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我正在尝试使用 minpack.lm 中的 nls.lm 函数将非线性模型拟合到来自心理物理学实验的某些数据。

我进行了搜索,但找不到很多关于包的信息,所以基本上复制了 nls.lm 帮助页面上给出的示例格式。不幸的是,我的脚本仍然无法运行,并且 R 抛出了这个错误:

Error in fn(par, ...) : 
unused argument (observed = c(0.1429, 0.2857, 0.375, 0.3846, 0.4667, 0.6154))

脚本似乎认为我想要拟合模型的数据是无关紧要的,这绝对是错误的。

我期望它适合模型并为备用参数 (w) 生成 0.5403 的值。

任何帮助是极大的赞赏。我正在从 Matlab 转移到 R,如果我的代码看起来很草率,我深表歉意。

这是脚本。

  install.packages("pracma")
  require(pracma)
  install.packages("minpack.lm")
  require(minpack.lm)

  # Residual function, uses parameter w (e.g. .23) to predict accuracy error at a given ratio [e.g. 2:1]
  residFun=function(w,n) .5 * erfc( abs(n[,1]-n[,2])/ ((sqrt(2)*w) * sqrt( (n[,1]^2) + (n[,2]^2) ) ) )

  # example for residFun
  # calculates an error rate of 2.59%
  a=matrix(c(2,1),1,byrow=TRUE)
  residFun(.23,a)

  # Initial guess for parameter to be fitted (w)
  parStart=list(w=0.2)

  # Recorded accuracies in matrix, 1- gives errors to input into residFun
  # i.e. the y-values I want to fit the model
  Acc=1-(matrix(c(0.8571,0.7143,0.6250,0.6154,0.5333,0.3846),ncol=6))

  # Ratios (converted to proportions) used in testing
  # i.e. the points along the x-axis to fit the above data to
  Ratios=matrix(c(0.3,0.7,0.4,0.6,0.42,0.58,0.45,0.55,0.47,0.53,0.49,0.51),nrow=6,byrow=TRUE)

  # non-linear model fitting, attempting to calculate the value of w using the Levenberg-Marquardt nonlinear least-squares algorithm 
  output=nls.lm(par=parStart,fn=residFun,observed=Acc,n=Ratios)

  # Error message shown after running
  # Error in fn(par, ...) : 
  #   unused argument (observed = c(0.1429, 0.2857, 0.375, 0.3846, 0.4667, 0.6154))
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1 回答 1

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该错误意味着您向函数传递了一个它不期望的参数。?nls.lm没有参数observed,所以它被传递给传递给的函数fn,在你的情况下,residFun。但是,residFun也不要期望这个论点,因此会出现错误。您需要像这样重新定义此功能:

# Residual function, uses parameter w (e.g. .23) to predict accuracy error at a given ratio [e.g. 2:1]
residFun=function(par,observed, n) {
  w <- par$w
  r <- observed - (.5 * erfc( abs(n[,1]-n[,2])/ ((sqrt(2)*w) * sqrt( (n[,1]^2) + (n[,2]^2) ) ) ))
  return(r)
}

它给出了以下结果:

> output = nls.lm(par=parStart,fn=residFun,observed=Acc,n=Ratios)
> output
Nonlinear regression via the Levenberg-Marquardt algorithm
parameter estimates: 0.540285874836135 
residual sum-of-squares: 0.02166
reason terminated: Relative error in the sum of squares is at most `ftol'.

为什么会这样:

您似乎受到了他文档中的这个示例的启发:

## residual function
residFun <- function(p, observed, xx) observed - getPred(p,xx)
## starting values for parameters
parStart <- list(a=3,b=-.001, c=1)
## perform fit
nls.out <- nls.lm(par=parStart, fn = residFun, observed = simDNoisy,
xx = x, control = nls.lm.control(nprint=1))

请注意,这observedresidFun这里的一个论点。

于 2015-05-01T10:44:06.517 回答