I want to split a list on 1 element and take the rest of the list in the second of the tuple and return that.
like this: *> split [1,2,3,2] gives [(1,[2,3,2]),(2,[1,3,2]),(3,[1,2,2]),(2,[1,2,3])]
I tried some code like this but it keeps giving me errors. Can someone help me out with this?
split :: [Int] -> [(Int,[Int])]
split [] = []
split (x:xs) = (map (x:) (map snd (split xs))) ++ [(x,[])]
Thx!
My answer to an older question goes via an answer to this one.
I call this operation picks because it tells you all the ways to pick one out of a list (seen as a bag).
picks :: [x] -> [(x, [x])]
picks [] = []
picks (x : xs) = (x, xs) : [(y, x : ys) | (y, ys) <- picks xs]
It's standard issue zipper-wrangling kit, about which I've written at great length in other answers, such as the one linked above, and this one, for the works.
This is possible with a list comprehension. Using zip you can create a tuple with an index and each element, then you can consturct a new tuple with your element and the original list minus the element at that index
splitter :: [a] -> [(a, [a])]
splitter xs = [(x, removeN n xs) | (n, x) <- zip [0..] xs]
That's using a function to remove the nth element
removeN :: Int -> [a] -> [a]
removeN n xs = take n xs ++ drop (n + 1) xs
example
*Main> splitter [1..4]
[(1,[2,3,4]),(2,[1,3,4]),(3,[1,2,4]),(4,[1,2,3])]
splitAt should give better constant factors.
(x:)onto[(Int, [Int])], since(Int, [Int])is not a list. 3. If you only usexsin the recursive call, you would always forget the start of the list. 4. Try to be more verbose. Write a function that, for a given index, splits the indexed value and the rest. Use this together withzipWith [0..]for a first prototype.