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我在 Haskell 中编写了一个程序，它必须加载和解析 UTF8 格式的大文本文件。该文件代表一个字典，每行都有键：值对。在我的程序中，我想要一个用于快速字典搜索的 Data.Map 容器。我的文件大约 40MB，但在将其加载到我的程序后，使用了 1.5 GB 的 RAM，并且从未释放。我做错什么了？内存使用量是预期的吗？

这是我的程序中的代码示例：

模块主要在哪里

import Engine

import Codec.Archive.Zip
import Data.IORef
import System.IO
import System.Directory
import qualified System.IO.UTF8 as UTF8
import qualified Data.ByteString.Lazy as B
import qualified Data.ByteString.UTF8 as BsUtf
import qualified Data.Map as Map

import Graphics.UI.Gtk
import Graphics.UI.Gtk.Glade

maybeRead :: Read a => BsUtf.ByteString -> Maybe a
maybeRead s = case reads $ BsUtf.toString s of
     [(x, "")] -> Just x
     _         -> Nothing    

parseToEntries :: [BsUtf.ByteString] -> [(BsUtf.ByteString, Int)]
parseToEntries [] = []
parseToEntries (x:xs) = let (key, svalue) = BsUtf.break (==':') x
                            value = maybeRead svalue
                        in case value of 
                            Just x -> [(key, x)] ++ parseToEntries xs 
                            Nothing -> parseToEntries xs 

createDict :: BsUtf.ByteString -> IO (Map.Map BsUtf.ByteString Int)
createDict str = do
    let entries = parseToEntries $ BsUtf.lines str
        dict = Map.fromList entries
    return (dict)

main :: IO ()
main = do

    currFileName <- newIORef ""

    dictZipFile <- B.readFile "data.db"    
    extractFilesFromArchive [] $ toArchive dictZipFile
    dictFile <- UTF8.readFile "dict.txt"
    dict <- createDict $ BsUtf.fromString dictFile

...

searchAccent :: Map.Map BsUtf.ByteString Int -> String -> Int
searchAccent dict word = let sword = BsUtf.fromString $ map toLower word
                             entry = Map.lookup sword dict
                         in case entry of
                            Nothing -> -1
                            Just match -> 0
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2 回答 2

7

快速回答。
主要问题是System.IO.UTF8.readFile将文件读入String.

假设瓶颈在这里：

dictFile <- UTF8.readFile "dict.txt"
dict <- createDict $ BsUtf.fromString dictFile

在处理 UTF-8 文本时，最好使用Data.Text而不是ByteString. 尝试这样的事情：

import qualified Data.Text.Lazy as LT
import qualified Data.Text.Lazy.Encoding as LT

...
dictFile <- B.readFile "dict.txt"
dict <- createDict $ LT.decodeUtf8 dictFile

另一个瓶颈是解析数字：您正在转换ByteString为String然后read它。最好使用Data.Text.Lazy.Read：

import qualified Data.Text.Lazy.Read as LT

maybeRead :: LT.Text -> Maybe Int
maybeRead s = case LT.decimal s of
    Left _  -> Nothing
    Right i -> Just i
于 2012-04-05T21:39:19.957 回答
4

HaskellString类型是一个间接（因为懒惰）的字符链表；这在空间上是非常浪费的。对于大量文本，您可能希望尝试Data.Text（来自http://hackage.haskell.org/package/text ）。

（现在编辑源代码我看到字符串是惰性ByteString的而不是String，所以这不相关。分析是下一步。）

于 2012-04-05T21:31:43.220 回答