Re: [Haskell-cafe] What is the best way to preserve the order of a list

Henk-Jan van Tuyl wrote:

On Wed, 01 Nov 2006 10:29:18 +0100, Stefan Holdermans wrote:

...

sortOnLab = sortBy (\(l, _) (m, _) -> compare l m)

The following does the same: sortOnLab = sort Not quite, sortOnLab preserves the order of tuples with the same first element while plain sort orders them by the second. Also, only sort requires that the second element of the tuple have an Ord instance (Haskell 98 requires that sort is stable).

Test.QuickCheck.test (\xs -> sort xs == sortOnLab xs) Falsifiable, after 1 tests: [(-3,3),(-3,1)]

sortOnLab [(-3,3),(-3,1)] ==> [(-3,3),(-3,1)]

but sort [(-3,3),(-3,1)] ==> [(-3,1),(-3,3)] Also,

map fst $ sortOnLab [(2,asTypeOf),(1,const)] [1,2]

map fst $ sort [(2,asTypeOf),(1,const)]

<interactive>:1:10: No instance for (Ord (a -> a -> a)) arising from use of `sort' at <interactive>:1:10-38 Possible fix: add an instance declaration for (Ord (a -> a -> a)) In the second argument of `($)', namely `sort [(1, asTypeOf), (1, const)]' In the expression: (map fst) $ (sort [(1, asTypeOf), (1, const)]) In the definition of `it': it = (map fst) $ (sort [(1, asTypeOf), (1, const)]) Brandon