替换 ByteString 中的换行符

Question

我想要一个接受 ByteString 并将换行符 \n 和 \n\r 替换为逗号的函数，但想不出一个好的方法。

import qualified Data.ByteString as BS
import Data.Char (ord) 
import Data.Word (Word8)

endlWord8 = fromIntegral $ ord '\n' :: Word8

replace :: BS.ByteString -> BS.ByteString

我想过使用 BS.map，但看不出如何使用，因为我无法在 Word8 上进行模式匹配。另一种选择是 BS.split 然后用 Word8 逗号连接，但这听起来很慢而且不优雅。有任何想法吗？

Answer 1

使用 Data.ByteString.Char8 摆脱讨厌的 Word8、Char 转换，否则您必须这样做。根据 Data.ByteString.Char8 first sentence 不应更改性能。

另外使用 B.span 而不是 B.split，因为您还想替换 \n\r 组合，而不仅仅是 \n.

我自己的（可能很笨拙）尝试这样做：

module Test where

import Data.Monoid ((<>))
import Data.ByteString.Char8 (ByteString)
import qualified Data.ByteString.Char8 as B
import qualified Data.ByteString.Builder as Build
import qualified Data.ByteString.Lazy as LB

eatNewline :: ByteString -> (Maybe Char, ByteString)
eatNewline string
  | B.null string = (Nothing, string)
  | B.head string == '\n' && B.null (B.tail string) = (Just ',', B.empty)
  | B.head string == '\n' && B.head (B.tail string) /= '\r' = (Just ',', B.drop 1 string)
  | B.head string == '\n' && B.head (B.tail string) == '\r' = (Just ',', B.drop 2 string)
  | otherwise = (Nothing, string)

replaceNewlines :: ByteString -> ByteString
replaceNewlines = LB.toStrict . Build.toLazyByteString . go mempty
  where
    go :: Build.Builder -> ByteString -> Build.Builder
    go builder string = let (chunk, rest) = B.span (/= '\n') string
                            (c, rest1)    = eatNewline rest
                            maybeComma    = maybe mempty Build.char8 c
                        in if B.null rest1 then
                             builder <> Build.byteString chunk <> maybeComma
                           else
                             go (builder <> Build.byteString chunk <> maybeComma) rest1

希望 Data.ByteString.Builder 的 mappend 与 mappend 已用于其中一个操作数的次数不是线性关系，否则，此处将出现二次算法.

替换 ByteString 中的换行符

Replace newlines in ByteString

haskell

replace

pattern-matching

bytestring