similar to: R.squared in Weighted Least Square using the Lm Function

Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the observations is calculated as if the weights are on the original scale: [...] f <- z$fitted.values w <- z$weights [...] m

Hi, This is more of a general question than a pure R one, but I hope that is OK. I want to combine one-tailed independent p-values using the weighted version of fisher's inverse chi-square method. The unweighted version is pretty straightforward to implement. If x is a vector with p-values, then I guess that this will do for the unweighted version: statistic <- -2*sum(log(x)) comb.p <-

Do you mean w <- z$residuals ? Type names(z) to see the list of item in your model. I ran your code on a lm and it work fine. You don't need the brackets around mss <- Michael Long On 04/07/2016 02:21 PM, Murray Efford wrote: > Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted

On 08 Apr 2016, at 12:57 , Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 07/04/2016 5:21 PM, Murray Efford wrote: >> Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the

Hi, If I used drop1 in a weighted lm fit, it seems to ignore the weights in the AIC calculation of the dropped terms, see the example below. Can this be right? Yan -------------------- library(car) > unweighted.model <- lm(trSex ~ (river+length +depth)^2- length:depth, dno2) > Anova(unweighted.model) Anova Table (Type II tests) Response: trSex Sum Sq Df F value

On 07/04/2016 5:21 PM, Murray Efford wrote: > Following some old advice on this list, I have been reading the code for summary.lm to understand the computation of R-squared from a weighted regression. Usually weights in lm are applied to squared residuals, but I see that the weighted mean of the observations is calculated as if the weights are on the original scale: > > [...] > f

Thanks for these perfectly consistent replies - I didn't understand the purpose of m = sum(w * f/sum(w)) and saw it merely as a weighted average of the fitted values. My ultimate concern is how to compute an appropriate weighted TSS (or equivalently, MSS) for PRESS-R^2 = 1 - PRESS/TSS = 1 - PRESS/ (MSS + PRESS). Do you think it then makes sense to substitute the vector of leave-one-out fitted

Dear all, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) This is the unweighted fit, in the code of 'nls' one can see that 'nls' generates a vector

>>>>> Murray Efford <murray.efford at otago.ac.nz> >>>>> on Fri, 8 Apr 2016 18:45:33 +0000 writes: > Thanks for these perfectly consistent replies - I didn't > understand the purpose of m = sum(w * f/sum(w)) and saw it > merely as a weighted average of the fitted values. My > ultimate concern is how to compute an appropriate

Martin - Thanks, but although hatvalues() is useful for calculating PRESS, I can't find anything directly relevant to my question in the influence help pages. After some burrowing in the literature I'm doubting there is an answer out there (PRESS R^2 is always presented in a fairly ad hoc way). This is a new topic, as you say, and perhaps better handled on a statistics list. Murray Efford

> On Apr 10, 2016, at 3:11 AM, Murray Efford <murray.efford at otago.ac.nz> wrote: > > Martin - > Thanks, but although hatvalues() is useful for calculating PRESS, I can't find anything directly relevant to my question in the influence help pages. After some burrowing in the literature I'm doubting there is an answer out there (PRESS R^2 is always presented in a fairly

> On Apr 10, 2016, at 9:38 AM, David Winsemius <dwinsemius at comcast.net> wrote: > >> >> On Apr 10, 2016, at 3:11 AM, Murray Efford <murray.efford at otago.ac.nz> wrote: >> >> Martin - >> Thanks, but although hatvalues() is useful for calculating PRESS, I can't find anything directly relevant to my question in the influence help pages. After

What is the exact formula used in R lm() for the Adjusted R-squared? How can I interpret it? There seem to exist several formula's to calculate Adjusted R-squared. Wherry’s formula [1-(1-R2)·(n-1)/(n-v)] McNemar’s formula [1-(1-R2)·(n-1)/(n-v-1)] Lord’s formula [1-(1-R2)(n+v-1)/(n-v-1)] Stein 1-(n-1/n-k-1)(n-2)/n-k-2) (n+1/n) Theil's formula (found here:

Dear all, I'm struggling with weighted least squares, where something that I had assumed to be true appears not to be the case. Take the following data set as an example: df <- data.frame(x = runif(100, 0, 100)) df$y <- df$x + 1 + rnorm(100, sd=15) I had expected that: summary(lm(y ~ x, data=df, weights=rep(2, 100))) summary(lm(y ~ x, data=rbind(df,df))) would be equivalent, but

I am simulating an ANOVA model and get a strange behavior from the summary function. To be more specific: please run the following code and see for yourself: the summary()[["r.squared"]] values of two identical models are quite different!! ## 3 x 3 ANOVA of two factors x and z on outcome y s.size <- 300 # the sample size p.z <- c(0.25, 0.5, 0.25) # the probabilities of factor z ##

Greetings, I'm having trouble finding a simple way to calculate a weighted histogram where there may be zero raw counts in a given interval. Given equal-length vectors of data 'data' and weights 'w', and breaks (intervals) for the histogram, I calculate a weighted histogram as follows (see MASS's 'truehist' for an unweighted histogram): bin <- cut(data,

Hello, I'm with a conceptual doubt regarding Rsquared of both lm() and postResample(library caret). I've got a multiple regression linear model (lets say mlr) with anR² value of 67.52%. Then I use this model pro make predictions with predict() function using the same data as input , that is, use the generated model to predict the value associated with data that I used as input. Next, if

Hello R folks, Reproducible code below - I'm trying to do a weighted mean by a factor and can't figure it out. Thanks in advance for your assistance. Mike data<-data.frame(c(5,5,1,1,1), c(10,8,9,5,3), c("A","A","A","B","B"))

I sent this to r-devel the other day but didn't get any takers. This may not be a bug but rather an inconsistency. I'm not sure if this is intentional. summary.lm stores weighted residuals whereas I think most users will want print.summary.lm to summarize unweighted ones as if saying summary(resid(fit)). > set.seed(1) > dat <- data.frame(y = rnorm(15), x = rnorm(15), w = 1:15)

I would like to apply network measures (such as betweenness centrality, upper boundedness, etc.) to a weighted graph with non-integer weights, defined by a euclidean distance matrix. The package sna provides the measures that I want to use, but seems only to operate on binary graphs. I have read work by Mark Newman (http://aps.arxiv.org/abs/cond-mat/0407503/), who suggests that a weighted graph