显示在haskell中重复的单词列表

Question

I need to be able to write a function that shows repeated words from a string and return a list of strings in order of its occurrence and ignore non-letters

eg at hugs prompt

repetitions :: String -> [String]

repetitions > "My bag is is action packed packed."
output> ["is","packed"]
repetitions > "My name  name name is Sean ."
output> ["name","name"]
repetitions > "Ade is into into technical drawing drawing ."
output> ["into","drawing"]

Answer 1

To split a string into words, use the words function (in the Prelude). To eliminate non-word characters, filter with Data.Char.isAlphaNum . Zip the list together with its tail to get adjacent pairs (x, y) . Fold the list, consing a new list that contains all x where x == y .

Someting like:

repetitions s = map fst . filter (uncurry (==)) . zip l $ tail l
  where l = map (filter isAlphaNum) (words s)

I'm not sure that works, but it should give you a rough idea.

Answer 2

I am new to this language so my solution could be a kind of ugly in the eyes of an Haskell veteran, but anyway:

let repetitions x = concat (map tail (filter (\x -> (length x) > 1) (List.group (words (filter (\c -> (c >= 'a' && c <= 'z') || (c>='A' && c <= 'Z') ||  c==' ') x)))))

This part will remove all non letters and non spaces from a string s :

filter (\c -> (c >= 'a' && c <= 'z') || (c>='A' && c <= 'Z') ||  c==' ') s

This one will split a string s to words and group the same words to lists returning list of lists:

List.group (words s)

When this part will remove all lists with less than two elements:

filter (\x -> (length x) > 1) s

After what we will concatenate all lists to one removing one element from them though

concat (map tail s)

Answer 3

This might be inelegent, however it is conceptually very simple. I'm assuming that its looking for consecutive duplicate words like the examples.

-- a wrapper that allows you to give the input as a String
repititions :: String -> [String]
repititions s = repititionsLogic (words s)
-- dose the real work 
repititionsLogic :: [String] -> [String]
repititionsLogic [] = []
repititionsLogic [a] = []
repititionsLogic (a:as) 
    | ((==) a (head as)) = a : repititionsLogic as
    | otherwise = repititionsLogic as

Answer 4

Building on what Alexander Prokofyev answered:

repetitions x = concat (map tail (filter (\\x -> (length x) > 1) (List.group (word (filter (\\c -> (c >= 'a' && c <= 'z') || (c>='A' && c <= 'Z') || c==' ') x)))))

Remove unnecessary parenthesis:

repetitions x = concat (map tail (filter (\\x -> length x > 1) (List.group (word (filter (\\c -> c >= 'a' && c <= 'z' || c>='A' && c <= 'Z' || c==' ') x)))))

Use $ to remove more parenthesis (each $ can replace an opening parenthesis if the ending parenthesis is at the end of the expression):

repetitions x = concat $ map tail $ filter (\\x -> length x > 1) $ List.group $ word $ filter (\\c -> c >= 'a' && c <= 'z' || c>='A' && c <= 'Z' || c==' ') x

Replace character ranges with functions from Data.Char, merge concat and map:

repetitions x = concatMap tail $ filter (\\x -> length x > 1) $ List.group $ word $ filter (\\c -> isAlpha c || isSeparator c) x

Use a section and currying in points-free style to simplify (\\x -> length x > 1) to ((>1) . length) . This combines length with (>1) (a partially applied operator, or section ) in a right-to-left pipeline.

repetitions x = concatMap tail $ filter ((>1) . length) $ List.group $ word $ filter (\\c -> isAlpha c || isSeparator c) x

Eliminate explicit "x" variable to make overall expression points-free:

repetitions = concatMap tail . filter ((>1) . length) . List.group . word . filter (\\c -> isAlpha c || isSeparator c)

Now the entire function, reading from right to left, is a pipeline that filters only alpha or separator characters, splits it into words, breaks it into groups, filters those groups with more than 1 element, and then reduces the remaining groups to the first element of each.