Java count occurrence of each item in an array

Question

Is there any method for counting the occurrence of each item on an array?

Lets say I have:

String[] array = {"name1","name2","name3","name4", "name5"};

Here the output will be:

name1 1
name2 1
name3 1
name4 1
name5 1

and if I have:

String[] array = {"name1","name1","name2","name2", "name2"};

The output would be:

name1 2
name2 3

The output here is just to demonstrate the expected result.

Solution 1

You could use a MultiSet from Google Collections/Guava or a Bag from Apache Commons.

If you have a collection instead of an array, you can use addAll() to add the entire contents to the above data structure, and then apply the count() method to each value. A SortedMultiSet or SortedBag would give you the items in a defined order.

Google Collections actually has very convenient ways of going from arrays to a SortedMultiset.

Solution 2

List asList = Arrays.asList(array);
Set<String> mySet = new HashSet<String>(asList);

for(String s: mySet){
 System.out.println(s + " " + Collections.frequency(asList,s));
}

Solution 3

With java-8, you can do it like this:

String[] array = {"name1","name2","name3","name4", "name5", "name2"};
Arrays.stream(array)
      .collect(Collectors.groupingBy(s -> s))
      .forEach((k, v) -> System.out.println(k+" "+v.size()));

Output:

name5 1
name4 1
name3 1
name2 2
name1 1

What it does is:

• Create a Stream<String> from the original array
• Group each element by identity, resulting in a Map<String, List<String>>
• For each key value pair, print the key and the size of the list

If you want to get a Map that contains the number of occurences for each word, it can be done doing:

Map<String, Long> map = Arrays.stream(array)
    .collect(Collectors.groupingBy(s -> s, Collectors.counting()));

For more informations:

• Stream
• Collectors

Hope it helps! :)

Solution 4

I wrote a solution for this to practice myself. It doesn't seem nearly as awesome as the other answers posted, but I'm going to post it anyway, and then learn how to do this using the other methods as well. Enjoy:

public static Integer[] countItems(String[] arr)
{
    List<Integer> itemCount = new ArrayList<Integer>();
    Integer counter = 0;
    String lastItem = arr[0];

    for(int i = 0; i < arr.length; i++)
    {
        if(arr[i].equals(lastItem))
        {
            counter++;
        }
        else
        {
            itemCount.add(counter);
            counter = 1;
        }
        lastItem = arr[i];
    }
    itemCount.add(counter);

    return itemCount.toArray(new Integer[itemCount.size()]);
}

public static void main(String[] args)
{
    String[] array = {"name1","name1","name2","name2", "name2", "name3",
            "name1","name1","name2","name2", "name2", "name3"};
    Arrays.sort(array);
    Integer[] cArr = countItems(array);
    int num = 0;
    for(int i = 0; i < cArr.length; i++)
    {
        num += cArr[i]-1;
        System.out.println(array[num] + ": " + cArr[i].toString());
    }
}

Solution 5

Using HashMap it is walk in the park.

main(){
    String[] array ={"a","ab","a","abc","abc","a","ab","ab","a"};
    Map<String,Integer> hm = new HashMap();

    for(String x:array){

        if(!hm.containsKey(x)){
            hm.put(x,1);
        }else{
            hm.put(x, hm.get(x)+1);
        }
    }
    System.out.println(hm);
}

View more solutions

Comments

[james.garriss]

The link to Bag is 404.

[akshayb]

from my experience, Collections.frequency is too slow.Avoid if performance is a concern.

[Tuan]

I used getCardinalityMap() from org.apache.commons.collections.CollectionUtils.

[jrenk]

You may want to provide some explanation to your code.

[Jorn Vernee]

This doesn't work, you'd have to use hash.put(s, hash.get(s) + 1);. Doing i++; doesn't update the integer inside the hashtable. There's also other typos.

[Damian]

Not really, if it is a HashSet or HashMap.

[Gopal00005]

It won't work for array like this: String[] array = { "Gates", "Michael","Gates","Peterson","Bush","Johnson","Johnson","Gat‌​es"};

[naveenTekkem]

We can also pass the sampleList above to constructor of TreeSet and then loop to get the result

[ikel]

this works, but could explain little about hm.put(x, hm.get(x)+1); part ? thx

[ikel]

I meant how is hm.get(x) actually act like a counter not item value?

[Deva44]

@ikel If key (x in this case) is already there in the hm, get the corresponding value, increment by one and store it back.