similar to: strange `nls' behaviour

dear list, I stumbled over the follwing strange behaviour/error when using `nls' which I'm tempted (despite the implied "dangers") to call a bug: I've written a driver for `nls' which allows specifying the model and the data vectors using arbitrary symbols. these are internally mapped to consistent names, which poses a slight complication when using `deriv' to

Hi there, I have some data which are convenient to enter as lists. For example: t1<-list(fname="animal1",testname="hyla",dspkr="left",res1=39.7,res2=15.0) t2<-list(fname="animal1",testname="bufo",dspkr="left",res1=14.4,res2=56.1)

Hi R helpers! I have a question. I'm trying to create a function for an exercise. Here are the arguments I should include: x and y are numeric z is a name ("plus","minus","multiply","divide") and swap is logical. Here is what the function should do: When z="plus", then x+y is performed and so on for the other z names. It should give a NA

my understanding of `eval' behaviour is obviously incomplete. my question: usually `eval(expr)' and `eval(expr, envir=parent.frame())' should be identical. why does the last `eval' (yielding `r3') in this code _not_ evaluate in the local environment of function `f' but rather in the global environment (if `f' is called from there)?

Dear R users, I have a data structure as follows: id two res1 res2 c1 c2 inter 1 -0.786093166 1 0 1 2 6 3 -0.308495749 1 0 0 1 2 5 -0.738033048 1 0 0 0 1 7 -0.52176252 1 0

I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"

Hi, dat1<- read.table(text=" id??????????????? t???????????????????? scores 2???????????????? 0??????????????????????? 1.2 2???????????????? 2???????????????????????? 2.3 2???????????????? 3??????????????????????? 3.6 2???????????????? 4??????????????????????? 5.6 2???????????????? 6??????????????????????? 7.8 3???????????????? 0??????????????????????? 1.6 3????????????????

Hi, I am new to clustering and was wondering why pvclust using "maximum" as distance measure nearly always results in p-values above 95%. I wrote an example programme which demonstrates this effect. I uploaded a PDF showing the results Here is the code which produces the PDF file: ------------------------------------------------------------------------------------- s <-

dear list members, my question concerns computation of confidence intervals in nonlinear fits with `nls' when weigthing the fit. the seemingly correct procedure does not work as (I) expected. I'm posting this here since: (A) the problem might suggest a modification to the `m' component in the return argument of `nls' (making this post formally OK for this list) and (B) I got no

Dear R-users, I would like to use optim( ) to minimize a function which depends on 4 parameters: 2 vectors, a scalar, and a matrix. And I have a hard to define the parameters at the beginning of the function, and then to call optim. Indeed, all the examples I have seen dont treat cases where parameters are not all real. Here is my code, it doesnt work but its just to show you where is exactly my

Hi, I use lme to fit models like R> res1 <- lme(y~A+B, data=mydata, random=~1|subject) R> res2 <- lme(y~B+A, data=mydata, random=~1|subject) (only difference between these two models are the sequence in which the indep variables are written in formula) where y is continuous and A, B, and subject are factors. To get ANOVA table I used R> anova(res1) R> anova(res2) and found

Dear R-users, I would like to maximize the function g above which depends on 4 parameters (2 vectors, 1 real number, and 1 matrix) using optim() and BFGS method. Here is my code: # fonction to maximize g=function(x) { x1 = x[1:ncol(X)] x2 = x[(ncol(X)+1)] x3 = matrix(x[(ncol(X)+2):(ncol(X)+1+ncol(X)*ncol(Y))],nrow=ncol(X),ncol=ncol(Y)) x4 = x[(ncol(X)+1+ncol(X)*ncol(Y)+1):length(x)]

Dear R-users I have a problem choosing initial points for the function arms() in the package HI I intend to implement a Gibbs sampler and one of my conditional distributions is nonstandard and not logconcave. Therefore I'd like to use arms. But there seem to be a strong influence of the initial point y.start. To show the effect I constructed a demonstration example. It is reproducible

Hi, It would be better if you provided the output of dput(dataset).? I am not sure about the structure of your dataset. Just from reading the data as is shown. dat1<- read.table(text=" separator,tissID >,>,2 ,2,1 ,6,5 ,11,13 >,>,4 ,4,9 ,6,2 ,7,3 ,21,1 ,23,58 ,25,9 ,26,4 >,>,11 ,1,12 >,>,21 ,4,1 ,11,3

Dear R-Users, I want to use mtable from package "memisc" to produce Latex-style estimation output. However, mtable() only gives me a p-value and not the corresponding test-statistic. Does anyone know how to extract it, either from a glm/anova object or mtable? Here is a short example: # Run this #################### install.packages("memisc") library(memisc) set.seed(1)

Dear all I have tried to estimate the confidence intervals for predicted values of a nonlinear model fitted with nls. The function predict gives the predicted values and the lower and upper limits of the prediction, when the class of the object is lm or glm. When the object is derived from nls, the function predict (or predict.nls) gives only the predicted values. The se.fit and interval aguments

Deal all, as MCMClogit does not allow for the specification of several chains, I have run my model 3 times with different random number seeds and differently dispersed multivariate normal priors. For example: res1 = MCMClogit(y~x,b0=0,B0=0.001,data=mydat, burnin=500, mcmc=5500, seed=1234, thin=5) res2 = MCMClogit(y~x,b0=1,B0=0.01,data=mydat, burnin=500, mcmc=5500, seed=5678, thin=5) res3 =

Hi, I am trying to run the command "forward.sel.par," however I receive the error message: "Error: could not find function 'simpleRDA2'." I have the vegan library loaded. The documentation on "varpart" has not helped me to understand why I cannot call this function. Maybe I am missing something obvious because I am still an 'R' novice. Below is a

I am trying to run separate regressions for different groups of observations using the lapply function. It works fine when I write the formula inside the lm() function. But I would like to pass formulae into lm(), so I can do multiple models more easily. I got an error message when I tried to do that. Here is my sample code: #generating data x1 <- rnorm(100,1) x2 <- rnorm(100,1) y <-

What is going on here? In the lines ending in #### the inputs and outputs are identical yet one gives a warning and the other does not. a1 <- `rownames<-`(anscombe[1:3, ], NULL) a2 <- anscombe[1:3, ] ix <- 5:8 # input arguments to #### are identical in both cases identical(stack(a1[ix]), stack(a2[ix])) ## [1] TRUE identical(a1[-ix], a2[-ix]) ## [1] TRUE res1 <-