# Maps
The java.util.Map interface (opens new window) represents a mapping between keys and their values. A map cannot contain duplicate keys; and each key can map to at most one value.
Since Map
is an interface, then you need to instantiate a concrete implementation of that interface in order to use it; there are several Map
implementations, and mostly used are the java.util.HashMap
and java.util.TreeMap
# Iterating Map Entries Efficiently
This section provides code and benchmarks for ten unique example implementations which iterate over the entries of a Map<Integer, Integer>
and generate the sum of the Integer
values. All of the examples have an algorithmic complexity of Θ(n)
, however, the benchmarks are still useful for providing insight on which implementations are more efficient in a "real world" environment.
- Implementation using
Iterator
(opens new window) withMap.Entry
(opens new window)
Iterator<Map.Entry<Integer, Integer>> it = map.entrySet().iterator();
while (it.hasNext()) {
Map.Entry<Integer, Integer> pair = it.next();
sum += pair.getKey() + pair.getValue();
}
- Implementation using
for
(opens new window) withMap.Entry
(opens new window)
for (Map.Entry<Integer, Integer> pair : map.entrySet()) {
sum += pair.getKey() + pair.getValue();
}
- Implementation using
Map.forEach
(opens new window) (Java 8+)
map.forEach((k, v) -> sum[0] += k + v);
- Implementation using
Map.keySet
(opens new window) withfor
(opens new window)
for (Integer key : map.keySet()) {
sum += key + map.get(key);
}
- Implementation using
Map.keySet
(opens new window) withIterator
(opens new window)
Iterator<Integer> it = map.keySet().iterator();
while (it.hasNext()) {
Integer key = it.next();
sum += key + map.get(key);
}
- Implementation using
for
(opens new window) withIterator
(opens new window) andMap.Entry
(opens new window)
for (Iterator<Map.Entry<Integer, Integer>> entries =
map.entrySet().iterator(); entries.hasNext(); ) {
Map.Entry<Integer, Integer> entry = entries.next();
sum += entry.getKey() + entry.getValue();
}
- Implementation using
Stream.forEach
(opens new window) (Java 8+)
map.entrySet().stream().forEach(e -> sum += e.getKey() + e.getValue());
- Implementation using
Stream.forEach
(opens new window) withStream.parallel
(opens new window) (Java 8+)
map.entrySet()
.stream()
.parallel()
.forEach(e -> sum += e.getKey() + e.getValue());
- Implementation using
IterableMap
(opens new window) from Apache Collections (opens new window)
MapIterator<Integer, Integer> mit = iterableMap.mapIterator();
while (mit.hasNext()) {
sum += mit.next() + it.getValue();
}
- Implementation using
MutableMap
(opens new window) from Eclipse Collections (opens new window)
mutableMap.forEachKeyValue((key, value) -> {
sum += key + value;
});
Performance Tests (Code available on Github (opens new window))
Test Environment: Windows 8.1 64-bit, Intel i7-4790 3.60GHz, 16 GB
Benchmark Score Error Units
test3_UsingForEachAndJava8 308 ± 21 ns/op
test10_UsingEclipseMutableMap 309 ± 9 ns/op
test1_UsingWhileAndMapEntry 380 ± 14 ns/op
test6_UsingForAndIterator 387 ± 16 ns/op
test2_UsingForEachAndMapEntry 391 ± 23 ns/op
test7_UsingJava8StreamAPI 510 ± 14 ns/op
