Hi Felipe,
Daniel mentions imputation is a disputed practice. There are
recommendations and rules of thumb for its use. I am not sure that
imputation is disputed. I would be interested to see some links to articles
recommending against its use.
Paul
- Original Message -
From: "
Thanks for reminding me. But, the problem still exists when I combine these two
'joy' levels.
Quoting Douglas Bates <[EMAIL PROTECTED]>:
> By the way, did you notice that the levels of Emotion include both
> "joy" and "joy ". You may want to correct that.
>
> On Sat, Jul 12, 2008 at 7:47 AM, Do
Yes, your version (func2) is quick, quickest for longer vectors:
> m <- matrix(rexp(6e6,rate=0.05), nrow=5) # 120 cols
> m[m<20] <- 20
> func1 <- function(v,cut=20) max(which(v>cut))
> func2 <- function(v,cut=20) {
+x <- which(v>cut)
+x[length(x)]
+ }
> func3 <- function(v,cut=20) tail
Yes, that Rblas.dll is known to be faulty, and the person who built it is
unable to re-build it. It needs to be removed from CRAN.
(I've also tried to build on Core 2 Duo, and my Cygwin installation has a
compiler crash during the build.)
On Tue, 8 Jul 2008, Law, Jason wrote:
I noticed a p
Søren Højsgaard wrote:
The tricky part is not getting it through the checks on my computer. It is when I upload to CRAN I
get the problems, because "their" computers need Rgraphviz as well...
("Suggests" does not seem to be the solution...)
Søren,
"CRAN" (which means my machine in the cas
Prof Brian Ripley wrote:
Yes, that Rblas.dll is known to be faulty, and the person who built it
is unable to re-build it. It needs to be removed from CRAN.
Whoops, I forgot to remove it and will do so this afternoon.
Uwe
(I've also tried to build on Core 2 Duo, and my Cygwin installation
[EMAIL PROTECTED] wrote:
From the R console I invoke:
install.packages("RWinEdt")
and get:
Warning in install.packages("RWinEdt") :
argument 'lib' is missing: using 'F:\Users\Kevin\Documents/R/win-library/2.7'
--- Please select a CRAN mirror for use in this session ---
trying URL
'http:/
Hi!
Well, if you look at the output:
shapiro.test(rnorm(5000))
>
>Â Â Â Â Shapiro-Wilk normality test
>
> data:Â rnorm(5000)
> W = 0.9997, p-value = 0.6205
You can see that the p-value is 0.6205 so you can't refuse the normality
hypotesis.
H0: normal data   vs H1: not normal
So shapiro
The reason for the empty levels was I did not put drop=TRUE on the
split to remove unused levels. Here is the revised script:
> set.seed(1) # start with a known number
> x <- data.frame(cat=sample(LETTERS[1:3],20,TRUE),a=sample(letters[1:4], 20,
> TRUE), b=runif(20))
> x
cat a b
1
This may do what you want:
> x <- read.table("/tempxx.txt", comment="", quote="", sep="|", header=TRUE,
> as.is=TRUE)
> # split out by name
> z <- lapply(seq(nrow(x)), function(.row){
+ .result <- NULL
+ # construct the data output
+ for (i in c('picnic', 'food', 'other')){
+
Marta Colombo wrote:
Hi!
Well, if you look at the output:
shapiro.test(rnorm(5000))
    Shapiro-Wilk normality test
data:Â rnorm(5000)
W = 0.9997, p-value = 0.6205
You can see that the p-value is 0.6205 so you can't refuse the normality hypotesis.
H0: normal data   vs H1: not n
On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
> [...]
> A large P-value means nothing more than needing more data. No
> conclusion is possible. Please read the classic paper Absence of
> Evidence is not Evidence for Absence.
Is that ironic, Frank, or is there really a "classic paper" with
t
http://www.bmj.com/cgi/content/full/311/7003/485
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Ted Harding
Sent: Sunday, July 13, 2008
G'day all,
On Sun, 13 Jul 2008 15:55:38 +0100 (BST)
(Ted Harding) <[EMAIL PROTECTED]> wrote:
> On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
> > [...]
> > A large P-value means nothing more than needing more data. No
> > conclusion is possible.
I would have thought that "we need more data
Uwe Ligges <[EMAIL PROTECTED]> [Sat, Jul 12, 2008 at 11:48:38AM CEST]:
>
>
> Johannes Huesing wrote:
>> I am still trying to build a package. At the moment I am stuck with a
>> file not found error message when processing R code from the tests
>> subdirectory. What would be the accurate relative pa
What is the least surprising way of initializing a factor with
predefined levels and with length 0?
as.factor(c("eins", "zwei", "drei"))[FALSE]
does the job but looks a bit weird.
--
Johannes Hüsing There is something fascinating about science.
One
Johannes Huesing wrote:
Uwe Ligges <[EMAIL PROTECTED]> [Sat, Jul 12, 2008 at 11:48:38AM CEST]:
Johannes Huesing wrote:
I am still trying to build a package. At the moment I am stuck with a
file not found error message when processing R code from the tests
subdirectory. What would be the accu
Many thanks to Berwin, and also to Charles Annis, for the
references. The're good!
Ted.
On 13-Jul-08 15:22:03, Berwin A Turlach wrote:
> G'day all,
>
> On Sun, 13 Jul 2008 15:55:38 +0100 (BST)
> (Ted Harding) <[EMAIL PROTECTED]> wrote:
>
>> On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
>> > [
I was not able to follow the solution posted. Could you demonstrate
this technique on an example
data set. Thanks!
dat <- data.frame(a = letters[1:3], b = LETTERS[1:3], c = 1:3, d = 3:1)
On Wed, Jul 2, 2008 at 1:13 PM, Charles C. Berry <[EMAIL PROTECTED]> wrote:
> On Wed, 2 Jul 2008, Ben Tupper w
"Erin Hodgess" <[EMAIL PROTECTED]> writes:
> Dear R People:
>
> I'm running Rmpi on a single machine and I have the following
> statement from the command line:
>
> mpirun -np 3 ./R --no-save < eek1.in >stuff4.out
All three versions of eek1.in write to the same location, over-writing
one another
(Ted Harding) wrote:
On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
[...]
A large P-value means nothing more than needing more data. No
conclusion is possible. Please read the classic paper Absence of
Evidence is not Evidence for Absence.
Is that ironic, Frank, or is there really a "clas
On Sun, 2008-07-13 at 18:47 +0200, Johannes Huesing wrote:
> What is the least surprising way of initializing a factor with
> predefined levels and with length 0?
> as.factor(c("eins", "zwei", "drei"))[FALSE]
> does the job but looks a bit weird.
>
Notice that one does not need to specify any
Disputed was probably not the correct wording for it. However, imputation
means that you make assumptions regarding the distribution of your missing
data dependent on the data that is available to you. Felipe had time series
data, and it is common to predict from the past to the future in such
mode
Dear all,
I am new to r and most happy for now:)
I would like to ask an issue about rm-anova.
I have data of an experiment with 24 subjects 3 treatment (8 replicates for
each treatment) and 8 sampling through time. data sheet is something like
that(just an example, not real).
sample id,response(t
While the earlier solutions involving expression() and paste() work
great, unfortunately Gabor's first suggestion doesn't display on the
OS X default quartz device, and Gabor's second suggestion displays on
quartz, but not to the pdf() device.
In any event, the first replies in this thread
Gavin Simpson <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at 08:18:37PM CEST]:
> On Sun, 2008-07-13 at 18:47 +0200, Johannes Huesing wrote:
[...]
> > as.factor(c("eins", "zwei", "drei"))[FALSE]
> > does the job but looks a bit weird.
> >
[...]
> > factor(levels = c("one","two","three"))
> factor(0)
>
Frank E Harrell Jr <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at 08:07:37PM CEST]:
> (Ted Harding) wrote:
>> On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
>>> [...]
>>> A large P-value means nothing more than needing more data. No
>>> conclusion is possible. Please read the classic paper Absence
On Sun, 13 Jul 2008, Juliet Hannah wrote:
I was not able to follow the solution posted. Could you demonstrate
this technique on an example
data set. Thanks!
dat <- data.frame(a = letters[1:3], b = LETTERS[1:3], c = 1:3, d = 3:1)
Using your example:
dat <- data.frame(a = letters[1:3], b = LE
Hello everyone,
I am using the following code to try to calculate the mean :
dat<-read.table(file="C:\\Documents and Settings.txt")
dat<-as.numeric(dat)
x1.m__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do r
What does 'str(dat)' show? the statement
dat <- as.numeric(dat)
says you are trying to make an entire dataframe numeric. This is
probably not what you want to do. What is it you want to do? Have
you tried
summary(dat)
e.g.,
> x <- data.frame(a=1:10, b=101:110, c=letters[1:10])
> summary(x)
On 13-Jul-08 19:53:47, Johannes Huesing wrote:
> Frank E Harrell Jr <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at
> 08:07:37PM CEST]:
>> (Ted Harding) wrote:
>>> On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
[...]
A large P-value means nothing more than needing more data. No
conclu
Hi,
I would like to make a stem and leaf plot and I want to edit the
category-names.
So, by doing this:
> x <- c(1,2,2,3,3,3,3,2,2,1)
> stem(x)
I get:
1 | 00
1 |
2 |
2 |
3 |
First Question: Why do I get gaps between the categories?
(like in line 2 and line 4)
And seco
Thank you I will try drop=TRUE.
In the mean time do you know how I can access the members (for lack of a better
term) of the results of a split? In the sample you provided below you have:
z <- split(x, list(x$cat, x$a), drop=TRUE)
Now I can print out 'z[1], z[2] etc' This is nice but what if I
Thank you, but this is not what i want exactly.. i would want to launch
function "myfun" with this script:
> table <- sample(LETTERS[1:5], 20,TRUE)
>
> name <- "A"
>
> myfun <- function(name) {
+ r <- grep (name[^0-9], table )
+ return (r) }
but if I do it ,R doesn't accept this.. i want this be
I checked and reapplied full access to the whole WinEdt directory. I am not the
administrator but I am a member of the administrators group. This couple with
the fact the I have given full control to the WinEdt directory suggests that
the problem is elsewhere. But this is not the first time that
Ted Harding <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at 10:59:21PM CEST]:
> On 13-Jul-08 19:53:47, Johannes Huesing wrote:
> > Frank E Harrell Jr <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at
> > 08:07:37PM CEST]:
> >> (Ted Harding) wrote:
> >>> On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
> [..
I know how to set graphic parameters by calling par(), but what I'd like
is a way to set the default values so that subsequent calls to par() use
my defaults. The reason to want this is that every time I create a new
graphic window (I'm using quartz on OSX, and so far no answers in the
Mac mai
On Sun, 13 Jul 2008, Fran100681 wrote:
Thank you, but this is not what i want exactly.. i would want to launch
function "myfun" with this script:
table <- sample(LETTERS[1:5], 20,TRUE)
name <- "A"
myfun <- function(name) {
+ r <- grep (name[^0-9], table )
...XX
On Sun, Jul 13, 2008 at 5:45 PM, <[EMAIL PROTECTED]> wrote:
> Thank you I will try drop=TRUE.
>
> In the mean time do you know how I can access the members (for lack of a
> better term) of the results of a split? In the sample you provided below you
> have:
>
> z <- split(x, list(x$cat, x$a), dr
or your function looks like this were you dynamically create the string:
myfun <- function(name) {
r <- grep (paste(name, "[^0-9]", sep=""), table )
return (r) }
On Sun, Jul 13, 2008 at 7:24 AM, Fran100681 <[EMAIL PROTECTED]> wrote:
>
> Thank you, but this is not what i want exactly.. i would wa
See at end.
On 13-Jul-08 21:42:19, Johannes Huesing wrote:
> Ted Harding <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at
> 10:59:21PM CEST]:
>> On 13-Jul-08 19:53:47, Johannes Huesing wrote:
>> > Frank E Harrell Jr <[EMAIL PROTECTED]> [Sun, Jul 13, 2008 at
>> > 08:07:37PM CEST]:
>> >> (Ted Harding) wrot
Thanks Jim,
I wish I were comfortable enough with the language for the fix needed to the
syntax to be obvious, but it is not yet. With your example, I get:
Error in strsplit(x[.row, ][[i]], ";#") : non-character argument
x appears to be filled properly, but z is not due to the error.
One way that I do it is to save the default parameters with the
following statement in my profile:
assign('Default.par', par(no.readonly=T))
An then I have a function which will reset them:
plotReset <-
function()
{# reset plotting window
par(Default.par)
windows(width=7.5,height=4.7
Hi,
I am trying to do a Moran's I test on an ordinal logistic regression model.
I have a simple spatial weights matrix listed below I would like to use.
Y=
10 0 0 0 0 0 0 0
01 1 0 0 0 0 1 1
Hi,
I am trying to do a Moran's I test on an ordinal logistic regression model.
I have a simple spatial weights matrix listed below I would like to use.
Y=
1
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
1
1
0
1
1
0
0
0
0
1
1
0
0
0
1
1
1
1
0
0
0
0
0
1
1
1
1
0
0
0
0
0
1
1
1
1
0
0
0
0
0
1
1
1
1
0
0
Is one of the rows NULL? Do an 'str(x)' show? The example you sent
seems to work with the code. Are you reading in a different set of
data? I think I know what happened. I shortened the names on your
example so it was easier to access. Here is the data I used:
ID|name|picnic|food|other
1|Yog
Is there a better or more efficent approach than this without the use of t() ?
> (m <- matrix(1:40, ncol=4)) [,1] [,2] [,3] [,4] [1,]1 11 21 31
> [2,]2 12 22 32 [3,]3 13 23 33 [4,]4 14 24 34
> [5,]5 15 25 35 [6,]6 16 26 36 [7,]
m <- matrix(1:40, ncol=4);
groups <- rep(1:2, each=2);
uGroups <- unique(groups);
mMeans <- matrix(NA, nrow=nrow(m), ncol=length(uGroups));
for (gg in seq(along=uGroups)) {
mMeans[,gg] <- rowMeans(m[,groups == uGroups[gg], drop=FALSE]);
}
(Preallocation of result matrix is more memory efficient
On 13/07/2008 5:40 PM, Jörg Groß wrote:
Hi,
I would like to make a stem and leaf plot and I want to edit the
category-names.
If you have a computer you can do much better histograms. But since you
have chosen to do this, one way is to edit the underlying C code. It's
available in
https
I'm trying to learn about the tcltk package and its uses. Floundering
around a bit ... Have discovered Peter Dalgaard's articles in R-News,
which should help. Also James Wettenhall's suite of examples look like
they might be enlightening, even though the indications are that they
are
Windoze
On 14/07/2008, at 1:45 PM, Duncan Murdoch wrote:
On 13/07/2008 5:40 PM, Jörg Groß wrote:
Hi,
I would like to make a stem and leaf plot and I want to edit the
category-names.
If you have a computer you can do much better histograms. But
since you have chosen to do this, one way is to ed
Instead of rowMeans(m[,groups == uGroups[gg], drop=FALSE]), I used
apply(m[,groups == uGroups[gg], drop=FALSE], 1, mean), which really slow down
the processing. I modified your codes slightly so that I can use sd, median,
and mad also. Better and efficient approach ?> Date: Sun, 13 Jul 2008 18:
Where is the link on www.r-project.org or CRAN to download source code for
development versions of packages? This is straightforward for BioConductor
packages but I can't seem to find it for R packages.
Mark
--
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University Schoo
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