Hello All, As usual, my first attempt could be better.. -- partitions a list according to an equivalence relation import Data.List (partition) partitionBy :: (a -> a -> Bool) -> [a] -> [[a]] partitionBy eq [] = [] partitionBy eq ls = x:(partitionBy eq y) where (x,y) = partition ((head ls) `eq`) ls example> partitionBy (\x y -> (last x) == (last y)) ["abc","bd","bdc","abd"] result> [["abc","bdc"],["bd","abd"]] Of course, any crits would be appreciated, but maybe this will save somebody some time.. Thanks, Hans van Thiel