Hi Murali,
>> I am interested in plotting my regression analysis results(regression
>> coefficients and
>> standard errors obtained through OLS and Tobit models) in the form of
>> graphs.
plot(obj$lm) will give you a set of diagnostic plots. What you seem to be
after is ?termplot. Also look at
Package dynlm allows easy specification of linear models (using lm as
usual) involving time lags and differences.
I can't say much more without knowing details of your model.
-Felix
On Sun, Jul 27, 2008 at 2:39 PM, Ashish Kumar <[EMAIL PROTECTED]> wrote:
> Hi,
>
>
>
> I would like to know how to
On Sun, 27 Jul 2008, S Ellison wrote:
Looking at the legend() source the filled box line colour is hardcoded :
if (mfill) {
if (plot) {
fill <- rep(fill, length.out = n.leg)
rect2(left = xt, top = yt + ybox/2, dx = xbox, dy = ybox,
col = fill, densi
You are right to be dissatisfied with your code.
I suspect you will get more response if you say
what the code is supposed to do, and give a
smaller example of what the answer should be.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide
How to conquer this?
Thanks
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented
2008/7/28 Rolf Turner <[EMAIL PROTECTED]>:
>
> On 28/07/2008, at 12:32 PM, oscar Linares wrote:
>
>> Hello,
>>
>> I am trying to convert english words to numeric equivalents, e.g., abc to
>> 123. Is there a function or library or package for doing this in R? If not,
>> can it be done easily in R?
>
[EMAIL PROTECTED]:
> Has anyone developed a defensible method of estimating percentiles from a
> univariate kernel density estimate? I am working on a problem in which
> the density estimate is of interest, but I would also like to estimate the
> value of the variable for which the distribution w
Here is my attempt at this (taking a specific understanding of the ill-
defined "equivalence" relation),
unletter <- function(word){
word.broken <- strsplit(word, NULL)
set.of.numbers <- sapply(word.broken[[1]], function(let) which(let ==
letters))
paste(set.of.numbe
> "CG" == Christophe Genolini <[EMAIL PROTECTED]>
> on Sun, 27 Jul 2008 09:27:34 +0200 writes:
CG> Martin Maechler <[EMAIL PROTECTED]> a écrit :
>>> "CG" == Christophe Genolini <[EMAIL PROTECTED]>
>>> on Sat, 26 Jul 2008 12:12:12 +0200 writes:
>>
CG> Mart
On 28-Jul-08 09:29:22, baptiste auguie wrote:
> Here is my attempt at this (taking a specific understanding of the ill-
> defined "equivalence" relation),
>
> unletter <- function(word){
>
> word.broken <- strsplit(word, NULL)
> set.of.numbers <- sapply(word.broken[[1]], functio
As you point out, my proposed function does not verify an equivalence,
and it is not restricted to english words. I'm probably too keen to
see questions that call for an answer with relaxed assumptions
(understand, that i can manage).
Cheers,
baptiste
On 28 Jul 2008, at 10:47, (T
How about this?
unletter <- function(word) {
gsub('-64',' ',paste(sprintf("%02d",utf8ToInt(tolower(word)) -
96),collapse=''))
}
unletter("abc")
[1] "010203"
unletter("Aw")
[1] "0123"
unletter("I walk to school")
[1] "09 23011211 2015 190308151512"
--Hans
Dear list,
i am using the following code to produce a lattice xyplot, but the
axis.break-function is seemingly not executed.
Date<-seq(as.Date("2006-08-29"), as.Date("2007-08-28"), by="2 weeks")
Period<-
var1<-rnorm(27, 9000, 3000)
var2<-rnorm(27, 5,25000)
var3<-rnorm(27, 10
On 28 Jul 2008, at 12:23, Hans-Joerg Bibiko wrote:
How about this?
unletter <- function(word) {
gsub('-64',' ',paste(sprintf("%02d",utf8ToInt(tolower(word)) -
96),collapse=''))
}
unletter("abc")
[1] "010203"
unletter("Aw")
[1] "0123"
unletter("I walk to school")
[1] "09 23011211 2015 1903
When using stop clause with a condition, its documented that "If a condition
object is supplied it should be the only argument, and further arguments
will be ignored, with a warning".
This will not be the case when running codes from Winedit or TINN?
When I do something like:
if(length(content[c
Henning Wildhagen gmx.de> writes:
>
> Dear list,
>
> i am using the following code to produce a lattice xyplot, but the
> axis.break-function is seemingly not executed.
axis.break (from plotrix) won't work at all on lattice
graphics -- you can look to see what it does and try to
replicate
On Mon, 2008-07-28 at 12:19 +0200, Henning Wildhagen wrote:
> Dear list,
>
> i am using the following code to produce a lattice xyplot, but the
> axis.break-function is seemingly not executed.
>
> Date<-seq(as.Date("2006-08-29"), as.Date("2007-08-28"), by="2 weeks")
> Period<-
> var1<-rnorm(27,
If you are using Tinn-R, then I am assuming that you are using the
'send selection' which basically minics the keyboard. On the
keyboard, after a 'stop' you can type in the next command without any
problems. Here is a script I was using with Tinn-R:
x <- 1
stop('error')
cat('continue processing'
On Mon, 28 Jul 2008, Paulo Cardoso wrote:
When using stop clause with a condition, its documented that "If a condition
object is supplied it should be the only argument, and further arguments
will be ignored, with a warning".
I see no condition object in your example.
This will not be the ca
Hi
I am using the boot function in boot package. I am facing a problem in getting
the values of bias and st.error in the output.
r<-36
n<-40
shape<-2
theta11<-exp(1) # (=2.718282)
theta21<-exp(.5) #( =1.648721)
m0<- function(Ti,Tj) #a function that generates the MLestimates
{
loglik<-function(ti
> I see no condition object in your example
Isn't if(length(content[content%in%folders])!=4) a condition?
> stop() breaks execution of a block of code that source()d -- maybe that is
what you are looking for?
Yes
> I don't know what you mean by 'TINN' and 'Winedit' (what exact program,
how ar
Hi. I have made 100 experiments of an M/M/1 queue, and for each one I have
calculated both, mean and variance of the queue size. Now, a professor has told
me that variance is usually chi-squared distributed. Is there a way in R that I
can find the parameter that best fits a chi-square to the var
Does this do what you want:
> x <- matrix(1:36,6)
> x
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]17 13 19 25 31
[2,]28 14 20 26 32
[3,]39 15 21 27 33
[4,]4 10 16 22 28 34
[5,]5 11 17 23 29 35
[6,]6 12 18 24 30 36
Try sourcing in the 'new.legend' function below. It's the legend
function with a new argument called 'box.col'. The argument will change
the color of the box surrounding the legend. If I understand what it is
you are looking for, this should work. Also, I didn't see a way to
change the axis bar
Dear All and Mark,
Given a dataset that I have called dat, I was hoping to speed up the
following loop:
for(i in 1:835353){
for(j in 1:86){
if (is.na(dat[i,j])==TRUE){dat[i,j]<-0 }}}
Actually I am also having a memory problem. I get the following:
Error: cannot allocate vector of size 3.2 Mb
Is it what you want ?
> x <- matrix(c(1:3, NA, NA, 4, 1:2, NA), 3, 3)
> x
[,1] [,2] [,3]
[1,]1 NA1
[2,]2 NA2
[3,]34 NA
> x[is.na(x)] <- 0
> x
[,1] [,2] [,3]
[1,]101
[2,]202
[3,]340
2008/7/28 Denise Xifara <[EMAIL PROTECT
Dear Denise,
It looks like you want to replace all NA with 0 in the dataset? The code
below should do that trick without loops. And it will be rather fast.
dat[is.na(dat)] <- 0
> dat <- matrix(rbinom(40, 1, 0.75), ncol = 4, nrow = 10)
> dat[dat == 0] <- NA
> dat
[,1] [,2] [,3] [,4]
[1,]
Martin Maechler <[EMAIL PROTECTED]> a écrit :
"CG" == Christophe Genolini <[EMAIL PROTECTED]>
on Sun, 27 Jul 2008 09:27:34 +0200 writes:
CG> Martin Maechler <[EMAIL PROTECTED]> a écrit :
>>> "CG" == Christophe Genolini <[EMAIL PROTECTED]>
>>> on Sat, 26 Jul 2008 12:12:
If your matrix is 835353x86, then if it is numeric, then it will take
about 550MB for a single copy. You should therefore have at least 2GB
(so you can have a couple of copies as part of some processing) of
real memory on your system. If you want to replace NAs with zero,
then this is how you mig
this works too:
n = 6 # number of rows
m = 4 # number of coloumns
nm = n*m
mat = matrix(1:nm,n) # your matrix
pf = function(Col){
ind = rep(1:(n/2),each=2)
out = tapply(Col,ind,prod)
out
}
# pf performs forall vecotr x: x[i]*x[i-1], i=2,4,6
hi there,
I want to write the plots in the pdfs and the details about the graph in a
seperate notepad.
plot(as.numeric(lapply(resultgenes,length)),
main= "Geneset.gene#.bias.test",xlab="Top.Ranked.Genesets",
ylab="gene.number.per.geneset")
lines(loess.smooth(c(1:1000),as.numeric(lapply(resultge
We were not told this was a matrix, rather a 'dataset'.
If it is matrix, logical indexing via is.na(x) is pretty good, although it
will create an index equal in size to the dataset (but logical).
If 'dataset' means a data frame, you will use less memory using a for()
loop over columns, e.g.
Hello all,
I am having trouble running a count function in R using RODBC to query a table
I created in Oracle. It may very well be that my SQL coding is incorrect; I
just started learning it. But if someone could point me in the right direction
or tell me if I am going about this the correct wa
?pdf
and copy and write whatever notes you want i a note pad
On Mon, Jul 28, 2008 at 10:54 AM, Rajasekaramya <[EMAIL PROTECTED]>wrote:
>
> hi there,
>
> I want to write the plots in the pdfs and the details about the graph in a
> seperate notepad.
>
> plot(as.numeric(lapply(resultgenes,length)),
on 07/28/2008 09:44 AM Megan J Bellamy wrote:
Hello all,
I am having trouble running a count function in R using RODBC to
query a table I created in Oracle. It may very well be that my SQL
coding is incorrect; I just started learning it. But if someone could
point me in the right direction or te
Xin Meng,
Ignore all the "you didn't tell us your operating system" comments--I
will show you a simple, although very tedious way to do what you want
that is totally free of any operating system limitations.
Began by formatting a very simple Excel file with colored cells, then
save the file as a
Dear fellow R-users
I am trying to do some imputation using K-NN with the yaImpute library. All
seems to be going well until I try to use
AsciiGridImpute. All my data are correctly formatted and I am able to run and
view the results of yai. Below is my code:
Given char vector delme2:
> str(delme2)
chr [1:1065] "30-1-08 8:48:21" "30-1-08 8:55:17" "30-1-08 9:00:22" ...
I do:
> delme3 <- strptime(delme2,format="%d-%m-%y %H:%M:%S")
But then:
> str(delme3)
POSIXlt[1:9], format: "2008-01-30 08:48:21" "2008-01-30 08:55:17" ...
> length(delme3)
[1] 9
wh
Hi,
I have a simple graph:
x <- c(1,2,3)
plot(x, pch=16,type="b")
I would like to add some notes just beside these 3 dots, and the notes are
stored in a vector:
a <- c(12,54,84)
So the result will be: there should be a "12" below the first dot (or next
to it, but not replacing the solid dot
rlearner309 gmail.com> writes:
> x <- c(1,2,3)
> plot(x, pch=16,type="b")
>
> I would like to add some notes just beside these 3 dots, and the notes are
> stored in a vector:
>
> a <- c(12,54,84)
>
> So the result will be: there should be a "12" below the first dot (or next
> to it, but not
HiI would like to use the parametric bootstrap confidence intervals. The
problem is that I am using a step stress model based on a gamma distributed
data. So, first I need to generate a spacial data and then find the MLE`s then
I must find the bootstrap confidence interval.I dont know how to us
It's your R syntax that is faulty: Use "" (not ') to delimit the R string
which contains single quotes for use by SQL.
On Mon, 28 Jul 2008, Megan J Bellamy wrote:
Hello all,
I am having trouble running a count function in R using RODBC to query a table
I created in Oracle. It may very well b
On Mon, 28 Jul 2008, Agustin Lobo wrote:
Given char vector delme2:
str(delme2)
chr [1:1065] "30-1-08 8:48:21" "30-1-08 8:55:17" "30-1-08 9:00:22" ...
I do:
delme3 <- strptime(delme2,format="%d-%m-%y %H:%M:%S")
But then:
str(delme3)
POSIXlt[1:9], format: "2008-01-30 08:48:21" "2008-01-30
Hi Megan,
Marc's hint is the R way, which is needed with your original SELECT.
OTOH, it may be more efficient to move the WHERE clause into the
SELECT. Something like
select plotnum, sampyear, sptype
from density
where sampyear=1995
AND plotnum=1
AND sptype IN ('S', 'H')
should work i
The current default for the box/frame colour is not "black" (it is
par("fg")), but otherwise I've committed something very similar to the
R-devel version of R earlier today.
On Mon, 28 Jul 2008, Nutter, Benjamin wrote:
Try sourcing in the 'new.legend' function below. It's the legend
function
Please find release 3.5.0 of the CRAN package "randomSurvivalForest"
now posted on CRAN. Thank you.
[EMAIL PROTECTED]
Udaya B. Kogalur, Ph.D.
Kogalur Shear Corporation
5425 Nestleway Drive, Suite L1
Clemmons, NC 27012
Greetings,
I have R 2.7.1 in MacOs and I believe UTF encoding is already installed.
At least:
> Sys.getenv()
shows several variables, including:
LANG "pt_PT.UTF-8"
I installed the Rstem and tm packages and when I try the following code:
> wordStem(c("aberração","aberrações"), language="por
> Hello,
>
> I want to install the package "multiv" which is not maintained any
> more (found in the archive: multiv_1.1-6.tar.gz from 16 July 2003). I
> have installed an older version of R 1.4.1 on my Windows Vista in
> order to match the older package.
>
> As no binary is distributed, is it n
Hi everyone,
Is there a way to perform a rollapply operation on a time series data matrix
and preserve the time frame? Currently, when I apply rollapply in its
standart form, the date column is no longer present in the o/p matrix.
Thanks,
rcoder
--
View this message in context:
http://www.nab
I'm pretty sure that this is not Paul's design and he usually doesn't want
to touch base graphics.
I've added arguments lwd.ticks and col.ticks in R-devel. The simplest way
to suppress the axis line become axis(..., lwd=0, lwd.ticks=1).
On Mon, 16 Jun 2008, Peter Dalgaard wrote:
Andrew Yee
I am trying to create a graph that puts three x-y plots (age vs rate for
three categories ) on a single plot. SAS does this with the overlay
function. I cannot find in the R documentation a method for doing this. Any
help would be appreciated.
Thanks,
Phill
Phillip R. Hunt, Sc.D.
24
Hello,
I need to retrieve data from the internet to R using SOAP protocol. I found
that SSOAP package (http://www.omegahat.org/SSOAP/) should do the work.
However, I cannot install it properly. It just does not work. I am using R
2.7.1 (this was the only version on which other required packages (
Hi all,
I am trying to convert geometric means in a matrix to cover classes. My
values are as such:
perc<-c(0,0.025136418, 0.316227766, 1.414213562,3.16227766, 7.071067812,
15.8113883, 35.35533906, 61.23724357, 84.40971508, 97.46794345)
cover<-c(0,1,2,3,4,5,6,7,8,9,10)
This is what I am tryin
Use '&' for vectors and '&&' for scalars. Ditto applies to the OR
operator(s). /Henrik
On Mon, Jul 28, 2008 at 10:36 AM, Wade Wall <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I am trying to convert geometric means in a matrix to cover classes. My
> values are as such:
>
> perc<-c(0,0.025136418, 0
Hi,
maybe the following code helps to achieve what you want?
It seems to me it is basically a 'recode' question.
set.seed(1234) # not neccessary but this ensures we have the same
#results
random.values <- runif(n=30, min=0, max=100)
newgrouping <- cut(x=random.values, breaks=c(0,0.1, 1, 2, 5, 10
This is EXACTLY what I needed. I suppose the only issue was that I needed to
isolate the numeric output of PrettyR::describe. Thanks very much!
Gabor Grothendieck wrote:
>
> Using the built in BOD data frame:
>
> library(prettyR)
> str(describe(BOD)) # note Numeric component
>
> BOD.descri
Thank you all for your input! I was able to create my descriptive table as
well as a few other things I was having trouble with.
Thanks again!
J.
jcarmichael wrote:
>
> Hello everyone. I am new to R, so please bear with me. I am trying to
> find an easy way to export descriptive statistics
Hi Paulo.
My development version has that warning turned off.
However, the Rstem package predates the encoding in R, AFAIR.
So when I call wordStem() with a string which has an Encoding()
of UTF-8, the resulting string has Encoding() "unknown".
I'll take a look and add see if I can add support
?lines
On Mon, Jul 28, 2008 at 12:57 PM, Phillip R. Hunt <[EMAIL PROTECTED]>wrote:
> I am trying to create a graph that puts three x-y plots (age vs rate for
> three categories ) on a single plot. SAS does this with the overlay
> function. I cannot find in the R documentation a method for doing
And you may want to consider using the 'cut' function. In your case,
something like
veg_mean <- cut(veg_mean, breaks=c(0,.1,1,2,5,10,25,50,75,95,100),
right=FALSE)
should do the trick (see ?cut for more options).
Best,
Stefan
On 28 Jul 2008, at 19:52, Henrik Bengtsson wrote:
Use '&'
HI
the code is suppose to generate a random value selected uniformly as a time
interval in seconds (that's the dT) from the range given fromdtmin<-969
dtmax<-9884. Once that value is selected the program will generate the
interval (ts1, ts2, and so on. for each sampling without getting out o
On 28-Jul-08 17:52:31, Henrik Bengtsson wrote:
> Use '&' for vectors and '&&' for scalars. Ditto applies to the OR
> operator(s). /Henrik
What's wrong with using "&" for scalars? Surely it gives the
correct answer? Maybe it's simply a bit slower, or something?
Ted.
> On Mon, Jul 28, 2008 at 1
Hello.
I am attempting to duplicate a negative binomial regression in R. SAS uses
generalized estimating equations for model fitting in the GENMOD procedure.
proc genmod data=mydata (where=(gender='F'));
by agegroup;
class id gender type;
model count = var1 var2 var3 /dist=NB link=log offset=l
Hi,
I use R at home, and am interested in using it at my work company (which is
in the Fortune 100). I began the request, and our legal team has given some
gruff about the open source license. Not boring you with the details here,
but I used some info on gnu.org as a rebuttal, and someone at the
See ?cut for creating a factor based on ranges of values.
Regards,
Wade Wall wrote:
>
> Hi all,
>
> I am trying to convert geometric means in a matrix to cover classes. My
> values are as such:
>
> perc<-c(0,0.025136418, 0.316227766, 1.414213562,3.16227766, 7.071067812,
> 15.8113883, 35.3
I have a vector of data (species names) interspersed with NA values
and I want a function to "fill in the blanks", replacing NA values
with whatever the last species name was.
For example the vector:
"A","B",NA,NA,"C",NA,NA,NA,NA,"D",NA,NA.
should evaluate to:
"A" "B" "B" "B" "C" "C" "C"
Hi, all,
Does anyone now of a way to put multiple plots on gap.plot?
Much appreciated,
Art
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
a
Hi.
On Mon, Jul 28, 2008 at 11:21 AM, Ted Harding
<[EMAIL PROTECTED]> wrote:
> On 28-Jul-08 17:52:31, Henrik Bengtsson wrote:
>> Use '&' for vectors and '&&' for scalars. Ditto applies to the OR
>> operator(s). /Henrik
>
> What's wrong with using "&" for scalars? Surely it gives the
> correct a
All this is included in the distribution in "doc" folder:
1. see FAQ: "2.11 Can I use R for commercial purposes?"
2. Specific GNU License is in file COPYING
HTH,
Jim Porzak
Responsys, Inc.
San Francisco, CA
http://www.linkedin.com/in/jimporzak
On Mon, Jul 28, 2008 at 11:32 AM, zerfetzen <[EMA
I'll leave it to someone else to answer the question
that you asked, but I don't mind answering the
question that you didn't ask:
Are there alternatives ways of getting R into the
company?
