A map from keys to values. A map cannot contain duplicate keys; each key can map to at most one value.

\(O(1)\) Construct an empty map.

\(O(1)\) Construct a map with a single element.

\(O(1)\) Return True if this map is empty, False otherwise.

\(O(n)\) Return the number of key-value mappings in this map.

\(O(\log n)\) Return True if the specified key is present in the map, False otherwise.

\(O(\log n)\) Return the value to which the specified key is mapped, or Nothing if this map contains no mapping for the key.

\(O(\log n)\) Return the value to which the specified key is mapped, or Nothing if this map contains no mapping for the key.

This is a flipped version of lookup.

Since: 0.2.11

\(O(\log n)\) Return the value to which the specified key is mapped, or the default value if this map contains no mapping for the key.

Since: 0.2.11

\(O(\log n)\) Return the value to which the specified key is mapped, or the default value if this map contains no mapping for the key.

DEPRECATED: lookupDefault is deprecated as of version 0.2.11, replaced by findWithDefault.

\(O(\log n)\) Return the value to which the specified key is mapped. Calls error if this map contains no mapping for the key.

\(O(\log n)\) Associate the specified value with the specified key in this map. If this map previously contained a mapping for the key, the old value is replaced.

\(O(\log n)\) Associate the value with the key in this map. If this map previously contained a mapping for the key, the old value is replaced by the result of applying the given function to the new and old value. Example:

insertWith f k v map
  where f new old = new + old

\(O(\log n)\) Remove the mapping for the specified key from this map if present.

\(O(\log n)\) Adjust the value tied to a given key in this map only if it is present. Otherwise, leave the map alone.

\(O(\log n)\) 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.

\(O(\log n)\) 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)

\(O(\log n)\) The expression (alterF f k map) alters the value x at k, or absence thereof.

alterF can be used to insert, delete, or update a value in a map.

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

Since: 0.2.10

\(O(n \log m)\) Inclusion of maps. A map is included in another map if the keys are subsets and the corresponding values are equal:

isSubmapOf m1 m2 = keys m1 `isSubsetOf` keys m2 &&
                   and [ v1 == v2 | (k1,v1) <- toList m1; let v2 = m2 ! k1 ]

Examples

>>> fromList [(1,'a')] `isSubmapOf` fromList [(1,'a'),(2,'b')]
True

>>> fromList [(1,'a'),(2,'b')] `isSubmapOf` fromList [(1,'a')]
False

Since: 0.2.12

\(O(n \log m)\) Inclusion of maps with value comparison. A map is included in another map if the keys are subsets and if the comparison function is true for the corresponding values:

isSubmapOfBy cmpV m1 m2 = keys m1 `isSubsetOf` keys m2 &&
                          and [ v1 `cmpV` v2 | (k1,v1) <- toList m1; let v2 = m2 ! k1 ]

Examples

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

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

Since: 0.2.12

\(O(n+m)\) The union of two maps. If a key occurs in both maps, the mapping from the first will be the mapping in the result.

Examples

>>> union (fromList [(1,'a'),(2,'b')]) (fromList [(2,'c'),(3,'d')])
fromList [(1,'a'),(2,'b'),(3,'d')]

\(O(n+m)\) The union of two maps. If a key occurs in both maps, the provided function (first argument) will be used to compute the result.

\(O(n+m)\) The union of two maps. If a key occurs in both maps, the provided function (first argument) will be used to compute the result.

Construct a set containing all elements from a list of sets.

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: \( O (n * \log(m)) \), 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 !?)

Since: 0.2.13.0

\(O(n)\) Transform this map by applying a function to every value.

\(O(n)\) Transform this map by applying a function to every value.

\(O(n)\) Perform an Applicative action for each key-value pair in a HashMap and produce a HashMap of all the results.

Note: the order in which the actions occur is unspecified. In particular, when the map contains hash collisions, the order in which the actions associated with the keys involved will depend in an unspecified way on their insertion order.

\(O(n)\). 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 there is no guarantee which of the associated values is chosen for the conflicting key.

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

Since: 0.2.14.0

\(O(n \log m)\) Difference of two maps. Return elements of the first map not existing in the second.

\(O(n \log m)\) 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.

\(O(n \log m)\) Intersection of two maps. Return elements of the first map for keys existing in the second.

\(O(n \log m)\) Intersection of two maps. If a key occurs in both maps the provided function is used to combine the values from the two maps.

\(O(n \log m)\) Intersection of two maps. If a key occurs in both maps the provided function is used to combine the values from the two maps.

\(O(n)\) Reduce the map by applying a function to each element and combining the results with a monoid operation.

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the right-identity of the operator).

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the left-identity of the operator).

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the right-identity of the operator). Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the left-identity of the operator). Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the right-identity of the operator). Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the left-identity of the operator). Each application of the operator is evaluated before using the result in the next application. This function is strict in the starting value.

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the right-identity of the operator).

\(O(n)\) Reduce this map by applying a binary operator to all elements, using the given starting value (typically the left-identity of the operator).

\(O(n)\) Filter this map by retaining only elements which values satisfy a predicate.

\(O(n)\) Filter this map by retaining only elements satisfying a predicate.

\(O(n)\) Transform this map by applying a function to every value and retaining only some of them.

\(O(n)\) Transform this map by applying a function to every value and retaining only some of them.

\(O(n)\) Return a list of this map's keys. The list is produced lazily.

\(O(n)\) Return a list of this map's values. The list is produced lazily.

\(O(n)\) Return a list of this map's elements. The list is produced lazily. The order of its elements is unspecified.

\(O(n)\) Construct a map with the supplied mappings. If the list contains duplicate mappings, the later mappings take precedence.

\(O(n \log n)\) Construct a map from a list of elements. Uses the provided function f to merge duplicate entries with (f newVal oldVal).

Examples

Given a list xs, create a map with the number of occurrences of each element in xs:

let xs = ['a', 'b', 'a']
in fromListWith (+) [ (x, 1) | x <- xs ]

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

Given a list of key-value pairs xs :: [(k, v)], group all values by their keys and return a HashMap k [v].

let xs = [('a', 1), ('b', 2), ('a', 3)]
in fromListWith (++) [ (k, [v]) | (k, v) <- xs ]

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

Note that the lists in the resulting map contain elements in reverse order from their occurences in the original list.

More generally, duplicate entries are accumulated as follows; this matters when f is not commutative or not associative.

fromListWith f [(k, a), (k, b), (k, c), (k, d)]
= fromList [(k, f d (f c (f b a)))]

\(O(n \log n)\) Construct a map from a list of elements. Uses the provided function to merge duplicate entries.

Examples

Given a list of key-value pairs where the keys are of different flavours, e.g:

data Key = Div | Sub

and the values need to be combined differently when there are duplicates, depending on the key:

combine Div = div
combine Sub = (-)

then fromListWithKey can be used as follows:

fromListWithKey combine [(Div, 2), (Div, 6), (Sub, 2), (Sub, 3)]
= fromList [(Div, 3), (Sub, 1)]

More generally, duplicate entries are accumulated as follows;

fromListWith f [(k, a), (k, b), (k, c), (k, d)]
= fromList [(k, f k d (f k c (f k b a)))]

Since: 0.2.11

\(O(n)\) Produce a HashSet of all the keys in the given HashMap.

>>> HashSet.keysSet (HashMap.fromList [(1, "a"), (2, "b")]
fromList [1,2]

Since: 0.2.10.0