One thing I noticed is the '2007' in there. Fedora also is stuck at
TexLive 2007 but that is getting old and some CRAN packages do not
work with it (and conversely some do despite errors in vignettes that
are uncoverd in TeXlive 2009).
So this may not apply to Erin, but developers are advised
Hello,
I'm new to R.
In Sapa package there's an example for function SDF like
data <- as.numeric(sunspots)
methods <- c("direct","wosa","multitaper","lag window")
S <- lapply(methods, function(x, data) SDF(data, method=x), data)
x <- attr(S[[1]], "frequency")[-1]
y <- lapply(S,function(x) decibel(
hi people,
I am a newbie to R and I would appreciate it very much if someone can help me. I
am used to use SPSS so I don’t have experience with writing a script. I have a
questionnaire (questions 1-12) and I would like to use the omega reliability.
Question 1-8 is factor 1 and question 9-12 factor
Try this:
> library(zoo)
> t(rollapply(zoo(t(mydata)), 3, na.pad = TRUE, align = "right", FUN = prod))
1 2 3 4 5 6
x.1 NA NA 861 52521 202581 499041
x.2 NA NA 1848 57288 213528 518568
x.3 NA NA 2967 62307 224847 538587
x.4 NA NA 4224 67584 236544 559104
x.5
Try this:
> as.data.frame.table(tapply(DF[,3], DF[2:1], c), responseName = names(DF)[3])
YEAR ID HEIGHT
1 2007 Harry 62
2 2008 Harry 62
3 2007 Jack 70
4 2008 Jack NA
5 2007 James 68
6 2008 James NA
7 2007 Jordan NA
8 2008 Jordan 72
9 2007 M
Hi everyone,
I'm fairly new to R (and this forum!), so first of all apologies if this
question has already been answered (I've tried searching the forum but
haven't come across anything which has solved the problem I'm having).
I have a data frame containing an "independent variable" and a serie
What is a "good" way to calculate the "moving product", for each row of a
dataframe, where I wish to be able to specify the length of the moving product?
Depending on my needs, I'd like to be able to specify the "length" over which
to calculate the moving average (in this example, length=3).
Fo
This was actually the one I was looking for, thx. It's actually essentially
a wrapper for %in%, can't imagine I missed that one. I tried it out but I
put the vectors in the wrong order (giving a vector of length 3). Guess it's
too late to think clearly.
Cheers
On Sat, Apr 3, 2010 at 12:55 AM, wr
Thanks for the link, very useful. A simple fix to the code using
normalizePath(), or full path name.
read.xls(normalizePath("ttt.xls"))
--- On Fri, 4/2/10, Gabor Grothendieck wrote:
> From: Gabor Grothendieck
> Subject: Re: [R] can't read excel file with read.xls()
> To: "array chip"
> Cc:
On Fri, Apr 2, 2010 at 11:43 PM, Erin Hodgess wrote:
> Dear R People:
>
> I'm building a packages on an Ubuntu Karmic Koala 9.10 system and am
> getting the following errors:
>
>
> * checking PDF version of manual ... WARNING
> LaTeX errors when creating PDF version.
> This typically indicates Rd
Read about the bugs and workarounds on this page:
http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows&s=excel
On Fri, Apr 2, 2010 at 6:56 PM, array chip wrote:
> Hi, I encountered a problem of not being able to read in an excel spreadsheet
> using read.xls() in the xlsReadWrite package
Has anyone authored a Nonparametric approach to Canonical Correlation?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, m
Hi, I encountered a problem of not being able to read in an excel spreadsheet
using read.xls() in the xlsReadWrite package. can anyone help? Here is an
example code
write.xls(matrix(1:9,nrow=3),"ttt.xls")
read.xls("ttt.xls")
Error in read.xls("ttt.xls") :
Unexpected error. Message: Can't fin
Hi Joris
Try Is.element function: is.element (x,y)
Regards
mohamed
Joris Meys a écrit :
Dear all,
I have a vector, and for each element I want to check whether it is equal to
any element from another vector. I want a vector of logical values with the
length of the first one as return. In R
Dear R People:
I'm building a packages on an Ubuntu Karmic Koala 9.10 system and am
getting the following errors:
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
LaTeX errors found:
! Font T1/ptm/m/n/10=ptmr8t at 10.0pt n
On 3/04/2010, at 11:35 AM, Joris Meys wrote:
> Dear all,
>
> I have a vector, and for each element I want to check whether it is equal to
> any element from another vector. I want a vector of logical values with the
> length of the first one as return. In R this would be :
>
>> x <- 1:10
>> sap
Dear all,
I have a vector, and for each element I want to check whether it is equal to
any element from another vector. I want a vector of logical values with the
length of the first one as return. In R this would be :
> x <- 1:10
> sapply(x,function(y){any(y==c("2","3","4"))})
[1] FALSE TRUE T
Hi,
Is there any R library/package that calculates tetrachoric correlations from
given marginals and Pearson correlations among ordinal variables?
Inputs to polychor function in polycor package are either contingency tables or
ordinal data themselves. I am looking for something that takes marg
David Winsemius wrote:
>
>
> On Apr 2, 2010, at 5:32 PM, HouseBandit wrote:
>
>>
>> my goal is to return the selected fitted values ...
>
> Which were never really "selected".
>
>> ... and then perform a sum of
>> squares calcuation with them. I have looked at 'list' etc but cant
>> retur
See topic
--
View this message in context:
http://n4.nabble.com/Arellano-how-to-generate-data-for-y-given-t-tp1749656p1749656.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailm
Thanks, Manuel!
> Subject: Re: lineplot.CI in "sciplot": option "ci.fun" can't be changed?
> From: manuel.a.mora...@williams.edu
> To: shi...@hotmail.com
> CC: r-help@r-project.org; mmora...@williams.edu
> Date: Fri, 2 Apr 2010 14:22:33 -0400
>
> For now,
On Apr 2, 2010, at 5:32 PM, HouseBandit wrote:
my goal is to return the selected fitted values ...
Which were never really "selected".
... and then perform a sum of
squares calcuation with them. I have looked at 'list' etc but cant
return
anything. Its either all of the fitted values or
Inline below:
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Steve Lianoglou
Sent: Friday, April 02, 2010 2:34 PM
To: Jay
Cc: r-help@r-project.org
Subject: Re: [R] Cross-validation for
Hi
I am doing PCA using prcomp and when I try to get predicted values for the
different PC's the number of data points is always one less than in my
original data set. This is a problem because it prevents me from doing any
post-hoc analysis due to the fact that my dependent variables are one en
Hi,
On Fri, Apr 2, 2010 at 9:14 AM, Jay wrote:
> If my aim is to select a good subset of parameters for my final logit
> model built using glm(). What is the best way to cross-validate the
> results so that they are reliable?
>
> Let's say that I have a large dataset of 1000's of observations. I
my goal is to return the selected fitted values and then perform a sum of
squares calcuation with them. I have looked at 'list' etc but cant return
anything. Its either all of the fitted values or just the first and last of
the sub set that I need.
Cheers
--
View this message in context:
http:
On Apr 2, 2010, at 5:14 PM, HouseBandit wrote:
I have a regression model that I have used on some data and to look
at its
accuracy compared to some other models I have been extracting the
same 10%
of the real data to perform a sum of squares calculation
this is what I have tried but it g
On 2/04/2010, at 4:37 AM, Scott Duke-Sylvester wrote:
> I have a simple problem: I need to load a ERSI shapefile of US states
> and check whether or not a set of points are within the boundary of
> these states. I have the shapefile, I have the coordinates but I'm
> having a great deal of difficu
I have a regression model that I have used on some data and to look at its
accuracy compared to some other models I have been extracting the same 10%
of the real data to perform a sum of squares calculation
this is what I have tried but it gives me the '0.9*length(t)' fitted value
and the 'length
On 2010-04-02 11:07, Michael Friendly wrote:
This is a nice example; thanks for providing it in this form. I tried
to trim it down to show fewer groups, but ran into the following errors
that I can't understand:
## keep species 1:6
> dataset <- subset(dataset, species < 7)
Warning message:
In
A very simple option, since you're only looking to efficiently store
and retrieve, is something like a key-value store.
