Question

我有三个阵列 - 第一，第二和第三。第二个包含第一个和第三个的唯一值包含新的替换值，首先映射到第二个，如下所示：

module SOQN = 

   open System

   let first  = [|"A"; "B"; "C"; "A"; "B"; "A"; "C"; "B"; "C"; "C"; "C"|]
   let second = [|"A"; "B"; "C"|]
   let third  = [|"1"; "2"; "3"|]

   let rplc (x:string[]) (y:string[]) (z:string[]) = 
      first
      // |> Array.map (fun w -> Array.iter2 (fun x y -> (if w = x then y)) second third)

   let fourth = 
      rplc first second third

   printfn ""
   printfn "fourth: %A" fourth

   // Expected Result: fourth: [|"1"; "2"; "3"; "1"; "2"; "1"; "3"; "2"; "3"; "3"; "3"|]
   // Actual Result:   fourth: [|"A"; "B"; "C"; "A"; "B"; "A"; "C"; "B"; "C"; "C"; "C"|]

我的评论行失败但我不确定为什么？

Answer 1

最简单的方法是从第二个和第三个数组创建一个查找表，而不是映射第一个数组的每个元素，并将其用作键。

let first  = [|"A"; "B"; "C"; "A"; "B"; "A"; "C"; "B"; "C"; "C"; "C"|]
let second = [|"A"; "B"; "C"|]
let third  = [|"1"; "2"; "3"|]

let lookupTbl = Map(Array.zip second third) //create a Map/Dictionary from the zipped values

first
|> Array.map (fun x -> lookupTbl.[x]) //Use the first array's values as keys
//val it : string [] = [|"1"; "2"; "3"; "1"; "2"; "1"; "3"; "2"; "3"; "3"; "3"|]

如果您不确定所有密钥是否存在，也可以使用TryFind，但在您的情况下，这似乎并不合适。

您的原始案例不起作用，因为您尝试使用if作为语句，因此它返回unit（因为如果x不等于w，会发生什么）。

如果你想更接近你的原始结构，你可以模式匹配而不是if，然后删除不匹配。 Array.collect将数组数组折叠为一个数组。 match表达式与代码中的if相同，但如果匹配则返回Some值，否则返回None。最后，我们用None删除Some和Array.choose选项包装器。

let rplc (x:string[]) (y:string[]) (z:string[]) = 
  first
  |> Array.collect (fun w -> 
                        Array.map2 (fun x y -> 
                                                match (w = x) with
                                                | true -> Some(y)
                                                | _ -> None ) second third)
  |> Array.choose id

let fourth = 
  rplc first second third

printfn ""
printfn "fourth: %A" fourth

// val fourth : string [] =
//   [|"1"; "2"; "3"; "1"; "2"; "1"; "3"; "2"; "3"; "3"; "3"|]
// val it : unit = ()

F＃三个阵列 - 用新值替换原始数组中的唯一值

1 个答案: