Hmmm, I did not follow this thread closely, sorry for that,
just want to share my 2c.
If it is about quality, then I create EPS files and use the
psfrag latex package to replace the PS fonts with TeX's fonts.
This has the following advantages:
1) The figures have the same font as the text its
On Wed, Jul 16, 2008 at 04:48:28AM -0500, Gabor Csardi wrote:
[...]
> I have a little script that automates this for .fig files (this is
> based on figtex, another script that I found somewhere online and
> can't find it any more)
[...]
Ok, it is called figfrag, and
Senthil,
you can try the 'igraph' package. Export your two-column Excel file
as a .csv, use 'read.csv' to read that into R, then 'graph.data.frame'
to create an igraph graph from it. Finally, call 'betweenness' on
the graph. It is really just three/four lines, something like this:
tab <- read.cs
I am attaching the Test.csv file for your
> experiments. Thank you very much again.
>
> Best regards,
> Senthil
> (909) 267-0799
>
> -Original Message-
> From: Gabor Csardi [mailto:[EMAIL PROTECTED]
> Sent: Monday, July 21, 2008 1:57 AM
> To: Senthil Purusho
cing values that are either smaller than
2 _OR_ larger than 3, no number is smaller than 2 _AND_ larger than 3,
at least if we consider the usual ordering on numbers.
Best,
Gabor
[...]
--
Csardi Gabor <[EMAIL PROTECTED]>UNIL DGM
__
R-help@r-pr
On Thu, Jul 24, 2008 at 09:30:54AM -0700, Nordlund, Dan (DSHS/RDA) wrote:
[...]
> > > a <- c(rep(seq(1,4),4),NA,NA)
> > > b <- c(rep(seq(1,2),7),NA,NA,1,2)
> >
> > Andreas,
> >
> > what is wrong with
> >
> > a[ (a < 2 | a > 3) & b==1 ] <- NA
> >
> > ? Isn't this what you want?
> >
[...]
>
>
On Thu, Jul 24, 2008 at 10:39:34AM -0700, Nordlund, Dan (DSHS/RDA) wrote:
[...]
> Yes, it does help. I was misunderstanding how logical values are
> used for indexing. I assumed incorrectly that a value would be
> returned only if the index expression evaluated as TRUE. It would
> seem that the
Jacob, yes, e.g.
rep( c(4,3,4,2,3,4,1,2,3,4), 1 ) :)
Seriously, your question reminds me of (rather silly) television
contents, when one should find _the_ rule in a series of numbers. :)
Here is a solution, assuming I've found the "correct" rule:
unlist(lapply(4:1, seq, 4))
Best,
Gabor
On Fri
On Thu, Jun 05, 2008 at 10:26:08AM -0200, Alberto Monteiro wrote:
>
>
>
> Uwe Ligges wrote:
> >
> > You probably want
> >
> > write.table(t(read.table(file.in)), file = file.out, row.names =
> > FALSE, col.names = FALSE)
> >
> Ok, almost there. I forgot to tell (because I didn't know)
> that
Baptiste,
the igraph ARPACK interface is quite experimental, and igraph includes
only the ARPACK files (converted to C) that it needs to calculate
some graph measures on sparse graphs. Btw. the development version
of igraph is a bit better in this respect, I can send you a link to
the developme
Maybe I'm missing something, but where is the 3D here?
My tip is hist3d in package rgl. But there might be others,
it might worth to search the archive, I remember seeing this
question once.
Gabor
On Fri, Jun 20, 2008 at 08:55:36AM -0400, Richardson, Patrick wrote:
> Try
>
> ?hist
>
> -a
Nina, these are not row NUMBERS, but row NAMES. Numbers are actually
reset, they always start with 1 and they are continuous. Just try
doing
T[1,]
on your table. If you want to reset row names, you can do this:
rownames(T) <- seq(length=nrow(T))
or you can even remove them:
rownames(T) <- N
Some clarifications.
R's license (GPL v2) is not about money,
you can charge anyone as much as you wish.
If you create an R program (and don't modify R itself), then
you can distribute that program according to any license you wish.
