Dear R Users
Does some has any idea about how to execute a scilab file(.sce file) from
the Terminal in R.
Any kind of guidance would be highly welcomed and appreciated.
--
View this message in context:
http://www.nabble.com/Execution-of-a-.sce-file-through-R-tp21227441p21227441.html
Sent from t
Hello,
I was using garchFit {fGarch} to fit some GARCH processes.
I noticed that the result contains "Log Likelihood" value (right above
"Description"), but when I use @fit$llh to retrieve Log Likelihood value,
the sign switched.
I am confused about which value I should choose to report...
An
That is not what the message says: the actual error is on the last line.
Try 'library(splines)' directly.
I suspect the issue is an encoding issue with mh1823, a package not known
to us. In particular, do you have any non-ASCII code in your package?
Check over the advice on portability issues
Hi all,
I did some analysis using package R2WinBUGS to call WinBUGS. I set the
iterations to 5 (fairly a large number, I think), but after the program
was done, the effective posterior samples contained only 7 draws. I don't
know why.
By the way, I checked posterior sample size by using bug
One can't tell for sure without seeing the function, but I'd guess
that you have a numerical issue. Here is an example to reflect upon:
f=function(x) (exp(x)-exp(50))*(exp(x)+exp(50))
uniroot(f,c(0,100))
$root
[1] 49.7
$f.root
[1] -1.640646e+39
$iter
[1] 4
$estim.prec
[1] 6.103516e-
I have a strange problem with uniroot() function. Here is the result :
> uniroot(th, c(-20, 20))
$root
[1] 4.216521e-05
$f.root
[1] 16.66423
$iter
[1] 27
$estim.prec
[1] 6.103516e-05
Pls forgive for not reproducing whole code, here my question is how "f.root"
can be 16.66423? As it is finding
Greetings R-Helpers:
I am trying to help a Chinese colleague who is experiencing difficulty
loading my package. He is running WindowsXP and the Chinese version of
R2.8.1.
As you can see from his screen-copy (below), my package is successfully
unpacked but cannot be loaded because splines b
XML is a good tool reading data from web within R. But I wonder how could get
the encoding correctly.
library(XML)
url <- 'http://www.szitic.com/docc/jz-lmzq.html'
xml <- htmlTreeParse(url, useInternal=TRUE)
q <- "//tbody/tr/td"
dat <- unlist(xpathApply(xml, q, xmlValue))
df <- as.data.frame(t(m
Thanks, Frank. I am an old newbie, so your advice was not
straightforward to me, but it nudged me to look at the structure of
the contents object. I did solve the problem by passing the contents
data frame in the contents object. Using the example dfr from Dieter,
> g <- contents(dfr)
> g
See CRAN --> Task Views --> Machine Learning.
i.e.
http://cran.r-project.org/web/views/MachineLearning.html
Particularly, 'Recursive Partitioning'.
HTH,
Chuck
On Tue, 30 Dec 2008, Juan Antonio Gil Pascual wrote:
Michael thank you very much, what about the options I have tried al
Try this:
> aggregate(z$mph, trunc(time(z), "hour"), mean)
(09/01/05 00:00:00) (09/01/05 01:00:00) (09/01/05 02:00:00)
9.2750010.08333 9.2
On Tue, Dec 30, 2008 at 6:30 PM, Sherri Heck wrote:
> Dear All-
>
> I have a dataset that is comprised of the followi
Prof Brian Ripley wrote:
Well, we don't see what you see. but if ? was hex a7, the message is
entirely correct. If you want to enter that, use "\xa7".
We see different things. I see a section sign (double s) symbol. From
the symptoms, I would suspect that the terminal is set to latin-1 or -1
Dear All-
I have a dataset that is comprised of the following (LST = yymmddhhMM):
LST in mphDeg DegF DegF2%volts Degmph2
w/m2
050901 0.007.8 216.9 45.1 -999 24.4 -999 -999
10.60.2
0509010005 0.008.6 206.6 45.1 -999 25.2
oh, I do... Just obfuscating the actual path for work compliance issues...
On Tue, Dec 30, 2008 at 4:50 PM, Duncan Murdoch wrote:
> On 30/12/2008 5:48 PM, James Yoo wrote:
>
>> Thanks for your reply.
>>
>> here is my path:
>> I:\Rtools\bin;
>> i:\Rtools\perl\bin;
>> I:\Rtools\MinGW\bin;
>> I:\R-2
On 30/12/2008 5:48 PM, James Yoo wrote:
Thanks for your reply.
here is my path:
I:\Rtools\bin;
i:\Rtools\perl\bin;
I:\Rtools\MinGW\bin;
I:\R-2.8.1\bin;
You shouldn't have "" in there. Just put in whatever the directory
actually is.
Duncan Murdoch
I downloaded my R source from the cran.
Thanks for your reply.
here is my path:
I:\Rtools\bin;
i:\Rtools\perl\bin;
I:\Rtools\MinGW\bin;
I:\R-2.8.1\bin;
I downloaded my R source from the cran.
the command I executed from within R-2.8.1\src\gnuwin32 is:
'make all recommended'
I'd love to just install the binary, but I'm trying to comp
On 30/12/2008 4:59 PM, James Yoo wrote:
lemme clarify... the last line is:
make[4]: *** No rule to make target `inst/BasicClasses.R', needed by
`/R-2.8.1/library/methods/inst'. Stop.
On Tue, Dec 30, 2008 at 3:23 PM, James Yoo wrote:
hello,
I'm trying compile R-2.8.1. I've got Rtools instal
lemme clarify... the last line is:
make[4]: *** No rule to make target `inst/BasicClasses.R', needed by
`/R-2.8.1/library/methods/inst'. Stop.
On Tue, Dec 30, 2008 at 3:23 PM, James Yoo wrote:
> hello,
>
> I'm trying compile R-2.8.1. I've got Rtools installed, paths setup, etc...
> the compile
hello,
I'm trying compile R-2.8.1. I've got Rtools installed, paths setup, etc...
the compile progress up to a point and then complains:
... DLL made
installing DLL
collecting R files
preparing package methods for lazy loading
dumping R code in package `methods'
initializing class and m
Michael thank you very much, what about the options I have tried all
without success.
Try again to send a message to the list by changing the subject line.
Merry Christmas
Juan
--
=
Juan Antonio Gil Pascual
Prof. Titular de Métodos de Inv
Prof Brian Ripley stats.ox.ac.uk> writes:
>
> I think it should work on any reasonable machine, but it is worth pointing
> out that there are 4 separate tiff() devices:
>
> Unix, X11
> Unix, cairo,
> Mac OS X, quartz,
> Windows, graphapp
>
> and once again we really did need the OS informatio
Thanks, in fact it's quite clean this way. I've added this tip to the
R-wiki,
http://wiki.r-project.org/rwiki/doku.php?id=guides:overview-data-manip
baptiste
On 30 Dec 2008, at 19:25, Gabor Grothendieck wrote:
Or even:
abind(foo, along = 3)
On Tue, Dec 30, 2008 at 1:24 PM, Gabor Gr
Hi, am not sure if this is the best platform to ask this question, so please
excuse it if it's tangential. I am writing an R function that also involves
some brief interfacing with SCILAB. I am doing this via the terminal in
LINUX, using the system() command. I have an .sce file that I wish to
exe
On 30/12/2008 1:18 PM, Stavros Macrakis wrote:
On Tue, Dec 30, 2008 at 12:53 PM, hadley wickham wrote:
On Tue, Dec 30, 2008 at 10:21 AM, baptiste auguie wrote:
I thought this was a good candidate for the plyr package, but it seems that
l*ply functions are meant to operate only on separate lis
Or even:
abind(foo, along = 3)
On Tue, Dec 30, 2008 at 1:24 PM, Gabor Grothendieck
wrote:
> Try:
>
> do.call(abind, c(foo, along = 3))
>
>
> On Tue, Dec 30, 2008 at 1:15 PM, baptiste auguie wrote:
>> In fact, when writing my post I tried to do exactly what you did in creating
>> a 3d array fro
Try:
do.call(abind, c(foo, along = 3))
On Tue, Dec 30, 2008 at 1:15 PM, baptiste auguie wrote:
> In fact, when writing my post I tried to do exactly what you did in creating
> a 3d array from the list, and I failed miserably! This is (imho) partly
> because the syntax is not very clean or strai
I believe you are looking for:
Test[1, 2:6, drop=TRUE]
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
glenn wrote:
Assuming I have dataframe ³test² with dim = (2,10) say;
And that I can choose some o
On Tue, Dec 30, 2008 at 8:44 AM, wrote:
> It is no homework. It is part of a project where a binary matrix, whose 1s
> represent the position of the highest DWT coefficients energy, is used as a
> template to extract signal features.
> The approach I am following requires each row of the binary
On Tue, Dec 30, 2008 at 12:53 PM, hadley wickham wrote:
> On Tue, Dec 30, 2008 at 10:21 AM, baptiste auguie wrote:
>> I thought this was a good candidate for the plyr package, but it seems that
>> l*ply functions are meant to operate only on separate list elements:...
>> Perhaps a new case to co
In fact, when writing my post I tried to do exactly what you did in
creating a 3d array from the list, and I failed miserably! This is
(imho) partly because the syntax is not very clean or straightforward
as compared to the apply and *ply family. A list of matrices with
equal dimensions is
-Messaggio originale-
Da: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Inviato: mar 30/12/2008 13.42
A: mau...@alice.it
Cc: r-help@r-project.org
Oggetto: Re: [R] I would appreciate some help with clustering
Is this homework? If so, you should discuss it with the instructor,
not us.
Reg
On Tue, Dec 30, 2008 at 10:21 AM, baptiste auguie wrote:
> I thought this was a good candidate for the plyr package, but it seems that
> l*ply functions are meant to operate only on separate list elements:
>
> Lists are the simplest type of input to deal with because they are already
> naturally
>
not sure what you want either but reshape package may help, melt function.
Sarah Goslee wrote:
> On Tue, Dec 30, 2008 at 11:57 AM, glenn wrote:
>> Assuming I have dataframe ³test² with dim = (2,10) say;
>>
>> And that I can choose some of the data;
>>
>> Test[1,2:6]
>>
>> How do I turn this data
> --begin included -
> My endogenous variable is not a time depending variable but percentages
> which naturally are censored in the interval [0,100]. Unfortunately many
> data points are 0 or 100 exactly. The rest of the data is asymmetrically
> distributed. So I would like to apply a two-li
Well, we don't see what you see. but if ? was hex a7, the message is
entirely correct. If you want to enter that, use "\xa7".
On Tue, 30 Dec 2008, Wijffels, Jan wrote:
Hi,
We recently switched from R2.7.0 to R2.8.1 but having problems tracking down
this 'invalid multibyte character' encod
People,
If Test 1 estimates the likelihood of Population A of being Type Z as
60% and if a completely separate and unrelated Test 2 estimates the
likelihood of Population A of being Type Z as 70%, is the likelihood
that Population A being Type Z raised above 70% ?
My gut feeling is that if t
I believe this does what you want:
m[-sample(which(m[,1]<8 & m[,2]>12),2),]
Analysis:
Get a boolean vector of rows fitting criteria:
m[,1]<8 & m[,2]>12
What are their indexes?
which(...)
Choose two among those indexes:
sample(...,2)
Choose all except the selected rows from the or
On Tue, Dec 30, 2008 at 11:57 AM, glenn wrote:
> Assuming I have dataframe ³test² with dim = (2,10) say;
>
> And that I can choose some of the data;
>
> Test[1,2:6]
>
> How do I turn this data into a list [which I would like to run a function
> across].
> Tried as.list but doesn¹t look or feel rig
Assuming I have dataframe ³test² with dim = (2,10) say;
And that I can choose some of the data;
Test[1,2:6]
How do I turn this data into a list [which I would like to run a function
across].
Tried as.list but doesn¹t look or feel right.
Thanks in advance, it is simple I am sure so apologies for
Is there some reason you can't reverse the loops? Then you'd have
only one device open at a time.
for ( i in length(obj.list) ) {
dev.set( device.list( j ) )
for( j in iters) {
# do plotting and other stuff for this list member
}
# close current device
}
Sarah
On Tu
I thought this was a good candidate for the plyr package, but it seems
that l*ply functions are meant to operate only on separate list
elements:
> Lists are the simplest type of input to deal with because they are
> already naturally
> divided into pieces: the elements of the list. For this
Patrick,
The reason I'm having a problem is because R only allows a maximum of 64
devices to be open at one time. If it allowed more than that, I wouldn't
have a problem. One way to get around that problem would be if I could close
a device and then append to it later, but there doesn't seem to be
Hi,
We recently switched from R2.7.0 to R2.8.1 but having problems tracking down
this 'invalid multibyte character' encoding issue. Can someone point us how to
solve this?
