Pairing adjacent list items in Haskell

Question

I have a chained list like

[\"root\", \"foo\", \"bar\", \"blah\"]

And I\'d like to convert it to a list of tuples, using adjacent pairs. Li

Answer 1

It's possible to generalize the zipAdj in the question to work with arbitrary Traversable containers. Here's how we'd do it if we wanted the extra element on the front end:

import Data.Traversable

pairDown :: Traversable t => a -> t a -> t (a, a)
pairDown x = snd . mapAccumL (\old new -> (new, (old,new))) x

*Pairing> take 10 $ pairDown 0 [1..]
[(0,1),(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)]
*Pairing> pairDown 0 [1..10]
[(0,1),(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)]

To stick the extra element on the end, we can use mapAccumR:

import Data.Traversable

pairUp :: Traversable t => t a -> a -> t (a, a)
pairUp xs x = snd $ mapAccumR (\old new -> (new, (new,old))) x xs

This effectively traverses the container backwards.

*Pairing> pairUp [0..10] 11
[(0,1),(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10),(10,11)]
*Pairing> take 10 $ pairUp [0..] undefined
[(0,1),(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)]

It's impossible to generalize the apparently-desired function in quite this fashion, but it's possible to generalize it a bit differently:

import Data.Foldable
import Prelude hiding (foldr)

pairAcross :: Foldable f => f a -> [(a,a)]
pairAcross xs = foldr go (const []) xs Nothing
  where
    go next r Nothing = r (Just next)
    go next r (Just prev) = (prev, next) : r (Just next)

This gives

*Pairing> pairAcross [1..10]
[(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)]