Hash join
The classic hash join algorithm for an inner join of two relations has the following steps:
- Hash phase : Creating a hash table for one of the two relations by applying a hash function to the join attribute of each row. Ideally we should create a hash table for the smaller relation. Thus, optimizing for creation time and memory size of the hash table.
- Join phase : Scanning the larger relation and finding the relevant rows by looking in the hash table created before.
The algorithm is as follows:
for each tuple s in S do let h = hash on join attributes s(b) place s in hash table Sh in bucket keyed by hash value h for each tuple r in R do let h = hash on join attributes r(a) if h indicates a nonempty bucket (B) of hash table Sh if h matches any s in B concatenate r and s place relation in Q
Implement the Hash Join algorithm in your programming language (optionally providing a test case as well).
Haskell
The ST monad allows us to utilise mutable memory behind a referentially transparent interface, allowing us to use hashtables (efficiently).
Our hashJoin function takes two lists and two selector functions.
Placing all relations with the same selector value in a list in the hashtable allows us to join many to one/many relations. <lang Haskell>{-# LANGUAGE LambdaCase, TupleSections #-} import qualified Data.HashTable.ST.Basic as H import Data.Hashable import Control.Monad.ST import Control.Monad import Data.STRef
hashJoin :: (Eq k, Hashable k) =>
[t] -> (t -> k) -> [a] -> (a -> k) -> [(t, a)]
hashJoin xs fx ys fy = runST $ do
l <- newSTRef [] ht <- H.new forM_ ys $ \y -> H.insert ht (fy y) =<< (H.lookup ht (fy y) >>= \case Nothing -> return [y] Just v -> return (y:v)) forM_ xs $ \x -> do H.lookup ht (fx x) >>= \case Nothing -> return () Just v -> modifySTRef' l ((map (x,) v) ++) readSTRef l
test = mapM_ print $ hashJoin
[(1, "Jonah"), (2, "Alan"), (3, "Glory"), (4, "Popeye")] snd [("Jonah", "Whales"), ("Jonah", "Spiders"), ("Alan", "Ghosts"), ("Alan", "Zombies"), ("Glory", "Buffy")] fst
</lang>
λ> test ((3,"Glory"),("Glory","Buffy")) ((2,"Alan"),("Alan","Zombies")) ((2,"Alan"),("Alan","Ghosts")) ((1,"Jonah"),("Jonah","Spiders")) ((1,"Jonah"),("Jonah","Whales"))
The task require hashtables; however, a cleaner and more functional solution would be to use Data.Map (based on binary trees): <lang Haskell>{-# LANGUAGE TupleSections #-} import qualified Data.Map as M import Data.List import Data.Maybe import Control.Applicative
mapJoin xs fx ys fy = joined
where yMap = foldl' f M.empty ys f m y = M.insertWith (++) (fy y) [y] m joined = concat . catMaybes . map (\x -> map (x,) <$> M.lookup (fx x) yMap) $ xs
test = mapM_ print $ mapJoin
[(1, "Jonah"), (2, "Alan"), (3, "Glory"), (4, "Popeye")] snd [("Jonah", "Whales"), ("Jonah", "Spiders"), ("Alan", "Ghosts"), ("Alan", "Zombies"), ("Glory", "Buffy")] fst
</lang>
λ> test ((1,"Jonah"),("Jonah","Spiders")) ((1,"Jonah"),("Jonah","Whales")) ((2,"Alan"),("Alan","Zombies")) ((2,"Alan"),("Alan","Ghosts")) ((3,"Glory"),("Glory","Buffy"))