Question

如何将未排序的单词列表转换为元组列表（单词，计数），按单词排序（fst）min-max不区分大小写？

["AAA","America", "England", "Spain"," Ireland"," aeroplane",
 "aeroplane", "Aeroplane", "AeRoPlAne", "helicopter", "HELICOPTER",
 "Zebra", "America", "Elephant", "abc", "A.B.C." , "haskell", "Haskell", "ZZZ", "zzz"]

预期产出：

[("a.b.c.",1),("aaa",1),("abc",1),("aeroplane",4),("america",2),("elephant",1)
,("england",1),("haskell",2),("helicopter",2),("ireland",1),("spain",1),("zebra",1),("zzz",2)]

[("a.b.c",1), ("zzz", 2)]

Answer 1

我已将其分解为应该使这更容易掌握的步骤。首先，对列表进行排序，然后将其缩减为（Word，Count）元组列表。

import Data.Char (toLower)

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []  
quicksort (x:xs) =
  (quicksort [a|a<-xs, a<=x]) ++ [x] ++ (quicksort [a|a<-xs, a>x])

addCount :: String -> [(String, Int)] -> [(String, Int)]
addCount a [] = [(a, 1)]
addCount a (x:xs)
  | a == fst x = (fst x, (snd x) + 1) : xs
  | otherwise  = x : (addCount a xs)

occurrences :: [(String, Int)] -> [String] -> [(String, Int)]
occurrences counts [] = counts
occurrences counts (x:xs) =
  occurrences countL xs
  where countL = addCount x counts

myList = ["AAA","America", "England", "Spain","Ireland", "aeroplane",
          "aeroplane", "Aeroplane", "AeRoPlAne", "helicopter", "HELICOPTER",
          "Zebra", "America", "Elephant", "abc", "A.B.C." , "haskell", "Haskell",
          "ZZZ", "zzz"]

main = do
  let downcased = map (map toLower) myList
  let sorted = quicksort downcased
  let occurencesL = occurrences [] sorted
  putStrLn . show $ occurencesL
  putStrLn . show $ [head occurencesL, last occurencesL]

这是非常详细，但我希望你能更容易理解这种情况。

更短路的一个例子：

  import Data.Char (toLower)
  import Data.List

  main = do
    let l = map (\a -> (head a, length a)) $ groupBy (\x y -> (map toLower x) == (map toLower y)) myList
    putStrLn . show $ l
    putStrLn . show $ [head l, last l]

Answer 2

这是一个单行：

    Data.List.sort $ Data.Map.toList $ Data.Map.fromListWith (+) [(map Data.Char.toLower word, 1) | word <- lst]

Answer 3

分组功能：

myGroup :: [[Char]] -> [([Char],Int)]
myGroup = fmap (\x -> (head x,length x)) . groupBy (==) . sort . fmap (fmap toLower . filter (not . isSpace))

请注意，我假设您要从String中删除空格。您可以通过更改过滤器轻松更改该部分。

然后边界功能：

headLast :: [t] -> [t]
headLast xs = [head xs,last xs]

快速排序并将其转换为元组列表（elem，count）

3 个答案: