On 17/09/11 01:19, Michael Friendly wrote:
Trying to interpret associations in complex loglinear models from
tables of parameter estimates is like trying to extract sunlight from
a cucumber. You have to squeeze very hard, and then are usually
unhappy with the quality of the sunlight.
For
On Sep 16, 2011, at 7:47 PM, markm0705 wrote:
Dear R helpers
With some help from R-help I've moved the x-scale to the top of a
lattice
plot using...'alternating'
scales = list(x = list(log = 10),alternating=2),
but now I cannot seem to find a way to move the x-axis label t
Hope you don't find this a rethorical question, but why did you prefer
to "google" the function instead of reading the fine manual?
Em 16/9/2011 16:07, song_gpqg escreveu:
Hi!
I am doing a permutation test on 7 data randomly selected into two groups,
one has 4 and the other 3. The purpose is t
The first thing I would do, especially before letting it run for 48
hours, is to take a small subset of the data and run it with the
profiler (Rprof) enable to see where time is being spent. I would
also put some print statements in the main loop to periodically output
the amount of CPU and memory
Hi Mike,
Look at ?options particularly something like:
options(width = 120)
80 is the default, I believe. On 1920 pixels I can comfortably get
around 220 (depending on the overhead of the program, full screen,
etc.). I imagine it would also be possible to run into limitations
from the termina
you want
options(width= )
you can edit your .Rprofile file and the .First function in there to set it
when you start R or in the console interactively
On Fri, Sep 16, 2011 at 12:48 PM, Mike P wrote:
> Hi,
>
> I want to apologize in advance if this has already been asked. I
> wasn't able to fin
Dear R helpers
With some help from R-help I've moved the x-scale to the top of a lattice
plot using...'alternating'
scales = list(x = list(log = 10),alternating=2),
but now I cannot seem to find a way to move the x-axis label to the top ?
Thanks
--
View this message in contex
Hi guyz,
I have serious problems with an algorithm I let run on a supercomputer:
You find the functions under the following URLs:
simuFunctionCaller: http://pastebin.com/6gw2fJFb
calls Function simuFunctionNBM (http://pastebin.com/QeJDUnqx)
after reading a csv-file ordered like the following:
I am trying to build a classifier using SVM and I have 23 independent
variables and 1 response.
I want to check the classification ac curacies of each independent
variables. I did this without any problem.
Now, I want to check the all possible combinations for any two variables
using following sc
I am processing grayscale TIF images that represent X-rays of a sand- and
gravel-filled tube. Each image represents a plane through the porous medium,
and I need to classify each pixel in the image as void or solid. For the most
part, voids are dark and solids are bright, but there are many gr
Hi!
I am doing a permutation test on 7 data randomly selected into two groups,
one has 4 and the other 3. The purpose is to list all the 35 possible
outcomes and calculate p-value. I googled the function perm.test() and not
sure if my usage is correct. So I set a=c(7 data), b=(1,1,1,1,2,2,2) which
Hi,
Thanks for your reply.
I converted using your idea and since I have big file I thought reading as a
table would help.
This is just snippet of my data.
I have around 150 samples so wrting s in that form would be tedious.
>m<-read.table("test.txt",sep='\t',header=TRUE,colClasses=c('character','in
Below is the message from the R workspace.
I have installed the BiodiversityR and rmcdr
How do I get to BiodiversityR GUI ?
Thanks,
Shreya
[Missouri S&T]
Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.
> loca
Hi,
I want to apologize in advance if this has already been asked. I
wasn't able to find any information, either on google or from local
list search.
I'm running an R shell from a linux command line, in an xterm window.
Whenever I print a data frame, only the first couple of columns are
printed s
Just a minor aside; I would have done
my.slopes <- numeric(100)
Note that:
> class(numeric(5))
[1] "numeric"
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 9/16/11 12:37 PM, "Luke Miller" wrote:
>Create an output
It's probably not a sensible thing to do, but I'm going to guess.
With a name like "days_to_tumor_recurrence", I might expect numeric
(integer) values. But "null" and numeric don't mix.
as.numeric(c('24','null',23')) will return 24, NA, 23.
