更快的ByteString构造技巧

Question

我是Haskell的新手,我一直坚持效率问题.

任务是:从4GB文本文件构建CSV文件,其中列具有恒定大小

列大小是已知的,例如[col1:4个字符宽,col2:2个字符宽,等等...
文件只能包含[A-Z0-9] ASCII字符,因此转义单元格没有意义

I have: 

$ cat example.txt 
AAAABBCCCC...
AAA1B1CCC1...
... (72 chars per line, usually 50 mln lines)


I need: 
$ cat done.csv
AAAA,BB,CCCC, ...
AAA1,B1,CCC1, ...
...

这是我在Haskell中最快的代码,大约需要2分钟来处理整个4G​​B文件.
我需要最多30秒

import qualified Data.ByteString.Lazy as BL
import qualified Data.ByteString as B
import qualified Data.ByteString.Unsafe as U
import Data.ByteString.Lazy.Builder
import Data.Monoid
import Data.List

col_sizes = intercalate [1] $ map (`replicate` 0) cs
  where
    cs = [4, 4, 4, 3, 5, 1, 1, 3, 3, 3, 3, 3, 3, 10, 3, 1, 1, 1, 2, 3, 10]

sp = char8 ',' -- column separator
nl = char8 '\n'

separator !cs !cl !xs !xl !ci !xi
  | c  == 1   = ps
  | xi == xl  = mempty -- at the end of bytestring, end recursion
  | cl == ci  = pr
  | otherwise = pc
  where
    c  = U.unsafeIndex cs ci         -- get column separation indicator
    w  = word8 . U.unsafeIndex xs    -- get char from BS at position
    p  = separator cs cl xs xl       -- partial recursion call
    pr = nl   <> p  0       (xi + 1) -- end of row, put '\n', reset counter, recur
    ps = sp   <> p (ci + 1)  xi      -- end of column, put column separator, recur
    pc = w xi <> p (ci + 1) (xi + 1) -- in the middle of column, copy byte, recur


main = do
  contents <- B.getContents
  BL.putStr . toLazyByteString $ init_sep sp_after_char contents


init_sep cs xs = separator cs (l cs) xs (l xs) 0 0
  where l = fromIntegral . B.length

sp_after_char = B.pack col_sizes

这是我在C http://pastebin.com/Kjz3Mugs中的实现 (要在此处粘贴它...)
大约需要5秒钟来处理同一个文件

所以我的Haskell代码是约.慢20倍.

因为Haskell ByteString过滤器和映射比我在C中的实现更快
(两个处理相同的文件只需要不到2秒做一些简单的修改)
我希望我的Haskell代码中有一些错误,我不会被迫使用C.

更新:测试数据生成器可在此处获取http://pastebin.com/aJ3RW3jG

在生产时,数据从一个二进制文件传送到另一个二进制文件,因此没有硬盘驱动器IO

测试解决方案我使用SSD驱动器,但我认为Ext4缓存了该文件无论如何在RAM中

time cat test.txt > /dev/null
cat test.txt > /dev/null  0,00s user 0,35s system 99% cpu 0,353 total

香草发生器:

time ./data_builder | head -50000000 > /dev/null
./data_builder  0,02s user 1,09s system 30% cpu 3,709 total
head -50000000 > /dev/null  2,95s user 0,76s system 99% cpu 3,708 total

我的C解决方案:

time ./tocsvc < test.txt > /dev/null 
./tocsvc < test.txt > /dev/null  5,35s user 0,35s system 100% cpu 5,689 total

与发电机

time ./data_builder | head -50000000 | ./tocsvc > /dev/null
./data_builder  0,02s user 1,18s system 18% cpu 6,460 total
head -50000000  3,15s user 1,19s system 67% cpu 6,459 total
./tocsvc > /dev/null  5,81s user 0,55s system 98% cpu 6,459 total

@GabrielGonzalez Haskell解决方案

time ./tocsvh1 < test.txt > /dev/null 
./tocsv < test.txt > /dev/null  19,56s user 0,41s system 100% cpu 19,950 total

与发电机

time ./data_builder | head -50000000 | ./tocsvh1 > /dev/null 
./data_builder  0,11s user 3,04s system 7% cpu 41,320 total
head -50000000  7,29s user 3,56s system 26% cpu 41,319 total
./tocsvh2 > /dev/null  33,01s user 2,42s system 85% cpu 41,327 total

我的Haskell解决方案

time ./tocsvh2 < test.txt > /dev/null 
./tocsvh2 < test.txt > /dev/null  128,63s user 2,95s system 100% cpu 2:11,45 total

与发电机

time ./data_builder | head -50000000 | ./tocsvh2 > /dev/null 
./data_builder  0,02s user 1,26s system 28% cpu 4,526 total
head -50000000  3,17s user 1,33s system 99% cpu 4,524 total
./tocsvh2 > /dev/null  129,95s user 3,33s system 98% cpu 2:14,75 total

@LukeTaylor解决方案

time ./tocsvh3 < test.txt > /dev/null 
./tocsv < test.txt > /dev/null  324,38s user 4,13s system 100% cpu 5:28,18 total

与发电机

time ./data_builder | head -50000000 | ./tocsvh3 > /dev/null 
./data_builder  0,43s user 4,46s system 1% cpu 5:30,34 total
head -50000000  5,20s user 2,82s system 2% cpu 5:30,34 total
./tocsv > /dev/null  329,08s user 4,21s system 100% cpu 5:32,96 total

Answer 1

通过使用原始指针操作，我能够将 C 的系数控制在 3 倍之内：

import Control.Monad (unless, when, void)
import Foreign.Safe hiding (void)
import System.IO
import Foreign.C.Types

bufInSize :: Int
bufInSize = n * (1024 * 1024 `div` n) where n = sum sizes0 + 1

bufOutSize :: Int
bufOutSize = n * (1024 * 1024 `div` n) where n = sum sizes0 + length sizes0

sizes0 :: [Int]
sizes0 = [4, 4, 4, 3, 5, 1, 1, 3, 3, 3, 3, 3, 3, 10, 3, 1, 1, 1, 2, 3, 10]

-- I also tried using the C memset using the FFI, but got the same speed
memcpy :: Ptr Word8 -> Ptr Word8 -> Int -> IO ()
memcpy dst src n = when (n > 0) $ do
    x <- peek src
    poke dst x
    memcpy (dst `plusPtr` 1) (src `plusPtr` 1) (n - 1)

main = do
    allocaArray bufInSize  $ \bufIn0  -> do
    allocaArray bufOutSize $ \bufOut0 -> do
    with (44 :: Word8)  $ \cm -> do
        let loop bufIn bufOut sizes suffixIn suffixOut = do
                let (bytesIn, bytesOut, sizes', copy) = case sizes of
                        []     -> (1, 1    , sizes0,    memcpy bufOut bufIn 1)
                        [s]    -> (s, s    , []    ,    memcpy bufOut bufIn s)
                        s:izes -> (s, s + 1, izes  , do
                            memcpy  bufOut              bufIn s
                            memcpy (bufOut `plusPtr` s) cm    1 )

                if suffixIn < bytesIn
                then do
                    eof <- hIsEOF stdin
                    if eof
                        then hPutBuf stdout bufOut0 (bufOut `minusPtr` bufOut0)
                        else do
                            suffixIn' <- hGetBuf stdin bufIn0 bufInSize
                            loop bufIn0 bufOut sizes suffixIn' suffixOut
                else if suffixOut < bytesOut
                then do
                    hPutBuf stdout bufOut0 (bufOut `minusPtr` bufOut0)
                    loop bufIn bufOut0 sizes suffixIn bufOutSize
                else do
                    copy
                    loop (bufIn  `plusPtr` bytesIn )
                         (bufOut `plusPtr` bytesOut)
                         sizes'
                         (suffixIn  - bytesIn )
                         (suffixOut - bytesOut)
        loop bufIn0 bufOut0 sizes0 0 bufOutSize

以下是time使用 1000000 行输入文件进行的一些粗略测量：

$ # The C Version
$ time ./a.out < in.dat > out.dat

real    0m0.189s
user    0m0.116s
sys 0m0.068s
$ # The Haskell version
$ time ./csv < in.dat > out2.dat

real    0m0.536s
user    0m0.428s
sys 0m0.104s
$ diff out.dat out2.dat
$ # No difference