Haskell - 是否存在替换函数？ [英] Haskell - Does a replace function exist?

问题描述

我不知道，是否有像其他语言一样的替换函数。我搜索了一下，但不幸的是没有成功： - （

）

所以我的尝试还很薄弱。 strong> 1st函数：
$ b

  replace :: String  - > String  - > String  - > String 
替换findStr replaceStr myText = replace（）?? 
  

函数：

  replace :: String  - > String  - > String  - > String 
替换[] old new = [] 
 
替换str old new = loop str 
其中
 loop [] = [] 
 loop str = 
 let （prefix，rest）= splitAt n str 
 in 
 if old == prefix  -  found a occurrence？
 then new ++ loop rest  -  yes：replace 
 
 else head str：loop（tail str） -  no：继续寻找
n = long old 
  

第二个功能：

  replaceBasedIdx ::字符串 - > [字符串]  - >字符串 - >字符串
 replaceBasedIdx findStr replaceStrList myText = replace（）??? 
  

该函数应该将myTxt中的第1个findStr替换为replaceStrList的第1个元素，第2个findStr与第二元素等等...

示例：

  replaceBasedIdxa[G，V，X]Haskell is a language
HGskell is V lXnguage
  
  
 
 
 第二种功能的方法：  
 
 
  replaceBasedIdx :: String  - > [字符串]  - >字符串 - > String 
 replaceBasedIdx findStr replaceStrList myText = replaceBasedIdxSub findStr replaceStrList myText 0 
 
 replaceBasedIdxSub :: String  - > [字符串]  - >字符串 - > Int  - > String 
 replaceBasedIdxSub findStr replaceStrList myText counter = loop myText 
其中
 loop [] = [] 
循环myText = 
 let（prefix，rest）= splitAt n myText $ 
中的b $ b如果findStr ==前缀 - 找到一个出现？ 
 then（replaceStrList !!（counter + 1））++循环休息 - 是：替换它
 
 else head myText：loop（tail myText） -  no：继续寻找
n = length findStr 
  
 我现在已经非常接近最终结果，计数器不会增加。 
 
 
 
您可以告诉我，我的错误在哪里？ 
我该如何修改第一个或第二个函数来获得第三个函数？ 
 
 
 第三个函数： 
  replaceBasedIdxMultiple :: [String]  - > [字符串]  - >字符串 - >字符串
 replaceBasedIdxMultiple findStrList replaceStrList myText = replace（）??? 
  
该函数应该将myTxt中的findStrList的每个元素替换为replaceStrList中的相应元素，与1.，2.与2.等等... 
 
 
 
 示例：  
 
 
  replaceBasedIdxMultiple [A，X，G] [N，Y，K]ABXMG
NBYMK 
  
你能帮我解决这个问题吗？一些提示和提示，如何开始它？
 
 
 
我真的不同： - （
 
 
 
谢谢很多提前 
 
 
亲切的问候！ 
 
解决方案
  replace  exists在 Data.List.Utils 中， MissingH 包的一部分。
 
 
实际上，这是一个非常简洁的实现： 
 
 
  replace :: Eq a => [a]  - > [ a]  - > [a]  - > [a] 
替换old new = join new。split old 
  
 
I have to make three functions for replacing of flat strings and in lists.

I don't know, whether there is a replace function like in other languages. I searched for that however unfortunately without success :-(

So my attempt is yet quite thin.

1st function:
replace  ::  String  ->  String  ->  String  ->  String
replace findStr replaceStr myText = replace()??
My approach for the 1st function:
replace :: String -> String -> String -> String
replace [] old new = []

replace str old new = loop str
  where
    loop [] = []
    loop str =
      let (prefix, rest) = splitAt n str
      in
        if old == prefix                -- found an occurrence?
        then new ++ loop rest           -- yes: replace

        else head str : loop (tail str) -- no: keep looking
    n = length old  
2nd function:
replaceBasedIdx ::  String  ->  [String]  ->  String  ->  String
replaceBasedIdx findStr replaceStrList myText = replace()???
This function should replace the 1st findStr in myTxt with the 1st element of replaceStrList, the 2nd findStr with the 2nd element and so on...

Example:
replaceBasedIdx   "a"  ["G","V","X"]  "Haskell is a language"
"HGskell is V lXnguage"
My approach for the 2nd function:
replaceBasedIdx    ::  String  ->  [String]  ->  String  ->  String
replaceBasedIdx    findStr replaceStrList myText = replaceBasedIdxSub findStr replaceStrList myText 0

replaceBasedIdxSub  ::  String  ->  [String]  ->  String  -> Int -> String
replaceBasedIdxSub findStr replaceStrList myText counter = loop myText
  where
    loop [] = []
    loop myText =
      let (prefix, rest) = splitAt n myText
      in
        if findStr == prefix                                -- found an occurrence?
        then (replaceStrList !! (counter+1)) ++ loop rest   -- yes: replace it

        else head myText : loop (tail myText)               -- no: keep looking
    n = length findStr
I'm now very near to the final result, however the counter doesn't increment.

Could you please tell me, where my mistake is?
And how could I modifey the 1st or 2nd function to get the 3rd function also?

3rd function:
replaceBasedIdxMultiple  ::  [String]  ->  [String]  ->  String  ->  String
replaceBasedIdxMultiple  findStrList replaceStrList myText = replace()???
This function should replace each element of findStrList in myTxt with the corresponding element from the replaceStrList, so 1. with 1., 2. with 2. and so on...

Example: 
replaceBasedIdxMultiple ["A","X","G"] ["N","Y","K"]  "ABXMG"
"NBYMK"
Could you help me with this? some tips and hints, how to begin with it?

I'm really disparate :-(

Thanks a lot in advance

Kind greetings!
解决方案
replace exists in Data.List.Utils, part of MissingH package.

Actually, it's a really concise implementation:
replace :: Eq a => [a] -> [a] -> [a] -> [a]
replace old new = join new . split old


                        
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