Hi Jorge
It worked!
lapply(1:length(mylist), function(i) quantile(mylist[[i]], probs =
B[i])) #like yours but without coma after i
[[1]]
99.62995%
0.7989808
[[2]]
55.33357%
-0.653456
[[3]]
72.30253%
-0.4872166
[[4]]
39.69968%
-1.138819
[[5]]
29.84813%
On 17 May 2010 22:55, RATIARISON Eric wrote:
> It's ok Arne, i've build the MS windows binary.
>
> And the result is ok (sandwich function runs)with this next modifications in
> my user specified functions
> But with the following functions (individual one):
>
>
> sc= as.vector( c(log(mean(v)),re
Sorry, my attempt wasn't quite good enough. I didn't
consider the possibility of a 'negative' value in a
character/factor column. To fix that, see inline below.
On 2010-05-17 14:32, Peter Ehlers wrote:
On 2010-05-17 12:54, Henrique Dallazuanna wrote:
Try this:
newData<- sapply(numdat, functi
"Lorenzo Cattarino" writes:
> Hi Jim,
>
> Thanks for your reply. Your codes does work but I was hoping to find a
> way to use lapply and avoid the for loop.
>
> Lorenzo
>
>
> -Original Message-
> From: Jim Lemon [mailto:j...@bitwrit.com.au]
> Sent: Monday, 17 May 2010 8:27 PM
> To: Loren
Monte Shaffer writes:
> Hello,
>
> I have programmed in PHP a lot, and wanted to know if anyone figured out
> Variable variables using R.
>
> For example, I have several dataframes of unequal sizes that relate to L
> treatments (1, 2, 3, 4, 5,6, L) ... in this case L=7
You should create a list c
Thanks for teaching me something new. :)
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-ldply-tp2219094p2220473.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://sta
Hello,
I have programmed in PHP a lot, and wanted to know if anyone figured out
Variable variables using R.
For example, I have several dataframes of unequal sizes that relate to L
treatments (1, 2, 3, 4, 5,6, L) ... in this case L=7
fData.1
unique.1
fit.nls.1
summary.nls.1
fit.var.1
summary.var
Dear R-help,
Let me know if I should email r-devel instead of this list. This
message is addressed to Professor Lumley or anyone familiar with the
survey package.
Does svycoxph() implement the method outlined in Binder 1992 as
referenced in the help file? That is, are weights incorporated in th
Hi all,
I am looking to create a rank variable based on a continuous variable
for subsets of the data. For example, for an R integrated data set
about US states this is how a loop could create what I want:
### Example with loop
data <- cbind(state.region,as.data.frame(state.x77))[,1:2] #
choo
Dear R-help,
I would like to compute the variance for the proportion of treatment
effect by a surrogate in a survival model (Lin, Fleming, and De
Gruttola 1997 in Statistics in Medicine). The paper mentioned that
the covariance matrix matches that of the covariance matrix estimator
for the margin
Hi Jim,
Thanks for your reply. Your codes does work but I was hoping to find a
way to use lapply and avoid the for loop.
Lorenzo
-Original Message-
From: Jim Lemon [mailto:j...@bitwrit.com.au]
Sent: Monday, 17 May 2010 8:27 PM
To: Lorenzo Cattarino
Cc: r-help@r-project.org
Subject: Re:
jim holtman writes:
> try:
>
> pattern="*result*\\.csv$"
Just for the record, that's not quite correct. The * doesn't behave like
in a shell glob. Instead, * says "0 or more copies of the previous
character". So the above pattern picks up resul.csv, which I don't think
was intended.
I don't kno
Hello,
I am a bit over my head on this issue. My colleagues and I are running
R off of our server. We all have admin rights and prior to yesterday
we all had our own libraries. Our main system administrator advised us
that we should have a shared library. So, I am trying to do this.
I have downl
Dear R-help,
Let me know if I should email r-devel instead of this list. This
message is addressed to Professor Lumley or anyone familiar with the
survey package.
Does svycoxph() implement the method outlined in Binder 1992 as
referenced in the help file? That is, are weights incorporated in th
Thank you very much - it's very helpful!
Dimitri
On Mon, May 17, 2010 at 9:39 PM, Dan Davison wrote:
> Dimitri Liakhovitski writes:
>
>> Thanks again - and one follow-up question.
>> When I do do.call(rbind, lapply(dir(patt = "\\.csv$"), read.csv))
>> What is the right way to speicify (probably
Dear All,
Since no one has answered my previous question, let me revise it a
little and ask again.
My data set contains about 10,000 women born in 60 months. The outcome
variable is a binary variable indicating whether one has certain
health problems. My hypothesis is that the 60 months in which
Dimitri Liakhovitski writes:
> Thanks again - and one follow-up question.
> When I do do.call(rbind, lapply(dir(patt = "\\.csv$"), read.csv))
> What is the right way to speicify (probably under "patt") that I only
> need to grab those .csv files that contain a certain string, e.g.,
> "result"?
I
try:
pattern="*result*\\.csv$"
On Mon, May 17, 2010 at 9:06 PM, Dimitri Liakhovitski
wrote:
> Thanks again - and one follow-up question.
> When I do do.call(rbind, lapply(dir(patt = "\\.csv$"), read.csv))
> What is the right way to speicify (probably under "patt") that I only
> need to grab thos
Thanks again - and one follow-up question.
When I do do.call(rbind, lapply(dir(patt = "\\.csv$"), read.csv))
What is the right way to speicify (probably under "patt") that I only
need to grab those .csv files that contain a certain string, e.g.,
"result"?
I tried a couple of things, like patt= "\\.
Sorry, I was talking nonsense.
The actual problem was in your panel function, where you extract the
state boundaries, and then draw them using panel.polygon: the order of
states from map() is arbitrary, and does not correspond to the color
palette that you set up.
I suggest using mapplot(). This
Dear All,
I am trying to do a Bayesian change point analysis. My dependent
variable is a proportion, so the natural choice will be a Poisson
model with offset. Now the two options for change point model with R
is MCMCpack and bcp, but neither seems to be able to handle such
model, am I right? Are
Hi,
My understanding is that Chebyshev polynomials solve the minimax
approximation problem. If this correct, what you need is an algorithm to
compute Chebyshev polynomial approximation. I have written an R function to
do this. See the attached code that contains the function and an example.
Is
Thanks a lot for your help Duncan,
message forwarded.
Jean-Christophe
On Wed, May 12, 2010 at 15:20, Duncan Murdoch wrote:
> On 12/05/2010 8:34 AM, Jean-Christophe Domenge wrote:
>
>> Dear R gurus,
>> some guy at my company would like to compile a script written in R.
>> In short he would like t
Dear R-users,
I learned today that there exists an interesting topic in numerical
analysis names "best polynomial approximation" (BSA). Given a function
f the BSA of degree k, say pk, is the polynomial such that
pk=arginf sup(|f-pk|)
Although given some regularity condition of f, pk is unique, p
On Mon, 17 May 2010, ivan popivanov wrote:
library(xts)
library(TTR)
ndx = getYahooData("^NDX")
aa = ndx$Close
bb = aggregate(aa, as.yearweek, tail, 1)
The last operation takes forever, and then the bb dates are messed up.
The problem is problem due to the as.yearweek() function which is ne
On May 17, 2010, at 5:01 PM, Adam November wrote:
I believe I'm working with the newest version of R (2.11.0) and I've
tried a few of the most recent versions... No luck yet.
Luck has nothing to do with it. What part of ERROR in the CRAN package
check for the current version of lme4 with th
Thanks for pointing that out, baptiste.
No need to use windows(). Width and height should be specified in pdf().
On Mon, May 17, 2010 at 3:28 PM, baptiste auguie
wrote:
> No, that's only true for lattice and ggplot2 graphics. The problem
> here is with this line,
>
> windows(width=5, height=5)
>
Greetings
does anyone know of an R package that can do quantile regression with
instrumental variables. I've found 'sem' for IV estimation and 'quantreg' for
quantile regression but I would like to find something that can do a quantile
regression with instrumental variables.
Cheers,
Neil
==
You could also put the call to system.time inside the function itself:
f = function(x) {
system.time({
... #function's code
ret_val = ...
}); flush.console();
return ret_val;
}
i s'pose you'd miss out on the time taken to jump to the function code,
return the value, et
R users,
I am trying to write some functions where names from a dataframe are
arguments. I know it is done in functions like lm(), but I haven't
been able to figure out how unless I attach the dataframe first or
wrap the names in expression() so they are not evaluated immediately.
How can I get a
I am working on a script with R to calculate pKa values. I want to be able
then to graph pKa values to Energy, basically two variables which will give
me titration curves. It is basically a graph with two variables and help
with just graphing it will be just fine. But if anyone is familiar with
ti
Version 4.39 of the caret package was sent to CRAN.
caret can be used to tune the parameters of predictive models using
resampling, estimate variable importance and visualize the results.
There are also various modeling and "helper" functions that can be
useful for training models. caret has wrapp
In R 2.11.0, the following code that appends data using gzfile in 'ab'
mode doesn't work anymore (it worked in the previous versions):
con <- gzfile('myfile.gz', open='wb')
write.table(c(1,2,3), file=con)
close(con)
con <- gzfile('myfile.gz', open='ab')
write.table(c(4,5), file=con)
close(con)
Thanks Cedric. I did take a look at RServe, but that doesn't seem to support
concurrency in windows environment. As ours is primarily a windows
environment, we are looking for some other ways. Is there any way in
rJava/JRI to call commands directly from java code?
Thank you.
On Sat, May 15, 2010
one way:
x <- rnorm(100)
y <- x+rnorm(100)
par(mfrow=c(2,2)) # this sets up the graphics window to expect a 2x2 layout
plot(x,y)
boxplot(x,y,names=c("var1","var2"))
hist(x)
hist(y)
mtext("fourplots on same page",side=3,outer=T,line=-1.5)
and you can adjust as needed to put more or less plots o
Thank you Bill, Peter and Barry.
2010/5/17 Barry Rowlingson
> On Mon, May 17, 2010 at 6:24 PM, Peter Ehlers wrote:
>
> > Try
> > system.time(y <- f(x))
> >
> > and see ?"=".
> >
> > -Peter Ehlers
>
> Ah ha. That explains the curly brackets I saw in a posting with
> system.time on stack ove
I believe I'm working with the newest version of R (2.11.0) and I've
tried a few of the most recent versions... No luck yet.
Thanks,
-Adam
On Mon, May 17, 2010 at 1:51 PM, David Winsemius wrote:
>
> On May 17, 2010, at 4:44 PM, Adam November wrote:
>
>> Hi All,
>> Just thought I'd bring attention
I fit a GAM to turtle growth data using mgcv:
>m1 <- gam(growth~s(mean.size,
bs="cr")+s(year,bs="cr",k=7)+s(cohort,bs="cr")+s(age,bs="cr"), data=grow,
family=quasi(link="identity"))
The errors are skewed (and seem to be correlated with age) (code and plots
here:
https://docs.google.com/fileview?
On May 17, 2010, at 4:44 PM, Adam November wrote:
Hi All,
Just thought I'd bring attention to the fact that lme4 is failing cran
checks on the mac platform, and I can't seem to install it from source
on 10.5 or 10.6, either ("ld: library not found for -lgfortran ") .
Any help getting this worki
Hi All,
Just thought I'd bring attention to the fact that lme4 is failing cran
checks on the mac platform, and I can't seem to install it from source
on 10.5 or 10.6, either ("ld: library not found for -lgfortran ") .
Any help getting this working?
Thanks!
(sorry for the non-plain-text...)
__
Hi All,
Just thought I'd bring attention to the fact that lme4 is failing cran
checks on the mac platform, and I can't seem to install it from source on
10.5 or 10.6, either ("ld: library not found for -lgfortran
") . Any help getting this working?
Thanks!
[[alternative HTML version delete
library(xts)
library(TTR)
ndx = getYahooData("^NDX")
aa = ndx$Close
bb = aggregate(aa, as.yearweek, tail, 1)
The last operation takes forever, and then the bb dates are messed up. The
following produces the desired result:
time(aa) = as.Date(time(aa))
bb = aggregate(aa, as.yearweek, tail, 1)
On 2010-05-17 12:54, Henrique Dallazuanna wrote:
Try this:
newData<- sapply(numdat, function(x)lapply(strsplit(as.character(x), '-'),
function(.x)mean(as.numeric(.x
There's a potential problem if numdat contains negative numbers.
It would be better to restrict the recoding to character or
No, that's only true for lattice and ggplot2 graphics. The problem
here is with this line,
windows(width=5, height=5)
which shouldn't be there.
HTH,
baptiste
On 17 May 2010 22:23, Jun Shen wrote:
> If you do plotting in a loop, then you need to print it to the device.
>
> print(plot(xj,y))
>
If you do plotting in a loop, then you need to print it to the device.
print(plot(xj,y))
On Mon, May 17, 2010 at 3:02 PM, Shirley Bao wrote:
> Thanks!
> I got an error message when opening the pdf file: "There was an error
> opening this document. This file cannot be opened because it has no pag
Thanks!
I got an error message when opening the pdf file: "There was an error
opening this document. This file cannot be opened because it has no pages."
Here is what I did in plotting and saving the file:
pdf(file="C:/figure.pdf")
for (j in 1:numColumns)
{
windows(width=5, height=5)
plot(xj,y)
1.Open pdf device
>pdf()
2.Do your plotting as many as you want, you won't see the plots on the
screen because they go directly to the pdf() device.
3.Turn off the pdf()
>dev.off()
Then you can review your plots in the pdf file. For more details see ?pdf
Jun
On Mon, May 17, 2010 at 2:41 PM, Shirl
I have created separate plots in multiple graphics windows using the
windows() function in R.
How do I save all the plots in one PDF file?
I tried savePlot("C:/rplot.pdf", type = "pdf"). However, it only saved the
plot in the current graphics window.
Thank you!
[[alternative HTML versio
That works great, thanks
--
View this message in context:
http://r.789695.n4.nabble.com/Display-Large-Matrix-as-an-Image-tp2220139p2220219.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.
My mistake.
cars2 should be initalized to have all the extra rows.
cars2 <- data.frame(matrix(rep(NA, prod(dim(cars)) + ncol(cars)),
nrow(cars)+1))
cars2[2:nrow(cars2),] <- cars
In this way, insertion at any row is possible.
Nikhil
On May 17, 2010, at 2:46 PM, Peter Ehlers wrote:
data(
Hi,
try this,
m = matrix(runif(2000*2400), nrow=2000)
library(grid)
grid.raster(m)
HTH,
baptiste
On 17 May 2010 20:35, tetonedge wrote:
>
> I have a matrix that is 2400x2000 and I would like to display it as an image,
> I have tried image(), but due to the size of the matrix the drawing of t
On Mon, May 17, 2010 at 6:24 PM, Peter Ehlers wrote:
> Try
> system.time(y <- f(x))
>
> and see ?"=".
>
> -Peter Ehlers
Ah ha. That explains the curly brackets I saw in a posting with
system.time on stack overflow just now:
system.time({y=f(x)})
works as expected since the {} pair make a
Try this:
newData <- sapply(numdat, function(x)lapply(strsplit(as.character(x), '-'),
function(.x)mean(as.numeric(.x
On Mon, May 17, 2010 at 3:29 PM, Juliet Hannah wrote:
> I am recoding some data. Many values that should be 1.5 are recorded
> as 1-2. Some example data and my solution is bel
I have a matrix that is 2400x2000 and I would like to display it as an image,
I have tried image(), but due to the size of the matrix the drawing of the
plot is extremely slow. Does anybody know of any fast ways to visualize such
a large matrix. Thanks
--
View this message in context:
http://r.7
I am recoding some data. Many values that should be 1.5 are recorded
as 1-2. Some example data and my solution is below. I am curious about
better approaches or any other suggestions. Thanks!
# example input data
myData <- read.table(textConnection("id, v1, v2, v3
a,1,2,3
b,1-2,,3-4
c,,3,4"),head
This is not an R question, though you may receive help from this list. But
you would probably do better posting on a statistical list, especially one
focused on ecology.
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help
I am trying to perform quantile regression (using quantreg package) and
I am particularly interested to know whether the technique requires
independence of observations.
I am an ecologist and, in particular, I collected data of abundance of a
species in 15 location around an island. In each l
Works with warnings for me. but your method is better.
Nikhil
On May 17, 2010, at 1:35 PM, Peter Ehlers wrote:
data(cars)
cars2 <- cars
cars2[2:nrow(cars)+1,] <- cars2[1:nrow(cars),]
cars2[1,] <- NA
__
R-help@r-project.org mailing list
https://sta
On May 17, 2010, at 1:36 PM, GL wrote:
I have the two loops listed below. The first executes perfectly and
creates a
series of density plots. The second does not produce any output,
however, if
I enter the exact bwplot command after the loop executes, I get
output for
the last value in t
Subsequently saw this in FAQs
See FAQ:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f
--
View this message in context:
http://r.789695.n4.nabble.com/bwplot-in-loop-tp2220020p2220034.html
Sent from the R help mailing list archive at Nabble.com.
__
I have the two loops listed below. The first executes perfectly and creates a
series of density plots. The second does not produce any output, however, if
I enter the exact bwplot command after the loop executes, I get output for
the last value in the list of services. Why am I not getting output
On 2010-05-17 11:14, Nikhil Kaza wrote:
Does this work?
data(cars)
cars2 <- cars
cars2[2:nrow(cars)+1,] <- cars2[1:nrow(cars),]
cars2[1,] <- NA
Not for me. Did you try it?
-Peter Ehlers
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail
Hi, Everyone,
I am trying to fit a ring specific gravity dataset using the linear
mixed-effects model (lme function) in R. However, the AIC value is extrememly
low (such as -8000) and the loglikehihhod value is extremely big (such as
4000), which seems not correct to me. What could the result i
On 2010-05-17 10:49, Gustave Lefou wrote:
Dear all,
I have a function f(x).
ptm<- proc.time()
y=f(x)
proc.time() - ptm
works fine for me.
However it is not possible to write
system.time(y=f(x))
and
system.time(f(x))
does not store the value of f(x).
Should I prefer the 3 lines with
Thank-you for the replies. I believe I figured out what the problem was.
When I installed the package on linux it ran smoothly, but I just need to
install a lot of accessory packages to make gplots work with Windows.
Thanks again,
Aaron
2010/5/17 Uwe Ligges
> Additionally, please give the
On 2010-05-17 9:28, ecvet...@uwaterloo.ca wrote:
I have a large data frame 48:2185 with different numbers.
I would like to add only one row at the very top of my data frame with
0's or NA's.
I don't know which approach to use. Should i create 2 different data
frames and merge them? Ive also tried
Does this work?
data(cars)
cars2 <- cars
cars2[2:nrow(cars)+1,] <- cars2[1:nrow(cars),]
cars2[1,] <- NA
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On May 17, 2010, at 11:28 AM, ecvet...@uwaterloo.ca wrote:
I have a large dat
Dear All,
I am executing my R script from PHP using the shell_exec.
$return = shell_exec("/usr/bin/R --slave --args ".$Domain.",".$Gender." <
/var/www/html/trends/newTrend.R");
In newTrend.R, I am plotting graph and save in pdf file using the below
code. It will create a pdf file with two line.
Dear all,
I have a function f(x).
> ptm <- proc.time()
> y=f(x)
> proc.time() - ptm
works fine for me.
However it is not possible to write
> system.time(y=f(x))
and
> system.time(f(x))
does not store the value of f(x).
Should I prefer the 3 lines with proc.time ?
Thank you very much,
Gusta
Lot of examples for one way pipes, but I need to create some named
pipes from R to another process, especially SQLite. I am look at the
R/SQLite packages for help. ANy pointers?
___
R-packages mailing list
r-packa...@r-project.org
https://stat.ethz.ch/
Try this:
> x <- data.frame(a=1:10, b=1:10)
> x
a b
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
> # add a row at the top
> x <- x[c(1,seq(nrow(x))),]
> x
a b
11 1
1.1 1 1
22 2
33 3
44 4
55 5
66 6
77 7
8
Hello.
In this post:
http://finzi.psych.upenn.edu/Rhelp10/2010-March/233815.html
Uwe Ligges suggests using BRugs rather than R2WinBUGS under windows. He
also notes that it is not in the main CRAN repository, but it is in
"extras" which is a default repository under windows.
I have OpenBUGS
The long-promised revision to bigmemory has arrived, with package
4.2.3 now on CRAN. The mutexes (locks) have been extracted and will
be available through package synchronicity (on R-Forge, soon to appear
on CRAN). Initial versions of packages biganalytics and bigtabulate
are on CRAN, and new ver
Version 2.2.0 of package bcp is now available. It replaces the
suggests of NetWorkSpaces (previously used for optional parallel MCMC)
with the dependency on package foreach, giving greater flexibility and
supporting a wider range of parallel backends (see doSNOW, doMC,
etc...).
For those unfamili
Hi Holger,
I would also highly recommend you look at the ?boxcox and ?logtrans
functions in the MASS package. There is also a very illuminating,
concise discussion about their use on Pages 170 - 172 of
Venables, W. N. and Ripley, B. D. (2002) Modern Applied Statistics
with S. Fourth edition.
with
I have a large data frame 48:2185 with different numbers.
I would like to add only one row at the very top of my data frame with
0's or NA's.
I don't know which approach to use. Should i create 2 different data
frames and merge them? Ive also tried the rbind command with no luck.
I would app
Hello to the list
I'm using a Rscript launched via the linux terminal as the title of this
mail explains. The following is the code of the Rscript :
myRscript.r
1 #!/usr/bin/Rscript
2
3 #Read data
4
data=read.table("/home/morisseau/stage/recherche/gwas/data/CFHLC
Assuming you really do have a table,
> tab <- as.table(c(Friday=119, Monday=173, Saturday=80,
+ Sunday=96, Thursday=167, Tuesday=193, Wednesday=194))
> tab
FridayMonday SaturdaySunday Thursday Tuesday Wednesday
119 1738096 167 193 194
w
= Summary =
Version 0.8.0 of the Rcpp package was released to CRAN today. This release
marks another milestone in the ongoing redesign of the package, and
underlying C++ library.
= Overview =
Rcpp is an R package and C++ library that facilitates integration of C++
code in R pac
Hello Giovanni
I made a minor modification to your function, which now allows to
compute the within R-sq in Twoways Within models (see below).
However I ran into an issue that I have already encountered before:
whenever I try to fit Twoways Within models on my unbalanced data, the
process is stran
Hello,
What about something like this?
week.table <- data.frame(Friday=119, Monday=173, Saturday=80,
Sunday=96, Thursday=167, Tuesday=193, Wednesday=194)
ordered <-
c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday")
week.table <- week.table[,ordered]
library(plotrix)
pie3D
On 13 May 2010 00:41, RATIARISON Eric wrote:
> Hi, here my new version:
> I submit you my case:
> ( Pseudo likehood for exponential family with offset )
>
> loglik <- function(param) {
> b<-param
> m=as.vector(of+z%*%b)
> ll <- sum(v*m-exp(m)) }
>
> gradlik <- function(param) {
> b<-p
On May 17, 2010, at 10:31 AM, someone wrote:
well, actually the script is run by crone job once in a while and id
hate to
do things manually...
The order of levels of a factor variable variable can be changed with
a call along the lines of
fac <- factor(fac, levels=c"(Monday", "Tuesda
well, actually the script is run by crone job once in a while and id hate to
do things manually...
--
View this message in context:
http://r.789695.n4.nabble.com/Change-order-of-columns-in-table-tp2219536p2219725.html
Sent from the R help mailing list archive at Nabble.com.
___
Hello,
the positioning of text elements in a graphics is nortmally
done via x/y coordin ates of the displayed data.
most of the time this is very helpful (and missing in other
programs).
But when I want to set some text into a graphics always at the same position,
this will not be easy...
