机器算法验证 - 如何阅读 R 的 nls 的拟合优度？ - 吾爱随笔录

我正在尝试解释 nls() 的输出。我已经阅读了这篇文章，但我仍然不明白如何选择最合适的。根据我的情况，我有两个输出：

> summary(m)

  Formula: y ~ I(a * x^b)

  Parameters:
  Estimate Std. Error t value Pr(>|t|)    
  a 479.92903   62.96371   7.622 0.000618 ***
  b   0.27553    0.04534   6.077 0.001744 ** 
  ---
  Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

  Residual standard error: 120.1 on 5 degrees of freedom

  Number of iterations to convergence: 10 
  Achieved convergence tolerance: 6.315e-06

和

> summary(m1)

  Formula: y ~ I(a * log(x))

  Parameters:
  Estimate Std. Error t value Pr(>|t|)    
  a   384.49      50.29   7.645 0.000261 ***
  ---
  Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

  Residual standard error: 297.4 on 6 degrees of freedom

  Number of iterations to convergence: 1 
  Achieved convergence tolerance: 1.280e-11

第一个有两个参数和较小的残差。第二个唯一的参数，但残差最差。哪个最合适？

> x <- 1:10 > y <- 2*x + 3 > yeps <- y + rnorm(length(y), sd = 0.01) > > > m1=nls(yeps ~ a + b*x, start = list(a = 0.12345, b = 0.54321)) > summary(m1) Formula: yeps ~ a + b * x Parameters: Estimate Std. Error t value Pr(>|t|) a 2.9965562 0.0052838 567.1 <2e-16 *** b 2.0016282 0.0008516 2350.6 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.007735 on 8 degrees of freedom Number of iterations to convergence: 2 Achieved convergence tolerance: 3.386e-09 > > > m2=nls(yeps ~ a + b*x+c*I(x^5), start = list(a = 0.12345, b = 0.54321,c=10)) > summary(m2) Formula: yeps ~ a + b * x + c * I(x^5) Parameters: Estimate Std. Error t value Pr(>|t|) a 3.003e+00 5.820e-03 516.010 <2e-16 *** b 1.999e+00 1.364e-03 1466.004 <2e-16 *** c 2.332e-07 1.236e-07 1.886 0.101 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.006733 on 7 degrees of freedom Number of iterations to convergence: 2 Achieved convergence tolerance: 1.300e-06 > > anova(m1,m2) Analysis of Variance Table Model 1: yeps ~ a + b * x Model 2: yeps ~ a + b * x + c * I(x^5) Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F) 1 8 0.00047860 2 7 0.00031735 1 0.00016124 3.5567 0.1013 >