在功能上按分隔符拆分字符串的最佳方法是什么？

Question

我尝试在 Haskell 中编写程序，该程序将采用逗号分隔的整数字符串，将其转换为整数列表并将每个数字递增 1。

例如 "1,2,-5,-23,15" -> [2,3,-4,-22,16]

下面是生成的程序

import Data.List

main :: IO ()
main = do
  n <- return 1
  putStrLn . show . map (+1) . map toInt . splitByDelimiter delimiter
    $ getList n

getList :: Int -> String
getList n = foldr (++) [] . intersperse [delimiter] $ replicate n inputStr

delimiter = ','

inputStr = "1,2,-5,-23,15"

splitByDelimiter :: Char -> String -> [String]
splitByDelimiter _ "" = []
splitByDelimiter delimiter list =
  map (takeWhile (/= delimiter) . tail)
    (filter (isPrefixOf [delimiter])
       (tails
           (delimiter : list)))

toInt :: String -> Int
toInt = read

对我来说最困难的部分是函数 splitByDelimiter 的编程，它接受一个字符串并返回字符串列表

"1,2,-5,-23,15" -> ["1","2","-5","-23","15"]

认为它有效，但我对它的编写方式不满意。有很多括号，所以看起来像 Lisp。该算法也有些人为：

1. 在字符串开头添加分隔符",1,2,-5,-23,15"

2. 生成所有尾部列表[",1,2,-5,-23,15", "1,2,-5,-23,15", ",2,-5,-23,15", .... ]

3. 过滤并只留下以分隔符[",1,2,-5,-23,15", ",2,-5,-23,15", .... ]开头的字符串

4. 删除第一个分隔符并取符号，直到遇到下一个分隔符["1", "2", .... ]

所以问题是：

如何改进功能splitByDelimiter？

我可以删除前置和删除分隔符并直接拆分字符串吗？

如何重写函数以减少括号？

可能是我遗漏了什么，并且已经有具有此功能的标准功能？

Answer 1

Data.List.Split.splitOn 不这样做吗？

Answer 2

splitBy delimiter = foldr f [[]] 
            where f c l@(x:xs) | c == delimiter = []:l
                             | otherwise = (c:x):xs

编辑：不是原作者的，但下面是一个更（过度？）冗长且不太灵活的版本（特定于Char/String），以帮助阐明其工作原理。使用上述版本，因为它适用于具有 Eq 实例的任何类型列表。

splitBy :: Char -> String -> [String]
splitBy _ "" = [];
splitBy delimiterChar inputString = foldr f [""] inputString
  where f :: Char -> [String] -> [String]
        f currentChar allStrings@(partialString:handledStrings)
          | currentChar == delimiterChar = "":allStrings -- start a new partial string at the head of the list of all strings
          | otherwise = (currentChar:partialString):handledStrings -- add the current char to the partial string

-- input:       "a,b,c"
-- fold steps:
-- first step:  'c' -> [""] -> ["c"]
-- second step: ',' -> ["c"] -> ["","c"]
-- third step:  'b' -> ["","c"] -> ["b","c"]
-- fourth step: ',' -> ["b","c"] -> ["","b","c"]
-- fifth step:  'a' -> ["","b","c"] -> ["a","b","c"]

Answer 3

这有点小技巧，但见鬼，它确实有效。

yourFunc str = map (+1) $ read ("[" ++ str ++ "]")

这是一个使用unfoldr的非hack版本：

import Data.List
import Control.Arrow(second)

-- break' is like break but removes the
-- delimiter from the rest string
break' d = second (drop 1) . break d

split :: String -> Maybe (String,String)
split [] = Nothing
split xs = Just . break' (==',') $ xs

yourFunc :: String -> [Int]
yourFunc = map ((+1) . read) . unfoldr split

Answer 4

只是为了好玩，下面是如何使用 Parsec 创建一个简单的解析器：

module Main where

import Control.Applicative hiding (many)
import Text.Parsec
import Text.Parsec.String

line :: Parser [Int]
line = number `sepBy` (char ',' *> spaces)

number = read <$> many digit

一个优点是它可以轻松创建一个灵活的解析器：

*Main Text.Parsec Text.Parsec.Token> :load "/home/mikste/programming/Temp.hs"
[1 of 1] Compiling Main             ( /home/mikste/programming/Temp.hs, interpreted )
Ok, modules loaded: Main.
*Main Text.Parsec Text.Parsec.Token> parse line "" "1, 2, 3"
Right [1,2,3]
*Main Text.Parsec Text.Parsec.Token> parse line "" "10,2703,   5, 3"
Right [10,2703,5,3]
*Main Text.Parsec Text.Parsec.Token> 

Answer 5

这是 HaskellElephant 对原始问题的回答的应用，略有改动

splitByDelimiter :: 字符 -> 字符串 -> [字符串] splitByDelimiter = 展开器。拆分单 splitSingle :: Char -> String -> Maybe (String,String) splitSingle _ [] = 没有 splitSingle 分隔符 xs = let (ys, zs) = break (== delimiter) xs in 只是（是的，下降1 zs）

函数 splitSingle 通过第一个分隔符将列表拆分为两个子字符串。

例如： "1,2,-5,-23,15" -> Just ("1", "2,-5,-23,15")

Answer 6

splitBy del str = helper del str []   
    where 
        helper _ [] acc = let acc0 = reverse acc in [acc0] 
        helper del (x:xs) acc   
            | x==del    = let acc0 = reverse acc in acc0 : helper del xs []  
            | otherwise = let acc0 = x : acc     in helper del xs acc0 

Answer 7

这段代码运行良好 使用：- 拆分“你的字符串”[] 并将 ',' 替换为任何分隔符

split [] t = [t]
split (a:l) t = if a==',' then (t:split l []) else split l (t++[a])

Answer 8

import qualified Text.Regex as RegExp

myRegexSplit :: String -> String -> [String]
myRegexSplit regExp theString = 
  let result = RegExp.splitRegex (RegExp.mkRegex regExp) theString
  in filter (not . null) result

-- using regex has the advantage of making it easy to use a regular
-- expression instead of only normal strings as delimiters.

-- the splitRegex function tends to return an array with an empty string
-- as the last element. So the filter takes it out

-- how to use in ghci to split a sentence
let timeParts = myRegexSplit " " "I love ponies a lot"

Answer 9

另一个没有进口的：

splitBy :: Char -> String -> [String]
splitBy _ [] = []
splitBy c s  =
  let
    i = (length . takeWhile (/= c)) s
    (as, bs) = splitAt i s
  in as : splitBy c (if bs == [] then [] else tail bs)