similar to: MASS: rlm, MM and errors in observations AND regressors

Hello, I'm learning about robust M-estimators right now and had settled on the "Huber Proposal 2" as implemented in MASS, but further reading made clear, that at least 2 further weighting functions (Hampel, Tukey bisquare) exist. In a post from B.D. Ripley going back to 1999 I found the following quote: >> 2) Would huber() give me results that are similar (i.e., close

Dear R-devel, There seems to be a discrepancy in the order in which lm and rlm evaluate their arguments. This causes rlm to sometimes produce an error where lm is just fine. Here is a little script that illustrate the issue: > library(MASS) > ## create data > n <- 100 > dat <- data.frame(x=rep(c(-1,0,1), n), y=rnorm(3*n)) > > ## call lm, works fine > summary(lm(y ~

Hi, I would appreciate of someone could explain how the residual standard error is computed for rlm models (MASS package). Usually, one would expect to get the residual standard error by > sqrt(sum((y-fitted(fm))^2)/(n-2)) where y is the response, fm a linear model with an intercept and slope for x and n the number of observations. This does not seem to work for rlm models and I am wondering

To make a long story short I was doing some in-sample testing in which some dynamically created regressors would end up either all true or all false based on the validation portion. In my case a new mainframe configuration (this is a crappy way to handle a level shift but I do what I can.) So here is the code snippet that finally let me pre-check my regressors and drop any of them that were all

The model matrix for the code at the end the email is shown below. Since the model matrix doesn't have -1, I think that it is made of dummy regressors rather than deviation regressors. I'm wondering how to make a model matrix using deviation regressors. Could somebody let me know? > model.matrix(aaov) (Intercept) A2 B2 B3 A2:B2 A2:B3 1 1 0 0 0 0 0 2

Hello everyone, Hope you all are doing great! I have been fitting arima models and performing forecasts pretty straightforwardly in R. However, I wanted to add a couple of regressors to the arima model to see if it could improve the accuracy of the forecasts but have had a hard time trying to do so. I used the following R function: arima(x, order = c(0, 0, 0), seasonal = list(order = c(0, 0,

Hi everybody, I have built a model that includes subject ID as a random effect, and has a continous variable (time) and I want to test whether the slope of this line differs between treatments (this is tested with the interaction between treatment and "time"). My question now is that I also want to include regressors that might explain variation in this slope between subjects (and of

Hi all, I've experienced this problem using step() and stepAIC() when a name of a regressors has blanks in between (R:R1.7.0, os: w2ksp4). Please look at the following code: "x" <- c(14.122739306734, 14.4831100207131, 14.5556459667089, 14.5777151911177, 14.5285815352327, 14.0217803203846, 14.0732571632964, 14.7801310180502, 14.7839362960477, 14.7862217992577)

Hi, Using arima on data that are in a data frame, especially when adding xreg, would be much easier if the arima function contained 1) a "data=" argument 2) the possibility to include the covariate(s) in a formula style. Ideally the call could be something like > arima(symptome, order=c(1,0,0), xreg=~trait01*mesure0, data=anxiete) ( or arima(symptome~trait01*mesure0,

Hi, All. I have a logistic regression model that I have run. The question came up: which of these regressors is more important than another? (I'm using Design) Logistic Regression Model lrm(formula = iconicgesture ~ ST + SSP + magnitude + Condition + Expertise, data = d) Coef S.E. Wald Z P Intercept -3.2688 0.2854 -11.45 0.0000 ST 2.0871 0.2730 7.64

dear r-experts: there is probably a very easy way to do it, but it eludes me right now. I have a large data frame with, say, 26 columns named "a" through "z". I would like to define "sets of regressors" from this data frame. something like myregressors=c("b", "j", "x") lm( l ~ myregressors, data=... ) is the best way to create new

Does anyone know if there is a way to specify regressors dynamically rather than explicitly? More specifically, I have a data set in "long format" that details a number of individuals and their responses to a question (which can be positive, negative, or no answer). Each individual answers as many questions as they want, so there are a different number of rows per individual. For each

##### BEGIN CODE ###### #!/usr/bin/perl ###### # # --start, -s = The date you would like to start generating regressors #--end, -e = When to stop generating holiday regressros # --scope, -c = D, W for Daily or Weekly respectively (e.g. Does this week have a particular holiday) # --file, -f = Ummm where to write the output silly! # # **NOTE** The EOM holiday is "End of Month" for

Hi everyone, this should be pretty basic but I need asking for help as I got stuck. I am running simple linear regression models on R with k regressors where k > 1. In order to automate my code I packed all the regressors in a matrix X so that lm(y~X) will always produce the results I want regardless of the variables in X. I am new to R but I found this advice somewhere so I guess it is

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

Hi all, I am reading the MASS book but it doesn't give examples about the diagnosis and model checking for rlm... My data is highly non-Gaussian so I am using rlm instead of lm. My questions are: 0. Are goodness-of-fit and model-checking using rlm completely the same as usual regression? 1. Please give me some pointers about how to do goodness-of-fit and residual diagnosis for

Hello, I have a simple xyplot with rlm lines. I would like to add the a and b coefficients (y=ax+b) of the rlm calculation in each panel. I know I can do it 'outside' the xyplot command but I would like to do all at the same time. I found some posts with the same question, but no answer. Is it impossible ? Thanks in advance for your help. Ptit Bleu. x11(15,12) xyplot(df1$col2 ~

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 =

I am absolutely new to R and I am aware of only a few basic command lines. I was running a robust regression in R, using the following command line library (MASS) rfmodel2 <- rlm (TotalEmployment_2005 ~ ALABAMA + MISSISSIPPI + LOUISIANA + TotalEmployment_2000 + PCWhitePop_2005 + UnemploymentRate_2005 + PCUrbanPop2000 + PCPeopleWithACollegeDegree_2000 +

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'