Haskell中的类型理解问题

Question

Hey guys so I'm suppose to takes a list of words and returns a list just like it but with the following substitutions made each time they appear as consecutive words.

One example is you and turn it into u

I'm given the following:

hep :: [Word] -> [Word]
type Word = String

now what is giving me problem is that I'm trying to use case expressions so that I won't have to repeat code but I get the following error

Couldn't match expected type `Char' with actual type `[Char]'
In the pattern: "You"
In a case alternative: "You" -> "u" : hep xs
In the expression: case a of { "You" -> "u" : hep xs }

from the following code

hep [] = []
hep [a:xs] = case a of 
    "You" -> "u":hep xs

Anyone tell me what the problem is?

Edit:

I have added the following code

hep [] = [[]]
hep (a:xs) = case a of 
    "you" -> "u":hep xs
    "are" -> "r":hep xs
    "your" -> "ur":hep xs
    "boyfriend" -> "bf":hep xs
    "girlfriend" -> "gf":hep xs
    "great" -> "gr8":hep xs
    a -> a:hep xs

Now how would I be able to add a case so that if the list contains 2 or 3 certain words in an order, I could turn that into an acronym?

Ex

["By","The","way"] = ["btw"]

Answer 1

You're trying to match against a list of a list of Strings, but the type of hep is [Word] -> [Word] , which contradicts that. The error message is referring to this fact.

But I'm guessing what you actually want is this?

hep [] = []
hep (a:xs) = case a of 
    "You" -> "u":hep xs
    a -> a:hep xs

Answer 2

Something like this?

"By" -> let (y:ys) = xs
        in if y=="The" && head ys=="Way"
              then "btw": hep (drop 2 xs)
              else a:hep xs

Although I wouldn't want to write that 50 times in a row. How about this?

import Data.List
import Data.Maybe

hep [] = []
hep (a:xs) = case a of 
              "you" -> "u":hep xs
              "are" -> "r":hep xs
              "your" -> "ur":hep xs
              "boyfriend" -> "bf":hep xs
              "girlfriend" -> "gf":hep xs
              "great" -> "gr8":hep xs
              "by" -> matchPhrase 3
              "see" -> matchPhrase 2
              a -> a:hep xs
            where matchPhrase num = 
                    let found = lookup (concat $ take num (a:xs)) phrase_list
                    in case found of
                        Just _ -> fromJust found : hep (drop num (a:xs)) 
                        Nothing -> a:hep xs

phrase_list = [("bytheway", "btw"), ("seeyou","cu")]