Yes. There are now some commercially supported
versions of R -- see 'What is R-plus?' in the R-FAQ
for a l
Try this:
library(zoo)
x <- c("A","B",NA,NA,"C",NA,NA,NA,NA,"D",NA,NA)
na.locf(x)
On Mon, Jul 28, 2008 at 2:10 PM, Owen Jones <[EMAIL PROTECTED]> wrote:
> I have a vector of data (species names) interspersed with NA values and I
> want a function to "fill in the blanks", replacing NA values wi
try function na.locf() from package 'zoo', i.e.,
library(zoo)
x <- c("A","B",NA,NA,"C",NA,NA,NA,NA,"D",NA,NA)
na.locf(x)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, B
On 7/28/2008 2:32 PM, zerfetzen wrote:
Hi,
I use R at home, and am interested in using it at my work company (which is
in the Fortune 100). I began the request, and our legal team has given some
gruff about the open source license. Not boring you with the details here,
but I used some info on g
yeah, just use points
twogrp<-c(rnorm(5)+4,rnorm(5)+20,rnorm(5)+5,rnorm(5)+22)
gpcol<-c(2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5)
gap.plot(twogrp,gap=c(8,12),xlab="Index",ylab="Group values",
main="Plot gap on Y axis",col=gpcol)
twogrp2<-c(rnorm(5)+4,rnorm(5)+20,rnorm(5)+5,rnorm(5)+22)
points(
maybe this is easiest way to do it:
x<-c("A","B",NA,NA,"C",NA,NA,NA,NA,"D",NA,NA);
x[is.na(x)]<- "D"; x
thanks
y
Owen Jones-3 wrote:
>
> I have a vector of data (species names) interspersed with NA values
> and I want a function to "fill in the blanks", replacing NA values
> with wha
Hello,
I am going to be on vacation and would like to temporary unsubscribe
from the list. Sending "unsubscribe" didn't help. Could you please
guide me on the appropriate step? Thanks a lot in advance!
Best regards,
Endre
--
Endre Domiczi, CEO
Sevana Oy, http://www.sevana.fi
Email : [EMAIL P
Jim Porzak wrote:
All this is included in the distribution in "doc" folder:
1. see FAQ: "2.11 Can I use R for commercial purposes?"
2. Specific GNU License is in file COPYING
Furthermore,
the start-up message of R can (hopefully) help you in your case.
It mentions to type in
license()
or
lice
on 07/28/2008 01:32 PM zerfetzen wrote:
Hi,
I use R at home, and am interested in using it at my work company (which is
in the Fortune 100). I began the request, and our legal team has given some
gruff about the open source license. Not boring you with the details here,
but I used some info on
(Ted Harding) wrote:
On 28-Jul-08 17:52:31, Henrik Bengtsson wrote:
Use '&' for vectors and '&&' for scalars. Ditto applies to the OR
operator(s). /Henrik
What's wrong with using "&" for scalars? Surely it gives the
correct answer? Maybe it's simply a bit slower, or something?
on 07/28/2008 02:36 PM Endre Domiczi wrote:
Hello,
I am going to be on vacation and would like to temporary unsubscribe
from the list. Sending "unsubscribe" didn't help. Could you please
guide me on the appropriate step? Thanks a lot in advance!
Best regards,
Endre
Go here:
https://stat.ethz
Just for the record, a less efficient but certainly simple and maybe
adequate way for you to do it in standard R (no zoo package required) is:
for(i in seq_along(x)[-1])if(is.na(x[i])) x[i] <- x[i-1]
I tried timing it on my not so fancy Windows desktop for a vector of 10,000
values, but it was in
Thanks to each of you for your excellent input. I have copied the file and
will read it tonight. I haven't run into any heat from IT, but if I do, it
will be in the near future. The exact legal issue was touched upon. There
was a concern that anything associated with R (my code, etc.) would ha
Hi,
I am new to R and am trying to do a loop but it seems not to run after one
turn.
What I want to do is subset my dataframe (extract one station and one day)
to calculate and store the maxima that can then be plotted.
I have an error message at the end of each loop:
Error: unexpected '}' in "}"
You are missing a right paren in this line:
>while (jhttp://www.functionaldiversity.org
__
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
Just for the fun of it, here is a recursive solution to the same problem...
rna <- function(z) {
y <- c(NA, head(z, -1))
z <- ifelse(is.na(z), y, z)
if (any(is.na(z))) Recall(z) else z }
> x
[1] "A" "B" NA NA "C" NA NA NA NA "D" NA NA
> rna(x)
[1] "A" "B" "B" "B
Hi Ileana,
See this thread:
http://www.nabble.com/R-package-install-td18636993.html
HTH, Mark.
Somesan, Ileana wrote:
>
>> Hello,
>>
>> I want to install the package "multiv" which is not maintained any
>> more (found in the archive: multiv_1.1-6.tar.gz from 16 July 2003). I
>> have install
This all got very serious, though I've read it with interest.
Zerfetzen's original post reminded me of something which happened
around the time "Y2K" was becoming an issue, bringing up a similar
issue from 2000 years earlier. Admin etc. does not change over time ...
I've copied the email exchange
On Mon, 28 Jul 2008, Megan J Bellamy wrote:
Hello all,
I am having trouble running a count function in R using RODBC to query a table
I created in Oracle. It may very well be that my SQL coding is incorrect; I
just started learning it. But if someone could point me in the right direction
or
Dear Jeroen,
It's true that the sem() function only handles quantitative endogenous
variables, but you could use it along with functions in the polycor package
to fit models with ordered or dichotomous observed variables. There's an
example of a confirmatory factor analysis model with ordinal indi
Dear Tom,
It is in principle possible for residuals to be autocorrelated even when the
series for the response variable is not. Moreover, the DW test should be
appropriate for the model you fit. On the other hand, a DW statistic of 0
suggests perfect positive autocorrelation, and so I would suspec
rcoder wrote:
>
> Hi everyone,
>
> Is there a way to perform a rollapply operation on a time series data
> matrix and preserve the time frame? Currently, when I apply rollapply in
> its standart form, the date column is no longer present in the o/p matrix.
>
> Thanks,
>
> rcoder
>
--
View
On Sun, Jul 27, 2008 at 9:06 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
> I continue to struggle with mixed models. The square zero version
> of the problem that I am trying to deal with is as follows:
>
> A number (240) of students are measured (tested; for reading comprehension)
> on 6 separate
I'm having a problem with the error and warning functions. I've tried this
on multiple machine so I'm fairly sure it's not machine dependent and I've
tried it on the latest versions 2.6.0-2.7.1. Whenever my program gets to an
error or warning it crashes the entire program rather than throwing the
rollapply along an index:
library(zoo)
z <- zoo(matrix(101:110, 5), 201:205)
tt <- time(z)
zz <- zoo(seq_along(tt), tt)
out <- rollapply(zz, 3, function(ix) list(z[ix,]))
str(out) # list of zoo objects
On Mon, Jul 28, 2008 at 5:03 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
>
> rcoder wrote:
>>
>> H
On 28/07/2008 6:41 PM, Andrew Redd wrote:
I'm having a problem with the error and warning functions. I've tried this
on multiple machine so I'm fairly sure it's not machine dependent and I've
tried it on the latest versions 2.6.0-2.7.1. Whenever my program gets to an
error or warning it crashes
Hello R help list
I have been using the smoothScatter function within the "geneplotter"
package to make some graphs using a Sweave Rnw script called via Rscript
in a DOS/Windows batch file. The Rscript will ultimately be called by a
web service with time-out constraints, hence things need to run
And then lapply over out.
On Mon, Jul 28, 2008 at 6:41 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> rollapply along an index:
>
> library(zoo)
> z <- zoo(matrix(101:110, 5), 201:205)
> tt <- time(z)
> zz <- zoo(seq_along(tt), tt)
> out <- rollapply(zz, 3, function(ix) list(z[ix,]))
> str(ou
Thanks for the response. I ***think*** I'm making a bit of
progress
On 29/07/2008, at 10:14 AM, Douglas Bates wrote:
On Sun, Jul 27, 2008 at 9:06 PM, Rolf Turner
<[EMAIL PROTECTED]> wrote:
What I *don't* understand is the correlation structure of the
estimates
produced
On 28/07/2008 6:42 PM, [EMAIL PROTECTED] wrote:
Hello R help list
I have been using the smoothScatter function within the "geneplotter"
package to make some graphs using a Sweave Rnw script called via Rscript
in a DOS/Windows batch file. The Rscript will ultimately be called by a
web service wi
On Mon, Jul 28, 2008 at 4:29 PM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 28/07/2008 6:42 PM, [EMAIL PROTECTED] wrote:
>>
>> Hello R help list
>>
>> I have been using the smoothScatter function within the "geneplotter"
>> package to make some graphs using a Sweave Rnw script called via Rscrip
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