Question
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?
Solution
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());
}
Solution
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:
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();
}
Using foreach and Map.Entry
long i = 0;
for (Map.Entry<Integer, Integer> pair : map.entrySet()) {
i += pair.getKey() + pair.getValue();
}
Using forEach from Java 8
final long[] i = {0};
map.forEach((k, v) -> i[0] += k + v);
Using keySet and foreach
long i = 0;
for (Integer key : map.keySet()) {
i += key + map.get(key);
}
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);
}
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();
}
Using the Java 8 Stream API
final long[] i = {0};
map.entrySet().stream().forEach(e -> i[0] += e.getKey() + e.getValue());
Using the Java 8 Stream API parallel
final long[] i = {0};
map.entrySet().stream().parallel().forEach(e -> i[0] += e.getKey() + e.getValue());
Using IterableMap of Apache Collections
long i = 0;
MapIterator<Integer, Integer> it = iterableMap.mapIterator();
while (it.hasNext()) {
i += it.next() + it.getValue();
}
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)
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
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
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)
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.
Solution
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!
Solution
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();
// ...
}
Solution
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.
Solution
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();
// ...
}
Solution
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.
Solution
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.
Solution
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));
}
Solution
With Java 8, you can iterate Map using forEach and lambda expression,
map.forEach((k, v) -> System.out.println((k + ":" + v)));
Solution
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.
Solution
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.
Solution
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.
Solution
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.
Solution
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.
Solution
In Map one can Iteration over keys and/or values and/or both (e.g., entrySet) depends on one's interested in_ Like:
Iterate through the keys -> keySet() of the map:
Map<String, Object> map = ...;
for (String key : map.keySet()) {
//your Business logic...
}
Iterate through the values -> values() of the map:
for (Object value : map.values()) {
//your Business logic...
}
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);
}
Solution
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());
//...
}
Solution
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,
Solution
Most compact with Java 8:
map.entrySet().forEach(System.out::println);
Solution
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());
}
}
Solution
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.
Map implementations promise a certain iteration order, others do not.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.
You can see there are four Map implementations maintaining an order:
TreeMapConcurrentSkipListMap LinkedHashMapEnumMapNavigableMap interfaceTwo 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.
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.
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.
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.
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.
In choosing a Map implementation, also consider:
Collections::synchronizedMap (less preferable). Both of these considerations are covered in the graphic table above.
Solution
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());
}
Solution
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();
}
}
}
Solution
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());
}
Solution
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");
}
});
Solution
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());
}
Solution
//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();
}
Solution
Use Java 8:
map.entrySet().forEach(entry -> System.out.println(entry.getValue()));
Solution
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();
}
Solution
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);
}