Juanchi - 9 months ago 75

R Question

Using this parametrization for a growth curve logistic model

I created some points with: K =0.7 ; y0=0.01 ; r =0.3

`df = data.frame(x= seq(1, 50, by = 5))`

df$y = 0.7/(1+((0.7-0.01)/0.01)*exp(-0.3*df$x))

Can someone tell me how can I have a fitting error if create the data with the model starters?

`fo = df$y ~ K/(1+((K-y0)/y0)*exp(-r*df$x))`

model<-nls(fo,

start = list(K=0.7, y0=0.01, r=0.3),

df,

nls.control(maxiter = 1000))

Error in nls(fo, start = list(K = 0.7, y0 = 0.01, r = 0.3), df, nls.control(maxiter = 1000)) :

number of iterations exceeded maximum of 1000

Answer Source

**Do not use ‘nls’ on artificial "zero-residual" data.**, as documented in `?nls`

.

```
set.seed(0)
x <- seq(1, 50, by = 5)
y <- 0.7 / (1 + ((0.7 - 0.01) / 0.01) * exp(-0.3 * x))
y <- y + rnorm(length(x), sd = 0.05) ## add Gaussian error!!
dat <- data.frame(x = x, y = y); rm(x, y)
with(dat, plot(x, y))
fit <- nls(y ~ K / (1 + ((K - y0) / y0) * exp(-r * x)), data = dat,
start = list(K = 0.7, y0 = 0.01, r = 0.3))
#Nonlinear regression model
# model: y ~ K/(1 + ((K - y0)/y0) * exp(-r * x))
# data: dat
# K y0 r
#0.70013 0.01841 0.27950
# residual sum-of-squares: 0.02851
#
#Number of iterations to convergence: 12
#Achieved convergence tolerance: 4.145e-06
```

Also, avoid using `$`

in model formula, otherwise you will get into trouble when using `predict`

later.