`nls`拟合错误:无论起始值如何,总是达到最大迭代次数

Question

将此参数化用于增长曲线逻辑模型

我创建了一些点: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))

有人能告诉我如果用模型启动器创建数据我怎么能有一个拟合错误？

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 1

不要在人工"零残留"数据上使用'nls'.,如中所述?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

另外,请避免$在模型公式中使用,否则predict以后使用时会遇到麻烦.