Michael gmail.com> writes:
> There is no way to signal to "aov" the A and B are random effects and
> C is fixed effect; or A is random and B and C are fixed?
It is possible to do blocking with aov and the Error() term, but I would
recommend against it, even if this method is still taught in univ
Michael wrote:
if I do:
yyy=aov(Y~A*B*C);
it seems that the three way ANOVA is based on all fixed-effects.
There is no way to signal to "aov" the A and B are random effects and
C is fixed effect; or A is random and B and C are fixed?
Just put the appropriate Error() term in the model formul
if I do:
yyy=aov(Y~A*B*C);
it seems that the three way ANOVA is based on all fixed-effects.
There is no way to signal to "aov" the A and B are random effects and
C is fixed effect; or A is random and B and C are fixed?
Moreover, I guess I will need the Expected Mean Squares in order to do
the F
R_LATEXCMD is defined in R_HOME/etc/Renviron (sourced by R CMD), and
conditionally re-defined in Rd2dvi.
On a sub-architecture build such as the CRAN MacOS distribution that will
be R_HOME/etc//Renviron, and on that distribution it is defined as
'false'. So you do need to define to a proper v
dennis campos wrote:
> hey can anybody help me? i have to simulate the richardson Arms race
model on R.. for my simulation class...
Hi Dennis,
this list is not intended for solving your homework, however, the
following may help you to go one step ahead.
You should have a look into the help
Dirk Eddelbuettel wrote:
On 9 November 2008 at 20:35, Alan Jackson wrote:
| There must be a simple answer to this.
|
| I'm running ubuntu gutsy, currently have 2.7.2 loaded, but the update
| tools refuse to update it, and don't tell me why. With apt-get I get :
|
| [668 ~]$ sudo apt-get -u -V
The tricks for removing columns specified by name from data frames such as
x$mycol <- NULL
(and others described here
http://wiki.r-project.org/rwiki/doku.php?id=tips:data-frames:remove_columns_by_name)
do not seem to work for a zoo object. Any suggestions as to how to do
this, or is my best bet
Hi All,
Let say
> df
Session_Setup DCT FwdDataVols_bin counts
761 0 1 1 87162
762 0 1 2 11495
763 0 1 3 3986
764 0 1 4 1583
765 0 1
I'm running R 2.8.0 under Fedora 8 (32-bit). I installed the gam
package. I can fit gam models, but I get error messages when I try to
use step.gam and plot.gam, even for examples:
> library(gam)
> ?plot.gam
> data(gam.data)
> gam.object <- gam(y ~ s(x,6) + z,data=gam.data)
> plot(gam.object,se=TR
hi: below almost gets you there except for the endpoints. see filter for
more information.
d=0.5
L=20
x=seq(20, by=1, length.out=20)
temp <- as.numeric(filter(x,filter=c(d,(1-2*d),d),sides=2))
print(temp)
On Sun, Nov 9, 2008 at 10:57 PM, stephen sefick wrote:
#Is there a way to vectorize th
Hi,
i try to plot my graph into different device using x11(),
but when i do this comes up:
> x11(print(plot(A5e$ECAB,A5e$EXPEND,type='p',main='Per capita expenditure
> against economic
+ ability index without
outliners',xlab='ECAB',ylab='EXPEND',xlim=c(0,150),ylim=c(150,400),
+ col='red',col.axi
#Is there a way to vectorize the for loop
#maybe a fancy indexing trick?
#thanks
d=0.5
L=20
x=seq(20, by=1, length.out=20)
reflecting <- function(pre, d, L){
r=L-1
x=rep(0, L)
for(j in 2:r){
x[j]=((1-(2*d))*pre[j])+(d*pre[(j+1)])+(d*pre[(j-1)])
}
x[1
On 9 November 2008 at 20:35, Alan Jackson wrote:
| There must be a simple answer to this.
|
| I'm running ubuntu gutsy, currently have 2.7.2 loaded, but the update
| tools refuse to update it, and don't tell me why. With apt-get I get :
|
| [668 ~]$ sudo apt-get -u -V --simulate dist-upgrade
| R
There must be a simple answer to this.
I'm running ubuntu gutsy, currently have 2.7.2 loaded, but the update
tools refuse to update it, and don't tell me why. With apt-get I get :
[668 ~]$ sudo apt-get -u -V --simulate dist-upgrade
Reading package lists... Done
Building dependency tree
Rea
On Sun, 9 Nov 2008, Yan (Daniel) Zhao wrote:
Suppose I have a function f(x) (say, f(x)=exp(sin(x)), how do I find the
number y such that the integral of f(x) on the interval of (-infinity, y) is
equal to a specified number, say, .8?
Thanks,
Yan
See ?uniroot
David Scott
___
Hi all,
I am using "lm" to fit some anova factor models with interactions.
The default setting for my unordered factors is "treatment". I
understand the resultant "lm" coefficients for one factors, but when
it comes to the interaction term, I got confused.
> options()$contrasts
unordered
Suppose I have a function f(x) (say, f(x)=exp(sin(x)), how do I find the
number y such that the integral of f(x) on the interval of (-infinity, y) is
equal to a specified number, say, .8?
Thanks,
Yan
[[alternative HTML version deleted]]
__
R-h
If I execute
R CMD Rd2dvi foo.Rd
I get messages of the form:
Converting Rd files to LaTeX ...
foo.Rd
Creating dvi output from LaTeX ...
Saving output to 'foo.dvi' ...
cp: .Rd2dvi4366/Rd2.dvi: No such file or directory
Done
xdvi-xaw: Fatal error: foo.dvi: No such file.
Indeed if I add
And in the non-log case, all the previously significant coefficients
now became insignificant...
On Sun, Nov 9, 2008 at 5:36 PM, Michael <[EMAIL PROTECTED]> wrote:
> This is a linear regression of Y onto factors...
>
> If I take log of Y, and regress onto the factors, I got:
>
> Multiple R-squared
This is a linear regression of Y onto factors...
If I take log of Y, and regress onto the factors, I got:
Multiple R-squared: 0.4023, Adjusted R-squared: 0.2731
If I don't take log of Y, and directly regress Y onto the factors, I got:
Multiple R-squared: 0.1807, Adjusted R-squared: -0.0
Dear Christy,
Take a look at
http://www.nabble.com/Efficient-way-to-fill-a-matrix-to20351720.html#a20351720
Here are three options to do what you want:
# Data set
mydata=read.table(textConnection("
point1point2distance
1 10
1 24
2
what should your matrix look like? do you mean something like this?
p1=c(1,1,2,2)
p2=c(1,2,1,2)
d=c(0,4,4,0)
cbind(p1,p2,d)
d2=matrix(d,length(d)/2,length(d)/2)
d2
Cheers,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Hello,
I am trying to convert list of pairwise distances to a distance matrix for
spatial analysis (kriging). For instance, I have something like this for
each pair pf points, and I want to convert it to a matrix:
point1point2distance
1 10
1 2
Peter Dalgaard wrote:
> Heinz Tuechler wrote:
>> At 13:26 09.11.2008, Leon Yee wrote:
>>> Hi, friends
>>>
>>>Is there any functions for object comparing? For example, I have two
>>> list objects, and I want to know whether they are the same. Since the
>>> the components of list are not necessar
On Sun, Nov 9, 2008 at 2:07 AM, Duncan Temple Lang
<[EMAIL PROTECTED]> wrote:
> ( You will first need to have libfftw3 installed. And there is no
>
For those curious, and on Gentoo, emerge sci-libs/fftw.
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
Do you know
I'm not surprised that strange things happen if you try to handle times in
the duplicated hour without specifying which time you mean. If your
current time zone setting is something like EST5EDT that includes both
daylight-saving and standard times then 2008-11-02 01:16:00 occured
twice, so
Dear Tania,
An option would be splitting the data by marker and the perform the
chi-square test selecting both the statistics and p-value for each marker.
Here is an example:
# Dummy data set
mydata=read.table(textConnection("
Marker Treatment Genotype1 Genotype2 Genotype3
1A 23
If your grouping is a list, then you can use 'sapply'; e.g.,
sapply(yourRanking, function(x) sum(x$rank)) # or whatever you want the sum of.
On Sun, Nov 9, 2008 at 4:00 PM, Swanton0822 <[EMAIL PROTECTED]> wrote:
>
> Hi,
> i have group all the data, but now if i would want to sum all the rank in
Sebastian P. Luque wrote:
Hi Roger,
On Sat, 08 Nov 2008 14:31:01 -0800,
Roger Levy <[EMAIL PROTECTED]> wrote:
While I have been able to install rjags on my Windows computer, oddly
I have been unable to install rjags successfully on my 64-bit Linux
compute server (etch, Linux kernel 2.6.18).
On Sat, 8 Nov 2008, Mark Kimpel wrote:
> I found the article the "Y of R" in the latest R news to be very
> interesting. It is certainly challenging me to learn more about how R works
> "under the hood" as the author states. What is less clear to me is whether
> this approach is primarily for teac
Hi,
I am new to R and I need to perform multiple chi-square tests. I
manage to perform one at a time, but is there a specific command
to do multiple tests?
For example, I have a table that looks like this:
Marker Treatment Genotype1 Genotype2 Genotype3
1A 2357
[EMAIL PROTECTED] wrote:
Quoting Prof Brian Ripley <[EMAIL PROTECTED]>:
On Sat, 8 Nov 2008, Roger Levy wrote:
While I have been able to install rjags on my Windows computer, oddly I
have
been unable to install rjags successfully on my 64-bit Linux compute server
(etch, Linux kernel 2.6.18).
Hi,
i have group all the data, but now if i would want to sum all the rank in
each group, how can i do it?
ie. i want to sum the rank in every group, not total. so there will be a sum
of rank for month Jan,Feb,.Dec, therefore there will be total of 12
vaule of ranking sum.
many thanks,
--
Vie
First search the web; there seems to be an abundance of material on
what the algorithm is and then implement the algorithm in R. Once you
have something, then if you are having problems with it, then ask a
question. We can not solve your homework for you.
On Sun, Nov 9, 2008 at 3:18 PM, dennis c
Sure, but we aren't going to do it for you... Start simulating and
then give use specific problems.
On Sun, Nov 9, 2008 at 3:18 PM, dennis campos <[EMAIL PROTECTED]> wrote:
> hey can anybody help me? i have to simulate the richardson Arms race model
> on R.. for my simulation class...
>
> ___
hey can anybody help me? i have to simulate the richardson Arms race
model on R.. for my simulation class...
__
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/po
On 8/11/2008, at 5:48 PM, RON70 wrote:
Sorry to be off-topic. Can somebody please explain me what is
Portmanteau
test? Why it's name is like that? When I would say, a particular
test is
portmanteau test? I did some googling but got no satisfactory
answer at all.
Please anybody help for
# I think this does what you want
d <- rbind(c(1,0,6,4),
c(2,5, 7,5),
c(3,6,8,6),
c(4,0,0,0))
f <- as.matrix(d)
f[-which(rowSums(f==0)>0),]
On Sun, Nov 9, 2008 at 8:30 AM, mentor_ <[EMAIL PROTECTED]> wrote:
>
> Have found a solution:
>
> matrix[rowSums
On 7/11/2008, at 11:33 PM, Christoph Scherber wrote:
Dear all,
I would like to get standard errors (or confidence intervals) for
*predicted* values from an nls fit.
I have tried to implement code from p.225 in MASS (bootstrapping a
nls fit), but this gives only the
confidence intervals o
Dear mentor_,
Try also
yourmat <- matrix(c(
1,0,6,4,
2,5, 7,5,
3,6,8,6,
4,0,0,0
),ncol=4,byrow=TRUE)
yourmat[apply(yourmat,1,function(x) sum(x==0)<1),]
HTH,
Jorge
On Sun, Nov 9, 2008 at 7:39 AM, mentor_ <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I h
# I think this does what you want
d <- rbind(c(1,0,6,4),
c(2,5, 7,5),
c(3,6,8,6),
c(4,0,0,0))
f <- as.matrix(d)
f[-which(rowSums(f==0)>0),]
On Sun, Nov 9, 2008 at 8:30 AM, mentor_ <[EMAIL PROTECTED]> wrote:
>
> Have found a solution:
>
> matrix[rowSums
Hi,
I have a further question about matrix manipulation.
Imagine the following two matrices:
> test
[,1] [,2] [,3] [,4]
[1,]1064
[2,]2575
[3,]3686
[4,]4000
> matrix(is.element(test,0), ncol=4)
[,1] [,2] [,3] [,4]
[1,]
Have found a solution:
matrix[rowSums(matrix == 0) == 0, ]
mentor_ wrote:
>
> Hi,
>
> I have a further question about matrix manipulation.
>
> Imagine the following two matrices:
>> test
> [,1] [,2] [,3] [,4]
> [1,]1064
> [2,]2575
> [3,]368
At 14:24 09.11.2008, Peter Dalgaard wrote:
Heinz Tuechler wrote:
Dear All!
If I try to compare the attributes of two
objects, I find a surprising behaviour of
attr.all.equal(). With identical attributes I
receive the answert NULL. If the attributes
differ, the answer is as expecxted and diff
On Nov 4, 2008, at 8:02 AM, Luca Mortarini wrote:
I am dealing with a 2d vector field on a
regular grid. The field has sparse NA values
(t near the domain borders and not isolated),
which I am trying to substitute with
interpolated values of the nearest point.
What I am looking for is to replac
Hi Mike,
the model you consider is a special case of the four-parameter logistic model
where the
lower and upper asymptotes are fixed at 0.5 and 1, respectively.
Therefore, this (dose-response) model can fitted using the R package 'drc':
library(drc)
xy.m <- drm(y~x, fct = L.4(fixed=c(NA,0.5,
?which
--- On Sun, 11/9/08, Meir Preiszler <[EMAIL PROTECTED]> wrote:
> From: Meir Preiszler <[EMAIL PROTECTED]>
> Subject: [R] Finding Indices of Vector
> To: r-help@r-project.org
> Received: Sunday, November 9, 2008, 10:56 AM
> Lets say I have a vector as follows
> xx<-rnorm(20)
>
> How coul
Hi Roger,
On Sat, 08 Nov 2008 14:31:01 -0800,
Roger Levy <[EMAIL PROTECTED]> wrote:
> While I have been able to install rjags on my Windows computer, oddly
> I have been unable to install rjags successfully on my 64-bit Linux
> compute server (etch, Linux kernel 2.6.18). I am required to specif
On 09/11/2008 10:56 AM, Meir Preiszler wrote:
Lets say I have a vector as follows
xx<-rnorm(20)
How could I find the indices of the vector<0?
which(xx<0)
Duncan Murdoch
Thanks
Meir
Meir Preiszler - Research Engineer
I t a m a r M e d i c a l
Lets say I have a vector as follows
xx<-rnorm(20)
How could I find the indices of the vector<0?
Thanks
Meir
Meir Preiszler - Research Engineer
I t a m a r M e d i c a l Ltd.
Caesarea, Israel:
Tel: +(972) 4 617 7000 ext 232
Fax: +(972) 4 627 5598
C
Ah, perfect! Thanks so much Ken.
In the meantime I played with developing an optim() driven ML search
(included below for posterity), but the glm() + mafc approach is faster and
apparently yields identical results.
# First generate some data
# Define a modified logistic functio
Mike Lawrence thatmike.com> writes:
> Where f(x) is a logistic function, I have data that follow:
> g(x) = f(x)*.5 + .5
> How would you suggest I modify the standard glm(..., family='binomial')
> function to fit this? Here's an example of a clearly ill-advised attempt to
> simply use the standard
Mike Lawrence wrote:
On Sat, Nov 8, 2008 at 3:59 PM, Rubén Roa-Ureta <[EMAIL PROTECTED]> wrote:
...
The fit is for grouped data.
...
As illustrated in my example code, I'm not dealing with data that can be
grouped (x is a continuous random variable).
Four points:
1) I've showed y
Peter Dalgaard wrote:
Heinz Tuechler wrote:
At 13:26 09.11.2008, Leon Yee wrote:
Hi, friends
Is there any functions for object comparing? For example, I have two
list objects, and I want to know whether they are the same. Since the
the components of list are not necessary atomic, this kind
Heinz Tuechler wrote:
Dear All!
If I try to compare the attributes of two objects, I find a surprising
behaviour of attr.all.equal(). With identical attributes I receive the
answert NULL. If the attributes differ, the answer is as expecxted and
differences are shown.
all.equal(attributes(), a
Heinz Tuechler wrote:
At 13:26 09.11.2008, Leon Yee wrote:
Hi, friends
Is there any functions for object comparing? For example, I have two
list objects, and I want to know whether they are the same. Since the
the components of list are not necessary atomic, this kind of comparison
should be
At 13:26 09.11.2008, Leon Yee wrote:
Hi, friends
Is there any functions for object comparing? For example, I have two
list objects, and I want to know whether they are the same. Since the
the components of list are not necessary atomic, this kind of comparison
should be recursive. Does this k
Hi, friends
Is there any functions for object comparing? For example, I have two
list objects, and I want to know whether they are the same. Since the
the components of list are not necessary atomic, this kind of comparison
should be recursive. Does this kind of function exist?
Thank you for
Dear All!
If I try to compare the attributes of two
objects, I find a surprising behaviour of
attr.all.equal(). With identical attributes I
receive the answert NULL. If the attributes
differ, the answer is as expecxted and differences are shown.
all.equal(attributes(), attributes()) instead
Dear all,
I'm writing a code that requires Bessel functions with complex argument.
Searching the list, I found the continuation of a thread I initiated a few
months ago:
http://tolstoy.newcastle.edu.au/R/e4/devel/08/03/0746.html
As I understand, the most promising option would be to use the fort
There is also some info in Q7 on http://batchfiles.googlecode.com which
also has scripts movedir.bat and copydir.bat for faciliting the moving or
copoying of the packages from one library to another (or from any folder
to another for that matter) on Windows. That can facilitate the copying/moving
Thanks for your help and for being so patient with me
Daniel Malter wrote:
>
> Do:
>
> length(variablename)
>
> where variablename is in {mats,time,quar}, i.e. do it for each of them.
> This
> will tell you. Btw: I think you might wanna pickup an introductory manual.
>
> Btw. Peter Dalgaard
Quoting Prof Brian Ripley <[EMAIL PROTECTED]>:
> On Sat, 8 Nov 2008, Roger Levy wrote:
>
> > While I have been able to install rjags on my Windows computer, oddly I
> have
> > been unable to install rjags successfully on my 64-bit Linux compute server
> > (etch, Linux kernel 2.6.18). I am require
Hi Roger,
Try this:
install.packages("rjags",configure.args="--with-jags-lib=/usr/local/lib")
The original sin here is that R is installed in the wrong place. By default,
R is installed into /usr/local/lib64 on 64-bit Linux. But this is a
non-standard location on Debian: it should go in /usr/loc
At 06:25 09.11.2008, Prof Brian Ripley wrote:
On Sat, 8 Nov 2008, Heinz Tuechler wrote:
At 08:01 08.11.2008, Prof Brian Ripley wrote:
We have no idea what you understood (you didn't tell us), but the help says
encoding: character vector. The encoding(s) to be assumed when 'file'
is
'estimated variance' of what?
See ?vcov, ?summary.lm (especially 'sigma') for two possible answers (and
there may be more senses).
On Sun, 9 Nov 2008, cruz wrote:
Hi,
What formula is appropriate for calculating the estimate of variance
from the parametric fit? I have a linear regression mod
We don't know this was for Windows, but a better answer is in the rw-FAQ
Q2.8.
Please (everyone) don't give answers to FAQs but refer to the official
FAQ. (If you think you know a better answer, check that out with the FAQ
maintainer(s) first.)
On Sat, 8 Nov 2008, Christian Schulz wrote:
On Sat, 8 Nov 2008, Roger Levy wrote:
While I have been able to install rjags on my Windows computer, oddly I have
been unable to install rjags successfully on my 64-bit Linux compute server
(etch, Linux kernel 2.6.18). I am required to specify the JAGS module
directory upon installation; whe
On Sat, 8 Nov 2008, RON70 wrote:
Still waiting for some input. Did my question void forum rule in any manner?
It did go against the advice of the posting guide:
Questions about statistics: The R mailing lists are primarily intended for
questions and discussion about the R software. However,
Hi,
What formula is appropriate for calculating the estimate of variance
from the parametric fit? I have a linear regression model i.e. lm(a~b)
Thanks,
cruz
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do rea
try this:
mat <- matrix(sample(90:2000, 1, TRUE), 1000, 10)
ind <- rowSums(mat > 100 & mat < 120) > 0
mat[ind, ]
I hope it helps.
Best,
Dimitris
mentor_ wrote:
Hi,
if I have a large matrix e.g. with the dimensions of 1000 rows and 10
columns.
How can I select rows comprising one or mo
mentor_ wrote:
>
> Hi,
>
> if I have a large matrix e.g. with the dimensions of 1000 rows and 10
> columns.
> How can I select rows comprising one or more values between a specific
> range of two values?
> So e.g. a row has one or more values between the range of 100 - 120.
>
No tested, but t
Hi,
in Windows install the new R-base Installer in a new folder , move or
copy all additional packages from old library folder to the new one
(..do not overwrite the packages included in the base installation!).
Now start the new version and type: update.packages(ask=F).
regards, Christian
Hi,
if I have a large matrix e.g. with the dimensions of 1000 rows and 10
columns.
How can I select rows comprising one or more values between a specific range
of two values?
So e.g. a row has one or more values between the range of 100 - 120.
Cheers
--
View this message in context:
http://www
I am using a rock clustering algorithm to cluster on 40 different subsets of
a data set with 60,000 points. Below is my macro, for some reason I keep
getting a run-time error. I am new to R, was wondering if anyone had any
ideas...
The function works by finding the clusters for each subset, and
We are trying to build a human respiration model.
Preliminary analysis of some breathing signals has shown that humans breathe
through switching among
a finite number of patterns.
Hidden Markov seems to be the right approach. Since most of our code is
written in R scripting language, finding an R p
Hello,
Before submitting a bug report, I thought I would ask here, as
recommended in http://www.r-project.org/posting-guide.html.
Using R 2.8.0 under Mac OS X, I find that edge labels are misplaced
when doing plot.dendrogram(..., horiz=TRUE), for example
> sessionInfo()
R version 2.8.0 (2
While I have been able to install rjags on my Windows computer, oddly I
have been unable to install rjags successfully on my 64-bit Linux
compute server (etch, Linux kernel 2.6.18). I am required to specify
the JAGS module directory upon installation; when doing this within R, I
get:
install
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