data3-sort(c(data1,data2))
See ?cbind, ? rbind ?append and ?merge for combining data
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
___
Hi R,
I am fetching Bloomberg data from R. The problem I face is that I get a
downloading error once pasted, but the same code run again will download
the data. So, I assure you that it is not the problem with the R code. I
will not be able to download the data due to some system
Hi!
Thanks in advance.
I am using R-2.4.0 on Windows XP. I am trying to create dll file.
My C code:
/* useC1.c */
void useC(int *i) {
i[6] = 100;
}
I have tried to create useC1.dll.
C:\R-2.4.0\binR CMD SHLIB useC1.c
'perl' is not recognized as an
Dear all,
Suppose I have a date variable:
c = 99/05/12
I want to extract the parts of this date like month number, year and day. I
can do it in SPSS. Is it possible to do this in R as well?
Rgd,
-
Heres a
Dear All,
A method for the estimation is univariate unimodal densities (with unknown
mode) is described in Statistical Inference under Order Restrictions by
Barlow et al.. Would anyone know whether there is an R-implementation
(preferably with reference) for the estimation of univariate
Hi,
I was wondering if someone could tell me more about this book, (if it's
a good or bad one).
I can't find it, as it seems that O'Reilly doesn't publish any more.
Thanks,
Ben
--
Benoit Ballester
__
R-help@stat.math.ethz.ch mailing list
format.Date(c, %d)
format.Date(c, %m)
format.Date(c, %y)
format.Date(c, %Y)
On 01/02/07, stat stat [EMAIL PROTECTED] wrote:
Dear all,
Suppose I have a date variable:
c = 99/05/12
I want to extract the parts of this date like month number, year and
day. I can do it in SPSS. Is it
nbsp; Hi, List ,I am searching any package on R which can do wavelet
filtering for mother wavelet morlet ,is anybody having any script for the same
?I am new to the RwAVELET ANALSSIS..THANKS IN ADVANCE ANIL KUMAR
ANIL KUMAR(nbsp;METEOROLOGIST)LRF SECTIONnbsp;NATIONAL
You need to install perl and MinGW, at least.
If you have them installed, then you need to properly set PATH environment
variable and, probably, restart your command line session.
See chapter 5 of the manual Writing R extensions (installed in
R_HOME/doc/manual)
and these two links
Dear all,
it is likely a stupid question but I cannot solve it.
I want to have a matrix of 100 elements.
Each element must be a vector of 500 elements.
If I do:
imp-array(dim=100)
imp[1]-vector(length=500)
it does not work. Warning message: number of items to replace is not a
multiple
Hello,
In a nutshell, I've got a data.frame like this:
assignation - data.frame(value=c(6.5,7.5,8.5,12.0),class=c(1,3,5,2))
assignation
value class
1 6.5 1
2 7.5 3
3 8.5 5
4 12.0 2
and a long vector of classes like this:
x - c(1,1,2,7,6,5,4,3,2,2,2...)
And would
Read the help desk article in R News 4/1 about dates and note the table at the
end of it, in particular.
On 2/1/07, stat stat [EMAIL PROTECTED] wrote:
Dear all,
Suppose I have a date variable:
c = 99/05/12
I want to extract the parts of this date like month number, year and day. I
can
Deb Midya wrote:
Hi!
Thanks in advance.
I am using R-2.4.0 on Windows XP. I am trying to create dll file.
My C code:
/* useC1.c */
void useC(int *i) {
i[6] = 100;
}
I have tried to create useC1.dll.
C:\R-2.4.0\binR CMD SHLIB useC1.c
stat stat wrote:
Dear all,
Suppose I have a date variable:
c = 99/05/12
I want to extract the parts of this date like month number, year and day. I
can do it in SPSS. Is it possible to do this in R as well?
Rgd,
Yes. One way is to use substr(), e.g.:
one way is the following:
assignation$value[match(x, assignation$class)]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax:
assignation$value[match(x,assignation$class)]
[1] 6.5 6.5 12.0 NA NA 8.5 NA 7.5 12.0 12.0 12.0
On 2/1/07, javier garcia-pintado [EMAIL PROTECTED] wrote:
Hello,
In a nutshell, I've got a data.frame like this:
assignation - data.frame(value=c(6.5,7.5,8.5,12.0),class=c(1,3,5,2))
Hi!
I'm having trouble with importing spss files containing non-ascii characters
(R 2.4.1, debian linux, i386). To reproduce:
Download the following file:
http://statmath.wu-wien.ac.at/data/spss/de/comphomeneu.sav
require (foreign)
Sys.setlocale (locale=C)
read.spss(comphomeneu.sav)$ARBEIT[1]
Dear R users,
I have a dataframe with two columns: first column is date data (e.g.
1/1/2000 with character format: daily data from 1/1/1970 till 31/12/2003)
and second column is temperature value. Now I'd like to calculate mean for
each month in a year (i.e. May 2001, June 1997) and mean for
Thomas Friedrichsmeier wrote:
Hi!
I'm having trouble with importing spss files containing non-ascii characters
(R 2.4.1, debian linux, i386). To reproduce:
Download the following file:
http://statmath.wu-wien.ac.at/data/spss/de/comphomeneu.sav
require (foreign)
Sys.setlocale (locale=C)
On 1/31/07, Jon Stearley [EMAIL PROTECTED] wrote:
I need to sum the columns of a sparse matrix according to a factor -
ie given a sparse matrix X and a factor fac of length ncol(X), sum
the elements by column factors and return the sparse matrix Y of size
nrow(X) by nlevels(f). The appended
For the case someone is interested in it, here it is the solution
somebody suggested me: to use a list.
imp - vector(list, 100)
imp[[1]] - im[1:500,]
names(imp[[1]]) = the list of labels of imp[1:500,]
Thanks!
Federico
Federico Abascal wrote:
Dear all,
it is likely a stupid question but I
Hi,
On 2/1/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Thu, 1 Feb 2007, Vaidotas Zemlys wrote:
Hi,
On 1/31/07, Professor Brian Ripley [EMAIL PROTECTED] wrote:
Two comments:
1) ls() does not list all the objects: it has all.names argument.
Yes, I tried it with all.names,
Doug is right, I think, that this would be easier with full indexing
using the matrix.coo classe, if you want to use SparseM. But
then the tapply seems to be the way to go.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
Dear R-community,
I have some trouble with index mappings for arrays. If, for example,
I have the array
R A - array(1:9, c(3,3,2))
and two index mappings both of same size
R x - c(2, 3)
R y - c(1, 2)
Now I want to access the elements (A[1, x[i], y[i]])_i of A, i.e.
A[1, x[1], y[1]] = A[1, 2,
Gabor Grothendieck ggrothendieck at gmail.com writes:
To get the best results you need to transfer it using vector
graphics rather than bitmapped graphics:
http://www.stc-saz.org/resources/0203_graphics.pdf
There are a number of variations described here (see
entire thread). Its for
Hi Peter,
generally speaking, wavelets are known to be good at extracting signal from
noisy data and are
adaptive but I am not familiar with any R implementation of wavelets.
A simple way of looking at changes would be to use CUSUM (strucchange
package).
I hope this helps.
Ansel.
On 1/30/07,
Wessel van Wieringen wrote:
A method for the estimation is univariate unimodal densities (with
unknown mode) is described in Statistical Inference under Order
Restrictions by Barlow et al.. Would anyone know whether there is an
R-implementation (preferably with reference) for the estimation
probably you want something like the following:
A[cbind(rep(1, length(x)), x, y)]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel:
On Thu, 2007-02-01 at 07:22 +0100, Jean lobry wrote:
Dear R-help,
I'm using prop.test() to compute a confidence interval for a proportion
under R version 2.4.1, as in:
prop.test(x = 340, n = 400)$conf
[1] 0.8103309 0.8827749
I have two questions:
1) from the source code my
On 2/1/07, Ben Bolker [EMAIL PROTECTED] wrote:
Gabor Grothendieck ggrothendieck at gmail.com writes:
To get the best results you need to transfer it using vector
graphics rather than bitmapped graphics:
http://www.stc-saz.org/resources/0203_graphics.pdf
There are a number of
Matt,
Some time back I didn't like the uniroot restriction either so I wrote a short
function manyroots that breaks an interval into many shorter intervals and
looks for a single root in each of them. This function is NOT guaranteed to
find all roots in an interval even if you specify many
dfr-data.frame(day=c(1/1/1970,5/1/1970,5/12/2003,31/12/2003),temperature=c(1,-1,2,0.5))
dfr
day temperature
1 1/1/1970 1.0
2 5/1/1970-1.0
3 5/12/2003 2.0
4 31/12/2003 0.5
On Thursday 01 February 2007 14:18, Peter Dalgaard wrote:
so you should convert it:
iconv(Im B\xfcro, from=latin1, to=UTF-8)
[1] Im Büro
iconv(Im B\374ro,from=latin1, to=UTF-8)
[1] Im Büro
I see. Thanks!
Any chances of adding something like this to read.spss()?
read.spss -
--- John Kane [EMAIL PROTECTED] wrote:
Date: Thu, 1 Feb 2007 09:07:11 -0500 (EST)
From: John Kane [EMAIL PROTECTED]
Subject: Re: [R] read.spss and encodings
To: Thomas Friedrichsmeier
[EMAIL PROTECTED]
Hi Thomas,
I am using R 2.4.1 on WindowsXP and I don't seem to
be
having any
days - seq(as.Date(1970/1/1), as.Date(2003/12/31), days)
temp - rnorm(length(days), mean=10, sd=8)
tapply(temp, format(days,%Y-%m), mean)
tapply(temp, format(days,%b), mean)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Majid Iravani
Sent:
Thank you Peter.
It was not my question but I was just about to start
the morning's work by searching Help and RSiteSearch()
for this exact question.
--- Peter Dalgaard [EMAIL PROTECTED] wrote:
stat stat wrote:
Dear all,
Suppose I have a date variable:
c = 99/05/12
Hello Demi
The trick for array indexing
on an array A, where length(dim(A))==n,
is to use an n-column matrix M to extract the elements
by rows of M.
If I understand correctly, the following should help:
A - array(1:18,c(3,3,2))
x - 2:3
y - 1:2
A[cbind(x,y,1)]
[1] 2 6
You may find
When I generate a LaTeX table using xtable I have been setting column names
to strings with LaTeX code in order to get features like subscripts in the
column headings. I recently had to reinstall xtable and discovered that all
my LaTeX column headings were printing out in LaTeX code rather than
You could also use aggregate with the zoo package. Using the same
input data that Vladimir used, create a zoo variable and aggregate it:
library(zoo)
z - zoo(dfr[,2], as.Date(dfr[,1], %d/%m/%Y))
aggregate(z, as.yearmon, mean)
Jan 1970 Dec 2003
0.00 1.25
zoo is described in the
Hi,
or otherwise you may try:
plot(c(1,5), c(1,10),type=l)
kindest regard, Rense
On Feb 1, 2007, at 8:14 , Petr Pikal wrote:
Hi
see ?segments
segments(1,10,5,10)
HTH
Petr
On 1 Feb 2007 at 14:21, XinMeng wrote:
From: XinMeng [EMAIL PROTECTED]
To:
Hi,
I apologize for this repeat posting, which I first posted yesterday. I would
appreciate any hints on solving this problem:
I have two matrices A (m x 2) and B (n x 2), where m and n are large
integers (on the order of 10^4). I am looking for an efficient way to
create another matrix,
Thank you very much, that works fine!
I now realize that I should have looked up not only the help pages for plot and
mca but also for plot.mca, which I did not think possible (unfortunately I am a
too sporadic user to know where to get the appropriate information).
Best regards,
Michael
Hi Jeff,
The way I do this is to place all the options that I want, along with
functions I've written that I always want available, into the
Rprofile.site file. R always loads this file upon startup. That file
is located in the etc/ folder. E.g., on my computer, it is at:
C:\Program
the following seems to be a first improvement:
m - 2000
n - 5000
A - matrix(rnorm(2*m), ncol=2)
B - matrix(rnorm(2*n), ncol=2)
W1 - W2 - matrix(0, m, n)
##
##
g1 - function(x, y){
theta - atan((y[2] - x[2]) / (y[1] - x[1]))
theta +
Hi
A - matrix(runif(10),ncol=2)
B - matrix(runif(10),ncol=2)
g - function(i4){theta - atan2( (i4[4]-i4[2]),(i4[3]-i4[1]))
+ return(theta + 2*pi*(theta0))}
apply(A,1,function(x){apply(B,1,function(y){g(c(x,y))})})
[,1] [,2] [,3][,4] [,5]
[1,] 1.1709326
On Thu, 1 Feb 2007, Thomas Friedrichsmeier wrote:
Hi!
I'm having trouble with importing spss files containing non-ascii characters
Peter has explained what is going on. It would be ideal for read.spss()
to do the translation to the current locale. This would require knowing
what encoding
I would use latex() in the Hmsic package. Here is a short example
tmp - matrix(1:12,4)
library(Hmisc)
tmp.latex - latex(tmp, colheads=c(abc$_1$,def$^{12}_4$,$g\\times h$))
## note the escaped \ in the above colheads vector
print.default(tmp.latex)
Copy the contents of the file referenced in
Hello all,
I'm trying to use the levelplot lattice function and can not adapt it to
my tastes concering colors:
dens - data.frame(x=c(), y=c(), z=c(), run=c())
for(l in levels(degCorrel$run)) {
ind - degCorrel$run == l
dk - kde2d(log10(degCorrel$correlFunc[ind]), log10(degCorrel
Hello,
I've got a data.frame like this:
assignation - data.frame(value=c(6.5,7.5,8.5,12.0),class=c(1,3,5,2))
assignation
value class
1 6.5 1
2 7.5 3
3 8.5 5
4 12.0 2
and a long vector of classes like this:
x - c(1,1,2,7,6,5,4,3,2,2,2...)
And
Thank you, Dimitris and Robin.
Dimitris - your solution(s) works very well. Although my g function is a
lot more complicated than that in the simple example that I gave, I think
that I can use your idea of taking the whole matrix inside the function and
working directly with it.
Robin - using
i get the message :
configure: WARNING: you cannot build info or html versions of the R manuals
how to deal with it ?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
One way would be to use merge, like this:
merge(assignation,data.frame(class=x),all.y=TRUE)
There might well be better ways...
On 01/02/07, javier garcia-pintado [EMAIL PROTECTED] wrote:
Hello,
I've got a data.frame like this:
assignation -
Hi,
xt - assignation$value[match(x,assignation$class)]
HTH
ido
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
a - data.frame(value=c(6.5,7.5,8.5,12.0),class=c(1,3,5,2))
x - c(1,1,2,7,6,5,4,3,2,2,2)
match(x, a$class)
[1] 1 1 4 NA NA 3 NA 2 4 4 4
a[match(x, a$class), value]
[1] 6.5 6.5 12.0 NA NA 8.5 NA 7.5 12.0 12.0 12.0
-- Tony Plate
javier garcia-pintado wrote:
Hello,
Dear Dimitris,
I implemented your solution on my actual problem. I was able to generate my
large transition matrix in 56 seconds, compared to the previous time of
around 27 minutes. Wow!!!
I thank you very much for the help. R and the R-user group are truly
amazing!
Best regards,
Ravi.
On 2/1/07, xiaopeng hu [EMAIL PROTECTED] wrote:
i get the message :
configure: WARNING: you cannot build info or html versions of the R
manuals
how to deal with it ?
did you read the manual 'R Installation and Administration'?
see section 2.2 there.
it is mentioned there that you need
Hello
My name is Aida Eslami. I am a M.S.c student of statistics at Shahid
Beheshti University , Tehran, Iran. The subject of my thesis is Analysis of
Masked Data. I have some problems in
Hi, there
I'm currently trying to figure out how to keep my factor levels for a
variable when moving it from one data frame or matrix to another.
Example below:
vec1-(rep(10,5))
vec2-(rep(30,5))
vec3-(rep(80,5))
vecs-c(vec1, vec2, vec3)
resp-rnorm(2,15)
dat-as.data.frame(cbind(resp, vecs))
Michael Rennie wrote:
Hi, there
I'm currently trying to figure out how to keep my factor levels for a
variable when moving it from one data frame or matrix to another.
Example below:
vec1-(rep(10,5))
vec2-(rep(30,5))
vec3-(rep(80,5))
vecs-c(vec1, vec2, vec3)
resp-rnorm(2,15)
On Thu, 2007-02-01 at 12:13 -0500, Michael Rennie wrote:
Hi, there
I'm currently trying to figure out how to keep my factor levels for a
variable when moving it from one data frame or matrix to another.
Example below:
vec1-(rep(10,5))
vec2-(rep(30,5))
vec3-(rep(80,5))
vecs-c(vec1,
a.eslami a.eslami at Mail.sbu.ac.ir writes:
Hello
My name is Aida Eslami. I am a M.S.c student of statistics at Shahid
Beheshti University , Tehran, Iran. The
subject of my thesis is Analysis of Masked Data. I have some problems in
writing of my program
(optimization). Would you
Greetings.
For R gurus this may be a no brainer, but I could not find pointers to
efficient computation of this beast in past help files.
Background - I wish to implement a Cramer-von Mises type test statistic
which involves double sums of max(X_i,Y_j) where X and Y are vectors of
differing
Jeff,
Here is something which is a little faster:
sum1 - sum(outer(x, x, FUN=pmax))
sum3 - sum(outer(x, y, FUN=pmax))
Best,
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Well, a reproducible example would be nice =)
not tested:
x = rnorm(10)
y = rnorm(20)
mymax - function(t1, t2) apply(cbind(t1, t2), 1, max)
sum(outer(x, y, mymax))
is this sth like what you need?
b
On Feb 1, 2007, at 1:18 PM, Jeffrey Racine wrote:
Greetings.
For R gurus this may be a no
Benoit Ballester [EMAIL PROTECTED] writes:
Hi,
I was wondering if someone could tell me more about this book, (if it's
a good or bad one).
I can't find it, as it seems that O'Reilly doesn't publish any more.
I've never seen a copy so I can't comment about its quality (has
anyone seen a
XLSolutions Corporation (www.xlsolutions-corp.com) is proud to
announce our R/S-plus Fundamentals and Programming Techniques :
www.xlsolutions-corp.com/Rfund.htm
*** Washington DC / March 1-2, 2007
*** San Francisco / March 15-16, 2007
*** Princeton / Week of Feb 26 (dates coming soon)
Should
javier garcia-pintado jgarcia at ija.csic.es writes:
Hello,
I've got a data.frame like this:
assignation - data.frame(value=c(6.5,7.5,8.5,12.0),class=c(1,3,5,2))
assignation
value class
1 6.5 1
2 7.5 3
3 8.5 5
4 12.0 2
and a long
On Thu, 2007-02-01 at 10:45 -0800, Seth Falcon wrote:
Benoit Ballester [EMAIL PROTECTED] writes:
Hi,
I was wondering if someone could tell me more about this book, (if it's
a good or bad one).
I can't find it, as it seems that O'Reilly doesn't publish any more.
I've never seen a
Jeff,
you can do
sum1: \sum_i\sum_j max(X_i,X_j)
sum2: \sum_i\sum_j max(Y_i,Y_j)
sum(x * (2 * rank(x) - 1))
sum3: \sum_i\sum_j max(X_i,Y_j)
sum(outer(x, y, pmax))
Probably, the latter can be speeded up even more...
Z
__
R-help@stat.math.ethz.ch
Consider the following lines of code:
plot(function(x) sin(cos(x)*exp(-x/2)), from=-8,to=7,xlim=c(-5,5))
Uses integral points (integers from -5 to 5) to draw the plot, instead
of the usual default of n= 101 equally spaced points (from
?plot.function).
plot(function(x)
Hi,
I was wondering if there is a direct approach for lining up 2-column
matrices according to the values of the first column. An example and a
brute-force approach is given below:
x - cbind(1:10, runif(10))
y - cbind(5:14, runif(10))
z - cbind((-4):5, runif(10))
xx - seq(
On Thu, 2007-02-01 at 15:05 -0500, Christos Hatzis wrote:
Hi,
I was wondering if there is a direct approach for lining up 2-column
matrices according to the values of the first column. An example and a
brute-force approach is given below:
x - cbind(1:10, runif(10))
y - cbind(5:14,
Marc Schwartz wrote:
On Thu, 2007-02-01 at 10:45 -0800, Seth Falcon wrote:
Benoit Ballester [EMAIL PROTECTED] writes:
Hi,
I was wondering if someone could tell me more about this book, (if it's
a good or bad one).
I can't find it, as it seems that O'Reilly doesn't publish any
Hi.
I have the following code in a loop. It splits a vector into subvectors of
equal size. But if the size of the original vector is not an exact multiple
of the desired subvector size, then the first few subvectors have one more
element than the last few. I know that the cut function could
What is 'Solaris 11'? According to www.sun.com, the latest Solaris
version is 10, and my sysadmins have not heard of Solaris 11.
You seem to be missing the Solaris compilation tools, ar in this case.
In Solaris = 10 they are in /usr/ccs/bin, not in the path by default.
On Wed, 31 Jan 2007,
Thanks Marc and Phil.
My dataset actually consists of 50+ individual files, so I will have to do
this one column at a time in a loop...
I might look into SQL and outer joints as an alternative to avoid looping.
Thanks again.
-Christos
-Original Message-
From: Marc Schwartz
On Thu, 2007-02-01 at 21:32 +0100, Peter Dalgaard wrote:
Marc Schwartz wrote:
On Thu, 2007-02-01 at 10:45 -0800, Seth Falcon wrote:
Benoit Ballester [EMAIL PROTECTED] writes:
Hi,
I was wondering if someone could tell me more about this book, (if it's
a good or bad one).
Does anyone know a good introductory book or tutorial about time series
analysis? (time
series for a beginner).
Thank you so much.
John Lamak
_
Descubra como mandar Torpedos SMS do seu Messenger para o celular dos seus
amigos.
John --
Well, as a start, have a look at Modern Applied Statistics with S, by
Venables and Ripley, both of which names you will recognize if you read
this list often. There is a 30-page chapter on time series (with
suggestions for other readings), obviously geared to S and R, that is a
good
On Thu, 2007-02-01 at 15:45 -0500, Christos Hatzis wrote:
Thanks Marc and Phil.
My dataset actually consists of 50+ individual files, so I will have to do
this one column at a time in a loop...
I might look into SQL and outer joints as an alternative to avoid looping.
Thanks again.
I need to generate autocorrelated binary data. I've found references to
the IEKS package but none of the web pages currently exist. Does anyone
know where I can find this package or suggest another package?
Rick B.
__
R-help@stat.math.ethz.ch mailing
Christos,
Haccording to the Value section in ?merge:
A data frame. The rows are by default lexicographically sorted on the
common columns, but for sort=FALSE are in an unspecified order.
Looking at the code, while there is a lot of time spent on matching
things, the key sort() code
[Sorry I meant to reply to the list]
Thanks, Marc.
That's what I have done.
However, there seems to be a penalty from using merge repeatedly as it
appears to internally re-sort the datasets. In my case the datasets are
long (~35K rows) and already sorted so this step adds considerable and
Hi,
I am trying to read in my Affymetrix CEL files (48 files, total ~600 MB) but
I keep getting memory errors. Can somebody please help me with this. Or is
therea remote server I can send my data to for computation?
Any help is much appreciated.
Thanks
Dr. Tristan Coram
This might do what you want:
# test data
x - 1:43
nb - 5 # number of subsets
# create vector of lengths of subsets
ns - rep(length(x) %/% nb, nb)
# see if we have to adjust counts of initial subsets
if ((.offset - length(x) %% nb) != 0) ns[1:.offset] = ns[1:.offset] + 1
# create the
Anil Kumar,
it seems there isn't packages for continuous wavelet
transforms in R. Anyway, take a look at the packages
waveslim, wavethresh, wavelets or rwt. Maybe one of
them can be useful to you.
