修复要为x：y而不是仅1：y运行的函数

我定义了一个函数，该函数根据从2种出版物中提取的方程式来计算树木的高度（h）和直径（dbh）之间的关系。我的目标是使用论文1（Xiangtao）中建立的关系来预测论文2（Marechaux和Chave）中方程式的变量值。我想测试一下纸张2[x:y]的生成nls()曲线适合纸张1的直径范围。目前，我一直遇到错误（我相信plot()）

Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ

如果我用任何东西，除了X = 1[x:y]即dbh.min:dbh.max

我的功能如下：

# Plant.Functional.Type constants...
Dsb1 <- 2.09
Dsb2 <- 0.54
Db1 <- 0.93
Db2 <- 0.84
BDb1 <- 2.66
BDb2 <- 0.48
Eb1 <- 1.41
Eb2 <- 0.65
# # # # # # # # # # # # # # # # # # # # # # # # # # #
Generate.curve <- function(b1, b2, dbh.min, dbh.max){
# calculate Xiangtao's allometry...
  tmp_h <- c(dbh.min:dbh.max)
  for (dbh in dbh.min:dbh.max)
  {
    h = b1*dbh^(b2)
    tmp_h[dbh] = h
  }
# plot to check curve
  plot(dbh.min:dbh.max, tmp_h)

# define secondary function for Marechaux and Chave allometry
  h_fxn <- function(hlim,dbh,ah){
    h = hlim * (dbh / (dbh + ah))
    return(h)
  }

# use nonlinear least squares model to solve for ah and hlim
  # set model inputs
  start.ah <- 1 
  start.hlim <- 5
  tmp_v <- cbind(dbh.min:dbh.max,tmp_h)

tmp.fit <- nls(tmp_h ~ h_fxn(hlim,dbh.min:dbh.max,ah), start = list(hlim = start.hlim, 
                ah = start.ah), algorithm = "port", upper = list(hlim = 75, ah = 99))  
# seems to be no way of extracting ah and hlim from tmp.fit via subset
# extract manually and then check fit with
  # lines(dbh.min:dbh.max, hlim * (dbh.min:dbh.max/(dbh.min:dbh.max + ah)))
  # for equation h = hlim * (dbh / (dbh + ah)) from Marechaux and Chave
return(tmp.fit)
}
# # # # # # # # # # # # # # # # # # # # # # # # # # #

这非常适合

Generate.curve(Dsb1,Dsb2,1,100)
lines(1:100, 36.75 * (1:100/(1:100 + 52.51)))

但是我也希望能够检查诸如此类范围内的曲线拟合[80:100]。我一直在尝试找出为什么现在Generate.curve(Dsb1,Dsb2,80,100)返回错误大约3天。谢谢你的帮助。