nls2 treats vectored initial values as single elements, unlike nls

試著忘記壹切 提交于 2019-12-13 03:08:50

问题


I noticed that I can fit parameter vector, as below, using nls. This let's me decide the number of parameters I want to fit. As in the example below; where I am fitting y = k + a_1 x^2 + a_2 x^3 + a_3 x^3. I can simply change the number of initial values, which change the number of co-efficients to estimate.

But, this approach doesn't work with nls2. It just treats fits the y = k + a_1 * x, three times!

My questions is how to get nls2 to determine the number of parameters to fit based on the initial values - or something similar - as in the case with nls.

I don't have a lot of experience with nls or similar packages. So, I am trying to mend that. I am guessing nls2 has more capabilities than nls...

Data and model to fit

x <- c(32,64,96,118,126,144,152.5,158)  
y <- c(99.5,104.8,108.5,100,86,64,35.3,15)


model_fun <- function(x, int_sep, para) {
  int_sep +  rowSums(sapply(1:length(para), function(i) para[i] * x^i))
}

With nls package

mod_nls <- nls(y ~ model_fun(x, int_sep, para), 
             start = list(int_sep = 0, para=c(1, 1, 1)))

mod_nls
# Nonlinear regression model
# model: y ~ model_fun(x, int_sep, para)
# data: parent.frame()
# int_sep      para1      para2      para3 
# 1.269e+02 -1.626e+00  2.910e-02 -1.468e-04 
# residual sum-of-squares: 65.87
# 
# Number of iterations to convergence: 1 
# Achieved convergence tolerance: 1.732e-07

With nls2 package

 mod_nls2 <- nls2(y ~ model_fun(x, int_sep, para), 
                start = list(int_sep = 0, para=c(1, 1, 1)))

mod_nls2
# Nonlinear regression model
# model: y ~ model_fun(x, int_sep, para)
# data: parent.frame()
# int_sep     para 
# 143.0438  -0.5966 
# residual sum-of-squares: 3661
# 
# Number of iterations to convergence: 1 
# Achieved convergence tolerance: 7.602e-09

(edit: I am not interested in this particular model - it seems like an easy example)


回答1:


nls2 converts the start= argument to a data frame internally so if you provide it in a form such that as.data.frame(as.list(start)) works (where in the example "works" means it creates a data frame with 1 row and 2 columns, one column for each of the two parameters -- note that data frame columns can hold complex objects) then you should be OK:

nls2(y ~ model_fun(x, int_sep, para), 
  start = list(int_sep = 0, para = I(t(c(1, 1, 1)))))



回答2:


You don't need to guess regarding the capabilities of nls2. Read the documentation. You could report this issue on the package bug tracker, but I suspect @G.Grothendieck won't fix it unless you provide the fix.

Anyway, I would change your function.

model_fun <- function(x, int_sep, ...) {
  para <- c(...)
  #I prefer matrix algebra over a loop:
  int_sep +  c(tcrossprod(para, outer(x, seq_along(para), "^")))
}

library(nls2)
mod_nls2 <- nls2(y ~ model_fun(x, int_sep, alpha1, alpha2, alpha3), 
                 start = list(int_sep = 0, alpha1 = 1, alpha2 = 1, alpha3 = 1))

mod_nls2
#Nonlinear regression model
#  model: y ~ model_fun(x, int_sep, alpha1, alpha2, alpha3)
#   data: <environment>
#   int_sep     alpha1     alpha2     alpha3 
# 1.269e+02 -1.626e+00  2.910e-02 -1.468e-04 
# residual sum-of-squares: 65.87
#
#Number of iterations to convergence: 1 
#Achieved convergence tolerance: 1.732e-07


来源:https://stackoverflow.com/questions/46986535/nls2-treats-vectored-initial-values-as-single-elements-unlike-nls

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!