Displaying 20 results from an estimated 300 matches similar to: "Question about lmRob"
2007 Sep 04
1
Robust linear models and unequal variance
Hi all,
I have probably a basic question, but I can't seem to find the answer in
the literature or in the R-archives.
I would like to do a robust ANCOVA (using either rlm or lmRob of the
MASS and robust packages) - my response variable deviates slightly from
normal and I have some "outliers". The data consist of 2 factor
variables and 3-5 covariates (fdepending on the model).
2007 Nov 27
1
Difference between AIC in GLM and GLS - not an R question
Hi,
I have fitted a model using a glm() approach and using a gls() approach
(but without correcting for spatially autocorrelated errors). I have
noticed that although these models are the same (as they should be), the
AIC value differs between glm() and gls(). Can anyone tell me why they
differ?
Thanks,
Geertje
~~~~
Geertje van der Heijden
PhD student
Tropical Ecology
School of Geography
2007 Oct 26
1
Help needed on calculation of Moran's I
Hi,
I am trying to calculate Moran's I test for the residuals for a
regression equation, but I have trouble converting my coordinates into
nb format.
I have used the dnearneigh() funtion now with an arbitrarily high upper
distance to make it include all plots. However, when I do the
lm.morantest() I get a Moran's I value which is the same as the expected
value and a P-value of 1.
I
2007 Nov 20
1
Problem with code for bootstrapping chi square test with count data
Hi,
I'd like some advice on bootstrapping in R.
I have a species x with 20 individuals and a factor containing 0 and 1's
(in this case 5 zeros and 15 ones). I want to compare the frequency of
the occurrence of 1 with a probability value. This code seems to work to
do this in R.
attach(test)
p <- c(0.5272, (1-0.5272))
sp1_1 <- length(subset(x, x==1))
sp1_0 <- length(subset(x,
2007 Oct 22
3
Spatial autocorrelation
Hi,
I have collected data on trees from 5 forest plots located within the
same landscape. Data within the plots are spatially autocorrelated
(calculated using Moran's I). I would like to do a ANCOVA type of
analysis combining these five plots, but the assumption that there is no
autocorrelation in the residuals is obviously violated. Does anyone have
any ideas how to incorporate these spatial
2009 Mar 12
1
zooreg and lmrob problem (bug?)
Hi all and thanks for your time in advance,
I can't figure out why summary.lmrob complains when lmrob is used on a
zooreg object. If the zooreg object is converted to vector before
calling lmrob, no problems appear.
Let me clarify this with an example:
>library(robustbase)
>library(zoo)
>dad<-c(801.4625,527.2062,545.2250,608.2313,633.8875,575.9500,797.0500,706.4188,
2011 Jul 28
1
Problem with anova.lmRob() "robust" package
Dear R users,
I'd like to known your opinion about a problem with anova.lmRob() of "Robust" package that occurs when I run a lmRob() regression on my dataset.
I check my univariate model by single object anova as anova(lmRob(y~x)).
If I compare my model with the null model (y~1), I must obtain the same results,
but not for my data.
Is it possible?
My example:
2018 Mar 03
2
lmrob gives NA coefficients
Dear list members,
I want to perform an MM-regression. This seems an easy task using the
function lmrob(), however, this function provides me with NA coefficients.
My data generating process is as follows:
rho <- 0.15 # low interdependency
Sigma <- matrix(rho, d, d); diag(Sigma) <- 1
x.clean <- mvrnorm(n, rep(0,d), Sigma)
beta <- c(1.0, 2.0, 3.0, 4.0)
error <- rnorm(n = n,
2010 Dec 13
1
Wrong contrast matrix for nested factors in lm(), rlm(), and lmRob()
This message also reports wrong estimates produced by lmRob.fit.compute()
for nested factors when using the correct contrast matrix.
And in these respects, I have found that S-Plus behaves the same way as R.
Using the three available contrast types (sum, treatment, helmert)
with lm() or lm.fit(), but just contr.sum with rlm() and lmRob(),
and small examples, I generated contrast matrices for
2018 Mar 03
0
lmrob gives NA coefficients
> On Mar 3, 2018, at 3:04 PM, Christien Kerbert <christienkerbert at gmail.com> wrote:
>
> Dear list members,
>
> I want to perform an MM-regression. This seems an easy task using the
> function lmrob(), however, this function provides me with NA coefficients.
> My data generating process is as follows:
>
> rho <- 0.15 # low interdependency
> Sigma <-
2018 Mar 04
2
lmrob gives NA coefficients
Thanks for your reply.
I use mvrnorm from the *MASS* package and lmrob from the *robustbase*
package.
To further explain my data generating process, the idea is as follows. The
explanatory variables are generated my a multivariate normal distribution
where the covariance matrix of the variables is defined by Sigma in my
code, with ones on the diagonal and rho = 0.15 on the non-diagonal. Then y
2018 Mar 04
0
lmrob gives NA coefficients
What is 'd'? What is 'n'?
On Sun, Mar 4, 2018 at 12:14 PM, Christien Kerbert <
christienkerbert at gmail.com> wrote:
> Thanks for your reply.
>
> I use mvrnorm from the *MASS* package and lmrob from the *robustbase*
> package.
>
> To further explain my data generating process, the idea is as follows. The
> explanatory variables are generated my a
2009 Apr 08
1
predict "interval" for lmRob?
lm's "predict" function offers an "interval" parameter to choose between 'confidence' and 'prediction' bands. In the package "robust" and for "lmRob", there is also a "predict" but it lacks such a parameter, and the documented "type" parameter has only "response" offerred. Is there some way of obtaining
2018 Mar 04
1
lmrob gives NA coefficients
d is the number of observed variables (d = 3 in this example). n is the
number of observations.
2018-03-04 11:30 GMT+01:00 Eric Berger <ericjberger at gmail.com>:
> What is 'd'? What is 'n'?
>
>
> On Sun, Mar 4, 2018 at 12:14 PM, Christien Kerbert <
> christienkerbert at gmail.com> wrote:
>
>> Thanks for your reply.
>>
>> I use
2011 Mar 16
0
cross validation? when rlm, lmrob or lmRob
Dear community,
I have fitted a model using comands above, (rlm, lmrob or lmRob). I don't
have new data to validate de models obtained. I was wondering if exists
something similar to CVlm in robust regression. In case there isn't, any
suggestion for validation would be appreciated.
Thanks, user at host.com
--
View this message in context:
2008 Jan 11
0
Behaviour of standard error estimates in lmRob and the like
I am looking at MM-estimates for some interlab comparison work. The
usual situation in this particular context is a modest number of results
from very expensive methods with abnormally well-characterised
performance, so for once we have good "variance" estimates (which can
differ substantially for good reason) from most labs. But there remains
room for human error or unexpected chemistry
2011 Jul 28
0
R: Re: Problem with anova.lmRob() "robust" package
I'm sorry, maybe the question was bad posed.
Ista has well described my problem.
Thanks
Massimo
>----Messaggio originale----
>Da: izahn at psych.rochester.edu
>Data: 28/07/2011 17.52
>A: "David Winsemius"<dwinsemius at comcast.net>
>Cc: "m.fenati at libero.it"<m.fenati at libero.it>, <r-help at r-project.org>
>Ogg: Re: [R]
2013 Apr 03
0
Help with lmRob function
Hi,
I am fairly new to R and have encountered an issue with the lmRob function that I have been unable to resolve. I am trying to run a robust regression using the lmRob function which runs successfully, but the results are rather strange. I'm not sure it's important, but my model has 3 dichotomous categorical variables and 2 continuous variables in it. When I look at a summary of my
2008 May 14
1
rlm and lmrob error messages
Hello all,
I'm using R2.7.0 (on Windows 2000) and I'm trying do run a robust
regression on following model structure:
model = "Y ~ x1*x2 / (x3 + x4 + x5 +x6)"
where x1 and x2 are both factors (either 1 or 0) and x3.....x6 are numeric.
The error code I get when running rlm(as.formula(model), data=daymean) is:
error in rlm.default(x, y, weights, method = method, wt.method =
2018 Mar 04
0
lmrob gives NA coefficients
Hard to help you if you don't provide a reproducible example.
On Sun, Mar 4, 2018 at 1:05 PM, Christien Kerbert <
christienkerbert at gmail.com> wrote:
> d is the number of observed variables (d = 3 in this example). n is the
> number of observations.
>
> 2018-03-04 11:30 GMT+01:00 Eric Berger <ericjberger at gmail.com>:
>
>> What is 'd'? What is