Hi Thomas,
Yes I have the xlibmesa-gl, libglu1-mesa, libglu1-mesa-dev, etc, libraries.
But no way. You mentioned TLS (transport Layer Security?), is this OK? I'm
working on a Toshiba laptop with graphics driver I852GM
Cheers,
Antonio
-Mensaje original-
De: [EMAIL PROTECTED]
Hi Thomas
antonio rodriguez schrieb:
Hi Thomas,
Yes I have the xlibmesa-gl, libglu1-mesa, libglu1-mesa-dev,
etc, libraries.
But no way. You mentioned TLS (transport Layer Security?), is
this OK? I'm
working on a Toshiba laptop with graphics driver I852GM
No, just look at /usr/share
Hi Manuel,
Look at your memory.limit() and run R with '--mem-max=' option
Try also to use the simpler netCDF library
Antonio
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de Manuel Gutierrez
Enviado el: viernes, 21 de enero de 2005 10:03
Para:
Nicolas Degallier wrote:
Hi!
I am trying to install in my R environment the rhdf5 package and
library but it seems to have vanished from either the CRAN or
BioConductors sites.
Can you tell me where it would be possible to find it or any R
library (or function) able to read hdf
Hi All,
I have 2 arrays:
dim(a1)
[1] 3 23 23
dim(a2)
[1] 3 23 23
And I want a new array, to say, a3, where:
dim(a3)
[1] 6 23 23
where the first dimension is supposed to be (months) so the resultating
array would start in jan and finish in june.
Best regards,
Antonio
Hi All,
I have 2 arrays:
dim(a1)
[1] 3 23 23
dim(a2)
[1] 3 23 23
And I want a new array, to say, a3, where:
dim(a3)
[1] 6 23 23
where the first dimension is supposed to be (months) so the resultating
array would start in jan and finish in june.
Best regards,
Antonio
--
=
Por favor,
Arnau Mir Torres wrote:
Uwe Ligges wrote:
Arnau Mir Torres wrote:
Hello.
I am an nxn data frame in the variable frame.
I want to make a contour plot with it. That is, I want to plot a square
of dimensions [1,n]x[1,n] with the gray level of square [i,i+1]x[j,j+1]
equal to
[EMAIL PROTECTED] wrote:
Hi,
I've extracted netcdf data in R, and am trying to output it in a matrix
format. I can make the matrix, and I have the data, but I can't get the
matrix to include the data...
I'm not very experienced at using R, so please forgive the crudeness of
these
?spectrum
?spec.ar
Martin Albers wrote:
Is there any way to apply a spectra or periodogramm to my time series.
My first problem is, that I do not know how to get a spectra/periodogramm for
my data.
I have found time series tools but nothing for spectral analysis.
My second problem:
Have you look at clim.pact()
Sorin Cheval wrote:
Hello,
I have some problems in using R for specific
climatological aplications. I would appreciate a lot
if anybody could help me a bit.
All the best,
Sorin Cheval
__
Hi Neil,
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de [EMAIL PROTECTED]
Enviado el: lunes, 03 de mayo de 2004 23:22
Para: [EMAIL PROTECTED]
Asunto: [R] Factor loadings and principal component plots
Hi- Can anyone tell me the command(s) to
Many, many thanks to everybody. It helped a lot
Cheers
Antonio
---
__
[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
in one
file all your newly created individuals files (jan.gif, feb.gif,...), and
you will get an animated plot.
Saludos!
Antonio Rodriguez
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de Jesus Fernandez
Galvez
Enviado el: lunes, 10 de noviembre de 2003 17
Hi,
Is it possible to overlay over a world region drawn by function world(),
a filled.contour for that specific region?
I've managed to do the inverse, add world to a filled.contour, but the
appearence of the map is not the one I would like. With contour() is
easy to do the task, but with
Hi Feng,
AFIK SVD analysis provides a one-step method for computing all the
components of the eigen value problem, without the need to compute and
store big covariance matrices. And also the resulting decomposition is
computationally more stable and robust.
Cheers,
Antonio Rodriguez
Hi Peter,
Thanks it works fine now with this little trick
Cheers
Antonio
-Mensaje original-
De: Peter Dalgaard BSA [mailto:[EMAIL PROTECTED]
Enviado el: martes, 27 de mayo de 2003 10:04
Para: [EMAIL PROTECTED]
CC: R Help
Asunto: Re: [R] Rcmdr on Debian
[EMAIL PROTECTED] writes:
Hi,
Hi,
Running R 1.7.0 on Win XP I'm getting the following error message:
library(netCDF)
a-read.netCDF(c:/data/fnmoc/individuales/sst_rey.nc)
Error in .C(open_netcdf, filename, id = as.integer(verbose)) :
C/Fortran function name not in load table
Don't know what is going wrong. With
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de Duncan Murdoch
Enviado el: lunes, 16 de junio de 2003 14:43
Para: antonio rodriguez
CC: R-help; Thomas Lumley
Asunto: Re: [R] Error in .C(open_netcdf
On Mon, 16 Jun 2003 13:38:28 +0200, you wrote:
Hi
Hi,
The problem was the way I installed the netCDF package. Within the R
environment and invoking Install package from CRAN makes (at least for me)
that the netCDF library doesn't work. But if I download it and Install from
local zip file the problems are gone.
Cheers
Antonio Rodríguez
---
Hi,
go to your C:/R/rw1080/etc directory and edit the Rprofile file adding
(e.g.):
library(foreign)
library(survival)
cheers,
antonio rodriguez
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de imap
Enviado el: martes, 20 de enero de 2004 16:16
Para
-Mensaje original-
De: Janus Larsen [mailto:[EMAIL PROTECTED]
Enviado el: martes, 03 de febrero de 2004 11:36
Para: antonio rodriguez
Asunto: RE: [R] filled maps
Hi Antonio,
Thanks for your prompt answer - but I can't see how that will solve my
problem (which is overlaying
Hi,
Some time ago, Roger Peng posted this solution, which I found very useful:
junk.mat - matrix(rnorm(1600), 16, 100)
contour.mat - ifelse(junk.mat 2, 0, junk.mat)
filled.contour(junk.mat, color = terrain.colors,
plot.axes = contour(contour.mat, levels = 1,
Trés bien!!
Antonio
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de Shigeru Mase
Enviado el: jueves, 04 de marzo de 2004 12:36
Para: r-help
Asunto: [R] Statistiques avec R
Dear R users,
I want to share my joy with you. Please see the following
Hi Eduardo,
Probably a good idea is to do a multivariate analysis through spectral
analysis. With function spectrum you can look for coherence and phase
between several time series. In your case with different lengths, you can
use the option: na.action=na.omit
And in:
http://zoonek2.free.fr/UNIX/48_R/all.html
an excellent starting point (in french)
Salut
Antonio
-Mensaje original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] nombre de Stephane DRAY
Enviado el: lunes, 22 de marzo de 2004 17:33
Para: mathieu chausson ; Aide R
Asunto:
Hi,
Having several daily wind speed time series I want to extract those
consecutive days over and below certain values (i.e. 5 x 8) Don't
know which funtion to use (aggregate, lapply?) and how to do it.
Thanks in advance
Antonio
__
PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of antonio rodriguez
Sent: Tuesday, October 24, 2006 1:57 PM
To: r-help@stat.math.ethz.ch
Subject: [R] extract certain values from a ts
Hi,
Having several daily wind speed time series I want to extract those
consecutive days over and below
.
-Original Message-
From: antonio rodriguez [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 24, 2006 2:22 PM
To: Leeds, Mark (IED); R-Help
Subject: Re: [R] extract certain values from a ts
Leeds, Mark (IED) escribió:
I don't think it matters whether it's a ts or a vector
Leeds, Mark (IED) escribió:
maybe subset works. I'm not familiar with that.
neither me
-Original Message-
From: antonio rodriguez [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 24, 2006 2:49 PM
To: Leeds, Mark (IED); R-Help
Subject: Re: [R] extract certain values from a ts
-
From: antonio rodriguez [mailto:[EMAIL PROTECTED]
Sent: Tuesday, October 24, 2006 2:49 PM
To: Leeds, Mark (IED); R-Help
Subject: Re: [R] extract certain values from a ts
Leeds, Mark (IED) escribió:
take your ts object and convert it to a zoo object ( you need to have
the zoo package
Gabor Grothendieck escribió:
There was still a typo in this so here it is again:
# Consider the builtin in nhtemp ts series
# Then the runs below above and below the mean are
r - rle(as.vector(nhtemp mean(nhtemp)))
r
# with the first and last indexes of each
# runs being given by:
Hi,
Having an zoo object I can subset it to obtain the days where I have the
values within some range:
is.zoo(z)
TRUE
subset(z[,1], z[,1]=5 z[,1]= 10) #Yields: Year(day)
1988(13) 1988(14) 1988(16) 1988(20) 1988(21) 1988(22) 1988(25)
1988(26)
7.973946 9.933518 7.978227 7.512960
Gabor Grothendieck escribió:
On 10/28/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Hi,
Having an zoo object I can subset it to obtain the days where I have the
values within some range:
is.zoo(z)
TRUE
subset(z[,1], z[,1]=5 z[,1]= 10) #Yields: Year(day)
1988(13) 1988(14) 1988(16
Gabor Grothendieck escribió:
Try this:
# test data
x - c(1:4, 6:8, 10:14)
z - zoo(x, as.Date(x))
# idx is 1 for first run, 2 for second run, etc.
idx - cumsum(c(1, diff(z) != 1))
# starts replaces each time with the start time of that run
# ends is similar but for ends
starts -
in a form easily pastable by someone else into their session.
Thanks Gabor! Now it's OK (nice the output of dput. I was reluctant to
paste all data because is 6575 days long!)
Best regards
Antonio
On 10/29/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Gabor Grothendieck escribió:
Try
Gabor Grothendieck escribió:
On 10/29/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Gabor Grothendieck escribió:
Thanks Gabor! Now it's OK (nice the output of dput. I was reluctant to
paste all data because is 6575 days long!)
You can write the following, say, to display just a bit
Hi,
Having a matrix F.zoo (6575,189) with NA's in some columns I'm trying to
extract from each column the percent of days within an specific range,
so I've wrote this procedure:
length(subset(F.zoo[,86],(F.zoo[,86]=5) (F.zoo[,86]=
Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
[EMAIL PROTECTED]
On Wed, 1 Nov 2006, antonio rodriguez wrote:
Hi,
Having a matrix F.zoo (6575,189) with NA's in some columns
Gabor Grothendieck escribió:
Try this where m is the matrix:
100 * colMeans(m 5 m 9, na.rm = TRUE)
Dear Gabor,
Just perfect!
Thanks a lot,
Antonio
On 11/1/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Hi,
Having a matrix F.zoo (6575,189) with NA's in some columns I'm trying
understand now. Many thanks!!
Best regards,
Antonio
On Wed, 1 Nov 2006, antonio rodriguez wrote:
Phil Spector escribió:
Antonio -
When you're operating on each column of a matrix, you really should
consider the apply() function, which was written for the task.
Also, it's usually
) will output variable v
in a form easily pastable by someone else into their session.
On 10/29/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Gabor Grothendieck escribió:
Try this:
# test data
x - c(1:4, 6:8, 10:14)
z - zoo(x, as.Date(x))
# idx is 1 for first run, 2 for second run, etc
Brandt, T. (Tobias) escribió:
I think the following does what you want:
(d - structure(c(6586, 6586, 6589, 6593, 6593, 6593, 6598, 6598,
6598, 6598), class = Date))
[1] 1988-01-13 1988-01-13 1988-01-16 1988-01-20 1988-01-20
1988-01-20 1988-01-25
[8] 1988-01-25 1988-01-25 1988-01-25
or read.zoo and operated on by aggregate.zoo though most
other zoo operations will not accept them.
On 11/2/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Brandt, T. (Tobias) escribió:
I think the following does what you want:
(d - structure(c(6586, 6586, 6589, 6593, 6593, 6593, 6598, 6598
Hi,
Having a matrix F(189,6575) I want to do this:
z1-subset(F[,1], F[,1] = 5 F[,1] = 10)
.
.
.
z189-subset(F[,189], F[,189] = 5 F[,189] = 10)
I would prefer to have an empty matrix, say 'z' in order to fill its
columns with the output of subsetting F. But each of the subsets can
differ in
) and a
pretty good processor (P4, 2.8)
Antonio
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of antonio rodriguez
Sent: Tuesday, November 07, 2006 2:01 PM
To: R-Help
Subject: [R] subsetting a matrix and filling other
Hi,
Having a matrix F
antonio rodriguez escribió:
Christos Hatzis escribió:
Use a list to store the partial matrices:
z - vector(list, 189)
for(i in 1:189)
z[[i]] - subset(...)
Hi Christos,
I've tried it but it died unexpectedly with a (Killed) message and the
linux console prompt again. I think I
Hi,
Having 3 dataframes with different row numbers, but equal column names
(see below) I want to merge them by Var1 so I've tried:
merge(j1,j2,j3,by=Var1)
merge(j,j1,j2,by=names(Var1))
But always got the same message:
Erro en fix.by(by.x, x) : 'by' must specify column(s) as numbers, names
or
(or omit all=TRUE if you only want columns which are in all 3 data
frames).
On Wed, 22 Nov 2006, antonio rodriguez wrote:
Hi,
Having 3 dataframes with different row numbers, but equal column names
(see below) I want to merge them by Var1 so I've tried:
merge(j1,j2,j3,by=Var1)
merge(j
Hi,
Having a dataframe 'l1' (dput output is below):
dim(l1)
1274 2
l1[1:12,]
Var1 Freq
1 1988-01-131
2 1988-01-161
3 1988-01-203
4 1988-01-252
5 1988-01-301
6 1988-02-015
7 1988-02-084
8 1988-02-141
9 1988-02-161
10 1988-02-184
Marc Schwartz escribió:
Using just the example data:
table(substr(DF$Var1, 1, 7))
1988-01 1988-02 1988-03
5 6 1
This is just what I wanted!
Many thanks Marc!
BR
Antonio
See ?substr for more information.
HTH,
Marc Schwartz
and time(out) are the corresponding year-months.
See ?yearmon
On 11/22/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Hi,
Having a dataframe 'l1' (dput output is below):
dim(l1)
1274 2
l1[1:12,]
Var1 Freq
1 1988-01-131
2 1988-01-161
3 1988-01-203
4 1988-01-25
James J. Roper escribió:
Dear all,
I often use dates and times in analyses. I just can't figure out how to
format my date or time column in R. So, apparently R sees the date as
something other than date (character). Let's say I am opening a CSV
file, one of the columns of which is a
output.
HTH
Petr
Dear Petr,
It works great!!
Many Thanks,
Antonio
On 23 Nov 2006 at 21:06, antonio rodriguez wrote:
Date sent:Thu, 23 Nov 2006 21:06:18 +0100
From: antonio rodriguez [EMAIL PROTECTED]
To: R-Help r-help@stat.math.ethz.ch
Hi,
I have a list of 20 elements, each of them of variable length and with a
structure like this:
lasker[[1]][1:10,]
Var1 Freq
1 1988-023
2 1988-031
3 1988-041
4 1988-052
5 1988-063
6 1988-071
7 1988-081
8 1988-091
9 1989-031
10 1989-04
(zz)] - 0
Suggest you read the zoo vignette:
vignette(zoo)
Gabor,
Once again, thanks!
BR
Antonio
On 11/26/06, antonio rodriguez [EMAIL PROTECTED] wrote:
Hi,
I have a list of 20 elements, each of them of variable length and with a
structure like this:
lasker[[1]][1:10
Hi,
I'm trying to do a filled.contour plot where some points are labelled as
NA. How do I could plot this kind of graphics, so NA points are coloured
black, keeping the levels of remaining points. NA's values represent
land points (meaningless), and what I want to plot is the levels of a
Hi Gustaf
I'm having the same issue myself. What I've ended up doing is
replacing NA's with a big negative value, define levels as one color
for negative values, and a regular scale above.
How to define 'levels' as one color for negative values and a regular
scale above? I don't know how
Hi,
I have a object 'zoo':
dim(zz)
[1] 720 5551
where some columns only have NA's values (representing land data in a
sea surface temperature dataset) I find straightforward the use of
'na.approx' for individual columns from the zz matrix, but when applied
to the whole matrix:
# TRUE for each column that has more than 1 non-NA
idx - colSums(!!z, na.rm = TRUE) 1
idx
[1] TRUE FALSE FALSE TRUE
z[,idx] - na.approx(z[,idx])
z
1 1 3 NA 19
2 2 NA NA 20
3 3 NA NA 21
4 4 NA NA 22
5 5 NA NA 23
6 6 NA NA 24
On 5/27/07, antonio rodriguez [EMAIL PROTECTED] wrote
Hi,
I have this dataset where columns z1.3 and z1.4 are full of NA's. I want
to perform some calculations in the remaining columns, but after doing
this, I want to recontruct the original matrix. I can with:
out - which( colMeans( is.na( z ) ) == 1 )
gd-z[, - out]
select the columns full of
Hi,
I swear I have read almost all the posted messages about this issue, but
it's evident I couldn't find an answer (surely esay) to my problem. What
I want is the following:
Make 8 days aggregates from a daily series like this (dput output):
structure(c(6.91777181625366, 0.79051125049591,
Achim Zeileis escribió:
On Tue, 26 Jun 2007, antonio rodriguez wrote:
Hi,
I swear I have read almost all the posted messages about this issue, but
it's evident I couldn't find an answer (surely esay) to my problem. What
I want is the following:
Make 8 days aggregates from a daily
Or,
z-mydata #zoo object
new.time - as.Date(7 * floor(as.numeric(time(z))/7) + 7)
z2 - aggregate(z, new.time, mean)
Henrique Dallazuanna escribió:
Hi,
Perhaps you can try:
df
Date Amount
1 2007-06-01 1
2 2007-06-01 1
3 2007-06-04 2
4 2007-06-05
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