similar to: cvTools for 2 models not working

Dear R-experts, I guess I have a problem with my fast function (fast tau estimator) here below. Indeed, zero errors look highly suspicious. I guess there is a bug in my R code. How could I correct my R code ? # install.packages( "robustbase" ) # install.packages( "MASS" ) # install.packages( "quantreg" ) # install.packages( "RobPer" ) #

Dear R-experts, Doing cross-validation for 2 robust regressions (HBR and fast Tau). I can't get the 2 errors rates (RMSE and MAPE). The problem is to predict the response on the testing data. I get 2 error messages. Here below the reproducible (fictional example) R code. #install.packages("MLmetrics") # install.packages( "robustbase" ) # install.packages(

On 5/8/2018 12:26 PM, varin sacha via R-help wrote: > > Dear R-experts, > > Here below the reproducible example. I am trying to get the average of the 100 results coming from the "lst" function. I have tried lst$mean and mean(lst). It does not work. > Any help would be highly appreciated > > #################### > > ?## R script for getting MedAe and

Dear R-experts, Here below the reproducible example. I am trying to get the average of the 100 results coming from the "lst" function. I have tried lst$mean and mean(lst). It does not work. Any help would be highly appreciated. #################### ?## R script for getting MedAe and MedAeSQ from HBR model on Testing data install.packages("robustbase") install.packages(

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

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

I am needing some expertise with regard to the S-Plus command lmRobMM and its R counterpart rlm(formula,data,method="MM") I have used lmRobMM(formula,data) in S-Plus on the Stackloss data and obtained for my residuals 6.217777 1.150717 6.427946 8.174019 -0.6713005 -1.248641 -0.4236203 0.5763797 -1.057899 0.3593823 11 12 13 14 15 16

Hi all, I'm trying to fit a nonlinear regression by "nlrob": model3=nlrob(y~a1*x^a2,data=transient,psi=psi.bisquare, start=list(a1=0.02,a2=0.7),maxit=1000) However an error message keeps popping up saying that the function psi.bisquare doesn't exist. I also tried psi.huber, which is supposed to be the default for nlrob: model3=nlrob(y~a1*x^a2,data=transient,psi=psi.huber,

Hi, everyone: I ask for help about translating a stata program into R. The program perform cross validation as it stated. #1. Randomly divide the data set into 10 sets of equal size, ensuring equal numbers of events in each set #2. Fit the model leaving out the 1st set #3. Apply the fitted model in (2) to the 1st set to obtain the predicted probability of a prostate cancer diagnosis. #4. Repeat

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

Dear all, I am not sure if this mail is for R-help or should be sent to R-devel instead, and therefore post to both. While using nlrob from package 'robustbase', I ran into the following problem: For psi-functions that can become zero (e.g. psi.bisquare), weights in the internal call to nls can become zero. Example: d <- data.frame(x=1:5,y=c(2,3,5,10,9)) d.nlrob <-

Hi, I've been using the Matlab robustfit function for linear regressions where I suspect some data points are outliers. Is there an equivalent function in R? Take care, Darren PS, This is the Matlab help on robustfit: >> help robustfit ROBUSTFIT Robust linear regression B = ROBUSTFIT(X,Y) returns the vector B of regression coefficients, obtained by performing robust

I want to estimate the center of a distribution with lots of outliers in one tail, and thought I would use a function such as S-plus's location.m() with psi.fun=bisquare (as per MASS 3 p. 131). However, R seems not have such a function, so my questions are: 1) Is there an R equivalent to location.m()? 2) Would huber() give me results that are similar (i.e., close enough)? Thanks.

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

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 all, I have been using rlm() for robust regression. Could someone please suggest an appropriate measure of goodness-of-fit [1]? All I've found after trawling the web, literature databases, and previous r-help posts, is the "robust R^2" on pp. 362-363 of the S-plus manual, which is available at http://web.mit.edu/afs/athena/software/splus_v7.0/www/statman1.pdf (7.57 MB)

hi guys and gals ... How are you all ... i have to do something in robust regression by R programm , and i have some problems as following: *the first :* suppose w(r) =1/(1 r^2) and r <- c(7.01,2.07,7.061,5.607,8.502,54.909,12.222) and i want to exclude some values from r so that (abs(r)>4.9 )... after ,i want to used (w) to get on coefficients beta0 and beta1 (B1 <-

Dear R users: Recently, I tried to write a program to calculate cross-validated predicted value. My sources are as follows. However, the R reported an error. Could you please check the sources? Thanks. set.seed(100) x<-rnorm(100) y<-sample(rep(0:1,50),replace=T) dat<-data.frame(x,y) library(rms) fito<-lrm(y~x) preo<-predict(fito) pre<-matrix(NA,nrow=100,ncol=200) for (i in