search for: nls1

  Hi, I fitted tree growth data with Chapman-Richards growth function using nls. summary(fit.nls) Formula: Parameters: Estimate Std. Error t value Pr Signif. codes: 0 ''***'' 0.001 ''**'' 0.01 ''*'' 0.05 ''.'' 0.1 '' '' 1 Residual standard error: 1.879 on 713 degrees of freedom Algorithm

...res regression with AIC and anova. The simplest model is one nonlinear curve, and in the more complex model I have a categorical variable (producing parameter estimates for four curves). Both models run fine, but when I try to produce an AIC value for the second model I get the error: > AIC(pow.nls1) [1] 114408.3 > AIC(pow.nls2) Error in 1:p : NA/NaN argument And similarly, when I try to compare the models with ANOVA, I get: anova(pow.nls1, pow.nls2) Error in if (df[i] > 0) { : missing value where TRUE/FALSE needed It seemed as if this could be a problem with the dataset, but I have c...

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 start values, and I heeded advice from other threads and ammended my code to i...

Potential bug: I mistyped weights in the call ('weigths') and it did not produce any error= message. The coefs were exactly the same like without weights, so I was su= spicious and when weights(nls1) gave NULL, I saw my typo. Usually the function will say "Unused arguments", which shows you the error= , but not nls. Regards Stephen [[alternative HTML version deleted]]

...wanted to use the nls() module to solve a Problem from Sen & Srivastava (1990, p.209). Here is the (basic) code used to perform the estimation: library(SenSrivastava) library(nls) data(E9.8) # Use Linear Least Square for estimating start values lm.obj <- lm(R.1 ~ I.1 + S.1, data = E9.8) nls1.obj <- nls(R.1 ~ b.0 + b.1*(I.1^a.1-1)/a.1 + b.2*(S.1^a.2-1)/a.2, data = E9.8, start = list(b.0 = coef(lm.obj)[1], b.1 = coef(lm.obj)[2], b.2 = coef(lm.obj)[3], a.1 = 1, a.2 = 1) ) unfortunately, I rece...

I'd like to use linear and quadratic trend analysis in order to find out a change in slope. Basically, I need to solve a similar problem as discussed in http://www.gseis.ucla.edu/courses/ed230bc1/cnotes4/trend1.html My subjects have counted dots: one dot, two dots, etc. up to 9 dots. The reaction time increases with increasing dots. The theory is that 1 up to 3 or 4 points can be counted

...er is as follows: nlmr1<-nlmer(feed~grModel(temp,MW,Tmin,Tmax,Topt,kopt,m)~kopt| flat,data=egi,start=c(Tmin=0.2,Tmax=21.1,Topt=14.6,kopt=2.5,m=.51)) 'flat' is a factor with 11 levels. My start values are from nls that fits the model with no errors. The call I have used is as follows: nls1<-nls(feed~grModel(temp,MW,Tmin,Tmax,Topt,kopt,m),data=egi,start=c(Tmin=1,Tmax=20,Topt=13,kopt=3,m=.7)) The examples given on the help page of nlmer run with no errors. I would appreciate any hints and help to track the problem indicated by the error message. Thanks, Kari