similar to: Multiple comparisons with a glm

Dear R experts, Did someone implemented French Curve yet? Or can anyone point me some papers that I can follow to implement it? thanks in advance for your help. Paul

What does 'with ties in' mean? with some identical elements (par ex., au moins une paire ex-equo) HTH ____________________ Ken Knoblauch Inserm U371, Cerveau et Vision Department of Cognitive Neurosciences 18 avenue du Doyen Lepine 69675 Bron cedex France tel: +33 (0)4 72 91 34 77 fax: +33 (0)4 72 91 34 61 portable: 06 84 10 64 10 http://www.lyon.inserm.fr/371/

Hi, I'm trying to calculate the value of the variable, dp, below, in the argument to the integral of dnorm(x-dp) * pnorm(x)^(m-1). This corresponds to the estimate of the sensitivity of an observer in an m-alternative forced choice experiment, given the probability of a correct response, Pc, a Gaussian assumption for the noise and no bias. The function that I wrote below gives me an error:

Your response nicely clarifies a question that I've had for a long time, but which I've dealt with by giving each subject a unique label. Unless I'm missing something, both techniques should work as the toy example below gives exactly the same output in all 3 cases below (forgetting about the convergence problem). Would there be a reason to prefer labeling the levels one way or

Can anyone explain to me why the AIC values are so different when using glm.nb and glm with a negative.binomial family, from the MASS library? I'm using R 1.8.1 with Mac 0S 10.3.4. >library(MASS) > dfr <- data.frame(c=rnbinom(100,size=2,mu=rep(c(10,20,100,1000),rep(25,4))), + f=factor(rep(seq(1,4),rep(25,4)))) > AIC(nb1 <- glm.nb(c~f, data=dfr)) [1] 1047 >

Sorry, some things got garbled in that request for help that might make my request unintelligible, probably it being too late at night when I posted, here is the corrected function call for which I would like to learn how to add horizontal grid lines. Any suggestions would be appreciated. Apologies for the posting error xyplot(est ~ STE, data = all, groups = AREA, type = "b",

Hi, I'm fitting a standard nonlinear model to the luminances measured from the red, green and blue guns of a TV display, using nls. The call is: dd.nls <- nls(Lum ~ Blev + beta[Gun] * GL^gamm, data = dd, start = st) where st was initally estimated using optim() st $Blev [1] -0.06551802 $beta [1] 1.509686e-05 4.555250e-05 7.322720e-06 $gamm [1] 2.511870 This works fine but I

Hi! Is it possible to easily cut & paste data from an Excel spreadsheet to an R edit( ) grid or to variable? It seems that R cannot handle the cell delimiters Excel hands over. Regards, Werner

I'm struggling to find a way to add a line (abline, lline or lmline) to a single panel of a lattice plot. In one panel, there is an outlier and I'd like to include a second lmline fit to the data without this point. I've looked through nine pages of the archives on lattice and it wasn't obvious (to me) what additional search keyword to add to hone in on this, although I would

Hello Anyone that now of any function in R that can calculate copulas? Or if anyone have any code avaible I would be more than interested. Thank you in advance /Thomas ______________________________________________ R-help at stat.math.ethz.ch mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Try this: > ?convolve > x<-rnorm(1000) > y<-rep(1,20) > z<-convolve(x,y,type="filter") > plot(x,type="l") > str(z) num [1:981] 6.31 7.28 8.16 7.39 4.65 ... > lines(c(rep(0,10),z,rep(0,10)),col="yellow",lwd=3) > lines(c(rep(0,10),z,rep(0,10))/length(y),col="red",lwd=3) #running mean You wrote: Does anyone know an easy way

I've been struggling with this for a while and cannot get it to work. Is there a simple way to add grid lines ---- I'd like to only add the horizontal ones --- when using the groups argument in xyplot? I can get the grid lines to plot, or the points/lines, but not both. Here is a simple example of the sort of thing that I'm working with. Thanks, in advance. >all est STE

Dear List Members, I need a bit help about fitting some theoretical distributions (such as geometric, exponential, lognormal or weibull distribution) to the following *dry spell*, *wet spell*, *cycles (Wet-Dry or Dry-Wet)* from my meteorological (daily rainfall) data http://www.angelfire.com/ab5/get5/R.rainfall.txt only for rainy seasen (july - september) of 14 years only:

Hi, I would to get the first letter of a word like: > title_cat TitleCat 1 Training I would like T from Training! Thnaks a lot for your help Sabine --------------------------------- T侀l侀chargez le ici ! [[alternative HTML version deleted]]

Dear R users, I want to generate two random variables (X1, X2) from uniform distribution (-0.5, 0.5) with a specified correlation coefficient r. Does anyone know how to do it in R? Many thanks! Menghui

# R for Windows will not send your bug report automatically. # Please copy the bug report (after finishing it) to # your favorite email program and send it to # # r-bugs at r-project.org # ###################################################### This report is joint from Richard Heiberger <rmh at temple.edu> and Burt Holland <bholland at temple.edu>. Burt Holland is the coauthor

hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for "holm", "hochberg", bonferroni", but not "Tukey". Is it not possbile to get adjusted p-values after

Hi, When I compile the construction \begin{Scode}{eval=FALSE} ?HSP \end{Scode} with Sweave and latex, it outputs in the pdf as, > `?` (HSP) which is not incorrect but a bit more formal than I wanted for demonstrating the use of the help shortcut. I would like the output to look like, > ?HSP but I can't seem to make this work. I have also tried the results=verbatim argument.

I'm using scan in a script to record a series of responses of the user as a function of some graphs that I put up on the screen. A toy version would be, y <- rep(NA, 3) for (ix in seq( length(y) ) ) { y[ix] <- scan( n = 1 ) } However, if I include any code after this loop, for example, y <- rep(NA, 3) for (ix in seq( length(y) ) ) { y[ix] <- scan( n = 1 ) } y I get an error

Hi, I was taken off guard by the following behavior in a lattice plot. I frequently want to add a predicted curve defined at more points than in the formula expression of xyplot. There have been numerous examples of how to do this on r-help, but I still often struggle to make this work. I just realized that specifying one of the axes on a log scale does not guarantee that the added data for a