Haskell正則表達式庫替換/替換

是否有一個高級API用於在Haskell中使用正則表達式進行搜索和替換？特別是，我正在查看Text.Regex.TDFA或Text.Regex.Posix包。我真的很喜歡類型的東西：Haskell正則表達式庫替換/替換

f :: Regex -> (ResultInfo -> m String) -> String -> m String

因此，舉例來說，以取代「狗」與「貓」你可以寫

runIdentity . f "dog" (return . const "cat") -- :: String -> String

或做更多先進的東西與單子一樣，計數發生次數等。

Haskell文檔很缺乏。一些低級的API說明是here。

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2012-01-30 gatoatigrado

我不知道，創建此功能的任何現有的功能，但我認爲我會最終使用像AllMatches [] (MatchOffset, MatchLength) instance of RegexContent東西來模擬它：

replaceAll :: RegexLike r String => r -> (String -> String) -> String -> String 
replaceAll re f s = start end 
    where (_, end, start) = foldl' go (0, s, id) $ getAllMatches $ match re s 
     go (ind,read,write) (off,len) = 
      let (skip, start) = splitAt (off - ind) read 
       (matched, remaining) = splitAt len matched 
      in (off + len, remaining, write . (skip++) . (f matched ++)) 

replaceAllM :: (Monad m, RegexLike r String) => r -> (String -> m String) -> String -> m String 
replaceAllM re f s = do 
    let go (ind,read,write) (off,len) = do 
     let (skip, start) = splitAt (off - ind) read 
     let (matched, remaining) = splitAt len matched 
     replacement <- f matched 
     return (off + len, remaining, write . (skip++) . (replacement++)) 
    (_, end, start) <- foldM go (0, s, return) $ getAllMatches $ match re s 
    start end

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2012-01-30 23:30:14 rampion

如何在包裝文字的subRegex .Regex？

Prelude Text.Regex> :t subRegex 
subRegex :: Regex -> String -> String -> String 

Prelude Text.Regex> subRegex (mkRegex "foo") "foobar" "123" 
"123bar"

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2012-01-31 09:47:08 Simon

也許這種方法適合你。

import Data.Array (elems) 
import Text.Regex.TDFA ((=~), MatchArray) 

replaceAll :: String -> String -> String -> String   
replaceAll regex new_str str = 
    let parts = concat $ map elems $ (str =~ regex :: [MatchArray]) 
    in foldl (replace' new_str) str (reverse parts) 

    where 
    replace' :: [a] -> [a] -> (Int, Int) -> [a] 
    replace' new list (shift, l) = 
     let (pre, post) = splitAt shift list 
     in pre ++ new ++ (drop l post)

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2013-06-08 04:30:14 Anton

基於@風鈴草的答案，但有固定的錯字所以它不只是<<loop>>：

replaceAll :: Regex -> (String -> String) -> String -> String 
replaceAll re f s = start end 
    where (_, end, start) = foldl' go (0, s, id) $ getAllMatches $ match re s 
     go (ind,read,write) (off,len) = 
      let (skip, start) = splitAt (off - ind) read 
       (matched, remaining) = splitAt len start 
      in (off + len, remaining, write . (skip++) . (f matched ++))

來源

2013-06-17 14:31:54

您可以使用replaceAll從Data.Text.ICU.Replace module。

Prelude> :set -XOverloadedStrings 
Prelude> import Data.Text.ICU.Replace 
Prelude Data.Text.ICU.Replace> replaceAll "cat" "dog" "Bailey is a cat, and Max is a cat too." 
"Bailey is a dog, and Max is a dog too."

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2016-09-03 12:44:34

Haskell正則表達式庫替換/替換

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