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Haskell: splitting a list into a tuple of two new lists

I find it difficult to understand how to split the Ints list into a tuple containing two new lists, so that each element (starting from the first) goes to the first list and each other element to the second.

Same:

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 am trying to do this recursively (with protection) and only using a single xs argument

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))
+5
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7 answers

split , ++ split. , ++ , . , fst snd - , .

, . , , 2, , 2 , .

a Int .

[]:

split :: [a] -> ([a], [a])
split [] = ([], [])
split [x] = ([x], [])
split (x:y:xys) = (x:xs, y:ys) where (xs, ys) = split xys
+14

- . :

split xs = (odds xs, evens xs)

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

evens xs = odds (drop 1 xs)
+5
split :: [a] -> ([a], [a])
split xs | null xs = ([], [])
         | otherwise = (head xs : snd pair, fst pair)
  where pair = split (tail xs)

:

split :: [a] -> ([a], [a])
split = foldr (\x (ys, zs) -> (x : zs, ys)) ([], [])
+3

Haskell Blow Your Mind wiki, :

-- 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))
+2
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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..]
+1
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There is a mistake in the last sentence. You should get the results from the recursive call, and then add the 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)
0
source

If you are looking for an alternative 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))
0
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