similar to: Wrong contrast matrix for nested factors in lm(), rlm(), and lmRob()

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 =

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:

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

You need to pay attention to the documentation more closely. If you don't know what something means, that is usually a signal that you need to study more... in this case about the difference between an input variable and a design (model) matrix. This is a concept from the standard linear algebra formulation for regression equations. (Note that I have never used RobPer, nor do I regularly

I keep coming back to this problem of singular fits in rlm (MASS library), but cannot figure out a good solution. I am fitting a linear model with a factor variable, like lm( Y ~ factorVar) and this works fine. lm knows to construct the contrast matrix the way I would expect, which puts the first factor as the baseline level. But when I try rlm( Y ~ factorVar) I get the message "'x'

R-experts, I have fitted many different lines. The fast-tau estimator (yellow line) seems strange to me?because this yellow line is not at all in agreement with the other lines (reverse slope, I mean the yellow line has a positive slope and the other ones have negative slope). Is there something wrong in my R code ? Is it because the Y variable is 1 vector and should be a matrix ? Here is the

Dear R-experts, Here below my reproducible R code. I want to add many straight lines to a plot using "abline" The last fit (fast Tau-estimator, color yellow) will not appear on the plot. What is going wrong ? Many thanks for your reply. ########## Y=c(2,4,5,4,3,4,2,3,56,5,4,3,4,5,6,5,4,5,34,21,12,13,12,8,9,7,43,12,19,21)

On 31/03/2018 11:57 AM, varin sacha via R-help wrote: > Dear R-experts, > > Here below my reproducible R code. I want to add many straight lines to a plot using "abline" > The last fit (fast Tau-estimator, color yellow) will not appear on the plot. What is going wrong ? > Many thanks for your reply. > It's not quite reproducible: you forgot the line to create

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

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,

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:

Dear All, When I run rlm to obtain robust standard errors, my output does not include p-values. Is there any reason p-values should not be used in this case? Is there an argument I could use in rlm so that the output does include p-values? Thanks in advance, Celso [[alternative HTML version deleted]]

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,

> 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 <-

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

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

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).

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

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

Dear Group, I am having trouble with using rlm on multivariate data sets. When I call rlm I get Error in lm.wfit(x, y, w, method = "qr") : incompatible dimensions lm on the same data sets seem to work well (see code example). Am I doing something wrong? I have already browsed through the forums and google but could not find any related discussions. I use Windows XP and R