If you modify R itself _and_ distribute the modified version,
Hmmm, this is not very good:
> Vec <- c(10:1,1)
> Vec[ table(Vec) == 1 ]
[1] 9 8 7 6 5 4 3 2 1
and these are obviously not the unique values.
This one is better:
Vec [ ! duplicated(Vec) & ! duplicated(Vec, fromLast=TRUE) ]
Gabor
On Wed, Jun 25, 2008 at 11:29:31AM -0500, Marc Schwartz wrote
I'm sorry to say, but this one is wrong, too.
Maybe coffee really helps, I just had one. :)
> Vec <- c(20:30,20)
> which(table(Vec) == 1)
21 22 23 24 25 26 27 28 29 30
2 3 4 5 6 7 8 9 10 11
You would actually need the names, but that would involve
some numberic -> character -> numeric
Wow, that is smart, although is seems to be overkill.
I guess 'duplicated' is better than O(n^2), is it really?
Gabor
On Wed, Jun 25, 2008 at 05:43:30PM +0100, Prof Brian Ripley wrote:
> On Wed, 25 Jun 2008, Marc Schwartz wrote:
>
>> on 06/25/2008 11:19 AM Daren Tan wrote:
>>>
>>> unique(c(1
Wanding,
I'm the maintainer of igraph, but missed your previous email.
Yes, currently the released version of igraph fails to compile
with gcc 4.3.x. I made the required modifications to fix this,
but these are still in the igraph development tree, as there has been
no release since that.
Yo
paste(sep="", "graf", 1:250, ".jpg")
See ?paste,
G.
On Mon, Jun 30, 2008 at 11:58:51AM -0300, Leandro Marino wrote:
> Hi list,
>
> I want to make a lot of graphics to my end course project. So, i was using
> this sintax:
>
>
> jpeg(filename = "graf01.jpg", width = 1024, height = 1024,
> u
I think there are many simple solutions, here is one:
lapply(1:92, function(x) c(2*x-1, 2*x))
Gabor
On Tue, Jul 01, 2008 at 02:46:07PM +0200, Boks, M.P.M. wrote:
> Dear experts,
>
> For the makeGenotype function I need a list as in the example. However,
> since my list needs to be 184 long the
A data frame is a special list:
> d <- data.frame( A=numeric(), B=logical(), C=character() )
> lapply(d, class)
$A
[1] "numeric"
$B
[1] "logical"
$C
[1] "factor"
Gabor
On Tue, Jul 01, 2008 at 03:50:18PM +0200, Dong-hyun Oh wrote:
> Dear UseRs,
>
> I would like to know the way to find classes o
If this is about more than a handful files, then it is really painful
to do it with OpenOffice.org or LyX, I guess.
You can use imagemagick, this is fairly standard on Linux. Then it is
something like this, assuming you have bash:
for f in *.png; do convert $f ${f%png}pdf; done
for f in *.jpg; d
Ooops, please ignore my previous mail, I did not read the
question carefully enough.
Gabor
On Tue, Jul 08, 2008 at 02:27:51AM -0700, Mark Difford wrote:
>
> Hi Daren,
>
> Can R (out)do Emacs? I think you just need to ?Sweave a little.
>
> Mark.
>
>
> Daren Tan wrote:
> >
> >
> > I ha
E.g.
> plot(1:10,1:10,xlab=NA)
> title(xlab=expression(mu*"mol"/10^6*" cells"))
Gabor
On Wed, Jul 09, 2008 at 11:21:46AM +0200, Dani Valverde wrote:
> Hello,
> I am creating a plot and I would like to know how to put this expression
> to the y axis
>
Why don't you write it for yourself, it takes less time than writing
an email:
mysummary <- function(x) {
require(plotrix)
require(e1071)
c(Mean=mean(x), Std.Error=std.error(x), Std.Deviation=sd(x),
Kurtosis=kurtosis(x))
}
Gabor
On Wed, Jul 09, 2008 at 08:15:00AM -0700, nmarti wrote:
>
It is called 'rev', see ?rev.
> rev(1:10)
[1] 10 9 8 7 6 5 4 3 2 1
G.
On Thu, Jul 10, 2008 at 01:56:58PM +0200, Zroutik Zroutik wrote:
> Dear R-users,
>
> I'd like to turn a vector so it starts with it's end. For better
> understanding, this set of commands will do what I need:
>
> i
I'm sure this is possible with 'network', but i'm not very familiar
with that package. In case you don't get an answer on how
to do it with network, here is how to do it with the 'igraph' package:
library(igraph)
M <- matrix(runif(100)*2-1, 10, 10)
M[ lower.tri(M, diag=TRUE) ] <- 0
M[ abs(M) <
is it possible to show the names in
> the nodes of the graph (currently it just shows the row number)?
>
> Your help is much appreciated
>
> Kind regards
>
> Jonathan
>
> -Original Message-
> From: Gabor Csardi [mailto:[EMAIL PROTECTED]
> Sent:
indexing
> > gdata2
> Vertices: 10
> Edges: 4
> Directed: FALSE
> Edges:
>
> [0] 1 -- 5
> [1] 2 -- 6
> [2] 3 -- 7
> [3] 4 -- 7
> >
> -Original Message-
> From: Gabor Csardi [mailto:[EMAIL PROTECTED]
> Sent: 11 July 2008 14:54
>
Maybe there is a simpler way, but this works fine:
> l1 <- 1
> l2 <-2
> m <-10
> ls()
[1] "l1" "l2" "m"
> rm(list=grep("^l.*", ls(), value=TRUE))
> ls()
[1] "m"
>
You can supply a regular expression to grep.
Gabor
On Mon, Jul 14, 2008 at 10:45:13AM +0200, Oehler, Friderike (AGPP) wrote:
> Dea
What about reading the very last four lines of any email
you get from the list? Like this one.
G.
On Wed, Mar 19, 2008 at 12:09:29PM -0400, ablukacz wrote:
> Dear All,
>
> Can someone please give me instruction on how to unsubscribe from this
> list. I do not have the original emial that arrive
On Wed, Mar 19, 2008 at 12:56:13PM -0700, jeffreya wrote:
>
> Hi.
>
> I'm looking to create a user-friendly program built around some R methods
> I've written. The program should be as easy to install and use as possible
> and will be built around a GUI. This program will be cross-platform; that'
If you do
help.search("download")
you find
?download.file
G.
On Thu, Mar 20, 2008 at 04:51:22PM -0500, gilbert feng wrote:
> Hi, everyone
>
> I want to download a XML webpage and save it as a file in my local machine.
> Is there any way to do it in R?
>
> Thanks a lot
>
> Gilbert
>
>
Read R FAQ 7.31 ?
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
Gabor
On Fri, Mar 21, 2008 at 04:17:28PM +0100, John Lande wrote:
> dear all,
>
> I report a problem very simple, that I does non know how to handle.
>
> look at the following co
May i ask what was the problem with symbols()?
G.
On Sun, Mar 23, 2008 at 04:10:38AM -0700, ermimi wrote:
>
> Thank you very much for the help!!!
>
> Felix Andrews wrote:
> >
> > help.search("circle")
> >
> > should point you to grid.circle in the grid package, at least in my
> > R version 2.
On Sun, Mar 23, 2008 at 11:06:05AM -0400, Mark Leeds wrote:
> In an earlier post, a person wanted to divide each of the rows of
>
> rawdata by the row vector sens so he did below but didn't like it and
>
> asked if there was a better solution.
>
>
>
> rawdata <- data.frame(rbind(c(1,2,2), c
Yes, it is exactly 'apply', and its friends. E.g. you can collect the
objects into a list and then do
sapply(mylist, is.matrix)
G.
On Wed, Apr 09, 2008 at 11:52:08AM -0400, Mon Mag wrote:
> I would like to apply a simple function, like
> is.matrix
> to more than one data.frame
> How can I call
On Mon, Apr 14, 2008 at 08:32:55PM +0800, Ng Stanley wrote:
> Hi,
>
> Two questions:
>
> A) Assuming OB is an object, how do I store 20 of OB in a vector or list ?
replicate(20, OB, simplify=FALSE)
> B) Does R has something similar associative array to Perl ? For example,
> %our_friends = ('bes
On Mon, Apr 14, 2008 at 08:52:36PM +0800, Ng Stanley wrote:
> Hi,
>
> Didn't make myself clear on A). The twenty OBs are all different, how to
> store them in a vector or list ?
bigOB <- list(OB1, OB2, OB3, ..., OB20)
G.
