如何测试哪个迭代组合和＆amp;置换计算器是最有效的？

Question

我试图测试这两个函数中的哪一个在计算组合或排列的总和时更快。

更新：Option Explicit Sub ReturnActions() Dim itemnumber As String Dim finalrow As Integer Dim i As Integer Sheet1.Range("B35:N100").ClearContents itemnumber = Sheet1.Range("E8").Value finalrow = Sheet3.Range("G10").End(xlUp).Row Worksheets("Total List").Activate For i = 2 To finalrow If Cells(i, 7) = itemnumber Then Sheet3.Range(Cells(i, 7), Cells(i, 12)).Copy Sheet1.Range("B50").End(xlUp).Offset(1, 0).PasteSpecial xlPasteFormulasAndNumberFormats End If Next i Worksheets("Action Entry").Activate Range("E8").Select End Sub 包似乎是我原始问题的答案，但我想知道为什么microbenchmark中的切换比combo_beast便宜...

combo_unicorn

第一个函数使用嵌套开关：

> microbenchmark(combo_beast(5,'yes','yes'))
Unit: microseconds
                         expr   min    lq    mean median    uq    max neval
 combo_beast(5, "yes", "yes") 8.019 8.354 9.94767  8.688 9.022 32.412   100
> microbenchmark(combo_unicorn(5,'yes','yes'))
Unit: microseconds
                           expr   min    lq     mean median    uq    max neval
 combo_unicorn(5, "yes", "yes") 8.354 8.688 10.63269  9.022 9.357 41.768   100

第二个功能使用if / else if方法：

combo_beast <- function(n, order, rep){
  switch(order,
         yes = switch(rep,
                      yes = {
                        y <- 0
                        for(i in 1:n){
                          x <- n^i
                          y <- x + y}
                        return(y)
                      },
                      no = {
                        y <- 0
                        for(i in 1:n){
                          x <- factorial(n + i - 1)/(factorial(i)*factorial(n-1))  
                          y <- x + y}
                        return(y)
                      }
                     ),
         no = switch(rep,
                     yes = {
                        y <- 0
                        for(i in 1:n){
                          x <- factorial(n)/(factorial(n - i))  
                          y <- x + y}
                        return(y)
                      },
                      no = {
                        y <- 0
                        for(i in 1:n){
                          x <- factorial(n)/(factorial(n-i)*factorial(i))
                          y <- x + y}
                        return(y)
                      }
                     )
  )
}

我已尝试将combo_unicorn <- function(n, order, rep){ if(order == 'yes' & rep == 'yes'){ y <- 0 for(i in 1:n){ x <- n^i y <- x + y} return(y) } else if (order == 'yes' & rep == 'no'){ y <- 0 for(i in 1:n){ x <- factorial(n + i - 1)/(factorial(i)*factorial(n-1)) y <- x + y} return(y) } else if (order == 'no' & rep == 'no'){ y <- 0 for(i in 1:n){ x <- factorial(n)/(factorial(n-i)*factorial(i)) y <- x + y} return(y) } else { y <- 0 for(i in 1:n){ x <- factorial(n)/(factorial(n - i)) y <- x + y} return(y) } } 放在任一函数的开头，并在最后添加ptm <- proc.time()，但无论我做多大proc.time() - ptm，它都会保持返回0。