glenn g1enn.robe...@btinternet.com writes:
Is there a function (before I try and write it !) that allows the input of a
covariance or correlation matrix to calculate PCA, rather than the actual
data as in princomp()
Yes, there is: princomp(). :-)
--
Bjørn-Helge Mevik
Hi Paul,
Have you ever seen a drawing of the regions of an R plot with the
terminology that is used for parts?
From what I can remember, several documents on CRAN cover this. The one that
springs to mind is Alzola Harrell's An Introduction to S and the Hmisc
and Design Libraries,” which you
Hi James,
What you really need to do is to check out the many freely available pdfs
for R beginners. Here is a good place to start
http://cran.r-project.org/other-docs.html
If I am right interpreting what you want, I think you need to create a blank
plot with no axes, axis labels etc. Try
The PCA is just a singular value decomposition on a sample covariance/
correlation matrix. Do a search for ?svd and get the eigenvalues and
vectors from that function.
On Feb 14, 10:30 am, glenn g1enn.robe...@btinternet.com wrote:
Hi All, would appreciate an answer on this if you have a moment;
Dear R-Experts,
seek your help.
There are two parts I want to deal with.
1)
I want to create a time interval of say, 30 minutes starting from 00:00:00
hrs
Thus at the end, I want to create sequence:
00:00:00
00:30:00
01:00:00
01:30:00
..
..
How to do so ?
Later, I want to change the
Dear list,
I am trying to follow an example that estimates a 2x2 markov transition
matrix across several periods from aggregate data using restricted least
squares.
I seem to be making headway using solve.QP(quadprog) as the unrestricted
solution matches the example I am following, and I can
Hi Glen, Andrew,
The PCA is just a singular value decomposition on a sample covariance/...
I believe that Bjørn-Helge Mevik's point was that __if you read the
documentation__ you will see the argument covmat to princomp(). This,
really, is much more straightforward and practical than Andrew's
[reposting as a plain text - apologies for the double posting]
Dear list,
I am trying to follow an example that estimates a 2x2 markov
transition matrix across several periods from aggregate data using
restricted least squares.
I seem to be making headway using solve.QP(quadprog) as the
Dear listers,
I would like to compare the levels of a factor with 8 age categories
(0,10] (10,20] (20,30] (30,40] (40,50] (50,60] (60,70] (70,90] (however,
the factor has not been ordered yet). The default in glm is
cont.treatment (for unordered factors) and that leads to compare each
level to
Hello R-help
I am having trouble getting gls to find the R objects that comprise a linear
model when the data=named.object option(option!) is not specified. In the
gls() help it states data is an optional data frame containing the variables
named in model, correlation, weights, and subset. By
princomp uses the raw data and calculates the correlation or covariance matrix
on the way to the PC's, so that doesn't use a correlation matrix itself. You
do, however, get the choice.
However, PC's are the eigenvectors of the correlation (or covariance) matrix,
so in principle calling eigen()
Hi,
On 2/2/2009 3:10:07 AM, you were invited to join srinivasa
raghavan's UNYK address book so he/she would always have access to your contact
info and you to his/hers.
To accept his/her request, Click here.
sqrt(svd(x)$d) maybe 2 more operations than princomp(covmat=x), but it
is hardly a chore.
On Feb 16, 9:15 pm, Mark Difford mark_diff...@yahoo.co.uk wrote:
Hi Glen, Andrew,
The PCA is just a singular value decomposition on a sample covariance/...
I believe that Bjørn-Helge Mevik's point was
On Mon, 2009-02-16 at 10:45 +, S Ellison wrote:
princomp uses the raw data and calculates the correlation or
covariance matrix on the way to the PC's, so that doesn't use a
correlation matrix itself. You do, however, get the choice.
That *isn't* what princomp() does. If you supply a valid
Try this (and see R News 4/1 for more).
library(chron)
tt - times(0:47/48)
tt
[1] 00:00:00 00:30:00 01:00:00 01:30:00 02:00:00 02:30:00 03:00:00
03:30:00 04:00:00 04:30:00 05:00:00 05:30:00 06:00:00 06:30:00
07:00:00 07:30:00
[17] 08:00:00 08:30:00 09:00:00 09:30:00 10:00:00 10:30:00 11:00:00
Thank you very much for the precise response.
Regards,
Suresh
Suresh_FSFM wrote:
Dear R-Experts,
seek your help.
There are two parts I want to deal with.
1)
I want to create a time interval of say, 30 minutes starting from
00:00:00 hrs
Thus at the end, I want to create
Hi,
Please could somebody has any information about the following package:
IlluminaGUI, published here:
http://bioinformatics.oxfordjournals.org/cgi/content/abstract/btm101v1
The link given in the article is dead and authors doesn't reply !
Is there someone who uses it ?
Thank you very much
Dear Tanveer and Johannes,
it *is* indeed possible to estimate dynamic panels by GMM with plm. As
Johannes observes, ?pgmm is a good start. Please see also the package
vignette or its close cousin, this paper on JSS
http://www.jstatsoft.org/v27/i02, section 5.4.
Johannes, if you had problems
You're looking in the wrong repository !
It's on www.bioconductor.org
Wolfgang
nabler a écrit :
Hi,
Please could somebody has any information about the following package:
IlluminaGUI, published here:
http://bioinformatics.oxfordjournals.org/cgi/content/abstract/btm101v1
The link given in
Hi, I am trying to create a vector of length 10 (say), wherein each element
will be average of random sample of size 100, from a distribution, say
Normal. Can anyone please tell me without creating a for loop, how I can
do that?
Regards,
--
View this message in context:
Try this:
replicate(10, mean(rnorm(100)))
On Mon, Feb 16, 2009 at 8:59 AM, megh megh700...@yahoo.com wrote:
Hi, I am trying to create a vector of length 10 (say), wherein each element
will be average of random sample of size 100, from a distribution, say
Normal. Can anyone please tell me
There are three columns, I was just careless.
I succeed to merge the two tables. But:
1) in the first table there were rows, which were not present in the second
table, these rows were deleted from the merged table too, but I need them.
2) If i want to import from the second table just a certain
Hi All,
I am trying to assemble a system that will allow me to work with large
datasets (45-50 million rows, 300-400 columns) possibly amounting to
10GB + in size.
I am aware that R 64 bit implementations on Linux boxes are suitable
for such an exercise but I am looking for configurations that R
megh wrote:
Hi, I am trying to create a vector of length 10 (say), wherein each element
will be average of random sample of size 100, from a distribution, say
Normal. Can anyone please tell me without creating a for loop, how I can
do that?
Homework? Then please ask you course material or
Vincze Orsolya wrote:
There are three columns, I was just careless.
I succeed to merge the two tables. But:
1) in the first table there were rows, which were not present in the second
table, these rows were deleted from the merged table too, but I need them.
2) If i want to import from the
Many apologies for the poor steer; you are quite right.
'fraid I hit 'send' before double-checking the help page myself. Next time...
S
Gavin Simpson gavin.simp...@ucl.ac.uk 16/02/2009 10:59
On Mon, 2009-02-16 at 10:45 +, S Ellison wrote:
princomp uses the raw data and calculates the
There are a series of coercion functions, all beginning with as.
which are designed to return (when possible) objects of specified
types. Since you offer three examples, only one of which I recognize
as a valid R type, I am wondering what R text you have been using?
--
David Winsemius
On
It would be useful to have some sample input data and then what you
expect as output. You can always convert the characters to factors
and then use the integer values assigned.
On Mon, Feb 16, 2009 at 2:00 AM, Arup arup.pramani...@gmail.com wrote:
I have a data frame in R where all the
?zoo
On Mon, Feb 16, 2009 at 1:16 AM, miya ontiveros_pal...@yahoo.com wrote:
Hello everyone.
I am trying to plot data from a time series in R and have run across some
problems. I was wondering if someone could help.
I have data taken every 15 minutes of a list of rankings of 25 titles. I
No, it is not homework. I obviously could do that using a for-loop, and that
I already did. However I thought whether there could be a better approach as
it was looking very messy and unprofessional.
Uwe Ligges-3 wrote:
megh wrote:
Hi, I am trying to create a vector of length 10 (say),
I would also add:
1. Chapter 12 in An Introduction to R
2. Chapter 3 in Paul's R Graphics book:
http://www.stat.auckland.ac.nz/~paul/RGraphics/rgraphics.html
Note that the figures and code used for the graphics in the above
chapter are available here:
Given a cox model:
library(Hmisc); library(survival); (library(Design);
cox.model=cph(Surv(futime, fustat) ~ age, data=ovarian, surv=T)
str(cox.model)
What I need is the total estimated time until failure (death), not the
probability of failing at a given time (survival
Hi
x - seq(0,1,.01)
y - ifelse(abs(x-.5)=0.3,0,
+ ifelse(abs(w-.5)=0.4,-1,
+ifelse((0.1w w0.2),10*x-2,-10*x+8)))
Error in storage.mode(test) - logical : object w not found
what is w?
Why did you use ?
and indicate logical AND and | and || indicate logical
What happens if you type
runif(1)
before loading `mgcv'?
best,
Simon
On Sunday 15 February 2009 17:06, Veerappa Chetty wrote:
Hi ,When I try to load the 'mgcv package, often, but not always, get this
error message. What am I doing wrong? I even tried reinstalling a few
times.
Hello all,
suppose I have a time-stamp: 16-02-2009 00:20:00
and other array that stores lot of time values.
My tolerance limit = +5 minutes
I would like to find values from this array matching with the value:
16-02-2009 00:20:00 + tolerance
Before I write some function, I would like to know:
Given a cox model:
library(Hmisc); library(survival); (library(Design);
cox.model=cph(Surv(futime, fustat) ~ age, data=ovarian, surv=T)
str(cox.model)
What I need is the total estimated time until failure (death), not the
probability of failing at a given time (survival probability), or
Eleni Rapsomaniki wrote:
Given a cox model:
library(Hmisc); library(survival); (library(Design);
cox.model=cph(Surv(futime, fustat) ~ age, data=ovarian, surv=T)
str(cox.model)
What I need is the total estimated time until failure (death), not the probability of failing at a given time
megh wrote:
No, it is not homework. I obviously
For some value of obvious as you has not given a single line of code
as the posting guide suggests.
You probably want:
replicate(10, mean(rnorm(100)))
Uwe Ligges
could do that using a for-loop, and that
I already did. However I thought
ahh, didn't see Gabor's solution, that works much better :-)
On 16 Feb, 10:00, Suresh_FSFM suresh.ghals...@gmail.com wrote:
Dear R-Experts,
seek your help.
There are two parts I want to deal with.
1)
I want to create a time interval of say, 30 minutes starting from 00:00:00
hrs
Thus at
G'day Selwyn,
On Mon, 16 Feb 2009 10:11:20 +
Selwyn McCracken selwyn.mccrac...@gmail.com wrote:
I am trying to follow an example that estimates a 2x2 markov
transition matrix across several periods from aggregate data using
restricted least squares.
I seem to be making headway using
Dear r-helpers,
I want to show that time is flowing CCW in the following:
require(circular)
len - 8
labl - as.character(c(0, 1, 1, 1, 0, 0, 1, 0))
r - circular(2*pi* (rep(c(1, 3, 6), each = 200)/len + rnorm(600, 0,
0.025)))
r.dens - density(r, bw = 25, adjust = 4, kernel = 'vonmises')
plot(r,
?all.equal has a tolerance argument.
On Mon, Feb 16, 2009 at 8:41 AM, Suresh_FSFM suresh.ghals...@gmail.com wrote:
Hello all,
suppose I have a time-stamp: 16-02-2009 00:20:00
and other array that stores lot of time values.
My tolerance limit = +5 minutes
I would like to find values from
Hi everybody,
I'm trying to fit a Garch(1,1) process to the DAX returns. My data
consists of about 2300 10day-logreturns in chronologically descending
order (see attachment). But if I use the garch function I get a very
high alpha_1 and a quite low beta, which doesn't make that much sense. I
Hi,
I can't apply the summary function when I use de armaFit (fArma).
