I have a code histogram
, which shows the number of all characters in String.
histogram :: String -> [(Char, Int)]
histogram = foldr help [] where
help x [] = [(x,1)]
help x ((y,z):yzs)
| x == y = (y,z+1) : yzs
| otherwise = (y,z) : help x yzs
But the result is shown in a chaotic order. I want to sort the result by the second element of the list (ie by numbers). For this I implemented ssort
code but it sorts only by the first element (char).
import Data.List (minimum, delete)
ssort :: Ord t => [t] -> [t]
ssort [] = []
ssort xs = x : ssort (remove x xs) where
x = minimum xs
remove :: Eq a => a -> [a] -> [a]
remove _ [] = []
remove y (x:xs) | y == x = xs
| otherwise = x : remove y xs
Input: ssort [('a',1), ('g', 6), ('o',2), ('f',0)]
Output: [('a',1),('f',0),('g',6),('o',2)] (instead of [('f',0),('a',1),('o',2),('g',6)])
I also don't want to use any other builtins, as sort, sortOn and sortBy, I want to implement my own function.
How can I make it so that only the second elements are compared?
You may provide your own definition of minimumBy
:
minimumBy2 :: Ord t2 => [(t1, t2)] -> (t1, t2)
minimumBy2 (h:t) = go h t where
go x [] = x
go x (h0:t0) = go (min x h0) t0
minimumBy2 [] = error "minimumBy2: Empty list"
You can then use it in place of minimum
in ssort
and it should work.
Assuming you don't want to use minimumBy:: Foldable t => (a -> a -> Ordering) -> ta -> a
as well.
You can take advantage on how are tuples compared – first by the first element and then by the second. Therefore you can first swap all of the elements of your histogram, then sort and finally swap back:
ssortOn2 :: (Ord t1, Ord t2) => [(t1, t2)] -> [(t1, t2)]
ssortOn2 = map swap . ssort . map swap
(leaving all other functions unchanged)
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