如何将Seq.sum用于任何.Net值类型?

时间:2015-06-26 19:25:01

标签: f# type-parameter

我正在寻找一个处理任何数值类型数据(int,float,double)的seq的函数,通过映射对它进行一些计算,然后对这些计算值进行求和。我遇到的问题是Seq.sum(或者通常只是'(+)')会导致类型参数为整数,或者只是给出一个平坦的错误。似乎应该有一种方法可以通过使用类型约束来完成这项工作,但我似乎无法得到它。

type ValueWithComputation<'v> = {Value: seq<'v>; Computation: 'v -> 'v}

let calculateAndCombine (x: ValueWithComputation<'v>) = 
    x.Value
    |> Seq.map x.Computation
    |> Seq.sum // sometimes gives error: "Could not resolve the ambiguity inherent in the use of operator '(+)'

let x = {Value= {1..10}; Computation= (fun x->x*2)}
let y = {Value= {(1.0)..(10.0)}; Computation= (fun x->x*x)}

let totalX = calculateAndCombine x //this causes the code force 'v to be int
let totalY = calculateAndCombine y //gives an error since this isn't an int

这似乎与F# generics / function overloading syntax类似,但它并没有真正解释如何让它适用于所有值类型。

1 个答案:

答案 0 :(得分:5)

我让它像这样工作。阅读Foggy Finder链接的答案。然后,您还需要静态成员Zero才能使sum正常工作。

type ValueWithComputation< ^T when ^T: (static member (+): ^T * ^T -> ^T) and ^T: (static member Zero: ^T)> = 
  { Value: seq< ^T>
    Computation: ^T -> ^T }

let inline calculateAndCombine x = 
    x.Value
    |> Seq.map x.Computation
    |> Seq.sum

let x = {Value= {1..10}; Computation= (fun x->x*2)}
let y = {Value= {(1.0)..(10.0)}; Computation= (fun x->x*x)}

let totalX = calculateAndCombine x
let totalY = calculateAndCombine y