similar to: least median squares

Dear List-members, I found a strange behaviour in the lqs function. Suppose I have the following data: y <- c(7.6, 7.7, 4.3, 5.9, 5.0, 6.5, 8.3, 8.2, 13.2, 12.6, 10.4, 10.8, 13.1, 12.3, 10.4, 10.5, 7.7, 9.5, 12.0, 12.6, 13.6, 14.1, 13.5, 11.5, 12.0, 13.0, 14.1, 15.1) x1 <- c(8.2, 7.6,, 4.6, 4.3, 5.9, 5.0, 6.5, 8.3, 10.1, 13.2, 12.6, 10.4, 10.8, 13.1, 13.3, 10.4, 10.5, 7.7, 10.0, 12.0,

Hi I am looking for suitable packages in R that do regression analyses using least median squares method (or better). Additionally, I am also looking for packages that implement algorithms/methods for detecting outliers that can be discarded before doing the regression analyses. Although some websites refer to "lms" method under package "lps" in R, I am unable to find such a

Hi, trying to do a robudt regression of a two-way linear model, I keep getting the following error: > lqs(obs ~ y + s -1,method="lms", contrasts=list(s=("contr.sum"))) Error: lqs failed: all the samples were singular Robust regression with M-estimators works (also regular least square fits, of course): rlm.formula(formula = obs ~ y + s - 1, method = "M",

I am using "nls" to fit dose-response curves but am not sure how to approach more robust regression in R to get around the problem of the my error showing increased variance with increasing dose. My understanding is that "rlm" or "lqs" would not be a good idea here. 'Fairly new to regression work, so apologies if I'm missing something obvious.

Dear all; This must have a rather simple answer but haven't been able to figure it out: I have a data frame with say 2 groups (group 1 & 2). I want to select from group 1 say "n" rows and calculate the mean; then select "m" rows from group 2 and calculate the mean as well. So far I've been using a for loop for doing it but when it comes to a large data set is

Your understanding isn't similar to mine. Mine says robust/resistant methods are for data with heavy tails, not heteroscedasticity. The common ways to approach heteroscedasticity are transformation and weighting. The first is easy and usually quite effective for dose-response data. The second is not much harder. Both can be done in R with nls(). Andy From: Quin Wills > > I am

Here are several ways to get there, but your original loop is fine once it is corrected: > for (i in 1:2) smean[i] <- mean(toy$diam[toy$group==i][1:nsel[i]]) > smean [1] 0.271489 1.117015 Using sapply() to hide the loop: > smean <- sapply(1:2, function(x) mean((toy$diam[toy$group==x])[1:nsel[x]])) > smean [1] 0.271489 1.117015 Or use head() > smean <- sapply(1:2,

Dear all; Is there any way to make this to work?: .x<-rnorm(50,10,3) .y<-.x+rnorm(50,0,1) X<-data.frame(.x,.y) colnames(X)<-c("Xvar","Yvar") Ytext<-"Yvar" lm(Ytext~Xvar,data=X) # doesn't run lm(Yvar~Xvar,data=X) # does run The main idea is to use Ytext as input in a function, so you just type "Yvar" and the model should fit....

Hi Pedro, This may not be much of an improvement, but it was a challenge. selvec<-as.vector(matrix(c(nsel,unlist(by(toy$diam,toy$group,length))-nsel), ncol=2,byrow=TRUE)) TFvec<-rep(c(TRUE,FALSE),length.out=length(selvec)) toynsel<-rep(TFvec,selvec) by(toy[toynsel,]$diam,toy[toynsel,]$group,mean) Jim On 4/3/16, Pedro Mardones <mardones.p at gmail.com> wrote: > Dear all; >

Dear List-Members, I have a problem with the function lqs() from package MASS. In some cases it produces different results for the same settings and needs a random seed to be set, in other cases not. I really cannot understand, why this happens. As well I do not understand what exactly you need the random seed for. Is it a starting point for iterations? Or do different results occur because of

Dear R users; Is it possible to get the row and column number of a particular entry in a dist object? Let's say that I want to find the position of the value 1.1837 (the last entry on the dist object below), that is [6,3]. Can I get those values without transforming the object to a matrix?, i.e. working with the dist object only. 1 2 3 2 0.23935864

Dear all; A very basic question. I have the following data: ************************************************************************************ A <- 1/1000*c(347,328,129,122,18,57,105,188,57,257,53,108,336,163, 62,112,334,249,45,244,211,175,174,26,375,346,153,32, 89,32,358,202,123,131,88,36,30,67,96,135,219,122, 89,117,86,169,179,54,48,40,54,568,664,277,91,290,

Dear r-help, Is there any available description of the components of lqs objects found in the package "lqs"? > names(slts) [1] "crit" "sing" "coefficients" "bestone" [5] "fitted.values" "residuals" "scale" "terms" [9] "call"

I tried building R-1.6.1 under Mac OS X 10.2.2 and experienced the following build failure: <...stuff omitted...> cc1: warning: changing search order for system directory "/usr/local/include" cc1: warning: as it has already been specified as a non-system directory gcc -bundle -flat_namespace -undefined suppress -L/usr/local/lib -o ctest.so ansari.o chisqsim.o d2x2xk.o

Dear r-devel members, I tried to build R packages on a PC running Windows XP but experience problems. However, it is ok when there is no inst directory in a package. Any help would be appreciated. The following is an example, C:\work>R CMD check VR_7.2-19.tar.gz * checking for working latex ... OK * using log directory 'C:/work/VR.Rcheck' * using R version 2.1.1, 2005-06-20 *

Hi, I have been using the lqs function with method='lms'. However the results I get are a little different from the results noted by Rousseeuw & Leroy (Robust Regression and Outlier Detection) and I was wondering how to use these results for outlier detection. I'm using the stackloss dataset, for which the original Rousseeuw et al. program points out that observations 1,2,3,4

Dear list, (this is a follow up from a previous query) Why does R CMD INSTALL write most of its messages to stderr? If it wrote to stdout, then we could capture its output within an R session when calling sink("stdout.txt", type="output") install.packages("MASS", type="source") sink() As it stands, the stderr messages can't be captured via

Dear all; Does anyone know how to add grid lines to a persp plot? I've tried using lines(trans3d..) but the lines of course are superimposed into the actual 3d surface and what I need is something like the plot shown in the following link: http://thermal.gg.utah.edu/tutorials/matlab/matlab_tutorial.html I'll appreciate any ideas Thanks PM

Dear all; I have a function written in R that returns as a list of values as output that has associated some user defined attributes to it. How can hide these attributes when printing the output on screen? I'm using R-2.8.1 on WinXP....it's like hiding the attr of the output from the scale function.... Thanks in advance PM

Dear list member, by running Rcmd check on a package where a customised 'CITATION' file should be included instead of the automatically produced one, the following error occurs: adding build stamp to DESCRIPTION installing NAMESPACE file and metadata installing R files installing inst files FIND: Parameter format not correct make[2]: *** [c:/R/packages/urca/inst] Error 2 make[1]: