similar to: step factor below minimum

This is maybe not directly an R problem but I have used R to try to solve it so I think somebody may be able to help. I have a mixture model with three components and a quadratic Scheffe polynomial p1x1+p2x2+p3x3+p12x1x2+p13x1x3+p23x2x3 fitted to the response. Now I'd like to compute the mixture corresponding the maximum response. Model for Y1 has the parameters p1=124.02 p2=60.973 p3=41.479

I've sometimes thought it would be useful to have another mail list called "no I haven't read the documentation but someone can save me a lot of time," or perhaps just "newbie" for short. This can be very useful for people just getting started in R and with school starting soon I expect the help list may become overwhelmed. Of course, I think that people should read all

Dear list, I'm brand new to R (started using it few days ago...), so sorry for possibly stupid question. Anyways, I'm using R to cluster my data. I do have the dissimilarity matrix as a text file, numbers separated by space. It's at its best something like 2300x2300 matrix. Now, it seems to me, that the process of importing the matrix into R is rather slow. For the peak size of

I  am trying to learn nls using a simple simulation. I assumed that the binomial prob varies linearly as 0.2 + 0.3*x in  x {0,1}, and the objective is to recover the known parameters a=0.2, b=0.3 ..data frame d has 1000 rows... d$x<-runif(0,1)               d$y<-rbinom(1000,1,0.2+0.3*d$x)  table(d$y,cut(d$x,breaks=5));   (-0.000585,0.199] (0.199,0.399] (0.399,0.599] (0.599,0.799]

Is the project on creating R GUIs using QT interfaces still on? Any plans of using PyQT Regards Ajay Ohri Websites- http://decisionstats.com http://dudeofdata.com Linkedin- www.linkedin.com/in/ajayohri On Tue, Nov 9, 2010 at 8:33 PM, Kari Ruohonen <kari.ruohonen at utu.fi> wrote: > Hi, > I wonder if someone could help. I needed to transfer (copy) a workspace > file that

--- samba@samba.org wrote: > SAMBA Digest 2218 > > For information on unsubscribing see > http://samba.org/listproc/ > Topics covered in this issue include: > > 1) > by "Brandon Schnell" <brandonschnell@hotmail.com> > 2) AutoCAD R14 and Eagle Point > by Hank Burton <wburton@literati.com> > 3) Can't locate point and

Hi I'm trying to read SAS-data on VAX/VMS to Windows R using foreign pakage read.xport and experience some problems. Following lines are used in SAS to create XPORT file LIBNAME a ''xxx; LIBNAME b XPORT ''; PROC COPY IN=a OUT=b; RUN; and I succeed in getting file that looks like correct xport file.However when typing in R following

I have a couple of data sets that I want to combine into one data frame. One set contains a number of records on individual observations and includes a geographic descriptor called dacode. The dacode is not unique in that table. The other table contains a number of socio-economic variables for each of the geographic areas identified in the other table. This second table also includes a variable

Dear all, I am trying to fit a non linear regression model to time series data. If I do this: reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal)) I get this error message (translated to English from French): Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = aux[1], : le pas 0.000488281 became inferior to 'minFactor' of 0.000976562 I then tried to set

Dear all, I did a non-linear least square model fit y ~ a * x^b (a) > nls(y ~ a * x^b, start=list(a=1,b=1)) to obtain the coefficients a & b. I did the same with the linearized formula, including a linear model log(y) ~ log(a) + b * log(x) (b) > nls(log10(y) ~ log10(a) + b*log10(x), start=list(a=1,b=1)) (c) > lm(log10(y) ~ log10(x)) I expected coefficient b to be identical

Hi I want to change a control parameter for an nls () as I am getting an error message "step factor 0.000488281 reduced below 'minFactor' of 0.000976562". Despite all tries, it seems that the control parameter of the nls, does not seem to get handed down to the function itself, or the error message is using a different one. Below system info and an example highlighting the

Hi ! I have been using BMA (bayesian model Averaing) package for modeling purposes, but was faced with a problem of incorporating offset term in the Poisson regression of disease rates. It looks like bic.glm does not accept offset keyword like glm ? Any ways to solve the problem using wt option in BMA? Janne Pitk?niemi -- Department of Public Health P.O.Box 41 (Mannerheimintie 172) 00014

All - I have data: TL age 388 4 418 4 438 4 428 5 539 10 432 4 444 7 421 4 438 4 419 4 463 6 423 4 ... [truncated] and I'm trying to fit a simple Von Bertalanffy growth curve with program: #Creates a Von Bertalanffy growth model VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4, start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE) #Scatterplot of the data plot(TL~age, data=box5.4,

Hi, I have some data I want to fit with a non-linear function using nls, but it won't solve. > regresjon<-nls(lcfu~lN0+log10(1-(1-10^(k*t))^m), data=cfu_data, > start=(list(lN0 = 7.6, k = -0.08, m = 2))) Error in nls(lcfu ~ lN0 + log10(1 - (1 - 10^(k * t))^m), data = cfu_data, : step factor 0.000488281 reduced below 'minFactor' of 0.000976562 Tried to increase minFactor

Is there some way to change the line widths of plot borders? I couldn't find any parameters for that purpose. Atte Tenkanen University of Turku, Finland

Hi, What's wrong here?: > v=c(`-`,`+`,1,`^`,`^`,NA,NA,"X",9,"X",2) > i2=16 > v[i2] [[1]] NULL > is.null(v[i2]) [1] FALSE Is it a bug or have I misunderstood something? Atte Tenkanen University of Turku, Finland

Is there any possibility to divide too long text in a plot to two or more lines, when using labels-parameter in the text()-command? Here is an example picture: http://users.utu.fi/attenka/253.jpeg My example script is something like this: text(1,0.7,labels=Chordnames[fnid(pcs%%12)]) # according to Larry Solomon's table http://solomonsmusic.net/pcsets.htm Chordnames is a long vector with

Hi, How do you get the most common row from a matrix? If I have a matrix like this array(1:3,dim=c(4,5)) [,1] [,2] [,3] [,4] [,5] [1,] 1 2 3 1 2 [2,] 2 3 1 2 3 [3,] 3 1 2 3 1 [4,] 1 2 3 1 2 in which rows 1 and 4 are similar, I want to find that vector c (1,2,3,1,2). Atte Tenkanen University of Turku, Finland

Hi, Can somebody tell me, which is the fastest way to make comparisons between all rows in a matrix (here A) and put the results to the new symmetric matrix? I have here used cosine distance as an example, but the comparison function can be any other, euclidean dist etc. A=rbind(c(2,3),c(4,5),c(-1,2),c(5,6)) M=matrix(nrow=length(A[,1]),ncol=length(A[,1])) for(i in 1:length(A[,1])) { for(j in

I'd like to pick every imbricated five character long subsets from a vector. I guess there is some efficient way to do this without loops... Here is a for-loop-version and a model for output: VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6); ADDRESSES=c(); for(i in 1:(length(VECTOR)-4)){ ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse="") } > ADDRESSES [1] "14265"