A Map from keys k to values a.

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'

Same as difference.

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(log n). Find largest key smaller than the given one and return the corresponding (key, value) pair.

 lookupLT 3 (fromList [(3,'a'), (5,'b')]) == Nothing
 lookupLT 4 (fromList [(3,'a'), (5,'b')]) == Just (3, 'a')

O(log n). Find smallest key greater than the given one and return the corresponding (key, value) pair.

 lookupGT 4 (fromList [(3,'a'), (5,'b')]) == Just (5, 'b')
 lookupGT 5 (fromList [(3,'a'), (5,'b')]) == Nothing

O(log n). Find largest key smaller or equal to the given one and return the corresponding (key, value) pair.

 lookupLE 2 (fromList [(3,'a'), (5,'b')]) == Nothing
 lookupLE 4 (fromList [(3,'a'), (5,'b')]) == Just (3, 'a')
 lookupLE 5 (fromList [(3,'a'), (5,'b')]) == Just (5, 'b')

O(log n). Find smallest key greater or equal to the given one and return the corresponding (key, value) pair.

 lookupGE 3 (fromList [(3,'a'), (5,'b')]) == Just (3, 'a')
 lookupGE 4 (fromList [(3,'a'), (5,'b')]) == Just (5, 'b')
 lookupGE 6 (fromList [(3,'a'), (5,'b')]) == Nothing

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(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). The implementation uses the efficient hedge-union algorithm.

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

O(n+m). Union with a combining function. The implementation uses the efficient hedge-union algorithm.

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

O(n+m). Union with a combining function. The implementation uses the efficient hedge-union algorithm.

 let f key left_value right_value = (show key) ++ ":" ++ left_value ++ "|" ++ right_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+m). Difference of two maps. Return elements of the first map not existing in the second map. The implementation uses an efficient hedge algorithm comparable with hedge-union.

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

O(n+m). Difference with a combining function. When two equal keys are encountered, the combining function is applied to the values of these keys. If it returns Nothing, the element is discarded (proper set difference). If it returns (Just y), the element is updated with a new value y. The implementation uses an efficient hedge algorithm comparable with hedge-union.

 let f al ar = if al == "b" then Just (al ++ ":" ++ ar) else Nothing
 differenceWith f (fromList [(5, "a"), (3, "b")]) (fromList [(5, "A"), (3, "B"), (7, "C")])
     == singleton 3 "b:B"

O(n+m). Difference with a combining function. When two equal keys are encountered, the combining function is applied to the key and both values. If it returns Nothing, the element is discarded (proper set difference). If it returns (Just y), the element is updated with a new value y. The implementation uses an efficient hedge algorithm comparable with hedge-union.

 let f k al ar = if al == "b" then Just ((show k) ++ ":" ++ al ++ "|" ++ ar) else Nothing
 differenceWithKey f (fromList [(5, "a"), (3, "b")]) (fromList [(5, "A"), (3, "B"), (10, "C")])
     == singleton 3 "3:b|B"

O(n+m). Intersection of two maps. Return data in the first map for the keys existing in both maps. (intersection m1 m2 == intersectionWith const m1 m2). The implementation uses an efficient hedge algorithm comparable with hedge-union.

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

O(n+m). Intersection with a combining function. The implementation uses an efficient hedge algorithm comparable with hedge-union.

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

O(n+m). Intersection with a combining function. The implementation uses an efficient hedge algorithm comparable with hedge-union.

 let f k al ar = (show k) ++ ":" ++ al ++ "|" ++ ar
 intersectionWithKey f (fromList [(5, "a"), (3, "b")]) (fromList [(5, "A"), (7, "C")]) == singleton 5 "5:a|A"

O(n+m). A high-performance universal combining function. This function is used to define unionWith, unionWithKey, differenceWith, differenceWithKey, intersectionWith, intersectionWithKey and can be used to define other custom combine functions.

Please make sure you know what is going on when using mergeWithKey, otherwise you can be surprised by unexpected code growth or even corruption of the data structure.

