Aha!!! Now it's working. Just had to compile with -O2 :D :D
Now I'm over twice as fast for a list of 2 million! With better length
based analysis of how many buckets should be used, this number can be
improved.
You can feel free to use my code however you like. I've attached the final
version.
That was fun :D
Timothy
---------- Původní zpráva ----------
Od: timothyho...@seznam.cz
Datum: 3. 9. 2012
Předmět: Re: [Haskell-cafe] hstats median algorithm
"
Thanks for the advice. After taking most of it it is faster. But it is
still many times slower than it ought to be! This algorithm should be much
faster than simply sorting the list, and yet it is more than twice as slow!
One note, you said:
""""
> Increment length.
>
>> modifySTRef
>> lengthRef
>> (+1)
This will create a big thunk for the length, you should use
oldLength <- readSTRef lengthRef
writeSTRef lengthRef $! oldLength + 1
(I'm not sure if a strict modifySTRef exists somewhere...)""""
Actually, replacing modifySTRef with that code is just a hair slower... Not
sure why.
I'm attaching the super optimized version. Along with the profiler output.
I just can't understand what is slow here :(
Timothy
---------- Původní zpráva ----------
Od: Felipe Almeida Lessa <felipe.le...@gmail.com>
Datum: 3. 9. 2012
Předmět: Re: [Haskell-cafe] hstats median algorithm
"On Mon, Sep 3, 2012 at 11:18 AM, Felipe Almeida Lessa
<felipe.le...@gmail.com> wrote:
> Ditto for oldLen here. Also, you can simplify this lambda a lot:
>
> import Control.Applicative ((<$>))
>
> \(oldLen, oldVal) ->
> let newLen = oldLen + 1
> newVal = (number:) <$> oldVal
> in newLen `seq` newVal `seq` (newLen, newVal)
Or, using BangPatterns,
\(oldLen, oldVal) ->
let !newLen = oldLen + 1
!newVal = (number:) <$> oldVal
in (newLen, newVal)
Cheers,
--
Felipe."
"
NO COPYRIGHT.
Public domain where applicable, and CC_Zero elsewhere.
>{-# LANGUAGE BangPatterns #-}
So I've been playing with the median problem today. Not sure why, but it stuck in my head.
>import Data.List
>import Control.Monad.ST
>import Data.STRef
>import Control.Monad
>import Control.Applicative ((<$>))
>import qualified Data.Vector.Mutable as V
>import System.Environment
I've been using the hashing algorithm that I described last night, but it's quite slow. I must be missing something obvious. It really shouldn't be slow at all!
So I have a median function:
>median
First I assume you know the range of your data.
> (min,max)
Then I ask you to figure out how many buckets you want to create, and which of those buckets you actually want to fill. You should only fill the buckets near the middle, based on an educated guess of the distribution of your data, otherwise, you will end up wasting memory. I you guessed the middle correctly the function will return more quickly than if you didn't.
> numBuckets guessedMiddle
Then you are to pass it your list.
> list =
> let
First I seperate out the values into a list of buckets, each one holding values which are near to eachother. I can figure out the length of the list at the same time, since I have to go through the whole thing anyways.
> (myBuckets,length) =
> buckets
> (min,max)
> numBuckets
> guessedMiddle
> list
> halfLength = length `div` 2
The buckets are set up, so that the first bucket has the lowest values, and the last bucket has the highest values. Each bucket, is a tuple, of it's length and it's contents. We fold across the list of buckets, accumulating the "index so far", until we find the bucket in which the median must reside.
> Right ((_,medianBucketVals),stubLen,medianBucketIx) =
foo :: Either (Int,Int) ((a,Maybe [b]),a,Int)
foo =
> foldr
> (\thisBucket@(thisBucketLen,_)
> eitheriOrMedianBucket ->
> case eitheriOrMedianBucket of
> Left (i,bucketIx) ->
> let
> nextI = i + thisBucketLen
> in
> if nextI > halfLength
> then Right (thisBucket, thisBucketLen-(halfLength - i),(bucketIx-1))
> else (Left (nextI, (bucketIx + 1)))
> _ -> eitheriOrMedianBucket)
> (Left (0,0))
> myBuckets
> in
We then sort the bucket in which the median must reside, and we then find the median the normal way. This should be faster, since we only had to sort one bucket, rather than the entire list. It is not, so there must be something horrible going on.
> case medianBucketVals of
> Just medianBucket' ->
> sort medianBucket' `genericIndex` stubLen
> Nothing ->
If the user guessed the middle incorrectly, just start over now that we know where the middle really is.
> median
> (min,max)
> numBuckets
> (medianBucketIx,medianBucketIx)
> list
Here is my actual function for seperating the values out into buckets.
>buckets
> (min,max)
> numBuckets
> (guessedMiddleStart,guessedMiddleEnd)
> list =
> runST $ do
> lengthRef <- newSTRef 0
First we create a list of empty buckets. Since it would be a waste of memory to actually fill the buckets near the edges of our distribution(where we are not likely to find our median), our buckets contain Maybe lists, and the buckets which are outside of our guessed bucket range will be filled with Nothing.
> buckets' <- V.new numBuckets
> mapM
> (\n->
> V.write
> buckets'
> n
> (0,
> if n >= guessedMiddleStart && n <= guessedMiddleEnd
> then Just []
> else Nothing))
> [0..numBuckets-1]
Then we go through the buckets, figuring out which bucket to put a given value into. We calculate the length at the same time.
> let
> bottom =
> (fromIntegral (max - min))
> /
> (fromIntegral (numBuckets-1))
> forM_ list $ \number -> do
> let
Figure out which bucket to put this into.
> bucket =
> whichBucket
> (min,max)
> bottom
> numBuckets
> number
Increment length.
> modifySTRef
> lengthRef
> (+1)
However tests show that's not true.
Put the value into the appropriate bucket.
> (oldLen,oldVal) <- V.read
> buckets'
> bucket
> let !newLen = oldLen + 1
> !newVal = (number:) <$> oldVal
> V.write
> buckets'
> bucket
> (newLen, newVal)
> filledBuckets <-
> mapM
> (\index->V.read buckets' index)
> [0..numBuckets-1]
> length <- readSTRef lengthRef
> return (filledBuckets,length)
>whichBucket (min,max) bottom numBuckets number =
> floor((fromIntegral (number - min))
> /
> bottom)
So I created a little test scenario.
>someListNumBuckets = 100
>someListGuessedMiddle = (45,55)
>someListLength = genericLength someList
And found that this:
>realMedian = sort someList `genericIndex` (someListLength `div` 2)
Is actually faster ^_^ :O
*Main> realMedian
500
(7.18 secs, 2031543472 bytes)
Than this:
>someListMedian =
> median
> (someListMin,someListMax)
> someListNumBuckets
> someListGuessedMiddle
> someList
*Main> someListMedian
Just 500
(37.77 secs, 15376209200 bytes)
someListMin :: Integer
>someListMin = 0
someListMax :: Integer
>someListMax = 1000
someList :: [Integer]
>someList =
> concatMap
> (\n->intersperse n [someListMin..someListMax])
> [someListMax,someListMax-1..someListMin]
>main :: IO ()
>main = do
> args <- System.Environment.getArgs
> case args of
> ("n":_) ->
> print realMedian
> ("f":_) ->
> print someListMedian
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