similar to: strange `nls' behaviour

I initially thought, this should better be posted to r-devel but alas! no response. so I try it here. sory for the lengthy explanation but it seems unavoidable. to quickly see the problem simply copy the litte example below and execute f(n=5) which crashes. called with n != 5 (and of course n>3 since there are 3 parameters in the model...) everything is as it should be. in detail:

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

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)?

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

Hi, I guess you meant this: dat2<- read.table(text=" patient_id????? t???????? scores 1????????????????????? 0??????????????? 1.6 1????????????????????? 1??????????????? 2.6 1????????????????????? 2???????????????? 2.2 1????????????????????? 3???????????????? 1.8 2????????????????????? 0????????????????? 2.3 2?????????????????????? 2???????????????? 2.5 2?????????????????????

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"

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

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????????????????

I noticed Doug only circulated the following on S-news, but it may be of interest to R users who don't follow S-news. His findings are certainly consistent with my own, where my problems often force me into the element by element type of situation. Paul Gilbert _______ At the risk of beating this example to death, I went back and compared the execution time for the element-by-element method

Dear R-help, I'd like ask for your opinion on choosing the "right" strategy for a particular dataset. We conducted 24-hour electric field measurements on 90 subjects. They are grouped by job (2 categories) and location (3 categories). There are four exposure metrics assigned to each subject. An excerpt from the data: n job location M OA UE all 0 job1 dist_200 0.297 0.072 0.171

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 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, Using c11<- 0.01 c12<- 0.01 c1<- 0.10 c2<- 0.10 One possible problem is that: dim(res5) #[1] 513? 20 res6<-aggregate(.~m1+n1+m+n,data=res5[,c(1:6,9:12,21:24)] ,max) #Error in `[.data.frame`(res5, , c(1:6, 9:12, 21:24)) : ?# undefined columns selected A.K. ________________________________ From: Joanna Zhang <zjoanna2013 at gmail.com> To: arun <smartpink111 at

Hello I'm trying to solve two sets of equations (each set has four equations and all of them share common parameters) with nlminb procedure. I minimize one set and use their parameters as initial values of other set, repeating this until their parameters become very close to each other. I have several parameters (say,param1, param2) and their constraints are given as inequality and depend

If you wanted to do a t.test res1<-do.call(cbind,lapply(seq_len(nrow(m)),function(i) do.call(rbind,lapply(split(rbind(m[i,-1],n),1:nrow(rbind(m[i,-1],n))), function(x) {x1<- rbind(x,m[i,-1]); t.test(x1[1,],x1[2,])$p.value})))) ?res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i) c(c(tail(res1[seq(1,i,1),i],-1),1),res1[-c(1:i),i]))) ?attr(res2,"dimnames")<-NULL ?res2

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 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

Greetings. I try to set up cross subnet browsing but I've encountered some difficulties. Let me introduce my situation 10.1.2.0 10.1.1.0 A_N1 -+ +- A_N1 | | B_N1 -+--[R1]=======[R2]--+- B_N2 . | VPN accross | . . | the Internet | . Z_N1 -+ +- Z_N A_N1 is 10.1.2.4, PDC A_N2 is 10.1.1.1, BDC

Hi, I'm trying to use pgamma() within my C application but I'm getting strange results. I have a Debian system with r-mathlib 1.4.1-1 and gcc 2:2.95.4-9. Here are 3 example pgamma() calls executed by R: $ R -q --vanilla < gamma.R > pgamma(98, 100, 1) [1] 0.4333105 > pgamma(21, 100, 5) [1] 0.7002453 > pgamma(2, 100, 69) [1] 0.9997046 > and here is what the small C