There may have been such a conversion in the preparatio
On 2011-09-16 16:27, andrewH wrote:
Dear Bill--
Wow. This is very clever and I learned a lot from it. I've never seen the
...() trick before, and on a Google code search, I could not find anyone
else who had used it. And I've never used the "..." feature, which, BTW,
though mentioned in every int
Dear Bill--
Wow. This is very clever and I learned a lot from it. I've never seen the
...() trick before, and on a Google code search, I could not find anyone
else who had used it. And I've never used the "..." feature, which, BTW,
though mentioned in every intro to R text, has no help page I can f
You can also do the regressions "in parallel" as follows:
x = -25:25
y = 0.05*x^2 + 2* x - 4
y1 = y + rcauchy(51)
y2 = y + rcauchy(51)^2
y3 = y/10 - 5 + rnorm(51)
Y = cbind(y, y1, y2, y3)
m = lm(Y~x)
print(coef(m))
Hope this helps,
Michael
On Fri, Sep 16, 2011 at 3:37 PM, Luke Miller wrote:
On 11-09-16 4:48 PM, Tarca, Adi wrote:
Hi all,
I am trying to run R CMD check on a package which passes R CMD INSTALL.
The check stops because of a parsing problem in the example of a given function
at this line:
return(res[res$ID %in% list$targetGeneSets,])
The code is ok, since it runs if I
wget worked for me.
Thanks Willian and Rainer.
-M
On Fri, Sep 16, 2011 at 4:18 PM, William Dunlap wrote:
> Wrap the call that may abort with try() or tryCatch().
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
> > -Original Message-
> > From: r-help-boun...@r-project.or
Hi all,
I am trying to run R CMD check on a package which passes R CMD INSTALL.
The check stops because of a parsing problem in the example of a given function
at this line:
return(res[res$ID %in% list$targetGeneSets,])
The code is ok, since it runs if I paste it in R.
Is this a known parsing i
Wrap the call that may abort with try() or tryCatch().
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Rainer
> Schuermann
> Sent: Friday, September 16, 2011 1:09 PM
> T
I haven't tested it thoroughly but what worked here is replacing
> download.file(url, destfile, quiet = FALSE)
with
sys_call <- paste( "wget", url, ">", destfile, sep=" " )
system( sys_call )
Program execution continues, whether or not the download from url was
successful. However, wget is, I be
Already answered on the Mac mailing list. You are asked not to cross-
post.
On Sep 16, 2011, at 1:57 PM, Wei Wang wrote:
Dear list,
I ran into problem trying to load rjags on my Macbook Air running OS
X 10.7.1.
snip
[[alternative HTML version deleted]]
and you are asked to post in
Create an output vector to hold your slopes before starting the loop, then
use your index i to place each new slope in the appropriate position in your
vector.
y1<-rnorm(100, mean=0.01, sd=0.001)
y2<-rnorm(100, mean=0.1, sd=0.01)
x<-(c(10,400))
my.slopes = vector("numeric",100) # initialize a
Hello,
I would like to write a loop to 1) run 100 linear regressions, and 2)
compile the slopes of all regression into one vector. Sample input data
are:
y1<-rnorm(100, mean=0.01, sd=0.001)
y2<-rnorm(100, mean=0.1, sd=0.01)
x<-(c(10,400))
#I have gotten this far with the loop
for (i in 1:10
Yes, you are confusing, because that is exactly what your original message had
in it.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/B
I am planning to download a large number of files from some website. I am
using the following script.
files2down = c('aaa', 'bbb', )
for (i in 1: len)
{
print(paste('downloading file', i, ' of total ', len));
url = paste(urlPrefix, files2down[i], sep='')
destfile = paste (dest, 'in
Dear list,
I ran into problem trying to load rjags on my Macbook Air running OS
X 10.7.1.
I installed JAGS 3.1.0 through binary and it seems OK.
$ jags
Welcome to JAGS 3.1.0 on Fri Sep 16 13:27:03 2011
JAGS is free software and comes with ABSOLUTELY NO WARRANTY
Loading module: basemod: ok
L
Johann Hibschman wrote:
I'm having trouble calling randomForest::partialPlot programmatically.
It tries to use name of the (R) variable as the data column name.
You may want to consider looking at plotmo (in the plotmo package) which
doesn't have the above issue.
library(randomForest)
library
On Sep 16, 2011, at 11:18 AM, Marion Wenty wrote:
hello,
does anyone know how I can write several matrices from R into one
exel-file
using different sheets for the different matrices?
You should describe what sort of searching efforts you have attempted.
When I use what I think are the
Look at the XLConnect package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Marion Wenty
> Sent: Friday, S
Try the XLConnect package.
On Fri, Sep 16, 2011 at 11:18 AM, Marion Wenty wrote:
> hello,
>
> does anyone know how I can write several matrices from R into one exel-file
> using different sheets for the different matrices?
>
> thank you very much in advance for your help.
>
> Marion
>
> [[
Thanks Jim,
That is exactly what I was looking for.
Marlene
2011/9/16 jim holtman
> Will this work for you:
>
> > dat <-
> +
> matrix(c(5.4,4.8,5.6,4.8,NA,4.4,4.6,3.4,NA,NA,4,2.4,NA,NA,NA,2),byrow=TRUE,ncol=4)
> > k <- apply(dat, 1, function(x) which(x == min(x, na.rm = TRUE)))
> > k
> [[1]
Will this work for you:
> dat <-
+
matrix(c(5.4,4.8,5.6,4.8,NA,4.4,4.6,3.4,NA,NA,4,2.4,NA,NA,NA,2),byrow=TRUE,ncol=4)
> k <- apply(dat, 1, function(x) which(x == min(x, na.rm = TRUE)))
> k
[[1]]
[1] 2 4
[[2]]
[1] 4
[[3]]
[1] 4
[[4]]
[1] 4
> sapply(k, paste, collapse = ', ')
[1] "2, 4" "4"
Hello to all the people in the forum,
I'm trying to perform a Scheirer-Ray-Hare test in R, and despite I've found
some posts talking about this test, there no one in which I can found how to
do so.
Can anyone explain me how to do the test?
Thanks
Pablo
--
View this message in context:
http:/
Dear R-members,
My name is Pedro Latorre Carmona, program co-chair of the 2012
"International Conference on Pattern Recognition Applications and
Methods" (ICPRAM 2012) http://www.icpram.org
Please let me send you below the text version of the "Open Call for
papers for Special Sessions" in
I'm having trouble with the mediate() and medsens() functions from the
mediation package.
My treatment, mediator and outcomes variables are all continuous and
scaled to the 0-1 interval. The data is observational not
experimental. I am thus using lm() for the initial outcome and
mediation regressi
hello,
does anyone know how I can write several matrices from R into one exel-file
using different sheets for the different matrices?
thank you very much in advance for your help.
Marion
[[alternative HTML version deleted]]
__
R-help@r-projec
hello jim,
thank you for your reply.
Do you know a solution without having to use a loop?
Marion
2011/9/16 jim holtman
> for (i in names(kinderrechte)){
>q1 <- kinderrechte[, i]
>i 1 <- get(i)
> }
>
> On Fri, Sep 16, 2011 at 4:47 AM, Marion Wenty
> wrote:
> > Hello,
> >
> > does anyo
Not that anything more needs to be said ... it doesn't, not really ...
But I think that in cases like this it is helpful to use the digits arg to
print().
It probably would have shown in a simple way that the two numeric versions
aren't really equal
-- and demonstrated a little about R's default p
Thanks a lot, guys - very helpful!
On Fri, Sep 16, 2011 at 12:52 PM, Bert Gunter wrote:
> This is easy to do (e.g. divide by the appropriate power of 10, use floor,
> multiply the power of 10), but you are going about it in the wrong way.
>
> See the "xaxs" component of the ?par argument list f
This is easy to do (e.g. divide by the appropriate power of 10, use floor,
multiply the power of 10), but you are going about it in the wrong way.
See the "xaxs" component of the ?par argument list for the right way.
-- Bert
On Fri, Sep 16, 2011 at 9:28 AM, Dimitri Liakhovitski <
dimitri.liakh
On 16.09.2011 18:27, Martin Morgan wrote:
On 09/16/2011 04:21 AM, Søren Højsgaard wrote:
Thanks Uwe,
The problem, however, is not solved. If I do:
download.file("http://gbi.agrsci.dk/statistics/courses/2011-ISMLS-course/data/LiSciData_0.0-03.tar.gz";,
tempdir()) Error in
download.file("http
Just for fun:
roundDown <- function(x, digits = 2)
{
currplaces <- nchar(as.character((floor(x
x <- x * 10 ^ (digits - currplaces)
x <- floor(x)
x <- x / (10 ^ (digits - currplaces))
x
}
> roundDown(1.98, 2)
[1] 1.9
> roundDown(1.98, 1)
[1] 1
> roundDown(1.98, 3)
[1] 1.98
>
>
> roundDown(2
In other words, I am looking for something like floor, but more
flexible - as it should allow me to:
turn 1.98 into 1.9 or
turn 298 into 290,
etc.
Dimitri
On Fri, Sep 16, 2011 at 12:28 PM, Dimitri Liakhovitski
wrote:
> Hello!
>
> What function would allow me to "round" down, rather than up?
> For
I'll freely admit that I know little about chron and didn't have a
computer with R in
front of me at the time.
But the original querent converted the chron times to numeric and then used ==,
and thus it became a 7.31 issue. Converting to character is a different way to
approach the problem; which
On 09/16/2011 04:21 AM, Søren Højsgaard wrote:
Thanks Uwe,
The problem, however, is not solved. If I do:
download.file("http://gbi.agrsci.dk/statistics/courses/2011-ISMLS-course/data/LiSciData_0.0-03.tar.gz";,
tempdir()) Error in
download.file("http://gbi.agrsci.dk/statistics/courses/2011-ISMLS
Hello!
What function would allow me to "round" down, rather than up?
For example, x<-1.98
I'd like to get 1.9 - rather than 2.0.
Reason - I am creating a minimum for an axis for a plot, and I need it
to be lower than x (which, in turn, is the lowest number already).
Thank you!
--
Dimitri Liakho
Sorry, maybe I was not clear in what I want. I need the two indexes.
k
2 4
4
4
4
Marlene
2011/9/16 Jean-Christophe BOUËTTÉ
> Hi there,
> Assuming you would prefer to have only one min per row:
>
> apply(dat,1,which.min)
>
> there is a link to this function in the help page for ?min.
> JC
>
> 2
Hi there,
Assuming you would prefer to have only one min per row:
apply(dat,1,which.min)
there is a link to this function in the help page for ?min.
JC
2011/9/16 marlene marchena :
> Hi,
>
> I need to repor the index of a min value of each row in a matrix, but I
> don't know how to do that when
Your coded answer looks right, and is represented in a syntactically correct
manner. Your "desired" answer looks ambiguous, and is certainly not
syntactically correct R. If you want ambiguous and vague answers, perhaps you
need to keep a human in the loop.
---
Hi all,
I've made some progress today. I've managed to move the set seed part of the
function outside of the loop. The only problem I have now is that it says
" Error in results[k, ] <- lapply(1:nsim, function(k) { :
number of items to replace is not a multiple of replacement length"
Give
On Sep 16, 2011, at 11:49 AM, Ben Ganzfried wrote:
> Thanks David. That works perfectly! Although, I'm still not sure I
> understand entirely why the input file says the character value
> "null" and yet, it actually is NA...
Unless you show us the file as it would appear in a text editor, a
Hi,
I need to repor the index of a min value of each row in a matrix, but I
don't know how to do that when I have more than one min value.
Here is my example
> dat <-
matrix(c(5.4,4.8,5.6,4.8,NA,4.4,4.6,3.4,NA,NA,4,2.4,NA,NA,NA,2),byrow=TRUE,ncol=4)
> dat
[,1] [,2] [,3] [,4]
[1,] 5.4 4.8
This is want you want:
op <- as.vector(t(test))
JC
2011/9/16 Chris82 :
> Dear R users,
>
> I guess, I have a quit simple problem, but I'm not getting the solution.
>
> I create a matrix like this:
>
> test <- matrix(seq(1,3,1),ncol=5,nrow=3)
>
> then I want to create one vector, consisting of eve
On 16.09.2011 17:43, Chris82 wrote:
Dear R users,
I guess, I have a quit simple problem, but I'm not getting the solution.
I create a matrix like this:
test<- matrix(seq(1,3,1),ncol=5,nrow=3)
then I want to create one vector, consisting of every row of the matrix.
