similar to: 4PL algorithm

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 0.5000000 4 20.000 2.9957323

Hello, nls library provides 6 self-starting models, among them: SSfp, a four parameters logistic function. Its self-starting procedure involves several steps. One of these steps is: pars <- as.vector(coef(nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xydata, start = list(lscal = 0), algorithm = "plinear"))) which assumes an initial value of lscal equal to 0. If lscal

Thank you Martin. If I understand correctly, OP could do wheat.list <- nlsList(Prop ~ SSfpl(end, A, B, xmid, scal), data=wlg) or add some small value to all zeroes wlg$prop < -wlg$Prop+1e-7 wheat.list <- nlsList(prop ~ SSlogis(end,Asym, xmid, scal), data=wlg) which gives fairly reasonable results. plot(augPred(wheat.list)) Am I correct? Cheers Petr > -----Original Message-----

Full_Name: Madeleine Yeh Version: 1.9.1 OS: AIX 5.2 Submission from: (NULL) (151.121.225.1) After compiling R-1.9.1 on AIX 5.2 using the IBM cc compiler, I ran the checks. One of them failed. Here is the output from running the check solo. root@svweb:/fsapps/test/build/R/1.9.1/R-1.9.1/tests/Examples: ># ../../bin/R --vanilla < stats-Ex.R R : Copyright 2004, The R

>>>>> PIKAL Petr <petr.pikal at precheza.cz> >>>>> on Fri, 20 Oct 2017 06:33:36 +0000 writes: > Hi > Keep your messages in the list, you increase your chance to get some answer. > I changed your data to groupedData object (see below), but I did not find any problem in it. > plot(wlg) > gives reasonable picture and I am

Hi R help, I am trying to determine how nls() generates a function based on the self-starting SSlogis and what the formula for the function would be. I've scoured the help site, and other literature to try and figure this out but I still am unsure if I am correct in what I am coming up with. ************************************************************************** dat <-

Folks I have modified an existing function to calculate 'ec/ld/lc' 50 values and their associated Fieller's confidence limits. It is based on EC50.calc (writtien by John Bailer) - but also borrows from the dose.p (MASS) function. My goal was to make the original EC50.calc function flexible with respect to 1) probability at which to calculate the expected dose, and 2) the link

Hello, I have this data: Time AMP 0 0.2000000 10 0.1958350 20 0.2914560 40 0.6763628 60 0.8494534 90 0.9874526 120 1.0477692 where AMP is the concentration of this metabolite with time. If you plot the data, you can see that it could be fitted using a logistic regression. For this purpose, I used this code: AMP.nls <- nls(AMP~SSlogis(Time,Asym, xmid, scal), data

Dear list, I am running some linear regressions through lapply, >lapply(c('EMAX','EC50','KOUT','GAMMA'),function(x)confint(lm(get(x)~RR0,dataset2))) I got results like [[1]] 2.5 % 97.5 % (Intercept) 0.6595789212 0.8821691261 RR0 -0.0001801771 0.0001489083 [[2]] 2.5 % 97.5 % (Intercept) -63.83694930

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? >

Dear R users I'm a new user of R and I have a basic question about the 4-parameter logistic model. According to the information from Pinheiro & Bates the model is: y(x)=theta1+(theta2-theta1)/(1+exp((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

Hi, In Splus7 this statement xlrmN1 <- sample(c(0,1,2),400 ,prob=c(0.02 ,0.93 ,0.05 )) worked fine, but in R the interpreter reports that the length of the vector to chose c(0,1,2) is shorter than the size of many times I want to be selected from the vector c(0,1,2). Any good reason? See below the error. > xlrmN1 <- sample(c(0,1,2),400 ,prob=c(0.02 ,0.93 ,0.05 )) Error in

Hi I am having some problems running a covariate analysis with my colleage using R with the NonMem program we are using for a graduate school project. R and NonMem run fine without adding in the covariates, but the program is giving us a problem when the covariate analysis is added. We think the problem is with the R code to run the covariate data analysis. We have the control stream, R code

The error msg is telling you that R cannot evaluate the loss function, so you should not expect answers. You might try examining the data -- Are there NA or Inf entries? Or prepare a dataframe with just X and Y, sort by X and graph. Then check the nls computations by sampling, say, every 100 X's to give you a dataset with about 160 observations. If that doesn't work, it is at least

I have uploaded a datafile that contains the following two variables: time (X value) and response (Y value). This is a fairly extensive file (with > 16000 entries). I have two questions: 1. I want to use the following equation to regress Y on X: Y-hat = min + (max-min)/(1 + (X/EC50)^Hillslope). Here is my R command: nlsout <- nls(Y ~ (0 - (100-0)/(1 + (X/EC50)^hill)), start=c(EC50=125,

I am wondering how to run 'garch' function of 'tseries' package in R2.6.1. I installed R2.3.1 and R2.6.1 in my PC (Windows XP Home) and run a following simple GARCH function in both versions: >garch(dSP[1:300], order = c(1,1)) where 'dSP' is daily return series of a stock index. R2.6.1 can not finish calculation and also I can not stop the

Hello, We use drc to fit dose-response curves, recently we discovered that there are quite different standard error values returned for the same dataset depending on the drc-version / R-version that was used (not clear which factor is important) On R 2.9.0 using drc_1.6-3 we get an IC50 of 1.27447 and a standard error on the IC50 of 0.43540 Whereas on R 2.7.0 using drc_1.4-2 the IC50 is

Dear all, I want to do a F test, which involves calculation of the degrees of freedom for the residuals. Now say, I have a nlme object "mod.nlme". I have two questions 1.How do I extract the degrees of freedom? 2.How is this degrees of freedom calculated in an nlme model? Thanks. Jun Shen Some sample code and data =================================================================

Greetings, We are performing a meta-analysis of mink pup survival data versus chemical concentration. We have modeled percent survival successfully using nls as shown below and the plot. What we need to do is construct a confidence interval on the concentration at which we get 50% survival (aka the EC50, although we may want other percent survivals in the future). My first question is, what seems

Howdee, I am able to fit a 4-parameter logistic growth curve to a dataset which comprise many individuals (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 ~