Paul Sorenson vision-bio.com> writes:
:
: The solution I came up with myself was simply to coerce the integer back to
POSIXct:
:
: class(ev$date) = "POSIXct"
:
: Can't say it is the right way to do it but it seem to work.
That is not entirely correct, although it many cases it will wor
Benjamin M. Osborne uvm.edu> writes:
:
: I have a set of time-series climate data with missing entries. I need to add
: rows for these missing entries to this data set. The only way I know to do
: this is unsing a foreloop, but this won't work on a list. I've tried to
: convert the list to a
I'm surprised nobody has mentioned Alpha. It has highlighting,
indenting, parenthesis matching, excellent integration with R (or
that other commercial version of R). There are versions for Classic
(Alpha8), OS X (Alphax), as well as *NIX and Windows (AlphatTk).
Alpha is shareware, based on the
The solution I came up with myself was simply to coerce the integer back to
POSIXct:
class(ev$date) = "POSIXct"
Can't say it is the right way to do it but it seem to work.
A second related problem I haven't been able to solve as yet is how to include
"incidents" columns (those not in '
Perhaps an easier way would be to throw away the offending text at the
end of the strings, rather than matching all possible numeric
formulations at the beginning of the string, that is:
sub("\\.*[[:alpha:]]+$", "", x)
Easier to read, if nothing else, and it allows for 2e-7 as a valid
number. Thi
"Carsten Steinhoff" <[EMAIL PROTECTED]> writes:
> Hello,
>
>
>
> maybe that my Question is a "beginner"-Question, but up to now, my research
> didn't bring any useful result.
>
>
>
> I'm trying to fit a distribution (e.g. lognormal) to a given set of data
> (ML-Estimation). I KNOW about my
On Mon, 2005-01-31 at 16:06 -0500, Uri wrote:
> Dear R community,
>
> I'm currently using R to analyze functional Magnetic Resonance Imaging
> data. Each analysis involves running ~120,000 repeated-measures
> ANOVAs.
>
> I would like to know if there is any automatic way to access the F-
> and P
(Ted Harding) <[EMAIL PROTECTED]> writes:
> On 31-Jan-05 R user wrote:
> > You could use something like
> >
> > y <- gsub('([0-9]+(.[0-9]+)?)?.*','\\1',x)
> > as.numeric(y)
> >
> > But maybe there's a much nicer way.
> >
> > Jonne.
>
> I doubt it -- full marks for neat regexp footwork!
Hmm, I
A data.frame is a list, which may be why "mode(mansNew)" was "list".
Have you tried something like the following:
> tstList <- list(a=1:3, b=c(NA, 4, 6))
> DF <- as.data.frame(tstList)
> mode(DF)
[1] "list"
> class(DF)
[1] "data.frame"
> DF
a b
1 1 NA
2 2 4
3 3 6
> DF2 <- DF[c(1,1
What R commands were used to produce this estimate of the required
transformation? Did the command used produce a confidence interval for
the power transformation, as, e.g., "boxcox" in library(MASS) described
by Venables and Ripley (2002) Modern Applied Statistics with S, 4th ed.
(Spring
On Mon, 2005-01-31 at 12:42 -0700, Tae-Hoon Chung wrote:
> Hi, All;
>
> When I generate a "special" paper postscript image larger than "a4" or
> "letter" using R, I can only see one-page portion of all image, of course.
> What will be the simple solution for this? Is there any way I can set the
>
Dear Uri,
You can save the object produced by summary() (i.e., assign it to a
variable, not print it and dump it to a file) and extract what you want from
that. You can use str() to see the structure of the object.
I hope this helps,
John
John Fox
Department of
Bert,
Your question below prompted me to run the same code to the postscript device
and sure enough, the file created displays perfectly. So I guess the problem is
specific to my system and I should not try to fix it by changing my code.
My system is as follows:
platform i686-pc-linux-gnu
arch
On Mon, 31 Jan 2005, Carsten Steinhoff wrote:
maybe that my Question is a "beginner"-Question, but up to now, my research
didn't bring any useful result.
