search for: fits

Dear List: I am having difficulties with the fitted values at different levels of a multilevel model. My data set is a series of student test scores over time with a total of 7,280 observations, 1,720 students nested witin 60 schools. The data set is not balanced. The model was fit using eg.model.1<-lme(math~year, random=~year|schoolid/childid, data=single). When I call the random

I would like to fit a Pareto Distribution and I am using the following codes. I thought the fitted (fit1) should be the fitted value for the data, is it correct? As the result of the "fitted" turns out to be a single value for all. fit=vglm(ycf1 ~ 1, pareto1(location=alpha), trace=TRUE, crit="c") fitted(fit) The result is fitted(fit) [,1] [1,] 0.07752694

Dear R-team As I didn't get any answer to my bug-report last week I have taken the effort and extracted a minimal data set from my data (see below) where the following bug occurs: > glm(SKR.ein.aus ~ ., family = binomial, data = bugdata, na.action = na.omit) Error in names<-.default(*tmp*, value = ynames) : names attribute must be the same length as the vector In addition: Warning

to clean up some code I would like to make a list of arbitrary length to store?various objects for use in a loop sample code: ############ BEGIN SAMPLE ############## # You can see the need for a loop already linearModel1=lm(modelSource ~ .,mcReg) linearModel2=step(linearModel1) linearModel3=lm(modelSource ~ .-1,mcReg) linearModel4=step(linearModel3) #custom linearModel5=lm(modelSource ~ .

...5334e+00 7.5863272e-04 134 0 0.62 -2.6994067e+00 2.2881359e-01 135 1 1.00 -3.4141496e+01 1.2481102e-14 136 1 1.00 -1.6016066e+01 7.4169082e-07 137 0 0.00 1.0315917e+00 1.6787319e-02 138 0 0.16 -1.4051577e+00 1.1015032e-01 It is somehow strange to me that e.g. the second observation fits almost perfectly (actually it is 0.99999773), but that it has such a high residual. Obviously this is somehow compensated by the extremely low weight, but I don't understand actually what happens. Are the problems due to the possible perfect fit as it results in S-Plus? Thanks in advance for t...

Dear R-helpers: May I ask a question related to storing a number of lmer model fit into a list. Basically, I have a for-loop (see towards the bottom of this email) in the loop, I am very sure that the i-th model fit (i.e.,fit_i) is successfully generated and the character string (i.e., tmp_i) is created correctly. The problem stems from the following line in the for-loop #trouble making line

Dear All, I have been trying to simulate data from a fitted glm using the simulate() function (version details at the bottom). This works for lm() fits and even for lmer() fits (in lme4). However, for glm() fits its output does not make sense to me -- am I missing something or is this a bug? Consider the following count data, modelled as gaussian, poisson and binomial responses: counts <- data.frame( count = c(8, 3, 0, 3, 0, 9, 0...

Hallo! I want to understand / recalculate what is done to get the CI of the logistic regression evaluated with lrm. As far as I came back, my problem is the variance-covariance matrix fit$var of the fit (fit<-lrm(...), fit$var). Here what I found and where I stucked: ----------------- library(Design) # data D<-c(rep("a", 20), rep("b", 20)) V<-0.25*(1:40) V[1]<-25

Dear R-help, Here's a simple example of nonlinear curve fitting where nls seems to get the answer wrong on a very simple exponential fit (my R version 2.7.2). Look at this code below for a very basic curve fit using nls to fit to (a) a logarithmic and (b) an exponential curve. I did the fits using self-start functions and I compared the results with a more simple fit using a straight lm() command. The logarithmic fit works 100% correctly below. The problem is with the exponential fit: the nls command gives the wrong values and I have completely failed to see why this should happen. I...

Hey everybody, i am using the rugarch-package and its great! I have a pretty easy problem, but i just dont get it, so thanks if you can help me. Normally i use: / data(DATANAME) spec = ugarchspec() fit = ugarchfit(data = x[,1], spec = spec) fit slotNames(fit) names(fit at fit) coef(fit) infocriteria(fit) likelihood(fit) nyblom(fit) signbias(fit) head(as.data.frame(fit)) head(sigma(fit))

I imitated predict.glm, my thing worked, now I need to revise. It would help me very much if someone would explain predict.glm line 28, which says object$na.action <- NULL # kill this for predict.lm calls I want to know 1) why does it set the object$na.action to NULL 2) what does the comment after mean? Maybe I need a pass by value lesson too, because I can't see how changing that

Hey, all. I had a quick question about fitting new glm values and then looking at the error around them. I'm working with a glm using a Gamma distribution and a log link with two types of treatments. However, when I then look at the predicted values for each category, I find for the one that is close to 0, the error (using se.fit=T with predicted) actually makes it overlap 0.

Dear R-team There seems to be a bug in glm.fit - I got the following error message: > > > + Error in names<-.default(*tmp*, value = ynames) : names attribute must be the same length as the vector In addition: Warning messages: 1: fitted probabilities of 0 or 1 occurred in: (if (is.empty.model(mt)) glm.fit.null else glm.fit)(x = X, y = Y, 2: fitted probabilities of 0 or 1 occurred

Hi, Although my doubt is pretty,as i m not from stats background i am not sure how to proceed on this. Currently i am doing a forecasting.I used ARIMA to forecast and time series was volatile i used garchFit for residuals. How to use the output of Garch to correct the forecasted values from ARIMA. Here is my code: ###delta is the data fit<-arima(delta,order=c(2,,0,1)) fit.res <-

Dear all, predict.lme() throws an error when the fixed part consists of only an intercept and using newdata. See the reproducible example below. I've tracked the error down to asOneFormula() which returns in this case NULL instead of a formula. Changing NULL instead of ~1 in that function (see below) solves the problem in the case of an intercept only model (m1). It does not solve the problem

Hello, I'm still a newbie user and struggling to automate some analyses from SigmaPlot using R. R is a great help for me so far! But the following problem makes me go nuts. I have two spectra, both have to be fitted to reference data. Problem: the both spectra are connected in some way: the stoichiometry of coefficients "cytf.v"/"cytb.v" is 1/2. {{In the SigmaPlot

Dear all, I would like to use predict.lm() with an existing lm object but with new arbitrary coefficients. I modify 'fit$coef' (see example below) "by hand" but the actual model in 'fit' used for prediction does not seem to be altered (although fit$coef is!). Can anyone please help me do this properly? Thanks in advance, J?r?mie > dat <-

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,

I would really appreciate it if someone can give suggestions on how to do spectra fitting in R using ordinary least square fitting and non-negativity constraints. The lm() function works well for ordinary least square fitting, but how to specify non-negativity constraints? It wouldn't make sense if the fitting coefficients coming out as negative in absorption spectra deconvolution. Thanks.

I have some data which consists of time series for a number of sites. It appears that there is not much autocorrelation in the data and I have fitted a cubic for each site using lm. I would like to obtain a lattice plot with one panel for each site and showing the original data, and the fitted cubic. The closest I have got to doing what I want is: fit <- fitted(paraslm1) temp <-