> sessionInfo()
R version 2.8.1 (2008-12-22)
x86_64-unknown-linux-gnu
locale:
LC_CTYPE=en_US.UTF-8;LC_NUMER
Hi Marc,
thanks, that works great! With yet another apply(), I can also make it
accept psych::winsor() and trimming:
some.sort.of.apply <- function ( foo, FUN, ... ) {
matrix(apply(sapply(foo, c),1,FUN,...), dim(foo[[1]]))
}
some.sort.of.apply ( foo, FUN=winsor,trim=0.3 )
Thank you all for
J Michael Dean wrote:
Thanks, Dieter, for your response shown below.
library(Hmisc)
dfr <- data.frame(x=rnorm(20),y=sample(c('male','female'),20,TRUE))
cnt <- contents(dfr)
latex(cnt,label="tab:mytab",caption="This is a caption")
The problem is that I only want to display cnt[1] - I am simply
Hi,
The approach below uses a function. The nice thing about it is that you can
define the cutoff values dynamically (i.e. what is 8 and 12 in your
example). The functions extract a row index to remove. Be aware that there
is no warning if both return the same row index. You might have to adjust
I think it should work on any reasonable machine, but it is worth pointing
out that there are 4 separate tiff() devices:
Unix, X11
Unix, cairo,
Mac OS X, quartz,
Windows, graphapp
and once again we really did need the OS information asked for in the
posting guide. As I recall Ben uses one of
on 12/30/2008 08:33 AM Stephan Kolassa wrote:
> Dear useRs,
>
> I have a list, each entry of which is a matrix of constant dimensions.
> Is there a good way (i.e., not using a for loop) to apply a mean to each
> matrix entry *across list entries*?
>
> Example:
>
> foo <- list(rbind(c(1,2,3),c(4,
Thanks, Dieter, for your response shown below.
library(Hmisc)
dfr <- data.frame(x=rnorm(20),y=sample(c('male','female'),20,TRUE))
cnt <- contents(dfr)
latex(cnt,label="tab:mytab",caption="This is a caption")
The problem is that I only want to display cnt[1] - I am simply making
a table that s
Hi Phil,
thanks, that already helps: Reduce() gets me the means I need!
Unfortunately, Reduce() apparently won't help me with trimming or
winsorizing the means, judging from its "successively" philosophy... Any
other ideas out there?
Best,
Stephan
Phil Spector schrieb:
Stephan -
If you
Assuming your values aren't always in such neat order, you could use
something like:
valtoremove1 <- sample((1:nrow(m))[m[,1] < 8], 1)
valtoremove2 <- sample((1:nrow(m))[m[,1] > 12], 1)
Sarah
On Tue, Dec 30, 2008 at 9:59 AM, Guillaume Chapron
wrote:
> Hello all,
>
> I create the following matri
Hello all,
I create the following matrix:
m <- matrix(1:20, nrow = 10, ncol = 2)
which looks like:
[,1] [,2]
[1,]1 11
[2,]2 12
[3,]3 13
[4,]4 14
[5,]5 15
[6,]6 16
[7,]7 17
[8,]8 18
[9,]9 19
[10,] 10 20
Then, I want to re
Try this (last line only needed if you want times as Date class):
library(zoo)
z <- zoo(as.matrix(dat2), time(dat2))
time(z) <- as.Date(time(z)) # optional
or you can use as.zoo.timeSeries from the devel version of zoo:
source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/
Dear useRs,
I have a list, each entry of which is a matrix of constant dimensions.
Is there a good way (i.e., not using a for loop) to apply a mean to each
matrix entry *across list entries*?
Example:
foo <- list(rbind(c(1,2,3),c(4,5,6)),rbind(c(7,8,9),c(10,11,12)))
some.sort.of.apply(foo,FU
At 17:43 29/12/2008, Juan Antonio Gil Pascual wrote:
Quería saber si existe alguna función de R para
realizar el algoritmo CHAID de árbol de
clasificación, existen de J48 por ejemplo pero
no encuentro nada de este algorítmo.
If you try one of the various R search facilities you will find some
Hi Ista,
I think you looking for: n.rgroup? See help(latex).
Also, try posting your example in a copy and past form, like:
descriptives<-data.frame(n=..., mean=...) etc.
The easier it is for others to reproduce your initial results the
quicker the problem will be solved.
HTH,
Jon Loehrk
bioshaw qq.com> writes:
>
> Hi
> I'm want to obtain a plot of 1200dpi tiff format, but I met some difficulty.