[...]
--
Csardi Gabor <[EMAIL PROTECTED]>UNIL DGM
_
On Mon, Apr 14, 2008 at 09:47:49PM +0800, Ng Stanley wrote:
> Hi,
>
> Two questions:
>
> A) I need to initialize many variables to NULL. So I created variable_names
> <- c("a1", "a2"). What can I do to variable_names so that variable a1 is
> NULL and a2 is NULL ?
for (n in variable_names) assign
force the assignment ?
>
> > RG[["ABC"]] <- c("a", "b")
> Error in RG[["ABC"]] <- c("a", "b") :
> more elements supplied than there are to replace
>
>
> On Mon, Apr 14, 2008 at 9:53 PM, Gabor Csardi
I'm sure you'll get a friendlier answer, but... see
?"="
?"=="
Introduction to R
G.
On Tue, Apr 15, 2008 at 05:28:53AM -0700, Linn wrote:
>
> Hi
> Could anyone please explain to me the difference between the = and the ==?
> I'm quite new to R and I've tried to find out but didn't get any wiser
next
break
Another 'Introduction to R', or even ?"for" question
G.
On Fri, Apr 18, 2008 at 04:55:01PM +0800, Ng Stanley wrote:
> Hi,
>
> Is there any function to skip a loop in a for loop ?
>
> Thanks
> Stanley
>
> [[alternative HTML version deleted]]
>
> __
Hmm, my understanding is different,
m <- matrix(sample(10*10), ncol=10)
m2 <- rbind( m[1:5,], 1:10, m[6:10,] )
m3 <- cbind( m[,1:8], 1:10, m[,9:10] )
G.
On Sun, Apr 20, 2008 at 10:21:47AM -0300, Henrique Dallazuanna wrote:
> If I understand:
>
> m <- matrix(sample(10*10), ncol=10)
> m[5:6, 8:9]
On Sun, Apr 20, 2008 at 08:16:11PM +, David Winsemius wrote:
> Gabor Csardi <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]:
>
> > Hmm, my understanding is different,
> >
> > m <- matrix(sample(10*10), ncol=10)
> > m2 <- rbind( m[1:5,], 1:
On Mon, Apr 21, 2008 at 12:50:08PM +, David Winsemius wrote:
[...]
>
> Am I correct in assuming that after the creation of m by way of a
> temporary matrix that the temporary matrix would then be available for
> garbage collection, whereas if both m and m2 were created, there would
> be mor
See ?apply
M2 <- M[ apply(M!=0, 1, any), , drop=FALSE]
Gabor
On Tue, Apr 22, 2008 at 11:52:08AM +0200, Patrick Zimmermann wrote:
> Dear R-community,
> I have matrices/tables of different sizes which may contain rows with
> only zeros. Now I would like to delete these zero lines or create new
> m
I would rather not comment on matlab (where is
your matlab code by the way?), but your function
could be simplified a bit:
grw.permute <- function(v) {
cbind( rep(v, each=length(v)), rep(v, length(v)) )
}
> system.time(tmp <- f( 1:300))
user system elapsed
0.020 0.000 0.019
This is
But please consider that this benchmark is five years old, and i believe
that R has changed quite a lot since version 1.9.
Gabor
On Wed, Apr 30, 2008 at 04:21:51PM -0400, Wensui Liu wrote:
> Hi, ZD,
> Your comment about speed is too general. Here is a benchmark
> comparison among several langua
On Wed, Apr 30, 2008 at 06:59:38PM -0400, esmail bonakdarian wrote:
>
> This has been an interesting discussion, and brings up two questions
> for me:
>
> Is there a good collection of hints/suggestions for R language idoms in terms
> of efficiency? For instance I read not to use for-loops, so I
Solomon, if i understand two-mode networks properly (they're bipartite, right?),
then this is not hard to do with igraph. Basically, for each vertex create an
order=2 neighborhood, and then create a graph from the adjacency list,
it is something like this:
two.to.one <- function(g, keep) {
neis
See ?cbind and ?matrix.