My first idea was to do th
Thanks David. That works perfectly! Although, I'm still not sure I
understand entirely why the input file says the character value "null" and
yet, it actually is NA...
On Fri, Sep 16, 2011 at 11:35 AM, David Winsemius wrote:
>
> On Sep 16, 2011, at 11:08 AM, David Winsemius wrote:
>
>
>> On Sep
Dear R users,
I guess, I have a quit simple problem, but I'm not getting the solution.
I create a matrix like this:
test <- matrix(seq(1,3,1),ncol=5,nrow=3)
then I want to create one vector, consisting of every row of the matrix.
My first idea was to do this with a for loop.
op <- vector()
f
On Sep 16, 2011, at 11:08 AM, David Winsemius wrote:
On Sep 16, 2011, at 10:36 AM, Ben Ganzfried wrote:
Hi guys,
snipped
Second, I changed the first line of my code to:
tmp <-
ifelse(uncurated
$days_to_tumor_recurrence==NA,"norecurrence","recurrence")
I do not know why your original d
Hi Samir,
to me, autocorrelation is the cross-correlation of a signal with
itself. Which is why I don't really understand the meaning of your
question.
Are you looking for cross-correlation, for example the ccf function
documented in the same help page as acf ?
JC
2011/9/16 Samir Benzerfa :
> Be
You have to give more specifications as to what you want to do with
the data that is being processed. You have values being put into 'q1'
and 'i1' that don't seem to be used in the loop. Do you just want
these values returned? You can easily do it "without a loop" using
lapply:
result <- lapply
On 16/09/2011 10:53 AM, John Clark wrote:
My question was not related to embedding R codes in powerpoint / word,
rather I need to execute some of the commands in sequential way in R, so
that I do not need to type them. This is similar to what you get when you
use the following demo:
require(latt
I use NotePad++ and keep the source in there and then can highlight
and copy over to the R session by pressing a key. You can easily
extend this with the use of AutoHotKey to make it dependent on a mouse
click if you wanted.
On Fri, Sep 16, 2011 at 10:53 AM, John Clark wrote:
> My question was n
looking at the code of example I found that:
oldask <- grDevices::devAskNewPage(ask = TRUE)
x <- rnorm(1000)
hist(x)
plot(x)
plot(x^2)
grDevices::devAskNewPage(oldask)
which will do what you want in terms of displaying plots with a pause.
2011/9/16 John Clark :
> My question was not related to e
On Sep 16, 2011, at 10:36 AM, Ben Ganzfried wrote:
Hi guys,
My code (next 2 lines below) isn't doing what I'm expecting it to:
tmp <-
ifelse(uncurated
$days_to_tumor_recurrence=="null","norecurrence","recurrence")
curated$recurrence_status <- tmp
I want the column "recurrence_status" in my
It seems that your original data has NA values in it (which you should have
said in your original post) and that's where they are coming from in tmp.
E.g.,
R> NA == "null"
[1] NA
To identify them, you need to use is.na() rather than "==NA"
E.g., tmp[is.na(tmp)] <- "norecurrence"
and similarly
On Sep 16, 2011, at 10:08 AM, B77S wrote:
I'm sure Sarah's solution works (and she knows more about R than
myself), but
I ran into a similar problem and used:
as.character(start.time)==as.character(expected_start.time)
What you did was an implicit rounding of the values to the digits
leve
My question was not related to embedding R codes in powerpoint / word,
rather I need to execute some of the commands in sequential way in R, so
that I do not need to type them. This is similar to what you get when you
use the following demo:
require(lattice);
demo(lattice)
So I want to put my co
I think changing nrow() to NROW() will get around that error. However, this,
as you can see, this goes a little astray at the end. Perhaps something like
this will work for you:
z.ext <- rbind(z,array(0,c(9,3)))
z.out <- array(0, c(3,3,9))
for(i in 1:9){
z.out[,,i] <- cbind(z.ext[c(i,i+3,i+6),
Hi Michael,
I mean the character string "null".
Thanks,
Ben
On Fri, Sep 16, 2011 at 10:40 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> Do you actually mean the character string "null" or do you mean the NULL it
> doesn't exist thing? If that's the case, you need to write is.n
Do you actually mean the character string "null" or do you mean the NULL it
doesn't exist thing? If that's the case, you need to write is.null(),
something like is.NA()...
Michael Weylandt
On Fri, Sep 16, 2011 at 10:36 AM, Ben Ganzfried wrote:
> Hi guys,
>
> My code (next 2 lines below) isn't do
Hi,
maybe I did not get the point, but get() does exactly what you want:
whatIwant = 2
nomdeplume = "whatIwant"
get(nomdeplume)
cheers
Am 16.09.2011 16:09, schrieb Timothy Bates:
> On 16 Sep 2011, at 1:49 AM, andrewH wrote:
>>> I'm trying to make a function that takes column names then
>>> usin
Hi guys,
My code (next 2 lines below) isn't doing what I'm expecting it to:
tmp <-
ifelse(uncurated$days_to_tumor_recurrence=="null","norecurrence","recurrence")
curated$recurrence_status <- tmp
I want the column "recurrence_status" in my output file to have values
"recurrence" (if the input val
On 16 Sep 2011, at 1:49 AM, andrewH wrote:
>> I'm trying to make a function that takes column names then
>> using the variable colName, containing the name of
>> the column, to refer to the column itself
rmail...@justemail.net said
> What am I missing about your inquiry: It seems like x[ , colNam
I'm sure Sarah's solution works (and she knows more about R than myself), but
I ran into a similar problem and used:
as.character(start.time)==as.character(expected_start.time)
good luck regardless.
-BS
Sarah Goslee wrote:
>
> Sounds like a case for FAQ 7.31, or, yet another machine precision
Page 7 in my version of formatting.odt (to be sure I have the right
version I downloaded the latest odfWeave from CRAN) discusses
registering style definitions and "Examples of Changing Styles for
Tables, Paragraphs, Bullets and Pages" which has nothing to do with my
question (as far as I
Would not something like
V = apply(P,2,acf, plot=FALSE)
V = sapply(V,`[[`,"acf")
work? I'm pretty sure acf() doesn't return any sort of residuals so you'll
have to calculate those on your own.
Hope this helps,
Michael Weylandt
On Fri, Sep 16, 2011 at 9:20 AM, Jean-Christophe BOUËTTÉ <
jcboue..
On Mon, Sep 12, 2011 at 8:54 PM, Mikkel Grum wrote:
> I've created a chart with times that employees have entered data on named
> tasks as in the following example:
>
> Employee <- c(rep("Tom", 127),
> rep("Dick", 121),
> rep("Sally", 130)
> )
> Time <- c(seq(as.POSIXct("2011-09-12 07:00:00"), as
formatting.odf, page 7. The results are in formattingOut.odt
On Thu, Sep 15, 2011 at 2:44 PM, Jan van der Laan wrote:
> Max,
>
> Thank you for your answer. I have had another look at the examples (I
> already had before mailing the list), but could find the example you
> mention. Could you perhap
On 15/09/2011 10:22 PM, John Clark wrote:
R experts:
I want to prepare a demo (that will play sequentially in clicks) for a
presentation tutorial...Can somebody help me how can I write a demo, suppose
the following are steps in the demo..
#start
set.seed(1345)
x1<- sample(letters[1:10], 5)
x1
s
On Fri, Sep 16, 2011 at 2:17 AM, Cram Rigby wrote:
> I have a problem with lattice log scales that I could use some help with.
>
> I'm trying to print log y-axis scales without exponents in the labels.
> A similar thread with Deepayan' recommendation is here:
> http://tolstoy.newcastle.edu.au/R/e
Hi,
you did not supply a reproducible example. We do not know what your
data nor your code looks like.
Please follow the recommandations found at the bottom of this email!
You're more likely to get a quick and meaningful reply.
JC
2011/9/16 Samir Benzerfa :
> Hi everyone,
>
>
>
> I've got a questi
Hi Yana
Trying to interpret associations in complex loglinear models from tables
of parameter estimates is like trying to extract sunlight from a
cucumber. You have to squeeze very hard, and then are usually unhappy
with the quality of the sunlight.
Instead, you can visualize the association
I've found that while cutree() and rect.hclust() make the same classes
for a given height in the dendrogram, the actual labeling of the classes
is different. For example, both produce the same 4 classes but
class 1 according to cutree() is class 4 according to rect.hclust().
Would it be possible
Hi,
it is not clear to me what your exact request is.
If you want to integrate R code in a presentation, using sweave with
the beamer class is an option.
2011/9/15 John Clark :
> R experts:
>
> I want to prepare a demo (that will play sequentially in clicks) for a
> presentation tutorial...Can some
did not see the "_l" so changes below:
On Fri, Sep 16, 2011 at 8:33 AM, jim holtman wrote:
> for (i in names(kinderrechte)){
> q1 <- kinderrechte[, i]
> i 1 <- get(paste(i, "_l", sep = ""))
> }
>
> On Fri, Sep 16, 2011 at 4:47 AM, Marion Wenty wrote:
>> Hello,
>>
>> does anyone know the so
for (i in names(kinderrechte)){
q1 <- kinderrechte[, i]
i 1 <- get(i)
}
On Fri, Sep 16, 2011 at 4:47 AM, Marion Wenty wrote:
> Hello,
>
> does anyone know the solution to this problem:
>
> I imported a dataframe from Spss into R and saved it as an object called
> "kinderrechte".
>
> I hav
try this:
> str <- c("14.XYZ", "15.ABCDE", "16.dkieowo", "120.EIDKAI")
> x <- gsub("[^0-9]", "", str)
> x
[1] "14" "15" "16" "120"
> as.integer(x)
[1] 14 15 16 120
>
On Fri, Sep 16, 2011 at 5:13 AM, Karl Weinmayer wrote:
> Dear all,
>
>
>
> I have a vector, which looks about like this:
>
Hi Karl,
both strategies are possible using gsub:
gsub("([0-9]*).*","\\1",str) #extract numbers
gsub("[^0-9]","",str) #remove all non-numeric characters
wrap it nicely in an "as.numeric" call to get numeric values.
cheers.
Am 16.09.2011 11:13, schrieb Karl Weinmayer:
> Dear all,
>
>
>
> I h
Sounds like a case for FAQ 7.31, or, yet another machine precision issue.
Try all.equal() instead of ==
Sarah
On Fri, Sep 16, 2011 at 7:36 AM, mebstyne wrote:
> I have two local variables: startTime and expectedStartTime. Both are chron
> related objects.
> When I look at the class for the obje
Hello Folks,
I am trying to do linear optimization using linear constraints on the
decision variables. The problem is it is treating some of the constraints as
soft constraints(violating couple of constraints condition), instead of
throwing infeasible solution. Is there an option in lpsolveapi
Szeptember 12-től 26-ig irodán kívül vagyok, és az emailjeimet nem érem el.
Sürgős esetben kérem forduljon Kárpáti Edithez (karpati.e...@gyemszi.hu).
Üdvözlettel,
Mihalicza Péter
I will be out of the office from 12 till 26 September with no access to my
emails.
In urgent cases please contact
I have two local variables: startTime and expectedStartTime. Both are chron
related objects.
When I look at the class for the objects I can see they are of class
"times".
When I print them to the console, they both read: "09:30:00"
When I print them as.numeric(), they both read: 0.3958333
When I t
Thanks David, I'll give it a go!
Cheers Nikki
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Hi everyone,
I've got a question concerning the function acf(.) in R for calculating the
autocorrelation in my data.
I have a table with daily returns of several stocks over time and I would
like to calculate the autocorrelation for all the series (not only for one
time series). How can I d
Hi,
I'm using ade4 R package to perform some clustering operations.
I've got a problem with the representation of my data with rect.hclust
Currently, I do a tree plot, and I'd like to add color boxes on it to show
where the classes are.
As the order of the classes is never the same, I wanted to u
Hello,
does anyone know the solution to this problem:
I imported a dataframe from Spss into R and saved it as an object called
"kinderrechte".
I have got the following command lines:
q1 <- kinderrechte[,"q0007_0001"]
l1 <- q0007_0001_l
now i would like to be able to change the column name in t
Dear all,
I have a vector, which looks about like this:
str <- c("14.XYZ", "15.ABCDE", "16.dkieowo", "120.EIDKAI")
I need to extract the numerical characters out of the string, so I receive
in the end a normal vector containing:
vec <- c(14, 15, 16, 120)
I need a generic solutio
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