A Map from keys k to values 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(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(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). 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). 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). 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(1). Is the map empty?

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

O(m*log(n/m + 1)), m <= n. 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(m*log(n/m + 1)), m <= n. Difference of two maps. Return elements of the first map not existing in the second map.

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

O(m*log(n/m + 1)), m <= n. Intersection of two maps. Return data in the first map for the keys existing in both maps. (intersection m1 m2 == intersectionWith const m1 m2).

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

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)