i tried
pkgs <- c("pkg1", "pkg2", "pkg3")
install.packages(pkgs, repos = NULL,
+ contriburl =
"file:///C:/R/packages/bin/windows/contrib/2.11",
+ dependencies = TRUE)
i have put the zip files in C:/R/packages/bin/windows/contrib/2.11
*** and extracted the package
Gustaf Rydevik wrote:
> Oh, I forgot to mention that the workaround of using as.double (or
> as.numeric) works fine, and I've done that.
> It's just that it can take quite a while (as in several hours) to
> figure out that the reason for the error is that you have to force
> difftime objects to be
Hi all!
This is an R workspace memory processing question.
There is a method which from the R will control 10GB data at 500MB units?
my work environment :
R version : 2.11.1
OS : WinXP Pro sp3
Thanks and best regards.
Park, Young-Ju
from Korea.
[1][rKWLz
Hej Gregory,
Thanks for your reply and sorry for late reply, actually I was busy with
some other stuff but It helped me alot.
Thanks again sir.
Thanks to both of you.
Thank you
Jeet
On Wed, Jun 16, 2010 at 2:27 AM, Greg Snow wrote:
> OK, I did not see the website before, here is some code to
2010/6/23 Seth W Bigelow :
>> "with(alt.data[[which.packet()[1]]]" statements. It all graphs out
>> beautifully, but none of the usual tricks to get the proper legend to
> plot
>> are working, i.e., using auto.key, key, etc.
You haven't told us exactly what you mean by "proper legend", but it
is g
On Tue, Jun 22, 2010 at 7:46 PM, Yi wrote:
> Hi, folks,
>
> As I understand, Least-squares Estimate (second-moment assumption) and the
> Method of Maximum Likelihood (full distribtuion assumption) are used for
> linear regression.
>
> I do >?lm, but the help file does not tell me the model employe
Hi Tim,
I had solved the problem myself.
I found that the problem was caused by VMware. After re-installing
vmware-tools, the problem disappeared...
Furthermore, the issue has nothing to do with xz software. I had uninstalled
xz on my system and the "make check" still passed all test.
If you a
Say X is your data matrix with the variable, then you could do :
X <- matrix(rnorm(2100),300,7)
S <- var(X)
dist <- as.dist(
apply(X,1,function(i){
mahalanobis(X,i,S)
}
)
)
Cheers
Joris
On Tue, Jun 22, 2010 at 11:41 PM, yoo hoo wrote:
> I am a new R user. i have
There is:
1) read the help files
2) read the vignette (see ?vignette )
2) look up at www.rseek.org
3) check google
4) look at the references mentioned in the help files (or do first if
you're not familiar with the method)
5) ask here
If still nothing :
use another function/method.
Cheers
Joris
O
Apologize for not being clearer earlier. I would like to ask again. Thank
Joris and Markleeds for response.
Two examples:
1. Function 'var'. In R, it is the sum of square divided by (n-1) but not by
n. (I know this in R class)
2. Function 'lm'. In R, it is the residual sum of square divied by (n
Try this:
x <- ftable(a.df$DATE, paste(a.df$GENDER, a.df$Co, sep = "-"))
as.data.frame(`class<-`(`dimnames<-`(x, list(attr(x, 'row.vars')[[1]],
attr(x, 'col.vars')[[1]])), 'matrix'))
On Tue, Jun 22, 2010 at 6:50 PM, Erin Hodgess wrote:
> Dear R People:
>
> I have the following data set with the
It's normally always specified, unless in the case of least squares
linear regression (lm), where it is considered obvious.
Cheers
Joris
On Wed, Jun 23, 2010 at 1:46 AM, Yi wrote:
> Hi, folks,
>
> As I understand, Least-squares Estimate (second-moment assumption) and the
> Method of Maximum Like
anyone can give me a detailed instuction about how to install IPA in R and do
vowel plot? Thx.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the po
Dear All,
I´ve been trying to find an option to scatter3D from rcmdr to remove the
individual points from the plots but to no help so far. Removing the residuals
is easy, but I cannot find a similar point option. Is there such an option that
can be set to FALSE?
Best,
//M
__
I am a new R user. i have a question about Mahalanobis distance.actually i
have 300 rows and 7 columns. columns are different measurements, 300 rows are
genes. since genes can
classify into 4 categories. i used dist() with euclidean distance and cmdscale
to do MDS plot. but find out Mahalanobi
Here is the code for my recent query:
The first part just sets up the data. It's the last part
titled "individual depth graph" where the actual graphics code comes in
Thx
--Seth
library(lattice)
setwd("c:/sethdocs/3 Light and PLAS txs/8 Figures/Fg 6 Soil wetness")
d <- read.table("2007 data.t
On 06/20/10 02:31 PM, Muenchen, Robert A (Bob) wrote:
Hi All,
I've been fiddling around with various ways to estimate the popularity
of R, SAS, SPSS, Stata, JMP, Minitab, Statistica, Systat, BMDP, S-PLUS,
R-PLUS and Revolution R. It's not an easy task. You can see what I've
come up with so far a
Hi, folks,
As I understand, Least-squares Estimate (second-moment assumption) and the
Method of Maximum Likelihood (full distribtuion assumption) are used for
linear regression.
I do >?lm, but the help file does not tell me the model employed in R. But
in the book 'Introductory Statistics with R'
Rachel -
Not exactly a reproducible example, but maybe
t1 <- subset(nih2009, ic_name %in% levels(nih2009$ic_name)[c(27,51)])
- Phil Spector
Statistical Computing Facility
Hi All,
I have a factor variable with 52 levels -with long, annoying names. I want to
keep only rows with some variables. I can do this using this code:
test1 <- subset(nih2009,ic_name %in% c('NATIONAL EYE INSTITUTE','Veterans
Affairs'))
dim(test1)
[1] 2396 38
But this doesn't work:
t1 <
I have a moderately complex graph with three panels. There are data points
plotted, and fitted lines are added using a panel function, which includes
"with(alt.data[[which.packet()[1]]]" statements. It all graphs out
beautifully, but none of the usual tricks to get the proper legend to plot
ar
Erin -
Perhaps this reproducible example will help:
df = data.frame(a=sample(1:20,1,replace=TRUE),
b=sample(c('M','F'),1,replace=TRUE),
c=sample(LETTERS[1:8],1,replace=TRUE))
tt = as.data.frame(table(df))
tt$bc = paste(tt$b,tt$c,sep='.')
tt$b=NULL
tt
Regarding Newton-Raphson, I would add that there are certain kinds
of function for which the method *cannot* work.
N-R is essentially a root-finding technique, and when used for
finding a minimum (or maximum) it is applied to finding the root
of the deriviative.
As an example of a class of functi
Dear R People:
I have the following data set with the columns DATE, GENDER, and Co.
Co has 8 possible options.
> a.df[1:10,]
DATE GENDER Co
1 2009-04-16 F Rash
2 2009-04-16 F Other
3 2009-04-16 M Botulinic
4 2009-04-16 M
On Sun, Jun 20, 2010 at 12:46 PM, Ivan Calandra
wrote:
> Bob,
>
> I have no idea whether it is realistic, but if you look for the papers that
> used R or SAS (or anything), you might get better results by searching for
> the way R and SAS are cited.
I wonder what the effects of different format
Jim, I just checked your help out and everything works fine except the axis
labels, I should probably have been more specific. I had already been able
to draw the rownames() as labels, what I was looking for was the capacity to
define major ticks, and hence draw only every nth rowname on the grap
Try this:
library(lattice)
bwplot(values ~ TimePeriod | ind, cbind(stack(my.data), TimePeriod =
my.data$TimePeriod))
On Tue, Jun 22, 2010 at 1:45 PM, Shawn Morrison <
shawn.morri...@dryasresearch.com> wrote:
> Hi all,
>
> I have a dataset with several variables, each of which is a separate
> col
On Tue, Jun 22, 2010 at 1:00 AM, Samuel Okoye wrote:
> Hi,
>
> I have the following data
>
> data1 <- data.frame(count = c(0,1,1,2,4,5,13,16,14), weeks = 1:9,
> treat=c(rep("1mg",3),rep("5mg",3),rep("10mg",3)))
> and I am using
>
> library(splines)
>
> to fit
>
> glm.m <- glm(c
On Tue, Jun 22, 2010 at 9:30 AM, Michael Friendly wrote:
> Consider the following plot that shows separate regression lines ~ age for
> each subject in the Pothoff-Roy Orthodont data,
> with separate panels by Sex:
>
> library(nlme)
> #plot(Orthodont)
> xyplot(distance ~ age|Sex, data=Orthodont, t
John,
thanks for the comments, very useful.
Just three short additions specific to the ZINB case:
1. zeroinfl() doesn't uses BFGS (by default) and not Newton-Raphson
because we have analytical gradients but not an analytical Hessian.
2. If the default starting values do not work, zeroin
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Ivan Calandra
Sent: Sunday, June 20, 2010 3:47 PM
To: r-help@r-project.org
Subject: Re: [R] Popularity of R, SAS, SPSS, Stata...
Bob,
I have no idea whether it is realistic, but if
Hello,
I was going to use lmer() on this data, but it seemed easier -- and
more importantly, more meaningful -- to just analyze smaller sections
of it individually. I'd like to ask for help to see if I'm analyzing
the separate parts correctly. Each part is the same, and they all
look like this:
Excellent, thanks. This solved it.
ottar
On 22 June 2010 11:39, Hadley Wickham wrote:
> Hi Ottar,
>
> It's a bug in the code that automatically figures out the axis ticks.
> I have some code to fix it, but it's waiting for a release of another
> package. In the meantime, you can work aroun
I have not included the previous postings because they came out very strangely on my mail
reader. However, the question concerned the choice of minimizer for the zeroinfl()
function, which apparently allows any of the current 6 methods of optim() for this
purpose. The original poster wanted to u
Downey, Patrick wrote:
Hello,
I'm running a very long for loop that usually takes hours. For my own piece
of mind, it would be nice if I could check periodically and see which
iteration the loop is on. A line of code that told R to print the iteration
number every 100 or 200 iterations would be
If you want to test the null that all datasets come from the same distribution,
then I would probably do this as a permutation test. Find the largest K-S
distance between 2 groups, then randomly permute the data into a new set of k
groups (with same sample sizes) and find the largest K-S distan
Thanks Josh,
I do want to see each plot. I took your code and modified it (below) and
it appears to do what I wanted:
my.data <- data.frame(var1=rnorm(1000), var2=rnorm(1000),
TimePeriod=factor(rep((LETTERS[1:4]), 250)))
str(my.data)
lapply(names(my.data[ , 1:2]), function(y) {
quartz()
bo
Hello Shawn,
Does this do what you want? I'm assuming you want to look at each
plot, so I added a call to par().
###
my.data <- data.frame(var1=rnorm(1000), var2=rnorm(1000),
TimePeriod=factor(rep((LETTERS[1:4]), 250)))
str(my.data)
lapply(names(my.data[ , 1:2]), function(y) {
old.par <-
Many thanks Phil, that does help.
If I could ask a follow-up, how do I put each plot in its own device
window? Right now, all I get is the boxplot for var2 (var1 gets
overwritten?). I tried putting quartz() before the boxplot command but
got an error message.
Cheers,
Shawn
On 22/06/10 11:13
Hi,
I have a number of multilevel models and I would like some Latex outputs of
them. I usually use the "apsrtable" package, but it does not accept "lme"
outputs. Neither does the "mtable" function in the "memisc" package. Is
there any good alternative that I am missing?
Thanks, Jonas
See the aggregate= argument to read.zoo which will aggregate them as
it reads them in like this:
> library(zoo)
> Lines <- "02/10/1995 4925.5
+ 30/10/1995 4915.9
+ 23/01/1996 4963.5
+ 23/01/1996 5009.2
+ 04/03/1996 5031.9 # here
+ 04/03/1996 5006.5 # here
+ 03/04/19
Hello,
I am curious if anyone has had any success with finding a R version of a
k-sample Kolmogorov-Smirnov test. Most of the references that I have able to
find on this are fairly old and I am wondering if this type of analysis has
fallen out of favour. If so, how do people tend to compare distr
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
> Sent: Tuesday, June 22, 2010 11:24 AM
> To: johan...@huesing.name
> Cc: r-help@r-project.org
> Subject: Re: [R] Replacing elements of a list over a certain thres
PCA and factor analysis is implemented in the core R distribution, no
extra packages needed. When using princomp, they're in the object.
pr.c <- princomp(USArrests,scale=T)
pr.c$scores # gives you the scores
see ?princomp
When using factanal, you can ask for regression scores or Bartlett
scors
Hello,
I am curious if anyone has had any success with finding a R version of a
k-sample Kolmogorov-Smirnov test. Most of the references that I have able to
find on this are fairly old and I am wondering if this type of analysis has
fallen out of favour. If so, how do people tend to compare di
On 22.06.2010 19:30, Subramanian S wrote:
i am able to install zip files of packages from local zip files using the
file.choose() for the package argument in install.packages() function in R
console and also from the Packages menu>install from local zip files. But
if there are many packages to
On Jun 22, 2010, at 2:14 PM, Johannes Huesing wrote:
Jim Hargreaves [Mon, Jun 21, 2010 at 12:34:01PM
CEST]:
Dear List,
I have a list of length ~1000 filled with numerics. I need to
replace the elements of this list that are above a certain numerical
threshold with the value of the threshold
So the original email message was in error and you meant to type
"is.numeric" in the first line and not "as.numeric"?
(You might also note that sending duplicate messages to r-help and r-
devel is not considered proper etiquette. My reply did not go to r-
devel.)
On Jun 22, 2010, at 2:01 PM
Jim Hargreaves [Mon, Jun 21, 2010 at 12:34:01PM CEST]:
> Dear List,
>
> I have a list of length ~1000 filled with numerics. I need to
> replace the elements of this list that are above a certain numerical
> threshold with the value of the threshold.
>
> e.g
> example=list(1, 2, 3, 4, 5, 6, 7, 8,
On Tue, Jun 22, 2010 at 7:50 PM, David Winsemius wrote:
>
> On Jun 22, 2010, at 1:33 PM, Gustaf Rydevik wrote:
> Cannot help you there, but have you looked at the help page for difftime?
>
> "The as.double method returns the numeric value expressed in the specified
> units. Using units = "auto" m
Hey,
Thanks a lot everyone.
Taking care of the NA's and setting right the type-casting I could apply
lapply.
I have read the documentation for R and it is mentioned that statistical and
mathematical methods could be applied across data frames but when lapply
didn't work (due to the above said error
On Jun 22, 2010, at 1:33 PM, Gustaf Rydevik wrote:
On Thu, Jun 10, 2010 at 3:39 PM, Gustaf Rydevik
wrote:
Hi all,
The recent change in 2.11 that made as.numeric() return false on
difftime-objects broke some of my code that calculated age classes of
individuals using cut(). While this was no
Okay, Subramanian S, try this:
pkgs <- c("pkg1", "pkg2", "pkg3")
install.packages(pkgs, repos = NULL,
contriburl = "file:///c:/r/repos/2.12",
dependencies = TRUE)
where I've got zipfiles in c:/r/repos/2.12.
Comments:
1. Note the 3 slashes after "file:".
2. You
On Thu, Jun 10, 2010 at 3:39 PM, Gustaf Rydevik
wrote:
> Hi all,
>
> The recent change in 2.11 that made as.numeric() return false on
> difftime-objects broke some of my code that calculated age classes of
> individuals using cut(). While this was no big thing to fix for me, it
> might be wise
> t
Oops, I was too fast. I meant :
i <- length(x)- match(unique(index(x)),rev(index(x)))+1
(one has to reverse the indices again)
But in any case, the aggregate-way is indeed far cleaner. Thx for
pointing that out.
Cheers
Joris
On Tue, Jun 22, 2010 at 7:24 PM, Achim Zeileis wrote:
> On Tue, 22 Ju
i am able to install zip files of packages from local zip files using the
file.choose() for the package argument in install.packages() function in R
console and also from the Packages menu>install from local zip files. But
if there are many packages to install (say >15 or 20), then i have to factor
On Tue, 22 Jun 2010, Joris Meys wrote:
Try this :
x.Date <- as.Date("2003-02-01") + c(1, 3, 7, 7, 14) - 1
x <- zoo(1:5, x.Date)
x
2003-02-01 2003-02-03 2003-02-07 2003-02-07 2003-02-14
1 2 3 4 5
i <- match(unique(index(x)),rev(index(x)))
x
One approach is to use the within function:
> within(list(), {a<-1; b<-a})
$b
[1] 1
$a
[1] 1
Now, whether that is simpler or more compact than other options is for you (or
other users) to decide.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcar
Shawn -
Does this example help? (Please don't use cbind when creating
a data frame, since it first creates a matrix, which means everything
must be of the same mode.)
var1 = rnorm(1000)
var2 = rnorm(1000)
TimePeriod = rep((LETTERS[1:4]), 250)
my.data = data.frame(var1,var2,TimePeriod)
lapply
Duncan and I had nearly identical code examples in our replies. This is
becoming a theme. Great minds think alike??? :-)
UTF-8 is backward compatible with ASCII. Hence, if your "|" separated
table contains only ASCII, the UTF-8 encoded file may be
indistinguishable from an ASCII encoded file. Mayb
On Jun 22, 2010, at 12:43 PM, Jim Silverton wrote:
Hello,
Is there anywhere I can find some Bayesian Code for 2 by 2 tables or
even
the non-central hypergeometric distribution in R? Packages would be
helpful
but the actual coding in R is much better.
page 49 of ...
https://home.comcast.
Hello, everyone.
Recently, I started using the 64-bit version of R from the command
line in Mac OS X 10.6.4. Usually, I launch R with the command "R64
--save." I am using R version 2.11.0.
When I install a new package, an X11 window loads, but it takes a
short while to do this. However, the X11
Hi all,
I have a dataset with several variables, each of which is a separate
column. For each variable, I want to produce a boxplot and include the
name of the variable (ie, column name) on each plot.
I have included a sample dataset below. Can someone tell me where I am
going wrong?
Thank
Michael,
Look at the ancova function in the HH package.
## install.packages("HH") ## if not there yet.
library(HH)
library(nlme) ## for the Orthodont data
ancova(distance ~ age*Sex, data=Orthodont)
ancova(distance ~ age+Sex, data=Orthodont)
ancova(distance ~ age, groups=Sex, data=Orthodont)
anc
Hello,
Is there anywhere I can find some Bayesian Code for 2 by 2 tables or even
the non-central hypergeometric distribution in R? Packages would be helpful
but the actual coding in R is much better.
Thanks,
Thanks Jim
[[alternative HTML version deleted]]
Try this :
> x.Date <- as.Date("2003-02-01") + c(1, 3, 7, 7, 14) - 1
> x <- zoo(1:5, x.Date)
> x
2003-02-01 2003-02-03 2003-02-07 2003-02-07 2003-02-14
1 2 3 4 5
> i <- match(unique(index(x)),rev(index(x)))
> x[i]
2003-02-01 2003-02-03 2003-02-07 200
On Tue, Jun 22, 2010 at 8:57 AM, Muenchen, Robert A (Bob)
wrote:
> Interesting! I had no idea there were R-help lists in other languages. I
> don't see it on http://www.r-project.org/mail.html, but then that's in
> English! Is there a list of such sites?
It is listed here:
https://stat.ethz.ch/
Consider the following plot that shows separate regression lines ~ age
for each subject in the Pothoff-Roy Orthodont data,
with separate panels by Sex:
library(nlme)
#plot(Orthodont)
xyplot(distance ~ age|Sex, data=Orthodont, type='r', groups=Subject,
col=gray(.50),
main="Individual linear
Hi people of R forum,
I´m trying to install de "Event" Package, but I can´t find it.
Where I can get it?
Thanks in advance.
Best Regards
Pedro Mota Veiga
--
View this message in context:
http://r.789695.n4.nabble.com/EVENT-PACKAGE-tp2264357p2264357.html
Sent from the R help mailing list archive
Read the posting guide please. What's your data structure? That's
quite important. As I see it, I can easily get a matrix with what you
want by :
x1 <- rep("a","b",each=3)
x2 <- rep("c","d","f",times=2)
x3 <- rep("g",6)
x4 <- rep("h",6)
result <- rbind(x1,x2,x3,x4)
But that's not what you want pro
On 22/06/2010 9:29 AM, dbeest wrote:
Hi All,
I'm having difficulty making a contour plot a would like some help.
A standard contourplot can be made by having an x,y, and some matrix (shape
x*y) with contents:
x = 1:10
y = 1:10
cont <- matrix(runif(100,min=1,max=2),nrow=10,ncol=10)
filled.conto
open the R console.
type:
?install.packages.
press enter.
read.
say "Doh, that's easy."
do what you read.
cheers
Joris
On Tue, Jun 22, 2010 at 5:38 PM, Amy Li wrote:
> Hi I'm a new user. I need to install some new packages. Can someone show me?
>
> Amy
>
> [[alternative HTML version de
I'm probably being dense but I can't work out how to add the second
line:
...
...
I can get the first line using:
saveXML(xml$value(), prefix='',
encoding="ISO-8859-1")
Any ideas?
Thanks.
__
R-help@r-project.org mailing list
https://stat.ethz.
On Tue, 22 Jun 2010, Research wrote:
Hello,
I have a zoo time series read from an excel file which has some dates the
same, such as the following example:
02/10/1995 4925.5
30/10/1995 4915.9
23/01/1996 4963.5
23/01/1996 5009.2
04/03/1996 5031.9 # here
04/03/1996 5
I'm probably being dense but I can't work out how to add the second
line:
...
...
I can get the first line using:
saveXML(xml$value(), prefix='',
encoding="ISO-8859-1")
Any ideas?
Thanks.
__
R-help@r-project.org mailing list
https://stat.ethz.ch
Dear R users
I am analysing data from a group of twins and their siblings. The measures
that we are interested in are all correlated within families, with the
correlations being stronger between twins than between non-twin siblings.
The measures are all calculated from survey answers and by defini
for (i in 1:1)
{
if( (i - 100) %%100 == 0)
print(i);
}
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
__
Dear expeRts,
sorry for such a newbie question -
in PCA/factor analysis e.g. in SPSS it is possible to save scores from
the factors. Is it analogously possible to "save" the implied scores
from the latent variables in a measurement model or structural model
e.g. using the sem or lavaan package
Dear all,
I have a very large amount of dive files that need to be double checked
after the Zoccing process. i would like to load these file sequentially but
automatically into R, plot depth against duration with a simple scatterplot
with labels (DiveNumber) for each individual point and proceed to
Hi All,
I'm having difficulty making a contour plot a would like some help.
A standard contourplot can be made by having an x,y, and some matrix (shape
x*y) with contents:
x = 1:10
y = 1:10
cont <- matrix(runif(100,min=1,max=2),nrow=10,ncol=10)
filled.contour(x,y,cont)
Looks very nice.
Somet
On Jun 22, 2010, at 11:38 AM, Amy Li wrote:
Hi I'm a new user.
Please read the Posting Guide.
I need to install some new packages. Can someone show me?
?install.packages
Or use a GUI package installer, which would then of course depend on
what OS you are using
Amy
[[alternat
Hi Ottar,
It's a bug in the code that automatically figures out the axis ticks.
I have some code to fix it, but it's waiting for a release of another
package. In the meantime, you can work around it by specifying where
you want the breaks:
ggplot(data, aes(time, PU) ) +
geom_line() +
scale_x
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of phani kishan
Sent: Tuesday, June 22, 2010 8:10 AM
To: David Winsemius
Cc: r-help@r-project.org
Subject: Re: [R] Applying forecast fun
Hi I'm a new user. I need to install some new packages. Can someone show me?
Amy
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.
On Jun 22, 2010, at 11:10 AM, phani kishan wrote:
Hey!
On Tue, Jun 22, 2010 at 8:11 PM, David Winsemius > wrote:
On Jun 22, 2010, at 10:32 AM, Henrique Dallazuanna wrote:
Try this:
lapply(DF, auto.arima)
I am getting the following error:
"Error in if (PVAL == min(tablep)) warning("p-value
Hey!
On Tue, Jun 22, 2010 at 8:11 PM, David Winsemius wrote:
>
> On Jun 22, 2010, at 10:32 AM, Henrique Dallazuanna wrote:
>
> Try this:
>>
>> lapply(DF, auto.arima)
>>
> I am getting the following error:
"Error in if (PVAL == min(tablep)) warning("p-value smaller than printed
p-value") else wa
You could try this:
embed(a, k)
On Tue, Jun 22, 2010 at 3:35 AM, Ron Shefi wrote:
>
> Hi,
> I would like to generate a list of all subsequence of length k from an
> array with length n (k < n). The result should be organized in a matrix.
> So the matrix should look like the following whereas e
On Jun 22, 2010, at 4:57 AM, Jeff08 wrote:
Okay, it crosstab seems to work when I clear out a bunch of my
variables,
went from 800 mb Vcol to 100 mb
Can anyone explain how memory works in R, because my lack of
understanding
with memory was clearly the problem.
Making the assumption th
On Jun 22, 2010, at 2:35 AM, Ron Shefi wrote:
Hi,
I would like to generate a list of all subsequence of length k from
an array with length n (k < n). The result should be organized in a
matrix.
So the matrix should look like the following whereas each row is one
of a subsequence of len
On 22/06/2010 10:39 AM, venkata kirankumar wrote:
Hi Murdoch,
first of all thanks allot for your suggestion.
I tried the way you suggested but while opening in notepad and checking its
showing that the file is in ANSI
can any thing else I can do for this
What makes you think notepad is able
On Jun 22, 2010, at 10:32 AM, Henrique Dallazuanna wrote:
Try this:
lapply(DF, auto.arima)
I cannot comment on the relative merits of auto.arima or the forecast
function, but I did notice that earlier today that Gabor Grothendieck
(who, I suspect, could comment knowledgeably) mentioned t
Hi Murdoch,
first of all thanks allot for your suggestion.
I tried the way you suggested but while opening in notepad and checking its
showing that the file is in ANSI
can any thing else I can do for this
Thanks in advance
kiran
On Tue, Jun 22, 2010 at 7:02 PM, Duncan Murdoch wrote:
> On 22/06/2
Hello,
I have a zoo time series read from an excel file which has some dates
the same, such as the following example:
02/10/1995 4925.5
30/10/1995 4915.9
23/01/1996 4963.5
23/01/1996 5009.2
04/03/1996 5031.9 # here
04/03/1996 5006.5 # here
03/04/1996 5069.2
Try this:
lapply(DF, auto.arima)
On Tue, Jun 22, 2010 at 11:26 AM, phani kishan wrote:
> Hey,
> I have a list of 30 odd time-series (products) in columns of a data-frame.
> I want to apply time-series forecasting functions across all the columns of
> the data-frame in order to determine which is
Hey,
I have a list of 30 odd time-series (products) in columns of a data-frame.
I want to apply time-series forecasting functions across all the columns of
the data-frame in order to determine which is the best model to use.
How do I go about this?
Phani
[[alternative HTML version deleted
On Jun 21, 2010, at 10:28 PM, Bernhard Pollner wrote:
Hello,
With one procedure, I get a strangely warped text in the mtext-area
of a quartz-window, while with an other the text comes out OK - see
attached PDF.
(Number of Trials is the x-Axis label, which always comes out OK)
The two pr
Presuming you're talking about Perl, I might be able to help with
specific issues, but read the posting guide before you actually call
upon our help.
http://www.R-project.org/posting-guide.html
One book I definitely can recommend is Phil Spectors Data Manipulation with R
http://www.springer.com/
Yes. When my young son told someone that I was a "doctor", he had to
further explain I was a "number doctor", not a "people doctor".
Barry Rowlingson
Sent by: r-help-boun...@r-project.org
06/22/2010 09:45 AM
To
Marc Schwartz
cc
r-help@r-project.org, Patrick Burns
Subject
Re: [R] Popular
To me R is just another programming language. In fact it seems to share quite a
lot of the characteristics of "fashionable" languages such as python - for
example dynamic typing amonst others. The fact it happens to be good for
statistics and other mathematical stuff is a bonus.
David
--
On Jun 22, 2010, at 7:15 AM, sta...@quo.org wrote:
Hi Barbara,
You could use:
samples = matrix(rep(0,(5000*40)),ncol=40)
for(i in 1:5000) {
samples[i,] = sample(1:40,40)
}
Might also be a good time to use replicate():
samples = matrix(replicate(5000, sample(1:40) ), ncol=40, byrow=TRUE)
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