On Tue, 1 Dec 2009 14:48:04 +1100 Remko Duursma
wrote:
> any(duplicated(c(1,2,2)))
or
anyDuplicated(c(1,2,2))
which is slightly more efficient.
--
Karl Ove Hufthammer
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On Tue, 1 Dec 2009 14:10:17 +1300 Rolf Turner
wrote:
> Consider the following:
>
> > set.seed(42)
> > ff <- factor(sample(c(1,3,5),42,TRUE),levels=1:5)
> > x <- runif(42)
> > tapply(x,ff,sum)
> 12345
> 3.675436 NA 7.519675 NA 9.094210
>
>
Dear R helpers
If for a given data, I need to calculate Mean, Standard Deviation, Mode,
Median, Skewness, Kurtosis, is there any package in R, which will calculate
these moments?
Individually I can calculate these, but if there is any function which will
calculate these at a stretch, please le
On Dec 1, 2009, at 12:42 AM, johannes rara wrote:
Hi,
Is there a way to get different colour based on group when plotting
stripchart?
mydata <- data.frame(val1 = c(1.1, 3.2, 4.1, 2.5, 6.2, 5.3, 4.5, 2.2,
4.7, 2.7), val2 = c(4.2, 5.3, 3.4, 2.6, 5.3, 6.2, 7.7, 4.8, 3.4, 2.1),
group = rep(0:1
On Mon, 30 Nov 2009 16:47:41 + Henry Thorogood
wrote:
> The scale function seems to have tackled the skew, just looking at the
> boxplots for the data.
No, the 'scale' function does absolutely nothing to change the skew;
it's only a linear transformation.
--
Karl Ove Hufthammer
_
On Mon, 30 Nov 2009 14:35:45 -0800 (PST) DispersionMap
wrote:
> Is this book worth its dollar? If so, why?, if not, why not?
Yes, it's an excellent book, IMHO. It teaches you both the 'how' and the
'why' of ggplot2 in a nice, well-structured way, with lots of examples,
and information on vario
Peng Yu wrote:
>
> I see the follow explanation in help(cbind). I don't understand what
> the dispatch mechanism is. Could you point me what document or
> reference I should read?
>
> The method dispatching is _not_ done via 'UseMethod()', but by
> C-internal dispatching. Therefore, t
Hi,
The computer I'm using now doesn't have R on it, so this is a naive guess:
lines (density (y) )
regards
--
Charlotte Maia
http://sites.google.com/site/maiagx/home
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On Nov 30, 2009, at 11:07 PM, Peng Yu wrote:
x=list(a=c(1,2),b=c(3,4,5))
y=list(a=c(1,2),b=c(3,4,5))
lapply(seq(along=x),function(i){cbind(x[[i]],y[[i]])})
mapply("cbind", x, y)
I need to apply some function on two lists. I have to use the index to
do as shown above. But I feel more natur
Hello and thanks in advance
I am running a glm in R the code is as follows with residual diagnostic code
below
model4<-glm(Biomass~(Treatment+Time+Site)^2, data=bobB,
family=quasi(link="log", variance="mu"))
par(mfrow=c(2,2))
plot(model2)
to test the effect of grazing exclusion of feral ho
Hi,
Is there a way to get different colour based on group when plotting stripchart?
mydata <- data.frame(val1 = c(1.1, 3.2, 4.1, 2.5, 6.2, 5.3, 4.5, 2.2,
4.7, 2.7), val2 = c(4.2, 5.3, 3.4, 2.6, 5.3, 6.2, 7.7, 4.8, 3.4, 2.1),
group = rep(0:1, 5))
mydata.stack <- stack(mydata, select=-group)
strip
why don't you try this tmp=list(X=x,Y=y) and then lapply(tmp,fun(..))
Peng Yu wrote:
>
> x=list(a=c(1,2),b=c(3,4,5))
> y=list(a=c(1,2),b=c(3,4,5))
> lapply(seq(along=x),function(i){cbind(x[[i]],y[[i]])})
>
>
> I need to apply some function on two lists. I have to use the index to
> do as shown
Sorry, I just posted the email below but realised I did not give a name or
details, would it be possible to adjust before posting and send what is
below, sorry again, first time user...
From: Joanne Lenehan [mailto:jlene...@une.edu.au]
Sent: Tuesday, 1 December 2009 3:51 PM
To: 'r-help@r-proje
R 2.9
Windows XP
Is there an R function that will compute a measure of agreement(e.g. Kappa)
among three or more observers?
Thank you,
John
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
__
R-help@r
I see the follow explanation in help(cbind). I don't understand what
the dispatch mechanism is. Could you point me what document or
reference I should read?
The method dispatching is _not_ done via 'UseMethod()', but by
C-internal dispatching. Therefore, there is no need for, e.g.,
Suppose I run the following code in the R session. At the last prompt
'>', I want to retrieve the second command (staring with 'y'). But I
have to type up arrow many times, which is very inconvenient. I'm
wondering if there is a way to configure R to skip block of code in
the history?
> x=list(a=c
x=list(a=c(1,2),b=c(3,4,5))
y=list(a=c(1,2),b=c(3,4,5))
lapply(seq(along=x),function(i){cbind(x[[i]],y[[i]])})
I need to apply some function on two lists. I have to use the index to
do as shown above. But I feel more natural to refer to the lists
without using the index (using the syntax of somet
any(duplicated(c(1,2,2)))
r
-
Remko Duursma
Post-Doctoral Fellow
Centre for Plants and the Environment
University of Western Sydney
Hawkesbury Campus
Richmond NSW 2753
Dept of Biological Science
Macquarie University
North Ryde NSW 2109
Australia
See ?duplicated. But really I don't see anything wrong with your approach.
-Ista
On Mon, Nov 30, 2009 at 10:42 PM, Peng Yu wrote:
> length(unique(c(1,2,2)))==length(c(1,2,2))
>
> I use the above test to test if all the elements in a vector are
> unique. But I'm wondering if there is a convenient
Maybe you can just use a list?
mylis <- list("x", c("y","z"))
element 1 is then mylis[[1]], which is x, and so forth.
But it looks like you really need to 'map' some numbers to your list.
You could use separate vector that keeps track of the index numbers:
index <- c(0,1)
and then map like thi
On 12/01/2009 01:56 PM, Brock Tibert wrote:
Hi all,
I am new to R, but with the help I have received thus,I have already been able
to work through a bunch of questions on my own in just a few days. I imagine
my questions are relatively simple, but here she goes.
1) I am using a package where
length(unique(c(1,2,2)))==length(c(1,2,2))
I use the above test to test if all the elements in a vector are
unique. But I'm wondering if there is a convenient function to do so
in R library.
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It seems that an vector or other non elemental data type can not be
assigned to an element in the data.frame. I'm wondering what is the
walk around.
> li=data.frame(a=c(0,1), b=c('x','y'))
> li$b[[1]]= 'x'
> li$b[[2]]<- c('y','z')
Error in li$b[[2]] <- c("y", "z") :
more elements supplied than t
Hi all,
I am new to R, but with the help I have received thus,I have already been able
to work through a bunch of questions on my own in just a few days. I imagine
my questions are relatively simple, but here she goes.
1) I am using a package where one of the objects generated represents X/Y
Hi Harold,
Hans is right. You can see this if you use BB::spg
require(BB)
opt2 <- spg(startVal, fn) # this is fine
opt3 <- spg(startVal, fn,gradient) # this is not fine!
> opt3 <- spg(startVal, fn, gradient)
Gradient check details: max. relative difference in gradients= 0.001697913
It was arbitrary data and I made a mistake. Thanks for your help
nonetheless.
I solved the problem using matpoints, which does the job quite nicely:
> lx100=c(1,1,1,.8,.5,.4,.2,0)
> day100=c(0,1,2,3,4,5,6,7)
> 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)
> lx0=c(1,1,1,1,1
Hi,
i already read everything that is in the site that it is in English , i not
figuring out how to start.
I don t understand how to look at the data that where given to me and began
to relate them.
I m sorry for my english but it s the best i can do, im not american or
english so i write with
Consider the following:
> set.seed(42)
> ff <- factor(sample(c(1,3,5),42,TRUE),levels=1:5)
> x <- runif(42)
> tapply(x,ff,sum)
12345
3.675436 NA 7.519675 NA 9.094210
I got bitten by those NAs in the result of tapply(). Effectively
one is summi
Here is what you want:
xout <- c(1,5,10,25,50,100)
for(i in xout) { print(paste("Areal_Ppt_",i,"sqmi.txt", sep="")) }
Notice that 'i' will be assigned each value in xout; you do not have
to index into the vector. Notice that you second value is 50 which is
xout[5].
On Mon, Nov 30, 2009 at 7:49
Hi Douglas,
You were almost there. Just remember that your iterator is i and not xout.
R> xout <- c(1,5,10,25,50,100)
R> for(i in xout) print(paste("Areal_Ppt_", i,"sqmi.txt", sep=""))
[1] "Areal_Ppt_1sqmi.txt"
[1] "Areal_Ppt_5sqmi.txt"
[1] "Areal_Ppt_10sqmi.txt"
[1] "Areal_Ppt_25sqmi.txt"
[1] "A
On 30/11/2009 7:49 PM, Douglas M. Hultstrand wrote:
Hello,
I am trying to create subsets of grouped data (by area size), and use
the area size as part of the output name. The code below works for area
(xout) 1 and 50, the other files are given NA for an area.
A simple example:
xout <- c(1,5
Hello,
I am trying to create subsets of grouped data (by area size), and use
the area size as part of the output name. The code below works for area
(xout) 1 and 50, the other files are given NA for an area.
A simple example:
xout <- c(1,5,10,25,50,100)
for(i in xout) { print(paste("Areal_Pp
Phil and Peter, thanks - figured it was something obvious. I had
uninstalled splines to check the behavior of some other code. I just
reinstalled R 2.10.0 and it shows up. Thanks! Alan
Peter Ehlers wrote:
Have you searched your computer?
Rhome/library/splines
It's one of the packages autom
On Nov 30, 2009, at 6:47 PM, Alan Swanson wrote:
Dear R gurus,
I am using the gbm library under R version 2.10.0, which depends on
the splines library. I have searched the various repositories and
can't seem to locate any current version of the splines library.
This is strange because I
Have you searched your computer?
Rhome/library/splines
It's one of the packages automatically installed with R.
-Peter Ehlers
Alan Swanson wrote:
Dear R gurus,
I am using the gbm library under R version 2.10.0, which depends on the
splines library. I have searched the various repositories
Dear R gurus,
I am using the gbm library under R version 2.10.0, which depends on
the splines library. I have searched the various repositories and
can't seem to locate any current version of the splines library. This
is strange because I had previously installed splines (built under
2.10.0
I agree with Ista.
The book is very interesting. I put my 2cents too.
milton
On Mon, Nov 30, 2009 at 6:20 PM, Ista Zahn wrote:
> Depends on what you want. It's possible to find most of the
> information contained in the book in other places (the ggplto2
> website, Hadley's dissertation, the gg
On Mon, Nov 30, 2009 at 12:01:12PM +0100, Trafim wrote:
> Dear all,
>
> I cannot find a function which would allow drawing hist and density on the
> same graph.
>
> x <- seq(1,40,1)
> y <- 2*x+1+5*rnorm(length(x))
>
> hist(y,freq = FALSE)
> plot(density(y))
The package descr has the function hi
Depends on what you want. It's possible to find most of the
information contained in the book in other places (the ggplto2
website, Hadley's dissertation, the ggplot2 mailing list archive).
Personally I really appreciate having all that information in one
place, and in an easy to read format. It's
Thanks... that worked
is there a way to run r script?
for example open -n /Applications/R.app myscript.r
when I use this syntax... I get my script file opened but not executed :(
Thanks in advance
Ken Knoblauch wrote:
>
> chronos.phenomena gmail.com> writes:
>
>>
>>
>> This is really
Peter,
Thanks, layout() does exactly what I want. Now I also realize why I wasn't
able to get the legend where I wanted it. I didn't realize par(xpd=...) was
only for the last plot. That explains why I was getting weird results when
trying to move my legend around!
Aloha,
Tim
Tim Clark
D
Hello everyone:
I tried to fit a Beta distribution on a right-skewed dataset using:
fitdistr(temp,densfun="beta",start=list(shape1=3,shape2=2))
To assess the fit, I proceeded as follows:
Using distribution parameters from the sample resulting from fitdistr()
function, I generated 1000 samples
Is this book worth its dollar? If so, why?, if not, why not?
Cheers.
--
View this message in context:
http://n4.nabble.com/Ggplot2-Elegant-Graphics-for-Data-Analysis-Use-R-2009-Paperback-tp931702p931702.html
Sent from the R help mailing list archive at Nabble.com.
Hey!
Can anyone help me coding in R a normalized kernel matrix.
Basically, I want
K(x,y)/sqrt(*K*(*x, x*)*K*(*y, y*))
Anyone has a piece of code that you could share?
Many thanks,
Parmee
[[alternative HTML version deleted]]
__
R-help@r-proje
You should do your own homework.
cheers,
Rolf Turner
On 1/12/2009, at 10:37 AM, susan jacobs wrote:
Census from 1990 california in USA. It’s a data frame with data
from 20640
zones from Califórnia. Zones are chosen as being geograficly near,
wich one
with 1500 pe
Tim Clark wrote:
Dear List,
I am trying to plot a color.legend() in the right outer margin of my device region. I have read multiple threads on the subject and still can't get it right. I have stolen an example from one of the threads to demonstrate my problem. I have extended the outer marg
As Ben Bolker has indicated, I am working on various improvements to the
functionality of optim() along with others, esp. Ravi Varadhan and Kate
Mullen.
With relevance to the posts by Sebastien Bihorel and Ben Bolker about
output of point/function value information on each evaluation, I am
wo
Sorry, heres the message:
Hi
yes the freeze is rather a result of the kernelMatrix being non-positive
definite, have a look at the eigenvalues of kernelMatrix
with eigs and you will see even small negative values. This is usualy
the result of small numerical inacuracies in the computation of the
Census from 1990 california in USA. It’s a data frame with data from 20640
zones from Califórnia. Zones are chosen as being geograficly near, wich one
with 1500 persons (mean).
For each zone are provided several informations in the colums from data
frame:
- longitude e latitude from the geograf
Hi,
im working in R but honestly i don t know how to apply the formulas in my
problem, can someone give some help?
I already learn how we use the formulas but im not understanding the main
issue in my problem.
Johannes Huesing wrote:
>
> Esmail [Mon, Nov 30, 2009 at 01:07:15PM CET]:
>
>>
Hi there,
I am trying to use funcion metaMDS (vegan pakage) for Community Ecology
data, but I find no way to calculate the "expressed variance" of the first 2
axis? is there a way to do that?
Thanks a lot in advance,
Gian
--
View this message in context:
http://n4.nabble.com/how-to-calculate-
Data Analysis and Graphics Using R - An Example-Based Approach
John Maindonald and John Braun
2nd edn, Cambridge University Press, January 2007
susan jacobs wrote:
>
> Hi,
>
> im working in R but honestly i don t know how to apply the formulas in my
> problem, can someone give some help?
On Mon, 30 Nov 2009, Karin Groothuis wrote:
Hello,
I have installed R on my Apple Laptop. Next I wanted to install the package
Rcmdr which requires the package tcltk. But then I get errors like:
The downloaded packages are in
/var/folders/0p/0pD8fDrwHouNDsQ+k8dGmU+++TI/-Tmp-//RtmpSp
Hello,
I have installed R on my Apple Laptop. Next I wanted to install the
package Rcmdr which requires the package tcltk. But then I get errors
like:
The downloaded packages are in
/var/folders/0p/0pD8fDrwHouNDsQ+k8dGmU+++TI/-Tmp-//RtmpSp4q7p/
downloaded_packages
Loading required packag
Dear List,
I am trying to plot a color.legend() in the right outer margin of my device
region. I have read multiple threads on the subject and still can't get it
right. I have stolen an example from one of the threads to demonstrate my
problem. I have extended the outer margin using par(oma(
Esmail [Mon, Nov 30, 2009 at 01:07:15PM CET]:
> ps: Just checked, 'R in a Nutshell':
> http://oreilly.com/catalog/9780596801717
> release date dec 2009/jan 2010
You can't really judge a book by its cover, and in this case not even
the cover is presented to us. But judging by the author's
Hello again,
i got this message from the maintainer:
Am Montag, den 30.11.2009, 11:22 +0100 schrieb Heiko Strathmann:
> Ok, i reported it.
> thanks for trying it out again.
