Note that the use of read.table() does make a difference. If you did
x - scan(gzfile(xxx.gz), list(,,))
you would leave an unused connection, and showConnections(all=TRUE) would
show this. There is a finite pool of connections, and in general the
correct way to use them is
con -
On Tue, 3 Jul 2007, hadley wickham wrote:
Hi Stephane,
The problem is that the windows graphics device doesn't support
transparent colours. You can get around this in two ways:
It certainly does! Try col=transparent (and perhaps consult your
dictionary). It was news to me that the
On 7/3/07, Héctor Villalobos [EMAIL PROTECTED] wrote:
Hi all,
I want to retrieve the stats from a 'bwplot' with one factor. I have read
the help for 'panel'
function and I'm aware of the option 'stats' which defaults to
'boxplot.stats' but I didn't
understand it well and therefore I am
Hello,
I write us because I wanna know if it's possible to don't display
read item
when I execute a scan(textConnection())
thanks
_
[[alternative HTML version deleted]]
Hi all,
I am a freshman of R,but I am interested in it! Those days,I am
learning pages on NIST,with url
http://www.itl.nist.gov/div898/handbook/eda/section3/probplot.htm,
I am meeting a problem about probability plot and I don't know how to
plot a data set with R.
Could somebody tell me the
On 7/4/07, Prof Brian Ripley [EMAIL PROTECTED] wrote:
On Tue, 3 Jul 2007, hadley wickham wrote:
Hi Stephane,
The problem is that the windows graphics device doesn't support
transparent colours. You can get around this in two ways:
It certainly does! Try col=transparent (and perhaps
Hi,
I can't get this approach to work. When I first added the repsitory line to
/etc/apt/sources.list synaptic complained that it was a malformed line. I
fixed this
by adding main to end of the entry making it:
deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty main
However after
Dear all,
I am trying to shove a number of cmdscale() results into a single plot
(k=1 so I'm trying to get multiple columns in the plot). From ?par I
learned that I can/should set new=TRUE in either par() or the plot
function itself. However with the following reduced code, I get only a
plot
Dear R help,
I'm working with regular 8-daily time-series from 2000 up till now and
would like to be able to compare years with each other. E.g. by creating
a monthplot via the result of the stl() method it looks ok but I was
wondering whether there exist other methods to plot the different
I think fine tuning the function might be in order.
The function has just a single penalty for not meeting
the constraints no matter how close it is to meeting
them. A better approach is to have a penalty that
depends on the amount by which all of the constraints
are breached.
Patrick Burns
to end of the entry making it:
deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty main
However after this it still complains that it can't find packages.gz
Just a guess: have you replaced the my.favorite.cran.mirror by a mirror
which is close to you? If you're in UK it would
Hi
if you change your code
plot(1,10, xlim=range(0,3), ylim=range(0,10), type='n')
aa - rep(1,10)
bb - 1:10
plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE)
aa - rep(2,10)
par(new=T)
plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE)
you will get both columns plotted.
However you
Hi,
Thanks for the suggestion and I wish the solution was that obvious, but I
have changed it to really point at my favourite mirror.
Using your example Synaptic reports the following error when I try to update
the repositories:
The ksvm object is probably what you need to use.
[quote]
Dear R users,
I'm trying to run the Support Vector Regression by a general
quadratic programming function like ipop ( ) in kernlab or solve.QP ( )
in quadprog packages.
Since they are general, their application in Support Vector
On 7/4/07, RAVI VARADHAN [EMAIL PROTECTED] wrote:
Here is another approach: I wrote an R function that would generate interior
points as starting values for constrOptim. This might work better than the
LP approach, since the LP approach gives you a starting value that is on the
boundary of
Hi all, thank you very much.
livia wrote:
Hi, I would like to generate a series in the following form (0.8^1, 0.8^2,
..., 0.8^600)
Could anyone tell me how can I achieve that? I am really new to R.
--
View this message in context:
[EMAIL PROTECTED] wrote:
myYlabGrob -
function(..., main.ylab = ) ## ...is lab1, lab2, etc
{
## you can add arguments to textGrob for more control
## in the next line
labs - lapply(list(...), textGrob, rot=90)
main.ylab - textGrob(main.ylab, rot = 90)
nlabs -
Hi Petr,
On 7/4/07, Petr PIKAL [EMAIL PROTECTED] wrote:
par(new=T)
plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE)
So I need to activate the par(new=T) really just ahead of time when I
need it, not as sort of a general clause at the beginning of my
script?
However you can get
G'day Uwe,
On Tue, 03 Jul 2007 14:33:05 +0200
Uwe Ligges [EMAIL PROTECTED] wrote:
Pietrzykowski, Matthew (GE, Research) wrote:
I saw the new book,
The R Book, by Michael J. Crawley and wanted to know what R users
thoughts of it.
The author seems to be an expert in (almost?) all
Hi R Users,
Thanks in advance.
I am using R-2.5.1 on Windows XP.
I am trying to call C code (testCX1.C) from R. testCX1.c calls another C code
(funcC1.c) and returning a value to testCX1.c. I like to have this value in R.
My steps are below:
1. R CMD SHLIB testCX1.c
On Wed, Jul 04, 2007 at 04:39:18AM -0700, Deb Midya wrote:
Hi R Users,
Thanks in advance.
I am using R-2.5.1 on Windows XP.
I am trying to call C code (testCX1.C) from R. testCX1.c calls another C
code (funcC1.c) and returning a value to testCX1.c. I like to have this
msmith schrieb:
Hi,
Thanks for the suggestion and I wish the solution was that obvious, but I
have changed it to really point at my favourite mirror.
Using your example Synaptic reports the following error when I try to update
the repositories:
Stefan Grosse schrieb:
deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty main
Sorry, I copied a mistake there, it should be:
deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty/
Stefan
-=-=-
... The simple truth is that interstellar distances will not fit into
the human
Is this what you mean ?
---
mydata - c(1,2,3,4,5,7,5,4,3)
plot(mydata)
---
--- along zeng [EMAIL PROTECTED] wrote:
Hi all,
I am a freshman of R,but I am interested in it!
Those days,I am
learning pages on NIST,with url
Gabor,
Thank you very much for such a quick response.
As I am new to this area, will you please explain where can I put SEXP
func(SEXP a);
in my program.
Once again, thank you very much for your quick response.
Regards,
Deb
Gabor Csardi [EMAIL PROTECTED] wrote:
On Wed, Jul 04, 2007 at 05:15:15AM -0700, Deb Midya wrote:
Gabor,
Thank you very much for such a quick response.
As I am new to this area, will you please explain where can I put SEXP
func(SEXP a);
in my program.
Deb, anywhere before calling it. (Well outside a function
Hi, I would like to apply the following function for i between 1 and 12, and
then construct a list of the return series.
for (i in 1:12){
ewma[i] - emaTA(calm[[i]]^2,0.03)
standard[i]- calm[[i]]/sqrt(ewma[i])
standard - cbind(standard[i])
}
But it does not work. Could anyone give me some advice
Dear list,
if I do
smooth.spline(tmpSec, tmpT, all.knots=T)
with the attached data, I get this error-message:
Error in smooth.spline(tmpSec, tmpT, all.knots = T) :
smoothing parameter value too small
If I do
smooth.spline(tmpSec[-single arbitrary number], tmpT[-single arbitrary
Hi
In what way does it not work?
My guess is that you have not declared your values outside the for loop.
As they are local they will be lost on exit.
You need to declare them before:
ewma-vector(length=12)
standard-vector(length=12)
for ... {
}
John Seers
---
Hi, I
A more elegant way to do this is
standard - sapply(calm, function(calmi){calmi / sqrt(emaTA(calmi ^ 2,
0.03))})
Cheers,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for
On 7/4/07, RAVI VARADHAN [EMAIL PROTECTED] wrote:
My point is that it might be better to try multiple (feasible) starting
values for constrOptim to ensure that you have a good local minimum, since it
appears that constrOptim converges to a boundary solution where the gradient
is non-zero.
On 7/4/07, Paul Smith [EMAIL PROTECTED] wrote:
On 7/4/07, RAVI VARADHAN [EMAIL PROTECTED] wrote:
My point is that it might be better to try multiple (feasible) starting
values for constrOptim to ensure that you have a good local minimum, since
it appears that constrOptim converges to a
I am using R 2.5.1 for windows and my purpose is to estimate a clayton copula .
Since I have two time series marginals, I found that the most appropriate model
was an ARMA(1,0)+GARCH(1,1) model for both with sstd as conditional
distribution. Can anyone give me some tips about the code to
Hi R-ers:
I'm drawing a plot and have used different line types (lty) for
different race/ethnicity groups. I want a legend that explains what line
types correspond to the different race/ethnicity groups. I used the
following code:
legend( 1992 , 42 , c(Hispanic , non-Hispanic white (NHW) ,
Keith Alan Chamberlain Keith.Chamberlain at Colorado.EDU writes:
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') #
Cat - c('a','a','a','b','b','b','a','a','b')
C1 - ifelse(Cat == 'a', -1, 1)
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en
S is an array 1-dimensional, for example 1 X 10, and mean(S) is the mean of
these 10 elements.
So, I want to do:
minimize mean(S) with 0 b_func(S) 800.
That is, there are some boundaries on S according the b_funct
The function apply an iterative convergent criterion:
f_1=g(S), f_2=g(f_1),
C1 - rep(-1, length(Cat))
C1[Cat == b]] - 1
b
On Jul 4, 2007, at 9:44 AM, Keith Alan Chamberlain wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this,
but one
which can also handle
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1
On 04-Jul-07 13:44:44, Keith Alan Chamberlain wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values
from another vector? I'd like to call a pre-existing function for
this, but one which can also handle an arbitrarily large number
of categories. Any ideas?
Here are two ways. The second way is more than 10x faster.
set.seed(1)
C - sample(c(a, b), 10, replace = TRUE)
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
0.370.010.38
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.020.000.02
or
Cat - c('a','a','a','b','b','b','a','a','b')
C1 - (Cat=='a')*1
ONKELINX,
Thierry
Gabor Grothendieck wrote:
set.seed(1)
C - sample(c(a, b), 10, replace = TRUE)
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
0.370.010.38
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.020.000.02
system.time(s1 -
#Given
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
#and defining
coding-array(c(-1,1), dimnames=list(unique(Cat) ))
#(ie an array of values corresponding to your character array levels, and with
names set to those levels)
coding[Cat]
#does what you want.
Keith
I am new in R and I want to solve this problem;
I have a matrix X (with n-rows and p-colums) my problem is to obtain the
products of the vectors of rows and print out only the maximum value and
identify the row that gives the maximum value. Thanks
Oyeyemi, G.M
which.max(apply(mat, 1, prod))
Gabor
On Wed, Jul 04, 2007 at 09:18:41AM -0700, Yemi Oyeyemi wrote:
I am new in R and I want to solve this problem;
I have a matrix X (with n-rows and p-colums) my problem is to obtain the
products of the vectors of rows and print out only the maximum value
Dear Ted,
You are correct in that factors are probably what I had in mind since I
would be using them as predictors in a regression. I didn't know the syntax
to get R to do the arithmetic.
Many thanks to everyone who replied!
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair
In thinking about this a bit more I have found a slightly faster one still.
See s3. Also I have added s0, the original solution, to the timings.
set.seed(1)
C - sample(c(a, b), 100, replace = TRUE)
system.time({
+ s0 - vector(length = length(C))
+ for(i in seq_along(C)) s0[i] - if (C[i]
Yemi,
Try
which.max(apply(X,1,prod))
(or possibly abs(apply(X,1,prod)) if you're only interested in unsigned product
max.
S
Yemi Oyeyemi [EMAIL PROTECTED] 04/07/2007 17:18:41
I am new in R and I want to solve this problem;
I have a matrix X (with n-rows and p-colums) my problem is to
This was in error since s3 was not set. The as.numeric in the calculation
of s3 can be omitted if its ok to have an integer rather than numeric result
and in that case its still faster yet.
set.seed(1)
C - sample(c(a, b), 100, replace = TRUE)
system.time({
+ s0 - vector(length =
Hi R hep group member,
I want to know that how I can call my data in R for princomp function.
I want to calculate PCA of 200 descriptors of 4000 molecule(I am using
Linux). How I can call this in R.
Thanking you.
--
Nitish Kumar Mishra
Junior Research Fellow
BIC, IMTECH, Chandigarh, India
Dear,
Thanks for your prompt assistant. The problem I posted is
just a bit of my problem. The whole problem is that;
1. I simulated a multivariate data set X (n by p)
2. Then draw sample without replacement from X (k less than n)
3. Compute the eigen values of the
Dear all,
first I would like to tell you that I've been using R for two days... (so,
you can predict my knowledge of the language!).
Yet, I managed to implement some stuff related with the Long-Tail model [1].
I did some tests with the data in table 1 (from [1]), and plotted figure 2
(from [1]).
[Keith Alan Chamberlain]
Is there a faster way than below to set a vector based on values
from another vector? I'd like to call a pre-existing function for
this, but one which can also handle an arbitrarily large number of
categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') #
User and System are a measure of the CPU time that was consumed. Elapsed
time is the wall clock and even though they are both measured in seconds,
they are not really the same units. The reason for the difference is any
idle time that they system may have waiting for I/O to complete which does
Whether you can use optim or not depends on the nature of the constraints on
S. If you have simple box constraints, you can use the L-BFGS-B method in
optim. If not, optim may not be directly applicable, unless you can somehow
transform your problem into an unconstrained minimization problem.
One other thing, in a multiprocessor configuration, if your application is
making use of the additional CPUs, then
User + System Elapsed
In some cases.
On 7/4/07, jim holtman [EMAIL PROTECTED] wrote:
User and System are a measure of the CPU time that was consumed. Elapsed
time is the wall
If the constraints on S are linear inequalities, then linear programming
methods would work. See function solveLP in package linprog or simplex in
boot or package lpSolve.
Ravi.
- Original Message -
From: domenico pestalozzi [EMAIL PROTECTED]
Date: Wednesday, July 4, 2007 11:26 am
?scan
Help page says to set quiet=TRUE
On 7/4/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote:
Hello,
I write us because I wanna know if it's possible to don't display
read item
when I execute a scan(textConnection())
thanks
Hi All,
A question from a newbie using R 2-5-0 on windows XP. Why is it that
kmeans clustering with apparently the exact same parameters behaves so
differently between the two following examples :
cl1 - kmeans(subset(pointsUXO1, select = c(2:4)), 10)
Takes about 2 seconds to deliver a
Hey all; I'm a beginner++ user of R, trying to use it to do some processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user id. as
I run through the transactions, I need to update a separate table for
I think you simply had your nls() syntax wrong. Works here:
## first a neat trick to read the data from embedded text
fmdata - read.csv(textConnection(
+ rank,cum_value
10, 17396510
32, 31194809
96, 53447300
420,100379331
1187, 152238166
24234, 432238757
91242, 581332371
mfrumin wrote:
Hey all; I'm a beginner++ user of R, trying to use it to do some processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user id. as
I run through the transactions, I need to update a
Hi everybody,
I'm still interesting the possibility to use R for genetic programming (see
https://stat.ethz.ch/pipermail/r-help/2007-April/128782.html)
and I'd like to know, how to express for instance this kind of functions
(x^2+3x+1 etc, see the picture) using some kind of tree structures and
mike,
try installing directly using apt-get instead of Synaptic.
in my /etc/apt/sources.list i added the line:
deb http://cran.R-project.org/bin/linux/ubuntu/ dapper/
and then i did:
bash$ sudo apt-get install r-base r-doc-info r-doc-pdf r-doc-html
r-mathlib r-base-html
Michael,
A hash provides constant-time access, though the resulting perl-esque
data structures (a hash of lists, e.g.) are not convenient for other
manipulations
n_accts - 10^3
n_trans - 10^4
t - list()
t$amt - runif(n_trans)
t$acct - as.character(round(runif(n_trans, 1, n_accts)))
uhash
i wish it were that simple. unfortunately the logic i have to do on
each transaction is substantially more complicated, and involves
referencing the existing values of the user table through a number of
conditions.
any other thoughts on how to get better-than-linear performance time?
is
__
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, minimal, self-contained, reproducible code.
I just got 203 hits from RSiteSearch(copula) and 60 from
RSiteSearch(copula, fun). Most if not all of the first 24 or the
hits in the latter referred to the 'copula' package. Have you reviewed
these? The 25th hit in the latter referred to an 'fgac' package for
'Generalized Archimedean
Hi R Gurus!
I'm trying to create a test package using the package.skeleton function.
I wanted to add some compiled code too.
In the src library, I put together a baby subroutine, compiled it and created
a test.dll
When I use the R cmd build, it works fine. But I get into trouble
with the R CMD
On 04/07/2007 6:43 PM, Edna Bell wrote:
Hi R Gurus!
I'm trying to create a test package using the package.skeleton function.
I wanted to add some compiled code too.
In the src library, I put together a baby subroutine, compiled it and created
a test.dll
When I use the R cmd build, it
Michael Frumin wrote:
i wish it were that simple. unfortunately the logic i have to do on
each transaction is substantially more complicated, and involves
referencing the existing values of the user table through a number of
conditions.
any other thoughts on how to get better-than-linear
Dear all,
In optim() all parameters of a function to be adjusted is stored in a single
vector, with lower/upper bounds can be specified by a vector of the same
length.
In nls(), is it true that if I want to specify lower/upper bounds, functions
must be re-written so that each parameter is
On 7/4/07, Martin Morgan [EMAIL PROTECTED] wrote:
Michael,
A hash provides constant-time access, though the resulting perl-esque
data structures (a hash of lists, e.g.) are not convenient for other
manipulations
n_accts - 10^3
n_trans - 10^4
t - list()
t$amt - runif(n_trans)
t$acct
Ana
You are estimating a random coefficient model on 5 individuals (mean
and variance). Are you sure this is wise?
Ross Darnell
- Original Message -
From: Ana Conesa [EMAIL PROTECTED]
Date: Thursday, July 5, 2007 1:21 am
Subject: [R] questions on lme function
Dear list,
I am
Hi,I have a fundamental questions that I'm a bit confused. If any guru from
this circle could help me out, I would really appreciate.I have a system of
equations in which some of the endogs appear on right hand sides of some
equations. To solve this, one needs a technique like 2SLS or FIML to
Hi gurus,
I have a doubt about multiclass classification SVM. The population in my data
includes a couple of class labels that have relatively small proportion of the
entire population compared to other classes. I would like SVM to pay more
attention to these classes. However, the question I
I am trying to use the crr function in the cmprsk package to analyze a large
patient dataset (45000 +), The model has 100 + covariates and 5 competing
risks. I am finding that R seems to get bogged down and even if I let it run
for several hours I don't get anything back. Am I expecting too much,
Hi,
Sorry that I have many questions today. I am using svm function on about
180,000 points of training set. It takes very long time to run. However, I
would like it to spit out something to make sure that the run is not dead in
between. Would you please suggest anyway to do so?
And is
Dear R users,
I'm trying to fit stable distribution and hyperbolic distribution to my data
using stableFit(), and hypFit() of fBasics.
However, there are some problems
This is the result
==
stableFit(lm, alpha = 1, beta = 0, gamma = 1, delta = 0, doplot =
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