Word frequency finder

Question

Challenge: find the top three most frequently used words in a string. Only imports from the base package are allowed (source: codewars)

I'm looking for feedback on (1) readability, and (2) performance

• Was folding the string into "frequency map" a good choice?
• I have several small functions that I compose together. Are these abstraction layers helping or hurting readability?
• How are my functions names?
• Does the base package have any libraries that would make this code significantly easier to write?
• Could I benefit from using more pointfree style?

import qualified Data.Map as Map
import Data.List
import Data.Function
import Data.Char
import Data.Maybe

top3 :: [Char] -> [[Char]]
top3 str =
  let
    wordFrequencyMap = foldr (Map.alter increment) Map.empty (normalizedWords str)
    sortedWordFrequencies = sortBy reverseBySecond (Map.toList wordFrequencyMap)
  in map fst (take 3 sortedWordFrequencies)
  where
    normalizedWords = filter containsAtLeastOneAlphaNumericChar . lowerCaseWords . map isLegalChar
    containsAtLeastOneAlphaNumericChar w = find isAlphaNum w /= Nothing
    lowerCaseWords = map (map toLower) . words
    isLegalChar c = if isAlphaNum c || c =='\'' then c else ' '
    reverseBySecond x y = if snd x < snd y then GT else LT
    increment Nothing = Just 1
    increment (Just x) = Just (x + 1)

Answer 1

I'd write

increment = (<|> (Just 1)) . fmap (+1)

You can probably also write

reverseBySecond = flip $ comparing snd

if you don't mind getting EQ for equal pairs.

Answer 2

i had already answered this from within my http://lynx.browser.org/ unbloated hypertext browser within superfast http://www.washington.edu/alpine/ mail commander but http://stackexchange.com/ does not behave like written from humans for humans and rather unfortunately insists on a robot captcha unsolvable on a braille typewriter.

i will recreate my first intuitive answer which goes more lambda calculus and less hoare calculus than your code to try

top3=map head.take 3.reverse.sortBy(comparing length).group.sort.words