CODEWITHSUNDEEP
javadictionarycollectionsiteration
iMack
iMack

Reputation: 39213

How do I efficiently iterate over each entry in a Java Map?

If I have an object implementing the Map interface in Java and I wish to iterate over every pair contained within it, what is the most efficient way of going through the map?

Will the ordering of elements depend on the specific map implementation that I have for the interface?

Upvotes: 4021

Views: 3508983

Answers (30)

anand krish
anand krish

Reputation: 4415

These are all the possible ways of iterating HashMap.

HashMap<Integer,String> map = new HashMap<Integer,String>();
    map.put(1, "David"); // Adding elements to Map
    map.put(2, "John");
    map.put(4, "Samyuktha");
    map.put(3, "jasmin");
    System.out.println("Iterating Hashmap...");

    // way 1 (java 8 Method)
    map.forEach((key, value) -> {
        System.out.println(key + " : " + value);
    });

    // way 2 (java 7 Method)
    for (Map.Entry<Integer, String> entry : map.entrySet()) {
        System.out.println(entry.getKey() + " : " + entry.getValue());
    }

    // way 3 (java 6 Method)
    for (Integer key : map.keySet()) {
        System.out.println(map.get(key));
    }

    // way 4 (Legacy way to iterate HashMap)
    Iterator iterator = map.entrySet().iterator(); // map.keySet().iterator()
    while (iterator.hasNext())
    {
        Map.Entry<Integer, String> entry = (Map.Entry)iterator.next();
        System.out.println(entry.getKey() + " : " + entry.getValue());
    }   
}

Upvotes: 24

anandchaugule
anandchaugule

Reputation: 1137

An effective iterative solution over a Map is a for loop from Java 5 to Java 7. Here it is:

for (String key : phnMap.keySet()) {
    System.out.println("Key: " + key + " Value: " + phnMap.get(key));
}

From Java 8, you can use a lambda expression to iterate over a Map. It is an enhanced forEach

phnMap.forEach((k,v) -> System.out.println("Key: " + k + " Value: " + v));

If you want to write a conditional for lambda you can write it like this:

phnMap.forEach((k,v) -> {
    System.out.println("Key: " + k + " Value: " + v);
    if ("abc".equals(k)) {
        System.out.println("Hello abc");
    }
});

Upvotes: 17

Georgios Syngouroglou
Georgios Syngouroglou

Reputation: 19984

Method forEach is more concise (like mentioned in the comments), however you can still use Java Stream API with a lambda expression like below,

myMap.entrySet().stream().forEach((entry) -> {
    Object currentKey = entry.getKey();
    Object currentValue = entry.getValue();
});

For more information, follow this.

Upvotes: 38

Paul Tofunmi
Paul Tofunmi

Reputation: 530

Past answers are ok but I feel an explanation of each solution and time complexity will go long one to anyone searching for this.

  1. Using entrySet() and for-each loop:

     Map<String, Integer> myMap = new HashMap<>();
 myMap.put("Alice", 25);
 myMap.put("Bob", 30);
 myMap.put("Charlie", 20);

 // Using entrySet()
 for (Map.Entry<String, Integer> entry : myMap.entrySet()) {
     String key = entry.getKey();
     Integer value = entry.getValue();
     System.out.println(key + ": " + value);
 }

Explanation:

  • myMap.entrySet() returns a set of key-value pairs (Map.Entry objects).
  • The for-each loop iterates directly over these entries. Time Complexity:
  • Best/Average/Worst Case: O(n), where 'n' is the number of entries in the map. This is because it iterates through each entry once.

Overall Complexity:

  • Linear, making it very efficient.
  1. Using entrySet(), Iterator, and while loop:
    Map<String, Integer> myMap = new HashMap<>();
        myMap.put("Alice", 25);
        myMap.put("Bob", 30);
        myMap.put("Charlie", 20);
    // Using Iterator
    Iterator<Map.Entry<String, Integer>> iterator = myMap.entrySet().iterator();
    while (iterator.hasNext()) {
        Map.Entry<String, Integer> entry = iterator.next();
        String key = entry.getKey();
        Integer value = entry.getValue();
        System.out.println(key + ": " + value);
    }

Explanation:

  • myMap.entrySet().iterator() gets an iterator for the entry set.
  • iterator.hasNext() checks if there's another element.
  • iterator.next() gets the next entry and moves the iterator forward.

