Partial fold of a list in Haskell -


how apply fold functions finite number of times, instead of folding entire list? example, incomplete fold sum first 3 numbers, leaving rest untouched:

foldln 3 (+) [1..5] -- [6,4,5]

it seems simple fold on part of list, cons rest of list:

foldln :: int -> (a -> -> a) -> [a] -> [a] foldln n f xs = (foldl1 f $ take n xs) : drop n xs

result in repl:

*main> foldln 3 (+) [1..5] [6,4,5]

if worry taking , dropping same list double job, can rewrite splitat:

foldln n f xs = (foldl1 f $ fst t) : snd t   t = splitat n xs

note type of function, can't base fold on foldl, because can't have list [[1,2,3],4,5,6] in haskell.


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