Rogerio.
-- Cabeçalho original ---
De: [EMAIL PROTECTED]
Para:
My R search page at http://finzi.psych.upenn.edu/
which is also what you get with RSiteSearch()
has been slowing down the last few months (years?).
I thought this was because the archives were just getting too big.
But I discovered a simple fix. The technical term for the problem is
garbage. By
On Thu, 1 Feb 2007, Marc Schwartz wrote:
Christos,
Haccording to the Value section in ?merge:
A data frame. The rows are by default lexicographically sorted on the
common columns, but for sort=FALSE are in an unspecified order.
There is also a sort in the .Internal code. But I am
On Feb 1, 2007, at 6:22 AM, Douglas Bates wrote:
It turns out that in the sparse matrix code used by the
Matrix package the triplet representation allows for duplicate index
positions with the convention that the resulting value at a position
is the sum of the values of any triplets with
The zoo package has a multiway merge with optional zero fill.
Here are two ways:
library(zoo)
merge(x = zoo(x[,2], x[,1]),
y = zoo(y[,2], y[,1]),
z = zoo(z[,2], z[,1]),
fill = 0)
# or
library(zoo)
X - list(x = x, y = y, z = z)
merge0 - function(..., fill = 0) merge(..., fill =
The bioconductor mailing list is probably a better place to ask this
type of question.
[EMAIL PROTECTED]
But we also need to know what arrays are you working with, what the
errors are, what your sessionInfo() is
Let us know, ok?
b
On Feb 1, 2007, at 5:46 PM, Tristan Coram wrote:
Hi,
Tristan,
I have a soft spot for problems analyzing microarrays with R..
for the memory issue, there have been previous posts to this list..
But here is the answer I gave a few weeks ago.
If you need more memory, you have to move to linux or recompile R for
windows yourself..
.. But you'll still
I don't have a Linux system to try it with but
omitting both dev.control statements it worked for me
between two Windows XP sessions on the same
machine using this version of R:
R.version.string # Windows XP
[1] R version 2.4.1 Patched (2006-12-30 r40331)
It also successfully worked with:
On Thu, 2007-02-01 at 23:34 +, Prof Brian Ripley wrote:
On Thu, 1 Feb 2007, Marc Schwartz wrote:
Christos,
Haccording to the Value section in ?merge:
A data frame. The rows are by default lexicographically sorted on the
common columns, but for sort=FALSE are in an
Marc,
I don't think the issue is duplicates in the matching columns. The data
were generated by an instrument (NMR spectrometer), processed by the
instrument's software through an FFT transform and other transformations and
finally reported as a sequence of chemical shift (x) vs intensity (y)
On Thu, 2007-02-01 at 20:39 +, Prof Brian Ripley wrote:
What is 'Solaris 11'? According to www.sun.com, the latest Solaris
version is 10, and my sysadmins have not heard of Solaris 11.
That is the solaris express community release, or the pre-release of the
upcoming OpenSolaris
Thanks Gabor.
This is along the lines of what I was looking for. In fact the merge
function for zoo objects (ordered) turns out to be almost an order of
magnitude faster than the generic merge function for my problem:
system.time(
+ zz - merge( spec.1 = zoo(nmr.spectra.serum[[1]]$V2,
On Thu, 2007-02-01 at 22:46 -0500, Christos Hatzis wrote:
Marc,
I don't think the issue is duplicates in the matching columns. The data
were generated by an instrument (NMR spectrometer), processed by the
instrument's software through an FFT transform and other transformations and
finally
Marc,
The data structure is a list of data frames generated from read.table:
class(nmr.spectra.serum)
[1] list
class(nmr.spectra.serum[[1]])
[1] data.frame
dim(nmr.spectra.serum[[1]])
[1] 32768 2
Converting the data.frames to matrices does not have much of an effect on
timing.
Hi,
I am working on a regression tree in Rpart that uses a continuous response
variable that is ordered. I read a previous response by Pfr. Ripley to a
inquiry regarding the ability of rpart to handle ordinal responses in
2003. At that time rpart was unable to implement an algorithm to handle
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