Question

我是R的新手，我遇到了一个我不明白的问题。我已经找到了有关R中nls功能如何工作的文档或答案，但我找不到任何东西。

Test = data.frame(x = c(5, 10, 15, 25), y = c(5, 8, 12, 16))

fun = function(x, a, b) {
    cat("a =", a, ", b =", b, "\n")
    a*(x**b)
}

nls(y ~ fun(x, a, b), data = Test, start = list(a = 1.613, b = 0.718), trace = TRUE)

输出如下：

a = 1.613 , b = 0.718 
a = 1.613 , b = 0.718 
a = 1.613 , b = 0.718 
a = 1.613 , b = 0.718 
0.7964401 :  1.613 0.718
a = 1.612723 , b = 0.7177708 
a = 1.612723 , b = 0.7177708 
a = 1.612723 , b = 0.7177708 
0.7960992 :  1.6127232 0.7177708
a = 1.612698 , b = 0.7177761 
a = 1.612698 , b = 0.7177761 
a = 1.612698 , b = 0.7177762 
0.7960992 :  1.6126980 0.7177761
Nonlinear regression model
  model: y ~ fun(x, a, b)
   data: Test
     a      b 
1.6127 0.7178 
 residual sum-of-squares: 0.7961

Number of iterations to convergence: 2 
Achieved convergence tolerance: 1.43e-06

它表示需要两次迭代才能收敛，但看起来它已经运行了10次，并且只更改了两次参数值。是否有一个原因？我无法找到nls如何计算参数猜测的解释。

我试图将600个数据点放到一个大型函数中，因此这需要花费大量额外的不必要的时间。

Answer 1

在我看来，nls使用Gauss-Newton算法，它需要雅可比计算，因此您可以看到雅可比值的数值计算。

使用以下代码可以看到调用不完全相同：

Test = data.frame(x = c(5, 10, 15, 25), y = c(5, 8, 12, 16))

fun = function(x, a, b) {
    cat("a =", sprintf("%0.10f",a), ", b =", sprintf("%0.10f",b), "\n")
    a*(x**b)
}

nls(y ~ fun(x, a, b), data = Test, start = list(a = 1.613, b = 0.718), trace = TRUE)

它只打印更多小数。通过对所有参数进行微小的扰动来估计雅可比。

a = 1.6130000000 , b = 0.7180000000 
a = 1.6130000000 , b = 0.7180000000 
a = 1.6130000240 , b = 0.7180000000 
a = 1.6130000000 , b = 0.7180000107 
0.7964401 :  1.613 0.718
a = 1.6127231539 , b = 0.7177708205 
a = 1.6127231779 , b = 0.7177708205 
a = 1.6127231539 , b = 0.7177708312 
0.7960992 :  1.6127232 0.7177708
a = 1.6126979985 , b = 0.7177761452 
a = 1.6126980225 , b = 0.7177761452 
a = 1.6126979985 , b = 0.7177761559 
0.7960992 :  1.6126980 0.7177761
Nonlinear regression model
  model: y ~ fun(x, a, b)
   data: Test
     a      b 
1.6127 0.7178 
 residual sum-of-squares: 0.7961

Number of iterations to convergence: 2 
Achieved convergence tolerance: 1.43e-06

因此，这些调用对于nls计算jacobian是必须的。

nls每次迭代多次运行函数？

1 个答案: