在haskell中删除单词 [英] stemming words removal in haskell

问题描述

我的目标是从用户处获取一个字符串，并检查该字符串中是否存在可用的词干单词。如果是的话，我想删除词干并返回字符串。

 例如：如果输入是：他正在钓鱼和捕捉 
的输出是：他是鱼和捕捉
  

我可以在其他语言中执行此操作。但我不知道如何在Haskell中做到这一点。做任何一个知道如何使用Haskell解决这个问题..

  checkstemming :: [String]  - > [String] 
 checkstemming [] = [] 
 checkstemming xs = foldr（++）[]（[drop x | x < -  xs，x =ing]）
  

我试过了，但我知道它错了。任何人都可以帮助我解决这个问题，请.. ..
解决方案

一次解决一个字，这使事情变得更容易：

  removeStemming :: String  - > String 
 removeStemming [] = [] 
 removeStemming（x：ing）= [x] 
 removeStemming（x：xs）= x：removeStemming xs 
 
 checkStemming :: [String]  - > [String] 
 checkStemming = map removeStemming 
  

另一种方法是使用 isSuffixOf 来自 Data.List ：

  removeStemming :: String  - >字符串
 removeStemming xs 
 | ingisSuffixOf` xs = take（length xs  -  3）xs 
 |否则= xs 
  

am new to Haskell and functional programing..

my aim is to get a string from user and check whether there is any stemming words available in that string. if so i want to remove the stemming and return the string.

Eg: if input is :"He is fishing and catching"
    output is : "he is fish and catch"

i can do this in other languages. but i don't know how to do it in Haskell. do any one know how to solve this problem using Haskell..

checkstemming :: [String] -> [String]
checkstemming [] = []
checkstemming xs = foldr (++) [] ( [ drop x |x <- xs, x="ing"])

i tried, but i know its wrong. can anyone help me to sole this problem please..

解决方案

Tackle one word at a time, this makes things much easier:

removeStemming :: String -> String
removeStemming []        = []
removeStemming (x:"ing") = [x]
removeStemming (x:xs)    = x : removeStemming xs

checkStemming :: [String] -> [String]
checkStemming = map removeStemming

Another way to do this is to use isSuffixOf from Data.List:

removeStemming :: String -> String
removeStemming xs
  | "ing" `isSuffixOf` xs = take (length xs - 3) xs
  | otherwise             = xs

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