If you have to deal with missing values, you might be interested
in na.omit().
?na.omit
Regards,
Christoph
--
Christoph Buser <[EMAIL PROTECTED]>
Seminar fuer Statistik, LEO C13
ETH (Federal Inst. Technology) 8092 Zurich SWITZERLA
You can simply do
plot(xx)
plot(yy,add=T)
Aleš Žiberna
- Original Message -
From: "Mike Jones" <[EMAIL PROTECTED]>
To:
Sent: Thursday, June 02, 2005 9:53 PM
Subject: [R] overlaying plots
surely this not hard, but i can't figure it out. i'd like to overlay
the results of two ecdf's
gReetings,
I want to manipulate a formula object, containing the name "."
so that "." is replaced by a desired (arbitrary) expression.
What is a safe way to do this?
Adrian Baddeley
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R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/
Dear R users,
Could you please help me out. I am in trouble as I am unable to model graphs
to explain the GARCH (1 , 1) model, the Hill estimator (of alpha), and the
Pareto estimator.
I just got introduce to R. I am working on a paper which must be worked from
R.
You look at the difficult
By post hoc I guess you meant pairwise comparisons. You might want to check
out the npmc package on CRAN.
Andy
> From: [EMAIL PROTECTED]
>
> I´m looking for a program with a post hoc kruskall Wallis
> test like the Tukey-
> type non parametric test of ZAR(Biostatistical Analysis
> 2ºEdition).
"Jagarlamudi, Choudary" <[EMAIL PROTECTED]> writes:
> Hi All,
>
> I get the following error when i perform a t.test.
>
> studentt<-apply(tlr, 1, function(x) t.test(x[2:6],
> x[7:11],var.equal=TRUE)$p.value)
>
> # tlr is a table of 11 columns and 22500 rows. I am not able to post tlr due
>
They are calculated in the print() method for boot objects, but not
returned. boot:::print.boot has:
[...]
op <- cbind(t0, apply(t, 2, mean, na.rm = TRUE) -
t0, sqrt(apply(t, 2, function(t.st)
var(t.st[!is.na(t.st)]
dimnames(op) <- list(rn, c("original
Yes, you're correct. From the Green book ("Programming with Data"):
(p.323) "Strictly speaking, the [specific] method supplied [for a generic]
must have the exact same formal argument list as the generic ... However, if
you do supply a function with different arguments, setMethod will construct
a
Thanks a lot!
Both versions seem to work equally efficient.
N8,
Stefan
replace <- is.na(B)
B[replace] <- A[replace]
Duncan Murdoch
B[is.na(B)] <- A[is.na(B)]
Sarah
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Deepayan, Paul:
Thanks very much for the very useful leads. I just spent some time this
afternoon implementing your suggestions and my early test results are very
encouraging.
Paul, I understand the potential device re-sizing issue with the convert
family of functions in grid, and will be in touch
I tried the following (relevant excerpt only)
setMethod("likelihood",
signature(spec="Specification", covs="vector",
states="vector"),
function(spec, covs, states) {
setMethod("likelihood",
signature(model="Model", path="matrix"),
Hi All,
I get the following error when i perform a t.test.
studentt<-apply(tlr, 1, function(x) t.test(x[2:6],
x[7:11],var.equal=TRUE)$p.value)
# tlr is a table of 11 columns and 22500 rows. I am not able to post tlr due to
its size.
Error in if (stderr < 10 * .Machine$double.eps * max(abs
Hi
Deepayan Sarkar wrote:
On Wednesday 01 June 2005 05:43, Pikounis, Bill [CNTUS] wrote:
Hello,
I wish to customize the tick marks and labels of axes in panels produced by
high-level lattice functions, namely xyplot. I know I can use the scales
argument to specify values for rot, cex, etc. in
Please read the chapter on plotting in "An Introduction to R" where this is
explained.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
"The business of the statistician is to catalyze the scientific learning
process." - George E. P. Box
> -Original Message-
I´m looking for a program with a post hoc kruskall Wallis test like the Tukey-
type non parametric test of ZAR(Biostatistical Analysis 2ºEdition). SSPS
hasn´t got any non parametric post host test. I don´t know if R has an
appropiate post hoc non parametric test. I´m not sure if NDWD test of the
On 6/2/2005 4:19 PM, Stefan Mischke wrote:
Dear List
I have two large matrices A and B. Both have the same dimensions, let's
say 20k x 30k. About half the cells of B are missing. Now I'm looking
for an efficient way to merge them, so that the missing values in B are
replaced by the correspond
On 6/2/05, Stefan Mischke <[EMAIL PROTECTED]> wrote:
> Dear List
>
> I have two large matrices A and B. Both have the same dimensions, let's
> say 20k x 30k. About half the cells of B are missing. Now I'm looking
> for an efficient way to merge them, so that the missing values in B are
> replaced
Dear List
I have two large matrices A and B. Both have the same dimensions, let's
say 20k x 30k. About half the cells of B are missing. Now I'm looking
for an efficient way to merge them, so that the missing values in B are
replaced by the corresponding values of A.
Matrix A
[,1]
The symbol sizes, themselves, already encode the mean temperatures, so I'm
not exactly sure what you want to put in a legend. Perhaps the circles and
values for a few selected quantiles of the distribution to anchor the eye?
If so, legend() with pch and pt.cex does not appear capable of doing what
surely this not hard, but i can't figure it out. i'd like to overlay
the results of two ecdf's.
e.g. xx<--ecdf(x) and yy<--ecdf(y). i'd like to plot xx and yy on the
same graph.
thanks...mj
[[alternative HTML version deleted]]
__
R-he
On 6/2/2005 2:41 PM, Gabor Grothendieck wrote:
or alternately use new.env(parent = NULL) in which case you don't
need the inherits = FALSE on the get either.
On 6/2/2005 2:59 PM, Seth Falcon wrote:
>
> In terms of using R's environments as dictionary data structures I
> think what one wants m
On 6/2/2005 2:40 PM, Prof Brian Ripley wrote:
On Thu, 2 Jun 2005, Duncan Murdoch wrote:
Gabor Grothendieck wrote:
On 6/2/05, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on modification,
when
I am running a bootstrap
bootResult<-boot(...)
and would like to obtain the bias and std. error produced by the
bootstrap.
I can get the values to print to the screen, I can get the original
(non-bootstrap) estimates, OrgEst<-bootResult$t0
but I can not find any way to assign the bias and s
Brian Ripley wrote:
> But the real problem is more likely that Rolf has not passed
> model.matrix a model frame, so it calls model.frame() internally. The
> help page is a bit confused in that it says
>
> data: a data frame created with 'model.frame'.
>
> which the default for the argum
Not an answer, but a note. Encoding the mean temperature as the circle
radius is a bad idea: the eye perceives the area and so the radius^2, thus
the perceived effect is the square of the actual temperature effect (Howard
Wainer once referrred to this as "goosing up the effect by squaring the
eyeba
Duncan Murdoch wrote:
Gabor Grothendieck wrote:
On 6/2/05, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on
modification,
when any language would have to copy in this context.
Caveat: the
On 2 Jun 2005, [EMAIL PROTECTED] wrote:
> The original poster asked for caching of results. here is an example
> using new.env():
>
> memo <- function(fun) {
> mem <- new.env()
> function(x) {
> if (exists(as.character(x), envir=mem)) get(as.character(x),
> envir=mem, inherits=FALSE) else {
> va
I have created a symbol plot with circles that represent the mean temperature
at lat/lon locations over the United States. The radius of the circle
corresponds to the mean temperature. I would like to add a legend that
identifies a range of temperatures (e.g. 0-10, 10-20, etc) with circles of the
a
On 6/2/2005 2:14 PM, Kjetil Brinchmann Halvorsen wrote:
Prof Brian Ripley wrote:
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on
modification,
when any language would have to copy in this context.
