Isn't the binomial-log used to obtain risk ratios from the coefficients
rather than odds-ratios?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Thomas Lumley
Sent: Wednesday, September 10, 2008 4:36 PM
To: Prof Brian Ripley
Cc: r-help@r-project.org; Ben
on 09/10/2008 04:49 PM Rolf Turner wrote:
On 11/09/2008, at 9:29 AM, Marc Schwartz wrote:
on 09/10/2008 04:03 PM Rolf Turner wrote:
I tried to search for a string of words ``as one entity'' following the
example in the help file:
RSiteSearch({logistic regression})
and got the error
Hi Timur,
try
cor(log(a+1), log(b+1), method=pearson)
HTH,
miltinho
brazil
On Wed, Sep 10, 2008 at 6:04 PM, Timur Shtatland [EMAIL PROTECTED]wrote:
Hi,
How can I make the cor(x, y, method=spearman) call to produce an
error when the input to it (x, y) produces an error? Here is a simple
Hi,
I'm pretty new to R and I'm trying to use the functions
portfolioMarkowitz and portfolioMonteCarlo. I think that its in the
package fPortfolio but it doesn't seem to work with I install the
package and import the library. I think this may be because It only
existed in an old version of
On Wed, Sep 10, 2008 at 3:09 PM, Marc Schwartz
[EMAIL PROTECTED] wrote:
on 09/10/2008 04:49 PM Rolf Turner wrote:
On 11/09/2008, at 9:29 AM, Marc Schwartz wrote:
on 09/10/2008 04:03 PM Rolf Turner wrote:
I tried to search for a string of words ``as one entity'' following the
example in the
Hello,
I just had a lot of trouble compiling a version of R which uses a
recently-compiled version of ATLAS and LAPACK. ATLAS and LAPACK compiled
correctly and installed fine and passed all of their checks (and yes, I did
compile them all with -fPIC as is required for R). This was my
Hi R People!
I'm going to be putting together a completely online undergrad
business stats course (a second semester course) and was going to use
R and Rcmdr.
My question: has anyone else used R for an online course, please? If
so, did it go well, please?
Thanks in advance,
Sincerely,
Erin
You might try
apply(t(sapply(l,apply,2,sum)),2,sum)
--Adam
On Wed, 10 Sep 2008, Muhammad Azam wrote:
Dear R community
I have stored the results of arrays in a list consist of J-components (say
200 components). Each component containing same no of columns but may be
different no of rows.
Maybe, for vector v and list l
match(v,names(l))
?
--Adam
On Wed, 10 Sep 2008, Rajasekaramya wrote:
hi,
I have list of length 5453 and vector of length 14318.I need to compare the
vector with the list and return the list name if matched.I am thinking of
using an lapply but how to retrive
Try this:
lines(fitted(lm(x ~ y)), y, lty = 2)
On Wed, Sep 10, 2008 at 4:06 PM, David Epstein
[EMAIL PROTECTED] wrote:
Suppose x and y are numeric vectors of the same length.
plot(x,y) #scatterplot
lmObj1 - lm(y~x) # best fit line
abline(lmObj1) # good
lmObj2 - lm(x~y) #get best fit
On 11/09/2008, at 10:09 AM, Marc Schwartz wrote:
snip
As Prof. Ripley noted, try using debug(RSiteSearch) and then step
through the code to see where $HOME is being added to the URL
variable.
That can help pin it down further. If the URL is not corrupted by the
time browseURL() is
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gabor Grothendieck
Sent: Wednesday, September 10, 2008 3:45 PM
To: David Epstein
Cc: r-help@r-project.org
Subject: Re: [R] re flecting a line
Try this:
lines(fitted(lm(x ~ y)), y, lty = 2)
Hi Erika,
As mentioned, I haven't run the model before and I don't have access
to your data set, so you might want to post your reply to the list as well
(cc'd again).
As another guessing-without-trying-anything, I'd first make sure
that in fact pop and family are factor
Your code shows no errors only correct responses.
If you want an error when R is generating correct results, you will need to
check for the conditions yourself.
-Original Message-
From: Timur Shtatland [EMAIL PROTECTED]
To: r-help@r-project.org r-help@r-project.org
Sent: 9/10/08 3:51 PM
On Mon, Sep 8, 2008 at 1:06 PM, Erika Crispo
[EMAIL PROTECTED] wrote:
Hello,
I need to perform a mixed-model (with nesting) MANCOVA, using Type III sums
of squares. I know how to perform each of these types of tests individually,
but I am not sure if performing a mixed-model MANCOVA is
HI Steve,
I've checked the random forest package. The VariableImportance does exactly
what I need.
Thank you so much!
Gabriela
-Original Message-
From: Steven McKinney [mailto:[EMAIL PROTECTED]
Sent: Tuesday, September 09, 2008 8:48 PM
To: Gabriela Bucini; r-help@r-project.org
Thanks, Jim:
One of the purpose of this function is to try not to stop the program and
generate T/F result for further process. Actually find fix by myself with
tryCatch (but don't know why try doesn't work). The modified function handles
potential Date and yearmon in Zoo classes.
is.Date=
#
Hi my R buddies,
I'm trying to solve a specific group of convex optimization in R. The
admissible region is the inside and surface of a multi-dimensional
eclipse area and the goal function is the sum of absolution values of
the variables. Could any one please tell me whether there's a package
in
It does not stop the program. It generates an error message which
is caught by 'try' from which you can do a recovery. The purpose of
'stop' is to cause the 'try' to catch it and return the error message
as the result and you can test the class for being 'try-error'.
On Wed, Sep 10, 2008 at
Try this modification:
as.yearmon(paste(res[,1], res[,2], sep=-),format=%Y-%m)Andy
--- On Wed, 9/10/08, Megh Dal [EMAIL PROTECTED] wrote:
From: Megh Dal [EMAIL PROTECTED]
Subject: [R] Woring message in as.yearmon()
To: [EMAIL PROTECTED]
Date: Wednesday, September 10, 2008, 2:15 PM
I have
Dear Erika et al.,
I'm not sure exactly what you want to do, but you might take a look at the
Anova() function in the car package, which will compute type-II or
type-III tests for multivariate linear models, including models with
repeated measures; you can also get the traditional univariate
There is a series of data contains time in fixed step and energy
varying with time, how to test its periodicity?In R, it seems there is
no direct tools since I have search the R manual with periodic and I
have not found any related topic.
Thanks a lot
Well, I can speak from the receiving end. I'm enrolled in an online distance
masters degree in statistics at Texas AM University. I'm in my fourth course.
We are pretty much free to use any statistical software we want. The first
semester (distributions, goodness of fit, one- and two-
?spectrum
?acf
?ccf
library(wmtsa)
?wavCWT
library(sowas)
?wsp
you could also look at lagged plots to look for periodicity.
if you elaborate on the problem and include executable sample code you
will probably recieve more help.
On Wed, Sep 10, 2008 at 10:02 PM, yk [EMAIL PROTECTED] wrote:
There
I have taught myself R with undue suffering by this list over the last
year, and I see no reason why distance education wouldn't work. The
learning curve will be a little shallower than mine with a guide. I
would suggest using the Rcmdr to jump start everything because you can
see the code that
Hello,
The likelihood includes two parameters to be estimated: lambda
(=beta0+beta1*x) and alpha. The algorithm for the estimation is as
following:
1) with alpha=0, estimate lambda (estimate beta0 and beta1 via GLM)
2) with lambda, estimate alpha via ML estimation
3) with updataed alpha,
Your model is close, but not correct... there are no t's on the parameters
and the U's aren't lagged.
You can find an ARMAX example on our quick fix page:
http://www.stat.pitt.edu/stoffer/tsa2/R_time_series_quick_fix.htm . The
example is near the bottom and just above the spectral analysis
I have two variables (x,y) :
x : it takes all integer values from 0 to y and,
y : takes all values from 0, 10
I am looking for some R code to find all possible pairs of (x,y). Can anyone
please help me?
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The data I mentioned above is oscilating vs time，but there are not
obersevable fixed cycle if I just plot this data.
How to get the average cycle，or the most probable range of cycle with
statistical methods?
I don't know how to achieve it by R, is there any command?
On Sep 11, 10:52 am, stephen
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