I mi
7 data points? What's wrong with with doing it manually?
x<-c(173,193,194,167,119,80,96)
labs<-c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday")
pie3D(x,labels=labs)
-tgs
On Mon, May 17, 2010 at 8:46 AM, someone wrote:
>
> I'm an R noob and have a (maybe) stupid questi
Dear R-help,
just for the record a summary of responses to my problem:
> On 04/05/2010 9:41 AM, Rainer Scheuchenpflug wrote:
> a student of mine tries to use the Windows-Rconsole with screen
> reading software (she is blind), and cannot access the command line (Menus
are ok).
- Duncan Murdoch
I'm just guessing (since as David pointed out, no reproducible example
were given), but this perhaps could be the problem:
R> x <- 1:10
R> y <- rnorm(x)
R> fm <- loess(y ~ x)
R> predict(fm, data.frame(x=5:15))
[1] 0.1830450 0.2145826 -0.2158466 -0.3051978 -0.2635318 -0.1013985
[7] NA
Steve,
It sounds to me like you want to use the model to predict data in other
maps. If so, you should be using predict(), not getTree(). getTree()
shows the classification rules in _one_ of the trees in the forest
(default the first tree), which is rarely of interest if you've grown
hundreds of
For example if nombreC <- nombreC <- c("Juan", "Carlos", "Ana", "María","Mario")
I do not want as a result:
name index
1 Juan 1
2 Juan 5
3 Carlos 2
4Ana 3
5 María 4
6 Mario 0
I want:
name index
1 Juan 1
2 Juan 5
3 Carlos 2
4Ana 3
Please read the Posting Guide. Several of your questions will get
answered, including why the attachment did not arrive at any of out
mail clients.
We are having some difficulty understanding why you say your lecturer
"didn't teach us anything about how to use R". It appears from the
mang
On 17/05/2010 9:07 AM, APOLLOCHEROKEE wrote:
> Dear Sir or Madam:Hi,sorry for disturbing you. Currently, I'm a
> master student, and my lecturer didn't teach us anything about how to use R.
> But the assignment is all about R. So even a very stupid error,I still can't
> tell. So please f
Thank you David!
Le 5/17/2010 15:10, David Winsemius a écrit :
On May 17, 2010, at 8:07 AM, Ivan Calandra wrote:
Dear users,
I've noticed some time ago that some of the packages I use
(especially doBy) could not run (yet) on R 2.11. So I kept on using R
2.10.
But I think it should be fin
I'm working with the randomForest package and have successfully build a
model. I'd like to go one step further however, and use the output from
the model to construct a map using the output. My input data are spatial,
and I have an independent set of rasterize maps for each of the predictor
vari
Hi,
Try this,
saveMyWork <- .Last.value
HTH,
baptiste
On 17 May 2010 15:07, math_daddy wrote:
>
> Hello.
>
> I ran a simulation that took a few days to complete, and want to analyze the
> results, but have just realized that I (idiotically) did not assign the
> output to a variable when I int
See
.Last.value
On Mon, May 17, 2010 at 10:07 AM, math_daddy wrote:
>
> Hello.
>
> I ran a simulation that took a few days to complete, and want to analyze
> the
> results, but have just realized that I (idiotically) did not assign the
> output to a variable when I intitiated the simulation. Is
On May 17, 2010, at 8:22 AM, Moohwan Kim wrote:
Dear R Family,
I have an error message. I would like to learn how to deal with that.
The orginal series is as follows: I just pick up the first 10
observations.
[1:10]
[1] 0.0065880493 -0.0065880490 -0.0131743570 0.0197745715
0.0065889175
We still have a long way to go with the data we were given by some drug
discovery scientists.
The problem is to select the few variables (Collective Variables), from a set
of variables sampled during a
Molecular Dynamics simulation, which exhibit a consistent and coherent
relationship with the
Hello.
I ran a simulation that took a few days to complete, and want to analyze the
results, but have just realized that I (idiotically) did not assign the
output to a variable when I intitiated the simulation. Is there any way to
retrieve the last output produced by R so that these last few days
Dear Sir or Madam:Hi,sorry for disturbing you. Currently, I'm a master
student, and my lecturer didn't teach us anything about how to use R. But the
assignment is all about R. So even a very stupid error,I still can't tell. So
please forgive my innocence.
The following is the requiremen
1 - 100 of 135 matches
Mail list logo