test9_UsingApacheIterableMap 524 ± 8 ns/op
test4_UsingKeySetAndForEach 816 ± 26 ns/op
test5_UsingKeySetAndIterator 863 ± 25 ns/op
test8_UsingJava8StreamAPIParallel 5552 ± 185 ns/op
Benchmark Score Error Units
test10_UsingEclipseMutableMap 37606 ± 790 ns/op
test3_UsingForEachAndJava8 50368 ± 887 ns/op
test6_UsingForAndIterator 50332 ± 507 ns/op
test2_UsingForEachAndMapEntry 51406 ± 1032 ns/op
test1_UsingWhileAndMapEntry 52538 ± 2431 ns/op
test7_UsingJava8StreamAPI 54464 ± 712 ns/op
test4_UsingKeySetAndForEach 79016 ± 25345 ns/op
test5_UsingKeySetAndIterator 91105 ± 10220 ns/op
test8_UsingJava8StreamAPIParallel 112511 ± 365 ns/op
test9_UsingApacheIterableMap 125714 ± 1935 ns/op
Benchmark Score Error Units
test1_UsingWhileAndMapEntry 1184.767 ± 332.968 μs/op
test10_UsingEclipseMutableMap 1191.735 ± 304.273 μs/op
test2_UsingForEachAndMapEntry 1205.815 ± 366.043 μs/op
test6_UsingForAndIterator 1206.873 ± 367.272 μs/op
test8_UsingJava8StreamAPIParallel 1485.895 ± 233.143 μs/op
test5_UsingKeySetAndIterator 1540.281 ± 357.497 μs/op
test4_UsingKeySetAndForEach 1593.342 ± 294.417 μs/op
test3_UsingForEachAndJava8 1666.296 ± 126.443 μs/op
test7_UsingJava8StreamAPI 1706.676 ± 436.867 μs/op
test9_UsingApacheIterableMap 3289.866 ± 1445.564 μs/op
x: Size of Map
f(x): Benchmark Score (μs/op)
100 600 1100 1600 2100
---------------------------------------------------
10 | 0.333 1.631 2.752 5.937 8.024
3 | 0.309 1.971 4.147 8.147 10.473
6 | 0.372 2.190 4.470 8.322 10.531
1 | 0.405 2.237 4.616 8.645 10.707
Tests 2 | 0.376 2.267 4.809 8.403 10.910
f(x) 7 | 0.473 2.448 5.668 9.790 12.125
9 | 0.565 2.830 5.952 13.22 16.965
4 | 0.808 5.012 8.813 13.939 17.407
5 | 0.81 5.104 8.533 14.064 17.422
8 | 5.173 12.499 17.351 24.671 30.403
# Using Default Methods of Map from Java 8
Examples of using Default Methods introduced in Java 8 in Map interface
- Using getOrDefault
Returns the value mapped to the key, or if the key is not present, returns the default value
Map<Integer, String> map = new HashMap<>();
map.put(1, "First element");
map.get(1); // => First element
map.get(2); // => null
map.getOrDefault(2, "Default element"); // => Default element
- Using forEach
Allows to perform the operation specified in the 'action' on each Map Entry
Map<Integer, String> map = new HashMap<Integer, String>();
map.put(1, "one");
map.put(2, "two");
map.put(3, "three");
map.forEach((key, value) -> System.out.println("Key: "+key+ " :: Value: "+value));
// Key: 1 :: Value: one
// Key: 2 :: Value: two
// Key: 3 :: Value: three
- Using replaceAll
Will replace with new-value only if key is present
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.replaceAll((key,value)->value+10); //{john=30, paul=40, peter=50}
- Using putIfAbsent
Key-Value pair is added to the map, if the key is not present or mapped to null
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.putIfAbsent("kelly", 50); //{john=20, paul=30, peter=40, kelly=50}
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.remove("peter",40); //{john=30, paul=40}
If the key is present then the value is replaced by new-value. If the key is not present, does nothing.