There is a new rredis (redis) package on CRAN, as well as the
RBerkeley (Oracle Berkeley DB) package.
RBerkeley is as simple as db_put() and db_get() calls where you
specify a k
==
y=c(100,200,300,400,500)
treatment=c(1,2,3,3,4)
block=c(1,1,2,3,3)
summary(lm(y~as.factor(treatment)+as.factor(block)))
==
The aim is to find a model that can estimate
the comparison between treatment 1 with 2
and treatment 3 wit
On Apr 2, 2010, at 3:39 PM, Geoffrey Smith wrote:
Hello, I have an unbalanced panel data set that looks like:
ID,YEAR,HEIGHT
Tom,2007,65
Tom,2008,66
Mary,2007,45
Mary,2008,50
Harry,2007,62
Harry,2008,62
James,2007,68
Jack,2007,70
Jordan,2008,72
That is, James, Jack, and Jordan are missing a Y
#Hello, i have created this function, but optim doesnt maximize it, just
return the value at the inits
W<-function(l){
w<-rep(0,dim(D)[1])
for(i in 1:dim(D)[1]){
w[i]<-PAitk(D[i,],D[-i,],l)
}
return(prod(w))
}
#D is a matrix with entires in {0,1}, l is a vector which length(l)=
dim(D)[2]
#PAitk i
You can use identify() to obtain coordinates from plotted points but if you
want any coordinates you could use locator():
> plot(1:10)
> loc <- locator(n=3)
> str(loc)
List of 2
$ x: num [1:3] 2.3 5.4 8.29
$ y: num [1:3] 6.15 8.33 2.6
> points(loc$x, loc$y, col=2)
>
Walmes.
-
..ooo0
nmola...@gmail.com
--
Att: Nicolás Molano
[[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.R-project.org/posting-guide.html
and p
Are you thinking of ?identify ?
Nuno Prista wrote:
Hi,
I seem to recall coming across a function that allowed one to mouse-click on an
xy-plot and obtain x and y coordinates. Can anyone remind me its name?
Thanks,
Nuno
__
R-help@r-project.org mail
In an OpenOffice.org forum someone asked if it was possible to plot some raw
data and then add a line for the confidence interval.
Example at http://www.graphpad.com/help/Prism5/scatter%20-%20grouped.png
While it may be possible to do this in OOo's spreadsheet program it looks nasty
(both to
Hi,
I seem to recall coming across a function that allowed one to mouse-click on an
xy-plot and obtain x and y coordinates. Can anyone remind me its name?
Thanks,
Nuno
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Great, thanks for your help. I tried:
x <- 1:1
y <- lapply(1:1,function(t){t*runmean(x,t,alg="fast",endrule="trim")})
and it worked in about 90 sec.
Thanks again,
Andy
On Fri, Apr 2, 2010 at 3:43 PM, Gabor Grothendieck
wrote:
> There is also rollmean in the zoo package which might be
Hello, I have an unbalanced panel data set that looks like:
ID,YEAR,HEIGHT
Tom,2007,65
Tom,2008,66
Mary,2007,45
Mary,2008,50
Harry,2007,62
Harry,2008,62
James,2007,68
Jack,2007,70
Jordan,2008,72
That is, James, Jack, and Jordan are missing a YEAR.
Is there any command that will "fill in" the mis
hi List and Manuel,
I have encounter the following problem with the function "lineplot.CI". I'm
running R 2.10.1, sciplot 1.0-7 on Win XP. It seems like it's a scoping issue,
but I couldn't figure it out.
Thanks!
...Tao
> lineplot.CI(x.factor = dose, response = len, data = ToothGrowth)
Ah, I finally figured it out: I had asked
> In both of those cases, why is the [] needed?
