The map interface (the java™ tutorials collections interfaces)

8/30/2016 The Map Interface (The Java™ Tutorials > Collections > Interfaces)
https://docs.oracle.com/javase/tutorial/collections/interfaces/map.html 1/6
Trail: Collections
Lesson: Interfaces
The Map Interface
A Map is an object that maps keys to values. A map cannot contain duplicate keys: Each key can map to at most one value. It models the
mathematical function abstraction. The Map interface includes methods for basic operations (such as put, get, remove, containsKey,
containsValue, size, and empty), bulk operations (such as putAll and clear), and collection views (such as keySet, entrySet, and
values).
The Java platform contains three generalpurpose Map implementations: HashMap, TreeMap, and LinkedHashMap. Their behavior and
performance are precisely analogous to HashSet, TreeSet, and LinkedHashSet, as described in The Set Interface section.
The remainder of this page discusses the Map interface in detail. But first, here are some more examples of collecting to Maps using JDK 8 aggregate
operations. Modeling realworld objects is a common task in objectoriented programming, so it is reasonable to think that some programs might, for
example, group employees by department:
// Group employees by department
Map<Department, List<Employee>> byDept = employees.stream()
.collect(Collectors.groupingBy(Employee::getDepartment));
Or compute the sum of all salaries by department:
// Compute sum of salaries by department
Map<Department, Integer> totalByDept = employees.stream()
.collect(Collectors.groupingBy(Employee::getDepartment,
Collectors.summingInt(Employee::getSalary)));
Or perhaps group students by passing or failing grades:
// Partition students into passing and failing
Map<Boolean, List<Student>> passingFailing = students.stream()
.collect(Collectors.partitioningBy(s ‐> s.getGrade()>= PASS_THRESHOLD));
You could also group people by city:
// Classify Person objects by city
Map<String, List<Person>> peopleByCity
         = personStream.collect(Collectors.groupingBy(Person::getCity));
Or even cascade two collectors to classify people by state and city:
// Cascade Collectors
Map<String, Map<String, List<Person>>> peopleByStateAndCity
  = personStream.collect(Collectors.groupingBy(Person::getState,
  Collectors.groupingBy(Person::getCity)))
Again, these are but a few examples of how to use the new JDK 8 APIs. For indepth coverage of lambda expressions and aggregate operations see
the lesson entitled Aggregate Operations.
Map Interface Basic Operations
The basic operations of Map (put, get, containsKey, containsValue, size, and isEmpty) behave exactly like their counterparts in
Hashtable. The following program generates a frequency table of the words found in its argument list. The frequency table maps each word
to the number of times it occurs in the argument list.
import java.util.*;
Documentation
The Java™ Tutorials

public class Freq {
    public static void main(String[] args) {
        Map<String, Integer> m = new HashMap<String, Integer>();
        // Initialize frequency table from command line
        for (String a : args) {
            Integer freq = m.get(a);
            m.put(a, (freq == null) ? 1 : freq + 1);
        }
        System.out.println(m.size() + " distinct words:");
        System.out.println(m);
    }
}
The only tricky thing about this program is the second argument of the put statement. That argument is a conditional expression that has the effect
of setting the frequency to one if the word has never been seen before or one more than its current value if the word has already been seen. Try
running this program with the command:
java Freq if it is to be it is up to me to delegate
The program yields the following output.
8 distinct words:
{to=3, delegate=1, be=1, it=2, up=1, if=1, me=1, is=2}
Suppose you'd prefer to see the frequency table in alphabetical order. All you have to do is change the implementation type of the Map from
HashMap to TreeMap. Making this fourcharacter change causes the program to generate the following output from the same command line.
8 distinct words:
{be=1, delegate=1, if=1, is=2, it=2, me=1, to=3, up=1}
Similarly, you could make the program print the frequency table in the order the words first appear on the command line simply by changing the
implementation type of the map to LinkedHashMap. Doing so results in the following output.
8 distinct words:
{if=1, it=2, is=2, to=3, be=1, up=1, me=1, delegate=1}
This flexibility provides a potent illustration of the power of an interfacebased framework.
Like the Setand Listinterfaces, Map strengthens the requirements on the equals and hashCode methods so that two Map objects can be
compared for logical equality without regard to their implementation types. Two Map instances are equal if they represent the same keyvalue
mappings.
By convention, all generalpurpose Map implementations provide constructors that take a Map object and initialize the new Map to contain all the key
value mappings in the specified Map. This standard Map conversion constructor is entirely analogous to the standard Collection constructor: It
allows the caller to create a Map of a desired implementation type that initially contains all of the mappings in another Map, regardless of the other
Map's implementation type. For example, suppose you have a Map, named m. The following oneliner creates a new HashMap initially containing all
of the same keyvalue mappings as m.
Map<K, V> copy = new HashMap<K, V>(m);
Map Interface Bulk Operations
The clear operation does exactly what you would think it could do: It removes all the mappings from the Map. The putAll operation is the Map
analogue of the Collection interface's addAll operation. In addition to its obvious use of dumping one Map into another, it has a second, more
subtle use. Suppose a Map is used to represent a collection of attributevalue pairs; the putAll operation, in combination with the Map conversion
constructor, provides a neat way to implement attribute map creation with default values. The following is a static factory method that demonstrates
this technique.
static <K, V> Map<K, V> newAttributeMap(Map<K, V>defaults, Map<K, V> overrides) {
    Map<K, V> result = new HashMap<K, V>(defaults);
    result.putAll(overrides);
    return result;
}
Collection Views