> my code:
> ##
> tiff(file="shaw.tiff",width=8.6,height=8.6,units="cm",res=1200,pointsize=10)
> plot(bal100,type="l",pch=20,col="red",l
Abou,
I presume this may be a task you will need to perform with some
regularity. I took the liberty of turning Duncan Murdoch's suggested
code into a function for you (I hope you don't mind, Duncan). Good
luck.
On a personal note, it's good to see a familiar name. I hope all is
well with you.
Hi!
I am trying to estimate Engel curves using a big sample (>42,000) using lm and
taking heteroskedasticity into account by using the summaryHCCM posted here by
John Fox (Mon Dec 25 16:01:59 CET 2006).
Having used the SIC (with MASS stepAIC) to determine how many powers to use I
estimate the mo
Is this homework? If so, you should discuss it with the instructor,
not us.
Regardless, the methods you suggested are a reasonable place
to start, and you perhaps should have done so first before asking
here. You may well have gotten the results you needed without
the delay of waiting for an answe
Hi
I'm want to obtain a plot of 1200dpi tiff format, but I met some difficulty.
Could anybody show me some examples about it?
my code:
##
tiff(file="shaw.tiff",width=8.6,height=8.6,units="cm",res=1200,pointsize=10)
plot(bal100,type="l",pch=20,col="r
If cm is a similarity matrix, why are you taking its Euclidean distance?
(The usage I'm familiar with has similarity as a pairwise measure of
association.)
Otherwise, if you feel the stress is too high, that implies that a 2-dimensional
solution is inadequate for your data and you should consider
I have a binary vector whose length is known.
Such a vector contains an unspecified number of 1s.
My goal is
1. to generate as many clusters as the number of 1s
2. to place the 1 as much as possible at the center of its own cluster
Example. Say I have the following binary vector:
v <- c(0,0,1,0,0,
Still it seems that it is not downloading data properly. For example
observation for date "2008-12-09 " is missing :
> tail(dat2, 30)
2008-10-02 2008-10-06 2008-10-08 2008-10-16 2008-10-20 2008-10-21 2008-10-22
2008-10-24 2008-10-27 2008-10-29 2008-10-30
2.681252.368755.375001.937
Oh my mistake, I missed the argument nDaysBack = 366.
However pls let me know how to change the data to zoo object.
RON70 wrote:
>
> I was trying to dw data from Economagic
> [http://www.economagic.com/em-cgi/data.exe/libor/day-ussnon], using
> following code :
>
> library(fimport)
> dat2 = e
I was trying to dw data from Economagic
[http://www.economagic.com/em-cgi/data.exe/libor/day-ussnon], using
following code :
library(fimport)
dat2 = economagicSeries("libor/day-ussnon", frequency = "daily")
Here I see that data is not complete, downloaded data starts from
"2007-12-31 ", whereas
Try this:
do.call(rbind, lapply(lapply(l, unlist), "[", unique(unlist(c(sapply(l,
names))
On Mon, Dec 29, 2008 at 7:35 PM, Mark Heckmann wrote:
> thanks for the elegant solution!
>
> It worked out fine in my case, but when I tried to understand what it does
> I discovered one little issue.
On 29/12/2008 4:25 PM, AbouEl-Makarim Aboueissa wrote:
Dear ALL:
How I show the area under the normal curve for example for the following example:
Assume X has N(100,15). How to show the area corresponding to the probability:
P(90
With many thanks
range <- c(70, 130)
x <- seq(min(ran
Weijia You gmail.com> writes:
> I have two networks for the same group of the users.
> I want to compare individual's degree in different networks.
> But how could I get the degrere of the nodes according to its name?
>
> When I use degree(g1), I could only get a vector of the degree of each node
Utkarsh Singhal ambaresearch.com> writes:
> I am using nnet function of nnet package to fit neural networks. Now I want to
get a unique solution every time
> I run the function for the same data. If I give rang=0, it solves my problem
but I am not sure whether I am doing
> the right thing. Any co
J Michael Dean comcast.net> writes:
>
> I am trying to feed a list to latex to use with SWeave, and the list
> comes from contents(). Since it is a list, a caption and label are
> generated by latex.list - these are not overridden by setting the
> parameters (such as caption='myCaption',
Sarah E. McCormick student.gvsu.edu> writes:
>
> Hi, I just got a new computer with windows vista on it and I am having a
problem using Tinn R. The problem occurs
> when I select a chunk of code and try to run it all at once using the "R
send:selection (echo = TRUE)" button, I
> get this error:
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