Data.Map.Internal

A Map from keys k to values a.

The Semigroup operation for Map is union, which prefers values from the left operand. If m1 maps a key k to a value a1, and m2 maps the same key to a different value a2, then their union m1 <> m2 maps k to a1.

$O(\log n)$. 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'

$O(\log n)$. Find the value at a key. Returns Nothing when the element can not be found.

fromList [(5, 'a'), (3, 'b')] !? 1 == Nothing

fromList [(5, 'a'), (3, 'b')] !? 5 == Just 'a'

Since: containers-0.5.9

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")]

Note that adjust = alter . fmap.

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

Example:

interactiveAlter :: Int -> Map Int String -> IO (Map Int String)
interactiveAlter k m = alterF f k m where
  f Nothing = do
     putStrLn $ show k ++
         " was not found in the map. Would you like to add it?"
     getUserResponse1 :: IO (Maybe String)
  f (Just old) = do
     putStrLn $ "The key is currently bound to " ++ show old ++
         ". Would you like to change or delete it?"
     getUserResponse2 :: IO (Maybe String)

alterF is the most general operation for working with an individual key that may or may not be in a given map. When used with trivial functors like Identity and Const, it is often slightly slower than more specialized combinators like lookup and insert. However, when the functor is non-trivial and key comparison is not particularly cheap, it is the fastest way.

Note on rewrite rules:

This module includes GHC rewrite rules to optimize alterF for the Const and Identity functors. In general, these rules improve performance. The sole exception is that when using Identity, deleting a key that is already absent takes longer than it would without the rules. If you expect this to occur a very large fraction of the time, you might consider using a private copy of the Identity type.

Note: alterF is a flipped version of the at combinator from Control.Lens.At.

Since: containers-0.5.8

$O(m\log (\frac{n+1}{m+1})),m\le n$. 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")]

$O(m\log (\frac{n+1}{m+1})),m\le n$. Union with a combining function.

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

$O(m\log (\frac{n+1}{m+1})),m\le n$. Union with a combining function.

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")]

$O(m\log (\frac{n+1}{m+1})),m\le n$. 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"

$O(n+m)$. 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.

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.

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"

$O(m\log (\frac{n+1}{m+1})),m\le n$. 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"

$O(m\log (\frac{n+1}{m+1})),m\le n$. Intersection with a combining function.

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

$O(m\log (\frac{n+1}{m+1})),m\le n$. Intersection with a combining function.

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"

$O(m\log (\frac{n+1}{m+1})),m\le n$. Check whether the key sets of two maps are disjoint (i.e., their intersection is empty).

disjoint (fromList [(2,'a')]) (fromList [(1,()), (3,())])   == True
disjoint (fromList [(2,'a')]) (fromList [(1,'a'), (2,'b')]) == False
disjoint (fromList [])        (fromList [])                 == True

xs `disjoint` ys = null (xs `intersection` ys)

Since: containers-0.6.2.1

Relate the keys of one map to the values of the other, by using the values of the former as keys for lookups in the latter.

Complexity: $O(n\ast \log (m))$, where $m$ is the size of the first argument

compose (fromList [('a', "A"), ('b', "B")]) (fromList [(1,'a'),(2,'b'),(3,'z')]) = fromList [(1,"A"),(2,"B")]

(compose bc ab !?) = (bc !?) <=< (ab !?)

Note: Prior to v0.6.4, Data.Map.Strict exposed a version of compose that forced the values of the output Map. This version does not force these values.

Since: containers-0.6.3.1

A tactic for dealing with keys present in one map but not the other in merge.

A tactic of type SimpleWhenMissing k x z is an abstract representation of a function of type k -> x -> Maybe z .

Since: containers-0.5.9

A tactic for dealing with keys present in both maps in merge.

A tactic of type SimpleWhenMatched k x y z is an abstract representation of a function of type k -> x -> y -> Maybe z .

Since: containers-0.5.9

Along with zipWithMaybeAMatched, witnesses the isomorphism between WhenMatched f k x y z and k -> x -> y -> f (Maybe z).

Since: containers-0.5.9

Along with traverseMaybeMissing, witnesses the isomorphism between WhenMissing f k x y and k -> x -> f (Maybe y).

Since: containers-0.5.9

Merge two maps.

merge takes two WhenMissing tactics, a WhenMatched tactic and two maps. It uses the tactics to merge the maps. Its behavior is best understood via its fundamental tactics, mapMaybeMissing and zipWithMaybeMatched.

Consider

merge (mapMaybeMissing g1)
             (mapMaybeMissing g2)
             (zipWithMaybeMatched f)
             m1 m2

Take, for example,

m1 = [(0, 'a'), (1, 'b'), (3, 'c'), (4, 'd')]
m2 = [(1, "one"), (2, "two"), (4, "three")]

merge will first "align" these maps by key:

m1 = [(0, 'a'), (1, 'b'),               (3, 'c'), (4, 'd')]
m2 =           [(1, "one"), (2, "two"),           (4, "three")]

It will then pass the individual entries and pairs of entries to g1, g2, or f as appropriate:

maybes = [g1 0 'a', f 1 'b' "one", g2 2 "two", g1 3 'c', f 4 'd' "three"]

This produces a Maybe for each key:

keys =     0        1          2           3        4
results = [Nothing, Just True, Just False, Nothing, Just True]

Finally, the Just results are collected into a map:

return value = [(1, True), (2, False), (4, True)]

The other tactics below are optimizations or simplifications of mapMaybeMissing for special cases. Most importantly,

• dropMissing drops all the keys.
• preserveMissing leaves all the entries alone.

When merge is given three arguments, it is inlined at the call site. To prevent excessive inlining, you should typically use merge to define your custom combining functions.

Examples:

unionWithKey f = merge preserveMissing preserveMissing (zipWithMatched f)

intersectionWithKey f = merge dropMissing dropMissing (zipWithMatched f)

differenceWith f = merge preserveMissing dropMissing (zipWithMatched f)

symmetricDifference = merge preserveMissing preserveMissing (zipWithMaybeMatched $ \ _ _ _ -> Nothing)

mapEachPiece f g h = merge (mapMissing f) (mapMissing g) (zipWithMatched h)

Since: containers-0.5.9

When a key is found in both maps, apply a function to the key and values and maybe use the result in the merged map.

zipWithMaybeMatched :: (k -> x -> y -> Maybe z)
                    -> SimpleWhenMatched k x y z

Since: containers-0.5.9

When a key is found in both maps, apply a function to the key and values and use the result in the merged map.

zipWithMatched :: (k -> x -> y -> z)
               -> SimpleWhenMatched k x y z

Since: containers-0.5.9

Map over the entries whose keys are missing from the other map, optionally removing some. This is the most powerful SimpleWhenMissing tactic, but others are usually more efficient.

mapMaybeMissing :: (k -> x -> Maybe y) -> SimpleWhenMissing k x y

mapMaybeMissing f = traverseMaybeMissing (\k x -> pure (f k x))

but mapMaybeMissing uses fewer unnecessary Applicative operations.

Since: containers-0.5.9

Drop all the entries whose keys are missing from the other map.

dropMissing :: SimpleWhenMissing k x y

dropMissing = mapMaybeMissing (\_ _ -> Nothing)

but dropMissing is much faster.

Since: containers-0.5.9

Preserve, unchanged, the entries whose keys are missing from the other map.

preserveMissing :: SimpleWhenMissing k x x

preserveMissing = Merge.Lazy.mapMaybeMissing (\_ x -> Just x)

but preserveMissing is much faster.

Since: containers-0.5.9

Force the entries whose keys are missing from the other map and otherwise preserve them unchanged.

preserveMissing' :: SimpleWhenMissing k x x

preserveMissing' = Merge.Lazy.mapMaybeMissing (\_ x -> Just $! x)

but preserveMissing' is quite a bit faster.

Since: containers-0.5.9

Map over the entries whose keys are missing from the other map.

mapMissing :: (k -> x -> y) -> SimpleWhenMissing k x y

mapMissing f = mapMaybeMissing (\k x -> Just $ f k x)

but mapMissing is somewhat faster.

Since: containers-0.5.9

Filter the entries whose keys are missing from the other map.

filterMissing :: (k -> x -> Bool) -> SimpleWhenMissing k x x

filterMissing f = Merge.Lazy.mapMaybeMissing $ \k x -> guard (f k x) *> Just x

but this should be a little faster.

Since: containers-0.5.9

A tactic for dealing with keys present in one map but not the other in merge or mergeA.

A tactic of type WhenMissing f k x z is an abstract representation of a function of type k -> x -> f (Maybe z) .

Since: containers-0.5.9

A tactic for dealing with keys present in both maps in merge or mergeA.

A tactic of type WhenMatched f k x y z is an abstract representation of a function of type k -> x -> y -> f (Maybe z) .

Since: containers-0.5.9

An applicative version of merge.

mergeA takes two WhenMissing tactics, a WhenMatched tactic and two maps. It uses the tactics to merge the maps. Its behavior is best understood via its fundamental tactics, traverseMaybeMissing and zipWithMaybeAMatched.

Consider

mergeA (traverseMaybeMissing g1)
              (traverseMaybeMissing g2)
              (zipWithMaybeAMatched f)
              m1 m2

Take, for example,

m1 = [(0, 'a'), (1, 'b'), (3, 'c'), (4, 'd')]
m2 = [(1, "one"), (2, "two"), (4, "three")]

mergeA will first "align" these maps by key:

m1 = [(0, 'a'), (1, 'b'),               (3, 'c'), (4, 'd')]
m2 =           [(1, "one"), (2, "two"),           (4, "three")]

It will then pass the individual entries and pairs of entries to g1, g2, or f as appropriate:

actions = [g1 0 'a', f 1 'b' "one", g2 2 "two", g1 3 'c', f 4 'd' "three"]

Next, it will perform the actions in the actions list in order from left to right.

keys =     0        1          2           3        4
results = [Nothing, Just True, Just False, Nothing, Just True]

Finally, the Just results are collected into a map:

return value = [(1, True), (2, False), (4, True)]

The other tactics below are optimizations or simplifications of traverseMaybeMissing for special cases. Most importantly,

• dropMissing drops all the keys.
• preserveMissing leaves all the entries alone.
• mapMaybeMissing does not use the Applicative context.

When mergeA is given three arguments, it is inlined at the call site. To prevent excessive inlining, you should generally only use mergeA to define custom combining functions.

Since: containers-0.5.9

When a key is found in both maps, apply a function to the key and values, perform the resulting action, and maybe use the result in the merged map.

This is the fundamental WhenMatched tactic.

Since: containers-0.5.9

When a key is found in both maps, apply a function to the key and values to produce an action and use its result in the merged map.

Since: containers-0.5.9

Traverse over the entries whose keys are missing from the other map, optionally producing values to put in the result. This is the most powerful WhenMissing tactic, but others are usually more efficient.

Since: containers-0.5.9

Traverse over the entries whose keys are missing from the other map.

Since: containers-0.5.9

Filter the entries whose keys are missing from the other map using some Applicative action.

filterAMissing f = Merge.Lazy.traverseMaybeMissing $
  \k x -> (\b -> guard b *> Just x) <$> f k x

but this should be a little faster.

Since: containers-0.5.9

$O(n+m)$. An unsafe general combining function.

WARNING: This function can produce corrupt maps and its results may depend on the internal structures of its inputs. Users should prefer merge or mergeA.

When mergeWithKey is given three arguments, it is inlined to the call site. You should therefore use mergeWithKey only to define 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 Maps 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, filterWithKey f, or mapMaybeWithKey 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 m == 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

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

Since: containers-0.5.8

$O(n)$. 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 mapAccumRWithKey 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. The value at the greater of the two original keys is used as the first argument to 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.

Since: containers-0.5.4

$O(n)$. 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

$O(n)$. The set of all elements of the map contained in Args.

argSet (fromList [(5,"a"), (3,"b")]) == Data.Set.fromList [Arg 3 "b",Arg 5 "a"]
argSet 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

$O(n)$. Build a map from a set of elements contained inside Args.

fromArgSet (Data.Set.fromList [Arg 3 "aaa", Arg 5 "aaaaa"]) == fromList [(5,"aaaaa"), (3,"aaa")]
fromArgSet 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 key new_value old_value = show key ++ ":" ++ new_value ++ "|" ++ old_value
fromListWithKey f [(5,"a"), (5,"b"), (3,"b"), (3,"a"), (5,"c")] == fromList [(3, "3:a|b"), (5, "5:c|5:b|a")]
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

