Hi Armin,
Alternatively you could use premutation statistics. You could shuffle
your subjects between groups randomly under the Null hypothesis of no
differences between groups and each time claculating the lme model. I am
not sure, but if you do it at each time point of your repetition at
Hello R-user!
I am running R 2.7.0 on a Power Book (Tiger). (I am still R and
statistics beginner)
I try to find the most important variables to divide my dataset as
given in a categorical variable using randomForest.
Is randomForest() able to deal with count data?
Or is there no
Maybe a very, very basic question but how can I get a vector of values with the
specific format:
001,002,010,100
instead of:
1,2,10,100
Paulo
De: [EMAIL PROTECTED] em nome de Stephan Moratti
Enviada: seg 05-05-2008 11:43
Para: r-help@r-project.org
Assunto:
batchfiles 0.4-1 consists of a set of Windows Vista .bat and
other scripts used as front ends to R CMD ... and for other
purposes.
Whereas Version 0.4-0 of batchfiles eliminated the need to
set any paths when running R, version 0.4-1 now eliminates
the need to set any paths when building and
On May 5, 2008, at 7:03 AM, pecardoso wrote:
Maybe a very, very basic question but how can I get a vector of
values with the specific format:
001,002,010,100
instead of:
1,2,10,100
Not perfect, but might get you started:
sprintf(%03d,c(8:10,101))
or
formatC(c(8:10,101), width=3, flag=0)
Hello Tobias,
I am not sure what your wt variable is: I suspect a 'weight'. If it is a
nonnegative measure, then you want a positive density model, not a normal
density in the first place. I think you should try a Gamma GLM, and look at a
Gamma qqplot within each of your conditions. You could
Dear all:
I fitted Observed into a distribution as frequency. The predicted values
are calculated as predicted for frequency. bins is x. I plot observed,
predicted against x in a graph. the commond is here.
Try:
scan(textConnection(u), sep=,)
On Mon, May 5, 2008 at 12:59 AM, Georg Ehret [EMAIL PROTECTED] wrote:
Dear R Usergroup,
I have the following vector and I would like to split it on ,.
How can I do this?
u
[1]
Hi,
Is there a way in R to plot an envelope line from a cloud of points
(x, y data) ? That is, a smooth line that include all points, where
the points do not follow a strait linear pattern. Could somebody
redirect me to some package or function for this? Thanks in advance.
Juli
--
Dear R users,
I have just de-installed my old R 2.4.1 and installed R 2.7.0. I
downloaded this version from the South African mirror, as I am currently
working in South Africa.
When trying to paste any of my old analyses into the console (I keep
them in word-documents), I get an error when
smooth is in the eye of the beholder, but you could look at the
function convex.hull in the package tripack.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax:
Hello R-user!
I am running R 2.7.0 on a Power Book (Tiger). (I am still R and
statistics beginner)
I tried rfImpute (randomForest) and as far as I understood should it
replace NA`s using a proximity matrix:
set.seed(10)
Subset5Imputed-rfImpute(Sex~., data=Subset5)
ntree OOB
Since you are using WORD, you will have to turn off the smart quotes
since WORD is changing them as you type. You may just want to use
something like TINN-R to store your scripts.
On Mon, May 5, 2008 at 8:49 AM, Dennis Hansen [EMAIL PROTECTED] wrote:
Dear R users,
I have just de-installed my
The values NA and NA are different. The first is treated as missing;
the second is not. For example,
table(factor(c(NA,0,1,NA,NA)))
0 1 NA
1 1 2
I suspect you have NA where you want NA, and this is causing your problem.
James
--
James Reilly
Department of Statistics, University of
So, you are on Windows (but didn't tell us). Those 'quotes' in your email
are the so-called smart quotes -- only ASCII quotation marks are accepted
by R.
The change is that R 2.7.0 is actually reading what you pasted into the
clipboard, not some converted version (as previous versions of R
Goodmorning,
I'm looking for a package that can help me to calculate the most
probable number for a series of diluition.
Can you help me?
That you
Riccardo Rasera
---
Dott.Agr.Rasera Riccardo
Dipartimento
hi,
i have 2 lists of ranks for which i'd like to compute kendall tau.
there are ties in the ranks which (to the best of my knowledge) means i
cant use tau a but rather b or c.
how does R handle that? are ties automatically detected (using
corr.test()) and is tau b/c computed instead of tau a?
Xin [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Dear all:
I fitted Observed into a distribution as frequency. The
predicted values are calculated as predicted for frequency.
bins is x. I plot observed, predicted against x in a graph. the
commond is here.
Hi,
I have 2 questions about bwplot in R :
1) How to change the order of my different boxes in the graph ?
2) How to rename the names of the differents boxes ? because I know
how to do that with boxplot (using names) but I do not find the
equivalent parameter in bwplot.
thanks,
Cédric
Thank you James that you saved me from a huge mistake using NA as level.
I specified now
na.strings=NA
$ Sex: Factor w/ 2 levels
0,1: 2 2 2 2 2 2 2 2 2 2 ...
$ outLatTep_like_other : Factor w/ 2 levels
0,1: 2 2 2 2 2 2 2 1 2 2 ...
$
Hello,
Is there any way I can use the gregexpr functions (or a different function)
in a manner that will also return overlapping (i.e. non disjoint) regular
expressions?
For instance, when running gregexpr(AAA,AA), I get two matches, one
at position 1 and one at position 4. I'd like
Dear all,
Is there a less cumbersome way to rename a column by name (as opposed
to index) than --
names( X)[ names[ X] == bob]-sue
?
A semi-related question: how does one get the index of a column by
name, something along the lines of col.index( X, sue) ?
Chip Barnaby
Try this:
gregexpr(A(?=AA),AA, perl = TRUE)
Read about zero width lookahead assertions at ?regex
On Mon, May 5, 2008 at 10:16 AM, Schraga Schwartz
[EMAIL PROTECTED] wrote:
Hello,
Is there any way I can use the gregexpr functions (or a different function)
in a manner that will also
Chip Barnaby wrote:
Dear all,
Is there a less cumbersome way to rename a column by name (as opposed
to index) than --
names( X)[ names[ X] == bob]-sue
?
Not that I know of (notwithstanding that your code doesn't quite work as
it stands...). You might have thought that this would work, but
I have found that the schema.table syntax used in Postgresql (and
Oracle) does not work directly with RODBC.
This works
library(RODBC)
con-odbcConnect(mydb)
d-sqlQuery(con,select * from meso.trees)
However this does not.
d-sqlFetch(con,meso.trees)
Error in odbcTableExists(channel, sqtable) :
apply(matrix(int,nr=10),2,trbm)
?apply
-- Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of DAVID ARTETA GARCIA
Sent: Monday, May 05, 2008 8:19 AM
To: r-help@r-project.org
Subject: [R] tukey biweight of vector
Peter,
My method seems to work, see below. Is there
some reason it might not work in general? This
is the tersest way I have found to rename, but it
seems awkward, given that names( X) is repeated.
Chip Barnaby
--
head( airquality)
Ozone Solar.R Wind Temp Month
Chip Barnaby wrote:
Peter,
My method seems to work, see below. Is there some reason it might not
work in general? This is the tersest way I have found to rename, but
it seems awkward, given that names( X) is repeated.
Nono, this technique works fine. It just wasn't what you wrote (look
Yes, someone else pointed out my typo, sorry.
My issue here is that referring to columns by
index is risky. Hard-coded indices will be
buried in code and there will be trouble if
(when) the data organization changes. So I am
trying to learn how to work with names, but the
language does
Just having read the nice review article on boosting in the latest
Statistical Science, I would love to reproduce some of the plots inside
that article, but it is not clear to me how to create the partial
contribution plots for the Poisson regression.
Does anyone have example code for this ?
(The
Hi Cedric,
to change the names of different boxes, you could use levels on the
dataframe factor variable you're plotting- see this example:
library('lattice');
bwplot(voice.part ~ height, data=singer, xlab=Height (inches))
singer.m-singer
levels(singer.m$voice.part)- c(a,f,g,j,k,o,y,t)
On 5/5/2008 10:48 AM, Chip Barnaby wrote:
Dear all,
Is there a less cumbersome way to rename a column by name (as opposed
to index) than --
names( X)[ names[ X] == bob]-sue
I find this clearer:
X$sue - X$bob
X$bob - NULL
Duncan Murdoch
?
A semi-related question: how does one get the
Claire_6700 wrote:
Hello,
I am having really hard time finding a good article about simulating
genotypes of cases and controls at a disease locus using R.
if you guys can point me or guide me where i can find more information, it
will be helpful.
The popgen() package allows the
I agree there. I believe the same effect can be achieved with
'transform'.
transform(X, sue = bob, bob = NULL)
Perhaps there are performance reasons for using the names() - value
approach for very large data.frames?
Duncan Murdoch wrote:
On 5/5/2008 10:48 AM, Chip Barnaby wrote:
Dear
On Mon, May 5, 2008 at 12:08 PM, hadley wickham [EMAIL PROTECTED] wrote:
On Mon, May 5, 2008 at 10:54 AM, Chip Barnaby [EMAIL PROTECTED] wrote:
Yes, someone else pointed out my typo, sorry.
My issue here is that referring to columns by index is risky. Hard-coded
indices will be
On Mon, May 5, 2008 at 10:54 AM, Chip Barnaby [EMAIL PROTECTED] wrote:
Yes, someone else pointed out my typo, sorry.
My issue here is that referring to columns by index is risky. Hard-coded
indices will be buried in code and there will be trouble if (when) the data
organization changes. So
Hello,
I am using xYplot to plot lines with confidence bands (see test
example below). I would like to add a single point with
confidence bars to each graph (coordinates of the points below:
px, py, pxlow, etc...). I tried playing with
panel=panel.superpose, but failed to get it to work.
Duncan Murdoch wrote:
On 5/5/2008 10:48 AM, Chip Barnaby wrote:
Dear all,
Is there a less cumbersome way to rename a column by name (as opposed
to index) than --
names( X)[ names[ X] == bob]-sue
I find this clearer:
X$sue - X$bob
X$bob - NULL
...or, essentially the same:
X - within(X,
our group at Duke is currently attempting to connect two data sets fed
through a web-based system, one containing research questions (i.e.,
variables positioned with certain roles) and a database of statistical
methods. this connection is done informally in a number of R packages
and related
Question for the list members: Is anybody aware of any existing
ontologies computational focusing on this topic?
Have you look at zelig (http://gking.harvard.edu/zelig/) ? It unifies
a lot of statistical models within a framework that describes variable
types for input and output.
You
Hi everyone,
I'm working with the Epical library, specicatly using the power test in
proportions. I think this test is not working like in the book:
Biostatistical Analysis (4th Edition): Jerrold H. Zar
In the example 23.25. (I attach this Pic) It's not the same answer.
Using the follow command
Duncan Golicher writes:
[snip]
thread that I found led to an obvious solution. Brian Ripley pointed out
the issue with the dot syntax and suggested that you can set the schema
and then use unqualified names. Does anyone know how this can be done
for postgresql?
[snip]
The SQL-Syntax for
There is a rename function in the reshape package and also one in the
memisc package.
On Mon, May 5, 2008 at 10:48 AM, Chip Barnaby [EMAIL PROTECTED] wrote:
Dear all,
Is there a less cumbersome way to rename a column by name (as opposed to
index) than --
names( X)[ names[ X] == bob]-sue
?
Hi!
(a complete newby, but will not give up easily!)
I was wondering if there is any way to decouple the axis and tick mark
widths? As I understand they are both controlled by the lwd setting, and
cannot be controlled independently? For example I might want to create major
and minor ticks,
On 5/5/08, cfinet [EMAIL PROTECTED] wrote:
Hi,
I have 2 questions about bwplot in R :
1) How to change the order of my different boxes in the graph ?
2) How to rename the names of the differents boxes ? because I know how to
do that with boxplot (using names) but I do not find the
Hello,
I'm a new user of the R environment. I need to do some power analysis.
For this purpose, I installed the pwr package from the R window, but
unfortunately something went wrong. The installation of the package
was successful (I got a message saying so in R) but when I enter a
Hello Gurus:
When I plot a survival fit using rpart for the classification tree, for each
node, there is a decimal based number above the event/total. I tried to see if
it's the exponential ratio or logrithmics, neither seem to be the case. I'm
wondering if anyone knows what they are.
On 5/5/08, John Poulsen [EMAIL PROTECTED] wrote:
Hello,
I am using xYplot to plot lines with confidence bands (see test example
below). I would like to add a single point with confidence bars to each
graph (coordinates of the points below: px, py, pxlow, etc...). I tried
playing with
On 5/5/2008 3:02 PM, maiya wrote:
Hi!
(a complete newby, but will not give up easily!)
I was wondering if there is any way to decouple the axis and tick mark
widths? As I understand they are both controlled by the lwd setting, and
cannot be controlled independently? For example I might want to
For such simple expressions you can use 'beheading': remove the current
match with substr() and run regrepr again. That does not work for some
regexps which depend on e.g. word boundaries, though.
Alternatively you can modify the source code.
On Mon, 5 May 2008, Schraga Schwartz wrote:
See
?chull
juli pausas [EMAIL PROTECTED] 05/05/08 1:56 PM
Hi,
Is there a way in R to plot an envelope line from a cloud of points
(x, y data) ? That is, a smooth line that include all points, where
the points do not follow a strait linear pattern. Could somebody
redirect me to some package or
see ?library
You need to load the package before R can access its functions.
Kingsford Jones
On Mon, May 5, 2008 at 12:18 PM, [EMAIL PROTECTED] wrote:
Hello,
I'm a new user of the R environment. I need to do some power analysis. For
this purpose, I installed the pwr package from the R
On Mon, 5 May 2008, Chang Liu wrote:
Hello Gurus:
When I plot a survival fit using rpart for the classification tree,
I don't think that is a 'classification tree': no classification is going
on. I will assume you called rpart() with a survival response.
for each node, there is a
On Mon, 05-May-2008 at 03:18PM -0400, [EMAIL PROTECTED] wrote:
Hello,
I'm a new user of the R environment. I need to do some power analysis. For
this purpose, I installed the pwr package from the R window, but
unfortunately something went wrong. The installation of the package was
José Ignacio Bustos Melo wrote:
Hi everyone,
I'm working with the Epical library, specicatly using the power test in
proportions. I think this test is not working like in the book:
Biostatistical Analysis (4th Edition): Jerrold H. Zar
In the example 23.25. (I attach this Pic) It's not the same
Many thanks Albin,
Problem solved. This works fine for me now.
library(RODBC)
con-odbcConnect(mydb)
odbcQuery(con,SET search_path =meso, pg_catalog;)
d-sqlFetch(con,trees)
sqlSave(d,trees2)
Duncan
Albin Blaschka wrote:
Duncan Golicher writes:
[snip]
thread that I found led to an obvious
Hi
This is an ad-hoc fix..
d-rnorm(4)
plot(1:4,d)
g-3;
rug((1:4),ticksize=-.03,lwd=g,side=1)
rug((1:4),ticksize=0,lwd=(g-1),side=1,col=white)
just drawing another white axis on top of the old one.. it looks ok :)
y
maiya wrote:
Hi!
(a complete newby, but will not give up easily!)
Hi again R users. I was reviewing the manual of R about how to install
packages using the GUI and it's corresponding menu packages. My
question is how to download a specific set of packages in order to
install them in a computer that doesn't have internet access. Is there
a web page where I can
Dear Rodrigo,
I'm going to assume from your description that you're using a Windows OS. In
that case, you can download Windows binary packages for the current version
of R from the packages' pages on CRAN (e.g.,
http://cran.r-project.org/web/packages/car/index.html for the car package --
note,
Dear Rodrigo,
The packages are available in http://cran.r-project.org/web/packages/ What
you can do is download them to your flash drive in .zip format. Once you do
that, you can install them from R via Packages-Install package(s) from
local zip files and browsing them in your flash drive.
To
Try the download.packages function also:
download.packages(c(car, boot, MASS), destdir = C://)
On Mon, May 5, 2008 at 6:12 PM, Rodrigo Briceño [EMAIL PROTECTED]
wrote:
Hi again R users. I was reviewing the manual of R about how to install
packages using the GUI and it's corresponding menu
On Mon, 5 May 2008, John Fox wrote:
Dear Rodrigo,
I'm going to assume from your description that you're using a Windows OS. In
that case, you can download Windows binary packages for the current version
of R from the packages' pages on CRAN (e.g.,
Dear Helpers of R,
I was wondering if you can help me with a few question about the program. I
was trying create an expression file from my CEL files and I came upon an
error stating that I had not enough memory and to see help(memory.size).
From reading the help txt it seems that my max memory
I have numerous objects, each containing continuous data representing the
same variable, movement rate, yet each having a different number of rows.
e.g.
d1-as.matrix(rnorm(5))
d2-as.matrix(rnorm(3))
d3-as.matrix(rnorm(6))
How can I merge these three columns side-by-side in order to create a
you can do below but there's no way of getting arounf having NAs in your
final matrix.
d1-as.matrix(rnorm(5))
d2-as.matrix(rnorm(3))
d3-as.matrix(rnorm(6))
templist - list(d1,d2,d3)
maxnum - max(sapply(templist,length))
print(maxnum)
temp - lapply(templist,function(.mat) {
if (nrow(.mat)
Dear list,
I have a time-series Y of length n which has significant auto-correlation at
lag 1, as indicated by acf plots.
According to certain criteria, I have defined two groups of observations n1 and
n2. The objective is to estimate the difference of the means between these
groups and test
On 6/05/2008, at 11:19 AM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear list,
I have a time-series Y of length n which has significant auto-
correlation at lag 1, as indicated by acf plots.
According to certain criteria, I have defined two groups of
observations n1 and n2. The objective
thank you. i will take a look
claire,
Neil Shephard wrote:
Claire_6700 wrote:
Hello,
I am having really hard time finding a good article about simulating
genotypes of cases and controls at a disease locus using R.
if you guys can point me or guide me where i can find more
Try this also:
sapply(list(d1, d2, d3), [, 1:max(sapply(list(d1, d2, d3), length)))
On Mon, May 5, 2008 at 7:46 PM, T.D.Rudolph [EMAIL PROTECTED]
wrote:
I have numerous objects, each containing continuous data representing the
same variable, movement rate, yet each having a different number
Hi the list
Is it possible to add one argument to the arguments contain in ... ?
I would like to do :
f - function(ttt,...)[
if(... does not contain the args uuu)
[add uuu=3 to ...]
else[]
g(ttt,...)
]
More precisely, my function g is a callNextMethod()
Thanks
Christophe
Hey, all. I had a quick question about fitting new glm values and
then looking at the error around them. I'm working with a glm using a
Gamma distribution and a log link with two types of treatments.
However, when I then look at the predicted values for each category, I
find for the one
G'day Jarrett,
tapply(y,x,mean)
a b
1.00 25.58
On Mon, 5 May 2008 20:21:26 -0700
Jarrett Byrnes [EMAIL PROTECTED] wrote:
Hey, all. I had a quick question about fitting new glm values and
then looking at the error around them. I'm working with a glm using
a Gamma distribution
On Mon, 5 May 2008, qin gao wrote:
Dear Helpers of R,
I was wondering if you can help me with a few question about the program. I
was trying create an expression file from my CEL files and I came upon an
error stating that I had not enough memory and to see help(memory.size).
From reading the
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