When mergeWithKey is given three arguments, it is inlined to the call site. You should therefore use mergeWithKey only to define your custom combining functions. For example, you could define unionWithKey, differenceWithKey and intersectionWithKey as

 myUnionWithKey f m1 m2 = mergeWithKey (\k x1 x2 -> Just (f k x1 x2)) id id m1 m2
 myDifferenceWithKey f m1 m2 = mergeWithKey f id (const empty) m1 m2
 myIntersectionWithKey f m1 m2 = mergeWithKey (\k x1 x2 -> Just (f k x1 x2)) (const empty) (const empty) m1 m2

When calling mergeWithKey combine only1 only2, a function combining two IntMaps is created, such that

• if a key is present in both maps, it is passed with both corresponding values to the combine function. Depending on the result, the key is either present in the result with specified value, or is left out;
• a nonempty subtree present only in the first map is passed to only1 and the output is added to the result;
• a nonempty subtree present only in the second map is passed to only2 and the output is added to the result.

The only1 and only2 methods must return a map with a subset (possibly empty) of the keys of the given map. The values can be modified arbitrarily. Most common variants of only1 and only2 are id and const empty, but for example map f or filterWithKey f could be used for any f.

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). traverseWithKey f s == fromList $ traverse ((k, v) -> (,) k $ f k v) (toList m) That is, behaves exactly like a regular traverse except that the traversing function also has access to the key associated with a value.

 traverseWithKey (\k v -> if odd k then Just (succ v) else Nothing) (fromList [(1, 'a'), (5, 'e')]) == Just (fromList [(1, 'b'), (5, 'f')])
 traverseWithKey (\k v -> if odd k then Just (succ v) else Nothing) (fromList [(2, 'c')])           == Nothing

O(n). The function mapAccum threads an accumulating argument through the map in ascending order of keys.

 let f a b = (a ++ b, b ++ "X")
 mapAccum f "Everything: " (fromList [(5,"a"), (3,"b")]) == ("Everything: ba", fromList [(3, "bX"), (5, "aX")])

O(n). The function mapAccumWithKey threads an accumulating argument through the map in ascending order of keys.

 let f a k b = (a ++ " " ++ (show k) ++ "-" ++ b, b ++ "X")
 mapAccumWithKey f "Everything:" (fromList [(5,"a"), (3,"b")]) == ("Everything: 3-b 5-a", fromList [(3, "bX"), (5, "aX")])

O(n). The function mapAccumR threads an accumulating argument through the map in descending order of keys.

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 greatest of the original 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 using the given right-associative binary operator, such that foldr f z == foldr f z . elems.

For example,

 elems map = foldr (:) [] map

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

O(n). Fold the values in the map using the given left-associative binary operator, such that foldl f z == foldl f z . elems.

For example,

 elems = reverse . foldl (flip (:)) []

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

O(n). Fold the keys and values in the map using the given right-associative binary operator, such that foldrWithKey f z == foldr (uncurry f) z . toAscList.

For example,

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

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

O(n). Fold the keys and values in the map using the given left-associative binary operator, such that foldlWithKey f z == foldl (\z' (kx, x) -> f z' kx x) z . toAscList.

For example,

 keys = reverse . foldlWithKey (\ks k x -> k:ks) []

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

O(n). Fold the keys and values in the map using the given monoid, such that

foldMapWithKey f = fold . mapWithKey f

This can be an asymptotically faster than foldrWithKey or foldlWithKey for some monoids.

O(n). A strict version of foldr. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

O(n). A strict version of foldl. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

O(n). A strict version of foldrWithKey. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

O(n). A strict version of foldlWithKey. Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

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

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

O(n). Return all keys of the map in ascending order. Subject to list fusion.

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

O(n). An alias for toAscList. Return all key/value pairs in the map in ascending key order. Subject to list fusion.

 assocs (fromList [(5,"a"), (3,"b")]) == [(3,"b"), (5,"a")]
 assocs 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). Build a map from a set of keys and a function which for each key computes its value.

 fromSet (\k -> replicate k 'a') (Data.Set.fromList [3, 5]) == fromList [(5,"aaaaa"), (3,"aaa")]
 fromSet undefined Data.Set.empty == empty

O(n). Convert the map to a list of key/value pairs. Subject to list fusion.

 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.

If the keys of the list are ordered, linear-time implementation is used, with the performance equal to fromDistinctAscList.

 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 the map to a list of key/value pairs where the keys are in ascending order. Subject to list fusion.

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

O(n). Convert the map to a list of key/value pairs where the keys are in descending order. Subject to list fusion.

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

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")])

O(log n). The expression (split k map) is a pair (map1,map2) where the keys in map1 are smaller than k and the keys in map2 larger than k. Any key equal to k is found in neither map1 nor map2.

 split 2 (fromList [(5,"a"), (3,"b")]) == (empty, fromList [(3,"b"), (5,"a")])
 split 3 (fromList [(5,"a"), (3,"b")]) == (empty, singleton 5 "a")
 split 4 (fromList [(5,"a"), (3,"b")]) == (singleton 3 "b", singleton 5 "a")
 split 5 (fromList [(5,"a"), (3,"b")]) == (singleton 3 "b", empty)
 split 6 (fromList [(5,"a"), (3,"b")]) == (fromList [(3,"b"), (5,"a")], empty)

O(log n). The expression (splitLookup k map) splits a map just like split but also returns lookup k map.

 splitLookup 2 (fromList [(5,"a"), (3,"b")]) == (empty, Nothing, fromList [(3,"b"), (5,"a")])
 splitLookup 3 (fromList [(5,"a"), (3,"b")]) == (empty, Just "b", singleton 5 "a")
 splitLookup 4 (fromList [(5,"a"), (3,"b")]) == (singleton 3 "b", Nothing, singleton 5 "a")
 splitLookup 5 (fromList [(5,"a"), (3,"b")]) == (singleton 3 "b", Just "a", empty)
 splitLookup 6 (fromList [(5,"a"), (3,"b")]) == (fromList [(3,"b"), (5,"a")], Nothing, empty)

O(1). Decompose a map into pieces based on the structure of the underlying tree. This function is useful for consuming a map in parallel.

No guarantee is made as to the sizes of the pieces; an internal, but deterministic process determines this. However, it is guaranteed that the pieces returned will be in ascending order (all elements in the first submap less than all elements in the second, and so on).

Examples:

 splitRoot (fromList (zip [1..6] ['a'..])) ==
   [fromList [(1,'a'),(2,'b'),(3,'c')],fromList [(4,'d')],fromList [(5,'e'),(6,'f')]]

 splitRoot empty == []

Note that the current implementation does not return more than three submaps, but you should not depend on this behaviour because it can change in the future without notice.

O(n+m). This function is defined as (isSubmapOf = isSubmapOfBy (==)).

O(n+m). The expression (isSubmapOfBy f t1 t2) returns True if all keys in t1 are in tree t2, and when f returns True when applied to their respective values. For example, the following expressions are all True:

 isSubmapOfBy (==) (fromList [('a',1)]) (fromList [('a',1),('b',2)])
 isSubmapOfBy (<=) (fromList [('a',1)]) (fromList [('a',1),('b',2)])
 isSubmapOfBy (==) (fromList [('a',1),('b',2)]) (fromList [('a',1),('b',2)])

But the following are all False:

 isSubmapOfBy (==) (fromList [('a',2)]) (fromList [('a',1),('b',2)])
 isSubmapOfBy (<)  (fromList [('a',1)]) (fromList [('a',1),('b',2)])
 isSubmapOfBy (==) (fromList [('a',1),('b',2)]) (fromList [('a',1)])

O(n+m). Is this a proper submap? (ie. a submap but not equal). Defined as (isProperSubmapOf = isProperSubmapOfBy (==)).

