Get unique values from ArrayList in Java

Question

I have an ArrayList with a number of records and one column contains gas names as CO2 CH4 SO2, etc. Now I want to retrieve different gas names(unique) only without repeatation from the ArrayList. How can it be done?

Answer 1

You should use a Set. A Set is a Collection that contains no duplicates.

If you have a List that contains duplicates, you can get the unique entries like this:

List<String> gasList = // create list with duplicates...
Set<String> uniqueGas = new HashSet<String>(gasList);
System.out.println("Unique gas count: " + uniqueGas.size());

NOTE: This HashSet constructor identifies duplicates by invoking the elements' equals() methods.

Answer 2

You can use Java 8 Stream API.

Method distinct is an intermediate operation that filters the stream and allows only distinct values (by default using the Object::equals method) to pass to the next operation.
I wrote an example below for your case,

// Create the list with duplicates.
List<String> listAll = Arrays.asList("CO2", "CH4", "SO2", "CO2", "CH4", "SO2", "CO2", "CH4", "SO2");

// Create a list with the distinct elements using stream.
List<String> listDistinct = listAll.stream().distinct().collect(Collectors.toList());

// Display them to terminal using stream::collect with a build in Collector.
String collectAll = listAll.stream().collect(Collectors.joining(", "));
System.out.println(collectAll); //=> CO2, CH4, SO2, CO2, CH4 etc..
String collectDistinct = listDistinct.stream().collect(Collectors.joining(", "));
System.out.println(collectDistinct); //=> CO2, CH4, SO2

Answer 3

I hope I understand your question correctly: assuming that the values are of type String, the most efficient way is probably to convert to a HashSet and iterate over it:

ArrayList<String> values = ... //Your values
HashSet<String> uniqueValues = new HashSet<>(values);
for (String value : uniqueValues) {
   ... //Do something
}

Answer 4

you can use this for making a list Unique

ArrayList<String> listWithDuplicateValues = new ArrayList<>();
list.add("first");
list.add("first");
list.add("second");

ArrayList uniqueList = (ArrayList) listWithDuplicateValues.stream().distinct().collect(Collectors.toList());

Answer 5

Here's straightforward way without resorting to custom comparators or stuff like that:

Set<String> gasNames = new HashSet<String>();
List<YourRecord> records = ...;

for(YourRecord record : records) {
  gasNames.add(record.getGasName());
}

// now gasNames is a set of unique gas names, which you could operate on:
List<String> sortedGasses = new ArrayList<String>(gasNames);
Collections.sort(sortedGasses);

Note: Using TreeSet instead of HashSet would give directly sorted arraylist and above Collections.sort could be skipped, but TreeSet is otherwise less efficent, so it's often better, and rarely worse, to use HashSet even when sorting is needed.

Answer 6

ArrayList values = ... // your values
Set uniqueValues = new HashSet(values); //now unique

Answer 7

    public static List getUniqueValues(List input) {
      return new ArrayList<>(new LinkedHashSet<>(incoming));
    }

dont forget to implement your equals method first

Answer 8

When I was doing the same query, I had hard time adjusting the solutions to my case, though all the previous answers have good insights.

Here is a solution when one has to acquire a list of unique objects, NOT strings. Let's say, one has a list of Record object. Record class has only properties of type String, NO property of type int. Here implementing hashCode() becomes difficult as hashCode() needs to return an int.

The following is a sample Record Class.

public class Record{

    String employeeName;
    String employeeGroup;

    Record(String name, String group){  
        employeeName= name;
        employeeGroup = group;    
    }
    public String getEmployeeName(){
        return employeeName;
    }
    public String getEmployeeGroup(){
        return employeeGroup;
    }

  @Override
    public boolean equals(Object o){
         if(o instanceof Record){
            if (((Record) o).employeeGroup.equals(employeeGroup) &&
                  ((Record) o).employeeName.equals(employeeName)){
                return true;
            }
         }
         return false;
    }

    @Override
    public int hashCode() { //this should return a unique code
        int hash = 3; //this could be anything, but I would chose a prime(e.g. 5, 7, 11 )
        //again, the multiplier could be anything like 59,79,89, any prime
        hash = 89 * hash + Objects.hashCode(this.employeeGroup); 
        return hash;
    }

As suggested earlier by others, the class needs to override both the equals() and the hashCode() method to be able to use HashSet.

Now, let's say, the list of Records is allRecord(List<Record> allRecord).

Set<Record> distinctRecords = new HashSet<>();

for(Record rc: allRecord){
    distinctRecords.add(rc);
}

This will only add the distinct Records to the Hashset, distinctRecords.

Hope this helps.

Answer 9

If you have an array of a some kind of object (bean) you can do this:

List<aBean> gasList = createDuplicateGasBeans();
Set<aBean> uniqueGas = new HashSet<aBean>(gasList);

like said Mathias Schwarz above, but you have to provide your aBean with the methods hashCode() and equals(Object obj) that can be done easily in Eclipse by dedicated menu 'Generate hashCode() and equals()' (while in the bean Class). Set will evaluate the overridden methods to discriminate equals objects.