F # - splitting a list into tuples of odd-even lists (by element, not position)
Example: split [1;3;2;4;7;9];;
Output:([1;3;7;9], [2;4])
I'm new to F # and I can't seem to figure it out.
Cannot use built-in function partition
.
This is what I have so far:
let rec split xs =
match xs with
| [] -> [], []
| xs -> xs, []
| xh::xt -> let odds, evens = split xt
if (xh % 2) = 0 then xh::odds, xh::evens
else xh::odds, evens
Fixed code:
let rec split xs =
match xs with
| [] -> [], []
| xh::xt -> let odds, evens = split xt
if (xh % 2) = 0 then odds, xh::evens
else xh::odds, evens
* Thanks to @TheInnerLight for pointing out my mistakes: unreachable case and unnecessary change in odds
+3
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1 answer
You can use the built-in function List.partition
let splitOddEven xs =
xs |> List.partition (fun x -> x % 2 <> 0)
splitOddEven [1;3;2;4;7;9];;
val it : int list * int list = ([1; 3; 7; 9], [2; 4])
If you want a recursive implementation, I would probably go for a recursive tail implementation like this:
let splitOddEven xs =
let rec splitOddEvenRec oddAcc evenAcc xs =
match xs with
| [] -> oddAcc, evenAcc
| xh::xt ->
if (xh % 2) = 0 then splitOddEvenRec oddAcc (xh :: evenAcc) xt
else splitOddEvenRec (xh :: oddAcc) evenAcc xt
splitOddEvenRec [] [] xs
splitOddEven [1;3;2;4;7;9]
Note that this will give you two resulting lists in reverse order so you can undo them yourself.
+8
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