The problem is that: There is some rounding problems, for example
Library(chron)
any(times("4:00:01")==times("4:00:00")+times("00:00:01")))
False
But,it should be true
ggrothendieck-2 wrote:
>
> On Thu, Nov 6, 2008 at 12:10 PM, tedzzx <[EMAIL PROTECTED]> wrote:
>>
>> Hi,all
>>
>> I only got the
Dear All,
Encoding() goes beyond my understanding. See the
example. I would expect from reading the help for
Encoding() that strsplit preserves the encoding
for each resulting element, but for simple letters it gets lost.
Also it seems that an Encoding() cannot be
declared for simple letters.
Hi, before aplied, the function likfit, for one geodata class:
or.geo.lf<-likfit(or.geo,cov.model="matern",ini.cov.pars=c(2.5,1),kappa=3,fix.kappa=FALSE,nugget=1.2,lambda=0.019,fix.lambda=FALSE,hessian=TRUE)
got this with a summary
> summary(or.geo.lf)
Summary of the parameter estimation
-
Thank you two of you very much for your help,
I found the way to solve my problem.
Best wishes,
Tuan.
On Thu, Nov 6, 2008 at 4:14 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
>
> On 7/11/2008, at 9:26 AM, Hoang Trong Minh Tuan wrote:
>
> Hi all,
>> So far I only know one way to get the confide
Are you dealing with some rainfall data for your schoolwork?
Few days ago, somebody asked exactly the same question.
read this one to know how to calculate the H value (An Alternative
Formula for the Calculation of H)
http://faculty.vassar.edu/lowry/ch14a.html
You can calculate the rank by rank()
G'day Cruz,
On Fri, 7 Nov 2008 09:47:47 +0800
cruz <[EMAIL PROTECTED]> wrote:
> Hi,
>
> How do we get the value of a chi square as we usually look up on the
> table on our text book?
>
> i.e. Chi-square(0.01, df=8), the text book table gives 20.090
>
> > dchisq(0.01, df=8)
> [1] 1.036471e-08
>
> qchisq(0.01, df=8, lower.tail=FALSE)
[1] 20.09024
>
See ?dchisq
On Fri, 2008-11-07 at 09:47 +0800, cruz wrote:
> Hi,
>
> How do we get the value of a chi square as we usually look up on the
> table on our text book?
>
> i.e. Chi-square(0.01, df=8), the text book table gives 20.090
>
> > dc
> qchisq(0.01, df = 8, lower.tail = FALSE)
[1] 20.09024
cruz wrote:
Hi,
How do we get the value of a chi square as we usually look up on the
table on our text book?
i.e. Chi-square(0.01, df=8), the text book table gives 20.090
dchisq(0.01, df=8)
[1] 1.036471e-08
pchisq(0.01, df=8)
[1] 2.
On Fri, 7 Nov 2008, cruz wrote:
Hi,
How do we get the value of a chi square as we usually look up on the
table on our text book?
i.e. Chi-square(0.01, df=8), the text book table gives 20.090
dchisq(0.01, df=8)
[1] 1.036471e-08
pchisq(0.01, df=8)
[1] 2.593772e-11
qchisq(0.01, df=8)
[1] 1
Hi,
How do we get the value of a chi square as we usually look up on the
table on our text book?
i.e. Chi-square(0.01, df=8), the text book table gives 20.090
> dchisq(0.01, df=8)
[1] 1.036471e-08
> pchisq(0.01, df=8)
[1] 2.593772e-11
> qchisq(0.01, df=8)
[1] 1.646497
>
nono of them give me 20.
I am an R beginner and trying to run a market model using supply and
demand in R framework.
First, I conducted cointegration test. The results showed that there
were two ranks.
Now I need to test weak exogenity for each series of the seven
variables(my data have 7 variables).
Then I ne
Dear R Gurus:
How do you put together a 2^2 (or even 2^k) factorial problem, please?
Since you have 2 levels for A and B, do you put in "A+" and "A-" as
factors, please?
Thanks,
Edna Bell
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailm
Hi,
i have a question in R,
How and what command you need to do to run a kruskal-wallis test without the
built in command 'kruskal.test'?
many thanks.
_
[[alternative HTML version deleted]]
Hi,
Zitat von "Charles C. Berry" <[EMAIL PROTECTED]>:
> On Thu, 6 Nov 2008, Oliver Bandel wrote:
>
> > Hello Charles,
> >
>
> [snip]
>
> >> dim( as.matrix(replicate(10, sample(x, 3) ) ) )
> > [1] 3 10
> >> dim( as.matrix(replicate(10, sample(x, 2) ) ) )
> > [1] 2 10
> >> dim( as.matri
Dear Prof.Ripley!
Thank you very much for your attention. In the
given example Encoding(), or the encoding
parameter of read.csv solve the problem. I hope
your patch will solve also the problem, when I
read a spss file by spss.get(), since this
function has no encoding parameter and my real
Boy are you confused. This has nothing at all to do with substitution.
Instead do
test <- with(fc,ave.fc[match(diff_mirs_list,Probe)])
cheers,
Rolf Turner
On 7/11/2008, at 11:46 AM, Iain Gallagher wrote:
Hello list.
I have a vector of values:
eg
head(di
Thanks, I got to do it.,
2008/11/6 Pedro Mardones <[EMAIL PROTECTED]>
> try prcomp instead of princomp
>
> On Thu, Nov 6, 2008 at 3:20 PM, Lucke, Joseph F
> <[EMAIL PROTECTED]> wrote:
> > Neola
> > I'm a bit rusty on this, but I believe you can conduct on singular-value
> decomposition on the 4
Look at Encoding() on your two strings. The results are different, and
this seems to be the root of the problem. Adding encoding="latin1" to the
read.csv call is a workaround.
It looks like there is a problem in the use of the CHARSXP cache: if I
save the session then x0 == x becomes true wh
Hello list.
I have a vector of values:
eg
> head(diff_mirs_list)
[1] "hsa-miR-26b" "hsa-miR-26b" "hsa-miR-23a" "hsa-miR-27b" "hsa-miR-29a"
[6] "hsa-miR-29b"
and I would like to conditionally replace each value in this vector with a
number defined in a dataframe:
> fc
Probe ave.fc
try prcomp instead of princomp
On Thu, Nov 6, 2008 at 3:20 PM, Lucke, Joseph F
<[EMAIL PROTECTED]> wrote:
> Neola
> I'm a bit rusty on this, but I believe you can conduct on singular-value
> decomposition on the 436 by 518 matrix. The squares of your singular values
> (max of 436, 518-436 will
Hi All -
I'm running a faily long script that uses rpvm & snowFT to spawn off
multiple processes with the 'clusterApplyFT' function.
Specifically, what happens is that the head node generates a number of seed
clusters that are then spawned off to the pvm cluster (in this case, nodes
on a 4 dual-c
This is more of a statistical question. Let's say I have two numbers. One
is a lower bound and the other is a point estimate to the right of the lower
bound. Now, let's say I want to be able to estimate the mean and standard
deviation of a lognormal distribution, where 95% of the density falls w
Lu, Jiang wrote:
Thank you very much, Frank. I installed Design package and tried
survplot().
#R code
survplot(testfit,time.inc=365.25,xaxt='n',xlim=c(0,1826.25),ylim=c(0,1),conf='none',
fun=function(y)1-y,label.curves=list(keys=c('Med','Rev')),
abbrev.label=TRUE,n.risk=T
Hello,
I'm still a newbie user and struggling to automate some analyses from
SigmaPlot using R. R is a great help for me so far!
But the following problem makes me go nuts.
I have two spectra, both have to be fitted to reference data. Problem: the
both spectra are connected in some way: the stoi
On 7/11/2008, at 9:26 AM, Hoang Trong Minh Tuan wrote:
Hi all,
So far I only know one way to get the confidence limit for the
Poisson
distribution is to use the look-up table given by the 2 parameter (the
number of observation x and the confidence level, e.g. 95%) and the
table is
limit
On Thu, 6 Nov 2008, Hoang Trong Minh Tuan wrote:
Hi all,
So far I only know one way to get the confidence limit for the Poisson
distribution is to use the look-up table given by the 2 parameter (the
number of observation x and the confidence level, e.g. 95%) and the table is
limit by the maximu
Dear All!
Reading character strings containing an "umlaut"
from a csv-file I find a (to me) surprising
behaviour in R 2.8.0, that I did not notice in R 2.7.2.
A comparison by "==" results in FALSE, while grep does find the aggreement.
See the example below.
The crucial line is x=="div 1-2 Verä
Thank you very much, Frank. I installed Design package and tried survplot().
#R code
survplot(testfit,time.inc=365.25,xaxt='n',xlim=c(0,1826.25),ylim=c(0,1),conf='none',
fun=function(y)1-y,label.curves=list(keys=c('Med','Rev')),
abbrev.label=TRUE,n.risk=TRUE)
# End of R code
On Thu, 6 Nov 2008, Oliver Bandel wrote:
Hello Charles,
[snip]
dim( as.matrix(replicate(10, sample(x, 3) ) ) )
[1] 3 10
dim( as.matrix(replicate(10, sample(x, 2) ) ) )
[1] 2 10
dim( as.matrix(replicate(10, sample(x, 1) ) ) )
[1] 10 1
=
So, the
Hi all,
So far I only know one way to get the confidence limit for the Poisson
distribution is to use the look-up table given by the 2 parameter (the
number of observation x and the confidence level, e.g. 95%) and the table is
limit by the maximum number of observations (x <= 50).
I know the fo
Neola
I'm a bit rusty on this, but I believe you can conduct on singular-value
decomposition on the 436 by 518 matrix. The squares of your singular values
(max of 436, 518-436 will be zero) will be your eigenvalues, the same as in the
PC analysis. The post-eigenvectors will be your components.
Look at the rotate.wireframe function in the TeachingDemos package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> project.org] On Behalf Of Bill Sz
Hi all,
I'm noting the appearance of new versions of plotrix and prettyR as I
found a bug in the "brkdn" function that messed up the order of
"value.labels" if they had been imported from an SPSS data file. For
anyone using "brkdn", please upgrade to the new version of prettyR
(1.3-5) if you a
cruz wrote:
Thanks for all the responses, they are all very helpful:)
you don't need to assign dimension or classes to your objects.
It's easier if you do like this
this is something that really bothers me, when I need to define an
object which i will later fill with data, the dimens
Hi,
On Thu, Nov 6, 2008 at 2:12 AM, Yuri Volchik <[EMAIL PROTECTED]> wrote:
>
> Thanks for reply Henrik, seems obvious now.
> Can child class (B) access argument of the parent class, i.e. can i rewrite
> definition of the class B as
>
> setConstructorS3("ClassB", function() {
> extend(ClassA(), "
My matrix have 436 registers and 518 variables. I need to do a PCA analyst.
Usually I use princomp command to perform PCA analyst, but this time i can't
because of my variables are more than my registers.
2008/11/6 stephen sefick <[EMAIL PROTECTED]>
> would you please provide a dummy example tha
Hi Christabel,
Take a look at the function basicStats in the fBasics package. Here is an
example:
library(fBasics)
set.seed(123)
x=rnorm(20,24,2)
basicStats(x)
#x
#nobs 20.00
#NAs 0.00
#Minimum 20.066766
#Maximum 27.573826
#1. Quartile 23.012892
#3. Quartile 2
Hi Noela,
Take a loot at ?prcomp
HTH,
Jorge
On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez <[EMAIL PROTECTED]> wrote:
> I need perform PCA analyst with a matrix with more variables than units.
>
> The princomp command don't match with this matrix.
>
> Anybody knows a good command to do it?
>
?princomp refers you to prcomp for that case:
'princomp' only handles so-called R-mode PCA, that is feature
extraction of variables. If a data matrix is supplied (possibly
via a formula) it is required that there are at least as many
units as variables. For Q-mode PCA use 'p
would you please provide a dummy example that explains your problem.
Then maybe I can help you.
thanks
Stephen
On Thu, Nov 6, 2008 at 1:42 PM, Noela Sánchez <[EMAIL PROTECTED]> wrote:
> I need perform PCA analyst with a matrix with more variables than units.
>
> The princomp command don't match w
I need perform PCA analyst with a matrix with more variables than units.
The princomp command don't match with this matrix.
Anybody knows a good command to do it?