O(n+m). Is this a proper submap? (ie. a submap but not equal). The expression (isProperSubmapOfBy f m1 m2) returns True when m1 and m2 are not equal, all keys in m1 are in m2, and when f returns True when applied to their respective values. For example, the following expressions are all True:

 isProperSubmapOfBy (==) (fromList [(1,1)]) (fromList [(1,1),(2,2)])
 isProperSubmapOfBy (<=) (fromList [(1,1)]) (fromList [(1,1),(2,2)])

But the following are all False:

 isProperSubmapOfBy (==) (fromList [(1,1),(2,2)]) (fromList [(1,1),(2,2)])
 isProperSubmapOfBy (==) (fromList [(1,1),(2,2)]) (fromList [(1,1)])
 isProperSubmapOfBy (<)  (fromList [(1,1)])       (fromList [(1,1),(2,2)])

O(log n). Lookup the index of a key, which is its zero-based index in the sequence sorted by keys. The index is a number from 0 up to, but not including, the size of the map.

 isJust (lookupIndex 2 (fromList [(5,"a"), (3,"b")]))   == False
 fromJust (lookupIndex 3 (fromList [(5,"a"), (3,"b")])) == 0
 fromJust (lookupIndex 5 (fromList [(5,"a"), (3,"b")])) == 1
 isJust (lookupIndex 6 (fromList [(5,"a"), (3,"b")]))   == False

O(log n). Return the index of a key, which is its zero-based index in the sequence sorted by keys. The index is a number from 0 up to, but not including, the size of the map. Calls error when the key is not a member of the map.

 findIndex 2 (fromList [(5,"a"), (3,"b")])    Error: element is not in the map
 findIndex 3 (fromList [(5,"a"), (3,"b")]) == 0
 findIndex 5 (fromList [(5,"a"), (3,"b")]) == 1
 findIndex 6 (fromList [(5,"a"), (3,"b")])    Error: element is not in the map

O(log n). Retrieve an element by its index, i.e. by its zero-based index in the sequence sorted by keys. If the index is out of range (less than zero, greater or equal to size of the map), error is called.

 elemAt 0 (fromList [(5,"a"), (3,"b")]) == (3,"b")
 elemAt 1 (fromList [(5,"a"), (3,"b")]) == (5, "a")
 elemAt 2 (fromList [(5,"a"), (3,"b")])    Error: index out of range

O(log n). Update the element at index. Calls error when an invalid index is used.

 updateAt (\ _ _ -> Just "x") 0    (fromList [(5,"a"), (3,"b")]) == fromList [(3, "x"), (5, "a")]
 updateAt (\ _ _ -> Just "x") 1    (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "x")]
 updateAt (\ _ _ -> Just "x") 2    (fromList [(5,"a"), (3,"b")])    Error: index out of range
 updateAt (\ _ _ -> Just "x") (-1) (fromList [(5,"a"), (3,"b")])    Error: index out of range
 updateAt (\_ _  -> Nothing)  0    (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"
 updateAt (\_ _  -> Nothing)  1    (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"
 updateAt (\_ _  -> Nothing)  2    (fromList [(5,"a"), (3,"b")])    Error: index out of range
 updateAt (\_ _  -> Nothing)  (-1) (fromList [(5,"a"), (3,"b")])    Error: index out of range

O(log n). Delete the element at index, i.e. by its zero-based index in the sequence sorted by keys. If the index is out of range (less than zero, greater or equal to size of the map), error is called.

 deleteAt 0  (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"
 deleteAt 1  (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"
 deleteAt 2 (fromList [(5,"a"), (3,"b")])     Error: index out of range
 deleteAt (-1) (fromList [(5,"a"), (3,"b")])  Error: index out of range

O(log n). The minimal key of the map. Calls error if the map is empty.

 findMin (fromList [(5,"a"), (3,"b")]) == (3,"b")
 findMin empty                            Error: empty map has no minimal element

O(log n). The maximal key of the map. Calls error if the map is empty.

 findMax (fromList [(5,"a"), (3,"b")]) == (5,"a")
 findMax empty                            Error: empty map has no maximal element

O(log n). Delete the minimal key. Returns an empty map if the map is empty.

 deleteMin (fromList [(5,"a"), (3,"b"), (7,"c")]) == fromList [(5,"a"), (7,"c")]
 deleteMin empty == empty

O(log n). Delete the maximal key. Returns an empty map if the map is empty.

 deleteMax (fromList [(5,"a"), (3,"b"), (7,"c")]) == fromList [(3,"b"), (5,"a")]
 deleteMax empty == empty

O(log n). Delete and find the minimal element.

 deleteFindMin (fromList [(5,"a"), (3,"b"), (10,"c")]) == ((3,"b"), fromList[(5,"a"), (10,"c")])
 deleteFindMin                                            Error: can not return the minimal element of an empty map

O(log n). Delete and find the maximal element.

 deleteFindMax (fromList [(5,"a"), (3,"b"), (10,"c")]) == ((10,"c"), fromList [(3,"b"), (5,"a")])
 deleteFindMax empty                                      Error: can not return the maximal element of an empty map

O(log n). Update the value at the minimal key.

 updateMin (\ a -> Just ("X" ++ a)) (fromList [(5,"a"), (3,"b")]) == fromList [(3, "Xb"), (5, "a")]
 updateMin (\ _ -> Nothing)         (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"

O(log n). Update the value at the maximal key.

 updateMax (\ a -> Just ("X" ++ a)) (fromList [(5,"a"), (3,"b")]) == fromList [(3, "b"), (5, "Xa")]
 updateMax (\ _ -> Nothing)         (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"

O(log n). Update the value at the minimal key.

 updateMinWithKey (\ k a -> Just ((show k) ++ ":" ++ a)) (fromList [(5,"a"), (3,"b")]) == fromList [(3,"3:b"), (5,"a")]
 updateMinWithKey (\ _ _ -> Nothing)                     (fromList [(5,"a"), (3,"b")]) == singleton 5 "a"

O(log n). Update the value at the maximal key.

 updateMaxWithKey (\ k a -> Just ((show k) ++ ":" ++ a)) (fromList [(5,"a"), (3,"b")]) == fromList [(3,"b"), (5,"5:a")]
 updateMaxWithKey (\ _ _ -> Nothing)                     (fromList [(5,"a"), (3,"b")]) == singleton 3 "b"

O(log n). Retrieves the value associated with minimal key of the map, and the map stripped of that element, or Nothing if passed an empty map.

 minView (fromList [(5,"a"), (3,"b")]) == Just ("b", singleton 5 "a")
 minView empty == Nothing

O(log n). Retrieves the value associated with maximal key of the map, and the map stripped of that element, or Nothing if passed an

 maxView (fromList [(5,"a"), (3,"b")]) == Just ("a", singleton 3 "b")
 maxView empty == Nothing

O(log n). Retrieves the minimal (key,value) pair of the map, and the map stripped of that element, or Nothing if passed an empty map.

 minViewWithKey (fromList [(5,"a"), (3,"b")]) == Just ((3,"b"), singleton 5 "a")
 minViewWithKey empty == Nothing

O(log n). Retrieves the maximal (key,value) pair of the map, and the map stripped of that element, or Nothing if passed an empty map.

 maxViewWithKey (fromList [(5,"a"), (3,"b")]) == Just ((5,"a"), singleton 3 "b")
 maxViewWithKey empty == Nothing

O(n). Show the tree that implements the map. The tree is shown in a compressed, hanging format. See showTreeWith.

O(n). The expression (showTreeWith showelem hang wide map) shows the tree that implements the map. Elements are shown using the showElem function. If hang is True, a hanging tree is shown otherwise a rotated tree is shown. If wide is True, an extra wide version is shown.

  Map> let t = fromDistinctAscList [(x,()) | x <- [1..5]]
  Map> putStrLn $ showTreeWith (\k x -> show (k,x)) True False t
  (4,())
  +--(2,())
  |  +--(1,())
  |  +--(3,())
  +--(5,())

  Map> putStrLn $ showTreeWith (\k x -> show (k,x)) True True t
  (4,())
  |
  +--(2,())
  |  |
  |  +--(1,())
  |  |
  |  +--(3,())
  |
  +--(5,())

  Map> putStrLn $ showTreeWith (\k x -> show (k,x)) False True t
  +--(5,())
  |
  (4,())
  |
  |  +--(3,())
  |  |
  +--(2,())
     |
     +--(1,())

O(n). Test if the internal map structure is valid.

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