Julia, see
http://www.r-project.org/ -> Documentation -> Manuals (-> An introduction to R)
(or use: http://cran.r-project.org/manuals.html)
for a starting point. In addition, you might want to check the annotated list
of books to see which one might best fit your needs:
http://www.r-project.o
Thanks Jim, that is exactly what I was looking for!
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Sun, 11/29/09, Jim Lemon wrote:
> From: Jim Lemon
> Subject: Re: [R] Continuous legend colors
> To: "Tim Clark"
> Cc: r-help@r-project.org
> Date: Sunday, November 29, 2009,
Dear R helpers,
Almost 15 days back I have become member of this very active and wonderful
group. So far I have been only raising queries and in turn got them solved too
and I really thank for the spirit this group member show when it comes to the
guidance.
I wish to learn R language and I ha
So, I need some serious help with an R project. The project essentially
consists of some problems that have to be analyzed to decide the correct
statistical test to use, and then correctly use R to answer them. The use
of an instant messaging service for this would probably be great. I am
willi
i have a data frame and a numeric vector indexed as a subset of the rows in
the data.frame. what command can i use to assign the values in the vector
to the appropriate rows of the data.frame? here's my failed attempt. what
i would want is data[1,'z'] == 2, data[5,'z'] == -4, data[8,'z'] == -5,
Hi John,
Thanks A LOT for your reply and the code. What I want to do is to include the
ggobi display window to the widget window setup by me. I tried add before but
it said can not do it for GGobiScatterplotDisplay. Do you have any idea about
add displays?
Thanks again for your help.
Â
Thanks for the suggestion David. With mapply the lines are correctly
plotted but they are all red, the points are colored, but along the x
axis, not along the individual lines.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-
Yes, was wondering that other code line did not change so much :-)
Thanks a lot!
2009/11/30 Gabor Grothendieck :
>
>
> On Sun, Nov 29, 2009 at 9:16 AM, Gabor Grothendieck
> wrote:
>>
>> By the way, if you really do want to create the formula anyways then:
>>
>> ix <- 1:2
>> left <- paste(n
On 11/30/2009 03:22 PM, Tim Clark wrote:
Dear List,
I am trying to get a basic plot to show a continuous range of fill colors. It
is probably easiest to demonstrate. I would like a legend like in the
following example:
Satellite.Palette<-colorRampPalette(c("blue3","cyan","aquamarine","yello
On 11/30/2009 07:17 AM, Manuel Jesús López Rodríguez wrote:
Dear all,
I would like to know how could I execute a sequence or orders with just a
function, i.e, that just typing the function name, R gives me all the
parameters I want (for instance, if I want to see the summary, the standard
devi
This turns out to be easy. What you need are the
locations where to put the labels and those are given in
the som object. Using your code:
mysom <- som(wines.sc, grid=somgrid(5, 5, "hexagonal"))
plot(mysom, type="dist.neighbours" )
# check what's contained in mysom:
str(mysom)
# we need the $gr
Hi Rnewb,
Take a look at ?"%in%" .
HTH,
Jorge
On Sun, Nov 29, 2009 at 11:06 PM, Rnewb <> wrote:
>
> i have a data frame and a numeric vector indexed as a subset of the rows in
> the data.frame. what command can i use to assign the values in the vector
> to the appropriate rows of the data.fra
Sébastien Bihorel free.fr> writes:
>
> Dear R-users,
>
> I am trying to port to R something that I wrote in Matlab to perform model
> parameter optimization using the Nelder-Mead simplex method (fminsearch). I
> read the help on ?optim (which seems to be the way to go) as well as a bunch
> of p
Duncan Murdoch murdoch at stats.uwo.ca wrote:
> On 27/11/2009 3:36 PM, Alexander Søndergaard wrote:
> I'm new to R. Having a functional background, I was wondering what's
> the idiomatic way to iterate. It seems that for loops are the default
> given there's no tail-call optimization.
>
> I'm c
Dear List,
I am trying to get a basic plot to show a continuous range of fill colors. It
is probably easiest to demonstrate. I would like a legend like in the
following example:
Satellite.Palette
<-colorRampPalette(c("blue3","cyan","aquamarine","yellow","orange","red"))
require(fields)
image
i have a data frame and a numeric vector indexed as a subset of the rows in
the data.frame. what command can i use to assign the values in the vector
to the appropriate rows of the data.frame? here's my failed attempt. what
i would want is data[1,'z'] == 2, data[5,'z'] == -4, data[8,'z'] == -5,
Dear R-users,
I am trying to port to R something that I wrote in Matlab to perform model
parameter optimization using the Nelder-Mead simplex method (fminsearch). I
read the help on ?optim (which seems to be the way to go) as well as a bunch
of posts on the topic, but I would like to make sure abo
On Nov 29, 2009, at 9:30 PM, Frostygoat wrote:
I have 11 vectors representing insect survival probabilities in
response to different levels of toxins at 10 concentrations
lx100=c(1,1,1,.8,.5,.4,.2,0)
day100=c(0,1,2,3,4,5,6,7,8)
lx90=c(1,1,1,1,.9,.8,.6,.4,.2,.1,0)
day90=c(0,1,2,3,4,5,6,7,8,9,1
I have 11 vectors representing insect survival probabilities in
response to different levels of toxins at 10 concentrations
lx100=c(1,1,1,.8,.5,.4,.2,0)
day100=c(0,1,2,3,4,5,6,7,8)
lx90=c(1,1,1,1,.9,.8,.6,.4,.2,.1,0)
day90=c(0,1,2,3,4,5,6,7,8,9,10)
#...and so on10% and a zero (control) series
l
Many thanks, Peter, Gavin. That is exactly what it was...and old version. I
was running R 2.7 (needed it to integrate with SPSS 17).
I know I am a pain, but one last question. Is it possible to overlay a value
label onto the maps? In some of the SOMs we have worked through in class, we
saw
chronos.phenomena wrote:
>
> This is really annoying me... when I click R application icon it brings
> already opened session in the focus and it DOESN'T open new session
>
> any ideas?
>
I don't think this is possible with R.app on OS X. OS X applications do not
spawn separate processes
This is really annoying me... when I click R application icon it brings
already opened session in the focus and it DOESN'T open new session
any ideas?
--
View this message in context:
http://n4.nabble.com/How-do-I-run-to-or-more-R-consoles-on-Mac-OS-X-tp930979p930979.html
Sent from the R he
I tried out the code you wrote, it also works for me, but it lacks a
parameter i use in my code.
The problem (at this computer) seems to be this "cross"-parameter of
ksvm - if I, for example, add the parameter cross=10, i get the old
problem:
library("kernlab")
load("freeze_workspace.RDATA")
repl
You don't see the standard deviations because
only the final result (the output of summary() in
your case) is output by the function, not the
intermediate results (the results of the apply()
function in your case).
Try this:
resumen<-function(x) {
print( apply(x,2,sd,na.rm=TRUE))
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Manuel Jesús López Rodríguez
> Sent: Sunday, November 29, 2009 12:17 PM
> To: r-help@r-project.org
> Subject: [R] sequence of commands in R
>
> Dear all,
> I would like to know
On Sun, Nov 29, 2009 at 9:16 AM, Gabor Grothendieck wrote:
> By the way, if you really do want to create the formula anyways then:
>
>ix <- 1:2
>left <- paste(names(freeny)[ix], collapse = ",")
>fo <- as.formula(paste("cbind(", left, ") ~ ."))
>lm(fo, freeny)
>
> or possibly repla
Dear all,
I would like to know how could I execute a sequence or orders with just a
function, i.e, that just typing the function name, R gives me all the
parameters I want (for instance, if I want to see the summary, the standard
deviation, the number of valid cases, etc of a dataframe just with
Your function named 'gradient' is not the correct gradient. Take as an
example the following point x0, very near to the true minimum,
x0 <- c(-0.2517964, 0.4898680, -0.2517962, 0.4898681, 0.7500995)
then you get
> gradient(x0)
[1] -0.0372110470 0.0001816991 -0.0372102284 0.000182
On 11/26/2009 02:25 AM, Michael Hopgood wrote:
Hi Tom,
Thank you for the friendly and informative answer. It does explain a lot of
things, actually. As with any good answer, it inevitably leads to other
questions. In the first place, I need the arithmetic mean. It's what we
base our calculat
This works great.
Thanks for your help.
- Original Message
From: baptiste auguie
To: Jason Rupert
Cc: R-help
Sent: Thu, November 26, 2009 11:08:57 AM
Subject: Re: [R] Does nargin and nargout work with R functions?
Hi,
I think you can use match.call() to retrieve the number of ar
Peng Yu wrote:
>
> How to control what functions/classes are exported to a given namespace?
>
Namespace exports are set by package authors in the NAMESPACE file of an R
package. You could alter the NAMESPACE and rebuild the package yourself,
but an easier way would be to just use the ":::" o
Hi Jorge, Chuck and Kane,
thanks for your input!
The following code based on Jorge's answer did the trick to
standardize for subgroups within multiple columns:
# define a standardize function, but you could also define your custom
standardize function here
z.mean.sd <- function(data){
retu
How to control what functions/classes are exported to a given namespace?
On Sun, Nov 29, 2009 at 3:34 PM, baptiste auguie
wrote:
> Hi,
>
> They're not exported from the stats namespace,
>
> stats:::.Diag
> stats:::.asSparse
>
> ?":::"
>
> HTH,
>
> baptiste
>
> 2009/11/29 Peng Yu :
>> '.Diag' and
I don't find a good explanation on polynomial contrasts. Could
somebody recommend one to me?
__
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
an
I worry whether you understand what is happening when you lump all the
"unwanted levels" into a reference level. Be sure to watch the
intercept as you compare models. It will be some sort of adjusted mean
for whatever cases are in the reference levels of that and teh
reference levels of any
jerry83 yahoo.com> writes:
>
>
> Hi,
>
> I want to put a ggobi display into a GUI window setup by gWidgets, but error
> occur said it is not a S4 object.
>
> Does anyone have any idea about how to put it in or maybe it can not be put
> into a widget at all?
>
> Thanks A LOT!
To embed a GT
He is also the author of the rpad package and the email address on that
package on CRAN does correspond to a working web site.
On Sun, Nov 29, 2009 at 4:33 PM, Jason Rupert wrote:
> Dear Uwe,
>
> Thank you very much for your response.
>
> Maybe I should take the follow-up discussion about the ope
On 11/29/2009 4:23 PM, John Kane wrote:
> http://finzi.psych.upenn.edu/R/library/QuantPsyc/html/Make.Z.html
>
> Make.Z in the QuantPsych package may already do it.
For a single variable, you could use ave() and scale() together like this:
with(iris, ave(Sepal.Width, Species, FUN = scale))
T
Hi,
They're not exported from the stats namespace,
stats:::.Diag
stats:::.asSparse
?":::"
HTH,
baptiste
2009/11/29 Peng Yu :
> '.Diag' and '.asSparse' are defined in contrast.R. I'm wondering why I
> don't see them in my R session. Is it because that they start with
> '.'?
>
> ___
Dear Uwe,
Thank you very much for your response.
Maybe I should take the follow-up discussion about the open license issues off
line, as I have do not have a great deal of experience with such things,
especially with regards to the desired licenses for R packages and the existing
license for
http://finzi.psych.upenn.edu/R/library/QuantPsyc/html/Make.Z.html
Make.Z in the QuantPsych package may already do it.
--- On Sun, 11/29/09, Karsten Wolf wrote:
> From: Karsten Wolf
> Subject: [R] How to z-standardize for subgroups?
> To: r-help@r-project.org
> Received: Sunday, November 29, 20
'.Diag' and '.asSparse' are defined in contrast.R. I'm wondering why I
don't see them in my R session. Is it because that they start with
'.'?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting gu
My model has several independent and categorical variables. I would not like
to subset them as other variables in the data are useful. I just wanted to
set some coefficients for some levels in a single category.
A prototype of it can be something like y + constant *
(cat.variable1-Level1) ~ x1 + x2
Graham Williams wrote:
I don't have a Windows 7 to test this on yet - works on Vista and XP. Did
you install the GTK libraries (separately to R)?
I have manually tested on Windows Server 2008 64-bit where it works
(with most recent 32-bit binary versions of R, rattle and GTK).
Best wishes,
No "freezes" on a Mac using Ligges' 100 replication script. Tried it
twice. They took around a minute each time.
--
David.
--
David
On Nov 29, 2009, at 1:28 PM, Heiko Strathmann wrote:
Hello uwe,
Thanks for trying out.
the freeze happens after about 10 to 20 iterations. Did you try as
many
Jim,
Good catch! I know in my current problem there are objects with less than 3
rows, but will make sure to modify the function for future use.
Thanks,
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Sun, 11/29/09, Linlin Yan wrote:
> From: Linlin Yan
> Subject: Re: [R
I have constructed the function mml2 (below) based on the likelihood function
described in the minimal latex I have pasted below for anyone who wants to look
at it. This function finds parameter estimates for a basic Rasch (IRT) model.
Using the function without the gradient, using either nlminb
On Nov 29, 2009, at 12:41 PM, David Winsemius wrote:
On Nov 29, 2009, at 11:59 AM, DispersionMap wrote:
? like this do you mean...
Yes. Exactly. Unfortunately that data was passed though the summary
function which has done some odd things to the data, to wit:
> Weeks[Weeks==189]
2007-
Heiko Strathmann wrote:
Hello uwe,
Thanks for trying out.
the freeze happens after about 10 to 20 iterations. Did you try as many?
I just tried again:
library("kernlab")
load("freeze_workspace.RDATA")
replicate(100, ksvm(kernel="matrix", kernelMatrix, trainingDataYs,
type="C-svc", C=2))
a
Hi,
I've implemented the KhmaladzeTest for my linear quantile regression model for
the location-scale shift hypotesis as follow:
>formula=r ~ div + pe + por
>Ktest=KhmaladzeTest(formula,nullH="location-scale")
> Ktest
$nullH
[1] "location-scale"
$Tn
[1] 2.125804
$THn
div
Hello uwe,
Thanks for trying out.
the freeze happens after about 10 to 20 iterations. Did you try as many?
Am Sonntag, den 29.11.2009, 17:22 +0100 schrieb Uwe Ligges:
> I just tried
>
> ksvm(kernel="matrix", kernelMatrix, trainingDataYs, type="C-svc",
> cross=10, C=2)
>
> several times on both
Thomas Steiner gmail.com> writes:
>
> I have integers and I want R to give them back/output as Roman numerals:
> s=c(7,17)
> format(s,roman=T)
help.search("roman",agrep=FALSE)
?as.roman
as.roman(c(7,17))
__
R-help@r-project.org mailing list
https
I have integers and I want R to give them back/output as Roman numerals:
s=c(7,17)
format(s,roman=T)
is obviously wrong. Is there any other way/function to do this?
Thanks,
Thomas
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo
Late answer, but maybe still helpful (if I am correct this question is
still unanswered on R-help):
Clément Poirier wrote:
Dear useRs,
I'd like to plot a confidence interval on a periodogram. My problem is
that spec.pgram(sunspots,ci=0.95,log="yes") gives me a blue error bar on
the plot, b
On Nov 29, 2009, at 11:59 AM, DispersionMap wrote:
? like this do you mean...
Yes. Exactly. Unfortunately that data was passed though the summary
function which has done some odd things to the data, to wit:
> Weeks[Weeks==189]
2007-03-26 2007-07-09 2007-11-05 2008-02-25 2008-09-08 2009-0
See the Details section of ?lm where its all discussed.
On Sun, Nov 29, 2009 at 11:46 AM, Carl Witthoft wrote:
> Hi,
> A recent thread provided a (working) construct for lm:
>
> lm(as.matrix(freeny[ix]) ~., freeny[-ix])
>
>
> Can someone explain what is meant by the formula in that expression,
>
Sorry. Since . is commonly used but a matrix LHS less so I assumed you were
asking about the matrix part.
Dot means everything not on the LHS of the formula.
On Sun, Nov 29, 2009 at 12:18 PM, Carl Witthoft wrote:
> As others helpfully pointed out, the meaning of "." in a formula is
> provided i
I'm reading Section 2.4.1 Rules for Coding Expanded Formulas in
Statistical Models in S by Chambers and Hastie. It is a little
abstract for me. Could somebody point me some references that have
more examples and more explanations that can help me understand the
rules (in particular, the ones on pag
The help page for lm says:
"If ‘response’ is a matrix a linear model is fitted separately by
least-squares to each column of the matrix."
-Ista
On Sun, Nov 29, 2009 at 11:46 AM, Carl Witthoft wrote:
> Hi,
> A recent thread provided a (working) construct for lm:
>
> lm(as.matrix(freeny[ix])
On Nov 29, 2009, at 11:23 AM, sr danda wrote:
Hi,
I am a new R user. I am using it develop regression models with
categorical
variables.
Is there a way to force some regression coefficients to be zero for
some of
the values in a categorical variable (with 12 factor levels)?
I am recodin
As others helpfully pointed out, the meaning of "." in a formula is
provided in the Details section of ?formula. (But NOT in ?lm)
Ista Zahn wrote:
The help page for lm says:
"If ‘response’ is a matrix a linear model is fitted separately by
least-squares to each column of the matrix."
-I
? like this do you mean...
> dput(Weeks)
structure(c(370L, 342L, 333L, 317L, 308L, 298L, 289L, 269L, 265L,
257L, 254L, 253L, 252L, 249L, 243L, 243L, 239L, 239L, 236L, 234L,
233L, 232L, 230L, 230L, 229L, 229L, 229L, 228L, 227L, 226L, 225L,
222L, 218L, 217L, 216L, 215L, 215L, 214L, 214L, 214L, 2
On Nov 29, 2009, at 11:46 AM, Carl Witthoft wrote:
Hi,
A recent thread provided a (working) construct for lm:
lm(as.matrix(freeny[ix]) ~., freeny[-ix])
Can someone explain what is meant by the formula in that expression,
that is, what does "mymatrix~." do?
It doesn't say my matrix, it sa
Hi,
A recent thread provided a (working) construct for lm:
lm(as.matrix(freeny[ix]) ~., freeny[-ix])
Can someone explain what is meant by the formula in that expression,
that is, what does "mymatrix~." do? I couldn't find any such example
in the lm() or formula() help pages.
thanks
Carl
Jason Rupert wrote:
How do the R "powers that be" handle packages that are orphaned from CRAN?
>
Recently, I was looking for a function either part of the base functionality or an add-on package that mimicked the "poly" functionality from Octave (http://n4.nabble.com/Re-R-function-that-duplic
On 27/11/2009 3:36 PM, Alexander Søndergaard wrote:
Hi all,
I'm new to R. Having a functional background, I was wondering what's
the idiomatic way to iterate. It seems that for loops are the default
given there's no tail-call optimization.
I'm curious to know whether there is a way to transform
Hi,
I am a new R user. I am using it develop regression models with categorical
variables.
Is there a way to force some regression coefficients to be zero for some of
the values in a categorical variable (with 12 factor levels)?
I am recoding the values to the default value (1st in the order of d
I just tried
ksvm(kernel="matrix", kernelMatrix, trainingDataYs, type="C-svc",
cross=10, C=2)
several times on both workspaces and both returned some results after a
couple of seconds under the same versions (R version 2.10.0 and kernlab
0.9-9.) under Windows XP.
There mist be something el
How about a representation of the data that one could so something
with? By that I mean either by the "dump" method described in the
Posting Guide or by using dput:
> ttt <- c(1,2)
> dput(ttt)
c(1, 2)
> dump("ttt", stdout() )
ttt <-
c(1, 2)
Did you honestly expect anyone in their right mind
Антон Морковин wrote:
Dear all,
is there any functions which allow to calculate Student t-criterion using
means, their SE and sample size? I've seek for, but bulit-in t-criterion works
only with initial sample...
Not that I know, but then, you can easily calculate the t statistics in
a
David B. Thompson, Ph.D., P.E., D.WRE, CFM wrote:
Morning folks (at least here on the western side of the U.S.)...
This morning I constructed a contour plot of some bivariate
distributions I'm working with. When I attempted to add a second contour
to the plot using a dashed line (lty=2), R
Alexander Søndergaard wrote:
Hi all,
I'm new to R. Having a functional background, I was wondering what's
the idiomatic way to iterate. It seems that for loops are the default
given there's no tail-call optimization.
I'm curious to know whether there is a way to transform the following
toy sn
Hi,
I've implemented the KhmaladzeTest for my linear quantile regression model for
the location-scale shift hypotesis as follow:
>formula=r ~ div + pe + por
>Ktest=KhmaladzeTest(formula,nullH="location-scale")
> Ktest
$nullH
[1] "location-scale"
$Tn
[1] 2.125804
$THn
div p
Hi Karsten,
Let me assume your data is called d. If I understood what you are trying to
do, the following might help:
res <- apply(d, 2, tapply, d$group, scale)
res
See ?apply, ?tapply and ?scale for more information.
HTH,
Jorge
On Sun, Nov 29, 2009 at 10:41 AM, Karsten Wolf <> wrote:
> Hi f
Thanks again David,
Heres what happened:
> Weeks<-summary(cut(data$Raised.Date, breaks="weeks"))
> Weeks
2007-12-17 2009-01-05 2008-06-09 2008-12-08 2009-02-09 2008-12-01
370342333317308298
2008-05-12 2009-02-16 2007-01-22 2008-06-02 2007-01-29 20
Hi folks,
I have a dataframe df.vars with the follwing structure:
var1 var2 var3 group
Group is a factor.
Now I want to standardize the vars 1-3 (actually - there are many
more) by class, so I define
z.mean.sd <- function(data){
return.values <- (data - mean(data)) / (sd(dat
On Nov 29, 2009, at 10:09 AM, oscar linares wrote:
> Dear David,
>
> You are correct, I had a bug, here it is fixed. Note the times vector.
>
> # Need upright in same plots (e.g., dashed line)
> times <- seq(0,20, len=100)
> par(cex.lab=1.2,cex.axis=1.3)
>
> par(mfrow=c(2,1))
> plot(c(datasu$conc
On 2009.11.29 14:24:40, Prof Brian Ripley wrote:
> >> Windows 64-bit can certainly handle large memory spaces, but unless
> >> something has changed recently it my understanding Revolution
> >> Computing's 64-bit is the only 64-bit version of R available for
> >> Windows (due to the unavailability
Thank Jim! You are right. I didn't notice the case of none of rows
match the condition.
On Sun, Nov 29, 2009 at 10:10 PM, jim holtman wrote:
> One thing to be careful of is if no dataframe have less than 3 rows:
>
>> df1<-data.frame(letter=c("A","B","C","D","E"),number=c(1,2,3,4,5))
>> df2<-data.
On Nov 29, 2009, at 8:51 AM, oscar linares wrote:
Dear Wiz[R]ds,
I am deeply grateful for the help from Duncan Murdoch, Gray Calhoun,
and
others. We are almost there. For whatever reason, I can't change the
symbol
from a circle to a triangle in the upright posture plots. Any ideas?
I hav
On Sun, 29 Nov 2009, Jason Morgan wrote:
On 2009.11.28 21:50:09, Daniel Nordlund wrote:
- Is a Unix-like platform a better option than win-64? Again, would
this solve my memory limitation problems?
Possibly, but Win64 should provide plenty of memory (I believe Windows 7
Ultimate can use up to
On Nov 29, 2009, at 8:51 AM, oscar linares wrote:
Dear Wiz[R]ds,
I am deeply grateful for the help from Duncan Murdoch, Gray Calhoun,
and
others. We are almost there. For whatever reason, I can't change the
symbol
from a circle to a triangle in the upright posture plots.
Any ideas? I h
By the way, if you really do want to create the formula anyways then:
ix <- 1:2
left <- paste(names(freeny)[ix], collapse = ",")
fo <- as.formula(paste("cbind(", left, ") ~ ."))
lm(fo, freeny)
or possibly replace last line with:
eval(substitute(lm(fo, freeny))
which will cause th
Gabor Grothendieck a écrit :
Try this:
ix <- 1:2
lm(as.matrix(freeny[ix]) ~., freeny[-ix])
clean and clever!!! Thanks a lot!! You really simplified the code!!!
Just for curiosity, do you see why parse(eval)) was not working twice in
same formula?
thanks a lot!!
Matthieu
On Sun, Nov 29,
One thing to be careful of is if no dataframe have less than 3 rows:
> df1<-data.frame(letter=c("A","B","C","D","E"),number=c(1,2,3,4,5))
> df2<-data.frame(letter=c("A","B"),number=c(1,2))
> df3<-data.frame(letter=c("A","B","C","D","E"),number=c(1,2,3,4,5))
> df4<-data.frame(letter=c("A","B","C","
Jason Morgan wrote:
On 2009.11.28 21:50:09, Daniel Nordlund wrote:
- Is a Unix-like platform a better option than win-64? Again, would
this solve my memory limitation problems?
Possibly, but Win64 should provide plenty of memory (I believe Windows 7
Ultimate can use up to 192 GB of memory). Y
Try this:
ix <- 1:2
lm(as.matrix(freeny[ix]) ~., freeny[-ix])
On Sun, Nov 29, 2009 at 8:56 AM, Matthieu Stigler <
matthieu.stig...@gmail.com> wrote:
> Thanks for answering so fast!!
>
>>
>> lm(freeny)
>>
> :-)
> Ok that's working for the one equation case :-) Was example case...
>
> But now I
Hi Zi!
On 11/28/09, Yi Du wrote:
> I'd like to resort generalized methods of moment to estimate a regression
> model but I can't understand the help file of gmm package. If the regression
> model is y, the instrumental variable is x, how can I write the code?
>
> By the way, I use systemfit's 3sl
On 2009.11.28 21:50:09, Daniel Nordlund wrote:
> > > - Is a Unix-like platform a better option than win-64? Again, would
> > > this solve my memory limitation problems?
> >
> > Possibly, but Win64 should provide plenty of memory (I believe Windows 7
> > Ultimate can use up to 192 GB of memory). Yo
Thanks for answering so fast!!
lm(freeny)
:-)
Ok that's working for the one equation case :-) Was example case...
But now I want to have not only first column of freeny on the left but
both first? And I don't know their names a priori...
Thanks!
On Sun, Nov 29, 2009 at 8:49 AM, Matthie
On 28/11/2009 6:53 PM, Lars Bishop wrote:
Dear R users,
I’ve search the R site for help on this topic but it is hard to find a
precise answer for my questions.
Which are the best options to overcome the RAM memory limitation problems
when using R on “large” data sets (such as 2 or 3 million rec
Try this:
lm(freeny)
On Sun, Nov 29, 2009 at 8:49 AM, Matthieu Stigler <
matthieu.stig...@gmail.com> wrote:
> Hi
>
> My goal is to do a (multiple) regression, just knowing that my Y variables
> will be the say k first variables of a matrix/data frame. I thought I should
> do it with eval(pars
Hello again,
the freeze seems to depend on the kernel matrix.
With another kernel matrix of similiar size, gernerated with the same
kernel, but on another dataset, there is no freeze.
I have put a workspace with the working matrix and one with the freezing
matrix online for testing (see old email
Dear Wiz[R]ds,
I am deeply grateful for the help from Duncan Murdoch, Gray Calhoun, and
others. We are almost there. For whatever reason, I can't change the symbol
from a circle to a triangle in the upright posture plots. Any ideas? I have
included the problem in full.
# tritiated (3H)-Norepineph
Hi
My goal is to do a (multiple) regression, just knowing that my Y
variables will be the say k first variables of a matrix/data frame. I
thought I should do it with eval(parse)) but encounter a strange problem.
See:
lm(y~.-y, data=freeny) #that's what I want to do in the one equation case
#P
On Nov 29, 2009, at 7:52 AM, Linlin Yan wrote:
There is no year() function. Maybe you can try format() instead.
On Sun, Nov 29, 2009 at 8:44 PM, DispersionMap > wrote:
i have a column of dates in this format:
data[,"Raised.Date"] <- as.Date(data[,"Raised.Date"], "%d/%m/%Y");
data[1:10,"Rais
Hello,
I am using kernlab to do some binary classification on aminoacid
strings.
I am using a custom kernel, so i use the kernel="matrix" option of the
ksvm method.
My (normalized) kernel matrix is of size 1309*1309, my results vector
has the same length.
I am using C-svc.
My kernlab call is s
There is no year() function. Maybe you can try format() instead.
On Sun, Nov 29, 2009 at 8:44 PM, DispersionMap wrote:
>
> i have a column of dates in this format:
>
> data[,"Raised.Date"] <- as.Date(data[,"Raised.Date"], "%d/%m/%Y");
> data[1:10,"Raised.Date"]
> [1] "2006-07-07" "2006-07-07" "20
i have a column of dates in this format:
data[,"Raised.Date"] <- as.Date(data[,"Raised.Date"], "%d/%m/%Y");
data[1:10,"Raised.Date"]
[1] "2006-07-07" "2006-07-07" "2006-04-03" "2006-04-03" "2006-04-03"
"2006-04-03" "2006-04-03" "2006-04-03" "2006-04-03" "2006-04-03"
I can turn them into months
Arnab Maity wrote
>
>
>
>I like to call R from SAS. Could you please help me?
>
There are two methods that I know of:
1) Phil Rack has written a program called A Bridge to R see:
http://minequest.com/WordPress/?p=102
2) If you have SAS/IML licensed, you can link to R through IML studio. I
1 - 100 of 107 matches
Mail list logo