I know that the rest of the answers are shorter, but I recommend you to implement these lazy functions yourself before using the already Haskell implemented ones, so you understand how it works.
-- Order a list of tuples by the first item
orderBy1stTupleItem :: Ord a => (a, b1) -> (a, b2) -> Ordering
orderBy1stTupleItem tup1 tup2
| item1 > item2 = GT
| item1 < item2 = LT
| otherwise = EQ
where
item1 = fst tup1
item2 = fst tup2
-- Order a list of tuples by the second item
orderBy2ndTupleItem :: Ord a1 => (a2, a1) -> (a3, a1) -> Ordering
orderBy2ndTupleItem tup1 tup2
| item1 > item2 = GT
| item1 < item2 = LT
| otherwise = EQ
where
item1 = snd tup1
item2 = snd tup2
-- Wrapper Function: Order a list of tuples by the first item and later by the second item
orderTuplesBy1stThenBy2ndItem :: (Ord a1, Ord a2) => [(a2, a1)] -> [(a2, a1)]
orderTuplesBy1stThenBy2ndItem listTuples =
sortBy orderBy2ndTupleItem (sortBy orderBy1stTupleItem listTuples)
Example
let exampleListTuples = [(1,2),(0,8),(6,1),(3,6),(9,1),(7,8),(0,9)]
Then let's get the 1st list, ordered by the first item of each tuple:
> listOrderedByTuple1stItem = sortBy orderBy1stTupleItem exampleListTuples
> listOrderedByTuple1stItem
[(0,8),(0,9),(1,2),(3,6),(6,1),(7,8),(9,1)]
Now we order this result list by the second item of each tuple
> sortBy orderBy2ndTupleItem listOrderedByTuple1stItem
[(6,1),(9,1),(1,2),(3,6),(0,8),(7,8),(0,9)]
Or, you can just run the wrapper function orderTuplesBy1stThenBy2ndItem
as follows:
> sortBy orderTuplesBy1stThenBy2ndItem exampleListTuples