通过 sapply 和 uniroot 构建逆回归

Question

我有一个函数如下：

V <- seq(50, 350, by = 1)
> VK
    Voltage^0     Voltage^1     Voltage^2     Voltage^3 
-1.014021e+01  9.319875e-02 -2.738749e-04  2.923875e-07 
> plot(x = V, exp(exp(sapply(0:3, function(x) V^x) %*% VK)), type = "l"); grid()

现在我想对这个函数做一个逆回归。我看过 Solving for the inverse of a function in R :

inverse = function (f, lower = -100, upper = 100) { function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)1 }

square_inverse = inverse(function (x) x^2, 0.1, 100)

square_inverse(4)

我正在尝试根据我的目的对其进行如下调整：

certain_function <- function(x=V) { exp(exp(sapply(0:3, function(x) V^x) %*% VK)) }

inverse = function (f, lower = 50, upper = 350) {
  function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]
}

inverse_regression = inverse(certain_function, 50, 350)

inverse_regression(2)

不幸的是，这会产生：

 Error in uniroot((function(x) f(x) - y), lower = lower, upper = upper) : 
  f() values at end points not of opposite sign In addition: Warning messages:
1: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
  the condition has length > 1 and only the first element will be used
2: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
  the condition has length > 1 and only the first element will be used

据我了解：错误意味着根数多于一个根数（uniroot 只能处理一个根数）但不应该有多个根数，因为它是一个严格单调递增的函数。我不明白的警告..

编辑：我试图支持它。我删除了两个指数，产生了以下情节：

这仍然会产生以下错误：

> inverse_regression(0.1)

 Error in uniroot((function(x) f(x) - y), lower = lower, upper = upper) : 
  f() values at end points not of opposite sign In addition: Warning messages:
1: In if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
  the condition has length > 1 and only the first element will be used
2: In if (is.na(f.upper)) stop("f.upper = f(upper) is NA") :
  the condition has length > 1 and only the first element will be used

这是为什么？显然曲线在两个端点都有相反的符号。我猜端点是指根的左边和右边的点？

Answer 1

可能由于 "certain_function" 的定义，逆回归不起作用。这是一个矩阵向量乘积，结果又是一个向量。因此，我将其转换为常规函数 function(x) exp(exp(sum(x^0*VK[1],x^1*VK[2],x^2*VK[3],x^3*VK[4]))，现在它可以正常工作了。因此，一般情况下每次都要注意界限。

通过 sapply 和 uniroot 构建逆回归

Construct an inverse regression via sapply and uniroot

r

inverse