What's the most efficient way to cast String (or ByteString) to an integral type bitwise, for example:
smallEndianStringToInt32 :: ByteString -> Int32
smallEndianStringToInt32 str =
case str of
[a,b,c,d] -> shiftL d 24 .|. shiftL c 16 .|. shiftL b 8 .|. a
_ -> 0
In C we can simply do it like this, really low-cost:
char* some_string = "....";
int32_t x = *(int32_t*)some_string;
But the former implementation (smallEndianStringToInt32) doesn't look as fast as the C code.
How can this be done? (Or it actually IS as fast?)
I think this does what you're looking for, but like the other commenters, I'd recommend you clarify what you're actually trying to do, as this is very non-idiomatic Haskell:
{-# LANGUAGE OverloadedStrings #-}
import Data.ByteString.Internal
import Foreign.Storable
import Foreign.ForeignPtr
import Foreign.Ptr
import Data.Word (Word32)
main :: IO ()
main = withForeignPtr fptr $ \ptr -> do
i <- peek $ castPtr (ptr `plusPtr` off)
print (i :: Word32)
where
PS fptr off len = "A\0\0\0"
As should be painfully obvious, this code is quite susceptible to segfaults, for example.
Related
These two programs do the same thing, but one runs 10x faster.
This takes approx. 10 seconds on my machine:
import Control.Monad
import qualified Data.ByteString as B
import qualified Data.ByteString.Lazy as BL
theValueOne=B.singleton 1
main = replicateM_ 100000000 $ B.putStr theValueOne
The second version uses output-lazy IO. It is done in about 1 second (as fast as c):
import qualified Data.ByteString.Lazy as BL
main = BL.putStr $ BL.pack $ replicate 100000000 1
Question: Why is the non-lazy version so slow? More importantly, how can I make it fast? (I've tried recursion, forM, modifying the output buffer using hSetBuffering... Nothing has made a difference)
Note- This is more than just an academic question. The non-lazy version is an extremely simplified version of an executable my company uses in production, which is also slow in the same way. It would be nearly impossible to re-architect the larger program around the analogous lazy solution.
Updated: Added possible source of problem and a solution.
I don't think it has anything to do with lazy I/O. If you rewrite the strict I/O version to write two bytes at once:
theValueOne = B.singleton 1
main = replicateM_ 50000000 $ B.putStr (theValueOne <> theValueOne)
that halves the time. Write ten bytes at once:
theValueOne = B.singleton 1
main = replicateM_ 10000000 $ B.putStr (foldMap id (replicate 10 theValueOne))
and it's already faster than the lazy I/O version.
The issue is that there's a bit of overhead in a B.hPutStr call, much more than the overhead of a C fwrite call, and it's just not a particularly efficient way to write a single byte.
A good chunk of the overhead comes from the fact that Haskell I/O buffers have immutable metadata. Even though the buffer content itself is mutable, the pointers to valid data within the buffer are immutable, and so writing a single byte requires a heap allocation of a new GHC.IO.Buffer.Buffer structure which GHC can't optimize away
One solution is to use a hand-crafted buffering structure with a mutable pointer. The following works, and it's about twice as fast as the lazy I/O version in the original question.
{-# LANGUAGE RecordWildCards #-}
{-# OPTIONS_GHC -Wall #-}
import Control.Monad
import Data.IORef
import Data.Word
import Foreign.ForeignPtr
import Foreign.Ptr
import Foreign.Storable
import System.IO
data WriteBuffer = WriteBuffer
{ handle :: !Handle
, capacity :: !Int
, used :: !(IORef Int)
, content :: !(ForeignPtr Word8)
}
newBuffer :: Handle -> IO WriteBuffer
newBuffer h = do
hSetBinaryMode h True
hSetBuffering h NoBuffering
WriteBuffer h cap <$> newIORef 0 <*> mallocForeignPtrBytes cap
where cap = 4096
flushBuffer :: WriteBuffer -> IO ()
flushBuffer WriteBuffer{..} = do
n <- readIORef used
withForeignPtr content $ \p -> hPutBuf handle p n
writeIORef used 0
writeByte :: Word8 -> WriteBuffer -> IO ()
writeByte w buf#(WriteBuffer{..}) = do
n <- readIORef used
withForeignPtr content $ \p -> poke (plusPtr p n) w
let n' = n + 1
writeIORef used n'
when (n' == capacity) $
flushBuffer buf
main :: IO ()
main = do
b <- newBuffer stdout
replicateM_ 100000000 (writeByte 1 b)
flushBuffer b
Someone ironically, converting this to a version using an immutable counter and passing the WriteBuffer as state through foldM doubles the speed again, so it's about 4 times as fast as the lazy I/O version in the original question:
{-# LANGUAGE RecordWildCards #-}
{-# OPTIONS_GHC -Wall #-}
import Control.Monad
import Data.Word
import Foreign.ForeignPtr
import Foreign.Ptr
import Foreign.Storable
import System.IO
data WriteBuffer = WriteBuffer
{ handle :: !Handle
, capacity :: !Int
, used :: !Int
, content :: !(ForeignPtr Word8)
}
newBuffer :: Handle -> IO WriteBuffer
newBuffer h = do
hSetBinaryMode h True
hSetBuffering h NoBuffering
WriteBuffer h cap 0 <$> mallocForeignPtrBytes cap
where cap = 4096
flushBuffer :: WriteBuffer -> IO WriteBuffer
flushBuffer buf#WriteBuffer{..} = do
withForeignPtr content $ \p -> hPutBuf handle p used
return $ buf { used = 0 }
writeByte :: Word8 -> WriteBuffer -> IO WriteBuffer
writeByte w buf#(WriteBuffer{..}) = do
withForeignPtr content $ \p -> poke (plusPtr p used) w
let used' = used + 1
buf' = buf { used = used' }
if (used' == capacity)
then flushBuffer buf'
else return buf'
main :: IO ()
main = do
b <- newBuffer stdout
b' <- foldM (\s _ -> writeByte 1 s) b [(1::Int)..100000000]
void (flushBuffer b')
The reason this one is so fast seems to be that GHC is able to optimize away the WriteBuffer constructor entirely from the fold and just pass around unboxed pointers and integers in the loop. My guess is that if I modified the mutable version above to avoid boxing and unboxing the integer in the used IORef, it would be similarly fast.
In an ongoing endeavour to efficiently fiddle with bits (e.g. see this SO question) the newest challenge is the efficient streaming and consumption of bits.
As a first simple task I choose to find the longest sequence of identical bits in a bitstream generated by /dev/urandom. A typical incantation would be head -c 1000000 </dev/urandom | my-exe. The actual goal is to stream bits and decode an Elias gamma code, for example, i.e. codes that are not chunks of bytes or multiples thereof.
For such codes of variable length it is nice to have the take, takeWhile, group, etc. language for list manipulation. Since a BitStream.take would actually consume part of the bistream some monad would probably come into play.
The obvious starting point is the lazy bytestring from Data.ByteString.Lazy.
A. Counting bytes
This very simple Haskell program performs on par with a C program, as is to be expected.
import qualified Data.ByteString.Lazy as BSL
main :: IO ()
main = do
bs <- BSL.getContents
print $ BSL.length bs
B. Adding bytes
Once I start using unpack things should get worse.
main = do
bs <- BSL.getContents
print $ sum $ BSL.unpack bs
Suprisingly, Haskell and C show the almost same performance.
C. Longest sequence of identical bits
As a first nontrivial task the longest sequence of identical bits can be found like this:
module Main where
import Data.Bits (shiftR, (.&.))
import qualified Data.ByteString.Lazy as BSL
import Data.List (group)
import Data.Word8 (Word8)
splitByte :: Word8 -> [Bool]
splitByte w = Prelude.map (\i-> (w `shiftR` i) .&. 1 == 1) [0..7]
bitStream :: BSL.ByteString -> [Bool]
bitStream bs = concat $ map splitByte (BSL.unpack bs)
main :: IO ()
main = do
bs <- BSL.getContents
print $ maximum $ length <$> (group $ bitStream bs)
The lazy bytestring is converted to a list [Word8] and then, using shifts, each Word is split into the bits, resulting in a list [Bool]. This list of lists is then flattened with concat. Having obtained a (lazy) list of Bool, use group to split the list into sequences of identical bits and then map length over it. Finally maximum gives the desired result. Quite simple, but not very fast:
# C
real 0m0.606s
# Haskell
real 0m6.062s
This naive implementation is exactly one order of magnitude slower.
Profiling shows that quite a lot of memory gets allocated (about 3GB for parsing 1MB of input). There is no massive space leak to be observed, though.
From here I start poking around:
There is a bitstream package that promises "Fast, packed, strict bit streams (i.e. list of Bools) with semi-automatic stream fusion.". Unfortunately it is not up-to-date with the current vector package, see here for details.
Next, I investigate streaming. I don't quite see why I should need 'effectful' streaming that brings some monad into play - at least until I start with the reverse of the posed task, i.e. encoding and writing bitstreams to file.
How about just fold-ing over the ByteString? I'd have to introduce state to keep track of consumed bits. That's not quite the nice take, takeWhile, group, etc. language that is desirable.
And now I'm not quite sure where to go.
Update:
I figured out how to do this with streaming and streaming-bytestring. I'm probably not doing this right because the result is catastrophically bad.
import Data.Bits (shiftR, (.&.))
import qualified Data.ByteString.Streaming as BSS
import Data.Word8 (Word8)
import qualified Streaming as S
import Streaming.Prelude (Of, Stream)
import qualified Streaming.Prelude as S
splitByte :: Word8 -> [Bool]
splitByte w = (\i-> (w `shiftR` i) .&. 1 == 1) <$> [0..7]
bitStream :: Monad m => Stream (Of Word8) m () -> Stream (Of Bool) m ()
bitStream s = S.concat $ S.map splitByte s
main :: IO ()
main = do
let bs = BSS.unpack BSS.getContents :: Stream (Of Word8) IO ()
gs = S.group $ bitStream bs :: Stream (Stream (Of Bool) IO) IO ()
maxLen <- S.maximum $ S.mapped S.length gs
print $ S.fst' maxLen
This will test your patience with anything beyond a few thousand bytes of input from stdin. The profiler says it spends an insane amount of time (quadratic in the input size) in Streaming.Internal.>>=.loop and Data.Functor.Of.fmap. I'm not quite sure what the first one is, but the fmap indicates (?) that the juggling of these Of a b isn't doing us any good and because we're in the IO monad it can't be optimised away.
I also have the streaming equivalent of the byte adder here: SumBytesStream.hs, which is slightly slower than the simple lazy ByteString implementation, but still decent. Since streaming-bytestring is proclaimed to be "bytestring io done right" I expected better. I'm probably not doing it right, then.
In any case, all these bit-computations shouldn't be happening in the IO monad. But BSS.getContents forces me into the IO monad because getContents :: MonadIO m => ByteString m () and there's no way out.
Update 2
Following the advice of #dfeuer I used the streaming package at master#HEAD. Here's the result.
longest-seq-c 0m0.747s (C)
longest-seq 0m8.190s (Haskell ByteString)
longest-seq-stream 0m13.946s (Haskell streaming-bytestring)
The O(n^2) problem of Streaming.concat is solved, but we're still not getting closer to the C benchmark.
Update 3
Cirdec's solution produces a performance on par with C. The construct that is used is called "Church encoded lists", see this SO answer or the Haskell Wiki on rank-N types.
Source files:
All the source files can be found on github. The Makefile has all the various targets to run the experiments and the profiling. The default make will just build everything (create a bin/ directory first!) and then make time will do the timing on the longest-seq executables. The C executables get a -c appended to distinguish them.
Intermediate allocations and their corresponding overhead can be removed when operations on streams fuse together. The GHC prelude provides foldr/build fusion for lazy streams in the form of rewrite rules. The general idea is if one function produces a result that looks like a foldr (it has the type (a -> b -> b) -> b -> b applied to (:) and []) and another function consumes a list that looks like a foldr, constructing the intermediate list can be removed.
For your problem I'm going to build something similar, but using strict left folds (foldl') instead of foldr. Instead of using rewrite rules that try to detect when something looks like a foldl, I'll use a data type that forces lists to look like left folds.
-- A list encoded as a strict left fold.
newtype ListS a = ListS {build :: forall b. (b -> a -> b) -> b -> b}
Since I've started by abandoning lists we'll be re-implementing part of the prelude for lists.
Strict left folds can be created from the foldl' functions of both lists and bytestrings.
{-# INLINE fromList #-}
fromList :: [a] -> ListS a
fromList l = ListS (\c z -> foldl' c z l)
{-# INLINE fromBS #-}
fromBS :: BSL.ByteString -> ListS Word8
fromBS l = ListS (\c z -> BSL.foldl' c z l)
The simplest example of using one is to find the length of a list.
{-# INLINE length' #-}
length' :: ListS a -> Int
length' l = build l (\z a -> z+1) 0
We can also map and concatenate left folds.
{-# INLINE map' #-}
-- fmap renamed so it can be inlined
map' f l = ListS (\c z -> build l (\z a -> c z (f a)) z)
{-# INLINE concat' #-}
concat' :: ListS (ListS a) -> ListS a
concat' ll = ListS (\c z -> build ll (\z l -> build l c z) z)
For your problem we need to be able to split a word into bits.
{-# INLINE splitByte #-}
splitByte :: Word8 -> [Bool]
splitByte w = Prelude.map (\i-> (w `shiftR` i) .&. 1 == 1) [0..7]
{-# INLINE splitByte' #-}
splitByte' :: Word8 -> ListS Bool
splitByte' = fromList . splitByte
And a ByteString into bits
{-# INLINE bitStream' #-}
bitStream' :: BSL.ByteString -> ListS Bool
bitStream' = concat' . map' splitByte' . fromBS
To find the longest run we'll keep track of the previous value, the length of the current run, and the length of the longest run. We make the fields strict so that the strictness of the fold prevents chains of thunks from being accumulated in memory. Making a strict data type for a state is an easy way to get control over both its memory representation and when its fields are evaluated.
data LongestRun = LongestRun !Bool !Int !Int
{-# INLINE extendRun #-}
extendRun (LongestRun previous run longest) x = LongestRun x current (max current longest)
where
current = if x == previous then run + 1 else 1
{-# INLINE longestRun #-}
longestRun :: ListS Bool -> Int
longestRun l = longest
where
(LongestRun _ _ longest) = build l extendRun (LongestRun False 0 0)
And we're done
main :: IO ()
main = do
bs <- BSL.getContents
print $ longestRun $ bitStream' bs
This is much faster, but not quite the performance of c.
longest-seq-c 0m00.12s (C)
longest-seq 0m08.65s (Haskell ByteString)
longest-seq-fuse 0m00.81s (Haskell ByteString fused)
The program allocates about 1 Mb to read 1000000 bytes from input.
total alloc = 1,173,104 bytes (excludes profiling overheads)
Updated github code
I found another solution that is on par with C. The Data.Vector.Fusion.Stream.Monadic has a stream implementation based on this Coutts, Leshchinskiy, Stewart 2007 paper. The idea behind it is to use a destroy/unfoldr stream fusion.
Recall that list's unfoldr :: (b -> Maybe (a, b)) -> b -> [a] creates a list by repeatedly applying (unfolding) a step-forward function, starting with an initial value. A Stream is just an unfoldr function with starting state. (The Data.Vector.Fusion.Stream.Monadic library uses GADTs to create constructors for Step that can be pattern-matched conveniently. It could just as well be done without GADTs, I think.)
The central piece of the solution is the mkBitstream :: BSL.ByteString -> Stream Bool function that turns a BytesString into a stream of Bool. Basically, we keep track of the current ByteString, the current byte, and how much of the current byte is still unconsumed. Whenever a byte is used up another byte is chopped off ByteString. When Nothing is left, the stream is Done.
The longestRun function is taken straight from #Cirdec's solution.
Here's the etude:
{-# LANGUAGE CPP #-}
#define PHASE_FUSED [1]
#define PHASE_INNER [0]
#define INLINE_FUSED INLINE PHASE_FUSED
#define INLINE_INNER INLINE PHASE_INNER
module Main where
import Control.Monad.Identity (Identity)
import Data.Bits (shiftR, (.&.))
import qualified Data.ByteString.Lazy as BSL
import Data.Functor.Identity (runIdentity)
import qualified Data.Vector.Fusion.Stream.Monadic as S
import Data.Word8 (Word8)
type Stream a = S.Stream Identity a -- no need for any monad, really
data Step = Step BSL.ByteString !Word8 !Word8 -- could use tuples, but this is faster
mkBitstream :: BSL.ByteString -> Stream Bool
mkBitstream bs' = S.Stream step (Step bs' 0 0) where
{-# INLINE_INNER step #-}
step (Step bs w n) | n==0 = case (BSL.uncons bs) of
Nothing -> return S.Done
Just (w', bs') -> return $
S.Yield (w' .&. 1 == 1) (Step bs' (w' `shiftR` 1) 7)
| otherwise = return $
S.Yield (w .&. 1 == 1) (Step bs (w `shiftR` 1) (n-1))
data LongestRun = LongestRun !Bool !Int !Int
{-# INLINE extendRun #-}
extendRun :: LongestRun -> Bool -> LongestRun
extendRun (LongestRun previous run longest) x = LongestRun x current (max current longest)
where current = if x == previous then run + 1 else 1
{-# INLINE longestRun #-}
longestRun :: Stream Bool -> Int
longestRun s = runIdentity $ do
(LongestRun _ _ longest) <- S.foldl' extendRun (LongestRun False 0 0) s
return longest
main :: IO ()
main = do
bs <- BSL.getContents
print $ longestRun (mkBitstream bs)
Is there a standard Haskell equivalent to NumPy's argsort function?
I'm using HMatrix and, so, would like a function compatible with Vector R which is an alias for Data.Vector.Storable.Vector Double. The argSort function below is the implementation I'm currently using:
{-# LANGUAGE NoImplicitPrelude #-}
module Main where
import qualified Data.List as L
import qualified Data.Vector as V
import qualified Data.Vector.Storable as VS
import Prelude (($), Double, IO, Int, compare, print, snd)
a :: VS.Vector Double
a = VS.fromList [40.0, 20.0, 10.0, 11.0]
argSort :: VS.Vector Double -> V.Vector Int
argSort xs = V.fromList (L.map snd $ L.sortBy (\(x0, _) (x1, _) -> compare x0 x1) (L.zip (VS.toList xs) [0..]))
main :: IO ()
main = print $ argSort a -- yields [2,3,1,0]
I'm using explicit qualified imports just to make it clear where every type and function is coming from.
This implementation is not terribly efficient since it converts the input vector to a list and the result back to a vector. Does something like this (but more efficient) exist somewhere?
Update
#leftaroundabout had a good solution. This is the solution I ended up with:
module LAUtil.Sorting
( IndexVector
, argSort
)
where
import Control.Monad
import Control.Monad.ST
import Data.Ord
import qualified Data.Vector.Algorithms.Intro as VAI
import qualified Data.Vector.Storable as VS
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import Numeric.LinearAlgebra
type IndexVector = VU.Vector Int
argSort :: Vector R -> IndexVector
argSort xs = runST $ do
let l = VS.length xs
t0 <- VUM.new l
forM_ [0..l - 1] $
\i -> VUM.unsafeWrite t0 i (i, (VS.!) xs i)
VAI.sortBy (comparing snd) t0
t1 <- VUM.new l
forM_ [0..l - 1] $
\i -> VUM.unsafeRead t0 i >>= \(x, _) -> VUM.unsafeWrite t1 i x
VU.freeze t1
This is more directly usable with Numeric.LinearAlgebra since the data vector is a Storable. This uses an unboxed vector for the indices.
Use vector-algorithms:
import Data.Ord (comparing)
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Algorithms.Intro as VAlgo
argSort :: (Ord a, VU.Unbox a) => VU.Vector a -> VU.Vector Int
argSort xs = VU.map fst $ VU.create $ do
xsi <- VU.thaw $ VU.indexed xs
VAlgo.sortBy (comparing snd) xsi
return xsi
Note these are Unboxed rather than Storable vectors. The latter need to make some tradeoffs to allow impure C FFI operations and can't properly handle heterogeneous tuples. You can of course always convert to and from storable vectors.
What worked better for me is using Data.map, as it is subject to list fusion, got a speed up. Here n=Length xs.
import Data.Map as M (toList, fromList, toAscList)
out :: Int -> [Double] -> [Int]
out n !xs = let !a= (M.toAscList (M.fromList $! (zip xs [0..n])))
!res=a `seq` L.map snd a
in res
However this is only aplicable for periodic lists, as:
out 12 [1,2,3,4,1,2,3,4,1,2,3,4] == out 12 [1,2,3,4,1,3,2,4,1,2,3,4]
Hi have binaries of float data (single-precision 32-bit IEEE) that I would like to work on.
How can I best load this for further use, ideally as (IOArray Int Float).
bytesToFloats :: ByteString -> [Float]
bytesToFloatArray :: ByteString -> IOArray Int Float
If you've got bog standard single-precision floats, and you just want to work them over in Haskell, you can always be down and dirty about it:
import Data.ByteString.Internal as BS
import qualified Data.Vector.Storable as V
bytesToFloats :: BS.ByteString -> V.Vector Float
bytesToFloats = V.unsafeCast . aux . BS.toForeignPtr
where aux (fp,offset,len) = V.unsafeFromForeignPtr fp offset len
I think you might be happier with Data.Vector:
http://www.haskell.org/haskellwiki/Numeric_Haskell:_A_Vector_Tutorial#Parsing_Binary_Data
You could also use cereal library, for example:
import Control.Applicative
import Data.ByteString
import Data.Serialize
floatsToBytes :: [Float] -> ByteString
floatsToBytes = runPut . mapM_ putFloat32le
-- | Parses the input and returns either the result or an error description.
bytesToFloat :: ByteString -> Either String [Float]
bytesToFloat = runGet $ many getFloat32le
If you can convert 4 bytes to a Word32, you can use the function wordToFloat in the data-binary-ieee754 package to convert it to a float. You could then load this into any kind of list-like structure you want to manipulate it.
What is the best way to convert between Storable.Vector Word8 and a strict ByteString?
Of course a non-copying (no-op) way would be most appreciated.
Should I just unsafeCoerce or is there a library function for that (I couldn't find one)?
Also, will the approach be the same for an Unboxed.Vector Word8?
A simple unsafeCoerce will not work, as the layout of the data constructors is different:
data StorableArray i e = StorableArray !i !i Int !(ForeignPtr e)
vs.
data ByteString = PS {-# UNPACK #-} !(ForeignPtr Word8) -- payload
{-# UNPACK #-} !Int -- offset
{-# UNPACK #-} !Int -- length
You can import Data.Array.Storable.Internals and Data.ByteString.Internal to get access to the raw constructors and then construct one out of the other without copying the data:
> let bs = pack [1,2,3]
> bs
"\SOH\STX\ETX"
> let sa = case bs of (PS ptr 0 n) -> StorableArray 0 (n-1) n ptr
> :t sa
sa :: StorableArray Int GHC.Word.Word8
> Data.Array.MArray.readArray sa 1
2
> Data.Array.MArray.readArray sa 0
1
> Data.Array.MArray.readArray sa 3
*** Exception: Ix{Int}.index: Index (3) out of range ((0,2))
(I removed the rather long prompt of Prelude Data.Array.Storable.Internals Data.ByteString.Internal Data.ByteString>).
This will not work for Data.Vector.Unboxed, because here the data is on the Haskell heap and managed by the GHC runtime, while the other two manage the data outside the Haskell heap.
byteStringToVector :: (Storable a) => BS.ByteString -> V.Vector a
byteStringToVector bs = vec where
vec = V.unsafeFromForeignPtr (castForeignPtr fptr) (scale off) (scale len)
(fptr, off, len) = BS.toForeignPtr bs
scale = (`div` sizeOfElem vec)
sizeOfElem vec = sizeOf (undefined `asTypeOf` V.head vec)
http://hackage.haskell.org/packages/archive/spool/0.1/doc/html/Data-Vector-Storable-ByteString.html