On Tue, Nov 03, 2015 at 11:12:31AM +0100, Alexandre Delanoë wrote:
Hello, I am looking for such function:
function :: Ord a => [a] -> [a] -> [[a]] function [1,3,6] [0..6] == [[0,1],[2,3],[4,5,6]]
Question: does it exist already ?
``Data.List.Split`` (from package ``split``) may have what you are looking for.
Let:
partIt :: Ord a => a -> [a] -> [[a]] partIt m xs = [takeWhile (<= m) xs, dropWhile (<= m) xs] example: partIt 1 [0..5] == [[0,1],[2,3,4,5]]
Question: How to define recursively partIt 1) with a list as parameter ? 2) recursively on the second part of the list ?
With pattern matching: partIt :: Ord a => a -> [a] -> [[a]] partIt [] xs = xs partIt (m:ms) xs = let (a, b) = span (<= m) xs in (a : partIt ms b) λ> partIt [1,3,6] [0..6] [[0,1],[2,3],[4,5,6],[]] -- expected: splits on 6 too Learn You a Haskell has nice section on pattern matching. Did this answer the question?