similar to: test if all predictors in a glm object are factors

In http://tolstoy.newcastle.edu.au/R/e2/help/06/10/3064.html a method was given for converting a frequency table to an expanded data frame representing each observation as a set of factors. A slightly modified version was later included in the NCStats package, only on http://rforge.net/ (and it has too many dependencies to be useful). I've tried to make it more general, allowing an input

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates <- as.Date(data.df[,1]) selection<-which(dates>=as.Date("1986-1-1") & dates<=as.Date("2007-12-31")) dep <- ts(data.df[selection,c("dep")]) indep.ret1

Hello, Is there a better way to use paste such as: a = paste(colnames(list.indep)[1],colnames(list.indep)[2],colnames(list.indep)[3],colnames(list.indep)[4],colnames(list.indep)[5],sep="+") > a [1] "aa+dummy1+dummy2+bb+cc" I tried a = paste(colnames(list.indep)[1:5],sep="+") > a [1] "aa" "dummy1" "dummy2"

Hi useRs, I'm experiencing something quite weird with glm() and weights, and maybe someone can explain what I'm doing wrong. I have a dataset where each row represents a single case, and I run glm(...,family="binomial") and get my coefficients. However, some of my cases have the exact same values for predictor variables, so I should be able to aggregate up my data frame and

Hello R community, I have a question about the logistic regression function. Specifically, when the predictor variable has not just 0's and 1's, but also fractional values (between zero and one). I get a warning when I use the "glm(formula = ... , family = binomial(link = "logit"))" which says: "In eval(expr, envir, enclos) : non-integer #successes in a binomial

Hello R users, I am new to R and am wondering if anyone can help me out with the following issue: I wrote a function to build ts models using different inputs, but when R displays the call for a model, I cannot tell which variables it is using because it shows the arguments instead of the real variables passed to the function. (e.g Call: lm(formula = dyn(dep ~ lag(dep, -1) + indep)) --->

I would like to extract the names, modes [numeric/factor] and levels of variables needed in a data frame supplied as newdata= argument to predict.glm() Here is a small example illustrating my troubles; what I want from (both of) the glm objects is the vector c("x","f","Y") and an indication that f is a factor: library( splines ) dd <- data.frame( D =

I am using R on a Windows XP professional platform. The following code is part of a bigger one CODE press=function(y,x){ library(qpcR) models.press=numeric(0) cat("\n") dep=y print(dep) indep=log(x) print(indep) yfit=dep-PRESS(lm(dep~indep))[[2]] cat("\n yfit\n") print(yfit) yfit.orig=yfit presid=y-yfit.orig press=sum(presid^2)

Hi, Some try: > names(mi$xlevels) [1] "f" > all.vars(mi$formula) [1] "D" "x" "f" "Y" > names(mx$xlevels) [1] "f" > all.vars(mx$formula) [1] "D" "x" "f" When offset is indicated out of the formula, it does not work... Marc Le 07/03/2018 ? 06:20, Bendix Carstensen a ?crit?: > I would like

In a function, say foo.glm for glm objects I want to use the name of the object as a label for some output, but *only* if a glm object was passed as an argument, not a call to glm() producing that object. How can I distinguish these two cases? For example, I can use the following to get the name of the argument: foo.glm <- function(object) { oname <- as.character(sys.call())[2]

Could anyone tell me what am I doing wrong? > pro<-function(indep,dep){ + d<-data.frame(indep) + form<-formula(lm(dep~.,data=d)) + forward<-step(lm(dep~X1,data=d),scope=form,trace=0,direction='f') + return(forward) + } > pro(m,q) Error in inherits(x, "data.frame") : Object "d" not found Where q is a vector with the dependent variable's

Default arguments are evaluated in the function frame, not in the calling environment (nor in the same place as explicit arguments). > Which to me reads that a with statement as above is equivalent to > > > attach(data) ; aov.SS1(y=Obs) ; detach(data) > > Or is that just wishful thinking?? The latter. On Thu, 7 Oct 2004, RenE J.V. Bertin wrote: > Hello, > >

Hello R users, I have been trying to output all my results (text, plots, etc) into the same postscript file as one document, but have been unable to...Can anyone help me improve my code below so that I can accomplish this? Currently I have to output them separately then piece them back together into one document.. Thanks in Advance for any help! options (scipen=999, digits=7)

all.vars works fine, EXCEPT, it give a bit too much. I only want the regression variables, but in the following example I also get "k" the variable holding the chosen knots. Any machinery to find only "real" regression variables? cheers, Bendix library( splines ) y <- rnorm(100) x <- rnorm(100) k <- -1:1 ml <- lm( y ~ bs(x,knots=k) ) mg <- glm( y ~

Hello world, I'm trying to do an anova on data in data.set, dependent variable is a column named "dep.var", grouping variable is in a column called "indep.var", and is.factor(indep.var) is TRUE... why can't I just do aov(dep.var ~ indep.var, data = data.set)? What have I done to deserve this?! What gives? Am I missing something totlly obvious? R-base-1.0.0-1,

Below is a SIMEX object that was generated with the "simex" function from the "simex" package applied to a logistic regression fit. From this mountain of information I would like to extract all of the values summarized in this line: .. ..$ variance.jackknife: num [1:5, 1:4] 1.684 1.144 0.85 0.624 0.519 ... Can someone suggest how to go about doing this? I can extract the

Dear R Users, I am trying to plot a matrix (a Digital Elevation Model) using wireframe [lattice] and color that matrix based on a separate/independent matrix of the same resolution (both have the same number of rows and columns) as the DEM. More specifically I would like to plot a DEM using wireframe and color each cell/tile based on interpolated surface temperature measurements. Within the

I finally picked up a "real" modem (not a winmodem) and put it in my machine. I pulled down the hsfmodem driver from the source, got a license and installed the driver at full power. If I use wvdial, I can get out through the modem, so I know that works. However, sendfax is rejecting the fax capability (from /var/log/sendfax.log): 11/30 00:31:52 sendfax: interim release 1.1.33-Apr10

Since yesterday I've been getting the error below when building llvm-gcc on Ubuntu Hardy on x86. For some reason, several instances of autoconf are getting confused and failing to detect a stdlib.h. John /home/regehr/z/tmp/llvm-gcc-r62547-src/build/./prev-gcc/xgcc -B/home/regehr/z/tmp/llvm-gcc-r62547-src/build/./prev-gcc/