Can you help?
fit-armaFit(~arma(1,0),data=age55)
fit
Title:
ARIMA Modelling
Call:
armaFit(formula = ~arma(1, 0), data = age55)
Model:
ARIMA(1,0,0) with method: CSS-ML
Coefficient(s):
ar1 intercept
Or safer:
df - as.integer(round(...))
On Mon, Feb 16, 2009 at 10:54 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this:
On Mon, Feb 16, 2009 at 9:02 AM, Michael Friendly frien...@yorku.ca wrote:
For glm() models, I often find both the print() and summary() method
disappointing
Uwe Ligges wrote:
Vincze Orsolya wrote:
There are three columns, I was just careless.
I succeed to merge the two tables. But:
1) in the first table there were rows, which were not present in the
second
table, these rows were deleted from the merged table too, but I need
them.
2) If i
Before using lca,
try
data-as.matrix(data).
For example when I tried using lca lca(LSAT,2,niter=100)
for known LSAT data (library ltm), I took error messages.
But, when I use
data-as.matrix(LSAT)
lca(data,2,niter=100)
I took results for lca
King Regards,
Evgenia
Tryntsje Wesselius wrote:
Gabor Grothendieck ggrothendieck at gmail.com writes:
Or safer:
df - as.integer(round(...))
Did you try? I believe it is a problem of printCoefmat that has quite
a few options for special column, but none for df. Ask Martin Mächler.
Dieter
__
Thank you for the positve response.
Gabor Grothendieck wrote:
?all.equal has a tolerance argument.
On Mon, Feb 16, 2009 at 8:41 AM, Suresh_FSFM suresh.ghals...@gmail.com
wrote:
Hello all,
suppose I have a time-stamp: 16-02-2009 00:20:00
and other array that stores lot of time
On Mon, Feb 16, 2009 at 11:08 AM, Dieter Menne
dieter.me...@menne-biomed.de wrote:
Gabor Grothendieck ggrothendieck at gmail.com writes:
Or safer:
df - as.integer(round(...))
Did you try? I believe it is a problem of printCoefmat that has quite
a few options for special column, but none
I have data in a format like this:
namessexsex viewnum rating rt
ahl4f m f 56 -1082246
ahl4f m f 74 85 1444
ahl4f m f 52 151 1595
ahl4f m f 85 1 1447
Hi,
I am having difficulty entering a 'programmable' argument into the multinom
function from the nnet package. Interactively, I can get the function to work
fine by calling it this way:
z1=multinom(formula = class.ind(grp[-outgroup])~ (PC1 + PC2 + PC3),
data=data.frame(scores))
However I
Gabor Grothendieck ggrothendieck at gmail.com writes:
On Mon, Feb 16, 2009 at 11:08 AM, Dieter Menne
Yes, with as.integer(round(...)) It looks like this:
modelFit.glm(berk.mod2)
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
modelFit.glm(berk.mod2)
Analysis of
Peter Dalgaard wrote:
Uwe Ligges wrote:
Vincze Orsolya wrote:
There are three columns, I was just careless.
I succeed to merge the two tables. But:
1) in the first table there were rows, which were not present in the
second
table, these rows were deleted from the merged table too, but I
If the goal is to look professional, then
'replicate' probably suits. If the goal is to
compute as fast as possible, then that isn't
the case because 'replicate' is really a 'for'
loop in disguise and there are other ways.
Here's one other way:
function (size, replicates, distfun, ...)
{
On Mon, Feb 16, 2009 at 12:59 PM, megh megh700...@yahoo.com wrote:
Hi, I am trying to create a vector of length 10 (say), wherein each element
will be average of random sample of size 100, from a distribution, say
Normal. Can anyone please tell me without creating a for loop, how I can
do
On Mon, Feb 16, 2009 at 10:21 AM, William Simpson
william.a.simp...@gmail.com wrote:
I have data in a format like this:
namessexsex viewnum rating rt
ahl4f m f 56 -1082246
ahl4f m f 74 85 1444
ahl4f
Hello, everyone!
The code below allows me to produce the graph I want (I know - the
colors are strange, but it's just for the sake of an example).
After you run the plot- part and then do print(plot) - that's what I want.
However, when I run the bits of code below (with graphics devices) -
what
Hi R users,
I am doing cross correlation analysis on 2 time series (call them y-series
and x-series) where I need the use the model developed on the x-series to
prewhiten the yseries.. Can someone point me to a function/filter in R that
would allow me to do that?
Thanks in advance for any
Hi to all,
In other statistical software, such as Eviews, it is possible to
regress a model with the Least Squares method, assuming that the
residuals follow an AR(q) process.
For example the resulting regression is something like
y = 1.2154 + 0.2215 x + 0.251 AR(1)
How is it possible to do
Hi,
I am looking for two ways to speed up my computations:
1. Is there a function that efficiently computes the 'sandwich product' of
three matrices, say, ZPZ'
2. Is there a function that efficiently computes the determinant of a
positive definite symmetric matrix?
Thanks,
S.A.
Danger: More careful thought required.
Outliers (Title of a TECHNOMETRICS paper of a couple of decades ago)
are an artificial construct: there is NO SUCH THING in the abstract. They
exist only wrt to a model. So there is no such thing as software that tells
whether the changes are considered
Hi list! I have a large matrix which I'd like to partition into blocks
and for each block I'd like to compute the mean. Following a example
where each letter marks a block of the partition:
a a a d g g
a a a d g g
a a a d g g
b b b e h h
b b b e h h
c c c f i i
Dear Peter,Thanks for the answer, it's working very well. And again sorry,
about asking as a banal question, but I really didn't have who to ask to
help me.
Uwe: I don't have a teacher, I never learned R before. I'm just trying to
learn self-educating, because it's very helpful in data
Forget eval(parse(text = ))
See
?as.formula
?update.formula
and try out the example() s there.
HTH,
Chuck
On Mon, 16 Feb 2009, Crouch, Daniel wrote:
Hi,
I am having difficulty entering a 'programmable' argument into the multinom
function from the nnet package.
if you want to whiten the series why not just
new series - rnorm(num.obs)+series
On Mon, Feb 16, 2009 at 12:25 PM, Pele drdi...@yahoo.com wrote:
Hi R users,
I am doing cross correlation analysis on 2 time series (call them y-series
and x-series) where I need the use the model developed on
?gls
On Feb 16, 2009, at 12:28 PM, constantine wrote:
In other statistical software, such as Eviews, it is possible to
regress a model with the Least Squares method, assuming that the
residuals follow an AR(q) process.
For example the resulting regression is something like
y = 1.2154 +
Hi,
I don't know much about the RGL package, and I have read the
documentation and tried some parameters, with no luck... I would like
to generate a movie from a 3D object (code below), where the vortex A
is closer to the observer, and then the object rotates and the B
vortex gets closer. I would
I had a Murphy's law calendar a while back with many different laws in it. One
of those laws was along the lines of:
An easily understood, simple falsehood is often more useful than a complicated,
often misunderstood truth
(though the original was probably much better phrased than my memory).
do you mean:
f=function(x)
0*(abs(x-.5)=.3)-1*(abs(x-.5)=.4)+(10*x-2)*(x.1x.2)+(-10*x+8)*(x=.2x=.5)
f(x)
curve(f,0,1)
hope it helps.
Patrizio
2009/2/14 kathie kathryn.lord2...@gmail.com:
Dear R users,
From the code below, I try to compute y value. (In fact, y looks like a
trapezoid)
Hello
I tried a few searches on hist, histogram, equidist and space (space=0 was
mentioned in one contribution), but none of that so far worked. It also says
in the help ##-- For non-equidistant breaks, counts should NOT be graphed
unscaled: - which is precisely what I am looking for, but I
On Mon, Feb 16, 2009 at 01:45:52PM -0500, Stavros Macrakis wrote:
How are the blocks defined? As a priori index ranges? By factors? By
some property of i,j? Or...?
Ok, I should have been more specific.
The blocks are defined by factors. There's a factor for the columns and
a factor for the
On Mon, Feb 16, 2009 at 11:08:24AM -0800, Christian Langkamp wrote:
I could of course log the whole data
set, but then explaining that transformation within a presentation is
generally not a pleasant exercise.
You don't have to explain it. Just calculate the hist of the log and
label the axis
You will need
library(nlme)
first.
But not for ?arima, which seems the more obvious way to do this simple
example.
On Mon, 16 Feb 2009, Michael Kubovy wrote:
?gls
On Feb 16, 2009, at 12:28 PM, constantine wrote:
In other statistical software, such as Eviews, it is possible to
regress a
Patrick Burns wrote:
If the goal is to look professional, then
'replicate' probably suits. If the goal is to
compute as fast as possible, then that isn't
the case because 'replicate' is really a 'for'
loop in disguise and there are other ways.
Here's one other way:
function (size,
On Mon, 16 Feb 2009, Titus von der Malsburg wrote:
On Mon, Feb 16, 2009 at 01:45:52PM -0500, Stavros Macrakis wrote:
How are the blocks defined? As a priori index ranges? By factors? By
some property of i,j? Or...?
Ok, I should have been more specific.
The blocks are defined by factors.
David and Danel,
Indeed, the missing symbol is *. Now it works, although, with a change of
font in the brackets of the axis=label.
Thanks,
david
To: daviddou...@hotmail.com
Subject: Re: [R] superscript
From: davidcr...@charter.net
Date: Sat, 14 Feb 2009 17:49:11 -0600
Could it
Hi list,
I'm trying to install/compile rimage on ubuntu linux (i386) interpid.
However, the compilation hangs on:
gcc -std=gnu99 -I/usr/share/R/include -g -O2 -fpic -g -O2 -c
laplacian.c -o laplacian.o
laplacian.c: In function ‘laplacian’:
laplacian.c:14: warning: implicit declaration of
Assuming your matrix is:
mm - matrix(runif(6*6),6,6)
And your blocks are defined by integers or factors:
cfact - c(1,1,1,2,3,3)
rfact - c(1,1,1,2,2,3)
Then the following should do the trick:
matrix(tapply(mm, outer(rfact,cfact,paste), mean),
length(unique(rfact)))
Its not clear that the object returned from such an operation would be
a matrix, but if things remain very regular then perhapos you will
succeed with this:
markmtx - matrix(scan(textConnection(a a a d g g
+ a a a d g g
+ a a a d g g
+ b b b e h h
+ b b b e h h
+ c c c f
Hi Harsh,
The useR! 2008 site has useful information. E.g. talks by
Graham Williams:
http://www.statistik.uni-dortmund.de/useR-2008/slides/Williams.pdf
Dirk Eddelbuettel
http://www.statistik.uni-dortmund.de/useR-2008/tutorials/useR2008introhighperfR.pdf
and others
Wacek Kusnierczyk wrote:
Patrick Burns wrote:
If the goal is to look professional, then
'replicate' probably suits. If the goal is to
compute as fast as possible, then that isn't
the case because 'replicate' is really a 'for'
loop in disguise and there are other ways.
Here's one other way:
On 16 February 2009 at 21:09, Koen Hufkens wrote:
| Hi list,
|
| I'm trying to install/compile rimage on ubuntu linux (i386) interpid.
| However, the compilation hangs on:
|
| gcc -std=gnu99 -I/usr/share/R/include -g -O2 -fpic -g -O2 -c
| laplacian.c -o laplacian.o
| laplacian.c: In
You might want to look at the dev.copy function. I just tried with your
example and the postscript device, used the print(plot) command with the
postscript device open, then switched to the windows graphics device and did
dev.copy then closed the postscript file, the result had 2 pages, the
Dear R- Experts,
Seek your help.
I created a time sequence using:
x[i] -chron(dates, tt, format=c(dates=y-m-d, tt=h:m:s))
first element in the list is displayed as: (09-01-01 00:00:00)
Now, I want to store this value as date.
If I use: format.Date(x[1],%y-%m-%d %H:%M:%S), I expect following
One approach is to create your own contrasts matrix:
mycmat - diag(8)
mycmat[ row(mycmat) == col(mycmat) + 1 ] - -1
mycmati - solve(mycmat)
contrasts(agefactor) - mycmati[,-1]
Now when you use agefactor, the intercept will be the first age group and the
slopes will be the differences
Stavros Macrakis macra...@alum.mit.edu writes:
Assuming your matrix is:
mm - matrix(runif(6*6),6,6)
And your blocks are defined by integers or factors:
cfact - c(1,1,1,2,3,3)
rfact - c(1,1,1,2,2,3)
Then the following should do the trick:
matrix(tapply(mm,
I would love to fix it however my C++ skills are limited.
Hope someone fixes this before it disappears.
Koen
-Original Message-
From: Dirk Eddelbuettel [mailto:e...@debian.org]
Sent: Mon 16-2-2009 21:49
To: Hufkens Koen
Cc: r-help@r-project.org
Subject: Re: [R] rimage
On 16 February
Dear all,
I am trying to estimate the prediction from a fixed effects model and their
confidence intervals as well. Though I do not want to include in the
prediction and at the confidence intervals the intercept. For that reason I
used the argument incl.non.slopes=FALSE. But either if it is TRUE
On 17/02/2009, at 10:54 AM, dimitris kapetanakis wrote:
Dear all,
I am trying to estimate the prediction from a fixed effects model
and their
confidence intervals as well. Though I do not want to include in the
prediction and at the confidence intervals the intercept. For that
reason I
The Date class does not work with times and neither it nor
the chron dates and times classes use time zones so
that cannot be the problem if you are using those classes.
Much of this is discussed in R News 4/1.
On Mon, Feb 16, 2009 at 4:05 PM, Suresh_FSFM suresh.ghals...@gmail.com wrote:
Dear
Thanks, Gabor
No, that wasn't it at all. In print.anova, I found:
if (length(i - grep(Df$, cn)))
zap.i - zap.i[!(zap.i %in% i)]
so it only recognizes Df, not df as a column name prefix to print as
integers.
-Michael
Gabor Grothendieck wrote:
Or safer:
df - as.integer(round(...))
A couple of remarks on vQ's naive benchmark:
f.rep = function(n, m) replicate(n, rnorm(m))
I suppose you meant
f.rep = function(n, m) replicate(n, mean(rnorm(m)))
which doesn't make a substantial speed difference, though.
f.pat = function(n, m) colMeans(array(rnorm(n*m), c(n,
I am having difficulties getting the X-axis labels (dates) to be as
needed when plotting from chron
The help syntax from chron lists this example:
x - chron(dates = c(02/27/92, 02/27/92, 01/14/92, 02/28/92),
I have activity plots by time on the y-axis and the dates on the
x-axis. What I
On Mon, Feb 16, 2009 at 4:23 PM, Martin Morgan mtmor...@fhcrc.org wrote:
Stavros Macrakis macra...@alum.mit.edu writes:
matrix(tapply(mm, outer(rfact,cfact,paste), mean),
length(unique(rfact)))
or the variant
idx - outer(rfact, (cfact - 1) * max(rfact), +)
...
I suppose the clean way to do this would be to define a cartesian product of
two factors with the induced lexicographic order (is there a standard
function for doing this?):
Of course. ?interaction.
-- Bert Gunter
Genentch
__
On Feb 16, 2009, at 6:11 PM, Neotropical bat risk assessments wrote:
I am having difficulties getting the X-axis labels (dates) to be as
needed when plotting from chron
The help syntax from chron lists this example:
x - chron(dates = c(02/27/92, 02/27/92, 01/14/92, 02/28/92),
On my
I am trying to predict time series. I have several samples which
consist of measurement of 4 different variables over different
time periods, and other measurements over subsequent time
periods which I believe is predicted by the 4 variables.
I am thinking of using cluster analysis, to confirm
Hi,
I need to append my multiple plots in pdf files.
my problem is that I would want to run the R script a number times(closing
and opening) and still want to append. If i keep the dev.off() it wouldnt
let me see my plots while R is open.
any idea!!
Jorge Ivan Velez wrote:
Hi Ramya,
I am having difficulties getting the X-axis labels (dates) to be as
needed when plotting from chron
The help syntax from chron lists this example:
x - chron(dates = c(02/27/92, 02/27/92, 01/14/92, 02/28/92),
I have activity plots by time on the y-axis and the dates on the
x-axis. What I
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