Traversing a list is sometimes more difficult than it seems to be at the first glance. With "traversal" I mean to consume one or more lists and produce one or more new ones. Our goal is to do this efficiently and lazily.As a running example I use the
Its type signature is
partitionEithers :: [Either a b] -> ([a], [b])
and it does what you expect:
Prelude Data.Either> partitionEithers [Left 'a', Right False, Left 'z'] ("az",[False]) Prelude Data.Either> take 100 $ snd $ partitionEithers $ cycle [Left 'a', Right (0 :: Int)] [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
The second example is especially important because it shows that the input can be infinitely long and the output can be, too. That's the proof that the implementation is lazy. We will use this example as test for our implementations below.
1 First attempt - quadratic runtime, not lazy
In our first attempt we maintain a state containing two lists that we want to extend to the result lists step by step.
partitionEithers2 :: [Either a b] -> ([a], [b]) partitionEithers2 = let aux ab  = ab aux (as, bs) (Left a : es) = aux (as ++ [a], bs) es aux (as, bs) (Right b : es) = aux (as, bs ++ [b]) es in aux (, )
This implementation works for finite lists but fails for infinite ones. You will also notice that it is quite slow.The reason is that appending something to a list like
Since we do this repeatedly we end up with quadratic runtime.
2 Second attempt - linear runtime, still not lazy
We have learned that appending something to a list is expensive. However prepending a single element is very cheap, it needs only constant number of operations. Thus we will implement the following idea: We prepend new elements to the result list and since this reverses the order of elements, we reverse the result lists in the end.
partitionEithers1 :: [Either a b] -> ([a], [b]) partitionEithers1 xs = let aux ab  = ab aux (as, bs) (Left a : es) = aux (a : as, bs) es aux (as, bs) (Right b : es) = aux (as, b : bs) es (ys,zs) = aux (, ) xs in (reverse ys, reverse zs)
This implementation is much faster than the first onebut it cannot be lazy because
3 Third attempt - linear runtime and full laziness
In order to get linear runtime and full laziness we must produce the list in the same order as the input. However we must avoid appending to the end of the list. Instead we must prepend elements to lists that become known in the future. We must be very careful that the leading elements of the result lists can be generated without touching the following elements. Here is the solution:
partitionEithers :: [Either a b] -> ([a], [b]) partitionEithers  = (, ) partitionEithers (Left a : es) = let (as,bs) = partitionEithers es in (a:as, bs) partitionEithers (Right b : es) = let (as,bs) = partitionEithers es in (as, b:bs)
matches the top-most data constructor lazily. The following expressions would match strictly and thus would fail:
(\(as,bs) -> (a:as, bs)) $ partitionEithers es
case partitionEithers es of (as,bs) -> (a:as, bs)
Matching the pair constructor strictly meansthat the recursive call to
before the pair constructor of the result is generated. This starts a cascade that forces all recursive calls until the end of the input list.
This is different for lazy pattern matches.The above
let ~(as,bs) = partitionEithers es in (a:as, bs)
(\ ~(as,bs) -> (a:as, bs)) $ partitionEithers es
case partitionEithers es of ~(as,bs) -> (a:as, bs)
or without the tilde as syntactic sugar:
case partitionEithers es of ab -> (a : fst ab, snd ab)
contain strict pattern matches on the pair constructor but the key difference to above is that these matches happen inside the pair constructor of
That is, the outer pair constructor can be generatedbefore the evaluation of
4 Fourth attempt - expert solution
Now real experts would not recurse manuallybut would let
This allows for fusion. Additionally real experts would add the line
in order to generate the pair constructor of the result completely independent from the input. This yields maximum laziness.
partitionEithersFoldr :: [Either a b] -> ([a], [b]) partitionEithersFoldr = (\ ~(as,bs) -> (as,bs)) . foldr (\e ~(as,bs) -> case e of Left a -> (a:as, bs) Right b -> (as, b:bs)) (, )
5 Fifth attempt - your solution
If you are tired of all these corner cases that we need to respect in order to get full laziness then you might prefer to solve the problem by just combining functions that are known to be lazy. It is good style anyway to avoid explicit recursion. Of course, when combining lazy functions you must still take care that the combinators maintain laziness. Thus my exercise for you at the end of this articleis to implement
base before version 4.
turns out to be very useful.