Java count occurrence of each element in an integer array

Question

I've written the following snippet to count the number of occurrences of each element. Is it possible to achieve this in a much shorter way?

int[] arr = {1, 6, 2, 8, 5, 4, 7, 7, 5, 7};
Arrays.stream(arr)
        .collect(ArrayList::new, ArrayList::add, ArrayList::addAll)
        .stream()
        .collect(Collectors.groupingBy(s -> s))
        .forEach((k, v) -> System.out.println(k+" "+v.size()));

Also I would like to display only the elements which occur more than 1 time. So I tried modifying as below which resulted in an error.

.forEach((k, v) -> if(v.size() > 1) System.out.println(k+" "+v.size()));

What is the correct way to do this?

Eran · Accepted Answer · 2016-08-14 12:36:29Z

12

For the latter question, you have to change

.forEach((k, v) -> if(v.size() > 1) System.out.println(k+" "+v.size()));

to

.forEach((k, v) -> {if(v.size() > 1) System.out.println(k+" "+v.size());});

For the first part, it's not clear why you need the first collect followed by a second Stream pipeline. If the purpose was to convert an IntStream to a Stream<Integer>, use boxed():

Arrays.stream(arr)
      .boxed()
      .collect(Collectors.groupingBy(s -> s))
      .forEach((k, v) -> System.out.println(k+" "+v.size()));

As Dici suggested, you can also chain Collectors to group each number with its number of occurrences :

Map<Integer,Integer> occurrences = 
    Arrays.stream(arr)
          .boxed()
          .collect(Collectors.groupingBy(s -> s, Collectors.counting()));

edited Aug 14, 2016 at 12:36

answered Aug 14, 2016 at 12:05

Eran

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5 Comments

Dici Over a year ago

There's a slighlty better way using Collectors.groupingBy(Function.identity(), Collectors.counting()). With a bunch of static imports it looks nicer

DhiwaTdG Over a year ago

Also is it possible to return the result as a Map<Integer, Integer> instead of printing it?

Eran Over a year ago

@DhiwaTdG Yes, you can use Dici's suggestion.

DhiwaTdG Over a year ago

I tried that but it returns all the values whereas I wish to return only the elements which occur more than one time.

Eran Over a year ago

@DhiwaTdG You might have to run that Map through another Stream pipeline to filter out the numbers having a single occurrence. I'm not sure whether you can do that in the same Stream pipeline.

Donald Raab · Accepted Answer · 2020-12-08 18:52:16Z

3

If you are open to using a third-party library, Eclipse Collections has a Bag type which can be used as follows:

Bags.mutable.with(1, 6, 2, 8, 5, 4, 7, 7, 5, 7)
    .selectDuplicates()
    .forEachWithOccurrences((k, count) -> System.out.println(k+" "+count));

If you have to keep int[] arr variable as an int array, then you can use an IntBag as follows:

IntBags.mutable.with(arr)
    .selectDuplicates()
    .forEachWithOccurrences((k, count) -> System.out.println(k+" "+count));

IntBag is a primitive collection so does not box the int values into Integer wrappers.

Note: I am a committer for Eclipse Collections.

edited Dec 8, 2020 at 18:52

answered Aug 15, 2016 at 1:01

Donald Raab

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Comments

Shaido · Accepted Answer · 2021-07-02 03:13:05Z

2

List<Integer> numsList=Arrays.asList(1,2,3,5,2,4,3,1,2,2,4,4,5);
Map<Integer, Long> map=numsList.stream().collect(Collectors.groupingBy(Integer::intValue,Collectors.counting()));
map.forEach((k,v)->{System.out.println(k+" = "+v);});

edited Jul 2, 2021 at 3:13

Shaido

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answered Jul 1, 2021 at 6:31

ASR

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Comments

DhiwaTdG · Accepted Answer · 2016-08-14 13:00:50Z

0

I want to share my solution as well!!!

// Solution 1 [Improved from Eran's solution & suggestion]
int[] arr = {1, 6, 2, 8, 5, 4, 7, 7, 5, 7};
Map<Integer, Long> counts = Arrays.stream(arr)
    .boxed()
    .collect(collectingAndThen(groupingBy(n -> n, counting()),
        map -> map.entrySet().stream()
            .filter(n -> n.getValue() > 1)
            .collect(toMap(Entry::getKey, Entry::getValue))
));
System.out.println(counts.toString());

// Solution 2 [Improved from Dici's suggestion]
int[] arr = {1, 6, 2, 8, 5, 4, 7, 7, 5, 7};
Map<Object, Long> counts = Arrays.stream(arr)
    .collect(ArrayList::new, ArrayList::add, ArrayList::addAll)
    .stream()
    .collect(groupingBy(Function.identity(), counting()));
counts.values().removeIf(count -> count < 2);
System.out.println(counts.toString());

answered Aug 14, 2016 at 13:00

DhiwaTdG

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2 Comments

Tagir Valeev Over a year ago

Note that by groupingBy specification it's not guaranteed that returned map will be mutable. So such solution may fail in future Java (or non-OpenJDK-based) implementations.

Holger Over a year ago

You can easily solve the issue mentioned by @Tagir by using .collect(groupingBy( Function.identity(), HashMap::new, counting()));. Then, the result in guaranteed to be a mutable HashMap and you can use removeIf. Note that this also works together with collectingAndThen. By the way, you don’t need toString() when using System.out.println…

Eugene · Accepted Answer · 2016-08-15 13:10:48Z

0

Also might be done using frequency :

 List<Integer> list = ImmutableList.of(1, 2, 3, 4, 5, 6, 3, 4, 5);
 Map<Integer, Integer> result = list.stream().distinct().collect(Collectors.toMap(Function.identity(), token -> Collections.frequency(list, token)));

answered Aug 15, 2016 at 13:10

Eugene

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