similar to: Calculation of r.squared for linear model with offset

Hello all, I am using the function lm to do my weighted least square regression. model<-lm(Y~X1+X2, weight=w) What I am confused is the r.squared. It does not seem that the r.squared for the weighted case is an ordinary 1-RSS/TSS. What is that precisely? Is the r.squared measure comparable to that obtained by the ordinary least square? <I also notice that model$res is the unweighted

Hi all, Like a lot of people I noticed that I get different results when I use nls in R compared to the exponential fit in excel. A bit annoying because often the R^2 is higher in excel but when I'm reading the different topics on this forum I kind of understand that using R is better than excel? (I don't really understand how the difference occurs, but I understand that there is a

This is probably simple, but I just can't see it... I want to calculate the R^2s for a series of linear models where each term is dropped in turn. I can get the RSS from drop1(), and the r.squared from summary() for a given model, but don't know how to use the result of drop1() to get the r.squared for each model with one term dropped. Working example: library(vcd) # for

I've been working with a rather large data set (~10M rows), and while biglm works beautifully for generating coefficients, it does not report an r-squared. It does report RSS. Any idea on how one could coax an R-squared out of biglm? Thanks in advance for any help with this! Bryan Lim Lecturer Department of Finance University of Melbourne [[alternative HTML version deleted]]

I would appreciate help translating the following lme model to an lmer function. lme(lognrms ~ Group*Rotation*muscle*side*support*arms, random=~1|Subject/Stratum2/rep, data=Data) Many thanks Ross Darnell r.darnell at uq.edu.au

I am relatively new to R, so this may be a much simpler question than it seems to me. I am trying to create a function that includes two summations, and can't figure out how. I am attaching the equation as a pdf file. This function will then be optimized over a chosen range of values, but right now I just need help with the function. Thank you. Zack Darnell Duke University -- Zack

I have fitted a logistic regression model > failed.lr2$call lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long, na.action = na.omit) using the Design package functions and would like to generate a nomogram from this model. the datadist information is generated and stored in > ddist time.long$Age time.long$task2 Low:effect 45

I would like to create a page of two graphs (2 rows by 1 col) and then draw vertical lines (segments?) on both graphs from the minimum values to the corresponding maximum value. So I have tried # > y <- rnorm(3000) > par(mfrow=c(2,1)) > plot(y,type="l") > plot(cumsum(y),type="l") > segments(1000,min(cumsum(y)),1000,max(cumsum(y))) > par(mfg=c(1,1)) >

I would like to define the order of the levels of a factor. The relevel function would work but since I have 20 levels I would prefer to declare the order explicitly. Using a smaller example levels(oldfactor) "b1" "b2" "r1" "r2" nufactor <- order(oldfactor,order=c("b1","r1","b2","r2")) # my fabricated function

Maybe a useful addition to the predict functions would be to return the values of the predictor variables. It just (unless there are problems) requires an extra line. I have inserted an example below. "predict.glm" <- function (object, newdata = NULL, type = c("link", "response", "terms"), se.fit = FALSE,

Using R (version 0.63.3) for MS windows, I try the following command > shell(paste("cd ",getenv("RHOME"),sep="")) which replies with an error message Too many parameters - FILES\RW0633 which appears to suggest that the space if the path name is causing difficulties to the cd command. getenv returns > getenv("RHOME") RHOME

Dear R Help, I have noticed some inconsistent behaviour of add1 and drop1 with a weighted linear model, which affects the interpretation of the results. I have these data to fit with a linear model, I want to weight them by the relative size of the geographical areas they represent. _________________________________________________________________________________________ > example

John, I spoke with Level(3) last week regarding SIP termination. They quoted $0.01/minute, with an 11 Million Minute / Month minimum. Ugh! -dg -------------------------------------------------------------- Darnell Gadberry President binaryMedia darnell AT binmedia DOT com ------------ Date: Fri, 19 Dec 2003 21:12:22 -0500 To: asterisk-users@lists.digium.com From: John Todd

Greetings Listers! the R-squared value reported by summary of lm is calculated as 1 - RSS/RSS_m where RSS_m is the residual sum of squares of a minimal model. In most cases, the minimal model is simply y = mean(y), but when a constant is left out of the model, the minimal model is y = 0. However, if you manually add a constant, R still considers y = 0 the minimal model. This also causes

Hi, I'm trying to verify the calculation of coefficient of determination (r squared) for linear regression. I've done the calculation manually with a simple test case and using the definition of r squared outlined in summary(lm) help. There seems to be a discrepancy between the what R produced and the manual calculation. Does anyone know why this is so? What does the multiple r squared

I am using the Windows menu functions below which will work on the first pass, but if I repeat the same script I cannot get the WinMenuAddItem to work. This is a problem if I change the menu structure and reread the source code I am forced to quit and restart Rgui. "try.menu" <- function(){ OS <- .Platform$OS.type GUI <- .Platform$GUI if (!(OS == "windows" &

-----Original Message----- >From: Tommy Christensen [mailto:tommy.christensen@tpack.net] >Sent: Wednesday, October 08, 2003 10:09 AM >To: Christian Darnell >Cc: 'Linux 802.1Q VLAN'; Bart De Schuymer; netdev@oss.sgi.com; bridge >Subject: Re: [Bridge] RE: [VLAN] Re: [PATCH/RFC] Let {ip, arp}tables >"see" bridged VLAN tagged{I,AR}P packets > > > >This

Can anyone please tell me how to get the r-squared output from a piecewise (segmented) regression using the strucchange package? Here is the R code I have tried thus far. library(lmtest) library(strucchange) data <- ts(c(rnorm(30), runif(30)), frequency = 12, start = c(2005, 01)) bpts <- breakpoints(data ~ 1) print(bpts) summary(bpts) coeftest(bpts) [[alternative HTML version

Hi people, I want to obtain the adjusted r-square given a set of coefficients (without the intercept), and I don't know if there is a function that does it. Exist???????????????? I know that if you make a linear regression, you enter the dataset and have in "summary" the adjusted r-square. But this is calculated using the coefficients that R obtained,and I want other coefficients

Full_Name: Ross Darnell Version: 2.4.1 OS: Linux Submission from: (NULL) (130.102.133.33) rdarnell at pc:~/Data$ ls -al .RData -rwxrwxrwx 2 root root 13551 2006-12-06 08:58 .RData rdarnell at pc:~/Data$ R R version 2.4.1 (2006-12-18) Copyright (C) 2006 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to