# List traversal

### From HaskellWiki

(show five implementations of partitionEithers and their pros and cons) |
(link to lazy pattern match) |
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until the end of the input list. |
until the end of the input list. |
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− | This is different for lazy pattern matches. |
+ | This is different for [[lazy pattern match]]es. |

The above <hask>let</hask> can be rewritten equivalently to: |
The above <hask>let</hask> can be rewritten equivalently to: |
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<haskell> |
<haskell> |
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That is, the outer pair constructor can be generated |
That is, the outer pair constructor can be generated |
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before the evaluation of <hask>ab</hask> is started. |
before the evaluation of <hask>ab</hask> is started. |
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− | |||

== Fourth attempt - expert solution == |
== Fourth attempt - expert solution == |

## Latest revision as of 07:39, 7 September 2012

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 thesince `base-4.0`

.

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.

## Contents |

## [edit] 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 likeSince we do this repeatedly we end up with quadratic runtime.

## [edit] 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 one

but it cannot be lazy because

## [edit] 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 means

that the recursive call tobefore 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 abovelet ~(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 generated

before the evaluation of## [edit] 4 Fourth attempt - expert solution

Now real experts would not recurse manually

but would letThis 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)) ([], [])

## [edit] 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 article

is to implementsay, from `base`

before version 4.

turns out to be very useful.