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.replace("peter",50); //{john=20, paul=30, peter=50}
map.replace("jack",60); //{john=20, paul=30, peter=50}
- Using computeIfAbsent
This method adds an entry in the Map. the key is specified in the function and the value is the result of the application of the mapping function
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.computeIfAbsent("kelly", k->map.get("john")+10); //{john=20, paul=30, peter=40, kelly=30}
map.computeIfAbsent("peter", k->map.get("john")+10); //{john=20, paul=30, peter=40, kelly=30} //peter already present
- Using computeIfPresent
This method adds an entry or modifies an existing entry in the Map. Does nothing if an entry with that key is not present
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.computeIfPresent("kelly", (k,v)->v+10); //{john=20, paul=30, peter=40} //kelly not present
map.computeIfPresent("peter", (k,v)->v+10); //{john=20, paul=30, peter=50} // peter present, so increase the value
- Using compute
This method replaces the value of a key by the newly computed value
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
map.compute("peter", (k,v)->v+50); //{john=20, paul=30, peter=90} //Increase the value
- Using merge
Adds the key-value pair to the map, if key is not present or value for the key is null Replaces the value with the newly computed value, if the key is present Key is removed from the map , if new value computed is null
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("john", 20);
map.put("paul", 30);
map.put("peter", 40);
//Adds the key-value pair to the map, if key is not present or value for the key is null
map.merge("kelly", 50 , (k,v)->map.get("john")+10); // {john=20, paul=30, peter=40, kelly=50}
//Replaces the value with the newly computed value, if the key is present
map.merge("peter", 50 , (k,v)->map.get("john")+10); //{john=20, paul=30, peter=30, kelly=50}
//Key is removed from the map , if new value computed is null
map.merge("peter", 30 , (k,v)->map.get("nancy")); //{john=20, paul=30, kelly=50}
# Usage of HashMap
HashMap is an implementation of the Map interface that provides a Data Structure to store data in Key-Value pairs.
1. Declaring HashMap
Map<KeyType, ValueType> myMap = new HashMap<KeyType, ValueType>();
KeyType and ValueType must be valid types in Java, such as - String, Integer, Float or any custom class like Employee, Student etc..
For Example : Map<String,Integer> myMap = new HashMap<String,Integer>();
2. Putting values in HashMap.
To put a value in the HashMap, we have to call put
method on the HashMap object by passing the Key and the Value as parameters.
myMap.put("key1", 1);
myMap.put("key2", 2);
If you call the put method with the Key that already exists in the Map, the method will override its value and return the old value.
3. Getting values from HashMap.
For getting the value from a HashMap you have to call the get
method, by passing the Key as a parameter.
myMap.get("key1"); //return 1 (class Integer)
If you pass a key that does not exists in the HashMap, this method will return null
4. Check whether the Key is in the Map or not.
myMap.containsKey(varKey);
5. Check whether the Value is in the Map or not.
myMap.containsValue(varValue);
The above methods will return a boolean
value true or false if key, value exists in the Map or not.
# Iterating through the contents of a Map
Maps provide methods which let you access the keys, values, or key-value pairs of the map as collections. You can iterate through these collections. Given the following map for example:
Map<String, Integer> repMap = new HashMap<>();
repMap.put("Jon Skeet", 927_654);
repMap.put("BalusC", 708_826);
repMap.put("Darin Dimitrov", 715_567);
Iterating through map keys:
for (String key : repMap.keySet()) {
System.out.println(key);
}
Prints:
Darin Dimitrov
Jon Skeet
BalusC
keySet()
provides the keys of the map as a Set
(opens new window). Set
is used as the keys cannot contain duplicate values. Iterating through the set yields each key in turn. HashMaps are not ordered, so in this example the keys may be returned in any order.
Iterating through map values:
for (Integer value : repMap.values()) {
System.out.println(value);
}
Prints:
715567
927654
708826
values()
returns the values of the map as a Collection
(opens new window). Iterating through the collection yields each value in turn. Again, the values may be returned in any order.
Iterating through keys and values together
for (Map.Entry<String, Integer> entry : repMap.entrySet()) {
System.out.printf("%s = %d\n", entry.getKey(), entry.getValue());
}
Prints:
Darin Dimitrov = 715567
Jon Skeet = 927654
BalusC = 708826
entrySet()
returns a collection of Map.Entry
(opens new window) objects. Map.Entry gives access to the key and value for each entry.
# Add multiple items
We can use V put(K key,V value)
:
Associates the specified value with the specified key in this map (optional operation). If the map previously contained a mapping for the key, the old value is replaced by the specified value.
String currentVal;
Map<Integer, String> map = new TreeMap<>();
currentVal = map.put(1, "First element.");
System.out.println(currentVal);// Will print null
currentVal = map.put(2, "Second element.");
System.out.println(currentVal); // Will print null yet again
currentVal = map.put(2, "This will replace 'Second element'");
System.out.println(currentVal); // will print Second element.
System.out.println(map.size()); // Will print 2 as key having
// value 2 was replaced.
Map<Integer, String> map2 = new HashMap<>();
map2.put(2, "Element 2");
map2.put(3, "Element 3");
map.putAll(map2);
System.out.println(map.size());
Output:
3
To add many items you can use an inner classes like this:
Map<Integer, String> map = new HashMap<>() {{
// This is now an anonymous inner class with an unnamed instance constructor
put(5, "high");
put(4, "low");
put(1, "too slow");
}};
Keep in mind that creating an anonymous inner class is not always efficient and can lead to memory leaks so when possible, use an initializer block instead:
static Map<Integer, String> map = new HashMap<>();
static {
// Now no inner classes are created so we can avoid memory leaks
put(5, "high");
put(4, "low");
put(1, "too slow");
}
The example above makes the map static. It can also be used in a non-static context by removing all occurences of static
.
In addition to that most implementations support putAll
, which can add all entries in one map to another like this:
another.putAll(one);
# Merging, combine and composing Maps
Use putAll
to put every member of one map into another. Keys already present in the map will have their corresponding values overwritten.
Map<String, Integer> numbers = new HashMap<>();
numbers.put("One", 1)
numbers.put("Three", 3)
Map<String, Integer> other_numbers = new HashMap<>();
other_numbers.put("Two", 2)
other_numbers.put("Three", 4)
numbers.putAll(other_numbers)
This yields the following mapping in numbers
:
"One" -> 1
"Two" -> 2
"Three" -> 4 //old value 3 was overwritten by new value 4
If you want to combine values instead of overwriting them, you can use Map.merge
(opens new window), added in Java 8, which uses a user-provided BiFunction
to merge values for duplicate keys. merge
operates on individual keys and values, so you'll need to use a loop or Map.forEach
. Here we concatenate strings for duplicate keys:
for (Map.Entry<String, Integer> e : other_numbers.entrySet())
numbers.merge(e.getKey(), e.getValue(), Integer::sum);
//or instead of the above loop
other_numbers.forEach((k, v) -> numbers.merge(k, v, Integer::sum));
If you want to enforce the constraint there are no duplicate keys, you can use a merge function that throws an AssertionError
:
mapA.forEach((k, v) ->
mapB.merge(k, v, (v1, v2) ->
{throw new AssertionError("duplicate values for key: "+k);}));
# Composing Map<X,Y> and Map<Y,Z> to get Map<X,Z>
If you want to compose two mappings, you can do it as follows
Map<String, Integer> map1 = new HashMap<String, Integer>();
map1.put("key1", 1);
map1.put("key2", 2);
map1.put("key3", 3);
Map<Integer, Double> map2 = new HashMap<Integer, Double>();
map2.put(1, 1.0);
map2.put(2, 2.0);
map2.put(3, 3.0);
Map<String, Double> map3 = new new HashMap<String, Double>();
map1.forEach((key,value)->map3.put(key,map2.get(value)));
This yields the following mapping
"key1" -> 1.0
"key2" -> 2.0
"key3" -> 3.0
# Add an element
- Addition
Map<Integer, String> map = new HashMap<>();
map.put(1, "First element.");
System.out.println(map.get(1));
Output: First element.
- Override
Map<Integer, String> map = new HashMap<>();
map.put(1, "First element.");
map.put(1, "New element.");
System.out.println(map.get(1));
Output: New element.
HashMap
is used as an example. Other implementations that implement the Map
interface may be used as well.
# Clear the map
Map<Integer, String> map = new HashMap<>();
map.put(1, "First element.");
map.put(2, "Second element.");
map.put(3, "Third element.");
map.clear();
System.out.println(map.size()); // => 0
# Check if key exists
Map<String, String> num = new HashMap<>();
num.put("one", "first");
if (num.containsKey("one")) {
System.out.println(num.get("one")); // => first
}
# Maps can contain null values
For maps, one has to be carrefull not to confuse "containing a key" with "having a value". For example, HashMap
s can contain null which means the following is perfectly normal behavior :
Map<String, String> map = new HashMap<>();
map.put("one", null);
if (map.containsKey("one")) {
System.out.println("This prints !"); // This line is reached
}
if (map.get("one") != null) {
System.out.println("This is never reached !"); // This line is never reached
}
More formally, there is no guarantee that map.contains(key) <=> map.get(key)!=null
# Use custom object as key
Before using your own object as key you must override hashCode() and equals() method of your object.
In simple case you would have something like:
class MyKey {
private String name;
MyKey(String name) {
this.name = name;
}
@Override
public boolean equals(Object obj) {
if(obj instanceof MyKey) {
return this.name.equals(((MyKey)obj).name);
}
return false;
}
@Override
public int hashCode() {
return this.name.hashCode();
}
}
hashCode
will decide which hash bucket the key belongs to and equals
will decide which object inside that hash bucket.
Without these method, the reference of your object will be used for above comparison which will not work unless you use the same object reference everytime.
# Creating and Initializing Maps
# Introduction
Maps
stores key/value pairs, where each key has an associated value. Given a particular key, the map can look up the associated value very quickly.
Maps
, also known as associate array, is an object that stores the data in form of keys and values. In Java, maps are represented using Map interface which is not an extension of the collection interface.
/*J2SE < 5.0*/
Map map = new HashMap();
map.put("name", "A");
map.put("address", "Malviya-Nagar");
map.put("city", "Jaipur");
System.out.println(map);
/*J2SE 5.0+ style (use of generics):*/
Map<String, Object> map = new HashMap<>();
map.put("name", "A");
map.put("address", "Malviya-Nagar");
map.put("city", "Jaipur");
System.out.println(map);
Map<String, Object> map = new HashMap<String, Object>(){{
put("name", "A");
put("address", "Malviya-Nagar");
put("city", "Jaipur");
}};
System.out.println(map);
Map<String, Object> map = new TreeMap<String, Object>();
map.put("name", "A");
map.put("address", "Malviya-Nagar");
map.put("city", "Jaipur");
System.out.println(map);
//Java 8
final Map<String, String> map =
Arrays.stream(new String[][] {
{ "name", "A" },
{ "address", "Malviya-Nagar" },
{ "city", "jaipur" },
}).collect(Collectors.toMap(m -> m[0], m -> m[1]));
System.out.println(map);
//This way for initial a map in outside the function
final static Map<String, String> map;
static
{
map = new HashMap<String, String>();
map.put("a", "b");
map.put("c", "d");
}
//Immutable single key-value map
Map<String, String> singletonMap = Collections.singletonMap("key", "value");
Please note, that it is impossible to modify such map. Any attemts to modify the map will result in throwing the UnsupportedOperationException.
//Immutable single key-value pair
Map<String, String> singletonMap = Collections.singletonMap("key", "value");
singletonMap.put("newKey", "newValue"); //will throw UnsupportedOperationException
singletonMap.putAll(new HashMap<>()); //will throw UnsupportedOperationException
singletonMap.remove("key"); //will throw UnsupportedOperationException
singletonMap.replace("key", "value", "newValue"); //will throw UnsupportedOperationException
//and etc
# Remarks
A map (opens new window) is an object which store keys with an associated value for each key. A key and its value are sometimes called a key/value pair or an entry. Maps typically provide these features:
- Data is stored into the map in key/value pairs.
- The map may contain only one entry for a particular key. If a map contains an entry with a particular key, and you try to store a second entry with the same key, then the second entry will replace the first. In other words, this will change the value associated with the key.
- Maps provide fast operations to test whether a key exists in the map, to fetch the value associated with a key, and to remove a key/value pair.
The most commonly used map implementation is HashMap (opens new window). It works well with keys that are strings or numbers.
Plain maps such as HashMap are unordered. Iterating through key/value pairs may return individual entries in any order. If you need to iterate through map entries in a controlled fashion, you should look at the following: