similar to: predict.glm(..., type="response") loses names (was RE: [R] A sugg estion for predict function(s))

Dear all: "predict.glm" provides an example to perform logistic regression when the response variable is a tow-columned matrix. I find some paradox about the degree of freedom . > summary(budworm.lg) Call: glm(formula = SF ~ sex * ldose, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max -1.39849 -0.32094 -0.07592 0.38220 1.10375

Hi, I have recently been attempting to find the LD50 from two predicted fits (For male and females) in a Generalised linear model which models the effect of both sex + logdose (and sex*logdose interaction) on proportion survival (formula = y ~ ldose * sex, family = "binomial", data = dat (y is the survival data)). I can obtain the LD50 for females using the dose.p() command in the MASS

Here's a patch that should make predict.glm(..., type="response") retain the names. The change passes make check on our Opteron running SLES9. One simple test is: names(predict(glm(y ~ x, family=binomial, data=data.frame(y=c(1, 0, 1, 0), x=c(1, 1, 0, 0))), newdata=data.frame(x=c(0, 0.5, 1)), type="response")) which gives [1]

Dear R-Listers, to measure the psychometric curve of pitch discrimination, one sequentially presents two tones of slightly different pitch to an observer (animal will do), and asks "which is higher". The pschometric curve is the fraction of correct responses plotted against the pitch difference. It passes through 50% (pure guessing) at zero and normally approaches 100% at large

At useR 2006 I mentioned that it would be nice to have a way to specify binomial links that involved free parameters and described some experience with a Gosset link involving a free degrees of freedom parameter, and a Tukey-lambda link with two free parameters. My implementation of this involved some rather kludgey modifications of binomial, make.link and glm that (essentially) added a

I am working with Probit regression (I cannot switch to logit) can anybody help me in finding out how to obtain with R Finney's fiducial confidence intervals for the levels of the predictor (Dose) needed to produce a proportion of 50% of responses(LD50, ED50 etc.)? If the Pearson chi-square goodness-of-fit test is significant (by default), a heterogeneity factor should be used to calculate

Hi. I'm trying to plot the ratio of used versus unused bird houses (coded 1 or 0) versus a continuous environmental gradient (proportion of urban cover [purban2]) that I would like to convert into bins (0 - 0.25, 0.26 - 0.5, 0.51 - 0.75, 0.76 - 1.0) and I'm not having much luck figuring this out. I ran a logistic regression and purban2 ends up driving the probability of a box being

Hi - I am fitting a probit model using glm(), and the deviance and residual degrees of freedom are different depending on whether I use a binary response vector of length 80 or a two-column matrix response (10 rows) with the number of success and failures in each column. I would think that these would be just two different ways of specifying the same model, but this does not appear to be the case.

Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with

Dear R-help list members, I am currently trying to fit a generalized linear model using a binomial with the canonical link. The usual solution is to use the R function glm() in the package "stats". However, I run into problem when I want to fit a glm without an intercept. It is indicated that the solution is in changing the function glm.fit (also in "stats"), by specifying

I am not sure whether it is a design decision or just an oversight. When I ask for the standard errors of the predictions with predict(budwm.lgt,se=TRUE) where budwm.lgt is a logistic fit of the budworm data in MASS, I got Error in match.arg(type) : ARG should be one of response, terms If one is to construct a CI for the fitted binomial probability, wouldn't it be more natural to do

Dear R-helpers, I don't see a difference between the following two plots of effect objects, which I understand should be different. What am I missing? require(doBy) require(effects) data(budworm) m1 <- glm(ndead/20 ~ sex + log(dose), data=budworm, weight=ntotal, family=binomial) m1.eff <- all.effects(m1) plot(m1.eff, rescale.axis = FALSE, selection = 2, main = 'rescale =

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,

It really is unclear what is claimed to be a bug here. But see https://stat.ethz.ch/pipermail/r-devel/2007-May/045592.html for why the bug is not in R: your old and new data do not match. Your fit is to a category. [The problem with the web interface to R-bugs was reported last week: it is being worked on.] On Mon, 30 Apr 2007, r.darnell at uq.edu.au wrote: > This is a multi-part

Hello, I've got a program written in S-plus which I think is converted successfully to R with the exception of part of the opt.param function written. In S-plus it is: nlminb(start=x0, obj=negllgamma.f, scale=1, lower=c(0.01,0.0001), upper=c(10,0.9999), gamma=gamma, maxlik=maxlik, y=ldose, s=lse, max.iter = 1000, max.fcal = 1000)$par and so far with R I've got to: optim(par=x0,

Josef''s fs_mark test fs_mark -d /mnt/btrfs-test -D 512 -t 16 -n 4096 -F -S0 on a 2GB single metadata fs leaves about 400Mb of metadata almost unused. This patch reduces metadata chunk allocations by considering the proper metadata chunk size of 200MB in should_alloc_chunk(), not the default 256MB which is set in __btrfs_alloc_chunk(). Signed-off-by: Itaru Kitayama

Hi, I am trying to run two Non-Gaussian regressions: logistic and probit. I am receiving two different errors when I try to run these regressions and I am not sure what they mean or how to fix my syntax. Here is the logistic regression error: Error in family$linkfun(mustart) : Argument mu must be a nonempty numeric vector Here is the probit regression error: Error in pmax(eta, -thresh) :

This deadlock comes from xfstests 251. We''ll hold the chunk_mutex throughout the whole of a chunk allocation. But if we find that we''ve used up system chunk space, we need to allocate a new system chunk, but this will lead to a recursion of chunk allocation and end up with a deadlock on chunk_mutex. So instead we need to allocate the system chunk first if we find we''re

Hello all, Trying to get this piece of code to work on my data set. It is from http://www.itc.nl/personal/rossiter. logit.roc <- function(model, steps=100) { field.name <- attr(attr(terms(formula(model)), "factors"), "dimnames")[[1]][1] eval(parse(text=paste("tmp <- ", ifelse(class(model$data) == "data.frame", "model$data$",

Dear all, I get a strange error when I find polychoric correlations with the ML method, which I have been able to reproduce using randomly-generated data. What is wrong? I realize that the data that I generated randomly is a bit strange, but it is the only way that I duplicate the error message. > n<-100 > test.x<-rnorm(n, mean=0, sd=1) > test.c<-test.x + rnorm(n, mean=0,