An efficient implementation of maps from keys to values (dictionaries).

Since many function names (but not the type name) clash with Prelude names, this module is usually imported qualified, e.g.

  import Data.CompactMap (Map)
  import qualified Data.CompactMap as Map

The implementation of Map is based on size balanced binary trees (or trees of bounded balance) as described by:

• Stephen Adams, "Efficient sets: a balancing act", Journal of Functional Programming 3(4):553-562, October 1993, http://www.swiss.ai.mit.edu/~adams/BB/.
• J. Nievergelt and E.M. Reingold, "Binary search trees of bounded balance", SIAM journal of computing 2(1), March 1973.

Note that the implementation is left-biased -- the elements of a first argument are always preferred to the second, for example in union or insert.

Operation comments contain the operation time complexity in the Big-O notation http://en.wikipedia.org/wiki/Big_O_notation.

O(log n). Find the value at a key. Calls error when the element can not be found.

 fromList [(5,'a'), (3,'b')] ! 1    Error: element not in the map
 fromList [(5,'a'), (3,'b')] ! 5 == 'a'

O(1). Is the map empty?

 Data.Map.null (empty)           == True
 Data.Map.null (singleton 1 'a') == False

O(1). The number of elements in the map.

 size empty                                   == 0
 size (singleton 1 'a')                       == 1
 size (fromList([(1,'a'), (2,'c'), (3,'b')])) == 3

O(log n). Is the key a member of the map? See also notMember.

 member 5 (fromList [(5,'a'), (3,'b')]) == True
 member 1 (fromList [(5,'a'), (3,'b')]) == False

O(log n). Is the key not a member of the map? See also member.

 notMember 5 (fromList [(5,'a'), (3,'b')]) == False
 notMember 1 (fromList [(5,'a'), (3,'b')]) == True

O(log n). Lookup the value at a key in the map.

The function will return the corresponding value as (Just value), or Nothing if the key isn't in the map.

An example of using lookup:

 import Prelude hiding (lookup)
 import Data.Map

 employeeDept = fromList([("John","Sales"), ("Bob","IT")])
 deptCountry = fromList([("IT","USA"), ("Sales","France")])
 countryCurrency = fromList([("USA", "Dollar"), ("France", "Euro")])

 employeeCurrency :: String -> Maybe String
 employeeCurrency name = do
     dept <- lookup name employeeDept
     country <- lookup dept deptCountry
     lookup country countryCurrency

 main = do
     putStrLn $ "John's currency: " ++ (show (employeeCurrency "John"))
     putStrLn $ "Pete's currency: " ++ (show (employeeCurrency "Pete"))

The output of this program:

   John's currency: Just "Euro"
   Pete's currency: Nothing

O(log n). The expression (findWithDefault def k map) returns the value at key k or returns default value def when the key is not in the map.

 findWithDefault 'x' 1 (fromList [(5,'a'), (3,'b')]) == 'x'
 findWithDefault 'x' 5 (fromList [(5,'a'), (3,'b')]) == 'a'

O(1). The empty map.

 empty      == fromList []
 size empty == 0

O(1). A map with a single element.

 singleton 1 'a'        == fromList [(1, 'a')]
 size (singleton 1 'a') == 1

O(log n). Insert a new key and value in the map. If the key is already present in the map, the associated value is replaced with the supplied value. insert is equivalent to insertWith const.

 insert 5 'x' (fromList [(5,'a'), (3,'b')]) == fromList [(3, 'b'), (5, 'x')]
 insert 7 'x' (fromList [(5,'a'), (3,'b')]) == fromList [(3, 'b'), (5, 'a'), (7, 'x')]
 insert 5 'x' empty                         == singleton 5 'x'

O(log n). Insert with a function, combining new value and old value. insertWith f key value mp will insert the pair (key, value) into mp if key does not exist in the map. If the key does exist, the function will insert the pair (key, f new_value old_value).

