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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. The
Map
interface is shown below:The JDK contains two new general-purposepublic interface Map { // Basic Operations Object put(Object key, Object value); Object get(Object key); Object remove(Object key); boolean containsKey(Object key); boolean containsValue(Object value); int size(); boolean isEmpty(); // Bulk Operations void putAll(Map t); void clear(); // Collection Views public Set keySet(); public Collection values(); public Set entrySet(); // Interface for entrySet elements public interface Entry { Object getKey(); Object getValue(); Object setValue(Object value); } }Map
implementations.HashMap
, which stores its entries in a hash table, is the best-performing implementation.
TreeMap
, which stores its entries in a red-black tree, guarantees the order of iteration. Also,
Hashtable
has been retrofitted to implement
Map
. For more information on implementations, see the Implementations lesson.
If you've usedHashtable
, you're already familiar with the general flavor ofMap
. (Of courseMap
is an interface, whileHashtable
is a concrete implementation.) Here are the major differences:
Map
providesCollection
-views in lieu of direct support for iteration viaEnumeration
objects.Collection
-views greatly enhance the expressiveness of the interface, as discussed later in this lesson.Map
allows you to iterate over keys, values, or key-value pairs;Hashtable
did not provide the third option.Map
provides a safe way to remove entries in the midst of iteration;Hashtable
did not.Further,
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, as the key is the primary access mechanism for aHashtable
. The Map interface eliminates this source of confusion by renaming the methodcontainsValue
. Also, this improves the consistency of the interface:containsValue
parallelscontainsKey
nicely.
The basic operations (put
,get
,remove
,containsKey
,containsValue
,size
, andisEmpty
) behave exactly like their counterparts inHashtable
. Here'sa simple program to generate 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 thing even slightly tricky about this program is the second argument of theimport java.util.*; public class Freq { private static final Integer ONE = new Integer(1); public static void main(String args[]) { Map m = new HashMap(); // Initialize frequency table from command line for (int i=0; i<args.length; i++) { Integer freq = (Integer) m.get(args[i]); m.put(args[i], (freq==null ? ONE : new Integer(freq.intValue() + 1))); } System.out.println(m.size()+" distinct words detected:"); System.out.println(m); } }put
statement. It's 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. Let's run the program:Suppose you'd prefer to see the frequency table in alphabetical order. All you have to do is change the implementation type of the% java Freq if it is to be it is up to me to delegate 8 distinct words detected: {to=3, me=1, delegate=1, it=2, is=2, if=1, be=1, up=1}Map
fromHashMap
toTreeMap
. Making this four character change causes the program to generate the following output from the same command line:Are interfaces cool, or what?8 distinct words detected: {be=1, delegate=1, if=1, is=2, it=2, me=1, to=3, up=1}Like the
Set
and
List
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
objects 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 of the key-value mappings in the specifiedMap
. This standardMap
constructor is entirely analogous to the standard collection constructor forCollection
implementations. 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 copy = new HashMap(m);
Theclear
operation does exactly what you think it does: it removes all of 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 the standardMap
constructor, provides a neat way to implement attribute map creation with default values. Here's a static factory method demonstrating this technique:static Map newAttributeMap(Map defaults, Map overrides) { Map result = new HashMap(defaults); result.putAll(overrides); return result; }
TheCollection
-view methods allow aMap
to be viewed as aCollection
in three ways:The
keySet
: theSet
of keys contained in theMap
.values
: TheCollection
of values contained in theMap
. ThisCollection
is not aSet
, as 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
that is the type of the elements in thisSet
.Collection
-views provide the only means to iterate over aMap
. Here's an example illustrating the standard idiom for iterating over the keys in aMap
:The idiom for iterating over values is analogous. Here's the idiom for iterating over key-value pairs:for (Iterator i=m.keySet().iterator(); i.hasNext(); ) System.out.println(i.next());for (Iterator i=m.entrySet().iterator(); i.hasNext(); ) { Map.Entry e = (Map.Entry) i.next(); System.out.println(e.getKey() + ": " + e.getValue()); }When first presented with these idioms, many people worry that they may be slow because the
Map
has to create a newCollection
object each time aCollection
-view operation is called. Rest easy: This is not the case. There's no reason that aMap
can't always return the same object each time it is asked for a givenCollection
-view. This is precisely what all of the JDK'sMap
implementations do.With all three
Collection
-views, calling anIterator
'sremove
operation removes the associated entry from the backingMap
(assuming theMap
supports element removal to begin with). With theentrySet
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: theremove
,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, as the backingMap
'sput
andputAll
provide the same functionality.
When applied to theCollection
-views, the bulk operations (containsAll
,removeAll
andretainAll
) are a surprisingly potent tool. For starters, suppose you want to know whether oneMap
is a submap of another, that is, whether the firstMap
contains all of the key-value mappings in the second. The following idiom does the trick:Along similar lines, suppose that you want to know if twoif (m1.entrySet().containsAll(m2.entrySet())) { ... }Map
objects contain mappings for all the same keys:Suppose you have a map that represents a collection of attribute-value 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:if (m1.keySet().equals(m2.keySet())) { ... }Suppose you want to know all the keys common to twoboolean valid = true; Set attributes = attributeMap.keySet(); if(!attributes.containsAll(requiredAttributes)) { Set missing = new HashSet(requiredAttributes); missing.removeAll(attributes); System.out.println("Missing required attributes: "+missing); valid = false; } if (!permissibleAttributes.containsAll(attributes)) { Set illegal = new HashSet(attributes); illegal.removeAll(permissibleAttributes); System.out.println("Contains illegal attributes: "+illegal); valid = false; } if (valid) System.out.println("OK");Map
objects:A similar idiom gets you the common values, and the common key-value pairs. Extra care is needed if you get the common key-value pairs, as the elements of the resultingSet commonKeys = new HashSet(a.keySet()); commonKeys.retainAll(b.keySet());Set
, which areMap.Entry
objects, may become invalid if theMap
is modified.All the idioms presented thus far have been "non-destructive": 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:And what happens when you start mixing keys and values in the same bulk operation? Suppose you have am1.keySet().removeAll(m2.keySet());Map
calledmanagers
that maps each employee in a company to the employee's manager. We'll be deliberately vague about the types of the key and value objects. It doesn't matter, so long as they're the same. Now suppose you want to know who all the "individual contributors" are. (This is corporate-speak for employees who are not managers.) The following one-liner tells you exactly what you want to know:Suppose you want to fire all of the employees who report directly to some manager (we'll call him herbert):Set individualContributors = new HashSet(managers.keySet()); individualContributors.removeAll(managers.values());Note that this idiom makes use ofEmployee herbert = ... ; managers.values().removeAll(Collections.singleton(herbert));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 herbert's direct-reports were themselves managers). The following code tells you all of the employees whose manager no longer works for the company:
This example is a bit tricky. First it makes a temporary copy of theMap m = new HashMap(managers); m.values().removeAll(managers.keySet()); Set slackers = m.keySet();Map
. Then it removes from the temporary copy all entries whose (manager) value is a key in the originalMap
. Recall that the originalMap
contains an entry for every 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. If you stare at the example for a little while, this should all become clear. If it doesn't, now would be a good time to get a steaming hot mug of freshly brewed Java.There are many, 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 hard to come up with the right one when you need it.
A multimap is like a map, except it can map each key to multiple values. The 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 aMap
whose values areList
objects as a multimap. This technique is demonstrated in the next code example, which reads a dictionary containing one word per line (all lowercase) and prints out all of the permutation groups in the dictionary that meet a size criterion. A permutation 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: the name of the dictionary file, and the minimum size of permutation group to print out. Permutation groups containing fewer words than the specified minimum are not printed.There is a standard trick for finding permutation 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 of the words to which any given key maps form a permutation group. It's a simple matter to iterate over the keys in the multimap, printing out each permutation group that meets the size constraint.
The following program
is a straightforward implementation of this technique. The only tricky part is the
alphabetize
method, which returns a string containing the same characters as its argument, in alphabetical order. This routine (which has nothing to do with the Collections Framework) implements a slick bucket sort. It assumes that the word being alphabetized consists entirely of lowercase alphabetic characters.Running the program on an 80,000 word dictionary takes about 16 seconds on an aging UltraSparc 1. With a minimum permutation group size of eight, it produces the following output:import java.util.*; import java.io.*; public class Perm { public static void main(String[] args) { int minGroupSize = Integer.parseInt(args[1]); // Read words from file and put into simulated multimap Map m = new HashMap(); try { BufferedReader in = new BufferedReader(new FileReader(args[0])); String word; while((word = in.readLine()) != null) { String alpha = alphabetize(word); List l = (List) m.get(alpha); if (l==null) m.put(alpha, l=new ArrayList()); l.add(word); } } catch(IOException e) { System.err.println(e); System.exit(1); } // Print all permutation groups above size threshold for (Iterator i = m.values().iterator(); i.hasNext(); ) { List l = (List) i.next(); 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 the dictionary file we used. It is derived from the Public Domain ENABLE benchmark reference word list, at% java Perm dictionary.txt 8 9: [estrin, inerts, insert, inters, niters, nitres, sinter, triens, trines] 8: [carets, cartes, caster, caters, crates, reacts, recast, traces] 9: [capers, crapes, escarp, pacers, parsec, recaps, scrape, secpar, spacer] 8: [ates, east, eats, etas, sate, seat, seta, teas] 12: [apers, apres, asper, pares, parse, pears, prase, presa, rapes, reaps, spare, spear] 9: [anestri, antsier, nastier, ratines, retains, retinas, retsina, stainer, stearin] 10: [least, setal, slate, stale, steal, stela, taels, tales, teals, tesla] 8: [arles, earls, lares, laser, lears, rales, reals, seral] 8: [lapse, leaps, pales, peals, pleas, salep, sepal, spale] 8: [aspers, parses, passer, prases, repass, spares, sparse, spears] 8: [earings, erasing, gainers, reagins, regains, reginas, searing, seringa] 11: [alerts, alters, artels, estral, laster, ratels, salter, slater, staler, stelar, talers] 9: [palest, palets, pastel, petals, plates, pleats, septal, staple, tepals] 8: [enters, nester, renest, rentes, resent, tenser, ternes, treens] 8: [peris, piers, pries, prise, ripes, speir, spier, spire]http://personal.riverusers.com/~thegrendel/
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