R nls singular gradient

Question

I\'ve tried searching the other threads on this topic but none of the fixes are working for me. I have the results of a natural experiment and I want to show the number of c

Answer 1

1) linearize to get starting values You need better starting values:

# starting values
fm0 <- nls(log(y) ~ log(f(x, a, b)), dat2, start = c(a = 1, b = 1))

nls(y ~ f(x, a, b), dat2, start = coef(fm0))

giving:

Nonlinear regression model
  model: y ~ f(x, a, b)
   data: x
        a         b 
4214.4228   -0.8106 
 residual sum-of-squares: 2388

Number of iterations to convergence: 6 
Achieved convergence tolerance: 3.363e-06

1a) Similarly we could use lm to get the initial value by writing

y ~ a * exp(b * x)

as

y ~ exp(log(a) + b * x)

and taking logs of both to get a model linear in log(a) and b:

log(y) ~ log(a) + b * x

which can be solved using lm:

fm_lm <- lm(log(y) ~ x, dat2)
st <- list(a = exp(coef(fm_lm)[1]), b = coef(fm_lm)[2])
nls(y ~ f(x, a, b), dat2, start = st)

giving:

Nonlinear regression model
  model: y ~ f(x, a, b)
   data: dat2
       a        b 
4214.423   -0.811 
 residual sum-of-squares: 2388

Number of iterations to convergence: 6 
Achieved convergence tolerance: 3.36e-06

1b) We can also get it to work by reparameterizing. In that case a = 1 and b = 1 will work provided we transform the initial values in line with the parameter transformation.

nls(y ~ exp(loga + b * x), dat2, start = list(loga = log(1), b = 1))

giving:

Nonlinear regression model
  model: y ~ exp(loga + b * x)
   data: dat2
  loga      b 
 8.346 -0.811 
 residual sum-of-squares: 2388

Number of iterations to convergence: 20 
Achieved convergence tolerance: 3.82e-07

so b is as shown and a = exp(loga) = exp(8.346) = 4213.3

2) plinear Another possibility that is even easier is to use alg="plinear" in which case starting values are not needed for the parameters entering linearly. In that case the starting value of b=1 in the question seems sufficient.

nls(y ~ exp(b * x), dat2, start = c(b = 1), alg = "plinear")

giving:

Nonlinear regression model
  model: y ~ exp(b * x)
   data: dat2
        b      .lin 
  -0.8106 4214.4234 
 residual sum-of-squares: 2388

Number of iterations to convergence: 11 
Achieved convergence tolerance: 2.153e-06

Answer 2

Please check nlsLM function in minpack.lm package. This is a more robust version of nls and can handle data with zero residual sum of squares.

https://www.r-bloggers.com/a-better-nls/