It's because when on the left hand side of an assignment, the bracket
operator attempts to preserve the class and dimension of the object it's
subsetting. (Or at least, that's true when the object is a
For now, just change fun(x) to median(x) (or whatever) in your ci.fun()
below.
E.g.
lineplot.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun=
function(x) c(mean(x)-2*se(x), mean(x)+2*se(x)))
Otherwise, maybe the list members could help with a solution. An example
that illustrates
Thanks!
On Fri, Apr 2, 2010 at 11:35 AM, Erik Iverson wrote:
> Hello,
>
>
> Sam Albers wrote:
>
>> Hello there,
>>
>> I have a situation where I would like to select the first row of a
>> particular factor for a data frame (data example below). So that is, I
>> would
>> like to select the first
Hello
I wanted to compare two fingerprint images. How do you do with R?.
Is there a role for cross-correlation of images?
Thanks
--
=
Juan Antonio Gil Pascual
Prof. Titular de Métodos de Investigación en Educación
correo: j...@edu.uned.es
w
There is also rollmean in the zoo package which might be slightly
faster since its optimized for that operation.
k * rollmean(x, k)
e.g.
> 2 * rollmean(1:4, 2)
[1] 3 5 7
will give a rolling sum. runmean in the caTools package is even faster.
On Fri, Apr 2, 2010 at 2:31 PM, Jorge Ivan Velez
wrot
Hello,
Sam Albers wrote:
Hello there,
I have a situation where I would like to select the first row of a
particular factor for a data frame (data example below). So that is, I would
like to select the first entry when the factor1 =A and then the first row
when factor1=B etc. I have thousands of
Hi Andy,
Take a look at the rollapply function in the zoo package.
> require(zoo)
Loading required package: zoo
> x <- 1:4
> rollapply(zoo(x), 1, sum)
1 2 3 4
1 2 3 4
> rollapply(zoo(x), 2, sum)
1 2 3
3 5 7
> rollapply(zoo(x), 3, sum)
2 3
6 9
> rollapply(zoo(x), 4, sum)
2
10
# all at once
sappl
Hello there,
I have a situation where I would like to select the first row of a
particular factor for a data frame (data example below). So that is, I would
like to select the first entry when the factor1 =A and then the first row
when factor1=B etc. I have thousands of entries so I need some gene
Hello,
I'd like to take all possible sub-summands of a vector in the quickest and
most efficient way possible. By "sub-summands" I mean for each sub-vector,
take its sum. Which is to say: if I had the vector
x<-1:4
I'd want the "sum" of x[1], x[2], etc. And then the sum of x[1:2], x[2:3],
etc
Michael and others,
Here is my complete ancova example
http://astro.ocis.temple.edu/~rmh/HH/hotdog.pdf
This example, especially in Figure 6, places them in a context of a
Cartesian
product of models with the intercept having two levels and slope having
three levels.
It is based on the ancova c
The error message "F used instead of FALSE" is pretty clear to me ...:
Use FALSE rather than F in your code.
Uwe Ligges
On 30.03.2010 07:36, Dong H. Oh wrote:
Dear useRs,
I am trying to build my package (nonpareff) which deals with some models of
data envelopment analysis.
The building work
On 28.03.2010 21:20, Peter Ehlers wrote:
I haven't seen an answer to this yet.
Your problem may stem from having defined a variable T.
I can replicate your error messages with:
T <- "hello"
library(RMark)
So methinks that this probably indicates that there may be
a problem with using T for T
On Fri, Apr 2, 2010 at 1:01 PM, William Dunlap wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of e-letter
>> Sent: Friday, April 02, 2010 9:20 AM
>> To: Gabor Grothendieck
>> Cc: r-help@r-project.org
>> Subject: Re: [R]
Look at the serialize function, it may accomplish what you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf
Look at the my.symbols function in the TeachingDemos package. You can get line
segments using the ms.arrows function and setting the length argument to 0 (or
you can make your own plotting function by copying ms.arrows and replacing the
call to arrows with a call to segments).
Hope this helps
This is a nice example; thanks for providing it in this form. I tried
to trim it down to show fewer groups, but ran into the following errors
that I can't understand:
## keep species 1:6
> dataset <- subset(dataset, species < 7)
Warning message:
In Ops.factor(species, 7) : < not meaningful for
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of e-letter
> Sent: Friday, April 02, 2010 9:20 AM
> To: Gabor Grothendieck
> Cc: r-help@r-project.org
> Subject: Re: [R] Merge failure using zoo package
>
> On 02/04/2010, Gabor Gr
Hi R users,
I would like to construct a sort hybrid vector/scatter plot.
My data is in the following format: 3-column x,y,z data-frame in which
every row is a separate data-point.
The x & y columns are coordinates, and the z column contains orientation
data (range 0-180 degrees, with East=0 & N
Below is the format that was requested. This has the data followed by
the corrected code at the end. There are several things that were
wrong:
1. z1[,2] is wrong since z1 is a vector, not a 2d matrix. Ditto for
z2. Ideally zoo would have given an error message but in any case its
wrong. It sh
Google leads to some discussion on the Intel Sofware Network:
http://software.intel.com/en-us/forums/showthread.php?t=64585
Be warned: I haven't read the discussion.
-Peter Ehlers
On 2010-04-02 9:30, jacob wrote:
Dear Lists:
I recently ran quite annoyance problem while running R on Ubuntu
On 02/04/2010, Gabor Grothendieck wrote:
> The code does not run with the files. I need the requested
> information, namely a single file containing code and data and that I
> can just copy into a session without editing and see the result you
> see.
I don't understand how I can combine the four
On 02.04.2010 01:16, Jp2010 wrote:
Hi all,
Thanks for the wonderful forum with all the valuable help and comments here.
I have been a splus user for the past 7 to 8 years and now crossing the mind
of changing over to R. Have been doing a lot of reading and one of the main
reasons is being an
The code does not run with the files. I need the requested
information, namely a single file containing code and data and that I
can just copy into a session without editing and see the result you
see.
On Fri, Apr 2, 2010 at 11:27 AM, e-letter wrote:
> On 02/04/2010, Gabor Grothendieck wrote:
>
Dear Ping,
It is not possible to pass starting values for the fixed effects. It
doesn't make much sense to give starting values for the fixed effects
because they can be Gibbs sampled in a single pass conditional on the
latent variables and the (co)variance components - after a single
ite
Dear Lists:
I recently ran quite annoyance problem while running R on Ubuntu 9.10.
When running the program, the system suddenly exit from the R
session with the following warnings:
#
OMP: Hint: This may cause performance degradation and correctness
is
On 02/04/2010, Gabor Grothendieck wrote:
> The files only have one data column. What is the meaning of x[,2],
> etc. ? What is z1?
>
I only want to merge one column from one file with one column from
another file. With [x,2], I am trying to select the column of data.
> Please provide reproduci
Jp2010 yahoo.com> writes:
>From my understanding it is going to be difficult, is that my understanding
> right.?
Probably impossible ...
TIBCO, or whoever owns S-PLUS now (I don't pay much attention, so
it's hard for me to keep track) does try to achieve as much R-compatibility
as possible:
The files only have one data column. What is the meaning of x[,2],
etc. ? What is z1?
Please provide reproducible code and data all in a single file using
this style so its clear what is what. Also please cut down the size of
your data to the smallest size that will still illustrate the problem
On Apr 2, 2010, at 10:36 AM, Steve Murray wrote:
Dear all,
Thanks for the contributions so far. I've had a look at these and
the closest I've come to solving it is the following:
data_ave <- ave(data$rammday, by=c(data$month, data$year))
Warning messages:
1: In split.default(x, g) :
d
Jay
Unless I have misunderstood some statistical subtleties, you can use the
AIC in place of actual cross-validation, as the AIC is asymptotically
equivalent to leave-out-one cross-validation under MLE.
Joe
Stone, M.
An asymptotic equivalence of choice of model by cross-validation and
Akaike's
Data files test1, ...2, ...3, ...4 respectively.
time1,dataset1
01:01:00,0.73512097
01:01:30,0.34860813
01:02:00,0.61306418
01:02:30,0.01495898
01:03:00,0.27035612
01:03:30,0.69513898
01:04:00,0.46451758
01:04:30,0.61672569
01:05:00,0.82496122
01:05:30,0.34766154
01:06:00,0.69618714
01:06:30,0.390
Thank you, Ravi,
I have looked at that package, but I don't see any method to compare two ROC
curves. I beleive the method used by roccomp is based on Delong.
JoAnn
From: Ravi Kulkarni [via R] [ml-node+1748903-1261333028-216...@n4.nabble.com]
Sent: Friday, April
Beautiful. Thank you.
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Friday, April 02, 2010 10:59 AM
To: Doran, Harold
Cc: r-help@r-project.org
Subject: Re: [R] POSIX primer
On Fri, Apr 2, 2010 at 10:49 AM, Doran, Harold wrote:
> I have not used POSIX
Steve -
Take a closer look at the help page for ave(), especially
the ... argument. Try
data_ave <- ave(data$rammday, data$month, data$year,FUN=mean)
(Assuming you want to calculate the mean -- your example
didn't specify a function.)
- Phil Spector
On Fri, Apr 2, 2010 at 10:49 AM, Doran, Harold wrote:
> I have not used POSIX classes previously and now have a need to use them. I
> have sports data with times of some athletes
The main reason to use POSIXct is if you need time zones.
If you don't then you might be better off with chron. See
Readers,
Please refer to attached example data files. It seems that the merge
function fails for the latter section of the data set. Command
terminal output:
> library(chron)
> library(zoo)
> x<-read.zoo("test1.csv",header=TRUE,sep=",",FUN=times)
> y<-read.zoo("test2.csv",header=TRUE,sep=",",FUN=
Dear Steve,
Multiplying the mean with the number of observations is essentially the same as
summing the numbers.
Have a look at the plyr packages.
library(plyr)
ddply(data, c("month", "year"), function(x){
c(MeanMultiplied = mean(x$ramm) * nrow(x), Sum = sum(x$ramm))
})
--
I have not used POSIX classes previously and now have a need to use them. I
have sports data with times of some athletes after different events. I need to
perform some simple analyses using the times. I think I've figured out how to
do this. I just want to confirm with others who have more exper
If my aim is to select a good subset of parameters for my final logit
model built using glm(). What is the best way to cross-validate the
results so that they are reliable?
Let's say that I have a large dataset of 1000's of observations. I
split this data into two groups, one that I use for traini
Dear all,
Thanks for the contributions so far. I've had a look at these and the closest
I've come to solving it is the following:
> data_ave <- ave(data$rammday, by=c(data$month, data$year))
Warning messages:
1: In split.default(x, g) :
data length is not a multiple of split variable
2: In sp
> I'm using rpart function for creating regression trees.
> now how to measure the fitness of regression tree???
>
> thanks n Regards,
> Vibha
I read R-help as a digest so often come late to a discussion. Let me
start by being the first to directly answer the question:
> fit <- rpart(time ~ ag
On Apr 2, 2010, at 9:56 AM, Marshall Feldman wrote:
Hi,
I've been struggling with this problem the last few days and finally
discovered it's happening at a very fundamental level. Going through
Stephen Turner's tutorial on ggplot2, I entered these base graphics
commands:
with(diamonds, plot(
On Apr 2, 2010, at 9:26 AM, wenjun zheng wrote:
Dear R users,
can somebody give me some suggestions about how to build Mac
distribution on
my own Mac OS
It appears you have not read the most basic background information yet:
http://cran.r-project.org/doc/manuals/R-admin.pdf
--
David.
