r - 从预测的非线性模型数据中插值曲线，需要方程 R

Question

我需要定义一个方程来表示模型中的一系列点。由于它们是来自已经拟合的非线性回归的预测，因此噪声不应该成为真正的问题。我已经看到了许多允许对曲线进行预测的示例，但我仍然对如何从样条等各种函数中检索方程感到困惑。

根据我为 brms 中的非线性回归提供的公式（b1^b2，其中 b1 ~ 1 + x 和 b2 ~ 1），我希望这种关系类似于某种幂函数 [ax^b 或可能（int + ax)^b]。

我知道这可能只是一个近似值 - 只要它相对相似就可以（将与模型一起绘制以进行比较）。非常感谢任何见解，如果我可以提供更多信息，请告诉我。在下面提供了示例数据。

x <- c(7, 7.5, 8, 8.5, 9, 9.5, 10, 10.5, 11, 11.5, 
       12, 12.5, 13, 13.5, 14, 14.5, 15, 15.5, 
       16, 16.5, 17, 17.5, 18, 18.5, 19, 19.5, 
       20, 20.5, 21, 21.5, 22, 22.5, 23, 23.5, 
       24, 24.5, 25, 25.5, 26, 26.5, 27, 27.5, 
       28, 28.5, 29, 29.5, 30, 30.5, 31, 31.5, 
       32, 32.5, 33, 33.5, 34, 34.5, 35, 35.5, 
       36, 36.5, 37, 37.5, 38, 38.5, 39, 39.5, 
       40, 40.5, 41, 41.5, 42, 42.5, 43, 43.5, 
       44, 44.5, 45, 45.5, 46, 46.5, 47, 47.5, 
       48, 48.5, 49, 49.5, 50, 50.5, 51, 51.5, 
       52, 52.5, 53, 53.5, 54, 54.5, 55, 55.5, 
       56, 56.5, 57, 57.5, 58, 58.5, 59, 59.5, 
       60, 60.5, 61)

y <- c(38.0136100143096, 41.7377019494229, 45.0570784133659, 49.7413206704322, 
       54.1226793545035, 58.5143924911556, 64.0149939443995, 70.2996105120598, 
       74.5157778956172, 82.0845639640055, 88.1447395709772, 96.0901896008888, 
       104.331306746698, 112.587794035369, 116.793284692497, 129.660032344424, 
       139.148796580701, 154.021720779283, 164.352212638548, 174.265834929441, 
       186.345636353806, 199.537020067263, 217.599140020487, 233.877433057235, 
       246.132735977878, 267.47374365421, 283.863259012889, 301.899794004843, 
       322.769938301311, 351.430583957233, 383.849198025211, 390.736999633559, 
       423.249193342341, 454.638021905869, 485.240521462023, 509.31872397953, 
       560.804077944124, 580.52077607546, 623.750134417424, 663.267081364541, 
       698.778452300861, 737.091827116676, 796.938400847035, 856.419990917143, 
       906.017778910945, 965.168165262045, 1013.84413110147, 1065.85753522486, 
       1154.33958805326, 1204.13058748532, 1322.42312105971, 1385.41472565674, 
       1464.95636449756, 1526.04430007742, 1609.84006688626, 1738.45992222218, 
       1816.05261609804, 1896.50702990984, 2024.59349402727, 2142.55345863206, 
       2216.16848789472, 2384.32905391336, 2534.20713772314, 2599.97048189298, 
       2800.82684813604, 2956.9306593113, 3133.9558188069, 3320.82577045419, 
       3477.57994233654, 3602.69160098815, 3800.51701252841, 4062.6001499355, 
       4238.73826354165, 4645.61124406474, 4731.78190196944, 4947.75977778357, 
       5211.86923419648, 5619.65143369664, 5817.33879970144, 6079.00167282099, 
       6205.39929607976, 6681.42573862285, 7198.53801041513, 7267.01092369979, 
       7816.95660375773, 8133.793609116, 8261.13289172042, 9053.74531584895, 
       9473.33768555856, 10137.0504777871, 10400.797014819, 10966.3727089692, 
       11494.3384956937, 11920.4271066324, 12400.9643350626, 13096.7674700334, 
       13798.7159321491, 14358.3675127649, 15266.2675882016, 16072.0949530508, 
       17058.7362901159, 17663.6886724086, 18390.5282871649, 19307.9213621373, 
       20392.6725381775, 21431.4257870743, 21624.312458088, 23099.2069971926, 
       24090.3844352512)

data <- data.frame(x, y)


library(ggplot2)

ggplot(data, aes(x,y)) + geom_point()

编辑：回复@Bernhard

是的，我尝试使用这些输出，但没有发现它们非常接近曲线 - 也许我误解了如何使用它们。

因此，如果我的 brms 输出如下所示：

 Family: lognormal 
  Links: mu = identity; sigma = identity 
Formula: weight.g ~ b1^b2 
         b1 ~ 1 + dia.1.cm
         b2 ~ 1
   Data: CaesCombBot.data.wfrag (Number of observations: 2619) 
Samples: 4 chains, each with iter = 5000; warmup = 2000; thin = 5;
         total post-warmup samples = 2400

Population-Level Effects: 
             Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
b1_Intercept     6.61      0.81     5.08     8.24 1.00     1741     1787
b1_dia.1.cm      0.98      0.42     0.42     2.07 1.00     1738     1974
b2_Intercept     0.57      0.06     0.46     0.70 1.00     1727     1942

Family Specific Parameters: 
      Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
sigma     0.56      0.01     0.54     0.57 1.00     2156     2105

Samples were drawn using sampling(NUTS). For each parameter, Bulk_ESS
and Tail_ESS are effective sample size measures, and Rhat is the potential
scale reduction factor on split chains (at convergence, Rhat = 1).

我怎样才能用它来重新创建方程？

例如，当我尝试

curve((6.61 + 0.98*x)^0.57, from = 0, to = 50) 

只是做一个快速比较： x 值 40 (6.61 + 0.98*40)^0.57 = exp(8.846) [以对数正态分布] = 6946.547

我知道它看起来不一样（因为它也在对数正态分布内），但希望 exp 值与下图相匹配？（根据 add_predictions，x 值为 40 对应于 5256.29）

标签： rinterpolationcurve-fittingequationbrms