Caveat: the default is new.env(hash=FA
On 6/2/05, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> On 6/2/05, Kjetil Brinchmann Halvorsen <[EMAIL PROTECTED]> wrote:
> > Prof Brian Ripley wrote:
> >
> > > On Thu, 2 Jun 2005, hadley wickham wrote:
> > >
> > >>> An environment is a hash table, and copying occurs only on
> > >>> modification
On Thu, 2 Jun 2005, Duncan Murdoch wrote:
Gabor Grothendieck wrote:
On 6/2/05, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on modification,
when any language would have to copy in this contex
On 6/2/05, Kjetil Brinchmann Halvorsen <[EMAIL PROTECTED]> wrote:
> Prof Brian Ripley wrote:
>
> > On Thu, 2 Jun 2005, hadley wickham wrote:
> >
> >>> An environment is a hash table, and copying occurs only on
> >>> modification,
> >>> when any language would have to copy in this context.
> >>
> >
How might you fit a generalized linear model (glm) with variance =
mu+theta*mu^2 (where mu = mean of the exponential family random variable
and theta is a parameter to be estimated)?
This appears in Table 2.7 of Fahrmeir and Tutz (2001) Multivariate
Statisticial Modeling Based on General
Prof Brian Ripley wrote:
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on
modification,
when any language would have to copy in this context.
Caveat: the default is new.env(hash=FALSE), so an environment is a
hash table in one sense but
Martin Biuw wrote:
> Hello,
> I'm using the nls function and would like to increase the number of
> iterations. According to the documentation as well as other postings on
> R-help, I've tried to do this using the "control" argument:
>
> nls(y ~ SSfpl(x, A, B, xmid, scal), data=my.data,
> control=
> x <- readLines(...)
> tmp <- file()
> writeLines(x[substr(x, 83, 86) == "STD"], tmp)
> read.fwf(tmp, ...)
I wrapped this approach into function and it worked swimmingly. Thanks
all. -Andy
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https://stat.ethz.ch/m
Hello,
I'm using the nls function and would like to increase the number of
iterations. According to the documentation as well as other postings on
R-help, I've tried to do this using the "control" argument:
nls(y ~ SSfpl(x, A, B, xmid, scal), data=my.data,
control=nls.control(maxiter=200))
Gabor Grothendieck wrote:
On 6/2/05, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on modification,
when any language would have to copy in this context.
Caveat: the default is new.env(hash=FAL
On 6/2/05, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> On Thu, 2 Jun 2005, hadley wickham wrote:
>
> >> An environment is a hash table, and copying occurs only on modification,
> >> when any language would have to copy in this context.
>
> Caveat: the default is new.env(hash=FALSE), so an envi
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on modification,
when any language would have to copy in this context.
Caveat: the default is new.env(hash=FALSE), so an environment is a hash
table in one sense but not necessarily so in the str
On Thu, 2 Jun 2005, hadley wickham wrote:
An environment is a hash table, and copying occurs only on modification,
when any language would have to copy in this context.
Yes, I'm aware that copying only occurs on modification. However, it
was my understanding that
a <- list(a =1)
a$b <- 4
wo
Dear all R-helpers,
Thanks you very much for your help. I would like to thanks Sean Davis
and Gabor Grothendieck for their help.
Best wishes, Muhammad Subianto
On this day 6/2/2005 3:21 PM, Gabor Grothendieck wrote:
>
> Try this:
>
> names(prima) <- paste("xyz", names(prima), sep = ".")
>
On
> An environment is a hash table, and copying occurs only on modification,
> when any language would have to copy in this context.
Yes, I'm aware that copying only occurs on modification. However, it
was my understanding that
a <- list(a =1)
a$b <- 4
would create a new copy of a, whereas in Jav
On Thu, 2 Jun 2005, Dimitris Rizopoulos wrote:
I think you could bit that using (inside your function) something like this
old.o <- options(na.action = na.fail)
# old.o <- options(na.action = na.pass)
on.exit(old.o)
You can. But the real problem is more likely that Rolf has not passed
model
On Sun, 29 May 2005, John Fox wrote:
Dear Spencer,
-Original Message-
From: Spencer Graves [mailto:[EMAIL PROTECTED]
Sent: Sunday, May 29, 2005 4:13 PM
To: John Fox
Cc: r-help@stat.math.ethz.ch; 'Jacob van Wyk'; 'Eric-Olivier Le Bigot'
Subject: Re: [R] Errors in Variables
Hi, John:
On Thu, 2 Jun 2005, hadley wickham wrote:
cache results. Other languages usually offer a dictionary data type
which I can use as an efficient way to dynamically cache already
calculated results - what's the best way to do this in R?
It really depends on what sort of data you want to cache - i
AFAIK R does copy-on-modify, so if you call f(aList, ...) and inside f()
aList is not modified, then a copy is not really made.
Andy
> From: hadley wickham
>
> > cache results. Other languages usually offer a dictionary data type
> > which I can use as an efficient way to dynamically cache alre
On Thu, 2 Jun 2005, Peter Dalgaard wrote:
"Andy Bunn" <[EMAIL PROTECTED]> writes:
Hi all:
I have acquired a 100s of data files that I need to preprocess to get them
usable in R. The files are fixed width (to a point) and contain 1 to 3 lines
of header, followed by a variable number of fixed w
I think you could bit that using (inside your function) something like
this
old.o <- options(na.action = na.fail)
# old.o <- options(na.action = na.pass)
on.exit(old.o)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Cathol
> cache results. Other languages usually offer a dictionary data type
> which I can use as an efficient way to dynamically cache already
> calculated results - what's the best way to do this in R?
It really depends on what sort of data you want to cache - if it is
fundamentally 1D, use a vector,
Andy Bunn wrote:
Hi all:
I have acquired a 100s of data files that I need to preprocess to get them
usable in R. The files are fixed width (to a point) and contain 1 to 3 lines
of header, followed by a variable number of fixed width data lines (that I
can read with read.fwf). I want to read thro
"Andy Bunn" <[EMAIL PROTECTED]> writes:
> Hi all:
>
> I have acquired a 100s of data files that I need to preprocess to get them
> usable in R. The files are fixed width (to a point) and contain 1 to 3 lines
> of header, followed by a variable number of fixed width data lines (that I
> can read w
Ernesto Jardim wrote:
> Hi,
>
> I'm developing in S4 and I wanted to see the methods for a specific
> class using showMethods but I didn't succed. Can someone help ? See the
> example below.
>
> setClass("myclass",
> representation(
> name="character"
> )
> )
>
> setGeneric("
I have just been bitten by a quirk in the behaviour of model.matrix.
I used model.matrix inside a function, and passed to it a formula
that was built elsewhere.
The formula was of the form ``y ~ x + w + z''. Now, model.matrix
cheerfully accepts formulae of this form, although it only
***needs***
Hi all:
I have acquired a 100s of data files that I need to preprocess to get them
usable in R. The files are fixed width (to a point) and contain 1 to 3 lines
of header, followed by a variable number of fixed width data lines (that I
can read with read.fwf). I want to read through the files and r
Hi Mathias,
Try
abline(lm(z2 ~ z1))
Mikkel
..
Hi,
when I run
> plot.default(z1, z2, xlab = "x", ylab = "y", main =
"", pch = "+")
> abline(lm(z1 ~ z2))
then the plot is plotted perfectly (scatterplot),
however, the lm()
function doesnt appear on the p
Mathias Hunskår Furevik wrote:
Hi,
when I run
plot.default(z1, z2, xlab = "x", ylab = "y", main = "", pch = "+")
abline(lm(z1 ~ z2))
then the plot is plotted perfectly (scatterplot), however, the lm()
function doesnt appear on the plot. What could be wrong?
(Yesterday it worked perfec
Your function never uses the argument w. This argument is used to define
the bootstrap sample as described in the help for boot. You have choices
about how to do this for linear models. If you want to do case resampling
then you function should be
detefun3<-function (d,i) {
d <- d[i,]
On 6/2/05, Mathias Hunskår Furevik <[EMAIL PROTECTED]> wrote:
> Hi,
>
> when I run
>
> > plot.default(z1, z2, xlab = "x", ylab = "y", main = "", pch = "+")
> > abline(lm(z1 ~ z2))
I think you have the independent and dependent variables switched:
abline(lm(z2 ~ z1))
Sarah
--
Sarah Goslee
On 6/2/05, Muhammad Subianto <[EMAIL PROTECTED]> wrote:
> Dear R-helpers,
> First I apologize if my question is quite simple
> I have a large datasets which more 100 variables.
> For a research I need to change all name of variables with add one or
> more letters on each variables.
> For example,
>
Hi,
when I run
> plot.default(z1, z2, xlab = "x", ylab = "y", main = "", pch = "+")
> abline(lm(z1 ~ z2))
then the plot is plotted perfectly (scatterplot), however, the lm()
function doesnt appear on the plot. What could be wrong?
(Yesterday it worked perfectly, with the lm() line.)
Runnin
Hola,
Tengo un problema con el post hoc kruskal wallis. No encuentro en el SSPS
ningún test estadístico para el post hoc. En el libro de "Zar" proponen el
Turkey-type nonparametric multicomparison. Mi problema es que tengo muchos
datos y hacerlo a mano es complicado. El R es un programa que c
On Jun 2, 2005, at 01:23 pm, Dimitris Rizopoulos wrote:
try this:
h <- rep(3:2, c(4, 3))
n <- 10
matrix(rep(h, n), nrow = n, byrow = TRUE)
Hi
outer() also works:
> h <-rep(c(3,2),c(4,3))
> outer(rep(1,8),h)
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]333
NATALIA F TCHETCHERINA wrote:
Hello all,
I have data:
Genes time rep vart dye y trt
130911 sa1-d07 030min 1 col g 9.636244 o
145771 sa1-d07 030min 1 col r 8.107577 c
93335sa1-d07 030min 1 ler g 7.409566 o
94821sa1-d07 030min 1 ler r
try this:
h <- rep(3:2, c(4, 3))
n <- 10
matrix(rep(h, n), nrow = n, byrow = TRUE)
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/336899
Fax
"Friedrich, Andreas (dit)" <[EMAIL PROTECTED]> writes:
> Hallo,
>
> I'll need a matrix with n rows of the an identical vector.
>
>
>
> > h
> [1] 3 3 3 3 2 2 2
>
>
> The nmatrix should look like this:
>
> > x<-rbind(h,h,h)
> > x
> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
> h3333
Friedrich, Andreas (dit) wrote:
Hallo,
I'll need a matrix with n rows of the an identical vector.
h
[1] 3 3 3 3 2 2 2
The nmatrix should look like this:
x<-rbind(h,h,h)
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
h3333222
h3333222
Hallo,
I'll need a matrix with n rows of the an identical vector.
> h
[1] 3 3 3 3 2 2 2
The nmatrix should look like this:
> x<-rbind(h,h,h)
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
h3333222
h3333222
h3333222
Carlo Fezzi stat.unibo.it> writes:
>
> Dear R-helpers,
> Anybody knows which function can I use to comupute maximum likelihood
> standard errors?
>
> Using the function "nlm" I can get the estimate of the parameters of any
> likelihood that I want (for example now I am working on a jump diffus
look at mle() (package "stats4") and optim() (i.e., "optim(...,
hessian = TRUE)").
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/336899
Now I want to some simulation Impact of shrimp's releasing to genetic
structure of wild population,i think rmetasim can help me finish this
task. But on hand I have not the resource about it. Can anyone help me ?
Thanks!
luan_sheng
__
R-help@stat.math.
Hi!
How to calculate the correct SE of mean in a nested or spliplot anova?
Nested example:
-
m <- aov(Glycogen~Treatment+Error(Treatment/Rat/Liver))
> m
Call:
aov(formula = Glycogen ~ Treatment + Error(Treatment/Rat/Liver))
Grand Mean: 142.
Stratum 1: Treatment
Terms:
Dear R-helpers,
Anybody knows which function can I use to comupute maximum likelihood
standard errors?
Using the function "nlm" I can get the estimate of the parameters of any
likelihood that I want (for example now I am working on a jump diffusion
process) but what about the standard error?
Is t
See ?paste.
Something like below if you have 100 column names:
dimnames(pima.tr)[[2]] <-
paste(rep('xyz',100),dimnames(pima.tr)[[2]],sep=".")
You probably want to test the paste statement before setting the
dimnames, or operate on a copy of the data until you get the hang of
using paste.
Dear R-helpers,
First I apologize if my question is quite simple
I have a large datasets which more 100 variables.
For a research I need to change all name of variables with add one or
more letters on each variables.
For example,
> data(Pima.tr)
> Pima.tr[1:5,]
npreg glu bp skin bmi ped age t
Try unique(unlist(myList)).
Andy
> From: Sofyan Iyan
>
> Dear R-help again,
> I have a result something like this,
> [[1]]
> [1] "Game" "Internet"
>
> [[2]]
> [1] "Game" "Internet"
>
> [[3]]
> [1] "Game""Time"
>
> How could I make the result above like,
>
> [1] "Game","Internet",
Hi,
I'm developing in S4 and I wanted to see the methods for a specific
class using showMethods but I didn't succed. Can someone help ? See the
example below.
setClass("myclass",
representation(
name="character"
)
)
setGeneric("mymeth", function(obj, ...){
Dear R-help again,
I have a result something like this,
[[1]]
[1] "Game" "Internet"
[[2]]
[1] "Game" "Internet"
[[3]]
[1] "Game""Time"
How could I make the result above like,
[1] "Game","Internet","Time"
Regards, Sofyan
On 6/1/05, Sofyan Iyan <[EMAIL PROTECTED]> wrote:
> Dear R-he
Thoralf Mildenberger wrote:
I have been using the rgenoud package for a nonlinear least-squares
problem with lots of local minima, and it works very well but takes lots
of time. According to the article refrenced in the documentation, the
original GENOUD-software by the same authors seems to allo
Functions like arima() (in stats) may be what you want. Also you may
have a look at package dse1.
HTH, ft.
--
Fernando TUSELLe-mail:
Departamento de Econometría y Estadística [EMAIL PROTECTED]
Facultad de CC.EE. y Empresariales Tel: (+34)94.601.
You can use ´expand.grid'
» expand.grid(x=seq(42.2,45.2,by=1),y=seq(50.5,51.5,by=1))
xy
1 42.2 50.5
2 43.2 50.5
3 44.2 50.5
4 45.2 50.5
5 42.2 51.5
6 43.2 51.5
7 44.2 51.5
8 45.2 51.5
Eric
Eric Lecoutre
UCL / Institut de Statistique
Voie du Roman Pays, 20
1348 Louvain-la-Neuve
Belgi
Hello R-World,
i am trying to calculate data for a DEM (Digital Elavation Model) which i
also want to plot in R. i have the coordinates for the lower left corner
which look something like this:
x<-42,2
y<-50,5
besides i have the cellsize of the grid, which is:
z<-1,1
what i do is to calculate
Moin,
is there any possibility to choose the input variables which are
involved in the analysis,
" ..., and each column correspondends to a variable. ..."
(R-Reference-Manual)
or do I have to delete them from the data.frame?
An then I'd like to get the original dataset with the classification of
Hello!
I have an algorithm which performs lengthy operations and I would like to
cache results. Other languages usually offer a dictionary data type
which I can use as an efficient way to dynamically cache already
calculated results - what's the best way to do this in R?
Best wishes,
Sven C. Ko
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