As Rex commented, if you want this only for two cases, then you probably won't have any problem if you leave the code as it is. However, if you want to extract a common pattern, then you can write a function that splits a list into sub-list of a specified length (2 or 3 or any other number). Once you do that, you'll only use map
to turn each list of the specified length into Vector
.
The function for splitting list isn't available in the F# library (as far as I can tell), so you'll have to implement it yourself. It can be done roughly like this:
let divideList n list =
// 'acc' - accumulates the resulting sub-lists (reversed order)
// 'tmp' - stores values of the current sub-list (reversed order)
// 'c' - the length of 'tmp' so far
// 'list' - the remaining elements to process
let rec divideListAux acc tmp c list =
match list with
| x::xs when c = n - 1 ->
// we're adding last element to 'tmp',
// so we reverse it and add it to accumulator
divideListAux ((List.rev (x::tmp))::acc) [] 0 xs
| x::xs ->
// add one more value to 'tmp'
divideListAux acc (x::tmp) (c+1) xs
| [] when c = 0 -> List.rev acc // no more elements and empty 'tmp'
| _ -> failwithf "not multiple of %d" n // non-empty 'tmp'
divideListAux [] [] 0 list
Now, you can use this function to implement your two conversions like this:
seq { for [x; y] in floatList |> divideList 2 -> Vector2(x,y) }
seq { for [x; y; z] in floatList |> divideList 3 -> Vector3(x,y,z) }
This will give a warning, because we're using an incomplete pattern that expects that the returned lists will be of length 2 or 3 respectively, but that's correct expectation, so the code will work fine. I'm also using a brief version of sequence expression the ->
does the same thing as do yield
, but it can be used only in simple cases like this one.