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, 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
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?
> >
[...]
>
>
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
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
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
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
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
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
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
>
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:
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) <
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
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:
>
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
>
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
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
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
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
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
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
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
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
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
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,
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
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
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
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
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
>
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
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
> 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
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
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
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
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
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
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
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
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:
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]
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]]
>
> __
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
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
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
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 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
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 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
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.
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
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
>
>
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'
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
First 'c' and then 'paste' with 'collapse':
paste(collapse="", c(c("a", "b", "c"), "d"))
See ?paste
G.
On Mon, Mar 17, 2008 at 04:26:17PM +0530, Ajay Shah wrote:
> How does one convert objects c("a","b","c") and "d" into "abcd"?
>
>> paste(c("a","b","c"), "d")
> of course yields
>[1] "
On Sat, Mar 15, 2008 at 04:33:32PM +0100, Christophe Genolini wrote:
> Hi the list
>
> Is it possible to create an empty matrix ? I do not mean an matrix with
> a single value that is NA (which is not empty) but a real empty one,
> with length=0.
Sure:
> matrix(nrow=0, ncol=5)
[,1] [,2] [
Try
temp[1,,drop=FALSE]
See ?"[" for the explanation.
Gabor
On Fri, Mar 14, 2008 at 04:46:10PM -0400, Gregory Gentlemen wrote:
> Hi fellow R-users,
>
> I have run into a problem when trying to identify the number of rows in a
> matrix. Say we have an arbitrary 5 by 5 matrix called temp:
>
>
keep <- apply( DATA, 1, min ) >= 100
DATA <- DATA[ keep, ]
See ?apply for more.
Gabor
On Fri, Mar 14, 2008 at 01:26:49PM +, IAIN GALLAGHER wrote:
> Hi list.
>
> I have a numerical dataset 22,000 rows deep and 43 columns wide. I would like
> to remove those rows which contain only values le
Use %in%:
x [ x %in% y ]
G.
On Fri, Mar 14, 2008 at 12:37:45PM +0200, Rainer M Krug wrote:
> Hi
>
> Consider the following code
>
> > x <- rep(1:13, 13)
>
> > y <- 1:3
>
> I want to select all elements in x which are equal to 1, 2 or 3.
>
> I know that I could use
>
> > sel <- x==y[1] | x=
You want to do
thing <- list()# empty thing
for ( i in 1:100 ) {
thing[[i]] <- ?
}
But where is ? coming from? If you can index it with an integer
then it is exactly coming from the kind of object you want to create.
Chicken-egg problem. No?
G.
On Thu, Mar 13, 2008 at 09:04:11A
cumsum( mapply(function(i,j) sum(a$data[i:j]), x, y) )
Is this what you want?
Gabor
On Thu, Mar 13, 2008 at 06:02:13AM -0700, yoo wrote:
>
> Hi all, i have the following..
>
> a <- data.frame(data = seq(1,10))
>
> i have indices:
> x <- c(1, 5, 3, 9)
> y <- c(2, 7, 4, 10)
>
> I want the
M[ order(M[,1]), ]
G.
On Mon, Mar 10, 2008 at 01:56:45PM -0700, ermimi wrote:
>
> I have a matrix with 2 columns and n row. I need sort the matrix by the first
> column but the second row must be sort in the same order that the first
> column. Somebody know how I can sort this matrix.
> Thanks v
Jean,
On Fri, Mar 07, 2008 at 06:09:35PM +0100, Jean lobry wrote:
> Gabor,
>
> > this is nice, but
> > 1) the logo is a bitmap, it is ugly if you resize it
>
> Sure, it's a bitmap, but the naked eye resolution is only
> 100 $\mu$m so that a vectorized solution is overkilling
> in most common sit
Jean,
this is nice, but 1) the logo is a bitmap, it is ugly if you
resize it, 2) you don't need a pdf version for pdflatex, it
handles jpg (and maybe also png as well), so you can
just use the logos at the R developer site.
It would be really nice to have a non-bitmap version, though.
If it e
On Thu, Mar 06, 2008 at 06:54:41PM -0500, Charilaos Skiadas wrote:
> On Mar 6, 2008, at 1:49 PM, Mag. Ferri Leberl wrote:
>
> > Dear everybody!
> > Is there a command in \LaTeX to display the R-Logo or has anybody
> > made it up?
> > Thank you in advance.
>
> Isn't it just an image? Hence you w
Mark, graph.adjacency always preserves the order of the vertices,
so the vertex at row/column 1 will be vertex #0 in the igraph graph,
etc. I'll document this in a minute.
This means that you can always do
g <- graph.adjacency(A)
V(g)$name <- colnames(A)
But i completely agree that this should
On Wed, Mar 05, 2008 at 02:27:21AM -0500, Charilaos Skiadas wrote:
[...]
>
> Btw, you will likely want to take the betweenness call out, and call
> it once and store the result, instead of calling it twice (well,
> assuming the graph is largish). Or even better, use which.max:
>
> which.max(b
Bingo!
See ?exists
G.
On Tue, Mar 04, 2008 at 04:38:31PM +0100, Paul Hammer wrote:
> hi members,
>
> give it a function for requesting if a object, matrix, vector, variable
> or whatever already exists?
>
> e.g. if (*exists*(a) {print("yes") } else { print("no") }
>
> thanks
> paul
>
>
On Tue, Mar 04, 2008 at 01:18:43PM +0100, Martin Kaffanke wrote:
>
> Am Dienstag, den 04.03.2008, 12:34 +0100 schrieb Peter Dalgaard:
> > Martin Kaffanke wrote:
> > > Hi there!
> > >
> > > I use an gnome-terminal for using R. When I resize the termial to the
> > > maximum size, R uses only the le
options(width=120)
See ?options
G.
On Tue, Mar 04, 2008 at 12:04:54PM +0100, Martin Kaffanke wrote:
> Hi there!
>
> I use an gnome-terminal for using R. When I resize the termial to the
> maximum size, R uses only the left side of the window. Can I tell R to
> use the whole window somehow?
>
R FAQ 7.31. G.
On Mon, Mar 03, 2008 at 12:52:43PM -0500, Xuejun Qin wrote:
> Hi, there,
> I cannot get accurate value for calculation.
> for example:
> ld<-sqrt(1*0.05*0.95*0.05*0.95)
> 0.05*0.95-ld=-6.938894e-18
> 0.05*0.95-ld==0 is False.
>
> I met this problem in my program, how can I handl
I'm not a statistician, but do i remember well that among all
distributions with a given mean and variance, the normal distribution
has the highest entropy? This is good enough for me to call it
"normal"
Gabor
On Sun, Mar 02, 2008 at 10:10:21AM -0600, roger koenker wrote:
> A nice survey of
On Sat, Mar 01, 2008 at 12:54:56AM +0100, Louise Hoffman wrote:
> > If you still want to then read ?write.table, that can export your data
> > into a spreadsheet-like ascii format which can be used from GNUplot
> > easily.
>
> Very interesting.
>
> So if I e.g. write:
> ts.sim <- arima.sim(lis
I believe that R can export all formats that GNUplot can produce,
so i don't really see why you want to use GNUplot if you don't know it.
If you still want to then read ?write.table, that can export your data
into a spreadsheet-like ascii format which can be used from GNUplot
easily.
Btw, compar
colSums gives you exactly what you want, but the vector is *named*.
You can use it like a not-named numeric vector, without much trouble.
If you still want to get rid of the names see ?unname
G.
On Fri, Feb 29, 2008 at 12:02:46PM -0500, Jason Horn wrote:
> Does anyone know how to get a vector of
Simone, they are currently ignored. Just like edge direction.
Gabor
On Thu, Feb 28, 2008 at 03:17:24PM +0100, Simone Gabbriellini wrote:
> hello,
>
> I have a last question on cohesive blocks: if there are multiple links
> between some nodes in the graph, this is taken into account by
> cohe
It's in the 'limma' Bioconductor package.
Next time you can try
help.search("lmFit")
RSiteSearch("lmFit")
Gabor
On Sun, Feb 24, 2008 at 01:02:41PM -0800, Keizer_71 wrote:
>
> Hi Everyone,
>
> I am trying to use lmFit function; however, i cannot find it function
> anywhere.
>
> I have been try
Similar experience, with snow & MPI (LAM). Actually, plug and play.
G.
On Sat, Feb 23, 2008 at 10:57:22AM -0700, Eric W Anderson wrote:
> Hi Dan,
>
> I've had pretty good luck using Snow with with Rpvm. It's definitely
> not what you'd call "plug and play," but it does work. I'm using it on
>
It depends what you mean by 'hiding', you can start the function
names with a dot and then they are not listed by ls(), so this
is kind of hiding.
> .a <- function() TRUE
> ls()
character(0)
> .a
function() TRUE
Personally i would not do this though.
G.
On Sat, Feb 23, 2008 at 11:58:57AM +0100
igraph is a package for graphs and networks. It has a C core and
uses a simple and fast graph representation allowing millions
of vertices and edges.
NEW FEATURES:
- We use the ARPACK library for graph related eigenvalue problems,
like Page Rank calculation, Kleinberg's hub and authority sco
I think you need "-l":
R CMD check -l
Gabor
On Tue, Feb 19, 2008 at 02:00:38PM -, john seers (IFR) wrote:
[...]
> I think this is because I install my packages in "mylibrary" and use
> "R_LIBS=C:/PROGRA~1/R/mylibrary" in my Renviron.site file. If I move the
> package to the R main library
?RSiteSeach is useful. It gives you this:
http://tolstoy.newcastle.edu.au/R/e2/help/07/09/25536.html
and then keep clicking on "Next in thread".
Gabor
On Tue, Feb 19, 2008 at 12:33:21PM +0100, Alfonso Pérez Rodríguez wrote:
> Hello, I'm sure that this question is too simple, but, I'm begining w
Update: it is a Bioconductor package, so you need to do:
source("http://bioconductor.org/biocLite.R";)
biocLite("convert")
Gabor
On Mon, Feb 18, 2008 at 10:17:41AM +0100, Schmitt, Corinna wrote:
>
> Hallo,
>
> I am running R-2.6. on Windows. I have a code which uses
> library(convert). Can any
.. (Counting to ten.)
The package is called 'convert'. It seems that this package is not
on CRAN, however. I think you should ask the person whose code you're
running.
Gabor
On Mon, Feb 18, 2008 at 10:17:41AM +0100, Schmitt, Corinna wrote:
>
> Hallo,
>
> I am running R-2.6. on Windows. I
I don't know php very well, but perhaps you need
?sub
Gabor
On Fri, Feb 15, 2008 at 03:54:19PM +0100, Dong-hyun Oh wrote:
> Dear expeRt,
>
> I would like to know whether a function similar to str_replace() in
> php exists in R.
>
> Looking forward to hearing from you.
>
> Best,
>
>
>
>
RSiteSearch is your friend. E.g.:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/63365.html
and then click on 'Next in thread a couple of times
Gabor
On Thu, Feb 14, 2008 at 03:23:30PM -0600, Edna Bell wrote:
> Dear R Gurus:
>
> How do you get source for functions which say "UseMethod" wh
df2 = df
?
G.
On Thu, Feb 14, 2008 at 09:12:23AM -0800, joseph wrote:
>
>
> Hi
>
> I have a data frame df1 in which I would like to multiply col1
> by 2.
>
>
> The way I did it does not allow me to keep the old data
> frame.
>
>
> How can I do this and be able to create a new data frame
data <- data[ , !apply(is.na(data), 2, all)]
(or something like that)
G.
On Thu, Feb 14, 2008 at 12:59:46PM +, Martin Waller wrote:
> Hi,
>
> I guess this might be a FAQ or something, and there's probably a nice
> simple way to do it, but I can't think of it:
>
> Given a matrix, I want to
Personally, i would go for a pure R solution, except if the
files are very large. For that i'm afraid that you need to
read the whole file, manipulate it and write it back. Or implement
your own 'read' function instead of read.table.
Gabor
On Thu, Feb 14, 2008 at 09:57:14AM +, Romain Franco
It should be possible i think. You just supply all the arguments via
'...' and then cut off the last one. I don't see why this wouldn't work,
but maybe i'm missing something.
Gabor
On Tue, Feb 12, 2008 at 12:58:25PM -0600, Erik Iverson wrote:
> Alistair -
>
> I don't believe this is possible. T
I assume that data$Roman is character.
data[order(as.numeric(as.roman(data$Roman))),] should do it. Maybe
data[order(as.roman(data$Roman)), ] is enough too.
Gabor
On Tue, Feb 12, 2008 at 10:36:50AM +, Luis Ridao Cruz wrote:
> R-help,
>
> I have a data frame with one column containing roman
Because you need
test <= 0.2 | test >0.3
See ?"|"
Gabor
On Mon, Feb 11, 2008 at 09:12:57PM +0800, Stanley Ng wrote:
> That works beautfully. Why using test<=0.2 || test >0.3 gives error ?
>
> -Original Message-
> From: Gabor Csardi [mailto
which(apply(test<=0.2, 1, all))
See ?which, ?all, and in particular ?apply.
Gabor
On Mon, Feb 11, 2008 at 06:22:09PM +0800, Ng Stanley wrote:
> Hi,
>
> Given a simple example, test <- matrix(c(0.1, 0.2, 0.1, 0.2, 0.1, 0.1, 0.3,
> 0.1, 0.1), 3, 3)
>
> How to generate row indexes for which thei
sub("-", "--", v, fixed=TRUE)
See ?sub.
Gabor
On Sun, Feb 10, 2008 at 02:14:48PM -0500, Michael Kubovy wrote:
> Dear R-helpers,
>
> How do I transform
> v <- c('insd-otsd', 'sppr-unsp')
> into
> c('insd--otsd', 'sppr--unsp')
> ?
> _
> Professor Michael Kubovy
> Unive
It is a good idea to start with RSiteSearch("Excel")
G.
On Fri, Feb 08, 2008 at 03:49:29PM -0500, Christine Lynn wrote:
> This is the most basic question ever...I haven't used R in a couple years
> since college so I forget and haven't been able to find what I'm looking for
> in any of the manual
This is a "religion question" in some sense. Personally, i used
CVS and a bit Subversion too, but arch and bazaar look much better.
Especially if you're not always in online connection with the central
repository, or you don't really want a central repository at all.
Gabor
On Thu, Feb 07, 2008 a
You almost got it right. THe solution is
df[df$ind %in% subgr,]
See ?"%in%"
G.
On Tue, Feb 05, 2008 at 04:47:02PM +0100, Karin Lagesen wrote:
>
> Hi!
>
> I have a large dataframe that I want to extract a subset from. This
> subset has a certain column value that matches elements in a vector I
On Mon, Feb 04, 2008 at 03:21:10PM +0800, Ng Stanley wrote:
> Hi,
>
> Given a test matrix, test <- matrix(c(1,2,3,NA,2,3,NA,NA,2), 3,3)
>
> A) How to compute the counts of each column (excluding the NA) i.e., 3, 2, 1
apply(test, 2, function(x) sum(!is.na(x)))
> B) How to compute the counts of e
Derek,
the 0*m part zeros out everything in the matrix, expect for the NA's,
0*NA=NA by definition. If we add this to the y[ col(m) ] matrix, then
NA+anything=NA, but 0+anything=anything.
G.
ps. please answer to the list (as well)
On Fri, Feb 01, 2008 at 08:52:50AM -0500, Derek Cyr wrote:
>
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