>
> Am 30. November 2009 11:06 schrieb Uwe Ligges
> :
> My apologies, that must have been a copy&paste error and t
Hi
I have a ts from 1980:1 trough 2009:9 with a frequency of 12. I want to make
25 6 month forecast where the first 6 month forecast is between 2007:4 -
2007:9 and the last 2009:4 - 2009:9.
My script:
Forecast <- list()
for (i in 1:25){
j <- i + 326
Data <- window(omxr, end = time(omxr)[j])
Res
On Nov 30, 2009, at 2:24 PM, Henrique Dallazuanna wrote:
Dim argument must be a integer, see the return of:
Less true than you might expect:
> x <- 3.5
> y <- 5.2
> z <- array(0, dim=c(x,y))
> z
[,1] [,2] [,3] [,4] [,5]
[1,]00000
[2,]00000
[3,]
You can try something about like this:
yourColor <- c(250,235,215)
colors()[which.min(colSums(col2rgb(colors()) - yourColor
)[colSums(col2rgb(colors()) - yourColor) >= 0])]
On Mon, Nov 30, 2009 at 5:25 PM, Jorgy Porgee wrote:
> Hi all, I'm trying to convert a series of rgb codes into a color na
Fantastic. Thanks Duncan. Clearly my google key words were the wrong
ones, nothing of this sort came up. Hope my title is easier to find
somehow..
Regards,
George
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PLEASE
Try this:
> mycolor <- col2rgb("red"); mycolor
[,1]
red255
green0
blue 0
> colors()[sapply(lapply(colors(), col2rgb), identical, mycolor)]
[1] "red" "red1"
On Mon, Nov 30, 2009 at 2:25 PM, Jorgy Porgee wrote:
> Hi all, I'm trying to convert a series of rgb codes into a color
On 30/11/2009 2:25 PM, Jorgy Porgee wrote:
Hi all, I'm trying to convert a series of rgb codes into a color name.
What is my easiest option? So far I'm stuck with just converting to
hex using rgb() and I know R knows a number of colours() but a mapping
of the two has failed me.
Any help in this
In answer to your 4th question: you are trying to allocate an object
whose size is 1/4 of the addressable memory on your machine. You might
be lucky and have that much available in one piece, but I'm guessing you
don't, which is why the allocation fails. So don't worry about
questions 1,2, o
Hi all, I'm trying to convert a series of rgb codes into a color name.
What is my easiest option? So far I'm stuck with just converting to
hex using rgb() and I know R knows a number of colours() but a mapping
of the two has failed me.
Any help in this regard will be highly appreciated.
Regards,
Dim argument must be a integer, see the return of:
as.integer(slope*xdim)
Try this:
udim <- ceiling(max(slope*xdim,5))
On Mon, Nov 30, 2009 at 5:15 PM, Rupert Mazzucco wrote:
> Hello,
>
> I'm running into a very strange problem:
>
>> xrange <- c(-2.5,2.5)
>> xdim <- 100
>> mobility <- 0.1
>> s
Simply a manifestation of FAQ 7.31, i.e., a floating-point arithmetic issue.
On my machine,
> 1.16 * 100 == 116
[1] FALSE
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Rupert Mazzucco
> Sent: Monday, November 30, 2009 1:
Proably ole Faq 7.21 again. You are using floating point numbers and
expecting coercion to result in upward rounding.
My guess is that if you use trunc(udim) you will get a different number.
Yep:
> trunc(udim)
[1] 115
On Nov 30, 2009, at 2:15 PM, Rupert Mazzucco wrote:
Hello,
I'm running
Dear R-guru's,
I have some data that I'm writing to a mysql database. I need the database to
generate the primary key and it's not being generated as a unique ID. My
understanding was that primary keys should be unique. I submit a data.frame
with named columns. The only name that it's missing i
Hello,
I'm running into a very strange problem:
> xrange <- c(-2.5,2.5)
> xdim <- 100
> mobility <- 0.1
> slope <- 1.16
> urange <- slope*xrange
> udim <- max(slope*xdim,5)
> du <- (urange[2]-urange[1])/udim
> uvec <- urange[1]+(1:udim-0.5)*du
> # type dependent we
This seems to work quite well and gives an ~85% speed increase compared to
the for loop. Thanks a bunch!
Peter Ehlers wrote:
>
> Or maybe aperm() is faster:
>
> max(colSums(aperm(yourArray, c(2,1,3
>
> -Peter Ehlers
>
> Peter Ehlers wrote:
>> Does this
>>
>> max(apply(yourArray, 3,
This gives the correct result but appears to be slower than the for loop I
was using, I do appreciate the suggestion though.
Peter Ehlers wrote:
>
> Does this
>
> max(apply(yourArray, 3, rowSums))
>
> give you what you want?
>
> -Peter Ehlers
>
> Will Carr wrote:
>> Working with an NxMx
David Winsemius wrote:
>
> What are M, N, and O? All the same magnitude?
>
In this case, 4,13,10.
> What are you doing based on that if( ) determination?
>
Increasing a counter.
> Are you sure you are interpreting your profile results correctly?
>
No, but I think I am. I'm using proc.t
Let me begin stating that I read all help files and faq's on the subject
matter (there aren't more than about a dozen) but either did not find
solutions or found them not to work.
Here is the issue. I am trying to run a spatial regression on a
medium-sized dataset. Part of the functions in the s
One of the reasons we ask for a *reproducible* example, is that
it allows us to test our ideas and make sure that all the details
are taken care of. Here's a reproducible example that may help
solve your problem:
lx100=c(1,1,1,.8,.5,.4,.2,0)
day100=c(0,1,2,3,4,5,6,7)
lx90=c(1,1,1,1,.9,.8,.6,.4,.
On Nov 30, 2009, at 11:10 AM, Will Carr wrote:
Working with an NxMxO sized matrix, currently I can do this in my
code:
What are M, N, and O? All the same magnitude?
if (max(colSums(array)) >= number)
But to get an equivalent result using rowSums, I have to do:
for (i in 1:10)
{
if (max
Or maybe aperm() is faster:
max(colSums(aperm(yourArray, c(2,1,3
-Peter Ehlers
Peter Ehlers wrote:
Does this
max(apply(yourArray, 3, rowSums))
give you what you want?
-Peter Ehlers
Will Carr wrote:
Working with an NxMxO sized matrix, currently I can do this in my code:
if (max(co
On Nov 30, 2009, at 11:20 AM, Trafim wrote:
Sorry, maybe I confused everybody with what I want.
I have time process X_k, and corresponding returns process R_k
I am interesting in f(r|x), which I can find knowing f(r,x) and
f(x). I know
the latter I need the first.
How can I apploximate thi
Great! Have managed to shift my dataset to positive. Thanks for all your
help. Henry.
2009/11/30 David Winsemius
>
> On Nov 30, 2009, at 11:47 AM, Henry Thorogood wrote:
>
> The scale function seems to have tackled the skew, just looking at the
>> boxplots for the data.
>>
>> The boxcox functio
On Nov 30, 2009, at 11:47 AM, Henry Thorogood wrote:
The scale function seems to have tackled the skew, just looking at the
boxplots for the data.
The boxcox function I'm using is boxcox(), from MASS.
I'm puzzled. When I look at the boxcox function in MASS it says:
"Arguments
object a f
Does this
max(apply(yourArray, 3, rowSums))
give you what you want?
-Peter Ehlers
Will Carr wrote:
Working with an NxMxO sized matrix, currently I can do this in my code:
if (max(colSums(array)) >= number)
But to get an equivalent result using rowSums, I have to do:
for (i in 1:10)
{
if
Julia,
I can vouch for the 'Statistical Analysis in R'. Very good book, I learnt R
from it. The other one to look at is Spoetry. It solves a lot of problem for me.
Paul
http://www.burns-stat.com/pages/spoetry.html
On 30 Nov 2009, at 17:17, Ben Seligman wrote:
> Hi Julia,
>
> I'm also a newc
It appears that you were reading the number in as an integer and not
numeric. The value that you are seeing (-596864072) is the numeric
value trucated to 32 bit. The number would have been in hex
(6DC6C93B8) but dropping the leading '6' you will get the result as a
32 bit integer. Check your dat
Hi Julia,
I'm also a newcomer to R and to this listserv (in the past two weeks). One
book that was recommended to me, and has been extraordinarily helpful in
learning how to really make use of R, is "A Beginner's Guide to R", which
focuses on data entry/import, data manipulation, functions, and p
The scale function seems to have tackled the skew, just looking at the
boxplots for the data.
The boxcox function I'm using is boxcox(), from MASS. I've looked through
the help page, but I don't think (from what I can see) there's a way to make
the boxcox function handle the negative values, unlik
Hello,
I am trying to import data from an SQLite database to R.
Unfortunately, I seem to get wrong data when I try to import very large
numbers.
For example:
I look at the database via SQLiteStudio(v.1.1.3) and I see the following
values:
OrderID Day TimeToclose
1 2009-
On Nov 30, 2009, at 11:02 AM, Henry Thorogood wrote:
Hi,
I'm doing some work with linear models, and I've scaled my data
using the
scale(dataset) function. This was great at removing the skew, but I
now
can't perform the Box Cox transformation on the data set (using the
boxcox(dataset) fu
Hi Robert and Wai-Kuan,
I'd be interested as well. I see in the JMP Scripting Book that there appear
to be 3 possibilities (all in Chapter 11-at least for version 8):
1. using a JMP DataFeed object
2. using sockets
3. OLE automation
Sockets with something like RServe may be the way to go. I have
I used the archives, but I didn't use the FAQs. I removed the
compiled version- and all other R versions. I reinstalled through the
repository, and everything works fine. Sorry for the inconvenience.
regards,
Stephen
On Mon, Nov 30, 2009 at 10:14 AM, Dirk Eddelbuettel wrote:
>
> On 30 Novembe
Sorry, maybe I confused everybody with what I want.
I have time process X_k, and corresponding returns process R_k
I am interesting in f(r|x), which I can find knowing f(r,x) and f(x). I know
the latter I need the first.
How can I apploximate this f(r,x) in the such way that later I can find
f(X=
Hi,
I am building a regression tree (method=anova) by using rpart package
and as a final result I get the final leaves characterized by
different means and standard deviations for the dependent variable.
However, differently from the classification tree for categorical
variables I cannot find a way
Working with an NxMxO sized matrix, currently I can do this in my code:
if (max(colSums(array)) >= number)
But to get an equivalent result using rowSums, I have to do:
for (i in 1:10)
{
if (max(rowSums(array[,,i])) >= number)
}
I'm running both in a much larger loop that loops millions of tim
On 30 November 2009 at 15:12, Peter Dalgaard wrote:
| stephen sefick wrote:
| > Error in firstlib(which.lib.loc, package) :
| > Tcl/Tk support is not available on this system
| > Error : package 'tcltk' could not be loaded
| >
| > Ubuntu 9.10
| > I have both the 8.5 and 8.4 tcl and tk regular a
Also what if the grid matrix is not squared? How can I then find this kernel
density matrix/
Thanks a lot.
On Mon, Nov 30, 2009 at 4:56 PM, Trafim wrote:
> You are right.
> I brought this example just to see how to use its with two time series. In
> reality I have price process and returns, and
Hi,
I'm doing some work with linear models, and I've scaled my data using the
scale(dataset) function. This was great at removing the skew, but I now
can't perform the Box Cox transformation on the data set (using the
boxcox(dataset) function), as the scaling has returned negative values.
So my q
Dear Julia,
Welcome. It is good that you wish to learn more about R.
R has certainly become very vast in the last few years. Do you wish to
learn R for a particular reason (financial analyses, multivariate,
prediction/classification, genetics)? You might get more targeted
reading materials, b
Hi,
I am building a regression tree (method=anova) by using rpart package
and as a final result I get the final leaves characterized by
different means and standard deviations for the dependent variable.
However, differently from the classification tree for categorical
variables I cannot find a way
You are right.
I brought this example just to see how to use its with two time series. In
reality I have price process and returns, and I need conditional density.
Will highly appreciate your help.
I found function kde2d but cannot understand how to call for the values of
the matrix.
On Mon, No
Hi,
I am building a regression tree (method=anova) by using rpart package
and as a final result I get the final leaves characterized by
different means and standard deviations for the dependent variable.
However, differently from the classification tree for categorical
variables I cannot find a way
I've just realised that this is a very silly post as I can't read!!! The
output in the "anova" is the excluded variables - very sorry!
Laura
2009/11/30 Laura Bonnett
> Dear all,
>
> I have decided after much deliberation to use backward elimination and
> forward selection to produce a multivar
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