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AMap
is an object that maps keys to values. A map cannot contain duplicate keys: Each key can map to at most one value. TheMap
interface follows.The Java platform contains three general-purposepublic interface Map { //Basic operations V put(K key, V value); V get(Object key); V remove(Object key); boolean containsKey(Object key); boolean containsValue(Object value); int size(); boolean isEmpty(); //Bulk operations void putAll(Map<? extends K,? extends V> t); void clear(); // Collection Views public Set<K> keySet(); public Collection<V> values(); public Set<Map.Entry<K,V>> entrySet(); // Interface for entrySet elements public interface Entry { K getKey(); V getValue(); V setValue(V value); } }Map
implementations:HashMap
,TreeMap
, andLinkedHashMap
. Their behavior and performance are precisely analogous toHashSet
,TreeSet
, andLinkedHashSet
, as described in The Set Interface section. Also,Hashtable
was retrofitted to implementMap
.
If you've usedHashtable
, you're already familiar with the general basics ofMap
. (Of courseMap
is an interface, whileHashtable
is a concrete implementation.) The following are the major differences:Finally,
Map
providesCollection
views instead of direct support for iteration viaEnumeration
objects.Collection
views greatly enhance the expressiveness of the interface, as discussed later in this section.Map
allows you to iterate over keys, values, or key-value pairs;Hashtable
does not provide the third option.Map
provides a safe way to remove entries in the midst of iteration;Hashtable
did not.Map
fixes a minor deficiency in theHashtable
interface.Hashtable
has a method calledcontains
, which returnstrue
if theHashtable
contains a given value. Given its name, you'd expect this method to returntrue
if theHashtable
contained a given key, because the key is the primary access mechanism for aHashtable
. TheMap
interface eliminates this source of confusion by renaming the methodcontainsValue
. Also, this improves the interface's consistency containsValue
parallelscontainsKey
.
The basic operations ofMap
(put
,get
,containsKey
,containsValue
,size
, andisEmpty
) behave exactly like their counterparts inHashtable
. The followingprogram 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.The only tricky thing about this program is the second argument of theimport java.util.*; 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); } }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:The program yields the following output.java Freq if it is to be it is up to me to delegateSuppose you'd prefer to see the frequency table in alphabetical order. All you have to do is change the implementation type of the8 distinct words: {to=3, delegate=1, be=1, it=2, up=1, if=1, me=1, is=2}Map
fromHashMap
toTreeMap
. Making this four-character change causes the program to generate the following output from the same command line.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 to8 distinct words: {be=1, delegate=1, if=1, is=2, it=2, me=1, to=3, up=1}LinkedHashMap
. Doing so results in the following output.This flexibility provides a potent illustration of the power of an interface-based framework.8 distinct words: {if=1, it=2, is=2, to=3, be=1, up=1, me=1, delegate=1}Like the
Set
andList
interfaces,Map
strengthens the requirements on theequals
andhashCode
methods so that twoMap
objects can be compared for logical equality without regard to their implementation types. TwoMap
instances are equal if they represent the same key-value mappings.By convention, all
Map
implementations provide constructors that take aMap
object and initialize the newMap
to contain all the key-value mappings in the specifiedMap
. This standardMap
conversion constructor is entirely analogous to the standardCollection
constructor: It allows the caller to create aMap
of a desired implementation type that initially contains all of the mappings in anotherMap
, regardless of the otherMap
's implementation type. For example, suppose you have aMap
, namedm
. The following one-liner creates a newHashMap
initially containing all of the same key-value mappings asm
.Map<K, V> copy = new HashMap<K, V>(m);
Theclear
operation does exactly what you would think it could do: it removes all the mappings from theMap
. TheputAll
operation is theMap
analogue of theCollection
interface'saddAll
operation. In addition to its obvious use of dumping oneMap
into another, it has a second, more subtle use. Suppose aMap
is used to represent a collection of attribute-value pairs; theputAll
operation, in combination with theMap
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; }
TheCollection
view methods allow aMap
to be viewed as aCollection
in these three ways:The
keySet
theSet
of keys contained in theMap
.values
TheCollection
of values contained in theMap
. ThisCollection
is not aSet
, because multiple keys can map to the same value.entrySet
theSet
of key-value pairs contained in theMap
. TheMap
interface provides a small nested interface calledMap.Entry
, the type of the elements in thisSet
.Collection
views provide the only means to iterate over aMap
. This example illustrates the standard idiom for iterating over the keys in aMap
with afor-each
construct:and with anfor (KeyType key : m.keySet()) System.out.println(key);iterator
:The idiom for iterating over values is analogous. Following is the idiom for iterating over key-value pairs.//Filter a map based on some property of its keys. for (Iterator<Type> i=m.keySet().iterator(); i.hasNext(); ) if (i.next().isBogus()) i.remove();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 newCollection
instance each time aCollection
view operation is called. Rest easy: There's no reason that aMap
cannot always return the same object each time it is asked for a givenCollection
view. This is precisely what all theMap
implementations injava.util
do.With all three
Collection
views, calling anIterator
'sremove
operation removes the associated entry from the backingMap
, assuming that the backingMap
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 aMap.Entry
'ssetValue
method during iteration (again, assuming theMap
supports value modification to begin with). Note that these are the only safe ways to modify aMap
during iteration; the behavior is unspecified if the underlyingMap
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
, andclear
operations, as well as theIterator.remove
operation. (Yet again, this assumes that the backingMap
supports element removal.)The
Collection
views do not support element addition under any circumstances. It would make no sense for thekeySet
andvalues
views, and it's unnecessary for theentrySet
view, because the backingMap
'sput
andputAll
methods provide the same functionality.
When applied to theCollection
views, bulk operations (containsAll
,removeAll
, andretainAll
) are surprisingly potent tools. For starters, suppose you want to know whether oneMap
is a submap of another that is, whether the firstMap
contains all the key-value mappings in the second. The following idiom does the trick.Along similar lines, suppose you want to know whether twoif (m1.entrySet().containsAll(m2.entrySet())) { ... }Map
objects contain mappings for all the same keys.Suppose you have aif (m1.keySet().equals(m2.keySet())) { ... }Map
that represents a collection of attribute-value pairs, and twoSet
s 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.Suppose you want to know all the keys common to twostatic <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; }Map
objects.A similar idiom gets you the common values.Set<KeyType>commonKeys = new HashSet<KeyType>(m1.keySet()); commonKeys.retainAll(m2.keySet);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 the key-value pairs that oneMap
has in common with another.Suppose you want to remove from onem1.entrySet().removeAll(m2.entrySet());Map
all the keys that have mappings in another.What happens when you start mixing keys and values in the same bulk operation? Suppose you have am1.keySet().removeAll(m2.keySet());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.Suppose you want to fire all the employees who report directly to some manager, Simon.Set<Employee> individualContributors = new HashSet<Employee>(managers.keySet()); individualContributors.removeAll(managers.values());Note that this idiom makes use ofEmployee simon = ... ; managers.values().removeAll(Collections.singleton(simon));Collections.singleton
, a static factory method that returns an immutableSet
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 direct-reports were themselves managers). The following code will tell you all the employees whose manager no longer works for the company.
This example is a bit tricky. First, it makes a temporary copy of theMap<Employee, Employee> m = new HashMap<Employee, Employee>(managers); m.values().removeAll(managers.keySet()); Set<Employee> slackers = m.keySet();Map
, and it removes from the temporary copy all entries whose (manager) value is a key in the originalMap
. Remember that the originalMap
has an entry for each employee. Thus, the remaining entries in the temporaryMap
comprise all the entries from the originalMap
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.
A multimap is like aMap
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 touse aMap
whose values areList
instances as a multimap. This technique is demonstrated in the next code example, which reads a wordlist 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. The only tricky part is thealphabetize
method, which returns a string containing the same characters as its argument in alphabetical order. This routine, which has nothing to do with the Java Collections Framework, implements a slick bucket sort. It assumes that the word being alphabetized consists entirely of lowercase alphabetic characters.Running this program on a 173,000-word dictionary file takes about four seconds on a 1.8 GHz Pentium 4. With a minimum anagram group size of eight, it produces the following output.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])); String word; while(s.hasNext()) { String alpha = alphabetize(word = s.next()); 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) { int count[] = new int[256]; int len = s.length(); for (int i=0; i<len; i++) count[s.charAt(i)]++; StringBuffer result = new StringBuffer(len); for (char c='a'; c<='z'; c++) for (int i=0; i<count[c]; i++) result.append(c); return result.toString(); } }Many of these words seem a bit bogus, but that's not the program's fault; they're in the dictionary file. Here's the9: [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]dictionary file
we used. It is derived from the Public Domain ENABLE benchmark reference word list, at http://personal.riverusers.com/~thegrendel/
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