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I have some issue using nls() to estimate parameters. I have a following set of functions to explain some data in hand: 

funk1 <- function(a,x) { x^2*exp(-(l*(1-exp(-r*a))/r)) }

funk2 <- function(x) { sapply(x, function (s)
{ integrate(funk1, lower = 0, upper = s, x=s)$value }) }

I am trying to fit funk2 to y:

y <- sort(runif(100, 0, 10^8))

When I use nls(): 

nls(y ~ funk2(z1$days.post.bmt), data= z1, start=list(l=0.02, r=0.002), trace=T)

it shows me following error:

Error in f(x, ...) : object 'l' not found

Isn't the whole point of nls() to substitute different values for parameters l and r from parameter space to fit the function by minimizing SSR and give the parameter estimates? why it needs value of l for it to work? I am definitely missing something big here. Please help!

Thanks in advance!

4

1 回答 1

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您必须将参数l和r作为函数参数传递funk1和funk2。

funk1 <- function(a,x,l,r) {
  x^2*exp(-(l*(1-exp(-r*a))/r))
  }

funk2 <- function(x,l,r) {
  sapply(x, function (s) {
              integrate(funk1, lower = 0, upper = s, x=s, l=l, r=r)$value
              })
  }

我将生成一些数据进行测试：

z <- data.frame(days.post.bmt = 1:100,
                y = funk2(1:100, l = 1, r = 1) + rpois(100, 1:100))

nls(y ~ funk2(days.post.bmt,l,r), data = z, start = list(l = 0.5, r = 0.5))

#Nonlinear regression model
#  model: y ~ funk2(days.post.bmt, l, r)
#   data: z
#     l      r 
#0.9405 0.9400 
# residual sum-of-squares: 6709

#Number of iterations to convergence: 5 
#Achieved convergence tolerance: 2.354e-07

作为一个反例，请考虑：

bad_funk1 <- function(a,x) {
  x^2*exp(-(l*(1-exp(-r*a))/r))
  }

bad_funk2 <- function(x) {
  sapply(x, function (s) {
              integrate(funk1, lower = 0, upper = s, x=s)$value
              })
  }

nls(y ~ bad_funk2(days.post.bmt), data = z, start = list(l = 0.5, r = 0.5))
# Error in f(x, ...) (from #2) : argument "l" is missing, with no default
于 2016-07-20T05:51:57.393 回答