similar to: opposite estimates from zeroinfl() and hurdle()

Hello, I?m using zero-inflated Poisson regression via the zeroinfl() function to analyze data on seed-set of plants, but for some reason, I don?t seem to be getting the output for all three levels of my two categorical predictors. More about my data and model: My response variable is the number of viable seeds (AVInt), and my two categorical predictors are elevation (Elev) and Treatment

Hello all: I have a question regarding the fitted.values returned from the zeroinfl() function. The values seem to be nearly identical to those fitted.values returned by the ordinary glm(). Why is this, shouldn't they be more "zero-inflated"? I construct a zero-inflated series of counts, called Y, like so: b= as.vector(c(1.5, -2)) g= as.vector(c(-3, 1)) x <- runif(100) # x

I'm trying to fit a zero-inflated poisson model using zeroinfl() from the pscl library. It works fine for most models I try, but when I include either of 2 covariates, I get an error. When I include "PopulationDensity", I get this error: Error in solve.default (as.matrix(fit$hessian)) : system is computationally singular: reciprocal condition number = 1.91306e-34 When I

Hi all, I have been trying to use zeroinfl() with the pscl package with R version 2.1.1. and with the newest versions of the contrib packages compatible with R 2.1.1. I have read the examples, the vignette and all the posts relating to zeroinfl() but I am still confused as to how to structure the model. Here is a small example; the error message is the same for big data sets

Hi Just a quick problem that I hope is simple to resolve. I'm doing some work with zero inflated poisson models using the pscl package. I can build models using zeroinfl and get outputs fom them with no problem, but when I try to use the predict.zeroinfl function, I get Error: could not find function "predict.zeroinfl". I was using an older version of R, but still had the same

Hi! I have three models. In the first model, everything is fine. However, in the second and third models, I have NA's for standard errors: The hessians also have NaN's (same for m2 and m3). What should I do about it? It there a way to obtain the hessian without transforming my variables? I will greatly appreciate your help! -- View this message in context:

Dear R Users, When using zeroinfl() function to fit a Zero-Inflated Negative Binomial (ZINB) model to a dataset, the summary() gives an estimate of log(theta) and its standard error, z-value and Pr(>|z|) for the count component. Additionally, it also provided an estimate of Theta, which I believe is the exp(estimate of log(theta)). However, if I would like to have an standard error of Theta

Dear list, I'm wondering if anyone can help me calculate the deviance of either a zeroinfl or hurdle model from package pscl? Even if someone could point me to the correct formula for calculating the deviance, I could do the rest on my own. I am trying to calculate a pseudo-R-squared measure based on the R^{2}_{DEV} of [1], so I need to be able to calculate the deviance of the full and null

Dear all, I would like to use the biological symbols for male and female as plotting symbols in a scatterplot (ideally filled and non-filled). R does not seem to have these symbols using pch= in plot() nor are they implemented via expression() or at least I did not find them. I found that the symbols are e.g. available in the wasysym and the marvosym package in LaTeX. I have coded two very rough

Dear all, I would like to use the biological symbols for male and female as plotting symbols in a scatterplot (ideally filled and non-filled). R does not seem to have these symbols using pch= in plot() nor are they implemented via expression() or at least I did not find them. I found that the symbols are e.g. available in the wasysym and the marvosym package in LaTeX. I have coded two very rough

Hello, I have a data frame with 1425 observations, 539 of which are zeros. I am trying to fit the following ZINB: f3<-formula(Nbr_Abs~ Zone * Year + Source) ZINB2<-zeroinfl(f3, dist="negbin", link= "logit", data=TheData, offset=log(trans.area), trace=TRUE) Zone is a factor with 4 levels, Year a factor with 27 levels, and Source a factor with 3 levels. Nbr_Abs is counts

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,

I have an ANOVA model with 2 factors "Environment" and "Site", "Diameter" is the response variable. Site should be nested within Environment. Site is also a random factor while Environment is fixed. I can do this analysis using the "aov" function by using these commands: >model<-aov(Diam~Env+Error(Env%in%Site),data=environ) >summary(model)

Hi All. I am a new R user. Trying to do scatterplot. Not sure how to resolve this error message A<-subset (ErablesGatineau, station=="A") > B<-subset (ErablesGatineau, station=="B") > > plot(diam ~ biom) > abline(lm(diam ~ biom), col = "red") > > goodcases <- !(is.na(diam) | is.na(biom)) > lines(lowess(diam[goodcases] ~

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; >

I am attempting to fit a full zero-inflated Poisson model then use backward elimination to arrive at the best-fitting model. When I try to fit the model with zeroinfl() I get this error: Error in while (abs((ll_old - ll_new)/ll_old) > control$reltol) { : missing value where TRUE/FALSE needed Any suggestions? Thanks for your help! Paige Barlow MS Student Virginia Tech Dept Fish

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

Consider the following analysis of a class experiment done as a Latin Square: > spinner <- gl(4,4,16,label=c("Murray","Angela","Shasha","Stephen")) > order <- gl(4,1,16) > treat <- scan() 1: 1 2 4 3 5: 4 3 1 2 9: 3 4 2 1 13: 2 1 3 4 17: Read 16 items > coin <-

The list of your interest is R-help not R-sig-mac stefano Il giorno 26/gen/06, alle ore 01:20, Sylvain Charlat ha scritto: > Hi, > > Is there any simple way to get histogram for different levels of > factor? > > Say you have the following data set: > > Island Sp.diam > Moorea 1.21 > Moorea 1.27 > Moorea 1.28 > Moorea 1.22 > Moorea 1.28 > Rurutu

I would like to compute the MC test (rfm.test) available in the package MarkedPointProcess (for the data BITOEK for example) in order to test the dependence between the marks and their locations. Why the syntax of rfm.test is false here? I have the message : ****************************** ML WARNING! Forbidden values! -- if there are too many warnings try narrower lower and upper bounds for the