Gabor
On Sat, May 10, 2008 at 03:21:26PM -0700, Claire_6700 wrote:
>
> Hello,
>
> I have two data.
>
> x<-c(1, 2, 1, 3, 2)
> y<-c(3, 1, 2, 3, 5)
>
> How do i create matrix from this two.
>
> what i want is this
>
> x y
> 1 1 3
> 2 2 1
> 3 1 2
> 4 3
> a <- 1
> x <- "a"
> rm(list=x)
> a
Error: object "a" not found
See ?rm for details.
Gabor
On Tue, May 13, 2008 at 05:13:41PM +0530, Shubha Vishwanath Karanth wrote:
> Hi R,
>
>
>
> A simple question, but don't know the answer...
>
>
>
> x="a"
>
> a=5
>
>
>
> I need to remove the o
they're assumed to be
different types of nodes. Just create the graph, calculate the 'keep'
parameter, I assume that you know this from external information, and
then call the function.
G.
> Solomon
>
>
> >-Original Message-
> >From: Gabor Csardi [mai
Please stay on the list.
On Tue, May 13, 2008 at 06:05:15PM -0400, Messing, Solomon O. wrote:
> Gabor,
>
> By the way, this seems to work:
I'm a bit lost. So now you're converting your data frame
to a matrix? Why? Or you're doing the two-mode to one-mode
conversion here? It does not seem so to
ix
> g3 = df.to.nxn(df$actor, df$event)
> g4 = graph.adjacency(g3, mode = "undirected", diag = F)
> V(g4)$name = row.names(g3)
> g4
> ###
>
> This yields:
> > g4
> Vertices: 4
>
is.character(dd) && length(dd) == 0
should do it i think.
Gabor
On Thu, Nov 15, 2007 at 04:54:45PM -0500, Gang Chen wrote:
> I want to identify whether a variable is character(0), but get lost.
> For example, if I have
>
> > dd<-character(0)
>
> the following doesn't seem to serve as a good
Eleni, this question appears about every month on this list,
try using the RSiteSearch command before posting. Thanks.
RSiteSearch("replace NA")
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/109176.html
Gabor
On Wed, Nov 21, 2007 at 01:15:32PM +0200, Eleni Christodoulou wrote:
> Hi all!
>
>
John, what about
http://cran.r-project.org/src/contrib/PACKAGES.html
Isn't this good enough? You might also take a look at
http://cran.r-project.org/src/contrib/Views/
Gabor
On Wed, Nov 21, 2007 at 09:24:14AM -0500, John Sorkin wrote:
> Fellow Rers,
>
> Please forgive me if I have posted this
Please someone correct me if i'm wrong, but i think this is impossible
with the current R language. ':' is an ordinary function (see
get(":")) just like "[", so v[1:3] is actually the composition of two
functions, it is the same as "["(":"(1,3)). The ":" has no idea about
whether it'll be embedded
rm(list=ls())
G.
On Sun, Nov 25, 2007 at 01:12:19PM +0100, Jonas Stein wrote:
> Hi,
>
> as i am quite new to R i often play around with commands in R until my graph
> looks nice.
>
> After testing things there are often lots of variables left. How can i reset
> all vars in one command before i
> x <- c(1,4,3)
> freq <- c(3,2,4)
> rep(x, freq)
[1] 1 1 1 4 4 3 3 3 3
Gabor
On Mon, Nov 26, 2007 at 07:34:36PM +, Paul Smith wrote:
> Dear All,
>
> Suppose that you have the following data:
>
> X Frequency
> 1 3
> 4 2
> 3 4
>
> To get a vector with all observations, one c
Try some other mirror, or the main site:
http://cran.r-project.org/src/contrib/Descriptions/codetools.html
You can easily install it via
install.packages("codetools")
and then selecting an appropriate mirror.
Gabor
On Mon, Nov 26, 2007 at 08:19:39PM +, Martin Waller wrote:
> Hmm - I looked
Edna,
maybe there is a way, but Rd files are not that difficult,
writing an empty one requires less time than writing this email.
Then just copy your empty file as many times as you want.
Gabor
ps. perhaps you didn't install the manuals. You can always read
the up-to-date versions at the R home
Yes, it is indeed true for other systems as well, although
some configuration problems might arise, at least on Linux.
It is also true that there are several small Linux distributions
which easily fit into a flash drive, and then you can boot from the
flash drive. I used to use SLAX, this is mo
You can easily install ubuntu on it (although it might require
an external drive):
http://hup.hu/node/48116
So running R should not be problem.
G.
On Sun, Dec 16, 2007 at 12:56:39PM -0500, Burton Rothberg wrote:
> I'm thinking of getting one of these lightweight linux laptops for
> traveling. Doe
paste(rep(c("Factor", "Sign Factor"), 5), rep(1:5, each=2))
Replace '5' with the desired number,
Gabor
On Mon, Dec 17, 2007 at 03:08:09PM +0100, Jonas Malmros wrote:
> Hello,
>
> I have a vector of names, say :
>
> names <- c("Factor 1", "Factor 2", Factor 3")
>
> I am creating a dataframe and
miracle <- function(x) { lapply(seq(along=x), function(y) x[1:y]) }
Gabor
On Tue, Dec 18, 2007 at 11:40:37PM +0100, Johannes Graumann wrote:
> Hi all,
>
> What may be a smart, efficient way to get the following result:
>
> myvector <- c("A","B","C","D","E")
> myseries <- miracle(myvector)
> mys
On Wed, Dec 19, 2007 at 12:01:25AM +0100, Johannes Graumann wrote:
> Debugged version:
> lapply(1:length(myvector), function(.length) {
> myvector[1:.length]
> })
>
> Thanks for showing the direction!
>
> Joh
Note that this fails if length(myvector)==0.
Good to know the corner cases.
Gabor
>
Joh,
x <- c(1,2,3,4,7,8,9,10,12,13)
which(diff(x) != 1)
gives the indices of the 'jumps'.
Gabor
On Thu, Dec 20, 2007 at 10:43:05PM +0100, Johannes Graumann wrote:
> Hi all,
>
> Does anybody have a magic trick handy to isolate directly consecutive
> integers from something like this:
> c(1,2,3,
Well i don't know which Unix you have, but it shouldn't matter anyway.
If you're able to compile programs on the machine then download
the R source, compile it and install it into your local directory.
Gabor
On Fri, Dec 21, 2007 at 10:25:36AM -0500, Wensui Liu wrote:
> Good morning, Dear Lister,
l dir? do you mean dir under my user name with my profile?
I mean any directory you've write access to, on a disk where you've
enough space to install R.
Gabor
> On Dec 21, 2007 10:42 AM, Gabor Csardi <[EMAIL PROTECTED]> wrote:
> > Well i don't know which Unix y
Keith, are you looking for 'cumsum' ?
Gabor
On Sat, Jan 05, 2008 at 08:32:41AM -0600, Keith Jones wrote:
> Hi,
>
> Maybe I have not been looking in the right spot, but, I have not been
> able to fine a command to automatically calculate the running
> cumulative sum of a vector. Is there such
Lorenzo, why can't you actually generate the graph to find the
connection components? With the 'igraph' package this is something like:
g <- graph.adjacency( DIST < 0.5, mode="undirected" )
g <- simplify(g)
no.clusters(g)
assuming you have your distance matrix in 'DIST'. If N is too big
then you
if the sorted coordinates
are
x1
x2
x3
x4
then for finding the pairs of particle 1 you only need to check
particle 2 if x2-x1 Cheers
>
> Lorenzo
>
> On 07/01/2008, Gabor Csardi <[EMAIL PROTECTED]> wrote:
> > Lorenzo, why can't you actually generate the graph to f
names(alninc) <- c("F","V","G")
See ?names
Gabor
On Mon, Jan 07, 2008 at 01:35:59PM -0500, Wade Wall wrote:
> Hi all,
>
> I am trying to name list objects, but am having trouble doing so. At the
> moment I have list alninc that has 3 objects and I can refer to them as
> alninc[[1]] . . .alnin
Bram,
On Tue, Jan 08, 2008 at 03:11:07PM +0100, Bram Kuijper wrote:
> Hi all,
>
> As an R newbie, I wonder how I can store multiple matrices in a list()
> or vector() object without losing their structure. I should be able to
> retrieve the matrix from the list later on.
>
> If I just append()
On Wed, Jan 09, 2008 at 08:03:59AM +, Prof Brian Ripley wrote:
[...]
>
> You say you are using gcc, but that would be relevant to the Sun C++
> compiler (see the R-admin manual, which also says that configure adds it
> where needed). So which C++ compiler is this? The symptoms do look as i
Peter,
the changes you made should be ok. (No changes are needed for gcc/g++
anyway.) The problem is not the C compiler, which is (most likely)
gcc, but the c++ compiler, which is not the GNU
c++ compiler, but the Sun version. I guess that the GNU c++ compiler
is called g++ or gxx. If you found
Dear List,
i know there are some solutions for this in the archive,
but they're not very good for numeric matrices, since they
usually convert rows/columns to character strings. Is there
an easy way to do $subject for numeric matrices properly,
or i need to do it by hand?
Thanks,
Gabor
___
Hmmm, the only *perfect* way i know is to store the data internally
in a package, and implement all operations accessing it via
public functions. (Of course the data object itself is not exported
from the package.) This might be overkill, but it really works.
But there might be other ways, i'm not
rame
> Y <- data.frame( a=1:5, b=2:6, c=1:5)
> length(unique(t(Y)[,1]))
>
> --- Gabor Csardi <[EMAIL PROTECTED]> wrote:
>
> > Dear List,
> >
> > i know there are some solutions for this in the
> > archive,
> > but they're not very good for
On Sat, Jan 12, 2008 at 12:35:47PM -0500, John Kane wrote:
> I definately did not read it that way but that may
> have been my fault. That table approach is quite
> nice!
>
> Using it, you could just rebuild the vectors from the
> names. Does this do more or less what you want?
John, thanks. Sti
Chuck, thanks a lot, this is a very good starting point.
G.
On Sat, Jan 12, 2008 at 01:08:11PM -0800, Charles C. Berry wrote:
[...]
> Gabor,
>
> Try this. Order the matrix rows, conpare adjacent rows, and run length
> encode the logical vector of comparisons. Decode the rle() result to get
> t
Are you sure this is an R problem? I guess it is not, but
rather a pstoedit problem.
You can use the RSVGDevice package, or the Cairo package to
export your graphics directly into SVG from R.
Gabor
On Mon, Jan 14, 2008 at 07:10:13PM +0100, Mag. Ferri Leberl wrote:
> Dear everybody!
> I am maki
ce.
Just before you start installing Rtools.
Gabor
--
Csardi Gabor <[EMAIL PROTECTED]>UNIL DGM
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org
See ?outer
outer(Z, Z, function(x,y) x/y)
Gabor
On Thu, Jan 17, 2008 at 01:24:33PM -0300, Juan Pablo Fededa wrote:
> Dear Contributors:
>
> I have the next vector:
>
> "Z"
>
> 526
> 723
> 110
> 1110
> 34
> 778
> 614
> 249
> 14
>
> I want to generate a vector containing the ratios of all the
sked I chose for the Bamberg-mirror.
> Can you recognize a mistake? Should I chose for a different mirror?
> Thank you in advance.
> Yours, Mag. Ferri Leberl
>
>
>
>
> Am Montag, den 14.01.2008, 19:21 +0100 schrieb Gabor Csardi:
>
> Are you sure this is an
Eleni, download the package (I assume you know where it is),
on Linux you will need the source package. Then from R type
install.packages("", repos=NULL)
Gabor
On Tue, Jan 22, 2008 at 11:26:12AM +0200, Eleni Christodoulou wrote:
> Hi all,
>
> I am trying to install the package "GEOquery" in un
n't think it hurts if there are none).
>
> For Bioconductor packages, use biocLite()
>
> source("http://www.bioconductor.org/biocLite.R";)
> biocLite("GEOquery")
>
> Best,
>
> Jim
>
>
> Gabor Csardi wrote:
> >Eleni, download the p
It is definitely on CRAN:
http://cran.at.r-project.org/src/contrib/Descriptions/lattice.html
You can download and install it "by hand", but if you want to
make install.packages work, please give us at least an error message
or something.
G.
On Tue, Jan 22, 2008 at 03:29:19PM +0100, Armin Goralc
x
> ** building package indices ...
> * DONE (survival)
>
> The downloaded packages are in
> /tmp/RtmpHXe9Wk/downloaded_packages
> > install.packages('lattice')
> Warning in download.packages(unique(pkgs), destdir = tmpd, available =
> available, :
>
On Thu, Jan 24, 2008 at 03:03:22PM -0500, David Afshartous wrote:
>
> All,
>
> I'm trying to obtain a one-liner to generate a certain sequence of
> alternatign numbers.
>
> Consider:
> > unlist(rep(list(c(1,2), c(3,4)), each = 6))
> [1] 1 2 1 2 1 2 1 2 1 2 1 2 3 4 3 4 3 4 3 4 3 4 3 4
>
> I'd l
What about reading the last four lines of this email?
G.
On Fri, Jan 25, 2008 at 04:45:04PM +0100, [EMAIL PROTECTED] wrote:
> How can I join two lists? I have q1 and q2 and I want to merge them. I
[...]
> __
> R-help@r-project.org mailing list
> http
> library(help=lattice)
[...]
Built: R 2.6.0; i486-pc-linux-gnu; 2008-01-23 13:52:49; unix
[...]
G.
On Sat, Jan 26, 2008 at 11:44:53AM -0500, Charles Annis, P.E. wrote:
> Greetings, R-ians:
>
>
>
> I would like to know which version or R was used to create a given package.
> I think I
RSiteSearch("R asus")
G.
On Mon, Jan 28, 2008 at 03:31:45PM +0100, Dr. Walter H. Schreiber wrote:
> Dear list,
>
> I wonder if somebody has succeeded in installing R on an eeePC (Xandros
> desktop). Searching via Rseek (term eeePC) and in eeePC forums (term
> Cran) left me without proper hit
Even easier than you think... G.
> letters
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s"
[20] "t" "u" "v" "w" "x" "y" "z"
On Tue, Jan 29, 2008 at 04:39:11AM -0800, skestin wrote:
>
> I suppose it's very simple but I can't find the way to generate a sequence of
What exactly is the question? Selecting/permuting rows?
M[sample(length=nrow(M), count), ]
Selecting/permuting columns?
M[ , sample(length=ncol(M), count) ]
Permuting elements?
structure(sample(M), dim=dim(M))
Selecting elements?
sample(M, count)
Gabor
On Thu, Jan 31, 2008 at 09:10:11PM +0
I mean
On Thu, Jan 31, 2008 at 02:32:20PM +0100, Gabor Csardi wrote:
> What exactly is the question? Selecting/permuting rows?
>
> M[sample(length=nrow(M), count), ]
M[sample(seq(length=nrow(M)), count), ]
> Selecting/permuting columns?
>
> M[ , sample(length=ncol(M), cou
You don't need root access to compile a program (in mose cases).
Just a compiler and enough space.
Gabor
On Thu, Jan 31, 2008 at 03:45:51PM -0500, Wensui Liu wrote:
> I think the unix is SunOS.
> the secret is I don't have root priviledge. ^_^. So is there a possibility?
>
> Thanks.
>
> On Jan
?save
?load
Gabor
ps. although i'm not sure what an Rdata-project means, so maybe you
need something else
On Fri, Feb 01, 2008 at 08:24:32AM +0200, Atte Tenkanen wrote:
> Dear R-users,
>
> How do you save a big table or matrix as an independent object and attach it
> to your Rdata-project when
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