similar to: as.formula("x") error on C stack limit

Another example uses formula.character's other arguments: > as.formula("env") Error: object of type 'special' is not subsettable > as.formula("...") Error in eval(expr, envir, enclos) : '...' used in an incorrect context It may happen for the same reason that the following does not give an error: > y <- "response ~ pred1 + pred2" >

I have a function that, among other things, runs a linear model and returns r2. But, the number of predictor variables passed to the function changes from 1 to 3. How can I change the formula inside the function depending on the number of variables passed in? An example: get.model.fit <- function(response.dat, pred1.dat, pred2.dat = NULL, pred3.dat = NULL) { res <- lm(response.dat ~

could someone explain why the name of FPVAL gets " .value" concatenated onto it when the code below is run and temp is returned. I've been trying to figure this out for too long. It doesn't matter when I put the FPVAL in the return statement. It happens regardless of whether it's first or last. Thanks. f.lmmultenhanced <- function(response, pred1, pred2) {

hi, i'm working with mgcv packages and specially gam. My exemple is: >test<-gam(B~s(pred1)+s(pred2)) >plot(test,pages=1) when ploting test, you can view pred1 vs s(pred1, edf[1] ) & pred2 vs s(pred2, edf[2] ) I would like to know if there is a way to access to those terms (s(pred1) & s(pred2)). Does someone know how? the purpose is to access to equation of smooths terms

Say I fit a logistic model and want to calculate an odds ratio between 2 sets of predictors. It is easy to obtain the difference in the predicted logodds using the predict() function, and thus get a point-estimate OR. But I can't see how to obtain the confidence interval for such an OR. For example: model <- glm(chd ~age.cat + male + lowed, family=binomial(logit)) pred1 <-

I'm new to R and am trying to extract the factors of a dataframe using numeric indices (e.g. df[1]) that are input to a function definition instead of the other types of references (e.g. df$out). df[1] is a list(?) whose class is "dataframe". These indexed lists can be printed successfuly but are not agreeable to the plot() and lm() functions shown below as are their df$out

I'm fairly new to R, and I'm trying to write out a population model that satisfies the following; the system consists of s species, i= 1, 2,...,s network of interactions between species is specified by a (s x s) real matrix, C[i,j] x[i] being the relative population of the "ith" species (0 =< x[i] =< 1, sum(x[i]=1) the evolution rule being considered is as follows;

Hi, My questions concerns the ROCR package and I hope somebody here on the list can help - or point me to some better place. When evaluating a model's performane, like this: pred1 <- predict(model, ..., type="response") pred2 <- prediction(pred1, binary_classifier_vector) perf <- performance(pred, "sens", "spec") (Where "prediction" and

Dear R users I have a question related to the interpretation of results based on GAMMs using Simon Woods package gamm4. I have repeated measurements (hours24) of subjects (vpnr) and one factor with three levels (pred). The outcome (dv) is binary. In the first model I'd like to test for differences among factor levels (main effects only): gamm.11<-gamm4(dv ~ pred +s(hours24), random = ~

Full_Name: Renaud Lancelot Version: Version 2.3.0 (2006-04-24) OS: MS Windows XP Pro SP2 Submission from: (NULL) (82.239.219.108) I think there is a bug in predict.lme, when a polynomial generated by poly() is used as an explanatory variable, and a new data.frame is used for predictions. I guess this is related to * not * using, for predictions, the coefs used in constructing the orthogonal

Hello All,  I have used logistic regression glm in R and I am evaluating two models both learned with glm but with different predictors. model1 <- glm (Y ~ x4+ x5+ x6+ x7, data = dat, family = binomial(link=logit))model2 <- glm (Y~ x1 + x2 +x3 , data = dat, family = binomial(link=logit))  and I would like to compare these two models based on the prediction that I get from each model: pred1 =

Hi I'm trying to replicate Smith et al.'s (http://www.sciencemag.org/content/330/6008/1216.abstract) findings by fitting their Gompertz and logistic models to their data (given in their supplement). I'm doing this as I want to then apply the equations to my own data. Try as a might, I can't quite replicate them. Any thoughts why are much appreciated. I've tried contacting the

Dear friends of lme, After so many year with lme, I feel ashamed that I cannot get this to work. Maybe it's a syntax problem, but possibly a lack of understanding. We have growth curves of new dental bone that can well be modeled by a linear growth curve, for two different treatments and several subjects as random parameter. By definition, newbone is zero at t=0, so I tried to force the

Hi, I have found quite a few posts on normality checking of response variables, but I am still in doubt about that. As it is easy to understand I'm not a statistician so be patient please. I want to estimate the possible effects of some predictors on my response variable that is nº of males and nº of females (cbind(males,females)), so, it would be:

Hello all, How do I actually use the output of predict.lm(..., type="terms") to predict new term values from new response values? I'm a chromatographer trying to use R (2.15.1) for one of the most common calculations in that business: - Given several chromatographic peak areas measured for control samples containing a molecule at known (increasing) concentrations, first

Dear R-users, Some hours ago I successfully installed R 1.7.1 on my Linux computer (SuSE 8.1). Now I am trying to get my scripts work which I have written for R/Windows. * While with Windows I was able to use the vowels with the accents present in the Hungarian language: > d$crude.inc<-d$esetsz?m/d$?vk?zepi.l?leksz?m*1000000 > the same line leads to an error on Linux: >

Hi, I have a problem with library (rpart) (and/or library(tree)). I use a data.frame with variables "pnV22" (observation: 1, 0 or yes, no) "JTemp" (mean temperature) "SNied" (summer rain) I used function "rpart" to build a model: library(rpart) attach(data.frame) result <- rpart(pnV22 ~ JTemp + SNied) I got the following tree: n=55518 (50

THANKS so very much for your help (previous and future!). I have a two follow-up questions. 1) You say that dispersion = 1 by definition ....dispersion changes from 1 to 13.5 when I go from binomial to quasibinomial....does this suggest that I should use the binomial? i.e., is the dispersion factor more important that the 2) Is there a cutoff for too much overdispersion - mine seems to be

Hi - How can I 'manually' reproduce the results in 'pred1' below? My attempt is pred_manual, but is not correct. Any help is much appreciated. library(splines) set.seed(12345) y <- rgamma(1000, shape =0.5) age <- rnorm(1000, 45, 10) glm1 <- glm(y ~ ns(age, 4), family=Gamma(link=log)) dd <- data.frame(age = 16:80) mm <- model.matrix( ~ ns(dd$age, 4)) pred1 <-

Hi, May be you can try this: obj_name<- load("arun.RData") Pred1<- get(obj_name[1]) Actual1<- get(obj_name[2]) library(reshape2) dat<-cbind(melt(Pred1,id.vars="S1"),value2=melt(Actual1,id.vars="S1")[,3])? # to reshape to long form colnames(dat)[3:4]<- c("Predict","Actual") dat$variable<- as.character(dat$variable) #not that