similar to: R memory question (fwd)

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 ~

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, I am trying to build a series of rlm models. I have my data frame and the models will be built using various coulmns of the data frame. Thus a series of models would be m1 <- rlm(V1 ~ V2 + V3 + V4, data) m2 <- rlm(V1 ~ V2 + V5 + V7, data) m3 <- rlm(V1 ~ V2 + V8 + V9, data) I would like to automate this. Is it possible to use a string in place of the formula? I tried doing: fmla

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 ~

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'

Dear List, I used rlm to calculate two regression models for two data sets (rlm due to two outlying values in one of the data sets). Now I want to compare the two regression slopes. I came across some R-code of Spencer Graves in reply to a similar problem: http://www.mail-archive.com/r-help at stat.math.ethz.ch/msg06666.html The code was: > df1 <- data.frame(x=1:10, y=1:10+rnorm(10))

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 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 =

Hello, I'm trying to do a little rlm of some data that looks like this: UNIT COHORT perdo adjodds 1010 96 0.39890 1.06894 1010 97 0.48113 1.57500 1010 98 0.36328 1.21498 1010 99 0.44391 1.38608 It works fine like this: rlm(perdo ~ COHORT, psi=psisquare) But the problem is that I have about 100 UNITs, and I want to do a

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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

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

Dear all, When calling rlm with the following data, I get an error. (R v.1.8.1, WinXP Pro 2002 with service pack 1.) > d <- na.omit(data.frame(CPRATIO, HEIGHTZ, FAMILYID)) > c <- tapply(d$CPRATIO, d$FAMILYID, mean) > h <- tapply(d$HEIGHTZ, d$FAMILYID, mean) > c 1 2 3 6 7 9 10 11 6.000000 2.500000 3.250000

Dear r-bugs, I looked over the FAQ. Hope I'm reporting this correctly. I ran this on both solaris and windows. I've provided terminal snapshots which include how R was called from the command line, and the result of version at the R prompt. I have attached the .r file, and the data file and the output snapshots. Below also find everything except only a few lines of the data file. Note

Hei Group, I want to compare the relative importance of predictors in a multiple linear regression y~a+bx1+cx2... However, bptest indicates heteroskedasticity of my model. I therefore perform a robust regression (rlm), in combination with bootstrapping (as outlined in J. Fox, Bootstrapping Regression Models). Now I want to compare the relative importance of my predictors. Can I rely on the

How do we get 2-tailed p-values for the rlm summary? I'm using the following: > fit <- rlm(oatRT ~ oatoacData$erp, psi=psi.bisquare, maxit=100, na.action='na.omit') > fitsum <- summary(fit, cor=F) > print(fitsum) Call: rlm(formula = oatRT ~ oatoacData$erp, psi = psi.bisquare, maxit = 100, na.action = "na.omit") Residuals: Min 1Q Median

Hi all, I am also confused about the manual: a. The input arguments: wt.method are the weights case weights (giving the relative importance of case, so a weight of 2 means there are two of these) or the inverse of the variances, so a weight of two means this error is half as variable? w (optional) initial down-weighting for each case. init (optional) initial values for the

We have a choice when calculating the Huber location estimate: > set.seed(221205) > y <- 7 + 3*rt(30,1) > library(MASS) > huber(y)$mu [1] 5.9117 > coefficients(rlm(y~1)) (Intercept) 5.9204 I was surprised to get two different results. The function huber() works directly with the definition whereas rlm() uses iteratively reweighted least squares. My surprise is

On Thu, 16 Sep 2004, RenE J.V. Bertin wrote: > Dear all, > > A few quick questions about interpreting and evaluating the results of > linear regressions, to which I hope equally quick answers are possible. > > 1) The summary.lm method prints the R and R^2 correlation coefficients > (something reviewers like to see). It works on glm objects and (after > tweaking it to

I want to generate a lattice plot of a multiple linear regression. I'm using the code: xyplot(y ~ x1 + x2 | status, data=datam, xlab="Peak separation",ylab="G/W",main="G/W vs Fuzzy peak separation: Threshold=1.8", groups=Fuzzy.gw.t.score>1.8, subset=(status %in% c("control","patient","sibling")),