--
Noela
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Does this help
(mylist <- list(NULL))
(mylist[[3]] <- data.frame(a=1:4, b=letters[1:4]))
mylist
(mylist[[2]] <- matrix(1:12, nrow=4))
mylist
--- On Thu, 11/6/08, cruz <[EMAIL PROTECTED]> wrote:
> From: cruz <[EMAIL PROTECTED]>
> Subject: Re: [R] How to avoid "$ operator is invalid for atomi
cruz wrote on 11/06/2008 12:16 PM:
Thanks for all the responses, they are all very helpful:)
you don't need to assign dimension or classes to your objects.
It's easier if you do like this
this is something that really bothers me, when I need to define an
object which i will later fill with d
>
> Does that answer your question?
>
Thanks:)
I received one from Erin:
x <- NULL
__
R-help@r-project.org 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 provi
Lu, Jiang wrote:
Dear R helper,
I encountered a problem when I tried to plot the cumulative failure rate
(i.e. 1 - survival probability). I have used the following code to plot. The
scenario is that patients are randomized to different treatment arm (rev in
the code), the PCI revascularization w
?mean
kurtosis
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/110186.html
--- On Thu, 11/6/08, christabel_jane prudencio <[EMAIL PROTECTED]> wrote:
> From: christabel_jane prudencio <[EMAIL PROTECTED]>
> Subject: [R] mean computation for external data
> To: r-help@r-project.org
> Received: T
Thanks for all the responses, they are all very helpful:)
> you don't need to assign dimension or classes to your objects.
> It's easier if you do like this
this is something that really bothers me, when I need to define an
object which i will later fill with data, the dimension of this object
s
Hi Solveig,
Thanks very much, it's a nice solution and one I had not even begun to think
about. I MUST learn more about dates!
--- On Thu, 11/6/08, Solveig Mimler <[EMAIL PROTECTED]> wrote:
> From: Solveig Mimler <[EMAIL PROTECTED]>
> Subject: Re: Simple rep() question duplicating times and
Does this help?
library(chron)
tms<-c("19:30:23","18:39:10")
mytimes <- times(tms)
mytimes[1]-mytimes[2]
--- On Thu, 11/6/08, tedzzx <[EMAIL PROTECTED]> wrote:
> From: tedzzx <[EMAIL PROTECTED]>
> Subject: [R] How to manipulate the time data without the date?
> To: r-help@r-project.org
> Recei
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of anna freni
> sterrantino
> Sent: Thursday, November 06, 2008 10:00 AM
> To: cruz; r-help@r-project.org
> Subject: Re: [R] How to avoid "$ operator is invalid for
> atomic vectors"
>
> Hi Cruz
> you d
Dear R helper,
I encountered a problem when I tried to plot the cumulative failure rate
(i.e. 1 - survival probability). I have used the following code to plot. The
scenario is that patients are randomized to different treatment arm (rev in
the code), the PCI revascularization was monitored over 5
Hi Cruz
you don't need to assign dimension or classes to your objects.
It's easier if you do like this
> a=c(0,1,2,4,1,1)
> length(a)
[1] 6
> b=matrix(a,3,2,byrow=T)
> b
[,1] [,2]
[1,]01
[2,]24
[3,]11
of course you can change the colnames and assign what
you prefer
On Thu, Nov 6, 2008 at 12:10 PM, tedzzx <[EMAIL PROTECTED]> wrote:
>
> Hi,all
>
> I only got the time data such as:
> tms<-c("19:30:23","18:39:10".)
>
> I want to manipulate this time series data. For example, plus one second(or
> minute) or minus one second
>
> This data only has the time(h:m:
Why not give it an arbitrary date such 2008-01-01?
On Thu, 6 Nov 2008, tedzzx wrote:
Hi,all
I only got the time data such as:
tms<-c("19:30:23","18:39:10".)
I want to manipulate this time series data. For example, plus one second(or
minute) or minus one second
This data only has the tim
Thought I should copy the list with Matthieu's response.
H
-Original Message-
From: Matthieu Stigler [mailto:[EMAIL PROTECTED]
Sent: Wednesday, November 05, 2008 8:29 PM
To: Horace Tso; [EMAIL PROTECTED]
Subject: Re: [rkward-devel] questions on RKWard
some answer only for the third ques
Hi,all
I only got the time data such as:
tms<-c("19:30:23","18:39:10".)
I want to manipulate this time series data. For example, plus one second(or
minute) or minus one second
This data only has the time(h:m:s), without the date. I know that there are
chron package, ISOPix class and the tim
Try this:
library(alr3)
library(gsubfn)
example(delta.method) # defines m1
num <- 2
fn$delta.method(m1, "-b1/($num*b2)")
See gsubfn home page at:
http://gsubfn.googlecode.com
On Thu, Nov 6, 2008 at 12:13 PM, Christoph Scherber
<[EMAIL PROTECTED]> wrote:
> Dear all, Dear Keith,
>
> Well, of cou
I sent this message to the maintainer's email address listed on the
signal package, but it bounced. Perhaps somebody on this list has more
insight into the signal package than I do (or knows the maintainer's new
address):
Subject:
question about buttord function in R signal module
From:
Micha
> I am trying to use a background photo in a lattice plot. I
> am using the rimage and TeachingDemos packages to plot the
> photo and translate from the photo coordinates in pixels to
> geographic coordinates, which is what I want to use for
> plotting contours, lines, etc. The (unrunable) co
Dear all, Dear Keith,
Well, of course in fact the problem is more complicated than that. The example was just for
illustration.
I have several statistical models for which I want to retrieve predictions using delta.method (from
library(alr3)).
Now for this I need a character string such as
Hello Charles,
thank you for the hint.
Zitat von "Charles C. Berry" <[EMAIL PROTECTED]>:
[...]
> > This looks good (and correct to me).
>
> Look again .
>
> It is not the same as what you have above.
[...]
OK, yes, you are right!
I mixed the two parameters...
Now I get the same problem als
Hello -
cruz wrote:
Hi,
I am writing this in a wrong way, can someone please correct me?
A <- matrix()
length(A) <- 6
dim(A) <- c(3,2)
colnames(A) <- c("X","Y")
A
X Y
[1,] NA NA
[2,] NA NA
[3,] NA NA
A$X
Error in A$X : $ operator is invalid for atomic vectors
A[, "X"] may be what
On 06/11/2008 11:18 AM, BKMooney wrote:
I am new to R and am running into trouble with the function plot.
When I enter in the simple code:
x<-1:4
y<-5:8
plot(x,y)
I get a scatter plot with 4 points as expected.
However, with my own data, A and B are both vectors of length ~85, each
entry
Hi,
Firstly, I'd recommend using '<-' for assignment, rather than '='; it saves
confusion
Secondly, I don't think you want 'a*x+b' as a formula, I think you want an
expression.
Thirdly, your 'y' has only one term, a 9 character constant = "a * x + b"
Consider instead,
y <- expression(a*x+b)
a
Hello -
In you example, what are the classes of x and y?
x<-1:4
y<-5:8
plot(x,y)
class(x)
class(y)
In your 'real' data, what are the classes of A and B
class(A)
class(B)
One may be a factor?
How are you reading your data into R, read.table? Make sure your data
are numeric, then plot th
On Thu, 6 Nov 2008, Christoph Scherber wrote:
Dear all,
How can I replace text in objects that are of class "formula"?
y="a * x + b"
class(y)="formula"
grep("x",y)
y[1]
What exactly are you trying to accomplish??
And why did you assign 'formula' as the class of a character string?
'y' is
You need to provide more information. PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Show at least an 'str' of your data if you can not include it and the
commands that you were using.
On Thu, Nov 6,
Hi,
I am writing this in a wrong way, can someone please correct me?
> A <- matrix()
> length(A) <- 6
> dim(A) <- c(3,2)
> colnames(A) <- c("X","Y")
> A
X Y
[1,] NA NA
[2,] NA NA
[3,] NA NA
> A$X
Error in A$X : $ operator is invalid for atomic vectors
>
Thanks,
cruz
_
Assuming you mean data frame and not matrix and
depending on whether the empty spaces are intended
to represent NA or 0 we have:
> DF <- data.frame(V1 = LETTERS[1:3], V2 = LETTERS[24:26], V3 = 1:3)
> tapply(DF[[3]], DF[1:2], c)
V2
V1 X Y Z
A 1 NA NA
B NA 2 NA
C NA NA 3
> xtabs(V3 ~
Hi Phillippe,
Thanks for the pointer. It looks like a nice resource.
Hadley
On Thu, Nov 6, 2008 at 9:34 AM, Philippe Grosjean
<[EMAIL PROTECTED]> wrote:
> Hello Hadley,
>
> I have started this:
> http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:export.
>
> One solution that works
I am new to R and am running into trouble with the function plot.
When I enter in the simple code:
x<-1:4
y<-5:8
plot(x,y)
I get a scatter plot with 4 points as expected.
However, with my own data, A and B are both vectors of length ~85, each
entry a decimal in [0,1].
Using the same plo
On Thu, Nov 6, 2008 at 10:05 AM, Max Kuhn <[EMAIL PROTECTED]> wrote:
>> * svg: R output devices still experimental
>
> I've been using the svg device in the Cairo package for a while now.
> I've never had any issues with it and wouldn't characterize it as
> experimental (of course, others may have
On Thu, 6 Nov 2008, Oliver Bandel wrote:
Hello,
for a simulation I tried the following:
=
sampmeanvec <- function (from, n, repititions)
{
print( paste("samplesize n:", n, "repititions:", repititions) )
samples.mat <- as.matri
> * svg: R output devices still experimental
I've been using the svg device in the Cairo package for a while now.
I've never had any issues with it and wouldn't characterize it as
experimental (of course, others may have had issues).
I have had problems generating svg using some of the non-Cairo
The R wiki also discusses this:
http://wiki.r-project.org/rwiki/doku.php?id=tips:data-frames:sort&s=sort
On Wed, Nov 5, 2008 at 1:24 PM, Richardson, Patrick
<[EMAIL PROTECTED]> wrote:
> http://www.ats.ucla.edu/stat/R/faq/sort.htm
>
> A great tutorial about sorting data in R.
>
> HTH,
>
>
Try the 'rgl' package.
On Thu, Nov 6, 2008 at 10:38 AM, Bill Szkotnicki <[EMAIL PROTECTED]> wrote:
> I've been using lattice/wireframe succesfully to visualize some data.
> I have one question.
> I want to be able to change the viewpoint ( i.e. rotate the plotted figure a
> bit left or right or up
Am Donnerstag, den 06.11.2008, 10:38 -0500 schrieb Bill Szkotnicki:
> I've been using lattice/wireframe succesfully to visualize some data.
> I have one question.
> I want to be able to change the viewpoint ( i.e. rotate the plotted
> figure a bit left or right or up and down )
> Is there a way to
you just need pmax() or pmin(), e.g., check this:
set.seed(123)
M1 <- matrix(rnorm(20), 4, 5)
M2 <- matrix(rnorm(20), 4, 5)
M3 <- matrix(rnorm(20), 4, 5)
M1; M2; M3
pmax(M1, M2, M3)
pmin(M1, M2, M3)
I hope it helps.
Best,
Dimitris
Diogo André Alagador wrote:
Dear all,
I have 3 matrices w
#how about this
A <- rnorm(10)
B <- rnorm(10)
C <- rnorm(10)
D <- data.frame(A,B,C)
apply(D, MARGIN=1, FUN=min)
On Thu, Nov 6, 2008 at 10:00 AM, Diogo André Alagador
<[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I have 3 matrices with the same dimension, A,B,C and I would like to produce
> a matrix
Dear all,
I have 3 matrices with the same dimension, A,B,C and I would like to produce
a matrix D where in each position would retrieve the max(or min) value along
A,B,C taken from the same position.
I guess that apply functions should fit, but for matrices objects I am not
getting it.
thanks i
Dear all,
How can I replace text in objects that are of class "formula"?
y="a * x + b"
class(y)="formula"
grep("x",y)
y[1]
Suppose I would like to replace the "x" by "w" in the formula object "y".
How can this be done? Somehow, the methods that can be used in character objects do not work 1:1
Hi Ramya,
Sorry if I missed something, but unless I have problems reading your
message it seems that there is no line in your example matching both
your test conditions...
Hope this helps. Olivier
Hi there,
I have a dataframe length.unique.info
> length.unique.info
abc 12 345
def
I've been using lattice/wireframe succesfully to visualize some data.
I have one question.
I want to be able to change the viewpoint ( i.e. rotate the plotted
figure a bit left or right or up and down )
Is there a way to do that?
Or is there some other package around that could help?
Thanks.
__
Hello Hadley,
I have started this:
http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:export.
One solution that works not too bad in OpenOffice is to output the graph
in XFig format, and then use fig2dev from transfig to get an EMF file.
That one is rather well readable by OpenOf
Hi all,
I'm trying to write up some recommendations for what graphics formats
are most useful for inclusion into ms office and openoffice. There
have been a few discussions on the list in the past, but I haven't
seen a summary. These are the options I've seen so far, along with
there costs and b
Dear Prof Ripley,
Am I correct if I use the following code to get c.i.´s for predicted values of
the nls fit:
puro1<-nls(rate~a*conc/(b+conc), data=Puromycin[1:12,], start=list(a=200, b=1))
#set up nls model
# assume only one predicted value is obtained using
predict(puro1,list(conc=0.02)):
Release notes:
http://www.nabble.com/Bio7-1.3-Linux-released!-td20360723.html#a20360723
http://www.nabble.com/Bio7-1.3-Linux-released!-td20360723.html#a20360723
With kind regards
M.Austenfeld
--
View this message in context:
http://www.nabble.com/New-version-of-ecological-modeling-software-
Hello,
for a simulation I tried the following:
=
sampmeanvec <- function (from, n, repititions)
{
print( paste("samplesize n:", n, "repititions:", repititions) )
samples.mat <- as.matrix( replicate( repititions, sample(from, n)
Thanks Mark, it was exactly what I was looking for.
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Em nome de Mark Difford
Enviada em: quinta-feira, 6 de novembro de 2008 09:22
Para: r-help@r-project.org
Assunto: Re: [R] R Mixed Anova
Hi Rodrigo,
Here are two option
I have been learning how to use these functions and would like to know
the following as
I have so far been unable to find the answers in the documentation.
1) What stopping rules are used ?
2) Can the stopping rules be changed?
3) Can the results of each step be stored as objects in R and if so
Thank you for your prompt assistance, cruz and Bart.
Bart set me on the right track, and I modified his proposal to this:
f <- function(data){
m <- match(data$stop,data$start)
n <- min(length(m),which(is.na(m)))
data$stop[n]
}
by(data,data$id,f)
It also handles some spe
Hi John,
I had the same problem yesterday and solved it like following:
seq(ISOdate(Year,Month,Day,Hour), by="hour", length=24*365)
for example: seq(ISOdate(2005,1,1,0), by="hour", length=8760)
Regards,
Solveig
EIFER
Europäisches Institut für Energieforschung
Institut européen de recherche s
I have an external data (.txt) for
annual peak flood. The first column is the year, second column is the
observation date, and the last is the observed discharge. My task is to
calculate the mean, skewness and kurtosis of the said data. I was advised to use
read.table() to read the entire data.
christabel_jane prudencio wrote:
I have an external data (.txt) for
annual peak flood. The first column is the year, second column is the
observation date, and the last is the observed discharge. My task is to
calculate the mean, skewness and kurtosis of the said data. I was advised to use
I produced the 3D plot with "cloud" function (lattice package), but rgl package
works well: I can easily plot in 3D and annotate.
Thank you very much for your help.
Sebastian.
-
Merci de répondre à cette adresse e-mail et à [EMAIL PROTECTED]
Please reply both to this email
On Thu, 6 Nov 2008, Fán Lóng wrote:
Hi there,
I am intending to get the length of an UTF-8 string which contains
some Japanese characters (let's say, rstr) in R language.
I try to use the nchar(rstr) to get its length, however, it returns
the "NA" for it contains some multi-byte characters.
Is
Dear list,
I'm currently analyzing some count data using a hurdle model. I've used
the rcspline.eval function in the Hmisc-library to contruct the spline
terms for the regression model, and what I want in the end is the ability
to compute coefficients and confidence intervals for different chan
Abdou Ali agrhymet.ne> writes:
>
> Dear Sirs,
>
> Is there a R function to read an Idrisi image (*.rdc & *.rst)
Yes, please see the rgdal package, which can read this format. The "Spatial"
Task View (CRAN) might have got you the information you need without posting.
Depending on the size of
Hi,
I'm not sure how use curve(dexGAUS(
None of the following four works:
rt<- rexGAUS(100, mu=300, nu=100, sigma=35)
m1<-gamlss(rt~1, family=exGAUS)
curve(dexGAUS(rt=x, mu=300 ,sigma=35,nu=100), 100, 600, main = "The ex-
GAUS density mu=300 ,sigma=35,nu=100")
curve(dexGAUS(x=rt, mu=300 ,sigm
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