Time Complexity:

  • Best/Average/Worst Case: O(n), same as the for-each loop.

  • Overall Complexity: Linear. Compared to the earlier code written, this solution gives you the flexibility of using an iterator if you need to remove elements during the iteration safely.

  1. Using Java 8 Lambda function
 Map<String, Integer> myMap = new HashMap<>();
 myMap.put("Alice", 25);
 myMap.put("Bob", 30);
 myMap.put("Charlie", 20);
// Using forEach()
myMap.forEach((key, value) -> System.out.println(key + ": " + value));

Explanation

  • myMap.forEach() takes a lambda expression (a concise function) that's applied to each entry. Here, we simply print the key-value pair. Time Complexity:
  • Best/Average/Worst Case: O(n), the same as the other methods.
  • Overall Complexity: Linear. Provides a more concise syntax if you only need to process elements and don't require an iterator.

Conclusion All of these methods are effectively equally efficient. In most cases, the choice comes down to readability and whether you need the flexibility of an iterator:

  • For-each loop (entrySet): Clean syntax for simple iteration.
  • Iterator: Useful if you might need to remove entries while iterating.
  • forEach(): follows functional-style approach which is concise

Upvotes: 1

fechidal89
fechidal89

Reputation: 723

I copied the data of a map to another with this code:

HashMap product = (HashMap) shopping_truck.get(i);
HashMap tmp = new HashMap();
for (Iterator it = product.entrySet().iterator(); it.hasNext();) {
    Map.Entry thisEntry = (Map.Entry) it.next();
    tmp.put(thisEntry.getKey(), thisEntry.getValue());
}

Upvotes: 2

Syd Lambert
Syd Lambert

Reputation: 1495

If you want to iterate through the map in the order that the elements were added, use LinkedHashMap as opposed to just Map.

This approach has worked for me in the past:

LinkedHashMap<String, Integer> test = new LinkedHashMap();

test.put("foo", 69);
test.put("bar", 1337);

for (int i = 0; i < test.size(); i++) {
    System.out.println(test.get(test.keySet().toArray()[i]));
}

Output:

69
1337

Upvotes: 3

Badri Paudel
Badri Paudel

Reputation: 1630

You can search for the key and with the help of the key you can find the associated value of the map as the map has unique key, see what happens when the key is duplicated here or here.

Demo map :

Map<String, String> map = new HashMap();
map.put("name", "Name");
map.put("age", "23");
map.put("address", "NP");
map.put("faculty", "BE");
map.put("major", "CS");
map.put("head", "MDK");

To get the key only, you can use map.keySet(); like this :

for (String key : map.keySet()) {
    System.out.println(key);
}

To get value only, you can use map.values(); like this:

for (String value : map.values()) {
    System.out.println(value);
}

To get both the key and its value, you still can use map.keySet(); and get its corresponding value, like this:

// this prints the key + value pair
for (String k : map.keySet()) {
    System.out.println(k + " " + map.get(k) + " ");
}

map.get(key) gives the value pointed by that key.

Upvotes: 5

J.B.Vala
J.B.Vala

Reputation: 299

There are several ways to iterate a map. Please refer to the following code.

When you iterate a map using the Iterator Interface you must go with Entry<K,V> or entrySet().

It looks like this:

import java.util.*;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

public class IteratMapDemo {

    public static void main(String arg[]) {
        Map<String, String> mapOne = new HashMap<String, String>();
        mapOne.put("1", "January");
        mapOne.put("2", "February");
        mapOne.put("3", "March");
        mapOne.put("4", "April");
        mapOne.put("5", "May");
        mapOne.put("6", "June");
        mapOne.put("7", "July");
        mapOne.put("8", "August");
        mapOne.put("9", "September");
        mapOne.put("10", "Octomber");
        mapOne.put("11", "November");
        mapOne.put("12", "December");

        Iterator it = mapOne.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry me = (Map.Entry) it.next();
            // System.out.println("Get Key through While loop = " + me.getKey());
        }

        for (Map.Entry<String, String> entry:mapOne.entrySet()) {
            // System.out.println(entry.getKey() + "=" + entry.getValue());
        }

        for (Object key : mapOne.keySet()) {
            System.out.println("Key: " + key.toString() + " Value: " +
                               mapOne.get(key));
        }
    }
}

Upvotes: 5

Sajad NasiriNezhad
Sajad NasiriNezhad

Reputation: 779

//Functional Operations
Map<String, String> mapString = new HashMap<>();
mapString.entrySet().stream().map((entry) -> {
    String mapKey = entry.getKey();
    return entry;
}).forEach((entry) -> {
    String mapValue = entry.getValue();
});

//Intrator
Map<String, String> mapString = new HashMap<>();
for (Iterator<Map.Entry<String, String>> it = mapString.entrySet().iterator(); it.hasNext();) {
    Map.Entry<String, String> entry = it.next();
    String mapKey = entry.getKey();
    String mapValue = entry.getValue();
}

//Simple for loop
Map<String, String> mapString = new HashMap<>();
for (Map.Entry<String, String> entry : mapString.entrySet()) {
    String mapKey = entry.getKey();
    String mapValue = entry.getValue();

}

Upvotes: 14

Fadid
Fadid

Reputation: 1298

Iterator iterator = map.entrySet().iterator();
while (iterator.hasNext()) {
    Map.Entry element = (Map.Entry) it.next();
    LOGGER.debug("Key: " + element.getKey());
    LOGGER.debug("value: " + element.getValue());    
}

Upvotes: 17

Rupesh Yadav
Rupesh Yadav

Reputation: 12302

In Map one can Iteration over keys and/or values and/or both (e.g., entrySet) depends on one's interested in_ Like:

  1. Iterate through the keys -> keySet() of the map:

    Map<String, Object> map = ...;

for (String key : map.keySet()) {
    //your Business logic...
}

  2. Iterate through the values -> values() of the map:

    for (Object value : map.values()) {
    //your Business logic...
}

  3. Iterate through the both -> entrySet() of the map:

    for (Map.Entry<String, Object> entry : map.entrySet()) {
    String key = entry.getKey();
    Object value = entry.getValue();
    //your Business logic...
}

Moreover, there are 3 different ways to iterate through a HashMap. They are as below:

//1.
for (Map.Entry entry : hm.entrySet()) {
    System.out.print("key,val: ");
    System.out.println(entry.getKey() + "," + entry.getValue());
}

//2.
Iterator iter = hm.keySet().iterator();
while(iter.hasNext()) {
    Integer key = (Integer)iter.next();
    String val = (String)hm.get(key);
    System.out.println("key,val: " + key + "," + val);
}

//3.
Iterator it = hm.entrySet().iterator();
while (it.hasNext()) {
    Map.Entry entry = (Map.Entry) it.next();
    Integer key = (Integer)entry.getKey();
    String val = (String)entry.getValue();
    System.out.println("key,val: " + key + "," + val);
}

Upvotes: 32

Suresh Atta
Suresh Atta

Reputation: 122006

Yes, as many people agreed this is the best way to iterate over a Map.

But there are chances to throw NullPointerException if the map is null. Don't forget to put null .check in.

                                                 |
                                                 |
                                         - - - -
                                       |
                                       |
for (Map.Entry<String, Object> entry : map.entrySet()) {
    String key = entry.getKey();
    Object value = entry.getValue();
}

Upvotes: 12

tomaj
tomaj

Reputation: 1590

If your reason for iterating trough the Map, is to do an operation on the value and write to a resulting Map. I recommend using the transform-methods in the Google Guava Maps class.

import com.google.common.collect.Maps;

After you have added the Maps to your imports, you can use Maps.transformValues and Maps.transformEntries on your maps, like this:

public void transformMap() {
    Map<String, Integer> map = new HashMap<>();
    map.put("a", 2);
    map.put("b", 4);

    Map<String, Integer> result = Maps.transformValues(map, num -> num * 2);
    result.forEach((key, val) -> print(key, Integer.toString(val)));
    // key=a,value=4
    // key=b,value=8

    Map<String, String> result2 = Maps.transformEntries(map, (key, value) -> value + "[" + key + "]");
    result2.forEach(this::print);
    // key=a,value=2[a]
    // key=b,value=4[b]
}

private void print(String key, String val) {
    System.out.println("key=" + key + ",value=" + val);
}

Upvotes: 6

user1098063
user1098063

Reputation:

I like to concat a counter, then save the final value of the counter;

int counter = 0;
HashMap<String, String> m = new HashMap<String, String>();
for (int i = 0; i < items.length; i++)
{
    m.put("firstname" + i, items.get(i).getFirstName());
    counter = i;
}

m.put("recordCount", String.valueOf(counter));

Then when you want to retrieve:

int recordCount = Integer.parseInf(m.get("recordCount"));
for (int i = 0; i < recordCount; i++)
{
    System.out.println("First Name :" + m.get("firstname" + i));
}

Upvotes: 6

Ali Akram
Ali Akram

Reputation: 5327

Map.forEach

What about simply using Map::forEach where both the key and the value are passed to your BiConsumer?

map.forEach((k,v) -> {
    System.out.println(k + "->" + v);
});

Upvotes: 7

shivampip
shivampip

Reputation: 2144

Iterating a Map is very easy.

for (Object key : map.keySet()) {
   Object value = map.get(key);
   // Do your stuff
}

For instance, you have a Map<String, int> data;

for (Object key : data.keySet()) {
  int value = data.get(key);
}

Upvotes: 11

Fathah Rehman P
Fathah Rehman P

Reputation: 8761

public class abcd {
    public static void main(String[] args)
    {
        Map<Integer, String> testMap = new HashMap<Integer, String>();
        testMap.put(10, "a");
        testMap.put(20, "b");
        testMap.put(30, "c");
        testMap.put(40, "d");

        for (Integer key : testMap.keySet()) {
            String value = testMap.get(key);
            System.out.println(value);
        }
    }
}

OR

public class abcd {
    public static void main(String[] args)
    {
        Map<Integer, String> testMap = new HashMap<Integer, String>();
        testMap.put(10, "a");
        testMap.put(20, "b");
        testMap.put(30, "c");
        testMap.put(40, "d");

        for (Entry<Integer, String> entry : testMap.entrySet()) {
            Integer key = entry.getKey();
            String value = entry.getValue();
        }
    }
}

Upvotes: 21

abods
abods

Reputation: 301

Try this with Java 1.4:

for ( Iterator entries = myMap.entrySet().iterator(); entries.hasNext();) {

  Entry entry = (Entry) entries.next();

  System.out.println(entry.getKey() + "/" + entry.getValue());

  //...
}

Upvotes: 30

Nitin Mahesh
Nitin Mahesh

Reputation: 3852

Lambda Expression Java 8

In Java 1.8 (Java 8) this has become lot easier by using forEach method from Aggregate operations(Stream operations) that looks similar to iterators from Iterable Interface.

Just copy paste below statement to your code and rename the HashMap variable from hm to your HashMap variable to print out key-value pair.

HashMap<Integer,Integer> hm = new HashMap<Integer, Integer>();
/*
 *     Logic to put the Key,Value pair in your HashMap hm
 */

// Print the key value pair in one line.

hm.forEach((k, v) -> System.out.println("key: " + k + " value:" + v));

// Just copy and paste above line to your code.

Below is the sample code that I tried using Lambda Expression. This stuff is so cool. Must try.

HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
    Random rand = new Random(47);
    int i = 0;
    while(i < 5) {
        i++;
        int key = rand.nextInt(20);
        int value = rand.nextInt(50);
        System.out.println("Inserting key: " + key + " Value: " + value);
        Integer imap = hm.put(key, value);
        if( imap == null) {
            System.out.println("Inserted");
        } else {
            System.out.println("Replaced with " + imap);
        }               
    }

    hm.forEach((k, v) -> System.out.println("key: " + k + " value:" + v));
    
Output:

Inserting key: 18 Value: 5
Inserted
Inserting key: 13 Value: 11
Inserted
Inserting key: 1 Value: 29
Inserted
Inserting key: 8 Value: 0
Inserted
Inserting key: 2 Value: 7
Inserted
key: 1 value:29
key: 18 value:5
key: 2 value:7
key: 8 value:0
key: 13 value:11

Also one can use Spliterator for the same.

Spliterator sit = hm.entrySet().spliterator();

UPDATE

Including documentation links to Oracle Docs. For more on Lambda go to this link and must read Aggregate Operations and for Spliterator go to this link.

Upvotes: 39

Donald Raab
Donald Raab

Reputation: 6706

With Eclipse Collections, you would use the forEachKeyValue method on the MapIterable interface, which is inherited by the MutableMap and ImmutableMap interfaces and their implementations.

MutableMap<Integer, String> map = 
    Maps.mutable.of(1, "One", 2, "Two", 3, "Three");

MutableBag<String> result = Bags.mutable.empty();
map.forEachKeyValue((key, value) -> result.add(key + value));

MutableBag<String> expected = Bags.mutable.of("1One", "2Two", "3Three");
Assertions.assertEquals(expected, result);

The reason using forEachKeyValue with Eclipse Collections (EC) Map implementations will be more efficient than using entrySet is because EC Map implementations do not store Map.Entry objects. Using entrySet with EC Map implementations results in Map.Entry objects being generated dynamically. The forEachKeyValue method is able to avoid creating the Map.Entry objects because it can navigate the internal structure of the Map implementations directly. This is a case where there is a benefit of using an internal iterator over an external iterator.

Note: I am a committer for Eclipse Collections.

Upvotes: 44

ScArcher2
ScArcher2

Reputation: 87237

Map<String, String> map = ...
for (Map.Entry<String, String> entry : map.entrySet()) {
    System.out.println(entry.getKey() + "/" + entry.getValue());
}

On Java 10+:

for (var entry : map.entrySet()) {
    System.out.println(entry.getKey() + "/" + entry.getValue());
}

Upvotes: 5942

Tom Hawtin - tackline
Tom Hawtin - tackline

Reputation: 147154

Typical code for iterating over a map is:

Map<String,Thing> map = ...;
for (Map.Entry<String,Thing> entry : map.entrySet()) {
    String key = entry.getKey();
    Thing thing = entry.getValue();
    ...
}

HashMap is the canonical map implementation and doesn't make guarantees (or though it should not change the order if no mutating operations are performed on it). SortedMap will return entries based on the natural ordering of the keys, or a Comparator, if provided. LinkedHashMap will either return entries in insertion-order or access-order depending upon how it has been constructed. EnumMap returns entries in the natural order of keys.

(Update: I think this is no longer true.) Note, IdentityHashMap entrySet iterator currently has a peculiar implementation which returns the same Map.Entry instance for every item in the entrySet! However, every time a new iterator advances the Map.Entry is updated.

Upvotes: 153

Witold Kaczurba
Witold Kaczurba

Reputation: 10505

The ordering will always depend on the specific map implementation. Using Java 8 you can use either of these:

map.forEach((k,v) -> { System.out.println(k + ":" + v); });

Or:

map.entrySet().forEach((e) -> {
            System.out.println(e.getKey() + " : " + e.getValue());
        });

The result will be the same (same order). The entrySet backed by the map so you are getting the same order. The second one is handy as it allows you to use lambdas, e.g. if you want only to print only Integer objects that are greater than 5:

map.entrySet()
    .stream()
    .filter(e-> e.getValue() > 5)
    .forEach(System.out::println);

The code below shows iteration through LinkedHashMap and normal HashMap (example). You will see difference in the order:

public class HMIteration {


    public static void main(String[] args) {
        Map<Object, Object> linkedHashMap = new LinkedHashMap<>();
        Map<Object, Object> hashMap = new HashMap<>();

        for (int i=10; i>=0; i--) {
            linkedHashMap.put(i, i);
            hashMap.put(i, i);
        }

        System.out.println("LinkedHashMap (1): ");
        linkedHashMap.forEach((k,v) -> { System.out.print(k + " (#="+k.hashCode() + "):" + v + ", "); });

        System.out.println("\nLinkedHashMap (2): ");

        linkedHashMap.entrySet().forEach((e) -> {
            System.out.print(e.getKey() + " : " + e.getValue() + ", ");
        });


        System.out.println("\n\nHashMap (1): ");
        hashMap.forEach((k,v) -> { System.out.print(k + " (#:"+k.hashCode() + "):" + v + ", "); });

        System.out.println("\nHashMap (2): ");

        hashMap.entrySet().forEach((e) -> {
            System.out.print(e.getKey() + " : " + e.getValue() + ", ");
        });
    }
}

Output:

LinkedHashMap (1):
10 (#=10):10, 9 (#=9):9, 8 (#=8):8, 7 (#=7):7, 6 (#=6):6, 5 (#=5):5, 4 (#=4):4, 3 (#=3):3, 2 (#=2):2, 1 (#=1):1, 0 (#=0):0,
LinkedHashMap (2):
10 : 10, 9 : 9, 8 : 8, 7 : 7, 6 : 6, 5 : 5, 4 : 4, 3 : 3, 2 : 2, 1 : 1, 0 : 0,
HashMap (1):
0 (#:0):0, 1 (#:1):1, 2 (#:2):2, 3 (#:3):3, 4 (#:4):4, 5 (#:5):5, 6 (#:6):6, 7 (#:7):7, 8 (#:8):8, 9 (#:9):9, 10 (#:10):10,
HashMap (2):
0 : 0, 1 : 1, 2 : 2, 3 : 3, 4 : 4, 5 : 5, 6 : 6, 7 : 7, 8 : 8, 9 : 9, 10 : 10,

Upvotes: 27

Dubstep
Dubstep

Reputation: 147

Map<String, String> map = 
for (Map.Entry<String, String> entry : map.entrySet()) {
    MapKey = entry.getKey() 
    MapValue = entry.getValue();
}

Upvotes: 3

Younes El Ouarti
Younes El Ouarti

Reputation: 2354

Since Java 10, you can use local variable inference (a.k.a. "var") to make a lot of the already available answers less bloated. For example:

for (var entry : map.entrySet()) {
    System.out.println(entry.getKey() + " : " + entry.getValue());
}

Upvotes: 7

Taras Melnyk
Taras Melnyk

Reputation: 3265

With Java 8, you can iterate Map using forEach and lambda expression,

map.forEach((k, v) -> System.out.println((k + ":" + v)));

Upvotes: 58

Basil Bourque
Basil Bourque

Reputation: 339787

If I have an object implementing the Map interface in Java and I wish to iterate over every pair contained within it, what is the most efficient way of going through the map?

If efficiency of looping the keys is a priority for your app, then choose a Map implementation that maintains the keys in your desired order.

Will the ordering of elements depend on the specific map implementation that I have for the interface?

Yes, absolutely.

  • Some Map implementations promise a certain iteration order, others do not.
  • Different implementations of Map maintain different ordering of the key-value pairs.

See this table I created summarizing the various Map implementations bundled with Java 11. Specifically, notice the iteration order column. Click/tap to zoom.

Table of map implementations in Java 11, comparing their features

You can see there are four Map implementations maintaining an order:

  • TreeMap
  • ConcurrentSkipListMap
  • LinkedHashMap
  • EnumMap

NavigableMap interface

Two of those implement the NavigableMap interface: TreeMap & ConcurrentSkipListMap.

The older SortedMap interface is effectively supplanted by the newer NavigableMap interface. But you may find 3rd-party implementations implementing the older interface only.

Natural order

If you want a Map that keeps its pairs arranged by the “natural order” of the key, use TreeMap or ConcurrentSkipListMap. The term “natural order” means the class of the keys implements Comparable. The value returned by the compareTo method is used for comparison in sorting.

Custom order

If you want to specify a custom sorting routine for your keys to be used in maintaining a sorted order, pass a Comparator implementation appropriate to the class of your keys. Use either TreeMap or ConcurrentSkipListMap, passing your Comparator.

Original insertion order

If you want the pairs of your map to be kept in their original order in which you inserted them into the map, use LinkedHashMap.

Enum-definition order

If you are using an enum such as DayOfWeek or Month as your keys, use the EnumMap class. Not only is this class highly optimized to use very little memory and run very fast, it maintains your pairs in the order defined by the enum. For DayOfWeek, for example, the key of DayOfWeek.MONDAY will be first found when iterated, and the key of DayOfWeek.SUNDAY will be last.

Other considerations

In choosing a Map implementation, also consider:

  • NULLs. Some implementations forbid/accept a NULL as key and/or value.
  • Concurrency. If you are manipulating the map across threads, you must use an implementation that supports concurrency. Or wrap the map with Collections::synchronizedMap (less preferable).

Both of these considerations are covered in the graphic table above.

Upvotes: 24

Slava Vedenin
Slava Vedenin

Reputation: 60174

To summarize the other answers and combine them with what I know, I found 10 main ways to do this (see below). Also, I wrote some performance tests (see results below). For example, if we want to find the sum of all of the keys and values of a map, we can write:

  1. Using iterator and Map.Entry

    long i = 0;
Iterator<Map.Entry<Integer, Integer>> it = map.entrySet().iterator();
while (it.hasNext()) {
    Map.Entry<Integer, Integer> pair = it.next();
    i += pair.getKey() + pair.getValue();
}

  2. Using foreach and Map.Entry

    long i = 0;
for (Map.Entry<Integer, Integer> pair : map.entrySet()) {
    i += pair.getKey() + pair.getValue();
}

  3. Using forEach from Java 8

    final long[] i = {0};
map.forEach((k, v) -> i[0] += k + v);

  4. Using keySet and foreach

    long i = 0;
for (Integer key : map.keySet()) {
    i += key + map.get(key);
}

  5. Using keySet and iterator

    long i = 0;
Iterator<Integer> itr2 = map.keySet().iterator();
while (itr2.hasNext()) {
    Integer key = itr2.next();
    i += key + map.get(key);
}

  6. Using for and Map.Entry

    long i = 0;
for (Iterator<Map.Entry<Integer, Integer>> entries = map.entrySet().iterator(); entries.hasNext(); ) {
    Map.Entry<Integer, Integer> entry = entries.next();
    i += entry.getKey() + entry.getValue();
}

  7. Using the Java 8 Stream API

    final long[] i = {0};
map.entrySet().stream().forEach(e -> i[0] += e.getKey() + e.getValue());

  8. Using the Java 8 Stream API parallel

    final long[] i = {0};
map.entrySet().stream().parallel().forEach(e -> i[0] += e.getKey() + e.getValue());

  9. Using IterableMap of Apache Collections

    long i = 0;
MapIterator<Integer, Integer> it = iterableMap.mapIterator();
while (it.hasNext()) {
    i += it.next() + it.getValue();
}

  10. Using MutableMap of Eclipse (CS) collections

    final long[] i = {0};
mutableMap.forEachKeyValue((key, value) -> {
    i[0] += key + value;
});

Perfomance tests (mode = AverageTime, system = Windows 8.1 64-bit, Intel i7-4790 3.60 GHz, 16 GB)

  1. For a small map (100 elements), score 0.308 is the best

    Benchmark                          Mode  Cnt  Score    Error  Units
test3_UsingForEachAndJava8         avgt  10   0.308 ±  0.021  µs/op
test10_UsingEclipseMap             avgt  10   0.309 ±  0.009  µs/op
test1_UsingWhileAndMapEntry        avgt  10   0.380 ±  0.014  µs/op
test6_UsingForAndIterator          avgt  10   0.387 ±  0.016  µs/op
test2_UsingForEachAndMapEntry      avgt  10   0.391 ±  0.023  µs/op
test7_UsingJava8StreamApi          avgt  10   0.510 ±  0.014  µs/op
test9_UsingApacheIterableMap       avgt  10   0.524 ±  0.008  µs/op
test4_UsingKeySetAndForEach        avgt  10   0.816 ±  0.026  µs/op
test5_UsingKeySetAndIterator       avgt  10   0.863 ±  0.025  µs/op
test8_UsingJava8StreamApiParallel  avgt  10   5.552 ±  0.185  µs/op

  2. For a map with 10000 elements, score 37.606 is the best

    Benchmark                           Mode   Cnt  Score      Error   Units
test10_UsingEclipseMap              avgt   10    37.606 ±   0.790  µs/op
test3_UsingForEachAndJava8          avgt   10    50.368 ±   0.887  µs/op
test6_UsingForAndIterator           avgt   10    50.332 ±   0.507  µs/op
test2_UsingForEachAndMapEntry       avgt   10    51.406 ±   1.032  µs/op
test1_UsingWhileAndMapEntry         avgt   10    52.538 ±   2.431  µs/op
test7_UsingJava8StreamApi           avgt   10    54.464 ±   0.712  µs/op
test4_UsingKeySetAndForEach         avgt   10    79.016 ±  25.345  µs/op
test5_UsingKeySetAndIterator        avgt   10    91.105 ±  10.220  µs/op
test8_UsingJava8StreamApiParallel   avgt   10   112.511 ±   0.365  µs/op
test9_UsingApacheIterableMap        avgt   10   125.714 ±   1.935  µs/op

  3. For a map with 100000 elements, score 1184.767 is the best

    Benchmark                          Mode   Cnt  Score        Error    Units
test1_UsingWhileAndMapEntry        avgt   10   1184.767 ±   332.968  µs/op
test10_UsingEclipseMap             avgt   10   1191.735 ±   304.273  µs/op
test2_UsingForEachAndMapEntry      avgt   10   1205.815 ±   366.043  µs/op
test6_UsingForAndIterator          avgt   10   1206.873 ±   367.272  µs/op
test8_UsingJava8StreamApiParallel  avgt   10   1485.895 ±   233.143  µs/op
test5_UsingKeySetAndIterator       avgt   10   1540.281 ±   357.497  µs/op
test4_UsingKeySetAndForEach        avgt   10   1593.342 ±   294.417  µs/op
test3_UsingForEachAndJava8         avgt   10   1666.296 ±   126.443  µs/op
test7_UsingJava8StreamApi          avgt   10   1706.676 ±   436.867  µs/op
test9_UsingApacheIterableMap       avgt   10   3289.866 ±  1445.564  µs/op

Graphs (performance tests depending on map size)

Enter image description here

Table (perfomance tests depending on map size)

          100     600      1100     1600     2100
test10    0.333    1.631    2.752    5.937    8.024
test3     0.309    1.971    4.147    8.147   10.473
test6     0.372    2.190    4.470    8.322   10.531
test1     0.405    2.237    4.616    8.645   10.707
test2     0.376    2.267    4.809    8.403   10.910
test7     0.473    2.448    5.668    9.790   12.125
test9     0.565    2.830    5.952   13.220   16.965
test4     0.808    5.012    8.813   13.939   17.407
test5     0.810    5.104    8.533   14.064   17.422
test8     5.173   12.499   17.351   24.671   30.403

All tests are on GitHub.

Upvotes: 1622

Lova Chittumuri
Lova Chittumuri

Reputation: 3323

Using Java 7

Map<String,String> sampleMap = new HashMap<>();
for (sampleMap.Entry<String,String> entry : sampleMap.entrySet()) {
    String key = entry.getKey();
    String value = entry.getValue();

    /* your Code as per the Business Justification  */

}

Using Java 8

Map<String,String> sampleMap = new HashMap<>();

sampleMap.forEach((k, v) -> System.out.println("Key is :  " + k + " Value is :  " + v));

Upvotes: 6

akhil_mittal
akhil_mittal

Reputation: 24167

Java 8

We have got forEach method that accepts a lambda expression. We have also got stream APIs. Consider a map:

Map<String,String> sample = new HashMap<>();
sample.put("A","Apple");
sample.put("B", "Ball");

Iterate over keys:

sample.keySet().forEach((k) -> System.out.println(k));

Iterate over values:

sample.values().forEach((v) -> System.out.println(v));

Iterate over entries (Using forEach and Streams):

sample.forEach((k,v) -> System.out.println(k + ":" + v)); 
sample.entrySet().stream().forEach((entry) -> {
            Object currentKey = entry.getKey();
            Object currentValue = entry.getValue();
            System.out.println(currentKey + ":" + currentValue);
        });

The advantage with streams is they can be parallelized easily in case we want to. We simply need to use parallelStream() in place of stream() above.

forEachOrdered vs forEach with streams ? The forEach does not follow encounter order (if defined) and is inherently non-deterministic in nature where as the forEachOrdered does. So forEach does not guarantee that the order would be kept. Also check this for more.

Upvotes: 43

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