The Collection view methods allow a Map to be viewed as a Collection in these three ways:
keySet — the Set of keys contained in the Map.
values — The Collection of values contained in the Map. This Collection is not a Set, because multiple keys can map to the same
value.
entrySet — the Set of keyvalue pairs contained in the Map. The Map interface provides a small nested interface called Map.Entry, the
type of the elements in this Set.
The Collection views provide the only means to iterate over a Map. This example illustrates the standard idiom for iterating over the keys in a Map
with a foreach construct:
for (KeyType key : m.keySet())
    System.out.println(key);
and with an iterator:
// Filter a map based on some
// property of its keys.
for (Iterator<Type> it = m.keySet().iterator(); it.hasNext(); )
    if (it.next().isBogus())
        it.remove();
The idiom for iterating over values is analogous. Following is the idiom for iterating over keyvalue pairs.
for (Map.Entry<KeyType, ValType> e : m.entrySet())
    System.out.println(e.getKey() + ": " + e.getValue());
At first, many people worry that these idioms may be slow because the Map has to create a new Collection instance each time a Collection
view operation is called. Rest easy: There's no reason that a Map cannot always return the same object each time it is asked for a given Collection
view. This is precisely what all the Map implementations in java.util do.
With all three Collection views, calling an Iterator's remove operation removes the associated entry from the backing Map, assuming that the
backing Map supports element removal to begin with. This is illustrated by the preceding filtering idiom.
With the entrySet view, it is also possible to change the value associated with a key by calling a Map.Entry's setValue method during iteration
(again, assuming the Map supports value modification to begin with). Note that these are the only safe ways to modify a Map during iteration; the
behavior is unspecified if the underlying Map is modified in any other way while the iteration is in progress.
The Collection views support element removal in all its many forms — remove, removeAll, retainAll, and clear operations, as well as
the Iterator.remove operation. (Yet again, this assumes that the backing Map supports element removal.)
The Collection views do not support element addition under any circumstances. It would make no sense for the keySet and values views, and
it's unnecessary for the entrySet view, because the backing Map's put and putAll methods provide the same functionality.
Fancy Uses of Collection Views: Map Algebra
When applied to the Collection views, bulk operations (containsAll, removeAll, and retainAll) are surprisingly potent tools. For starters,
suppose you want to know whether one Map is a submap of another — that is, whether the first Map contains all the keyvalue mappings in the
second. The following idiom does the trick.
if (m1.entrySet().containsAll(m2.entrySet())) {
    ...
}
Along similar lines, suppose you want to know whether two Map objects contain mappings for all of the same keys.
if (m1.keySet().equals(m2.keySet())) {
    ...
}
Suppose you have a Map that represents a collection of attributevalue pairs, and two Sets representing required attributes and permissible attributes.
(The permissible attributes include the required attributes.) The following snippet determines whether the attribute map conforms to these constraints
and prints a detailed error message if it doesn't.
static <K, V> boolean validate(Map<K, V> attrMap, Set<K> requiredAttrs, Set<K>permittedAttrs) {
    boolean valid = true;
    Set<K> attrs = attrMap.keySet();

    if (! attrs.containsAll(requiredAttrs)) {
        Set<K> missing = new HashSet<K>(requiredAttrs);
        missing.removeAll(attrs);
        System.out.println("Missing attributes: " + missing);
        valid = false;
    }
    if (! permittedAttrs.containsAll(attrs)) {
        Set<K> illegal = new HashSet<K>(attrs);
        illegal.removeAll(permittedAttrs);
        System.out.println("Illegal attributes: " + illegal);
        valid = false;
    }
    return valid;
}
Suppose you want to know all the keys common to two Map objects.
Set<KeyType>commonKeys = new HashSet<KeyType>(m1.keySet());
commonKeys.retainAll(m2.keySet());
A similar idiom gets you the common values.
All the idioms presented thus far have been nondestructive; that is, they don't modify the backing Map. Here are a few that do. Suppose you want to
remove all of the keyvalue pairs that one Map has in common with another.
m1.entrySet().removeAll(m2.entrySet());
Suppose you want to remove from one Map all of the keys that have mappings in another.
m1.keySet().removeAll(m2.keySet());
What happens when you start mixing keys and values in the same bulk operation? Suppose you have a Map, managers, that maps each employee in
a company to the employee's manager. We'll be deliberately vague about the types of the key and the value objects. It doesn't matter, as long as
they're the same. Now suppose you want to know who all the "individual contributors" (or nonmanagers) are. The following snippet tells you exactly
what you want to know.
Set<Employee> individualContributors = new HashSet<Employee>(managers.keySet());
individualContributors.removeAll(managers.values());
Suppose you want to fire all the employees who report directly to some manager, Simon.
Employee simon = ... ;
managers.values().removeAll(Collections.singleton(simon));
Note that this idiom makes use of Collections.singleton, a static factory method that returns an immutable Set with the single, specified
element.
Once you've done this, you may have a bunch of employees whose managers no longer work for the company (if any of Simon's directreports were
themselves managers). The following code will tell you which employees have managers who no longer works for the company.
Map<Employee, Employee> m = new HashMap<Employee, Employee>(managers);
m.values().removeAll(managers.keySet());
Set<Employee> slackers = m.keySet();
This example is a bit tricky. First, it makes a temporary copy of the Map, and it removes from the temporary copy all entries whose (manager) value is
a key in the original Map. Remember that the original Map has an entry for each employee. Thus, the remaining entries in the temporary Map
comprise all the entries from the original Map whose (manager) values are no longer employees. The keys in the temporary copy, then, represent
precisely the employees that we're looking for.
There are many more idioms like the ones contained in this section, but it would be impractical and tedious to list them all. Once you get the hang of
it, it's not that difficult to come up with the right one when you need it.
Multimaps
A multimap is like a Map but it can map each key to multiple values. The Java Collections Framework doesn't include an interface for multimaps
because they aren't used all that commonly. It's a fairly simple matter to use a Map whose values are List instances as a multimap. This technique
is demonstrated in the next code example, which reads a word list containing one word per line (all lowercase) and prints out all the anagram groups
that meet a size criterion. An anagram group is a bunch of words, all of which contain exactly the same letters but in a different order. The program

takes two arguments on the command line: (1) the name of the dictionary file and (2) the minimum size of anagram group to print out. Anagram
groups containing fewer words than the specified minimum are not printed.
There is a standard trick for finding anagram groups: For each word in the dictionary, alphabetize the letters in the word (that is, reorder the word's
letters into alphabetical order) and put an entry into a multimap, mapping the alphabetized word to the original word. For example, the word bad
causes an entry mapping abd into bad to be put into the multimap. A moment's reflection will show that all the words to which any given key maps
form an anagram group. It's a simple matter to iterate over the keys in the multimap, printing out each anagram group that meets the size constraint.
The following program is a straightforward implementation of this technique.
import java.util.*;
import java.io.*;
public class Anagrams {
    public static void main(String[] args) {
        int minGroupSize = Integer.parseInt(args[1]);
        // Read words from file and put into a simulated multimap
        Map<String, List<String>> m = new HashMap<String, List<String>>();
        try {
            Scanner s = new Scanner(new File(args[0]));
            while (s.hasNext()) {
                String word = s.next();
                String alpha = alphabetize(word);
                List<String> l = m.get(alpha);
                if (l == null)
                    m.put(alpha, l=new ArrayList<String>());
                l.add(word);
            }
        } catch (IOException e) {
            System.err.println(e);
            System.exit(1);
        }
        // Print all permutation groups above size threshold
        for (List<String> l : m.values())
            if (l.size() >= minGroupSize)
                System.out.println(l.size() + ": " + l);
    }
    private static String alphabetize(String s) {
        char[] a = s.toCharArray();
        Arrays.sort(a);
        return new String(a);
    }
}
Running this program on a 173,000word dictionary file with a minimum anagram group size of eight produces the following output.
9: [estrin, inerts, insert, inters, niters, nitres, sinter,
     triens, trines]
8: [lapse, leaps, pales, peals, pleas, salep, sepal, spale]
8: [aspers, parses, passer, prases, repass, spares, sparse,
     spears]
10: [least, setal, slate, stale, steal, stela, taels, tales,
      teals, tesla]
8: [enters, nester, renest, rentes, resent, tenser, ternes,
     treens]
8: [arles, earls, lares, laser, lears, rales, reals, seral]
8: [earings, erasing, gainers, reagins, regains, reginas,
     searing, seringa]
8: [peris, piers, pries, prise, ripes, speir, spier, spire]
12: [apers, apres, asper, pares, parse, pears, prase, presa,
      rapes, reaps, spare, spear]
11: [alerts, alters, artels, estral, laster, ratels, salter,
      slater, staler, stelar, talers]

9: [capers, crapes, escarp, pacers, parsec, recaps, scrape,
     secpar, spacer]
9: [palest, palets, pastel, petals, plates, pleats, septal,
     staple, tepals]
9: [anestri, antsier, nastier, ratines, retains, retinas,
     retsina, stainer, stearin]
8: [ates, east, eats, etas, sate, seat, seta, teas]
8: [carets, cartes, caster, caters, crates, reacts, recast,
     traces]
Many of these words seem a bit bogus, but that's not the program's fault; they're in the dictionary file. Here's the dictionary file we used. It was
derived from the Public Domain ENABLE benchmark reference word list.
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