通过sapply和uniroot构造逆回归

Question

我的功能如下：

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()

现在我想对这个特定函数进行逆回归。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）

我试图根据我的目的调整它，如下所示：

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

据我所知：错误意味着有更多的根而不是只有一个（uniroot只能处理一个根）但是不应该有多个根，因为它是严格单调递增的函数。 警告我不明白..

编辑：我试图躲过它..我删除了两个指数，产生了以下情节：

，这仍会产生以下错误：

> 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。我猜终点意味着从根向左和向右的点？

Answer 1

由于“Certain_function”的定义，逆向回归可能不起作用。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BEGIN TRAN CREATE TABLE #TMP (Department NVARCHAR(50),Name NVARCHAR(50)) INSERT INTO #TMP SELECT 'Insights', 'Mike' UNION ALL SELECT 'Insights','Mike' UNION ALL SELECT 'Insights','Chris'UNION ALL SELECT 'Market' ,'Julie' UNION ALL SELECT 'Research', 'Will'UNION ALL SELECT 'Research', 'Sabrina'UNION ALL SELECT 'Research','Bryan' SELECT * FROM #TMP SELECT department, COUNT(DISTINCT NAME) FROM #TMP GROUP BY department ROLLBACK TRAN ，现在它正在运行。 因此，每次都应该意识到界限。