search for: ssfpl

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851...

Hi, I was hoping to get some advice on how to derive estimates of slopes from four parameter logistic models fit with SSfpl. I fit the model using: model<-nls(temp~SSfpl(time,a,b,c,d)) summary(model) I am interested in the values of the lower and upper asymptotes (parameters a and b), but also in the gradient of the line at the inflection point (c) which I assume tells me my rate of increase when it is steepest (?...

...ot;plinear"))) which assumes an initial value of lscal equal to 0. If lscal is very different to 0, the evaluation could fail (singular gradient,...), as it is the case with the dataset provided hereunder (see end of this message). As a workaround, I propose a modified self-start function for SSfpl that remedy to this: --------------------------------------- fpl <- function (input, A, B, xmid, scal) { .expr1 <- B - A .expr2 <- xmid - input .expr4 <- exp((.expr2/scal)) .expr5 <- 1 + .expr4 .expr8 <- 1/.expr5 .expr13 <- .expr5^2 .value <- A +...

Hi I am struggling to fit a non-linear trend using the likfit function in geoR. Specifically I want a sigmoidal function, something like SSfpl in the nls package to fit the trend. But it seems trend.spatial in geoR only works with lm or glm type models. Any ideas how I can specify the model to calculate the kriging parameters using REML, including the parameters of a sigmoidal trend function (such as SSfpl)? Thanks David

...xp((theta3-x)/theta4)) == y(x)=A+(B-A)/(1+exp((xmid-input)/scal)) from the graph in page 518 of the book of the same authors (mixed models in S) theta 1 corresponds to the horizontal asymptote as x goes to infinity and theta2 the horizontal asymptote as x goes to -infinity. When I use the function SSfpl(input,A,B,xmid,scal), I'm not sure why the value of A is the lower of the two asymptotes if according to the original function A should be equal to theta1 (upper asymptote).. or maybe I'm wrong. I'll appreciate any comment on this. Best Regards CM

...duals (using R v. 2.3.1). Yet, if I want to obtain the parameters for each individual (i.e., for each 'id') using nlsList, then I obtain an Error message which I have trouble interpreting. Any advice as to how I can solve this problem? Thanks for your time, Marc > reg <-nls(mass ~ SSfpl(age, A, B, xmid, scal), + na.action=na.omit, + data=data) > summary(reg) Formula: mass ~ SSfpl(age, A, B, xmid, scal) Parameters: Estimate Std. Error t value Pr(>|t|) A 1.42895 0.24472 5.839 5.43e-09 *** B 21.71508 0.06551 331.495 < 2e-16 *** xmid 5.93563 0.0...

...tarting values that don't result in singular gradients? I have had no luck with the "selfStart" models, e.g., "SSgompertz" -- the formula in "SSgompertz" is not the same as the one I need above, since it has three parameters instead of four. I've tried it and SSfpl thusly: > getInitial(y ~ SSfpl(x,a,b,c,d),data=data.frame(x=x,y=y)) 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 And this: > getInitial(y ~ SSgompertz(x,a,b,c),data=data.frame(x...

...t is possible to specify weights, possibly that is the way to do it, but I didn't manage to get it working the weights should then be something like: SUM (log(1+(resid(.)/quantile(all_residuals,0.68))^2)) / SUM (resid(.)) the test data I use : x<-seq(-5,-2,length=50) x<-rep(x,4) y<-SSfpl(x,0,100,-3.5,1) y<-y+rnorm(length(y),sd=5) y[sample(1:length(y),floor(length(y)/50))]<-200 # add 2% outliers at 200 thanks a lot Hans Vermeiren [[alternative HTML version deleted]]

HI I'm a student in chemical engineering, and i have to implement an algoritm about FIVE PARAMETERS INTERPOLATION for a calibration curve (dose, optical density) y = a + (c - a) /(1+ e[-b(x-m]) where x = ln(analyte dose + 1) y = the optical absorbance data a = the curves top asymptote b = the slope of the curve c = the curves bottom asymptote m = the curve X intercept Have you never seen

...kage but the self-starting functions are not in the derivitive tables. When I spellout the functions and supply starting values I repeatedly get an error message regardless of the starting values I use (see output below). Does anyone have a solution for this problem? > oh_fit2<-nls(MBB ~ SSfpl(AGE, a, b, c, d)) > oh_fit2 Nonlinear regression model model: MBB ~ SSfpl(AGE, a, b, c, d) data: parent.frame() a b c d 9.308 36.420 67.982 12.815 residual sum-of-squares: 46834 Number of iterations to convergence: 9 Achieved convergence tolerance: 8.727e-06 &...

...r of individuals using a 4-parameter logistic growth curve. 2. nlme does not converge with the random structure that I want to use. 3. nlsList does not converge for some individuals. 4. I decided to go around nlsList using: t(sapply(split(data, list(data$id)), function(subd){coef(nls(mass ~ SSfpl(age, A, B, xmid, scal), data = subd))})) 5. This does not converge either: 'Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, : singular gradient' 6. Would anyone know if I can get R to skip non-converging use of nls() so that I can at least obtain the par...

Dear all I am struggling to fit data which form something like CDF by lognorm. Here are my data: proc <- c(0.9, 0.84, 0.5, 0.16, 0.1) size <- c(0.144, 0.172, 0.272, 0.481, 0.583) plot(size, proc, xlim=c(0,1), ylim=c(0,1)) fit<-nls(proc~SSfpl(size, 1, 0, xmid, scal), start=list(xmid=0.2, scal=.1)) lines(seq(0,1,.01), predict(fit, newdata=data.frame(sito=seq(0,1,.01))), col=2) I tried to fit with SSfpl, which is let say close but probably not enough. I tried to find how to fit lognorm CDF to my data but did not find an easy solution. B...

...try to repeat 8.19 fig which is OK until I try: fit4 <- update(fit2, fixed = list(A+B~1,xmid~vykon.fac, scal~1), start = c(57, 100, 700, rep(0,3), 13)) I know I should provide an example but maybe somebody will be clever enough to point me to an explanation without it. nlme version 3.1-75 SSfpl model R 2.4.0dev (but is the same in 2.3.1), W2000. Thank you Best regards. Petr PikalPetr Pikal petr.pikal at precheza.cz

...> wrote: >Dear all > >I am struggling to fit data which form something like CDF by lognorm. > >Here are my data: > >proc <- c(0.9, 0.84, 0.5, 0.16, 0.1) >size <- c(0.144, 0.172, 0.272, 0.481, 0.583) >plot(size, proc, xlim=c(0,1), ylim=c(0,1)) >fit<-nls(proc~SSfpl(size, 1, 0, xmid, scal), start=list(xmid=0.2, >scal=.1)) >lines(seq(0,1,.01), predict(fit, >newdata=data.frame(sito=seq(0,1,.01))), col=2) > >I tried to fit with SSfpl, which is let say close but probably not >enough. I tried to find how to fit lognorm CDF to my data but did not &...

Hi, I am trying to fit a 4-parameter logistic model to my gradient data using nls. I tried to specify the model directly in the nls formula and also tried to use the self-start function SSfpl. For the following data, the first method worked, but the second didn't. I thought both ways were equivalent, can anyone tells me why? > test=data.frame(cbind(conc=c(13294,3940,1170,346,102,30.20,8.94,2.65,13294,3940,1170,346,102,30.20,8.94,2.65), signal=c(2609,487,110,35,17.5,16,11,12.5,26...

...) > data2<-cbind(c(8.05,6.50,5.03,4.37,4.03,3.92,3.87,3.89,7.84,6.27,4.74,4.14,3.76,3.69,3.69,3.71),c(8.07,6.94,5.59,4.43,3.66,3.00,2.64,2.40,8.09,6.90,5.56,4.44,3.50,2.71,2.48,2.08)) > par(mfrow=c(2,2)) > for (i in 1:2) { conc<-data1[,i] signal<-data2[,i] fit<-nls(signal~SSfpl(conc,A,B,xmid,scal)) p.conc<-data.frame(conc=(1:99)*(max(conc)-min(conc))/100+min(conc)) plot(as.numeric(p.conc[,1]),as.numeric(predict(fit,newdata=p.conc)),type='l',col=4) } When the above code was run in R, it worked very well, but when it was run in S-plus, it gave me the following...

...0-x[5])))^(1/x[3]))+rnorm(240,0,s derror) return(Y)})) params contains different parameters for 200 persons in 4 groups. Next I sample for each person five observations one at the start, one at the end and three randomly in between. Now I get the following error messages: testnlslist2<-nlsList(SSfpl,groupedData(Y~Time|group, datset)) Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, : step factor 0.000488281 reduced below 'minFactor' of 0.000976562 Error in nls(formula = formula, data = data, control = control) : singular gradient For the fir...

...an then be used to test parallelism using either Chi-square or F test. Now, fitting 2 separate curves are straight forward using nls(), but how to fit a global model to only allow "xmid" to be different among the 2 curves? BTW, is there a selfStart function for 5-parameter logistic like SSfpl for 4-parameter logistic function? Thanks very much, John

...8.74608021735751, 7.58120982619635, 6.361302477573 , 5.91889385427315, 5.63835466933375 , 5.80211837537706, 5.64897423816121, 8.6195692580331 , 8.70367275835886) ) i <- 1 while(i < 500) { # I usually hit an error within 50 iterations cat(i, "\n") nlrq(y ~ SSfpl(x, A, B, xmid, scal), data = test) i <- i + 1 } ### Errors occur with version 1.8.1 and 1.9.0 on both Windows (two different machines) and UNIX, but not on version 2.0 on a MAC (these are the only R version - OS permutations I was able to get reports on easily). Anyone understand what is h...

Hello, I'm using the nls function and would like to increase the number of iterations. According to the documentation as well as other postings on R-help, I've tried to do this using the "control" argument: nls(y ~ SSfpl(x, A, B, xmid, scal), data=my.data, control=nls.control(maxiter=200)) but no matter how much I increase "maxiter", I get the following error message: Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, : number of iterations exceeded maximum of 50 The...