similar to: nls.control( ) has no influence on nls( ) !

Dear friends. I use nls() and encounter the following puzzling problem: I have a function f(a,b,c,x), I have a data vector of x and a vectory y of realized value of f. Case1 I tried to estimate c with (a=0.3, b=0.5) fixed: nls(y~f(a,b,c,x), control=list(maxiter = 100000, minFactor=0.5 ^2048),start=list(c=0.5)). The error message is: "number of iterations exceeded maximum of

Dear Colleagues. I believe this should be a problem encountered by many: nls( ) is a very useful and efficient function to use if we are just to display the estimated value on screen. What if we need R to store the estimated parameter in a variable? For example: x=rnorm(10, mean=1000, sd=10) y=x^2+100+rnorm(10) a=nls(y~(x^2+para),control=list(maxiter = 1000, minFactor=0.5

Hi I?m trying to fit a nonlinear model to a derivative of the logistic function y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls) The derivative calculated with D function is: > logis<- expression(a/(1+exp((b-x)/c))) > D(logis, "x") a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2 So I enter this expression in the nls function:

Hello colleagues, I am attempting to determine the nonlinear least-squares estimates of the nonlinear model parameters using nls. I have come across a common problem that R users have reported when I attempt to fit a particular 3-parameter nonlinear function to my dataset: Error in nls(r ~ tlm(a, N.fix, k, theta), data = tlm.data, start = list(a = a.st, : step factor 0.000488281

Dear All, A couple of questions about the nls package. 1. I'm trying to run a nonlinear least squares regression but the routine gives me the following error message: step factor 0.000488281 reduced below `minFactor' of 0.000976563 even though I previously wrote the following command: nls.control(minFactor = 1/4096), which should set the minFactor to a lower level than the default

Hello, I've seen various threads on people reporting: step factor 0.000488281 reduced below `minFactor' of 0.000976563 While I know how to set the minFactor, what I'd like to have happen is for nls to return to me, the last or closest fitted parameters before it errors out. In other words, so I don't get convergence, I'd still like to acquire the values of the parameters

Subject: nls, step factor 0.000488281 reduced below 'minFactor' of 0.000976563 Hi there, I'm trying to conduct nls regression using roughly the below code: nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1)) I checked my start values by plotting the relationship etc. but I kept getting an error message saying maximum iterations exceeded. I have tried changing these

Dear friends. Every loop of my program will result in a list that is very long, with a structure similar to the one below: Lst <- list(name="Fred", wife="Mary", daily.incomes=c(1:850)) Please notice the large size of "daily.incomes". I need to store all such lists in a csv file so that I can easily view them in Excel. Excel cannot display a row of more than 300

When I run gnls I get the error: Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, : step factor 0.000488281 reduced below 'minFactor' of 0.000976563 My first thought was to decrease minFactor but gnlsControl does not contain minFactor nor nlsMinFactor (see below). It does however contain nlsMaxIter and nlsTol which I assume are the analogs of

Dear Friends. I have a very long expression and I use function D to find its derivative, which will be even longer. I save the resulting expression in a variavle, say bbb. But when I tried to display bbb on the screen, the R screen is not long enough for me to view it. Is there a way to save the expression to a file? Best Wishes Yuchen Luo [[alternative HTML version deleted]]

Hello. I am trying to do a non-linear fit using the 'nls' command. The data that I'm using is as follows pH k 1 3.79 34.21 2 4.14 25.85 3 4.38 20.45 4 4.57 15.61 5 4.74 12.42 6 4.92 9.64 7 5.11 7.30 8 5.35 5.15 9 5.67 3.24 with a transformation of pH to H <- 10^-pH When using the nls command for a set of parameters - a, b and c, I receive two sets of errors: >

Dear Friends. I found a puzzling phenomenon in R when you use 'if' within a function: # defining a function aaa aaa=function(a) {if (a==1) {aaa=1}; if (a!=1) {aaa=2} } # using the function: > b=20 > bbb=aaa(b) > bbb [1] 2 > typeof(bbb) [1] "double" > > > c=1 > ccc=aaa(c) > ccc NULL > typeof(ccc) [1] "NULL" It seems that only the last

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

I'm trying to fit a Gompertz sigmoid as follows: x <- c(15, 16, 17, 18, 19) # arbitrary example data here; y <- c(0.1, 1.8, 2.2, 2.6, 2.9) # actual data is similar gm <- nls(y ~ a+b*exp(-exp(-c*(x-d))), start=c(a=?, b=?, c=?, d=?)) I have been unable to properly set the starting value '?'s. All of my guesses yield either a "singular gradient" error if they

Dear Friends. Greetings! This should be a very common operation and I believe there should be a nice way in R to handle it. I couldn't find it in the manual or by searching on line. I am wondering if I could ask for some help in this community. I am trying to record the results of my program to a csv file in the hard drive so as to save memory space and also to read the results in excel

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 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

Hello all, This is the error message that I get. > hyp.res <- nls(log(y)~log(pdf.hyperb(theta,X)), data=dataModel, + start=list(theta=thetaE0), + trace=TRUE) 45.54325 : 0.1000000 1.3862944 -4.5577142 0.0005503 3.728302 : 0.0583857346 0.4757772859 -4.9156128701 0.0005563154 1.584317 : 0.0194149477 0.3444648833 -4.9365149150 0.0004105426 1.569333 :

I am using a "for" loop to read a table row by row and I have to specify how many records are there in the table. I need to read row by row because the table is huge and the memory not large enough for the whole table.: number.of.records=100 fp=file("abc.csv","r") pos=seek(fp, rw="read") for (i in 1:number.of.record){ current.row=scan(file=fp,

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