Start list of parameters for nls and rational function

Question

I would like to fit a rational function to my data:

data:

 [1] 2.000000 3.000000 2.333333 1.750000 2.000000 1.833333 2.416667  1.916667
 [9] 1.750000 2.166667 2.116667 1.916667 1.944444 1.611111 1.722222 1.777778
[17] 1.877778 1.944444 1.958333 1.833333 2.041667 2.020833 1.908333 1.916667
[25] 1.733333 1.833333 1.800000 1.933333 1.893333 1.866667 1.888889 1.805556
[33] 1.833333 1.847222 1.822222 1.805556 1.833333 1.904762 1.880952 1.833333
[41] 1.804762 1.809524 1.708333 1.708333 1.750000 1.708333 1.683333 1.687500
[49] 1.611111 1.666667 1.648148 1.611111 1.611111 1.611111 1.650000 1.600000
[57] 1.650000 1.625000 1.630000 1.616667 1.469697 1.560606 1.590909 1.651515
[65] 1.651515 1.651515 1.513889 1.555556 1.625000 1.638889 1.647222 1.652778
[73] 1.679487 1.717949 1.705128 1.698718.

The model I would like to fit is the following:

Model <- function(t, a, b, c, d) {
                 (a + b*t)/(1 + c*t + d*t^2)
 }
I know that I firstly have to give a starting list of a, b, c... for nls but I really don't know how to set the parameters. Since I'm not an expert I found in this http://www.css.cornell.edu/faculty/dgr2/teach/R/R_rat.pdf document a useful guide. But at some point it says:

"Given a set of ordered pairs (ti,yi) where in general there are repeated measurements at each value of t, the parameters of a rational function can be fitted by non-linear least-squares estimation, for example with the nls method in R. One we have the four parameters, we can compute the value of t at which this is maximized, by computing the first derivative....".

Although I don't report additional data I have another column that represents time (integers from 1:76 representing years).

Can anyone help me?

Best

E.

What do you mean with "nls"? Non-linear system? And do you mean you want to optimize that function to have good estimates for your parameters? — mickey, Nov 16 '18 at 18:23

G. Grothendieck · Accepted Answer · 2018-11-17T15:12:36.170

The model is not fully specified in the question but assuming the model in the code below and the data shown reproducibly in Note 2 below if we set c = d = 0 then it is a linear model so we can use the coefficients from a linear model fit as starting values:

fm1 <- lm(y ~ t)
st2 <- list(a = coef(fm1)[[1]], b = coef(fm1)[[2]], c = 0, d = 0)
fm2 <- nls(y ~ Model(t, a, b, c, d), start = st2)

giving:

> fm2
Nonlinear regression model
  model: y ~ Model(t, a, b, c, d)
   data: parent.frame()
        a         b         c         d 
2.5097712 0.6038808 0.3205409 0.0008663 
 residual sum-of-squares: 1.684

Number of iterations to convergence: 16 
Achieved convergence tolerance: 8.029e-06

Looking at the fit graphically:

# model is shown in red. See Note 1 for fm4 (blue) and fm5 (green) models.
plot(y ~ t)
lines(fitted(fm2) ~ t, col = "red")
lines(fitted(fm4) ~ t, col = "blue")
lines(fitted(fm5) ~ t, col = "green")
legend("topright", c("fm2", "fm4", "fm5"), col = c("red", "blue", "green"), lty = 1)

Note 1

The following is a different model which fits almost as well but only uses 3 parameters. See blue line on graph above.

fm3 <- lm(log(y) ~ log(t))
st4 <- list(a = coef(fm3)[[1]], b = 0, c = coef(fm3)[[2]])
fm4 <- nls(y ~ exp(a + b/t + c*log(t)), start = st4)

> fm4
Nonlinear regression model
  model: y ~ exp(a + b/t + c * log(t))
   data: parent.frame()
      a       b       c 
 0.9845 -0.1767 -0.1157 
 residual sum-of-squares: 1.685

Number of iterations to convergence: 4 
Achieved convergence tolerance: 2.625e-06

Also this model isn't too bad. It uses only two parameters, it is linear in them and it has a residual sum of squares of 1.728837 compared to 1.684 for the fm2 model and 1.685 for the fm4 model. See green line on graph above.

fm5 <- lm(y ~ log(t))

> fm5

Call:
lm(formula = y ~ log(t))

Coefficients:
(Intercept)       log(t)  
     2.4029      -0.1793  

> deviance(fm5)
[1] 1.728837

Note 2

y <- c(2, 3, 2.333333, 1.75, 2, 1.833333, 2.416667, 1.916667, 1.75, 
2.166667, 2.116667, 1.916667, 1.944444, 1.611111, 1.722222, 1.777778, 
1.877778, 1.944444, 1.958333, 1.833333, 2.041667, 2.020833, 1.908333, 
1.916667, 1.733333, 1.833333, 1.8, 1.933333, 1.893333, 1.866667, 
1.888889, 1.805556, 1.833333, 1.847222, 1.822222, 1.805556, 1.833333, 
1.904762, 1.880952, 1.833333, 1.804762, 1.809524, 1.708333, 1.708333, 
1.75, 1.708333, 1.683333, 1.6875, 1.611111, 1.666667, 1.648148, 
1.611111, 1.611111, 1.611111, 1.65, 1.6, 1.65, 1.625, 1.63, 1.616667, 
1.469697, 1.560606, 1.590909, 1.651515, 1.651515, 1.651515, 1.513889, 
1.555556, 1.625, 1.638889, 1.647222, 1.652778, 1.679487, 1.717949, 
1.705128, 1.698718)

t <- seq_along(y)

I had a similar thought. If b=C=d then using a=mean(y) would make sense too — Dason, Nov 16 '18 at 19:19
Thank you guys for your precious help. It is what I was looking for. @Grothendieck I don't know why but after running the lines of your answer it gives the following error after fm2: Error in data[, nm[i]] : incorrect number of dimensions. I don't know what is wrong. I used the y vector you use and the same with t. — NewUsr_stat, Nov 16 '18 at 20:05
God I made a mistake because I confounded the "Model" I defined with the "Model" function...a naive mistake. I apologize. Thank you again for your precious help! — NewUsr_stat, Nov 16 '18 at 20:21

score 0 · Answer 2 · answered Nov 16 '18 at 18:46

I believe you can do something like

nls(y ~ (a1 + b1*times) / (1 + c1*times + d1*times^2))

where y is the data you gave above and times=1:76. I added a 1 to the parameters, because nls did not recognize c as a parameter, but rather as the function c().

However, when I ran this, I get an error of singular gradient and suggests initialize the starting values to something other than 1 (the default). You can specify the starting values with the argument start = list("a1"=0.1, "b1"=0.1, "c1"=0.1, "d1"=0.1), but this didn't seem to help. Perhaps you have a better idea of what the starting values should be?

Please note that there is a more stable version of `nls` in the package *nlsr* by John Nash. Trying `nls` in Base R will likely rush you into the infamous "singular gradient" error, while `nlsr::nlsb` will give the same solution like the one above by Grothendieck for all reasonable starting values. — Hans W., Nov 16 '18 at 19:27

Start list of parameters for nls and rational function

2 Answers2

Note 1

Note 2