Question

嘿伙计们，所以我想要一个单词列表并返回一个类似的列表但是每次出现连续单词时都会进行以下替换。

一个例子是you并将其转换为u

我给出了以下内容：

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

现在给我的问题是我正在尝试使用case表达式，这样我就不必重复代码了但是我得到以下错误

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 }

来自以下代码

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

有人告诉我这是什么问题吗？

编辑：

我添加了以下代码

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

现在我如何能够添加案例，以便如果列表中包含2或3个特定单词，我可以将其转换为首字母缩略词？

实施例

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

Answer 1

您尝试匹配字符串列表的列表，但hep的类型为[Word] -> [Word]，这与此相矛盾。错误消息指的是这个事实。

但我猜你真正想要的是这个吗？

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

Answer 2

这样的东西？

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

虽然我不想连续写50次。怎么样？

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")]

了解haskell中类型的问题

2 个答案: