QuickSort 应用于 Haskell 中的元组列表 (Int,[Int])

Question

Good afternoon,

I'm a newb to Haskell, and I'm trying to adapt a QuickSort algorithm I made to sort lists of 'Int' with a list of tuples, but I quite can't figure out how to bind the tail to 'a' to make it work as I need it to, or if it's possible to reuse the code at all. Here's what I've been using to sort lists of 'Int':

quickSort :: Ord a => [a] -> [a]
quickSort [] = []  
quickSort (x:xs) =   
let smallSort  = quickSort [a | a <- xs, a < x]  
    biggerSort = quickSort [a | a <- xs, a > x]  
in  smallSort ++ [x] ++ biggerSort

And here is what I've tried to do with it so I could sort lists of tuples (Int,[Int]). I want to sort the tuples by the first element of the tuple, so if I get a lista like [(2,[1]),(1,[]),(3,[2,1])] it returns this [(1,[]),(2,[1]), (3,[2,1])].

quickSortTuplas ((x,(ks)) : []) = [(x,(ks))]
quickSortTuplas ((x,(ks)):ps) =   
let smallSort  = quickSort [a | a <- ps, a < x]  
    biggerSort = quickSort [a | a <- ps, a > x]  
in  smallSort ++ [(x,(ks))] ++ biggerSort

If I try to load this, I get the following error:

 Occurs check: cannot construct the infinite type: a ~ (a, [a1])
* In the second argument of `(>)', namely `x'
  In the expression: a > x
  In a stmt of a list comprehension: a > x
* Relevant bindings include
    a :: (a, [a1]) (bound at reNuevoHaskell.hs:60:37)
    biggerSort :: [(a, [a1])] (bound at reNuevoHaskell.hs:60:9)
    ps :: [(a, [a1])] (bound at reNuevoHaskell.hs:58:27)
    ks :: [a1] (bound at reNuevoHaskell.hs:58:22)
    x :: a (bound at reNuevoHaskell.hs:58:19)
    quickSortTuplas :: [(a, [a1])] -> [(a, [a1])]
      (bound at reNuevoHaskell.hs:57:1)

Many thanks in advance for any insight you could give me.

Answer 1

notice in the expresion [a | a <- ps, a < x] [a | a <- ps, a < x] . a is a tuple whereas x is an Int . Hence a < x makes no sense. Any case, because your quicksort works on Ord a , you can use it for ordering a list of tuples as well. Try it`!

quickSort :: Ord a => [a] -> [a]
quickSort [] = []  
quickSort (x:xs) =   
  let smallSort  = quickSort [a | a <- xs, a < x]  
      biggerSort = quickSort [a | a <- xs, a > x]  
   in smallSort ++ [x] ++ biggerSort

main = print $ quickSort [(1,[2,3,4]) , (0, [4,5,6]), (2,[1,2,3])] -- This works fine