I'm trying to fit a distribution (e.g. lognormal) to a given set of data
(ML-Estimation). I KNOW about my data that there is a truncation for all
Why not just save your current global environment and reload it after
running the offending code? See ?save.image to save, then rm(list =
ls()) to clear the global environment.
If a program tries to get or assign something to a global environment,
then either you rewrite the program to do otherwis
I guess setting par(mgp) would help.
Andy
> From: Wittner, Ben
>
> Hi,
>
> In the code below, the labels I put on the x-axis are too
> high (they cross the
> axis). Can anyone tell me how to move them down? I've tried
> adj=, padj=, mar=,
> and various other things, but cannot move them down.
Hi,
In the code below, the labels I put on the x-axis are too high (they cross the
axis). Can anyone tell me how to move them down? I've tried adj=, padj=, mar=,
and various other things, but cannot move them down.
Thanks.
-Ben
labs <- paste('sample', 1:10)
plot(1:10, xaxt='n', xlab='')
axis(1,
Dear R community,
I'm currently using R to analyze functional Magnetic Resonance Imaging
data. Each analysis involves running ~120,000 repeated-measures
ANOVAs.
I would like to know if there is any automatic way to access the F-
and P-value data that are associated with each of these 120,000
ANO
Will the following work for you?
Cheers, Pierre
> junk <- list()
> for(i in 1:5) junk[[paste("Item",i,sep="")]] <- runif(4)
> str(junk)
List of 5
$ Item1: num [1:4] 0.687 0.508 0.734 0.704
$ Item2: num [1:4] 0.5497 0.1777 0.4081 0.0129
$ Item3: num [1:4] 0.768 0.808 0.592 0.697
$ Item4: num
Hello,
maybe that my Question is a "beginner"-Question, but up to now, my research
didn't bring any useful result.
I'm trying to fit a distribution (e.g. lognormal) to a given set of data
(ML-Estimation). I KNOW about my data that there is a truncation for all
data below a well known thresh
Hi, I need to simulate the following two time series,
1. X=cos(omega1*t+phi1)+cos(omega2*t+phi2)+cos(omega3+phi3)
2. Y=cos(omega1*t+phi1)+cos(omega2*t+phi2)+cos(omega3+phi1+phi2)
where omega3=omega1+omega2 and phi1, phi2, phi3 are independent
uniform random variables from 0 to 2pi.
Question 1: w
All -
Each time through a loop I create a new dataset, which I would like to
append to a list object. Each item of the list should be the data matrix
created in that step of the loop; I would like the NAME (or tag) of that
list item to be assigned the value of a character string:
I've tried so
I have a set of time-series climate data with missing entries. I need to add
rows for these missing entries to this data set. The only way I know to do
this is unsing a foreloop, but this won't work on a list. I've tried to
convert the list to a data frame, but that won't happen, either.
I want
Hi, All;
When I generate a "special" paper postscript image larger than "a4" or
"letter" using R, I can only see one-page portion of all image, of course.
What will be the simple solution for this? Is there any way I can set the
bounding box information on the image? Or any other suggestions?
Tha
?time
dat <- ts(rnorm(10), start = c(1900), frequency = 1)
dat[time(dat) == 1905]
HTH, Andy
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of Ferrari Nicolas
> Sent: Monday, January 31, 2005 2:26 PM
> To: '[EMAIL PROTECTED]'
> Subject: [R] Help about ti
Hello,
When I create a ts object, I would like to get a particular value of this
time serie according to the date and not the rank.
However, it seems necessary to use the rank as if it were a simple vector.
I would be very grateful if you could help out of this.
I thank you in advance,
Nicolas F
Silvia
You may want to use the disk image instead. It is here (for italy):
http://microarrays.unife.it/CRAN/bin/macosx/R-2.0.1.dmg
Sean
On Jan 31, 2005, at 1:52 PM, silvia simoni wrote:
I have a problem in installing R-2.0.1, downloaded from the R web
site, on mac o sx version 10.3.7.
when i
I have a problem in installing R-2.0.1, downloaded from the R web site,
on mac o sx version 10.3.7.
when i launch the command ./configure i get the following error message:
checking for C compiler default output file name... configure: error:
C compiler cannot create executables
i've aleady in
Dr Carbon wrote:
Sorry, what I meant was that I want dat.sub[1] to line up with dat[7],
dat.sub[2] to line up with dat[19], dat.sub[3] with dat[12+12+7],
etc...
Sorry, I haven't read your mail carefully enough:
plot(dat)
plot(x = time(dat.sub, offset = 0.5), y = dat.sub,
type = "b", xy.lab
Dr Carbon gmail.com> writes:
:
: I'm probably apporaching this all wrong to start but
:
: Suppose I have a monthly time series and I want to compute the mean of
: months 6,7, and 8. I want to plot the original time series and the
: seasonal time series, one above the other. When I do that a
[EMAIL PROTECTED] wrote:
Dear All,
I'm trying to calculate the following:
Select the rows of the data.frame and just calculate what you already
have written:
g.rate<-(SL-int)/NO
For the lines you cited this means, e.g.:
X <- data.frame(CODE = c(123:125, 234),
SL = c(12.5, 11.
Sorry, what I meant was that I want dat.sub[1] to line up with dat[7],
dat.sub[2] to line up with dat[19], dat.sub[3] with dat[12+12+7],
etc...
On Mon, 31 Jan 2005 18:34:55 +0100, Uwe Ligges
<[EMAIL PROTECTED]> wrote:
> Dr Carbon wrote:
> > I'm probably apporaching this all wrong to start but
Dr Carbon wrote:
I'm probably apporaching this all wrong to start but
Suppose I have a monthly time series and I want to compute the mean of
months 6,7, and 8. I want to plot the original time series and the
seasonal time series, one above the other. When I do that as below the
time series don'
On Sun, 30 Jan 2005 17:47:31 -0500, you wrote:
<-SNIP
|=[:o) >
|=[:o) >
|=[:o) >
|=[:o) Why are you using a double square root transformation? Is the
|=[:o) transformation for the response variable? Transfromation is one way to
|=[:o) help insure that the error distri
Dear All,
I'm trying to calculate the following:
g.rate<-(SL-int)/NO
where SL and NO are individual length and counts for the same subject, and the
int value is a day-specific value i.e. for each DAY=88, 101..., 172 I'd like to
use a different values of int=9.32, 8.43, ..., 9.81. All variables a
I'm probably apporaching this all wrong to start but
Suppose I have a monthly time series and I want to compute the mean of
months 6,7, and 8. I want to plot the original time series and the
seasonal time series, one above the other. When I do that as below the
time series don't line up for re
On 31-Jan-05 R user wrote:
> You could use something like
>
> y <- gsub('([0-9]+(.[0-9]+)?)?.*','\\1',x)
> as.numeric(y)
>
> But maybe there's a much nicer way.
>
> Jonne.
I doubt it -- full marks for neat regexp footwork!
To comply fully with Mike's request, I'd wrap it in sort():
sort(as.
There are several institutions here which offer distance learning.
http://programs.gradschools.com/distance/statistics.html
Min-Han Tan
On Mon, 31 Jan 2005 14:13:38 +0100, CG Pettersson
<[EMAIL PROTECTED]> wrote:
> 30 Januari, Tilo Blenk wrote:
>
>to
> ... snip ...
>
>
> Dear Tilo
>
Hi,
see these papers or books (some are available on the
web):
Diego Kuonen, Introduction au data mining avec R :
vers la reconqu?te du `knowledge discovery in
databases' par les statisticiens. Bulletin of the
Swiss Statistical Society, 40:3-7, 2001.
http://www.statoo.com/en/publications/2001.R.
Paul Sorenson vision-bio.com> writes:
:
: I have a frame which contains 3 columns:
:
: "date" "defectnum" "state"
:
: And I want to get the most recent state change for a given defect number.
date is POSIXct.
:
: I have tried:
: aggregate(ev$date, by=list(ev$defectnum), max)
:
: Whic
Lic. Adrián Cecotto - FACEA wrote:
R people,
I need to know if is possible to make data mining with R. If so, is
Almost sure, depends on your definition of data mining.
There are several manuals and books, see CRAN.
Uwe Ligges
there any manual or somewhere/one to consult about that.
Thank you ver
R people,
I need to know if is possible to make data mining with R. If so, is
there any manual or somewhere/one to consult about that.
Thank you very much
Adrián
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
Thanks a lot!
Indeed, both implementations agree on the 'best' points. Your answer helped
me a great deal.
Rainer
> The two implementations use different consistency factors as well as
> different small sample correction factors.
>
> 1. The search parts of both implementations produce the same
Jérôme,
On Sat, 2005-01-29 at 14:14 -0500, Jérôme Lemaître wrote:
> Hello alls,
>
> I found in the literature a technique that has been evaluated as one of the
> more robust to assess statistically the significance of the loadings in a
> PCA: bootstrapping the eigenvector (Jackson, Ecology 1993,
On Mon, 31 Jan 2005, Prof Brian Ripley wrote:
> On Sun, 30 Jan 2005, Paul Roebuck wrote:
>
> > Section 3.5.3 The call stack (pg 23 of R-lang.pdf)
> > ...the computation the the currently active environment...
> > ^^^
>
> Which version of R is this? I think it has been corrected
30 Januari, Tilo Blenk wrote:
http://statmaster.sdu.dk/
Cheers
/CG
CG Pettersson, MSci, PhD Stud.
Swedish University of Agricultural Sciences
Dep. of Ecology and Crop Production. Box 7043
SE-750 07 Uppsala
Sweden
__
R-help@stat.math.ethz.ch mailing
Hi,
see this web page:
http://www.stats.ox.ac.uk/~vos/DataMining.html
you'll find something about classification with
examples:
4. Classification
Decision theory, Linear discriminant analysis,
Quadratic discriminant analysis, Fisher's LDA,
Logistic discrimination, Non-parametric classification
t
Eric Rodriguez wrote:
Hi,
I would like to have some advices about how to fit such function:
f(x) = a.x^b if x <= t
c.x - c.t + a.t^b otherwise
I have to minimize the least-squares error knowing x and f(x).
Do you have any propositions of what should I use to a
Hi,
I would like to have some advices about how to fit such function:
f(x) = a.x^b if x <= t
c.x - c.t + a.t^b otherwise
I have to minimize the least-squares error knowing x and f(x).
Do you have any propositions of what should I use to achieve it ?
Thank
You could use something like
y <- gsub('([0-9]+(.[0-9]+)?)?.*','\\1',x)
as.numeric(y)
But maybe there's a much nicer way.
Jonne.
On Mon, 2005-01-31 at 08:51 +, Mike White wrote:
> Hi
> Does anyone know if there is a function similar to as.numeric that will
> extract a numeric prefix from a
Hi
Does anyone know if there is a function similar to as.numeric that will
extract a numeric prefix from a string as in the following examples?
x<-c(3, "abc", 5.67, "2.4a", "6a", "6b", "2.4.a", 3, "4.2a")
df.x<-data.frame(Code=x)
x.str<-levels(df.x[,1])
# required function result
2.40 3.00 4.20 5
On Mon, 31 Jan 2005, Patrick Giraudoux H wrote:
Dear Listers,
We are organising practical trainings for students with R 2.0.1 under MacOS
X. I am used with R 2.0.1 under Windows XP and thus has been surprised not to
find functions in the MacOS X version of R providing vectorized chart outputs
to
Please ask programming-related questions on the R-devel list -- see the
advice in the posting guide.
On Mon, 31 Jan 2005, Faheem Mitha wrote:
Dear People,
Here is something I do not understand. Consider
*
foo.cc
*
On Mon, 2005-01-31 at 08:06 +0100, Patrick Giraudoux H wrote:
> Dear Listers,
>
> We are organising practical trainings for students with R 2.0.1 under MacOS
> X. I am used with R 2.0.1 under Windows XP and thus has been surprised not
> to find functions in the MacOS X version of R providing vec
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