Dear Arun,
Prof. Rob Hyndman's forecast package has automated arima.
You have to install 'forecast' library for that.
in linux go to r-prompt and type install.packages().
Hope this helps
nuncio
On Mon, Jun 13, 2011 at 12:10 PM, siddharth arun wrote:
> How we can call auto.arima in
How we can call auto.arima in R.
Is there any cran package we need to install for this function?
--
Siddharth Arun,
4th Year Undergraduate student
Industrial Engineering and Management,
IIT Kharagpur
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On 06/10/2011 02:29 PM, Richard M. Heiberger wrote:
> I am interested in running R commands asynchronously...
You can do this with my Rdsm package on CRAN.
Set up 2 Rdsm clients, which I'll call A and B. Use client A as you
main R session, where you do most of your work, but start your
asynchro
I didn't include code because I wasn't asking for help; I was merely
commenting that I had seen the same thing happen. But thank you for
mentioning dev.off, which I had not heard of before.
On Mon, Jun 13, 2011 at 3:55 PM, Prof Brian Ripley
wrote:
> Usually this happens when you forget to run de
Usually this happens when you forget to run dev.off(), as in that
example.
But we don't have the
commented, minimal, self-contained, reproducible code.
the posting guide and the footer of every R message asks for.
On Sun, 12 Jun 2011, Mark Seeto wrote:
Raptorista wrote:
Now, the graph
On 13/06/11 15:19, pdb wrote:
Is there an r function that will be able to identify the computer the code is
running on?
I have some common code that I run on several computers and each has a
database with a different server name - although the content is identical.
I need to set thisServer dep
Hi Weidong,
thank you very much. It really works fine.
Robert
2011/6/12 Weidong Gu :
> this may work.
> X<-data.frame(sapply(X,function(x) as.factor(x)))
> reg3=lm(Y~.,data=X)
> dummy.coef(reg3)
>
> Weidong Gu
>
> On Sun, Jun 12, 2011 at 4:55 PM, Robert Ruser wrote:
>> Hi,
>> but I want to get t
Not exactly R, but how about
> pcname <- system('uname -n',intern=T)
Tom
On Sun, Jun 12, 2011 at 11:19 PM, pdb wrote:
>
> Is there an r function that will be able to identify the computer the code is
> running on?
>
> I have some common code that I run on several computers and each has a
> data
Raptorista wrote:
>
> Now, the graph that appears is very nice: indeed it has a title, two
> axes with their labels and all the rest;
> but when I give commands
>
> postscript(file="plot.eps", onefile=FALSE)
> qqnorm (col)
>
> to save the graph to a file "plot.eps" to include it in a TeX, the f
?paste
something like...
paste (group, i, sep="_")
jiliguala wrote:
>
> Hello R users,
>
> I am new to R and am having difficulty with the output name of my "for"
> loops.
>
> here is the problem:
>
>
> for (i in c(1:100))
> {
> the name of the groups <- which(k1$cluster==i)
> }
>
> how
?paste
something like...
paste ("group", i, sep="_")
jiliguala wrote:
>
> Hello R users,
>
> I am new to R and am having difficulty with the output name of my "for"
> loops.
>
> here is the problem:
>
>
> for (i in c(1:100))
> {
> the name of the groups <- which(k1$cluster==i)
> }
>
> ho
Is there an r function that will be able to identify the computer the code is
running on?
I have some common code that I run on several computers and each has a
database with a different server name - although the content is identical.
I need to set thisServer depending on which machine the code
Hello R Community,
I'm continuing to work through logistic regression (thanks for all the help on
score test) and have come up against a new opposition.
I'm trying to compute Somers Dyx as some suggest this is the preferred method
to Somers Dxy (Demaris, 1992). I have searchered the [R] arc
Dear Sigrid,
At 12:46 PM -0700 6/12/11, Sigrid wrote:
I am new and self taught in R, so please bear with me.
I want to create two scatter plots side by side. The data set includes
measurements from two different countries with 7 treatments over a timeline
(x-axis).
Problem 1
I want to have eac
Thank you for very informative answers. This is great help. Peter, thanks
for correcting the model. That was exactly what I meant - apologies for the
typo.
I was able to run the analysis on some simulated data and the subset of data
that I had posted earlier. However, when I apply the analysis to
Lists are recursive and heterogenous in R. Just assign the values to elements
in lists and assign those lists to elements in other lists to build your tree.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#
Thanks very much. I did it using the do.call and compared it to my loop
method and the results were exactly the same. In future i think i will use
this function - a lot less typing and time consuming!
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Hi:
If you use RStudio, then you can use its manipulate package to figure
out starting values for the model visually through sliders. Walmes
Zeviani posted the template of how to do this last week; I've just
adapted it for the models under consideration. It's interesting to
play with this because
this may work.
X<-data.frame(sapply(X,function(x) as.factor(x)))
reg3=lm(Y~.,data=X)
dummy.coef(reg3)
Weidong Gu
On Sun, Jun 12, 2011 at 4:55 PM, Robert Ruser wrote:
> Hi,
> but I want to get the coefficients for every variables from x1 to x5.
> (x1 was an example)
>
> Robert
>
> 2011/6/12 Jorge
Hi!
Am 12.06.2011 21:43, schrieb bstudent:
Error in solve.default(Reduce("+", A2)) :
System ist für den Rechner singulär: reziproke Konditionszahl =
4.08048e-22
Error in solve.default(Reduce("+", A2)) :
System is singulary for the computer: reciprocal number of conditions =
4.08048e-22
Hello Everyone,
I am new to R and would like to create a quad tree in R. However the problem
is that I don't think R has pointers. Is there any way to create a tree in
R?
Thanks
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On Sun, 12 Jun 2011, peter dalgaard wrote:
On Jun 12, 2011, at 18:57 , Diviya Smith wrote:
Hello there,
I am trying to fit an exponential fit using Least squares to some data.
#data
x <- c(1 ,10, 20, 30, 40, 50, 60, 70, 80, 90, 100)
y <- c(0.033823, 0.014779, 0.004698, 0.001584,
On Jun 12, 2011, at 18:57 , Diviya Smith wrote:
> Hello there,
>
> I am trying to fit an exponential fit using Least squares to some data.
>
> #data
> x <- c(1 ,10, 20, 30, 40, 50, 60, 70, 80, 90, 100)
> y <- c(0.033823, 0.014779, 0.004698, 0.001584, -0.002017, -0.003436,
> -0.06
On Sun, 12 Jun 2011, Jorge Ivan Velez wrote:
Hi Diviya,
Take a look at the lrtest function in the lmtest package:
install.packages('lmtest)
require(lmtest)
?lrtest
Yes, when you have to nls() fits, say m1 and m2, you can do
lrtest(m1, m2)
However, I don't think that both m1 and m2 can be i
On Jun 12, 2011, at 20:25 , Diviya Smith wrote:
> Hello there,
>
> I am trying to use R function NLS to analyze my data and one of the examples
> in the documentation is -
>
> ## the nls() internal cheap guess for starting values can be sufficient:
>
> x <- -(1:100)/10
> y <- 100 + 10 * exp(x
Hi,
but I want to get the coefficients for every variables from x1 to x5.
(x1 was an example)
Robert
2011/6/12 Jorge Ivan Velez :
> Hi Robert,
>
> Try this:
> reg2 <- lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) +
> factor(x5) - 1, data = X )
> cof(ref2)
> HTH,
> Jorge
>
> On Sun, J
Hmm, did you shut the device down afterward (i.e., call dev.off() )?
I do not have any logic why that would induce the behavior you say you
are getting, but this works just fine for me:
postscript("tmp.eps", onefile = TRUE)
qqnorm(rnorm(20))
dev.off()
and creates the attached file (possibly not a
Hi Robert,
Try this:
reg2 <- lm( Y ~ factor(x1) + factor(x2) + factor(x3) + factor(x4) +
factor(x5) - 1, data = X )
cof(ref2)
HTH,
Jorge
On Sun, Jun 12, 2011 at 4:40 PM, Robert Ruser <> wrote:
> Prof. Ripley, thank you very much for the answer but wanted to get
> something else. There is an
Hi Nanami,
I am not sure exactly what you mean by "fits the peaks"...are any of
these plots what you want?
x <- 1:20
y <- c(19.4, 17.9, 8.1, 11.3, 7.8, 8, 5, 1.7, 3.9, 5.4, 7.5, 5.4,
4.7, 5, 4.9, 3.5, 2.9, 2.4, 1.4, 1.7)
## Find peaks
index <- which(c(NA, diff(sign(diff(y == -2)
dev.new()
la
Prof. Ripley, thank you very much for the answer but wanted to get
something else. There is an example and an explanation:
options(contrasts=c("contr.sum","contr.poly")) # contr.sum uses ‘sum
to zero contrasts’
Y <- c(6,3,5,2,3,1,1,6,6,6,7,4,1,6,6,6,6,1)
X <- structure(list(x1 = c(2L, 3L, 1L, 3L,
I am new and self taught in R, so please bear with me.
I want to create two scatter plots side by side. The data set includes
measurements from two different countries with 7 treatments over a timeline
(x-axis).
Problem 1
I want to have each plot to include the data from one of the countries wit
Hi Diviya,
Take a look at the lrtest function in the lmtest package:
install.packages('lmtest)
require(lmtest)
?lrtest
HTH,
Jorge
On Sun, Jun 12, 2011 at 1:16 PM, Diviya Smith <> wrote:
> Hello there,
>
> I want to perform a likelihood ratio test to check if a single exponential
> or a sum of
Hello,
although I searched for a solution related to my problem I didn´t find one,
yet. My skills in R aren´t very large, however.
For my Diploma thesis I need to run a GMM estimation on a dynamic panel
model using the "pgmm" - function in the plm-Package.
The model I want to estimate is: "Y(t) =
I'm using plotnetwork {spaa} in order to get a correlation network plot of my
data (e.g.: http://www.oga-lab.net/RGM2/func.php?rd_id=spaa:plotnetwork).
By default, 'interval' argument indicate the number of intervals by which the
range of input data is partitioned (the number of partitions betw
Hello there,
I am trying to use R function NLS to analyze my data and one of the examples
in the documentation is -
## the nls() internal cheap guess for starting values can be sufficient:
x <- -(1:100)/10
y <- 100 + 10 * exp(x / 2) + rnorm(x)/10
nlmod <- nls(y ~ Const + A * exp(B * x), trace=T
Hello there,
I want to perform a likelihood ratio test to check if a single exponential
or a sum of 2 exponentials provides the best fit to my data. I am new to R
programming and I am not sure if there is a direct function for doing this
and whats the best way to go about it?
#data
x <- c(1 ,10,
Hello everybody! This is my first mail so I'll write a couple of lines
of self-introduction.
My name is Massimiliano, I'm from Italy and I'm studying Mathematical
Engineering.
I started using R in my Statistics course and have to use it to make a
project which I'll discuss at the end of the course.
Hello there,
I am trying to fit an exponential fit using Least squares to some data.
#data
x <- c(1 ,10, 20, 30, 40, 50, 60, 70, 80, 90, 100)
y <- c(0.033823, 0.014779, 0.004698, 0.001584, -0.002017, -0.003436,
-0.06, -0.004626, -0.004626, -0.004626, -0.004626)
sub <- data.frame(x
Hello R users,
I am new to R and am having difficulty with the output name of my "for"
loops.
here is the problem:
for (i in c(1:100))
{
the name of the groups <- which(k1$cluster==i)
}
how can it automatically generate the name for 100 cluster? what should i
put in the bold letter place.
hello:
Does anyone know any R function which handles statisitcal model when the
independent variable is censored? I know survival package does the analysis
for censored dependnent variable.
thanks
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Hi all,
I have been struggling with some plotting issues (maybe not that
difficult but I am stuck).
If I have 2 vectors : 1st with average values of the columns of a data
frame, which, looks similar to : v:
(1.4,3,3,3,4.5,0,0,0,0,0,0,0,25.5,6,6,4.2)
2nd with t
Hi everyone,
I am given two vectors - for example -
x= c(0:20)
> x
[1] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
and
> y
[1] 19.4 17.9 8.1 11.3 7.8 8.0 5.0 1.7 3.9 5.4 7.5 5.4 4.7 5.0
4.9
[16] 3.5 2.9 2.4 1.4 1.7
> plot(x,y,xlim=range(x),ylim=range(y))
?dummy.coef
(NB: 'R' does as you tell it, and if you ask for the default contrasts
you get coefficients a2 and a3, not a1 and a2. So perhaps you did
something else and failed to tell us? And see the comment in
?dummy.coef about treatment contrasts.)
On Sun, 12 Jun 2011, Robert Ruser wrote
Prof Ripley
Thanks for your message - you are absolutely right. I should have provided
more information.
Meanwhile, I found out that loading an R workspace prevented parApply from
working properly. The workspace contains many objects and I could not
figure out which one was responsible for causin
Dear R Users,
Using lm() function with categorical variable R use contrasts. Let
assume that I have one X independent variable with 3-levels. Because R
estimate only 2 parameters ( e.g. a1, a2) the coef function returns
only 2 estimators. Is there any function or trick to get another a3
values. I
Yes, the basic problem is that you forgot to melt the data before
trying to cast it.
Hadley
On Thursday, June 9, 2011, misterbray wrote:
> Dennis, doing some more research, and it seems you actually can include the
> ... term directly in the formula: cf. page 8 of
> http://www.had.co.nz/reshape/
On Sun, 12 Jun 2011, Unger, Kristian, Dr. wrote:
Hi
I try parallelising some code using the snow package and the following lines:
cl <- makeSOCKcluster(8)
pfunc <- function (x) (if(x <= (-th)) 1 else 0) ###correlation coefficient
clusterExport(cl,c("pfunc","th"))
cor.c.f <- parApply(cl,tms,c(
for categorical independent variables, regression models in R will
generate dummy indicators based on your setting of contrasts (default
contr.treatment). Use model.matrix(your model) to see how R does this
internally.
Weidong Gu
On Sun, Jun 12, 2011 at 2:38 AM, setlist wrote:
> Hello, I wanted
Hi,
I still haven't found a solution for this problem. Is there a way in which I
can slice the objects in c based on the info in h? Thanks a lot!
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Sent from the R hel
Please see my previous note and please read the documentation. Also note the
difference between a character string literal in quotes and the name of a
vector, which is usually not quoted.
Frank
Chalikias George wrote:
>
> The actual dataframe that I have imported on R is
>
>Death adp
> 1
Hi
I try parallelising some code using the snow package and the following lines:
cl <- makeSOCKcluster(8)
pfunc <- function (x) (if(x <= (-th)) 1 else 0) ###correlation coefficient
clusterExport(cl,c("pfunc","th"))
cor.c.f <- parApply(cl,tms,c(1,2),FUN=pfunc)
The parApply results in the error m
On Jun 12, 2011, at 12:50 , eldor ado wrote:
> dear r community...
>
> it loks like i won't be able to reach homogenity of variance for my
> dataset, so i end up with welch anova instead of regular anova.
> documentation on this test is rather scarce, so maybe someone here can
> enlighten me a b
A good start might be to go to http://rseek.org and search for
partial correlation to get information on the packages and functions
that offer it.
>From there, if you have specific questions, describing your data,
what you've tried, what results you've gotten, and what results
you expect would be
dear r community...
it loks like i won't be able to reach homogenity of variance for my
dataset, so i end up with welch anova instead of regular anova.
documentation on this test is rather scarce, so maybe someone here can
enlighten me a bit:
- do i understand that no two-way implementation of th
how can I compute partial correlation
there's four variables;
income, age, educational level, race.
income is indenpendent variable
and race is nominal variable
I want to calculate partial correlation or semi-partial correlation between
income & race
help me out... thanks.
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Hello, I wanted to do the multiple regression on categorical predictor data
there's variable x1,x2,x3 and x3 is categorical one.
so i just used as.factor(x3) and then ran multiple regression
is it a good way to do the multiple regression on categorical predictor
data?
and how can I interpret the es
The actual dataframe that I have imported on R is
Death adp
1 0 58.00
2 1 18.70
3 0 21.75
4 1 25.35
5 0 20.55
6 1 28.05
7 0 50.15
8 1 31.25
9 1 32.75
10 1 28.95
11 1 15.10
12 0 45.05
13 1 19.95
14 0 32.95
15 0 22.60
16 0
Dear Lister,
I have collected data in 6 geographical areas on prevalence of a
parasite in humans and in foxes. The results are expressed as a number
of positive or negative cases in human and foxes in the following
data.frame:
Pvtab <-
structure(list(posHum = c(3, 5, 3, 17, 0, 4), negHum = c
On Fri, Jun 10, 2011 at 10:15:36PM +0200, Kairavi Bhakta wrote:
> Thanks for your answer. The reason I want the data to be uniform: It's the
> first step in a machine learning project I am working on. If I know the data
> isn't uniformly distributed, then this means there is probably something
> wr
On Jun 12, 2011, at 03:36 , Thomas Lumley wrote:
> On Sun, Jun 12, 2011 at 12:44 PM, Tiago Pereira
> wrote:
>
>> The test I am working on has an asymptotic 0.5*chi2(1)+0.5*chi2(2)
>> distribution, where numbers inside parenthesis stand for the degrees of
>> freedom. Is is possible to compute qu
On 12-Jun-11 01:36:00, Thomas Lumley wrote:
> On Sun, Jun 12, 2011 at 12:44 PM, Tiago Pereira
> wrote:
>
>> The test I am working on has an asymptotic 0.5*chi2(1)+0.5*chi2(2)
>> distribution, where numbers inside parenthesis stand for the
>> degrees of freedom. Is is possible to compute quickly i
On Jun 12, 2011, at 07:54 ,
wrote:
> The score test looks at the effect of adding extra columns to the model
> matrix. The function glm.scoretest takes the fitted model object as the
> first argument and the extra column, or columns, as the second argument.
> Your x2 argument has length o
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