Hi,
I've been struggling with this problem the last few days and finally
discovered it's happening at a very fundamental level. Going through
Stephen Turner's tutorial on ggplot2, I entered these base graphics
commands:
> with(diamonds, plot(carat,price))
> with(diamonds, plot(carat,
Here are a few ways:
Try this:
set.seed(123)
TS <- ts(1:25 + rnorm(25))
tt <- time(TS)
tt.pred <- end(tt)[1] + 1:10
both <- ts(c(TS, predict(lm(TS ~ tt), list(tt = tt.pred
ts.plot(both, TS, gpars = list(type = "o", col = 2:1, pch = 20))
and read ?ts, ?start, ?ts.plot and next time please pro
Dear R users,
can somebody give me some suggestions about how to build Mac distribution on
my own Mac OS
Thanks
--
Wenjun
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Please learn how to use `RsiteSearch' before posting questions to the list:
RSiteSearch("derivative smooth function")
This should have provided you with plenty of solutions.
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Di
Please keep in mind this question has absolutely nothing to do with
finance, and therefore needs to instead be directed to R-help.
Thanks in advance for keeping the R-finance list on topic.
Jeff
On Fri, Apr 2, 2010 at 3:36 AM, FMH wrote:
>
> Dear All,
>
> I've been searching for appropriate cod
On Fri, 2 Apr 2010, Duncan Murdoch wrote:
On 02/04/2010 7:01 AM, Tal Galili wrote:
Hi all,
The call to:
library(rJava)
Results in the following warning massage:
Warning message:
In inDL(x, as.logical(local), as.logical(now), ...) :
DLL attempted to change FPU control word from 8001f to 900
Hello,
I am using plot( ) function to plot time-series.
it takes time-series object as an argument
but i want to plot predicted data with training set, to compare them.
is there any function available?
Vibha
[[alternative HTML version deleted]]
___
Hi Muhammad,
The problem is that you set the par() options before creating your png.
I've tried, and it works if you do this:
... # x and y
png("image.png")
par(mar=c(5,5,5,7)+0.1) # inner margin
par(oma=c(3,3,3,7))# outer margin
... # rest of your code
HTH,
Ivan
Le 4/2/2010 12:12, Muhammad
Hello,
I have a problem. I am using the NLME library to fit a non-linear model. There
is a linear component to the model that has a couple parameter values that can
only be positive (the coefficients are embedded in a sqrt). When I try and
fit the model to data the search algorithm tries t
Hello,
I'm using earth function for Multivariate Adaptive Regression splines.
what is the significance of RSq in earth function??
following's the code.
printed value is of RSq.
> tr.wage<-sample(1:nrow(HCMwage), 0.8*nrow(HCMwage))
> tst.wage<- (1:nrow(HCMwage))[-tr.wage]
>
HCMwageModel<-earth(V
Thanks Duncan and Romain,
I'll go and do that.
with regards,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.
On 02/04/2010 7:01 AM, Tal Galili wrote:
Hi all,
The call to:
library(rJava)
Results in the following warning massage:
Warning message:
In inDL(x, as.logical(local), as.logical(now), ...) :
DLL attempted to change FPU control word from 8001f to 9001f
After some searching I found the follow
Thanks Ivan, Jim and Petr.
The output turns out as desired after I've taken your suggestions.
Muhammad
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
On Thu, Apr 1, 2010 at 7:08 PM, Gabor Grothendieck
wrote:
> Perhaps something like this:
>
> library(zoo)
> library(chron)
> # read in data
>
> Lines1 <- "date time level temp
> 2009/10/01 00:01:52.0 2.8797 18.401
> 2009/10/01 00:16:52.0 2.8769 1
Hi
r-help-boun...@r-project.org napsal dne 02.04.2010 12:12:02:
> Dear useRs,
>
> I'm having a slight problem with plotting on 2 axes. While the following
> code works alright on screen, the saved output does not turn out as
> desired i.e. the secondary y-axis does not display fully.
>
> Jus
1 - 100 of 109 matches
Mail list logo