On 09/06/2010 07:38 AM, Christian Lederer wrote:
> Dear R-Users,
>
> after the last upgrade from r-base 2.11.1-2hardy0 to 2.11.1-5hardy0
> i lost the tlctk package.
>
> I this a general problem under Ubuntu Hardy or should i search for
> a configuration error on my system?
Assuming that you didn
Dear R-Users,
after the last upgrade from r-base 2.11.1-2hardy0 to 2.11.1-5hardy0
i lost the tlctk package.
I this a general problem under Ubuntu Hardy or should i search for
a configuration error on my system?
Christian
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On Sep 5, 2010, at 11:48 PM, Charles Annis, P.E. wrote:
Requisite info: R version 2.11.1 (2010-05-31) running on a 64 bit HP
Windows
7 machine.
Doubt that makes much of a difference here.
I have used contour() for several years. Now I would like to
extract from a
contour plot the x, y c
sseq <- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste("arima", i, sep=""), arima(data.ts[sseq[i]:(sseq[i]+200)],
order=c(1,1,1)))
...but why would you want to do so?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Requisite info: R version 2.11.1 (2010-05-31) running on a 64 bit HP Windows
7 machine.
Greetings, R-ians:
I have used contour() for several years. Now I would like to extract from a
contour plot the x, y coordinates of a contour z=constant. This seems as
though it would be straight-forwa
On Sep 5, 2010, at 10:11 PM, Aks Ism wrote:
No, I mean dynamically. Like list.push_back
Not sure what you mean by "dynamically" and my Google search for the
unspecified function list.push_back uncovered a C++ function that is
at least as complex as what I believe would be the equivalent R
You also may want to look at auto.arima in the 'forecast' package,
which will return the "best" ARIMA model based on AIC/AICc/BIC values
?auto.arima
hth
c
On 09/05/2010 06:02 PM, Stephan Kolassa wrote:
Hi,
basically, you know 5 periods later. If you use a good error measure,
that is.
I a
No, I mean dynamically. Like list.push_back
On Mon, Sep 6, 2010 at 1:01 AM, David Winsemius wrote:
>
> On Sep 5, 2010, at 1:21 PM, Aks Ism wrote:
>
> Hi,
>>
>> I've looked at previous discussions and did not get anything. I want to be
>> able to append to a list in a loop. Is this possible?
>>
>
Hi,
I have a dataset in .dbf format. It contains Coordinates and Time.
The TIMESTAMP is as follows:
03/18/2006 13:30:37
I am not working with the TIMESTAMP column, but when I print out my
manipulated dataset using
*write.dbf* I get the value *390 *where the TIMESTAMP value should be. Can
Any
Hello,
I am now learning to use the arima.sim command to model an AR(1) process.
The AR(1) equations looks something like:
r_t = phi_0 + phi_1*r_t-1 + u_t
While simulating the AR(1) model, we specify the value of phi_1 in the
equation.
For example, arima.sim(list(c(1,0,0),ar=0.9),200)
Could y
arima1 = arima(data.ts[1:200], order = c(1,1,1))
arima2 = arima(data.ts[5:205], order = c(1,1,1))
arima3 = arima(data.ts[10:210], order = c(1,1,1))
arima4 = arima(data.ts[15:215], order = c(1,1,1))
arima5 = arima(data.ts[20:220], order = c(1,1,1))
arima6 = arima(data.ts[25:225], order = c(1,1,1))
colorRampPalette returns a function, not a list of colours. You might want to
try something like:
ascols <-
colorRampPalette(c("gray", "yellow", "darkgoldenrod1", "orange", "red"),
interpolate="spline")
x <- 10*(1:nrow(volcano))
y <- 10*(1:ncol(volcano))
image(x, y, volcano, col = ascols(30
Hello and thanks in advance,
Using the dataset volcano:
ascols =
colorRampPalette(c("gray","yellow","darkgoldenrod1","orange","red"),interpolate="spline")
x <- 10*(1:nrow(volcano))
y <- 10*(1:ncol(volcano))
image(x, y, volcano, col = ascols, axes = FALSE)
In the example above, how would I change
?mtext
mtext("secondary y-axis label", 4, line=-1)
you will have to adjust 'line" to get your spacing the way you want it.
On Sun, Sep 5, 2010 at 7:52 PM, Magali teurlai wrote:
> Hi all,
>
> We managed to plot two series on the same graph, with two y-axis,
> however, we haven't managed to add
Hi all,
We managed to plot two series on the same graph, with two y-axis,
however, we haven't managed to add a title to the second y-axis
x1=rnorm(25, mean=0, sd=1)
y1=dnorm(x1, mean=0, sd=1)
x2=rnorm(25, mean=0, sd=1)
y2=dnorm(x2, mean=0, sd=1)
plot(x1, y1, type='p', xlab='x', ylab='y')
par(new
CRAN (and crantastic) updates this week
New packages
* CFL (0.1)
Maintainer: Pablo Michel Marin Ortega
Author(s): Pablo Michel Marin Ortega, Kornelius Rohmeyer
http://crantastic.org/packages/CFL
The main results on this project are: a package to analyze the data
based on c
David Winsemius wrote:
>
> That is different than my understanding of AIC. I thought that the AIC
> and BIC both took as input the difference in -2LL and then adjusted
> those differences for the differences in number of degrees of freedom.
>
>
David! Your words make sense to me now. Sor
Hi,
basically, you know 5 periods later. If you use a good error measure,
that is.
I am a big believer in AIC for model selection. I believe that arima()
also gives you the AIC of a fitted model, or try AIC(arima1).
Other ideas include keeping a holdout sample or some such.
I'd recommend l
Hi David,
This rm .RData works perfectly.
Many Many Thanks
Chunhao
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R-
How do you evaluate the predictive models? For example if I have:
arima1 = arima(training, order = c(1,1,1))
arima2 = arima(training, order = c(0,0,0))
x.fore = predict(arima1, n.ahead=5)
x.fore1 = predict(arima2, n.ahead = 5)
How do I know which arima model is better for prediction?
--
View
I want to also choose the post optimal parameters in the order argument. How
could I easily do this?
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On Sep 5, 2010, at 12:50 PM, Chunhao wrote:
Hi R Users,
I was accidentally save R file when I quit R program. I try to
delete but I
am not sure the R saved file was successfuly delete or not. Now
everyt time
when I open the R, it always appears
[R.app GUI 1.34 (5589) i386-apple-darwin9.8
Depending on your operating system, the freed up memory might not be
returned to the operating system, but kept in the process space.
On Sat, Sep 4, 2010 at 10:17 PM, Hyunchul Kim wrote:
> Hi, all
>
> Thank you for your comments.
> I think that I misunderstood what gc() does because gc() is worki
x <- list()
for (i in whatever){
x[[length(x) + 1L]] <- someValuesForList
}
On Sun, Sep 5, 2010 at 1:21 PM, Aks Ism wrote:
> Hi,
>
> I've looked at previous discussions and did not get anything. I want to be
> able to append to a list in a loop. Is this possible?
>
> [[alternative H
Is this what you are looking for ...
mylist=list()
for (i in 1:3){
mylist[i] = rnorm(1)
}
print(mylist)
HTH
Pete
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On Sep 5, 2010, at 1:21 PM, Aks Ism wrote:
Hi,
I've looked at previous discussions and did not get anything. I want
to be
able to append to a list in a loop. Is this possible?
Of course:
?"c"
?"[["
--
David Winsemius, MD
West Hartford, CT
__
1. try the predict function
e.g. predict(arima1, n.ahead=10)
2. try the resid function
e.g. resid(arima1)
HTH
Pete
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Hello,
I am working with the igraph package and I am wondering if it is possible to
get the plot function to plot multiple edges (and multiple loops) as
separate lines. Thanks in advance for any help on this.
sasha
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I have a data file with a given time series of price data and I would like to
split the time series into a test set and training set. I would then like to
build an ARIMA model on the training set and apply this model on test set.
Below is some code:
[CODE]
data= read.table("A.txt",sep=",")
attach(
Hi,
I've looked at previous discussions and did not get anything. I want to be
able to append to a list in a loop. Is this possible?
[[alternative HTML version deleted]]
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How do I get the predicted values and the errors for each arima model?
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Hi R Users,
I was accidentally save R file when I quit R program. I try to delete but I
am not sure the R saved file was successfuly delete or not. Now everyt time
when I open the R, it always appears
[R.app GUI 1.34 (5589) i386-apple-darwin9.8.0]
[Workspace restored from /Users/ctu/.RData]
I h
great! Thanks
On Sun, Sep 5, 2010 at 9:31 PM, Joshua Wiley wrote:
> On Sun, Sep 5, 2010 at 8:36 AM, rajesh j wrote:
> > No, the examples dont cover it...if i have
> > dat <- list(INT = c("1","2","3"),
> > NUM = c("2.34","4.56","6.78"),
> > INT = c("4", "5", "6"),
> >
On Sep 5, 2010, at 11:41 AM, John Helly wrote:
Hi.
I'm trying to get 'mu' to show up as a Greek symbol but, despite
trying every example I could find, can't get it to work. Any
insights would be welcome.
This is what I'm using that works, but displays mu with the letter u.
plotTimeXMas
On Sun, Sep 5, 2010 at 8:36 AM, rajesh j wrote:
> No, the examples dont cover it...if i have
> dat <- list(INT = c("1","2","3"),
> NUM = c("2.34","4.56","6.78"),
> INT = c("4", "5", "6"),
> NUM = c("3.44"))
>
> I need to do a which on INT[0],NUM[0],INT[0] and NU
I am running a 3-parameter nonlinear fit using the default Gauss-Newton
method of nls.
initialValues.L = list(b=4,d=0.04,t=180);
fit.nls.L = nls(
myModel.nlm ,
fData.L,
start = initialValues.L,
control = nls.control(warnOnly = TRUE),
trace=T
);
summary.nls.L = summary(fit.nls.L);
I run the sam
Dr Shrink wrote:
>
> I need to include the repeated structure in our data set object,
> recall.sums.df, before using gls function.
> Thus I used groupedData.
> But I encountered error messages which may mean '*' is not not meaningful
> factor.
>
I know that when reading Pinheiro/Bates one has
Hi.
I'm trying to get 'mu' to show up as a Greek symbol but, despite trying every
example I could find, can't get it to work. Any insights would be welcome.
This is what I'm using that works, but displays mu with the letter u.
plotTimeXMastPAR <- qplot(DT,MastPAR, data=A, xlab = "", ylab = quo
Hi Team,
Can you please tell me the rules of assignment in R, by value or by
reference.
>From my about 3 months of experience of part time job of R, it seems most
>times it is by value, especially in function parameter and return values
>assignment; and it is by reference when referen
install.packages('sos') # if you haven't already
library(sos)
dch <- ???dirichlet
# found 288 matches = help pages on contributed packages
summary(dch)
# in 65 packages
dch
# lists the 288 matches in a table in a browser with links the help pages
# sorted to put the package with the most matches
Hello,
I'm sorry to just pop-up on the mailing list like this and ask a relatively
non-R related question but I had no idea whom else to contact on this matter.
I'm working on a completely different port of an application to OS X which
requires GTK and through Google'ing stumbled on a rather rec
John Helly wrote:
>
> I'm trying to get 'mu' to show up as a Greek symbol but, despite trying
> every example I could find, can't get it to work.
>
>
Try
library(ggplot2)
qplot(mpg, wt, data=mtcars, xlab = "", ylab = bquote(mu~(E ~m^-2~s^-1)),
geom="line")
(thanks to Baptiste and David i
Hi,
in a way similar to names(dat), can I address a particular index of every
vector?
like [0] of all the vectors? So that I could do something like,
ints<-which(=="x")
On Sun, Sep 5, 2010 at 8:34 PM, Joshua Wiley wrote:
> Hi Rajesh,
>
> This will work, unfortunately it seems like lapply() drops
this should work great. thanks
On Sun, Sep 5, 2010 at 8:34 PM, Joshua Wiley wrote:
> Hi Rajesh,
>
> This will work, unfortunately it seems like lapply() drops the names
> before it passes each element of the list which lead to my clumsy work
> around using which(). I'm sure there are other ways
Hi.
I'm trying to get 'mu' to show up as a Greek symbol but, despite trying every
example I could find, can't get it to work. Any insights would be welcome.
This is what I'm using that works, but displays mu with the letter u.
plotTimeXMastPAR <- qplot(DT,MastPAR, data=A, xlab = "", ylab = quo
No, the examples dont cover it...if i have
dat <- list(INT = c("1","2","3"),
NUM = c("2.34","4.56","6.78"),
INT = c("4", "5", "6"),
NUM = c("3.44"))
I need to do a which on INT[0],NUM[0],INT[0] and NUM[0]...ie the [0] index
of all the vectors...
and test if they
On Sun, Sep 5, 2010 at 8:16 AM, rajesh j wrote:
> Hi,
> in a way similar to names(dat), can I address a particular index of every
> vector?
> like [0] of all the vectors? So that I could do something like,
> ints<-which(=="x")
Not sure I follow you exactly, but you can use which() with any
logica
On Sep 5, 2010, at 10:53 AM, Donald Braman wrote:
Does anyone know of a package (or workaround) for fitting a dirichlet
distribution by maximum likelihood?
Searching:
RSiteSearch("fitting dirichlet distribution maximum likelihood")
(Rapidly) produces this candidate:
http://finzi.psych.upen
Hi Rajesh,
This will work, unfortunately it seems like lapply() drops the names
before it passes each element of the list which lead to my clumsy work
around using which(). I'm sure there are other ways.
dat <- list(INT = c("1","2","3"),
NUM = c("2.34","4.56","6.78"),
INT
On Sep 5, 2010, at 10:47 AM, rajesh j wrote:
I'm sorry. you seem to have misunderstood my data representation for
input.
Because you did not follow the Posting Guide's advice regarding
producing examples using valid R code. You give me text; I work on text.
#-
PLEASE do read the
Does anyone know of a package (or workaround) for fitting a dirichlet
distribution by maximum likelihood?
(I'm looking for something like this: http://repec.org/bocode/d/dirifit.html,
that allows for both dependent variables summing to 1 & predictive variables
of any sort.)
Don
--
Donald Brama
I'm sorry. you seem to have misunderstood my data representation for input.
Here's what I have
c<-list(INT=c("1","2","3"),NUM=c("2.34","4.56","6.78"))
I need
c<-list(INT=c(1,2,3),NUM=c(2.34,4.56,6.78))
On Sun, Sep 5, 2010 at 7:57 PM, David Winsemius wrote:
> So there is one item per line and the
On Sun, 5 Sep 2010, st...@wittongilbert.free-online.co.uk wrote:
David Winsemius wrote:
1. is glm the right thing to use before I waste my time
Yes, but if your outcome variable is binomial then the family argument
should be "binomial". (And if you thought it should be poisson,
then why
So there is one item per line and the task is to recognize the strings
"INT" and "NUM" and create variables with numeric type and
INT=c(2,3,4)
NUM=c(2.37, 4.56) # ???
I worry that is not a full description of the task if there be
more than just two variable names and if all the INTs ha
Dear Experts,
I need to include the repeated structure in our data set object,
recall.sums.df, before using gls function.
Thus I used groupedData.
But I encountered error messages which may mean '*' is not not meaningful
factor.
Please let me know what I have to do.
Thanks,
Jeong
> recall.sums.
The string vector actually comes as a part of a list, and the vector is
named "int", and the numbers are strings. I then have to make it a vector
that is still called "int" and has 4,5,6 etc. the types are either integer
or numeric. The number of items in the vector is unknown.
here's an example,
Calum-4 wrote:
>
> Hi I know asking which test to use is frowned upon on this list... so
> please do read on for at least a couple on sentences...
>
> I have some multivariate data slit as follows
>
> Tumour Site (one of 5 categories) #
> Chemo Schedule (one of 3 cats) ##
> Cycle (one of 3 ca
On Sep 5, 2010, at 9:22 AM, rajesh j wrote:
> for e.g., I get the following as a string vector
> "int" "4" "5" "6"
> after reading the first element, I have to convert this to a integer
> vector
But what is the right answer? And what number of items are possble
per line? And what are the o
for e.g., I get the following as a string vector
"int" "4" "5" "6"
after reading the first element, I have to convert this to a integer vector
On Sun, Sep 5, 2010 at 6:44 PM, David Winsemius wrote:
>
> On Sep 5, 2010, at 8:48 AM, rajesh j wrote:
>
> Hi,
>>
>> Is it possible to convert a string v
On Sep 5, 2010, at 8:48 AM, rajesh j wrote:
Hi,
Is it possible to convert a string vector to integer or numeric
vector? In
my situation I receive data in a string vector and have to convert
it based
on a given type.
Can you give an example? I don't understand either what sort of
conve
See
?numeric
?integer
and possible the colClasses argument of
read.table
-Ista
On Sun, Sep 5, 2010 at 12:48 PM, rajesh j wrote:
> Hi,
>
> Is it possible to convert a string vector to integer or numeric vector? In
> my situation I receive data in a string vector and have to convert it based
> on
Hi,
Is it possible to convert a string vector to integer or numeric vector? In
my situation I receive data in a string vector and have to convert it based
on a given type.
--
Rajesh.J
[[alternative HTML version deleted]]
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On Sep 5, 2010, at 6:06 AM, st...@wittongilbert.free-online.co.uk wrote:
David Winsemius wrote:
1. is glm the right thing to use before I waste my time
Yes, but if your outcome variable is binomial then the family
argument should be "binomial". (And if you thought it should
be pois
Thanks for the replies! Obviously I must have used to wrong search
terms - sorry.
@greg: I care about the levels after the subset, because if they are
not dropped, then they still appear in the subsequent heatmap I make
with ggplot (with my read data-set of course). Admittedly I am quite
green, a
Another way that I like is reshape::melt.list() because it keeps track
of the name of the original data.frames,
l = replicate(1e4, data.frame(x=rnorm(100),y=rnorm(100)), simplify=FALSE)
system.time(a <- rbind.fill(l))
# user system elapsed
# 2.482 0.111 2.597
system.time(b <- melt(l,id=1:2)
David,
your example clarify me the use of sink.
I really appreciate your help
Evgenia
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Thanks for the fast answers!
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David Winsemius wrote:
1. is glm the right thing to use before I waste my time
Yes, but if your outcome variable is binomial then the family argument
should be "binomial". (And if you thought it should be poisson,
then why below did you use gaussian???
Used gaussian below because it was
Thanks Ista!
Indeed this is the workaround I had found, but not very practical as it
fails on univariate series, so you should know before which class is the
object... I feel a head.mts would make sense, or did I misunderstand
something?
Thanks!
Matthieu
Le 04. 09. 10 16:09, Ista Zahn a éc
ya, unfortunately k was actually a dataframe and not a matrix when it was
returned
from the function, which explains why I got unexpected results
On Sun, Sep 5, 2010 at 12:21 AM, wrote:
> Is this all you want?
>
> > j <- matrix(nrow=10,ncol=10)
> > k <- matrix(seq(1:50), ncol=10)
> > row.names(
weird, I tried that but it didnt appear to work.. hmm. Thanks I try it again
On Sun, Sep 5, 2010 at 12:21 AM, wrote:
> Is this all you want?
>
> > j <- matrix(nrow=10,ncol=10)
> > k <- matrix(seq(1:50), ncol=10)
> > row.names(k) <- seq(2,10,by=2)
> >
> > row.names(j) <- 1:10
> > j[row.names(k),
Is this all you want?
> j <- matrix(nrow=10,ncol=10)
> k <- matrix(seq(1:50), ncol=10)
> row.names(k) <- seq(2,10,by=2)
>
> row.names(j) <- 1:10
> j[row.names(k), ] <- k
>
> j
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
1NA NA NA NA NA NA NA NA NANA
2 16
j[as.numeric(rownames(k)), ] <- k
> j
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] NA NA NA NA NA NA NA NA NANA
[2,]16 11 16 21 26 31 36 4146
[3,] NA NA NA NA NA NA NA NA NANA
[4,]27 12 17 22
j<-matrix(nrow=10,ncol=10)
k<-matrix(seq(1:50), ncol=10)
row.names(k) <- seq(2,10,by=2)
j
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] NA NA NA NA NA NA NA NA NANA
[2,] NA NA NA NA NA NA NA NA NANA
[3,] NA NA NA NA NA
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