how to say in Haskell...

[juan_gandhi]

pairwise :: [a] →  [(a, a)]
pairwise [] = []
pairwise (x:y:rest) = (x,y) : pairwise rest

is there something that does it already? Could not figure out.

Comments

[permea-kra.livejournal.com]

No, there is no standard way to do such things, i.e. there is no famous functions in base libraries that does something alike. You can build your one primitives to work with lists, however. Parser combinators library like ReadP is a good start point.

[rinver.livejournal.com]

Наверное так. Но что если этой функции подать список из нечётного количества элементов ? :) Или нефиг ? :)

[http://users.livejournal.com/_navi_/]

Nothing standard comes to my. There's a problem with your code btw: pairwise is not tail-recursive and will blow up the stack.

[nivanych.livejournal.com]

> is there something that does it already

А какая разница, если это пишется за полминуты?

[nivanych.livejournal.com]

Вот так будет лучше (и паттерн полный) ->
pairwise (x:y:rest) = (x,y) : pairwise rest
pairwise _ = []

[vyahhi.livejournal.com]

zip xs (tail xs)
и как-нибудь взять все элементы с позиций 0,2..

полет фантазии

[huzhepidarasa.livejournal.com]

pairwise xs = uncurry zip $ pairmap (map snd) $ partition fst $ zip (cycle [True, False]) xs
  where pairmap f (x,y) = (f x, f y)

хотел без рекурсии и не определяя никаких вспомогательных ф-й, но не вышло

[http://users.livejournal.com/_adept_/]

Как всегда, забыли unfoldr :)


splitInto n = unfoldr (chunk n)
where chunk _ [] = Nothing
chunk n lst = Just $ splitAt n lst