Haskell: Splitting list into tuple of two new lists

问题

I am having difficulty figuring out how to split a list of Ints into a tuple containing two new lists, such that every element (starting with first) goes into the first list and every other element in the second.

Like so:

split [] = ([],[])
split [1] = ([1],[])
split [1,2] = ([1],[2])
split [1,2,3] = ([1,3],[2])
split [1,2,3,4] = ([1,3],[2,4])

I'm trying to accomplish this recursively(with guards) and only using the single argument xs

This is my approach that keeps getting error messages:

split :: [Int] -> ([Int],[Int])
split xs | length(xs) == 0 = ([],[])
         | length(xs) == 1 = (xs !! 0 : [],[])
         | length(xs) == 2 = (xs !! 0 : [], xs !! 1 : [])
         | otherwise = (fst ++ xs !! 0, snd ++ xs !! 1) ++ split(drop 2 xs))    

回答1:

Your split function returns a pair, but in the last case you are using ++ on the result of split. That will be a type error, since ++ works on lists, not pairs. There is also a type error because fst and snd are functions to pick out the elements of a pair, but you are using them is a strange way.

Furthermore, use pattern matching instead of using length. Also, the case where you test if the length is 2 is not needed, since the general case removes 2 elements which takes you down to the base case of the empty list.

You can also make your function more general by using a type variable a instead of Int in the type.

[Edit]: Added code

split :: [a] -> ([a], [a])
split [] = ([], [])
split [x] = ([x], [])
split (x:y:xys) = (x:xs, y:ys) where (xs, ys) = split xys

回答2:

Another way to do this is with mutual recursion. It comes out very easy to read:

split xs = (odds xs, evens xs)

odds (x:xs) = x : evens xs
odds xs     = []

evens xs = odds (drop 1 xs)

回答3:

split :: [a] -> ([a], [a])
split xs | null xs = ([], [])
         | otherwise = (head xs : snd pair, fst pair)
  where pair = split (tail xs)

But you should be using a fold:

split :: [a] -> ([a], [a])
split = foldr (\x (ys, zs) -> (x : zs, ys)) ([], [])

回答4:

The Haskell Blow Your Mind wiki, has some one liners:

-- splitting in two (alternating)
-- "1234567" -> ("1357", "246")

-- the lazy match with ~ is necessary for efficiency, especially enabling
-- processing of infinite lists
foldr (\a ~(x,y) -> (a:y,x)) ([],[])

(map snd *** map snd) . partition (even . fst) . zip [0..]

transpose . unfoldr (\a -> toMaybe (null a) (splitAt 2 a))

回答5:

Two alternative versions:

split = conv . map (map snd) . groupWith (even.fst) . zip [0..] where
  conv [xs,ys] = (xs,ys)

split xs = (ti even xs, ti odd xs) where
  ti f = map snd . filter (f.fst) . zip [0..]

回答6:

There is a mistake in the last clause. You have to get results from recursive call and then add first and second elements to them.

split :: [Int] -> ([Int],[Int])
split xs | length(xs) == 0 = ([],[])
         | length(xs) == 1 = (xs !! 0 : [],[])
         | length(xs) == 2 = (xs !! 0 : [], xs !! 1 : [])
         | otherwise = let (fst, snd) = split(drop 2 xs) in
                     (xs !! 0 : fst, xs !! 1 : snd)

回答7:

In case you are looking for some alternate way to do this, below is one such implementation:

split xs = 
     let (a,b) = partition (odd . snd) (zip xs [1..]) 
     in ( (map fst a), (map fst b))

来源：https://stackoverflow.com/questions/7410989/haskell-splitting-list-into-tuple-of-two-new-lists