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

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

Since: containers-0.5.8

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

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

Since: containers-0.5.8

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

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

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

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

Since: containers-0.5.8

$O(n)$. 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"

$O(\log n)$. Take while a predicate on the keys holds. The user is responsible for ensuring that for all keys j and k in the map, j < k ==> p j >= p k. See note at spanAntitone.

takeWhileAntitone p = fromDistinctAscList . takeWhile (p . fst) . toList
takeWhileAntitone p = filterWithKey (k _ -> p k)

Since: containers-0.5.8

$O(\log n)$. Drop while a predicate on the keys holds. The user is responsible for ensuring that for all keys j and k in the map, j < k ==> p j >= p k. See note at spanAntitone.

dropWhileAntitone p = fromDistinctAscList . dropWhile (p . fst) . toList
dropWhileAntitone p = filterWithKey (\k _ -> not (p k))

Since: containers-0.5.8

$O(\log n)$. Divide a map at the point where a predicate on the keys stops holding. The user is responsible for ensuring that for all keys j and k in the map, j < k ==> p j >= p k.

spanAntitone p xs = (takeWhileAntitone p xs, dropWhileAntitone p xs)
spanAntitone p xs = partitionWithKey (\k _ -> p k) xs

Note: if p is not actually antitone, then spanAntitone will split the map at some unspecified point where the predicate switches from holding to not holding (where the predicate is seen to hold before the first key and to fail after the last key).

Since: containers-0.5.8

$O(m\log (\frac{n+1}{m+1})),m\le n$. Restrict a Map to only those keys found in a Set.

m `restrictKeys` s = filterWithKey (\k _ -> k `member` s) m
m `restrictKeys` s = m `intersection` fromSet (const ()) s

Since: containers-0.5.8

$O(m\log (\frac{n+1}{m+1})),m\le n$. Remove all keys in a Set from a Map.

m `withoutKeys` s = filterWithKey (\k _ -> k `notMember` s) m
m `withoutKeys` s = m `difference` fromSet (const ()) s

Since: containers-0.5.8

$O(n)$. 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.

Since: containers-0.5.4

$O(m\log (\frac{n+1}{m+1})),m\le n$. This function is defined as (isSubmapOf = isSubmapOfBy (==)).

$O(m\log (\frac{n+1}{m+1})),m\le n$. 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)])

Note that isSubmapOfBy (_ _ -> True) m1 m2 tests whether all the keys in m1 are also keys in m2.

$O(m\log (\frac{n+1}{m+1})),m\le n$. Is this a proper submap? (ie. a submap but not equal). Defined as (isProperSubmapOf = isProperSubmapOfBy (==)).

$O(m\log (\frac{n+1}{m+1})),m\le n$. Is this a proper submap? (ie. a submap but not equal). The expression (isProperSubmapOfBy f m1 m2) returns True when keys m1 and keys 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, 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.

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

$O(\log n)$. Take a given number of entries in key order, beginning with the smallest keys.

take n = fromDistinctAscList . take n . toAscList

Since: containers-0.5.8

$O(\log n)$. Drop a given number of entries in key order, beginning with the smallest keys.

drop n = fromDistinctAscList . drop n . toAscList

Since: containers-0.5.8

$O(\log n)$. Split a map at a particular index.

splitAt !n !xs = (take n xs, drop n xs)

Since: containers-0.5.8

$O(\log n)$. The minimal key of the map. Returns Nothing if the map is empty.

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

Since: containers-0.5.9

$O(\log n)$. The maximal key of the map. Returns Nothing if the map is empty.

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

Since: containers-0.5.9

$O(\log n)$. 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 empty                                      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 empty map.

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

Identity functor and monad. (a non-strict monad)

Since: base-4.8.0.0

Map covariantly over a WhenMissing f k x.

Since: containers-0.5.9

Map covariantly over a WhenMatched f k x y.

Since: containers-0.5.9

Map contravariantly over a WhenMissing f k _ x.

Since: containers-0.5.9

Map contravariantly over a WhenMatched f k _ y z.

Since: containers-0.5.9

Map contravariantly over a WhenMatched f k x _ z.

Since: containers-0.5.9

Map covariantly over a WhenMissing f k x, using only a 'Functor f' constraint.

Map covariantly over a WhenMatched f k x, using only a 'Functor f' constraint.