使用 nlsLM 拟合呈现肘/膝弯曲的数据集，并“强制”系数接近阈值

Question

由于需要使用 sestak berggren 模型（源自逻辑模型）拟合与二维扩散过程 D2 过程相关的数据集，我需要了解如何使用 nlsLM 当存在肘部/膝盖时，因为以下“简单的方法不起作用”

x=c(1.000000e-05, 1.070144e-05, 1.208082e-05, 1.456624e-05, 1.861581e-05, 2.490437e-05, 3.407681e-05, 4.696710e-05,
 6.474653e-05, 8.870800e-05, 1.206194e-04, 1.624442e-04, 2.172716e-04, 2.882747e-04, 3.794489e-04, 4.956619e-04,
 6.427156e-04, 8.275095e-04, 1.058201e-03, 1.344372e-03, 1.697222e-03, 2.129762e-03, 2.657035e-03, 3.296215e-03,
 4.067301e-03, 4.992831e-03, 6.098367e-03, 7.412836e-03, 8.968747e-03, 1.080251e-02, 1.295471e-02, 1.547045e-02,
 1.839960e-02, 2.179713e-02, 2.572334e-02, 3.024414e-02, 3.543131e-02, 4.136262e-02, 4.812205e-02, 5.579985e-02,
 6.449256e-02, 7.430297e-02, 8.533991e-02, 9.771803e-02, 1.115573e-01, 1.269824e-01, 1.441219e-01, 1.631074e-01,
 1.840718e-01, 2.071477e-01, 2.324656e-01, 2.601509e-01, 2.903210e-01, 3.230812e-01, 3.585200e-01, 3.967033e-01,
 4.376671e-01, 4.814084e-01, 5.278744e-01, 5.769469e-01, 6.284244e-01, 6.819947e-01, 7.371982e-01, 7.933704e-01,
 8.495444e-01, 9.042616e-01)

ynorm=c(
 1.000000e+00, 8.350558e-01, 6.531870e-01, 4.910995e-01, 3.581158e-01, 2.553070e-01, 1.814526e-01, 1.290639e-01,
 9.219591e-02, 6.623776e-02, 4.817180e-02, 3.543117e-02, 2.624901e-02, 1.961542e-02, 1.478284e-02, 1.123060e-02,
 8.597996e-03, 6.631400e-03, 5.151026e-03, 4.028428e-03, 3.171096e-03, 2.511600e-03, 2.001394e-03, 1.604211e-03,
 1.292900e-03, 1.047529e-03, 8.530624e-04, 6.981015e-04, 5.739778e-04, 4.740553e-04, 3.932255e-04, 3.275345e-04,
 2.739059e-04, 2.299339e-04, 1.937278e-04, 1.637946e-04, 1.389500e-04, 1.182504e-04, 1.009406e-04, 8.641380e-05,
 7.418032e-05, 6.384353e-05, 5.508090e-05, 4.762920e-05, 4.127282e-05, 3.583451e-05, 3.116813e-05, 2.715264e-05,
 2.368759e-05, 2.068935e-05, 1.808802e-05, 1.582499e-05, 1.385102e-05, 1.212452e-05, 1.061032e-05, 9.278534e-06,
 8.103650e-06, 7.063789e-06, 6.140038e-06, 5.315870e-06, 4.576585e-06, 3.908678e-06, 3.298963e-06, 2.732866e-06,
 2.189810e-06, 1.614149e-06)


dfxy <-  data.frame(x[1:length(ynorm)],ynorm)
fn=funSel <-"co*((1-x)^m)*(x^n)"
mod_fit <- nlsLM(ynorm~eval(parse(text=fn)),start=c(co=0.5,m=-1,n=0.5),data=dfxy)
plot(dfxy$x,dfxy$y,xlim=c(0,0.001))
plot(dfxy$x,(fitted(mod_fit))[1:length(dfxy$x)],xlim=c(0,0.001))

我找到的唯一解决方案是基于https://stackoverflow.com/a/54286595/6483091。因此，首先找到“肘部”，然后仅将回归应用于减少的数据集。一切都以这种方式工作，但我想知道是否还有其他解决方案（调整回归的参数而不是分两步制作，以某种方式让 nlsLM 使用动态一阶导数阈值“识别”曲线，但仍然强制fn 用于回归） “最大的问题是我已经知道参数的“范围””（即 使用“良好”起点应用回归（“基本事实” ynorm

“基本事实”

 plot(x,yrnom) yt <- 0.973*(1-x)^(0.425)*x^(-1.008)
 lines(x,yt/max(yt))

Answer 1

这是一个使用nls 和双曲线拟合的解决方案：

x=c(1.000000e-05, 1.070144e-05, 1.208082e-05, 1.456624e-05, 1.861581e-05, 2.490437e-05, 3.407681e-05, 4.696710e-05,
    6.474653e-05, 8.870800e-05, 1.206194e-04, 1.624442e-04, 2.172716e-04, 2.882747e-04, 3.794489e-04, 4.956619e-04,
    6.427156e-04, 8.275095e-04, 1.058201e-03, 1.344372e-03, 1.697222e-03, 2.129762e-03, 2.657035e-03, 3.296215e-03,
    4.067301e-03, 4.992831e-03, 6.098367e-03, 7.412836e-03, 8.968747e-03, 1.080251e-02, 1.295471e-02, 1.547045e-02,
    1.839960e-02, 2.179713e-02, 2.572334e-02, 3.024414e-02, 3.543131e-02, 4.136262e-02, 4.812205e-02, 5.579985e-02,
    6.449256e-02, 7.430297e-02, 8.533991e-02, 9.771803e-02, 1.115573e-01, 1.269824e-01, 1.441219e-01, 1.631074e-01,
    1.840718e-01, 2.071477e-01, 2.324656e-01, 2.601509e-01, 2.903210e-01, 3.230812e-01, 3.585200e-01, 3.967033e-01,
    4.376671e-01, 4.814084e-01, 5.278744e-01, 5.769469e-01, 6.284244e-01, 6.819947e-01, 7.371982e-01, 7.933704e-01,
    8.495444e-01, 9.042616e-01)

ynorm=c(
  1.000000e+00, 8.350558e-01, 6.531870e-01, 4.910995e-01, 3.581158e-01, 2.553070e-01, 1.814526e-01, 1.290639e-01,
  9.219591e-02, 6.623776e-02, 4.817180e-02, 3.543117e-02, 2.624901e-02, 1.961542e-02, 1.478284e-02, 1.123060e-02,
  8.597996e-03, 6.631400e-03, 5.151026e-03, 4.028428e-03, 3.171096e-03, 2.511600e-03, 2.001394e-03, 1.604211e-03,
  1.292900e-03, 1.047529e-03, 8.530624e-04, 6.981015e-04, 5.739778e-04, 4.740553e-04, 3.932255e-04, 3.275345e-04,
  2.739059e-04, 2.299339e-04, 1.937278e-04, 1.637946e-04, 1.389500e-04, 1.182504e-04, 1.009406e-04, 8.641380e-05,
  7.418032e-05, 6.384353e-05, 5.508090e-05, 4.762920e-05, 4.127282e-05, 3.583451e-05, 3.116813e-05, 2.715264e-05,
  2.368759e-05, 2.068935e-05, 1.808802e-05, 1.582499e-05, 1.385102e-05, 1.212452e-05, 1.061032e-05, 9.278534e-06,
  8.103650e-06, 7.063789e-06, 6.140038e-06, 5.315870e-06, 4.576585e-06, 3.908678e-06, 3.298963e-06, 2.732866e-06,
  2.189810e-06, 1.614149e-06)


dfxy <-  data.frame(x[1:length(ynorm)],ynorm)
plot(ynorm ~ x.1.length.ynorm.., data = dfxy)
mod <- nls(ynorm ~ a/x.1.length.ynorm.. + b, data = dfxy, start = list(a = 1, b = 0))
lines(x = dfxy$x.1.length.ynorm.., y = predict(mod, newdata = dfxy$x.1.length.ynorm..))

不过，合身并不完美。我想没有适合直角的连续函数...

根据您要使用回归的目的，您还可以使用黄土回归：

dfxy <-  data.frame(x[1:length(ynorm)],ynorm)
names(dfxy) <- c("x", "y")
plot(y ~ x, data = dfxy)
mod <- loess(y ~ x, data = dfxy, span = 0.1)
lines(x = dfxy$x, y = predict(mod, newdata = dfxy$x), col = "red")

导致：