Question

我遇到了F＃的麻烦，因为我正在学习。我有类似下一个代码的东西：

let A = [| [| [|"1";"Albert"|];[|"2";"Ben"|] |];[| [|"1";"Albert"|];[|"3";"Carl"|] |] |]

（A型：string[][][]）

我正在尝试将A转换为：

let B = [| [|"1"; "Albert" |] ; [| "2"; "Ben"|] ; [| "3"; "Carl"|] |]

（B型：string[][]）

我不知道该怎么做。我一直在尝试一些for和递归函数，但我没有得到它。

Answer 1

您可以使用Array.concat将string[][][]转换为string[][]，然后使用Seq.distinct删除重复的字符串数组。

let b = 
    [| [| [|"1";"Albert"|];[|"2";"Ben"|] |];[| [|"1";"Albert"|];[|"3";"Carl"|] |] |]
        |> Array.concat
        |> Seq.distinct
        |> Seq.toArray

Answer 2

以下是实现此功能的其他几个选项，可帮助您更好地概念化此类问题：

let A = [| [| [|"1";"Albert"|];[|"2";"Ben"|] |];[| [|"1";"Albert"|];[|"3";"Carl"|] |] |]

//Matthew's answer
//This is exactly what you were asking for.  
//It takes all the subarrays and combines them into one
A |> Array.concat
  |> Seq.distinct
  |> Seq.toArray

//This is the same thing except it combines it with a transformation step, 
//although in your case, the transform isn't needed so the transform 
//function is simply `id`
A |> Seq.collect id
  |> Seq.distinct
  |> Seq.toArray

//The same as the second one except using a comprehension.  
//This form makes it somewhat more clear exactly what is happening (iterate 
//the items in the array and yield each item).
//The equivalent for the first one is `[|for a in A do yield! a|]`
[for a in A do for b in a -> b] 
|> Seq.distinct
|> Seq.toArray

在f＃中将string [] [] []转换为string [] []

2 个答案: