How to count the number of occurrences of an element in a List

I have an ArrayList, a Collection class of Java, as follows:

ArrayList<String> animals = new ArrayList<String>();
animals.add("bat");
animals.add("owl");
animals.add("bat");
animals.add("bat");

As you can see, the animals ArrayList consists of 3 bat elements and one owl element. I was wondering if there is any API in the Collection framework that returns the number of bat occurrences or if there is another way to determine the number of occurrences.

I found that Google's Collection Multiset does have an API that returns the total number of occurrences of an element. But that is compatible only with JDK 1.5. Our product is currently in JDK 1.6, so I cannot use it.

I'm pretty sure the static frequency-method in Collections would come in handy here:

int occurrences = Collections.frequency(animals, "bat");

That's how I'd do it anyway. I'm pretty sure this is jdk 1.6 straight up.

In Java 8:

Map<String, Long> counts =
    list.stream().collect(Collectors.groupingBy(e -> e, Collectors.counting()));

Alternative Java 8 solution using Streams:

long count = animals.stream().filter(animal -> "bat".equals(animal)).count();

This shows, why it is important to "Refer to objects by their interfaces" as described in Effective Java book.

If you code to the implementation and use ArrayList in let's say, 50 places in your code, when you find a good "List" implementation that count the items, you will have to change all those 50 places, and probably you'll have to break your code ( if it is only used by you there is not a big deal, but if it is used by someone else uses, you'll break their code too)

By programming to the interface you can let those 50 places unchanged and replace the implementation from ArrayList to "CountItemsList" (for instance ) or some other class.

Below is a very basic sample on how this could be written. This is only a sample, a production ready List would be much more complicated.

import java.util.*;

public class CountItemsList<E> extends ArrayList<E> { 

    // This is private. It is not visible from outside.
    private Map<E,Integer> count = new HashMap<E,Integer>();

    // There are several entry points to this class
    // this is just to show one of them.
    public boolean add( E element  ) { 
        if( !count.containsKey( element ) ){
            count.put( element, 1 );
        } else { 
            count.put( element, count.get( element ) + 1 );
        }
        return super.add( element );
    }
    
    // This method belongs to CountItemList interface ( or class ) 
    // to used you have to cast.
    public int getCount( E element ) { 
        if( ! count.containsKey( element ) ) {
            return 0;
        }
        return count.get( element );
    }

    public static void main( String [] args ) { 
        List<String> animals = new CountItemsList<String>();
        animals.add("bat");
        animals.add("owl");
        animals.add("bat");
        animals.add("bat");

        System.out.println( (( CountItemsList<String> )animals).getCount( "bat" ));
    }
}

OO principles applied here: inheritance, polymorphism, abstraction, encapsulation.

Sorry there's no simple method call that can do it. All you'd need to do though is create a map and count frequency with it.

HashMap<String,int> frequencymap = new HashMap<String,int>();
foreach(String a in animals) {
  if(frequencymap.containsKey(a)) {
    frequencymap.put(a, frequencymap.get(a)+1);
  }
  else{ frequencymap.put(a, 1); }
}

There is no native method in Java to do that for you. However, you can use IterableUtils#countMatches() from Apache Commons-Collections to do it for you.

Simple Way to find the occurrence of string value in an array using Java 8 features.

public void checkDuplicateOccurance() {
		List<String> duplicateList = new ArrayList<String>();
		duplicateList.add("Cat");
		duplicateList.add("Dog");
		duplicateList.add("Cat");
		duplicateList.add("cow");
		duplicateList.add("Cow");
		duplicateList.add("Goat");    		
		Map<String, Long> couterMap = duplicateList.stream().collect(Collectors.groupingBy(e -> e.toString(),Collectors.counting()));
		System.out.println(couterMap);
	}

Output : {Cat=2, Goat=1, Cow=1, cow=1, Dog=1}

You can notice "Cow" and cow are not considered as same string, in case you required it under same count, use .toLowerCase(). Please find the snippet below for the same.

Map<String, Long> couterMap = duplicateList.stream().collect(Collectors.groupingBy(e -> e.toString().toLowerCase(),Collectors.counting()));

Output : {cat=2, cow=2, goat=1, dog=1}

I wonder, why you can't use that Google's Collection API with JDK 1.6. Does it say so? I think you can, there should not be any compatibility issues, as it is built for a lower version. The case would have been different if that were built for 1.6 and you are running 1.5.

Am I wrong somewhere?

Actually, Collections class has a static method called : frequency(Collection c, Object o) which returns the number of occurrences of the element you are searching for, by the way, this will work perfectly for you:

ArrayList<String> animals = new ArrayList<String>();
animals.add("bat");
animals.add("owl");
animals.add("bat");
animals.add("bat");
System.out.println("Freq of bat: "+Collections.frequency(animals, "bat"));

A slightly more efficient approach might be

Map<String, AtomicInteger> instances = new HashMap<String, AtomicInteger>();

void add(String name) {
     AtomicInteger value = instances.get(name);
     if (value == null) 
        instances.put(name, new AtomicInteger(1));
     else
        value.incrementAndGet();
}

To get the occurrences of the object from the list directly:

int noOfOccurs = Collections.frequency(animals, "bat");

To get the occurrence of the Object collection inside list, override the equals method in the Object class as:

@Override
public boolean equals(Object o){
	Animals e;
	if(!(o instanceof Animals)){
		return false;
	}else{
		e=(Animals)o;
		if(this.type==e.type()){
			return true;
		}
	}
	return false;
}

Animals(int type){
	this.type = type;
}

Call the Collections.frequency as:

int noOfOccurs = Collections.frequency(animals, new Animals(1));

To achieve that one can do it in several ways, namely:

Methods that return the number of occurrence of a single element:

Collection Frequency

Collections.frequency(animals, "bat");

Java Stream:

Filter

animals.stream().filter("bat"::equals).count();

Just iteration thought the list

public static long manually(Collection<?> c, Object o){
    int count = 0;
    for(Object e : c)
        if(e.equals(o))
            count++;
    return count;
}

Methods that create a map of frequencies:

Collectors.groupingBy

Map<String, Long> counts = 
       animals.stream()
              .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

merge

Map<String, Long> map = new HashMap<>();
c.forEach(e -> map.merge(e, 1L, Long::sum));

Manually

Map<String, Integer> mp = new HashMap<>();
        animals.forEach(animal -> mp.compute(animal, (k, v) -> (v == null) ? 1 : v + 1));

A running example with all the methods:

import java.util.*;
import java.util.function.Function;
import java.util.stream.Collectors;

public class Frequency {

    public static int frequency(Collection<?> c, Object o){
        return Collections.frequency(c, o);
    }

    public static long filter(Collection<?> c, Object o){
        return c.stream().filter(o::equals).count();
    }

    public static long manually(Collection<?> c, Object o){
        int count = 0;
        for(Object e : c)
            if(e.equals(o))
                count++;
        return count;
    }

    public static Map<?, Long> mapGroupBy(Collection<?> c){
        return c.stream()
                .collect(Collectors.groupingBy(Function.identity() , Collectors.counting()));
    }

    public static Map<Object, Long> mapMerge(Collection<?> c){
        Map<Object, Long> map = new HashMap<>();
        c.forEach(e -> map.merge(e, 1L, Long::sum));
        return map;
    }

    public static Map<Object, Long> manualMap(Collection<?> c){
        Map<Object, Long> map = new HashMap<>();
        c.forEach(e -> map.compute(e, (k, v) -> (v == null) ? 1 : v + 1));
        return map;
    }


    public static void main(String[] args){
        List<String> animals = new ArrayList<>();
        animals.add("bat");
        animals.add("owl");
        animals.add("bat");
        animals.add("bat");

        System.out.println(frequency(animals, "bat"));
        System.out.println(filter(animals,"bat"));
        System.out.println(manually(animals,"bat"));
        mapGroupBy(animals).forEach((k, v) -> System.out.println(k + " -> "+v));
        mapMerge(animals).forEach((k, v) -> System.out.println(k + " -> "+v));
        manualMap(animals).forEach((k, v) -> System.out.println(k + " -> "+v));
    }
}

The methods name should have reflected what those methods are doing, however, I used the name to reflect the approach being used instead (given that in the current context it is okey).

What you want is a Bag - which is like a set but also counts the number of occurances. Unfortunately the java Collections framework - great as they are dont have a Bag impl. For that one must use the Apache Common Collection link text

List<String> list = Arrays.asList("as", "asda", "asd", "urff", "dfkjds", "hfad", "asd", "qadasd", "as", "asda",
		"asd", "urff", "dfkjds", "hfad", "asd", "qadasd" + "as", "asda", "asd", "urff", "dfkjds", "hfad", "asd",
		"qadasd", "as", "asda", "asd", "urff", "dfkjds", "hfad", "asd", "qadasd");
	
	
	

Method 1:

Set<String> set = new LinkedHashSet<>();
set.addAll(list);

for (String s : set) {

	System.out.println(s + " : " + Collections.frequency(list, s));
}
	

Method 2:

int count = 1;
Map<String, Integer> map = new HashMap<>();
Set<String> set1 = new LinkedHashSet<>();
for (String s : list) {
	if (!set1.add(s)) {
		count = map.get(s) + 1;
	}
	map.put(s, count);
	count = 1;

}
System.out.println(map);

​If you use Eclipse Collections, you can use a Bag. A MutableBag can be returned from any implementation of RichIterable by calling toBag().

MutableList<String> animals = Lists.mutable.with("bat", "owl", "bat", "bat");
MutableBag<String> bag = animals.toBag();
Assert.assertEquals(3, bag.occurrencesOf("bat"));
Assert.assertEquals(1, bag.occurrencesOf("owl"));

The HashBag implementation in Eclipse Collections is backed by a MutableObjectIntMap.

Note: I am a committer for Eclipse Collections.

So do it the old fashioned way and roll your own:

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

void add(String name) {
     Integer value = instances.get(name);
     if (value == null) {
        value = new Integer(0);
        instances.put(name, value);
     }
     instances.put(name, value++);
}

Put the elements of the arraylist in the hashMap to count the frequency.

Java 8 - another method

String searched = "bat";
long n = IntStream.range(0, animals.size())
            .filter(i -> searched.equals(animals.get(i)))
            .count();

package traversal;

import java.util.ArrayList;
import java.util.List;

public class Occurrance {
	static int count;

	public static void main(String[] args) {
		List<String> ls = new ArrayList<String>();
		ls.add("aa");
		ls.add("aa");
		ls.add("bb");
		ls.add("cc");
		ls.add("dd");
		ls.add("ee");
		ls.add("ee");
		ls.add("aa");
		ls.add("aa");

		for (int i = 0; i < ls.size(); i++) {
			if (ls.get(i) == "aa") {
				count = count + 1;
			}
		}
		System.out.println(count);
	}
}

> Output: 4

If you are a user of my ForEach DSL, it can be done with a Count query.

Count<String> query = Count.from(list);
for (Count<Foo> each: query) each.yield = "bat".equals(each.element);
int number = query.result();

List<String> lst = new ArrayList<String>();

lst.add("Ram");
lst.add("Ram");
lst.add("Shiv");
lst.add("Boss");

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

for (String string : lst) {
	
	if(mp.keySet().contains(string))
	{
		mp.put(string, mp.get(string)+1);
		
	}else
	{
		mp.put(string, 1);
	}
}

System.out.println("=mp="+mp);

Output:

=mp= {Ram=2, Boss=1, Shiv=1}

I didn't want to make this case more difficult and made it with two iterators I have a HashMap with LastName -> FirstName. And my method should delete items with dulicate FirstName.

public static void removeTheFirstNameDuplicates(HashMap<String, String> map)
{
    
    Iterator<Map.Entry<String, String>> iter = map.entrySet().iterator();
    Iterator<Map.Entry<String, String>> iter2 = map.entrySet().iterator();
    while(iter.hasNext())
    {
        Map.Entry<String, String> pair = iter.next();
        String name = pair.getValue();
        int i = 0;

        while(iter2.hasNext())
        {

            Map.Entry<String, String> nextPair = iter2.next();
            if (nextPair.getValue().equals(name))
                i++;
        }

        if (i > 1)
            iter.remove();

    }

}

Map<String,Integer> hm = new HashMap<String, Integer>();
for(String i : animals) {
    Integer j = hm.get(i);
    hm.put(i,(j==null ? 1 : j+1));
}
for(Map.Entry<String, Integer> val : hm.entrySet()) {
    System.out.println(val.getKey()+" occurs : "+val.getValue()+" times");
}

You can use groupingBy feature of Java 8 for your use case.

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

public class Test {
    public static void main(String[] args) {
        List<String> animals = new ArrayList<>();

        animals.add("bat");
        animals.add("owl");
        animals.add("bat");
        animals.add("bat");

        Map<String,Long> occurrenceMap =
                animals.stream().collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
        System.out.println("occurrenceMap:: " + occurrenceMap);
    }
}

Output

occurrenceMap:: {bat=3, owl=1}

 Integer[] spam = new Integer[]  {1,2,2,3,4};
 List<Integer>	 list=Arrays.asList(spam);

System.out.println(list.stream().collect(Collectors.groupingBy(Function.identity(),Collectors.counting())));
System.out.println(list.stream().collect(Collectors.groupingBy(Function.identity(),HashMap::new,Collectors.counting())));
	

output

{1=1, 2=2, 3=1, 4=1}