nov 8, 2007 11:22 AM, Van Campenhout Bjorn
<[EMAIL PROTECTED]> wrote:
> On 11/6/07, Van Campenhout Bjorn <[EMAIL PROTECTED]> wrote:
> >> Hi all,
> >>
> >> I made a dotplot() with lattice, which comes out nice on the graphics
> >> device. I can save this as a eps using postscript() and inc
On 11/11/2007 11:49 AM, Alex Park wrote:
> R Help
>
> Feel sure there is a simple answer to this but answer has eluded me so far.
>
> NB. Using R 2.6.0.
>
> I am plotting a simple chart using plot():
>
> plot(df, ylim=c(as.numeric(min(df)), as.numeric(max(df))), col="OliveDrab",
> xlab="", ylab
U might look at Vector Auto regression model. Try library(mAr)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Giusy
Sent: Sunday, November 11, 2007 10:11 PM
To: r-help@r-project.org
Subject: [R] Multivariate time series
Hello to everyone!
I have a quest
You may want to have a look at the vars package
Frank
Giusy schrieb:
> Hello to everyone!
> I have a question for you..I need to predict multivariate time series, for
> example sales of 2 products related one to the other, having the 2 prices
> like inputs..
> Is there in R a function to do it? I
On Nov 11, 2007 11:23 AM, David Winsemius <[EMAIL PROTECTED]> wrote:
> It worked. But this student of R is left wondering why or how. In
> particular the construction [.][.] was puzzling. I doubt that
> matters greatly, but I am using R in both a Mac (OSX 10.2 v 2.0.1) and
> WinXP (v 2.4.1.)
>
> I
Hi - I follow some references and now implement my own state-space model
estimation. I have a question. In case, my equations are like this:
y(t) = Ax(t)+Bu(t)+eps(t) # observation eq
x(t) = Cx(t-1)+Du(t)+eta(t) # state eq
Using EM, after backward recursion, you will use the smoothed state estim
Hi to all,
I need to append/delete stopwords from the list that i can use from de
TM package. I use Portuguese stopwords.
When i see the list of stopwords using >stopwords("portuguese") I have
some words with special characters like this:
"verdadeiro""você" "vocês""vos"
I t
Hi Gabor,
I replaced multiple spaces with a single one and tried
the code you suggested. I got:
> library(sqldf)
Loading required package: RSQLite
Loading required package: DBI
Loading required package: gsubfn
Loading required package: proto
> source("http://sqldf.googlecode.com/svn/trunk/R/sqldf
I obtain the coefficients of a nonlinear function
thus.
--
m=nls(y ~ a + b*x + c*x^2 + d*x^3 +
e*exp(x)+f*log(x)+g*log2(x), start = list(a = 0, b =
1, c = 1, d=1, e=1, f=1, g=1))
c=coef(m)
---
However, when I look at
Peter Lauren wrote:
>
> I obtain the coefficients of a nonlinear function
> thus.
> --
> m=nls(y ~ a + b*x + c*x^2 + d*x^3 +
> e*exp(x)+f*log(x)+g*log2(x), start = list(a = 0, b =
> 1, c = 1, d=1, e=1, f=1, g=1))
>
> c=coef(m)
> -
I've been looking for ways to calculate a large number (100) of non-crossing
Nonparametric quantile regressions on large populations (1000+).
Can the quantreg package in R ensure the non-crossing property?
If not, do you know any alternative?
Thank you,
Paulbegc
--
View this message in con
On 11/11/2007, at 1:43 AM, Uwe Ligges wrote:
> Paul Smith wrote:
>> Dear All,
>>
>> Can R perform multivariate integration with infinite limits of
>> integration?
>
> No, R does numerical (not symbolical) calculations, hence it can never
> perform integration (not even univariate) with infinite
On Nov 11, 2007, at 2:20 PM, paulbegc wrote:
>
> I've been looking for ways to calculate a large number (100) of non-
> crossing
> Nonparametric quantile regressions on large populations (1000+).
>
> Can the quantreg package in R ensure the non-crossing property?
You can estimate a (reasonably)
On 10/11/2007, at 3:48 AM, Irene Mantzouni wrote:
> Dear all,
>
> probably this is quite clear for most of you but for me it is a
> headache...
>
> I am regressing response A against the continuous covariate B and
> the relationship is clearly quadratic.
> When I add a second covariate B, the
Hello,
I am trying to use the "tree" package in R with RSperl.
I have generated a regression tree using the "tree" package in R and saved it
in a file say "RegTree_2.Rdata"
If I use the predict function from R command line, it works fine. However, I
want to use it from within perl using RSperl
Hello,
I am trying to use the "tree" package in R with RSperl.
I have generated a regression tree using the "tree" package in R and saved it
in a file say "RegTree_2.Rdata"
If I use the predict function from R command line, it works fine. However, I
want to use it from within perl using RSperl
Dear list,
I am new to R and very inexperienced. Sorry for the trouble.
I have two txt files and want to merge them by taking the average.
More specifically, for example, the txt file1, with row names and column names,
consists of 238000 rows and 196 columns. Each column corresponds
to a sample. T
Hi,
maybe I don't understand your question correctly, but I don't see a reason
why there should be a significant interaction term at all.
To the question why you regression is linear in B and C:
your regressor B is linear (in B), i.e. B= 1B.
your regressor C is some quadratic function of B, i.e.
On 10 November 2007 at 13:41, Uwe Ligges wrote:
| [EMAIL PROTECTED] wrote:
[...]
| I'd recommend to compile your version of R yourself from sources. It's
| easy, just follow the R Installation and Administration manual.
I think you need a more nuanced view than to always recommend compilation
fr
Here is one way of doing it:
> x
[,1] [,2] [,3]
[1,]246
[2,]12 NA
[3,]46 NA
> y
[,1] [,2] [,3]
[1,] NA44
[2,]22 NA
[3,]122
> z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
> dim(z) <- dim(x)
> z
[,1] [,2] [,3]
On Nov 11, 2007 2:28 PM, affy snp <[EMAIL PROTECTED]> wrote:
> Hi Gabor,
>
> I replaced multiple spaces with a single one and tried
> the code you suggested. I got:
>
> > library(sqldf)
> Loading required package: RSQLite
> Loading required package: DBI
> Loading required package: gsubfn
> Loading
What is the most efficient alternative to x[order(x)][1:n] where
length(x)>>n?
I also need the positions of the mins/maxs perhaps by preserving names.
Thanks for any suggestions.
--
View this message in context:
http://www.nabble.com/Largest-N-Values-Efficiently--tf4788033.html#a13697535
Sent
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Rachana Jain wrote:
> Hello,
>
> I am trying to use the "tree" package in R with RSperl.
>
> I have generated a regression tree using the "tree" package in R and saved it
> in a file say "RegTree_2.Rdata"
>
> If I use the predict function from R
You can apply 1:n to order(x) so you don't wind up
subscripting x by every element in order(x).
o <- head(order(x), n) # positions
x[o]
A completely different approach, if X is a data frame with d as a data
column is this where the row names give the positions (don't know
about speed):
> library
Is it possible to run a linux syntax (i.e. like i was at a linux
terminal) from an R code?
Just one example:
Let say that I have an R code with
hist(rnorm(100))
dev.copy2eps(file="Dnormal.eps")
and then within the same R code, i want to convert the file
'Dnormal.eps' to 'Dnormal.pdf'. This can
?system
On 12/11/2007, at 2:28 PM, Christian Salas wrote:
> Is it possible to run a linux syntax (i.e. like i was at a linux
> terminal) from an R code?
>
> Just one example:
> Let say that I have an R code with
>
> hist(rnorm(100))
> dev.copy2eps(file="Dnormal.eps")
>
> and then within the same
thanks,
I got it!
---
Christian SalasE-mail:[EMAIL PROTECTED]
PhD studenthttp://environment.yale.edu/salas
School of Forestry & Environmental Studies Tel: +1-(203)-432 51
G'day Diana,
On Fri, 9 Nov 2007 11:48:15 -0800 (PST)
<[EMAIL PROTECTED]> wrote:
> #general least squares model works fine
>
> glm.F <- glm(modF, data = F2, family = gaussian)
glm() fits generalised linear models which are different from
general(ised) least squares models. The latter can be fi
Hi Jim,
Thanks a lot! I am wondering why I ended up getting the result as follows:
> x<-read.table(file="x.txt",header=TRUE,row.names=1,na.strings = "NA")
Warning message:
In read.table(file = "x.txt", header = TRUE, row.names = 1, na.strings = "NA") :
incomplete final line found by readTableHe
Hi,
I'm plotting 5 autocorrelation plots on one page. Using
par(mfrow=c(3,2)) everything comes out fine. However, for
each plot, it prints a title on top of each plot that says
Series followed by the variable name used in the plot. I
want to suppress those titles, but I also want a general
What did your text files look like? It would appear that there was
not a line feed on the last line of the file. Also what does 'str' of
x and y show? It appears that one is a data frame and one is a
matrix. That might be causing some of the problems.
On Nov 11, 2007 10:30 PM, affy snp <[EMAIL
Try:
sort(x, partial=n)[1:n]
On Nov 11, 2007 6:43 PM, David Katz <[EMAIL PROTECTED]> wrote:
>
> What is the most efficient alternative to x[order(x)][1:n] where
> length(x)>>n?
> I also need the positions of the mins/maxs perhaps by preserving names.
>
> Thanks for any suggestions.
> --
> View t
Hi,Jim. I created two txt files as:
x.txt
id b1 b2 b3
a1 246
a2 12 NA
a3 46 NA
y.txt
idb1 b2 b3
a1 NA44
a2 22 NA
a3 122
I tried it one more time but got different z:
> x<-read.table(file="x.txt",header=TRUE,row.names=1,na.strings = "NA")
Here is the way to read the data and convert it. Your data was a
dataframe with the first column being the id:
> x <- read.table(textConnection("id b1 b2 b3
+ a1 246
+ a2 12 NA
+ a3 46 NA"), header=TRUE)
> y <- read.table(textConnection("idb1 b2 b3
+ a1 NA44
Try this:
par(mfrow = 1:2)
acf(cbind(" " = 1:10))
acf(cbind(" " = 1:10))
par(mfrow=c(1,1))
par(mfrow=c(1,1), oma=c(0,0,1,0))
mtext("My Title", 3, outer = TRUE, cex = par("cex.main"))
Please provide reproducible code next time as requested in
last line of every message to r-help.
On Nov 11, 2007
Hi Jim,
However, the real file has 238000 row ids and 196 column ids.
How could I read the file first and then convert it to a matrix?
Allen
On Nov 11, 2007 11:02 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> Here is the way to read the data and convert it. Your data was a
> dataframe with the f
Should be the exact same way. Are the columns all numeric? I assume
they would have to be if you are taking averages. What I sent should
work as is if the first column is coming in as the 'id' and all the
rest are numeric.
If you want, you can send me a subset of the data (e.g., 20 rows of
each
Thanks Jim. It works. I need to specify row.names as missing and then
get rid of the first column to make it as a matrix.
x<-read.table(file="x.txt",header=TRUE,row.names=NULL,na.strings = "NA")
x <- as.matrix(x[,-1])
y<-read.table(file="y.txt",header=TRUE,row.names=NULL,na.strings = "NA")
y
Unfortunately it is not quite as simple as this. It seems (to me) that
the change to the behaviour of as.numeric() in the latest release means
that the advice given in FAQ 7.10 (and the help) is incorrect when the
factor levels are integers. The following example illustrates:
> x1 <- c(1:4,4,4,4
I have a vector whose length is nearly 70 thousand.
I need randomize it for 1000 times .
for randomizing , I mean ,the elements of the vector remain intact while
their order in the vector get changed randomly.
I have written a function which seems to be able to solve short vectors ,
but waste a l
What is 'x' here? What type? Does it contain NAs? Are there ties? R's
ordering functions are rather general, and you can gain efficiency by
ruling some of these out.
See ?sort, look at the 'partial' argument, including the comments in the
Details. And also look at ?sort.list.
sort.int(x)
?sample
On Sun, 11 Nov 2007, leffgh wrote:
>
> I have a vector whose length is nearly 70 thousand.
> I need randomize it for 1000 times .
> for randomizing , I mean ,the elements of the vector remain intact while
> their order in the vector get changed randomly.
> I have written a function which
I don't know what you mean by 'the latest release', but R does not behave
this way in a vanilla session in 2.6.0 or R-patched. Had you given the
'at a minimum' information asked for in the posting guide, we might have
been able to deduce what the problem is. I get
> x1 <- c(1:4,4,4,4)
> f <-
Why didn't you do this in R?
> help.search("random sample")
Help files with alias or concept or title matching 'random sample'
using fuzzy matching:
sample(base)Random Samples and Permutations
Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
>
See ?sample, e.g
On my system
>system.time(x1 <- sort(x,decreasing=TRUE)[1:1000])
user system elapsed
0.030.000.03
whereas
> system.time(x1 <- x[order(x)][1:1000])
user system elapsed
0.110.000.11
I.e. using sort is about 30 times faster.
Best regards
Frede Aakmann Tøgersen
oh, thank you ,
I used the "sample"function to generate new subordinates
but never think that it can directly act on the vector .
thank you very much.
Frede Aakmann Tøgersen wrote:
>
> Why didn't you do this in R?
>
>> help.search("random sample")
> Help files with alias or concept or titl
Dear all,
I'm using the "ape" package in R and want to draw a
phylogenetic tree with not only the tip labels but
also some labels for the edges. e.g. Mark the edge AB
as "m" in the tree ABC.
Couldn't find a way to do that. Can someone help?
Thanks,
Hua
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