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

Asked 2023-09-20 20:13:42 View 131,527

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?

Answers

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());
}

Answered   2023-09-20 20:13:42

  • If you do that, then it won't work as Entry is a nested Class in Map. java.sun.com/javase/6/docs/api/java/util/Map.html - anyone
  • you can write the import as "import java.util.Map.Entry;" and it will work. - anyone
  • @Pureferret The only reason you might want to use an iterator is if you need to call its remove method. If that is the case, this other answer shows you how to do it. Otherwise, the enhanced loop as shown in the answer above is the way to go. - anyone
  • I believe the form Map.Entry is clearer than importing the inner class into the current namespace. - anyone
  • Note that you can use map.values() or map.keySet() if you want to loop through values or keys only. - anyone

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.

Answered   2023-09-20 20:13:42

  • @Viacheslav : very nice answer. Just wondering how Java8 apis are hindered, in your benchmark, by capturing lambdas... (e.g. long sum = 0; map.forEach( /* accumulate in variable sum*/); captures the sum long, which may be slower than say stream.mapToInt(/*whatever*/).sum for example. Of course you can not always avoid capturing state, but that may be a reasonnable addition to the bench. - anyone
  • @ZhekaKozlov: look at the mindblowingly large error values. Consider that a test result of x±e implies that there were result within the interval from x-e to x+e, so the fastest result (1184.767±332.968) ranges from 852 to 1518, whereas the second slowest (1706.676±436.867) runs between 1270 and 2144, so the results still overlap significantly. Now look at the slowest result, 3289.866±1445.564, which implies diverging between 1844 and 4735 and you know that these test results are meaningless. - anyone
  • What about comparing the 3 main implementations : HashMap, LinkedHashMap and TreeMap ? - anyone
  • #1 and #6 are exactly the same. Using while vs. a for loop is not a different technique for iterating. And I am surprised they have such variation between them in your tests—which suggests that the tests are not properly isolated from external factors unrelated to the things you intend to be testing. - anyone
  • #8 is a terrible example, because of the parallel there's now a race condition when adding to i. - anyone

In Java 8 you can do it clean and fast using the new lambdas features:

 Map<String,String> map = new HashMap<>();
 map.put("SomeKey", "SomeValue");
 map.forEach( (k,v) -> [do something with key and value] );

 // such as
 map.forEach( (k,v) -> System.out.println("Key: " + k + ": Value: " + v));

The type of k and v will be inferred by the compiler and there is no need to use Map.Entry anymore.

Easy-peasy!

Answered   2023-09-20 20:13:42

  • Depending on what you want to do with a map, you can also use stream API on the entries returned by map.entrySet().stream() docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html - anyone
  • This won't work if you want to reference non-final variables declared outside your lambda expression from within the forEach()... - anyone
  • @Chris Correct. It won't work if you try to use effectively non-final variables from outside the lambda. - anyone

Yes, the order depends on the specific Map implementation.

@ScArcher2 has the more elegant Java 1.5 syntax. In 1.4, I would do something like this:

Iterator entries = myMap.entrySet().iterator();
while (entries.hasNext()) {
  Entry thisEntry = (Entry) entries.next();
  Object key = thisEntry.getKey();
  Object value = thisEntry.getValue();
  // ...
}

Answered   2023-09-20 20:13:42

  • Prefer for-loop than while.. for(Iterator entries = myMap.entrySet().iterator(); entries.hasNext(); ) {...} With this syntax the 'entries' scope is reduced to the for loop only. - anyone
  • @jpredham You are right that using the for construct as for (Entry e : myMap.entrySet) will not allow you to modify the collection, but the example as @HanuAthena mentioned it should work, since it gives you the Iterator in scope. (Unless I'm missing something...) - anyone
  • IntelliJ is giving me errors on Entry thisEntry = (Entry) entries.next();: doesn't recognize Entry. Is that pseudocode for something else? - anyone
  • @JohnK try importing java.util.Map.Entry. - anyone
  • This solution will not work if you have a integer key and String key. - anyone

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.

Answered   2023-09-20 20:13:42

  • EnumMap also has this peculiar behaviour along with IdentityHashMap - anyone
  • "LinkedHashMap will either return entries in [...] access-order [...]" ... so you access the elements in the order you access them? Either tautological, or something interesting which could use a digression. ;-) - anyone
  • @jpaugh Only direct accesses to the LinkedHashMap count. Those through iterator, spliterator, entrySet, etc., do not modify the order. - anyone
  • 1. thoughif? 2. The last paragraph might benefit from a brush-up. - anyone

Example of using iterator and generics:

Iterator<Map.Entry<String, String>> entries = myMap.entrySet().iterator();
while (entries.hasNext()) {
  Map.Entry<String, String> entry = entries.next();
  String key = entry.getKey();
  String value = entry.getValue();
  // ...
}

Answered   2023-09-20 20:13:42

  • You should put Iterator in a for loop to limit its scope. - anyone
  • @SteveKuo What do you mean by "limit its scope" ? - anyone
  • @StudioWorks for (Iterator<Map.Entry<K, V>> entries = myMap.entrySet().iterator(); entries.hasNext(); ) { Map.Entry<K, V> entry = entries.next(); }. By using that construct we limit the scope of (visibility of the variable) entries to the for loop. - anyone

This is a two part question:

How to iterate over the entries of a Map - @ScArcher2 has answered that perfectly.

What is the order of iteration - if you are just using Map, then strictly speaking, there are no ordering guarantees. So you shouldn't really rely on the ordering given by any implementation. However, the SortedMap interface extends Map and provides exactly what you are looking for - implementations will aways give a consistent sort order.

NavigableMap is another useful extension - this is a SortedMap with additional methods for finding entries by their ordered position in the key set. So potentially this can remove the need for iterating in the first place - you might be able to find the specific entry you are after using the higherEntry, lowerEntry, ceilingEntry, or floorEntry methods. The descendingMap method even gives you an explicit method of reversing the traversal order.

Answered   2023-09-20 20:13:42

FYI, you can also use map.keySet() and map.values() if you're only interested in keys/values of the map and not the other.

Answered   2023-09-20 20:13:42

The correct way to do this is to use the accepted answer as it is the most efficient. I find the following code looks a bit cleaner.

for (String key: map.keySet()) {
   System.out.println(key + "/" + map.get(key));
}

Answered   2023-09-20 20:13:42

  • This is not the best approach, it's much more efficient to use the entrySet(). Findbugs will flag this code (see findbugs.sourceforge.net/…) - anyone
  • @JeffOlson meh, not really. map lookup is O(1) so both loops behave the same way. admittedly, it will be slightly slower in a micro benchmark but i sometimes do this as well because i hate writing the type arguments over and over again. Also this will quite likely never be your performance bottleneck, so go for it if it makes the code more readable. - anyone
  • more in detail: O(1) = 2*O(1) is pretty much the definition of the big O notation. you're right in that it runs a bit slower, but in terms of complexity they're the same. - anyone
  • @Jeff Olson: the comments that the “Big O” complexity doesn’t change, when there is only a constant factor, is correct. Still, to me it matters whether an operation takes one hour or two hours. More important, it must be emphasized that the factor is not 2, as iterating over an entrySet() does not bear a lookup at all; it’s just a linear traversal of all entries. In contrast, iterating over the keySet() and performing a lookup per key bears one lookup per key, so we’re talking about zero lookups vs. n lookups here, n being the size of the Map. So the factor is way beyond 2… - anyone
  • @kornero: it might be worth noting that you don’t need the keys to have the same hashcode to have a collision; there’s already a collision when hashcode % capacity is the same. Starting with Java 8, the complexity for items having the same hashcode % capacity, but different hashcode or are Comparable falls back to O(log n) and only keys having the same hash code and not being Comparable impose O(n) complexity. But the statement that the complexity of a lookup can be more than O(1) in practice still holds. - anyone

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

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

Answered   2023-09-20 20:13:42

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.

Answered   2023-09-20 20:13:42

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.

Answered   2023-09-20 20:13:42

In theory, the most efficient way will depend on which implementation of Map. The official way to do this is to call map.entrySet(), which returns a set of Map.Entry, each of which contains a key and a value (entry.getKey() and entry.getValue()).

In an idiosyncratic implementation, it might make some difference whether you use map.keySet(), map.entrySet() or something else. But I can't think of a reason why anyone would write it like that. Most likely it makes no difference to performance what you do.

And yes, the order will depend on the implementation - as well as (possibly) the order of insertion and other hard-to-control factors.

[edit] I wrote valueSet() originally but of course entrySet() is actually the answer.

Answered   2023-09-20 20:13:42

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.

Answered   2023-09-20 20:13:42

Java 8:

You can use lambda expressions:

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

For more information, follow this.

Answered   2023-09-20 20:13:42

  • @injecteer: Seems the motive of lambda expressions - anyone
  • You don’t need a stream if you just want to iterate over a map. myMap.forEach( (currentKey,currentValue) -> /* action */ ); is much more concise. - anyone

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);
}

Answered   2023-09-20 20:13:42

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());

  //...
}

Answered   2023-09-20 20:13:42

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,

Answered   2023-09-20 20:13:42

Most compact with Java 8:

map.entrySet().forEach(System.out::println);

Answered   2023-09-20 20:13:42

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 me : map.entrySet()) {
        System.out.println(me.getKey() + " : " + me.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 me = (Map.Entry)iterator.next();
        System.out.println(me.getKey() + " : " + me.getValue());
    }   
}

Answered   2023-09-20 20:13:42

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.

Answered   2023-09-20 20:13:42

  • Late comment on an answer that's also late to the party (but very informative). +1 from me for mentioning EnumMap, since it's the first time I've heard of it. There's probably many cases where this might come in handy. - anyone

If you have a generic untyped Map you can use:

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

Answered   2023-09-20 20:13:42

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();
        }
    }
}

Answered   2023-09-20 20:13:42

    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());    
    }

Answered   2023-09-20 20:13:42

An effective iterative solution over a Map is a for loop from Java 5 through 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");
    }
});

Answered   2023-09-20 20:13:42

You can do it using generics:

Map<Integer, Integer> map = new HashMap<Integer, Integer>();
Iterator<Map.Entry<Integer, Integer>> entries = map.entrySet().iterator();
while (entries.hasNext()) {
    Map.Entry<Integer, Integer> entry = entries.next();
    System.out.println("Key = " + entry.getKey() + ", Value = " + entry.getValue());
}

Answered   2023-09-20 20:13:42

Use Java 8:

map.entrySet().forEach(entry -> System.out.println(entry.getValue()));

Answered   2023-09-20 20:13:42

           //Functional Oprations
            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();

            }

Answered   2023-09-20 20:13:42

There are a lot of ways to do this. Below is a few simple steps:

Suppose you have one Map like:

Map<String, Integer> m = new HashMap<String, Integer>();

Then you can do something like the below to iterate over map elements.

// ********** Using an iterator ****************
Iterator<Entry<String, Integer>> me = m.entrySet().iterator();
while(me.hasNext()){
    Entry<String, Integer> pair = me.next();
    System.out.println(pair.getKey() + ":" + pair.getValue());
}

// *********** Using foreach ************************
for(Entry<String, Integer> me : m.entrySet()){
    System.out.println(me.getKey() + " : " + me.getValue());
}

// *********** Using keySet *****************************
for(String s : m.keySet()){
    System.out.println(s + " : " + m.get(s));
}

// *********** Using keySet and iterator *****************
Iterator<String> me = m.keySet().iterator();
while(me.hasNext()){
    String key = me.next();
    System.out.println(key + " : " + m.get(key));
}

Answered   2023-09-20 20:13:42

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);
}

Answered   2023-09-20 20:13:42

  • Well, this is unnecessarily slow because first get the keys and then the entries. Alternative: Get the entrySets and then for each entryset the key and the value - anyone