 insertWith (++) 5 "xxx" (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "xxxa")]
 insertWith (++) 7 "xxx" (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a"), (7, "xxx")]
 insertWith (++) 5 "xxx" empty                         == singleton 5 "xxx"

O(log n). Insert with a function, combining key, new value and old value. insertWithKey f key value mp will insert the pair (key, value) into mp if key does not exist in the map. If the key does exist, the function will insert the pair (key,f key new_value old_value). Note that the key passed to f is the same key passed to insertWithKey.

 let f key new_value old_value = (show key) ++ ":" ++ new_value ++ "|" ++ old_value
 insertWithKey f 5 "xxx" (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "5:xxx|a")]
 insertWithKey f 7 "xxx" (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a"), (7, "xxx")]
 insertWithKey f 5 "xxx" empty                         == singleton 5 "xxx"

O(log n). Combines insert operation with old value retrieval. The expression (insertLookupWithKey f k x map) is a pair where the first element is equal to (lookup k map) and the second element equal to (insertWithKey f k x map).

 let f key new_value old_value = (show key) ++ ":" ++ new_value ++ "|" ++ old_value
 insertLookupWithKey f 5 "xxx" (fromList [(5,"a"), (3,"b")]) == (Just "a", fromList [(3, "b"), (5, "5:xxx|a")])
 insertLookupWithKey f 7 "xxx" (fromList [(5,"a"), (3,"b")]) == (Nothing,  fromList [(3, "b"), (5, "a"), (7, "xxx")])
 insertLookupWithKey f 5 "xxx" empty                         == (Nothing,  singleton 5 "xxx")

This is how to define insertLookup using insertLookupWithKey:

 let insertLookup kx x t = insertLookupWithKey (\_ a _ -> a) kx x t
 insertLookup 5 "x" (fromList [(5,"a"), (3,"b")]) == (Just "a", fromList [(3, "b"), (5, "x")])
 insertLookup 7 "x" (fromList [(5,"a"), (3,"b")]) == (Nothing,  fromList [(3, "b"), (5, "a"), (7, "x")])

O(log n). Delete a key and its value from the map. When the key is not a member of the map, the original map is returned.

 delete 5 (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"
 delete 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a")]
 delete 5 empty                         == empty

O(log n). Update a value at a specific key with the result of the provided function. When the key is not a member of the map, the original map is returned.

 adjust ("new " ++) 5 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "new a")]
 adjust ("new " ++) 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a")]
 adjust ("new " ++) 7 empty                         == empty

O(log n). Adjust a value at a specific key. When the key is not a member of the map, the original map is returned.

 let f key x = (show key) ++ ":new " ++ x
 adjustWithKey f 5 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "5:new a")]
 adjustWithKey f 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a")]
 adjustWithKey f 7 empty                         == empty

O(log n). The expression (update f k map) updates the value x at k (if it is in the map). If (f x) is Nothing, the element is deleted. If it is (Just y), the key k is bound to the new value y.

 let f x = if x == "a" then Just "new a" else Nothing
 update f 5 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "new a")]
 update f 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a")]
 update f 3 (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"

O(log n). The expression (updateWithKey f k map) updates the value x at k (if it is in the map). If (f k x) is Nothing, the element is deleted. If it is (Just y), the key k is bound to the new value y.

 let f k x = if x == "a" then Just ((show k) ++ ":new a") else Nothing
 updateWithKey f 5 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "5:new a")]
 updateWithKey f 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a")]
 updateWithKey f 3 (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"

O(log n). Lookup and update. See also updateWithKey. The function returns changed value, if it is updated. Returns the original key value if the map entry is deleted.

 let f k x = if x == "a" then Just ((show k) ++ ":new a") else Nothing
 updateLookupWithKey f 5 (fromList [(5,"a"), (3,"b")]) == (Just "5:new a", fromList [(3, "b"), (5, "5:new a")])
 updateLookupWithKey f 7 (fromList [(5,"a"), (3,"b")]) == (Nothing,  fromList [(3, "b"), (5, "a")])
 updateLookupWithKey f 3 (fromList [(5,"a"), (3,"b")]) == (Just "b", singleton 5 "a")

O(log n). The expression (alter f k map) alters the value x at k, or absence thereof. alter can be used to insert, delete, or update a value in a Map. In short : lookup k (alter f k m) = f (lookup k m).

 let f _ = Nothing
 alter f 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a")]
 alter f 5 (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"

 let f _ = Just "c"
 alter f 7 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "a"), (7, "c")]
 alter f 5 (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "c")]

O(log n*m). The expression (union t1 t2) takes the left-biased union of t1 and t2. It prefers t1 when duplicate keys are encountered, i.e. (union == unionWith const).

 union (fromList [(5, "a"), (3, "b")]) (fromList [(5, "A"), (7, "C")]) == fromList [(3, "b"), (5, "a"), (7, "C")]

O(log n*m). Union with a combining function.

 unionWith (++) (fromList [(5, "a"), (3, "b")]) (fromList [(5, "A"), (7, "C")]) == fromList [(3, "b"), (5, "aA"), (7, "C")]

O(log n*m). Union with a combining function.

 let f key new_value old_value = (show key) ++ ":" ++ new_value ++ "|" ++ old_value
 unionWithKey f (fromList [(5, "a"), (3, "b")]) (fromList [(5, "A"), (7, "C")]) == fromList [(3, "b"), (5, "5:a|A"), (7, "C")]

The union of a list of maps: (unions == foldl union empty).

 unions [(fromList [(5, "a"), (3, "b")]), (fromList [(5, "A"), (7, "C")]), (fromList [(5, "A3"), (3, "B3")])]
     == fromList [(3, "b"), (5, "a"), (7, "C")]
 unions [(fromList [(5, "A3"), (3, "B3")]), (fromList [(5, "A"), (7, "C")]), (fromList [(5, "a"), (3, "b")])]
     == fromList [(3, "B3"), (5, "A3"), (7, "C")]

The union of a list of maps, with a combining operation: (unionsWith f == foldl (unionWith f) empty).

 unionsWith (++) [(fromList [(5, "a"), (3, "b")]), (fromList [(5, "A"), (7, "C")]), (fromList [(5, "A3"), (3, "B3")])]
     == fromList [(3, "bB3"), (5, "aAA3"), (7, "C")]

O(n). Map a function over all values in the map.

 map (++ "x") (fromList [(5,"a"), (3,"b")]) == fromList [(3, "bx"), (5, "ax")]

O(n). Map a function over all values in the map.

 let f key x = (show key) ++ ":" ++ x
 mapWithKey f (fromList [(5,"a"), (3,"b")]) == fromList [(3, "3:b"), (5, "5:a")]

O(n*log n). mapKeys f s is the map obtained by applying f to each key of s.

The size of the result may be smaller if f maps two or more distinct keys to the same new key. In this case the value at the smallest of these keys is retained.

 mapKeys (+ 1) (fromList [(5,"a"), (3,"b")])                        == fromList [(4, "b"), (6, "a")]
 mapKeys (\ _ -> 1) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 1 "c"
 mapKeys (\ _ -> 3) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 3 "c"

O(n*log n). mapKeysWith c f s is the map obtained by applying f to each key of s.

The size of the result may be smaller if f maps two or more distinct keys to the same new key. In this case the associated values will be combined using c.

 mapKeysWith (++) (\ _ -> 1) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 1 "cdab"
 mapKeysWith (++) (\ _ -> 3) (fromList [(1,"b"), (2,"a"), (3,"d"), (4,"c")]) == singleton 3 "cdab"

O(n). mapKeysMonotonic f s == mapKeys f s, but works only when f is strictly monotonic. That is, for any values x and y, if x < y then f x < f y. The precondition is not checked. Semi-formally, we have:

 and [x < y ==> f x < f y | x <- ls, y <- ls] 
                     ==> mapKeysMonotonic f s == mapKeys f s
     where ls = keys s

This means that f maps distinct original keys to distinct resulting keys. This function has better performance than mapKeys.

 mapKeysMonotonic (\ k -> k * 2) (fromList [(5,"a"), (3,"b")]) == fromList [(6, "b"), (10, "a")]
 valid (mapKeysMonotonic (\ k -> k * 2) (fromList [(5,"a"), (3,"b")])) == True
 valid (mapKeysMonotonic (\ _ -> 1)     (fromList [(5,"a"), (3,"b")])) == False

O(n). Fold the values in the map, such that fold f z == foldr f z . elems. For example,

 elems map = fold (:) [] map

 let f a len = len + (length a)
 fold f 0 (fromList [(5,"a"), (3,"bbb")]) == 4

O(n). Fold the keys and values in the map, such that foldWithKey f z == foldr (uncurry f) z . toAscList. For example,

 keys map = foldWithKey (\k x ks -> k:ks) [] map

 let f k a result = result ++ "(" ++ (show k) ++ ":" ++ a ++ ")"
 foldWithKey f "Map: " (fromList [(5,"a"), (3,"b")]) == "Map: (5:a)(3:b)"

O(n). Return all elements of the map in the ascending order of their keys.

 elems (fromList [(5,"a"), (3,"b")]) == ["b","a"]
 elems empty == []

O(n). Return all keys of the map in ascending order.

 keys (fromList [(5,"a"), (3,"b")]) == [3,5]
 keys empty == []

O(n). The set of all keys of the map.

 keysSet (fromList [(5,"a"), (3,"b")]) == Data.Set.fromList [3,5]
 keysSet empty == Data.Set.empty

O(n). Return all key/value pairs in the map in ascending key order.

 assocs (fromList [(5,"a"), (3,"b")]) == [(3,"b"), (5,"a")]
 assocs empty == []

O(n). Convert to a list of key/value pairs.

 toList (fromList [(5,"a"), (3,"b")]) == [(3,"b"), (5,"a")]
 toList empty == []

O(n*log n). Build a map from a list of key/value pairs. See also fromAscList. If the list contains more than one value for the same key, the last value for the key is retained.

 fromList [] == empty
 fromList [(5,"a"), (3,"b"), (5, "c")] == fromList [(5,"c"), (3,"b")]
 fromList [(5,"c"), (3,"b"), (5, "a")] == fromList [(5,"a"), (3,"b")]

O(n*log n). Build a map from a list of key/value pairs with a combining function. See also fromAscListWith.

 fromListWith (++) [(5,"a"), (5,"b"), (3,"b"), (3,"a"), (5,"a")] == fromList [(3, "ab"), (5, "aba")]
 fromListWith (++) [] == empty

O(n*log n). Build a map from a list of key/value pairs with a combining function. See also fromAscListWithKey.

 let f k a1 a2 = (show k) ++ a1 ++ a2
 fromListWithKey f [(5,"a"), (5,"b"), (3,"b"), (3,"a"), (5,"a")] == fromList [(3, "3ab"), (5, "5a5ba")]
 fromListWithKey f [] == empty

O(n). Convert to an ascending list.

 toAscList (fromList [(5,"a"), (3,"b")]) == [(3,"b"), (5,"a")]

O(n). Build a map from an ascending list in linear time. The precondition (input list is ascending) is not checked.

 fromAscList [(3,"b"), (5,"a")]          == fromList [(3, "b"), (5, "a")]
 fromAscList [(3,"b"), (5,"a"), (5,"b")] == fromList [(3, "b"), (5, "b")]
 valid (fromAscList [(3,"b"), (5,"a"), (5,"b")]) == True
 valid (fromAscList [(5,"a"), (3,"b"), (5,"b")]) == False

O(n). Build a map from an ascending list in linear time with a combining function for equal keys. The precondition (input list is ascending) is not checked.

 fromAscListWith (++) [(3,"b"), (5,"a"), (5,"b")] == fromList [(3, "b"), (5, "ba")]
 valid (fromAscListWith (++) [(3,"b"), (5,"a"), (5,"b")]) == True
 valid (fromAscListWith (++) [(5,"a"), (3,"b"), (5,"b")]) == False

O(n). Build a map from an ascending list in linear time with a combining function for equal keys. The precondition (input list is ascending) is not checked.

 let f k a1 a2 = (show k) ++ ":" ++ a1 ++ a2
 fromAscListWithKey f [(3,"b"), (5,"a"), (5,"b"), (5,"b")] == fromList [(3, "b"), (5, "5:b5:ba")]
 valid (fromAscListWithKey f [(3,"b"), (5,"a"), (5,"b"), (5,"b")]) == True
 valid (fromAscListWithKey f [(5,"a"), (3,"b"), (5,"b"), (5,"b")]) == False

O(n). Build a map from an ascending list of distinct elements in linear time. The precondition is not checked.

 fromDistinctAscList [(3,"b"), (5,"a")] == fromList [(3, "b"), (5, "a")]
 valid (fromDistinctAscList [(3,"b"), (5,"a")])          == True
 valid (fromDistinctAscList [(3,"b"), (5,"a"), (5,"b")]) == False

O(n). Filter all values that satisfy the predicate.

 filter (> "a") (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"
 filter (> "x") (fromList [(5,"a"), (3,"b")]) == empty
 filter (< "a") (fromList [(5,"a"), (3,"b")]) == empty

O(n). Filter all keys/values that satisfy the predicate.

 filterWithKey (\k _ -> k > 4) (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"

O(n). Partition the map according to a predicate. The first map contains all elements that satisfy the predicate, the second all elements that fail the predicate. See also split.

 partition (> "a") (fromList [(5,"a"), (3,"b")]) == (singleton 3 "b", singleton 5 "a")
 partition (< "x") (fromList [(5,"a"), (3,"b")]) == (fromList [(3, "b"), (5, "a")], empty)
 partition (> "x") (fromList [(5,"a"), (3,"b")]) == (empty, fromList [(3, "b"), (5, "a")])

O(n). Partition the map according to a predicate. The first map contains all elements that satisfy the predicate, the second all elements that fail the predicate. See also split.

 partitionWithKey (\ k _ -> k > 3) (fromList [(5,"a"), (3,"b")]) == (singleton 5 "a", singleton 3 "b")
 partitionWithKey (\ k _ -> k < 7) (fromList [(5,"a"), (3,"b")]) == (fromList [(3, "b"), (5, "a")], empty)
 partitionWithKey (\ k _ -> k > 7) (fromList [(5,"a"), (3,"b")]) == (empty, fromList [(3, "b"), (5, "a")])

O(n). Map values and collect the Just results.

 let f x = if x == "a" then Just "new a" else Nothing
 mapMaybe f (fromList [(5,"a"), (3,"b")]) == singleton 5 "new a"

O(n). Map keys/values and collect the Just results.

 let f k _ = if k < 5 then Just ("key : " ++ (show k)) else Nothing
 mapMaybeWithKey f (fromList [(5,"a"), (3,"b")]) == singleton 3 "key : 3"

O(n). Map values and separate the Left and Right results.

 let f a = if a < "c" then Left a else Right a
 mapEither f (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])
     == (fromList [(3,"b"), (5,"a")], fromList [(1,"x"), (7,"z")])

 mapEither (\ a -> Right a) (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])
     == (empty, fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])

O(n). Map keys/values and separate the Left and Right results.

 let f k a = if k < 5 then Left (k * 2) else Right (a ++ a)
 mapEitherWithKey f (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])
     == (fromList [(1,2), (3,6)], fromList [(5,"aa"), (7,"zz")])

 mapEitherWithKey (\_ a -> Right a) (fromList [(5,"a"), (3,"b"), (1,"x"), (7,"z")])
     == (empty, fromList [(1,"x"), (3,"b"), (5,"a"), (7,"z")])