On 10 Apr 2008, at 07:43, Shubha Vishwanath Karanth wrote:
> So powerful, the gsub... But I really don’t understand the how the
> regular expressions like " *\\S+$", need to be used and how to make
> best use of it... Any article/material/links that I can go through?
A good starting point is:
Hi,
I am plotting 5 charts using p <- par(mfrow = c(3, 2), how can I place my
legend in the last region ? I don't wan to put it into the margin.
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Found the solution,
p <- par(mfrow = c(3, 2), oma = c(0, 0, 2.1, 0))
title(main = paste("Title is here"), outer = TRUE, cex.main = 1.5, font.main
= 4)
On Thu, Apr 10, 2008 at 11:20 AM, Ng Stanley <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a 3 by 2 plots per page. How do I specify a title at
>From the 'R Installation and Administration Manual':
Note that @env{TMPDIR} will be used to execute @command{configure}
scripts when installing packages, so if @code{/tmp} has been mounted as
@samp{noexec}, @env{TMPDIR} needs to be set to a directory from which
execution is allowed.
Hello everyone,
Is it possible to perform "Autoregressive Conditional Duration" (ACD) in R?
I'm a graduate student at U of Oregon and want to use ACD in my research
project.
Please advise.
Best,
Yuki
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On Wed, Apr 9, 2008 at 9:30 PM, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> I've never seen that. Is there something unusual about the filesystem so
> it does not recognize rgl/configure is executable? (We've seen that on SMB
> filesystems with the wrong mount options.)
>
In /etc/fstab, I ha
So powerful, the gsub... But I really don’t understand the how the regular
expressions like " *\\S+$", need to be used and how to make best use of it...
Any article/material/links that I can go through?
BR, Shubha
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalo
On 10.04.2008, at 05:18, Ng Stanley wrote:
> Also, is there any way to specify scientific notation for axes label ?
Scientific notation à la 3E-4 is set by default. Or did you mean
engineering notation?
Maybe this could help:
http://wiki.r-project.org/rwiki/doku.php?id=tips:data-
strings:forma
Hi Bill: Thanks for the reply! As you've no doubt guessed, I'm not a
statistician (I'm a social scientist). I hadn't given thought to
modeling the cluster-based covariance explicitly--interesting
possibility. My responses are drawn from some 60 discussion groups,
and a critic of my current
How to fit a nonlinear regression model with power exponentially distributed
errors? I know gnlm has a function gnlr3 that could work, but I would be
grateful if example R code is provided.
Daniel
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At 2:00 AM +0300 4/10/08, Donatas G. wrote:
>I am looking for simple introduction to cluster analysis using R, that would
>be understandable to a novice in statistics. Or, could someone perhaps help
>me understand how to proceed in my analysis? I am very new to both statistics
>and R, but am tryin
On Wed, 9 Apr 2008, Roger Levy wrote:
> I am developing a package and I want to be able to load an updated
> version of the package from within an active R session. Suppose, for
> example, I have a function f within a package X. In my active R
> session, I have already loaded X. Then I change t
Hi,
I have a 3 by 2 plots per page. How do I specify a title at the top centre
of each page ?
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Hi,
I have a 3 by 2 plots per page, and would like to place a legend on the last
region. How to do that ?
Also, is there any way to specify scientific notation for axes label ?
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I concur!
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Rolf Turner
Sent: Wednesday, April 09, 2008 10:22 PM
To: R-help List
Subject:
On 10/04/2008, at 2:02 PM, <[EMAIL PROTECTED]> wrote:
> In many cases a) often looks difficult, but on closer inspection turns
> out to be impossible
I nominate this for a fortune.
cheers,
Rolf Turner
optim is a general purpose optimiser. You don't reallly use it to
'analyze' data and you cannot get a variance matrix directly from the
result, even using vcov. If you ask, it will give you the hessian
matrix of the objective function at the optimum value, from which you
can get a variance matrix
Hi, John:
I just got the following error right after the attempt to use
'rmvnorm'.
Error: could not find function "rmvnorm"
I tried 'library(mvtnorm)', but the 'rmvnorm' in that package gave
me the following:
Error in rmvnorm(1, mean = c(3, -20, -10, 3, 2), sd = c(0.1, 15,
Le mer. 9 avr. à 18:00, Paul Johnson a écrit :
> On Wed, Apr 9, 2008 at 4:07 PM, Laura Bonnett <[EMAIL PROTECTED]
> > wrote:
>> Dear Sirs,
>>
>> I am using both the Bioconductor adds on (Affy, AffyPLM,...) and the
>> 'standard' R-package.
>>
>> I am trying to select a list of genes which all have
Dear list,
I am a little puzzled by computing time in connection with calling C functions.
With the function mysolve1 given below I solve Ax=B, where the actual matrix
operation takes place in mysolve2. Doing this 5000 times takes 3.51 secs.
However, if I move the actual matrix inversion part
I've used optim to analyze some data I have with good results, but
need to correct the var-cov matrix for possible effects of clustering
of observations (respondents) in small groups (non-independence). Is
there any function to adjust the matrix? I heard some time ago that
the vcovHC function wou
I'm solving the differential equation dy/dx = xy-1 with y(0) = sqrt(pi/2).
This can be used in computing the tail of the normal distribution.
(The actual solution is y(x) = exp(x^2/2) * Integral_x_inf {exp(-t^2/2) dt}
= Integral_0_inf {exp (-xt - t^2/2) dt}. For large x, y ~ 1/x, starting
around x~
I am looking for simple introduction to cluster analysis using R, that would
be understandable to a novice in statistics. Or, could someone perhaps help
me understand how to proceed in my analysis? I am very new to both statistics
and R, but am trying hard to avoid having to use SPSS as everyone
I am developing a package and I want to be able to load an updated
version of the package from within an active R session. Suppose, for
example, I have a function f within a package X. In my active R
session, I have already loaded X. Then I change the R source code of f
within X and rebuild
A student brought this question to me and I can't find any articles or
examples that are directly on point.
Suppose there are 2 ordinal logistic regression models, and one wants
to set them into a simultaneous equation framework. Y1 might be a 4
category scale about how much the respondent likes
On Wed, Apr 9, 2008 at 4:07 PM, Laura Bonnett <[EMAIL PROTECTED]> wrote:
> Dear Sirs,
>
> I am using both the Bioconductor adds on (Affy, AffyPLM,...) and the
> 'standard' R-package.
>
> I am trying to select a list of genes which all have expression values below
> a certain threshold.
> I hav
On Apr 9, 2008, at 1:27 PM, Hans-Jörg Bibiko wrote:
>
> On 09.04.2008, at 17:46, Shubha Vishwanath Karanth wrote:
>> To put it simple,
>>
>> C=c("My Dog", "Its really good", "Beautiful")
>>
>> Now,
>> SOMEFUNCTION(C) should give: c("My", "Its really", "")
>
> SOMEFUNCTION <- function(x) gsub(" *\
For your first problem, you can probably do it in 2 statements:
V3 = ifelse(V2==A,V2,V3)
V3 = ifelse(V2==B|V2==C,D,V3)
If you want to split V1 into (0,a],(a,b],(b,c],(c,1], you can do, quite simply
V1.factor = cut(V1, c(0,a,b,c,1))
Abhijit
On Wed, 9 Apr 2008 13:58:17 -0700
Chang Liu <[EMAIL PROTE
Dear Sirs,
I am using both the Bioconductor adds on (Affy, AffyPLM,...) and the
'standard' R-package.
I am trying to select a list of genes which all have expression values below
a certain threshold.
I have done this by creating a vector which has 0s where the expression is
greater than the thres
Hi Gurus:
If I have a large dataset of the form of:
> x <- data.frame(V1 = runif(10), V2 = sample(c('A','B','C'),10,T)) > x
> V1 V21 0.2691580 A2 0.8711267 B3 0.2674728 C4 0.3278876 A5
> 0.1809152 A6 0.2499651 C7 0.9155174 A8 0.8004974 B9 0.7885516 A10
> 0.9301630
Can you get fitted values out of Statistica? If the 2 models are
equivalent, just using different encodings of the categorical variables,
then the fitted values will be the same (within roundoff error) even
though the coefficient estimates may differ.
So as a first step you should compare the fit
You can randomize the order (permute) of a single column with something
like:
> x <- cbind( x=1:10, y=101:110, z=201:210 )
> rownames(x) <- letters[1:10]
> new.x <- x
> new.x[,1] <- sample(new.x[,1])
> new.x
x y z
a 8 101 201
b 10 102 202
c 5 103 203
d 9 104 204
e 3 105 205
f 1 106 206
On Wed, Apr 09, 2008 at 04:09:57PM -0400, Vidhu Choudhary wrote:
> kegg<-read.table("c:/IDs.tab",header =TRUE,quote= "'", sep="\t") *
> *It doesnot read from Alzheimer's disease. Due the single qoutes in it.
You have almost given the answer yourself: you are declaring the single
quote to be a q
On Wed, Apr 09, 2008 at 11:08:28AM -0700, Grant Gillis wrote:
> I would like to randomize data within columns.
I think the following line does what you want (minus preserving the row
labels):
apply(data, 2, sample)
Maybe someone smarter than me knows how to preserve the row names. I
would proba
Grant Gillis wrote:
> Hello,
>
> I have what I suspect might be an easy problem but I am new to R and
> stumped. I have a data set that looks something like this
>
> b<-c(2,3,4,5,6,7,8,9)
> x<-c(2,3,4,5,6,7,8,9)
> y<-c(9,8,7,6,5,4,3,2)
> z<-c(9,8,7,6,1,2,3,4)
> data<-cbind(x,y,z)
> row.names(dat
Hi,
* I am using read.table command as follow
kegg<-read.table("c:/IDs.tab",header =TRUE,quote= "'", sep="\t") *
Fragment of file is as follow:
ID Pathway
04916Melanogenesis
04920Adipocytokine signaling pathway
04930Type II diabetes mellitus
04940Type I diabetes mellitus
0
> I have a problem concerning discrepances between R (which I use) and
> Statistica (which uses my supervisor).
---
can this be submitted as a fortune?
any other programs around that use supervisors, it's always good to
have a couple of
those around
ingmar
It is still not clear what is your response (y) variable and exactly
what is your predictor (x) variable(s).
If you have a separate vector of length 10 that is the response and you
want to regress it with each of the 8 vectors mentioned, then here is
one way to do that:
> p<-1:80+rnorm(80)
> dim(
On 4/9/2008 3:22 PM, Thomas Hoffmann wrote:
> Dear listmembers
>
> I would like to create an expression that looks like
>
> labl = expression(10^1,10^2,10^3,10^4,10^5)
>
> using a for-loop. However
>
> for (i in 1:5){ labl[i]=expression(10^i) }
>
> does not do the right thing. Does anybody kno
On Wed, 9 Apr 2008, Ravi Varadhan wrote:
> Hi,
>
>
>
> I have a data file, certain lines of which are character fields. I would
> like to skip these rows, and read the data file as a numeric data frame. I
> know that I can skip lines at the beginning with read.table and scan, but is
> there a wa
Right, that shows that the rgl configure script is not being run and no
linking is being done to any GL libraries.
I've never seen that. Is there something unusual about the filesystem so
it does not recognize rgl/configure is executable? (We've seen that on
SMB filesystems with the wrong mou
Thank you all for your help.
I successfully used Richard's suggestion (much like John's) and
continued with my work, neglecting to respond to the list.
Sorry about that.
Thanx again, DaveT.
>-Original Message-
>From: John Kane [mailto:[EMAIL PROTECTED]
>Sent: April 9, 2008 03:16 PM
>To: T
Dear listmembers
I would like to create an expression that looks like
labl = expression(10^1,10^2,10^3,10^4,10^5)
using a for-loop. However
for (i in 1:5){ labl[i]=expression(10^i) }
does not do the right thing. Does anybody knows help?
Thanks
Thomas
This may be a bit simple minded but why not change
those commmands into a single function something like
this and run it rather than the actual plot command?
myfunction <- function(a) {
plot(a, axes=FALSE)
axis(1, tcl=0.5)
axis(2, tcl=0.5)
axis(3, tcl=0.5, labels=FALSE)
axis(4, tcl=0.5, labels
A few comments,
My first impression on reading that abstract was that it was complete nonsense.
After thinking a bit about it and skimming the full article I decided that it
was nonsense, but nonsense that is important to research and discuss (and
therefore the paper is useful).
Why is it non
On Wed, 9 Apr 2008, Liviu Andronic wrote:
> Dear R users,
>
> This is a follow-up of a recent discussion on building rgl on Gentoo
> Linux. Please read bellow.
>
> On Tue, Mar 18, 2008 at 7:24 PM, Charles C. Berry <[EMAIL PROTECTED]> wrote:
>> Below substitute 'nvidia-drivers' or whatever you use
Here are a couple of ways using base graphics:
> library(TeachingDemos)
> x <- y <- seq(0,1,length.out=100)
> tmp <- expand.grid(x=x,y=y)
> z <- matrix( tmp$x + tmp$y, 100 )
>
> plot( 0:1, 0:1, type='n', xlab='x', ylab='y' )
> subplot( image(x,y,z, col=topo.colors(100), axes=FALSE, xlab='',
ylab='
Read the file in as lines of text (readLines), 'grep' through the
character vector and delete the lines you want and then use:
read.table(textConntection(yourvector))
to read the corrected data in.
On 4/9/08, Ravi Varadhan <[EMAIL PROTECTED]> wrote:
> Hi,
>
>
>
> I have a data file, certain line
Dear Duncan and Brian,
thank you for the quick replies. Please see bellow.
On Wed, Apr 9, 2008 at 6:11 PM, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> We'll need to see the output from install.packages("rgl") to have much idea:
>
localhost liviu # R CMD INSTALL "/home/liviu/inst/dwn_R/rgl_0.77
Hi,
I have a data file, certain lines of which are character fields. I would
like to skip these rows, and read the data file as a numeric data frame. I
know that I can skip lines at the beginning with read.table and scan, but is
there a way to skip a specified sequence of lines (e.g., 1, 2, 1
On Wed, Apr 9, 2008 at 10:12 AM, Thompson, David (MNR)
<[EMAIL PROTECTED]> wrote:
> Hello,
>
> How would I make the default behaviour of my plots produce output such
> as the following (i.e. tick marks inside on all axes, labels only on two
> (arbitrary?) sides) without needing the five addition
Hi all,
I'm trying to get a simple, linear decision surface from e1071's svm.
I've run it like this:
svm(as.factor(slow) ~ SLICE.3 + PSGR.7 + SOLUTIONS.6 + DR.10, y,
kernel='linear', cost=1e6, class.weights=c('FALSE'=1, 'TRUE'=10))
According to the docs, kernel='linear' has a kernel u'v. Sinc
Hi,
I have a problem concerning discrepances between R (which I use) and
Statistica (which uses my supervisor). I can't say what is the origin
of these differences but unfortunately my supervisor doesn't know that
either.
Our response variable is number (or presence/absence) of parasites in
ro
Hello,
I have what I suspect might be an easy problem but I am new to R and
stumped. I have a data set that looks something like this
b<-c(2,3,4,5,6,7,8,9)
x<-c(2,3,4,5,6,7,8,9)
y<-c(9,8,7,6,5,4,3,2)
z<-c(9,8,7,6,1,2,3,4)
data<-cbind(x,y,z)
row.names(data)<-c('a','b','c','d','e','f','g','h')
wh
Hello Christoph,
Oops, I see that the factor.levels function is in the help menu for
strip.default. I had tried it, but not specified it correctly.
Thanks again for you help,
John
Christoph Meyer wrote:
> Hi John,
>
> a solution to your problem would be using strip.custom like this:
>
> ...
On 09.04.2008, at 17:46, Shubha Vishwanath Karanth wrote:
> To put it simple,
>
> C=c("My Dog", "Its really good", "Beautiful")
>
> Now,
> SOMEFUNCTION(C) should give: c("My", "Its really", "")
SOMEFUNCTION <- function(x) gsub(" *\\w+$", "", x)
But be aware that this won't work for instance for
Hi,
I would like to compare the hazard functions of two samples using the
Cox proportional hazards model. For sample 1 I have individual time-to-
event data. For sample 2 I don't have individual data, but grouped
data that allows to obtain a hazard function.
I am wondering if there is an R f
Hi John,
a solution to your problem would be using strip.custom like this:
...
with(dat,xYplot(Cbind(ycomb, y.up, y.low)~x1|factor(grp),
data=dat,type="l", method="bands",
scales=list(y=list(relation="free"),x=list(alternating=c(1,1,1))),
ylim=list(c(0,1200),c(0,1)),
strip=strip.custom(fact
We'll need to see the output from install.packages("rgl") to have much
idea:
R CMD ldd /usr/lib/R/library/rgl/libs/rgl.so
would also be informative.
On Wed, 9 Apr 2008, Duncan Murdoch wrote:
> On 4/9/2008 10:53 AM, Liviu Andronic wrote:
>> Dear R users,
>>
>> This is a follow-up of a recent di
Actually the last example was all but the first word,
not all but the last word. All but the last word is:
> sapply(strapply(C, "\\w+"), head, -1)
[[1]]
[1] "My"
[[2]]
[1] "Its""really"
[[3]]
character(0)
On Wed, Apr 9, 2008 at 11:56 AM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> See
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
See strapply in the gsubfn package and the gsubfn home page at
gsubfn.googlecode.com.
Here we extract the words, the last word and all but the last word:
> library(gsubfn)
> strapply(C, "\\w+")
[[1]]
[1] "My" "Dog"
[[2]]
[1] "Its""really" "good"
[[3]]
[1] "Beautiful"
> sapply(strapply(C, "
On 9 Apr 2008, at 17:44, Shubha Vishwanath Karanth wrote:
> Got all the answers using ?strsplit... Is there any way without
> using string split?... More specifically... How can I just extract
> the last word in all the strings without using ?strsplit ?
Oops, sorry.
gsub(" *\w+$", "", C)
sh
To put it simple,
C=c("My Dog", "Its really good", "Beautiful")
Now,
SOMEFUNCTION(C) should give: c("My", "Its really", "")
Thanks,
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-
I would like to apply a simple function, like
is.matrix
to more than one data.frame
How can I call on more than one data.frame? (are there any wildcards, etc?)
I am a true beginner and have tried to look this up in help files, but
cannot figure it out.
Thank you.
[[alternative HTML versi
Something like that?
gsub(" {1,}\w+$", "", C)
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and provide commented, minimal, self-contained,
Got all the answers using ?strsplit... Is there any way without using string
split?... More specifically... How can I just extract the last word in all the
strings without using ?strsplit ?
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London *
length(unlist(strsplit(C, ' ')))
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Shubha
> Vishwanath Karanth
> Sent: Wednesday, April 09, 2008 11:21 AM
> To: [EMAIL PROTECTED]
> Subject: [R] Number of words in a string
>
> Hi R,
>
>
>
> A qui
Exactly...this is what I wanted... Can we also extract/remove the last word of
the strings?
Example:
> C=c("My Dog", "Its really good", "Beautiful")
> sapply(strsplit(C, " "), length)
[1] 2 3 1
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * Lond
On 9 Apr 2008, at 17:29, Markus Gesmann wrote:
> Would this:
>
> sapply(strsplit(C, " "), length)
>
> work for?
or
length(unlist(strsplit(C, " ")))
--Hans
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PLEASE do rea
To put in more general,
C=c("My Dog", "Its really good", "Beautiful")
And the resultant output should be
[1] 2 3 1
...the number of words
Thank you...
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
> How would I make the default behaviour of my plots produce output such
> as the following (i.e. tick marks inside on all axes, labels only on two
> (arbitrary?) sides) without needing the five additional commands each
> time?
>
> plot(1:10, axes=FALSE)
> axis(1, tcl=0.5)
> axis(2, tcl=0.5)
> axi
Would this:
sapply(strsplit(C, " "), length)
work for?
Markus Gesmann │Associate Director│Libero Ventures Ltd, One Broadgate, London
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Asking the question usually makes one think about the workarounds J.
I reviewed the pscl source and found the code for pR2. Sourcing the
following internal functions allowed me to compute the pseudo R2 measures.
pR2Work <- function(llh,llhNull,n){
McFadden <- 1 - llh/llhNu
Hi All,
I tried to install AnnotationDBI
like so:
source("http://bioconductor.org/biocLite.R";)
biocLite("AnnotationDbi")
and got this error:
Loading required package: RSQLite
Error in dyn.load(file, ...) :
unable to load shared library '/RHEL3/local/lib64/R/library/
RSQLite/libs/RSQLite
Dear all,
I've been trying to find a package to simulate a proportional hazard
mixture cure model. I've had a look at the nltm package and the
references seem to include a PHMCM method but the package doesn't seem
to. I've also had a look at the package semicure (as per this thread
http://t
Hi R,
A quick question: How do we find the number of words in a string?
Example:
C="Have a nice day"
And the number of words should be 4. any built in function or?...
Thanks, Shubha
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo *
This is in regards to the suggested use of type="lpmatrix" in the
documentation for mgcv::predict.gam. Could one not get the same result more
simply by using type="terms" and interpolating each term directly? What is
the advantage of the lpmatrix approach for prediction outside R? Thanks.
--
View
On 4/9/2008 10:53 AM, Liviu Andronic wrote:
> Dear R users,
>
> This is a follow-up of a recent discussion on building rgl on Gentoo
> Linux. Please read bellow.
>
> On Tue, Mar 18, 2008 at 7:24 PM, Charles C. Berry <[EMAIL PROTECTED]> wrote:
>> Below substitute 'nvidia-drivers' or whatever you
Hello,
How would I make the default behaviour of my plots produce output such
as the following (i.e. tick marks inside on all axes, labels only on two
(arbitrary?) sides) without needing the five additional commands each
time?
plot(1:10, axes=FALSE)
axis(1, tcl=0.5)
axis(2, tcl=0.5)
axis(3, tcl=0
Here's a simplified example
p<-1:80+rnorm(80)
dim(p)<-c(2,2,2,10)
We could say that the 4-d array p consists of 2*2*2 = 8 vectors of length
10. So what I'm asking for is a fast way to perform a linear fit to all
those vectors.
I'm sorry if I'm causing you to have a headache with all those dimens
Dear R users,
This is a follow-up of a recent discussion on building rgl on Gentoo
Linux. Please read bellow.
On Tue, Mar 18, 2008 at 7:24 PM, Charles C. Berry <[EMAIL PROTECTED]> wrote:
> Below substitute 'nvidia-drivers' or whatever you use for
>
>
> emerge -D mesa
> revdep-r
Thanks in advance for your kind attention.
I am using R to fit empirical data to generalized linear models. AIC (Akaike
information criterion) is a measure of the goodness of fit returned by calls
to glm(). I would also like to calculate the coefficient of determination
R2, altho
Thanks for all the reply. I solved the task using apply(as suggested by
Hans).
The tips on S-Poetry and ?Logic are very handy as well.
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Well, could you provide a little bit more information regarding what
you are trying to do (e.g., reproducible example).
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/
Thank you all very much for replying. Of course you are absolutely right
but unfortunately I really deal with the case of a 4-d matrix so what you
said does not apply. I should have specified but being a new R user I
hadn't realized the difference between a matrix and an array.
So please tell me i
"tom soyer" <[EMAIL PROTECTED]> wrote:
> I have a 2D chart that is divided into four quadrants, I, II, III, IV:
>
> plot(1:10,ylim=c(0,10),xlim=c(0,10),type="n")
> abline(v=5,h=5)
> text(x=c(7.5,7.5,2.5,2.5),y=c(2.5,7.5,7.5,2.5),labels=c("I","II","III","IV"))
> I would like to fill each quadrant
> I have a matrix with data and a column indicating whether it is censored
> or not. Is there a way to apply weibull and exponential maximum
> likelihood estimation directly on the censored data, like in the paper:
> Backtesting Value-at-Risk: A Duration-Based Approach, P Chrisoffersen
> and D Pel
Hi
it is a work for do.call
try
Nc<-do.call(c,N)
xyplot(Nc ~
t,groups=i,type="o",key=simpleKey(as.character(1:3),x=0,y=1,lines=T))
Regards
Petr
[EMAIL PROTECTED]
[EMAIL PROTECTED] napsal dne 09.04.2008 14:53:41:
> I would like to graph cumulative counts
> and use tapply(...,cumsum) to get
>
Dear R-users,
An updated version of the energy package, energy 1.1-0, is now available on
CRAN.
This version has merged the dcov package (previously available from my
personal web page) into energy.
New functions include:
dcov (distance covariance)
dcor (distance correlation)
DCOR (four stati
I would like to graph cumulative counts
and use tapply(...,cumsum) to get
the cumulative counts.
For instance:
library(lattice)
t<-rep(1:4,each=3)#time
i<-rep(1:3,times=4) #categories
n<-rpois(length(t),t) #count
xyplot(n ~ t,groups=i,
type="o",key=simpleKey(as.charact
ravi wrote:
>
> d3<-merge(d1,d2,by.x=time1,by.y=time2)
> will work only for exact matching. One possible option is to match
> the times for the date and hour times only (by filtering away the
> minute data). But this is only a partial solution as I am not
> interested in data where the time di
Hello again,
I have been trying to add an expression to the strips in xYplot to no
avail. For example, in the code below, the text in the strips for each
panel is "Anls" and "Plts". However, I would like it to add "Anls km^2"
and "Plants km^2" with the exponents raised.
I tried resetting the
Hi Spencer,
Sorry for not producing code as a worked example.
Here's an example:
==
# setting the seed number
set.seed(0)
# creating a correlation matrix
corr <- diag(5)
corr[lower.tri(corr)] <- 0.5
corr[upper.tri(corr)] <- 0.5
# Data for the minimisation
mat <-
On 4/9/2008 6:55 AM, Costas Douvis wrote:
> Hi all,
>
> My question is not really urgent. I can write a loop and solve the
> problem. But I know that I'll be in a similar situation many more times so
> it would be useful to find out the answer
>
> Is there a fast way to perform linear fit to all
If you have the same design matrix then you can specify a matrix of
responses in lm(), e.g.,
Y <- matrix(rnorm(100*10), 100, 10)
x <- rnorm(100)
fit <- lm(Y ~ x)
fit
summary(fit)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic U
Here is one way of doing it. Create a new dataframe with the values,
look for the breaks in times between the two sequences and then create
a results dataframe:
> # convert time to POSIXct
> d1$time1 <- as.POSIXct(paste(d1[[1]], d1[[2]]))
> d2$time2 <- as.POSIXct(paste(d2[[1]], d2[[2]]))
> d1$ID1
Hi all,
My question is not really urgent. I can write a loop and solve the
problem. But I know that I'll be in a similar situation many more times so
it would be useful to find out the answer
Is there a fast way to perform linear fit to all the columns of a matrix?
(or in the one dimension of a m
Ng Stanley gmail.com> writes:
>
> Hi,
>
> Are there any methods for computing the correlation between two
> multi-dimensional matrices ? Will transforming the matrices into vectors and
> applying pearson be fine ? Any blind spots that I should be aware ?
>
If they were something like multi-
Congratulation Bill for this very clear and useful explanation.
Heinz
At 14:58 08.04.2008, [EMAIL PROTECTED] wrote:
>'mode' is a mutually exclusive classification of objects according to
>their basic structure. The 'atomic' modes are numeric, complex,
>charcter and logical. Recursive objects ha
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