Hello,
I have been getting the following intermittent error from kmeans:
>str(cavint.p.r)
num [1:1967, 1:13] 0.691 0.123 0.388 0.268 0.485 ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:1967] "6" "49" "87" "102" ...
..$ : chr [1:13] "HYD" "NEG" "POS" "OXY" ...
> set.seed(34)
> kmeans(cav
Dirk and Ray have provided two very clever solutions which
perform transformation and selection in one go
by returning NA and NULL respectively for unwanted elements
and then eliminating the NAs and NULLs.
I thought it would be worthwhile to bring them together and
make some further minor
On Sun, Nov 09, 2003 at 09:29:26PM -0600, Dirk Eddelbuettel wrote:
> On Sun, Nov 09, 2003 at 10:00:42PM -0500, Gabor Grothendieck wrote:
> > There are languages that do allow this. For example, Python
> > has list comprehensions which are things like this:
> >
> ># give me the squares of the
On Sun, Nov 09, 2003 at 10:00:42PM -0500, Gabor Grothendieck wrote:
> There are languages that do allow this. For example, Python
> has list comprehensions which are things like this:
>
># give me the squares of the even numbers from 1-10, in a list.
>>>> [ x*x for x in range(1,11) if x
This has already been answered but I think behind your
question may have been the thought of whether its
possible to do it all at once: transformation and selection.
There are languages that do allow this. For example, Python
has list comprehensions which are things like this:
# give me
Currently the function symbols() permits plotting of circles, squares, rectangles,
stars,
thermometers, and boxplots.
Can anybody suggest an approach for extending/cloning this function to handle
arbitrary shapes specified, perhaps, as bitmaps, or as postscript primitives?
Thanks, Steve Dutky
Hi Tomomi:
(B
(BI received the other e-mail where you sent rimage_0.5-3. It installed
(Bwithout any problems using fftw-2.1.5. Thanks for the quick response.
(B
(BRajiv
(B
(B- Original Message -
(BFrom: "Tomomi TAKASHINA" <[EMAIL PROTECTED]>
(BTo: "Rajiv Prasad" <[EMAIL PROTECTE
Hi, Rajiv:
(B
(BAt Sun, 9 Nov 2003 16:10:23 -0800, "Rajiv Prasad" wrote:
(B
(B> I tried downgrading to fftw-2.1.5, but that was no help either. The error
(B> message is still the same, at the same place during configuration.
(B>
(B> It seems to me that configure is checking for the existan
When I did it in S-Plus 6.1 and R 1.8.0, I didn't get NULL entries: I
got NAs. There is a difference between NULLs and NAs. The NAs can be
deleted using is.na, as follows:
> list.of.vectors <- list(a=1, b=1:3)
> x3 <- sapply(list.of.vectors, function(x)x[3])
> x3
a b
NA 3
> x3
a b
NA 3
>
Hi,
I'm trying to subset a list which contains variable length vectors.
What I want to do is extract (eg.) the 3rd item in each vector (with
length >= 3). At the moment I'm using sapply(list.of.vectors,
function(x) {x[3]}). The problem with this is that sapply returns a
list of the same len
Hi Dirk (and Tomomi):
Thanks for the quick reply. I tried using
'--configure-args="--with-fftw-lib=/usr/local/fftw-3.0.1"', but it did not
help. configure still fails at the same place:
...
checking for fftwnd_one in -lfftw... no
configure: error: Sorry, cant find fftw library. Please use --wit
Well!
Isn´t that what Copulas is all about.
Fitting multifactor models into one model
/T
Thomas Holm
Södra Hesperiagatan 32A9
00100 Helsingfors
[EMAIL PROTECTED]
-Ursprungligt meddelande-
Från: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] För Peter Dalgaard
Skickat: den 9 november 2
Prof Brian Ripley <[EMAIL PROTECTED]> writes:
> Well, it is one of those things
>
> -- it works in R but not in S
> -- it appears in the examples for help(":") but is not otherwise mentioned
>on the help page (why?)
> -- it does not give a numerical list of combinations, as asked for
> -- it
Hi
Laura Quinn wrote:
I am trying to plot positions on a grid where the x and y axis equate to
longitudinal and latitudinal co-prdinates respectively. As these
co-ordinates are southings and westings, i need the origin of my graph to
contain the two highest values of each co0ordinate, with the val
On Mon, Nov 10, 2003 at 10:02:00AM -0800, Rajiv Prasad wrote:
> I separately compiled and installed fftw-3.0.1. The problem seems to be that
> I need to provide "--with-fftw-lib" option to the configure script. How do I
> do this?
As options to R CMD INSTALL as per:
[EMAIL PROTECTED]:~/chibud
Hi folks:
I am trying to install the package "rimage" in R 1.8.0 on an Alpha Linux box.
"R CMD INSTALL rimage_0.5-1.tar.gz" fails with the following:
...
...
checking for stdint.h... yes
checking for unistd.h... yes
checking fftw.h usability... yes
checking fftw.h presence... yes
checking for f
On Sun, 9 Nov 2003, Karim Elsawy wrote:
> I'm using R 1.7.1 under linux redhat
> it seems that the eigen values produced by eigen() do not follow
> a consistant order; I mean either ascending or discending
This is strange, since the source code explicitly sorts them. Can you send
an actual exampl
On Sun, 9 Nov 2003, mpie wrote:
> Please forgive my ignorance, but is there such thing
> as a "multiple cross-correlation function", i.e., a
> cross-correlation function for more than two time
> series. If there is, is it implemented in R? I
> couldn't find it.
?acf
--
Brian D. Ripley,
Well, it is one of those things
-- it works in R but not in S
-- it appears in the examples for help(":") but is not otherwise mentioned
on the help page (why?)
-- it does not give a numerical list of combinations, as asked for
-- it does give unused levels, which in this application is disastr
On 9 Nov 2003 at 13:29, Prof Brian Ripley wrote:
> Factor3 <- factor(unclass(Factor1) + nlevels(Factor1)*(unclass(Factor2)-1))
>
Cannot this be done even easier by calculating the interaction?
> a <- factor(rep(1:3,rep(3,3)))
> b <- factor(rep(1:3,3))
> ab <- a:b
> ab
[1] 1:1 1:2 1:3 2:1 2:2 2:
Hi all,
Please forgive my ignorance, but is there such thing
as a "multiple cross-correlation function", i.e., a
cross-correlation function for more than two time
series. If there is, is it implemented in R? I
couldn't find it.
Thanks,
Mark
__
[EMAIL
I'm using R 1.7.1 under linux redhat
it seems that the eigen values produced by eigen() do not follow
a consistant order; I mean either ascending or discending
e.g
for one system:
eigenV<-eigen(V)
> print(eigenV$values)
[1] -7.706828e+13 -4.702980e+13 -3.267579e+13 -1.701297e+
Factor3 <- factor(unclass(Factor1) + nlevels(Factor1)*(unclass(Factor2)-1))
will give you the unique combinations, not labelled as you do but then I
don't think you need that.
On Sun, 9 Nov 2003, Scott Norton wrote:
> This might be easy but I'm very new to R and this question doesn't seem to
>
This might be easy but I'm very new to R and this question doesn't seem to
have any nice keywords that bring up relevant search results when I search
the CRAN search engine. Therefore, I'll plead (as I have in the recent
past) Newbie status.
I have a data frame with two factors (Factor 1 and 2
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