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Convert string [] [] [] to string [] [] in f #

I'm having trouble with F # because I'm learning. I have something like the following code:

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

(Type A: string[][][]

)

I am trying to convert A to:

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

(Type B: string[][]

)

I do not know how to do that. I've tried multiple for

and recursive function, but I don't get it.

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2 answers


Here are some other options for implementing this to better understand the concept of this type:



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

//Matthew 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
+4


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You can use Array.concat

to include string[][][]

in string[][]

and then Seq.distinct

to remove duplicate arrays of strings.



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