Intersection and union of ArrayLists in Java

Question

Are there any methods to do so? I was looking but couldn't find any.

Another question: I need these methods so I can filter files. Some are AND filters and some are OR filters (like in set theory), so I need to filter according to all files and the unite/intersects ArrayLists that holds those files.

Should I use a different data structure to hold the files? Is there anything else that would offer a better runtime?

Answer 1

Here's a plain implementation without using any third-party library. Main advantage over retainAll, removeAll and addAll is that these methods don't modify the original lists input to the methods.

public class Test {

    public static void main(String... args) throws Exception {

        List<String> list1 = new ArrayList<String>(Arrays.asList("A", "B", "C"));
        List<String> list2 = new ArrayList<String>(Arrays.asList("B", "C", "D", "E", "F"));

        System.out.println(new Test().intersection(list1, list2));
        System.out.println(new Test().union(list1, list2));
    }

    public <T> List<T> union(List<T> list1, List<T> list2) {
        Set<T> set = new HashSet<T>();

        set.addAll(list1);
        set.addAll(list2);

        return new ArrayList<T>(set);
    }

    public <T> List<T> intersection(List<T> list1, List<T> list2) {
        List<T> list = new ArrayList<T>();

        for (T t : list1) {
            if(list2.contains(t)) {
                list.add(t);
            }
        }

        return list;
    }
}

Answer 2

Collection (so ArrayList also) have:

col.retainAll(otherCol) // for intersection
col.addAll(otherCol) // for union

Use a List implementation if you accept repetitions, a Set implementation if you don't:

Collection<String> col1 = new ArrayList<String>(); // {a, b, c}
// Collection<String> col1 = new TreeSet<String>();
col1.add("a");
col1.add("b");
col1.add("c");

Collection<String> col2 = new ArrayList<String>(); // {b, c, d, e}
// Collection<String> col2 = new TreeSet<String>();
col2.add("b");
col2.add("c");
col2.add("d");
col2.add("e");

col1.addAll(col2);
System.out.println(col1); 
//output for ArrayList: [a, b, c, b, c, d, e]
//output for TreeSet: [a, b, c, d, e]

Answer 3

This post is fairly old, but nevertheless it was the first one popping up on google when looking for that topic.

I want to give an update using Java 8 streams doing (basically) the same thing in a single line:

List<T> intersect = list1.stream()
    .filter(list2::contains)
    .collect(Collectors.toList());

List<T> union = Stream.concat(list1.stream(), list2.stream())
    .distinct()
    .collect(Collectors.toList());

If anyone has a better/faster solution let me know, but this solution is a nice one liner that can be easily included in a method without adding a unnecessary helper class/method and still keep the readability.

Answer 4

list1.retainAll(list2) - is intersection

union will be removeAll and then addAll.

Find more in the documentation of collection(ArrayList is a collection) http://download.oracle.com/javase/1.5.0/docs/api/java/util/Collection.html

Answer 5

Unions and intersections defined only for sets, not lists. As you mentioned.

Check guava library for filters. Also guava provides real intersections and unions

 static <E> Sets.SetView<E >union(Set<? extends E> set1, Set<? extends E> set2)
 static <E> Sets.SetView<E> intersection(Set<E> set1, Set<?> set2)

Answer 6

You can use CollectionUtils from apache commons.

Answer 7

The solution marked is not efficient. It has a O(n^2) time complexity. What we can do is to sort both lists, and the execute an intersection algorithm as the one below.

private  static ArrayList<Integer> interesect(ArrayList<Integer> f, ArrayList<Integer> s) { 
    ArrayList<Integer> res = new ArrayList<Integer>();

    int i = 0, j = 0; 
    while (i != f.size() && j != s.size()) { 

        if (f.get(i) < s.get(j)) {
            i ++;
        } else if (f.get(i) > s.get(j)) { 
            j ++;
        } else { 
            res.add(f.get(i)); 
            i ++;  j ++;
        }
    }


    return res; 
}

This one has a complexity of O(n log n + n) which is in O(n log n). The union is done in a similar manner. Just make sure you make the suitable modifications on the if-elseif-else statements.

You can also use iterators if you want (I know they are more efficient in C++, I dont know if this is true in Java as well).

Answer 8

One-liners since JAVA 8

Union

if there are no duplicates:

  return concat(a.stream(), b.stream()).collect(toList());

union and distinct:

  return concat(a.stream(), b.stream()).distinct().collect(toList());

union and distinct if Collection/Set return type:

  return concat(a.stream(), b.stream()).collect(toSet());

Intersect

if no duplicates:

  return a.stream().filter(b::contains).collect(toList());

PERFORMANCE: If collection b is huge and not O(1), then pre-optimize the filter performance by adding 1 line before return: Copy to HasSet (import java.util.Set;):

... b = Set.copyOf(b);

intersect and distinct:

  return a.stream().distinct().filter(b::contains).collect(toList());

- imports

import static java.util.stream.Stream.concat;
import static java.util.stream.Collectors.toList;
import static java.util.stream.Collectors.toSet;

Answer 9

Here is a way how you can do an intersection with streams (remember that you have to use java 8 for streams):

List<foo> fooList1 = new ArrayList<>(Arrays.asList(new foo(), new foo()));
List<foo> fooList2 = new ArrayList<>(Arrays.asList(new foo(), new foo()));
fooList1.stream().filter(f -> fooList2.contains(f)).collect(Collectors.toList());

An example for lists with different types. If you have a realtion between foo and bar and you can get a bar-object from foo than you can modify your stream:

List<foo> fooList = new ArrayList<>(Arrays.asList(new foo(), new foo()));
List<bar> barList = new ArrayList<>(Arrays.asList(new bar(), new bar()));

fooList.stream().filter(f -> barList.contains(f.getBar()).collect(Collectors.toList());

Answer 10

You can use commons-collections4 CollectionUtils

Collection<Integer> collection1 = Arrays.asList(1, 2, 4, 5, 7, 8);
Collection<Integer> collection2 = Arrays.asList(2, 3, 4, 6, 8);

Collection<Integer> intersection = CollectionUtils.intersection(collection1, collection2);
System.out.println(intersection); // [2, 4, 8]

Collection<Integer> union = CollectionUtils.union(collection1, collection2);
System.out.println(union); // [1, 2, 3, 4, 5, 6, 7, 8]

Collection<Integer> subtract = CollectionUtils.subtract(collection1, collection2);
System.out.println(subtract); // [1, 5, 7]

Answer 11

I think you should use a Set to hold the files if you want to do intersection and union on them. Then you can use Guava's Sets class to do union, intersection and filtering by a Predicate as well. The difference between these methods and the other suggestions is that all of these methods create lazy views of the union, intersection, etc. of the two sets. Apache Commons creates a new collection and copies data to it. retainAll changes one of your collections by removing elements from it.

Answer 12

• retainAll will modify your list
• Guava doesn't have APIs for List (only for set)

I found ListUtils very useful for this use case.

Use ListUtils from org.apache.commons.collections if you do not want to modify existing list.

ListUtils.intersection(list1, list2)

Answer 13

In Java 8, I use simple helper methods like this:

public static <T> Collection<T> getIntersection(Collection<T> coll1, Collection<T> coll2){
    return Stream.concat(coll1.stream(), coll2.stream())
            .filter(coll1::contains)
            .filter(coll2::contains)
            .collect(Collectors.toSet());
}

public static <T> Collection<T> getMinus(Collection<T> coll1, Collection<T> coll2){
    return coll1.stream().filter(not(coll2::contains)).collect(Collectors.toSet());
}

public static <T> Predicate<T> not(Predicate<T> t) {
    return t.negate();
}

Answer 14

If the objects in the list are hashable (i.e. have a decent hashCode and equals function), the fastest approach between tables approx. size > 20 is to construct a HashSet for the larger of the two lists.

public static <T> ArrayList<T> intersection(Collection<T> a, Collection<T> b) {
    if (b.size() > a.size()) {
        return intersection(b, a);
    } else {
        if (b.size() > 20 && !(a instanceof HashSet)) {
            a = new HashSet(a);
        }
        ArrayList<T> result = new ArrayList();
        for (T objb : b) {
            if (a.contains(objb)) {
                result.add(objb);
            }
        }
        return result;
    }
}

Answer 15

I was also working on the similar situation and reached here searching for help. Ended up finding my own solution for Arrays. ArrayList AbsentDates = new ArrayList(); // Will Store Array1-Array2

Note : Posting this if it can help someone reaching this page for help.

ArrayList<String> AbsentDates = new ArrayList<String>();//This Array will store difference
      public void AbsentDays() {
            findDates("April", "2017");//Array one with dates in Month April 2017
            findPresentDays();//Array two carrying some dates which are subset of Dates in Month April 2017

            for (int i = 0; i < Dates.size(); i++) {

                for (int j = 0; j < PresentDates.size(); j++) {

                    if (Dates.get(i).equals(PresentDates.get(j))) {

                        Dates.remove(i);
                    }               

                }              
                AbsentDates = Dates;   
            }
            System.out.println(AbsentDates );
        }

Answer 16

Intersection of two list of different object based on common key - Java 8

 private List<User> intersection(List<User> users, List<OtherUser> list) {

        return list.stream()
                .flatMap(OtherUser -> users.stream()
                        .filter(user -> user.getId()
                                .equalsIgnoreCase(OtherUser.getId())))
                .collect(Collectors.toList());
    }

Answer 17

public static <T> Set<T> intersectCollections(Collection<T> col1, Collection<T> col2) {
    Set<T> set1, set2;
    if (col1 instanceof Set) {
        set1 = (Set) col1;
    } else {
        set1 = new HashSet<>(col1);
    }

    if (col2 instanceof Set) {
        set2 = (Set) col2;
    } else {
        set2 = new HashSet<>(col2);
    }

    Set<T> intersection = new HashSet<>(Math.min(set1.size(), set2.size()));

    for (T t : set1) {
        if (set2.contains(t)) {
            intersection.add(t);
        }
    }

    return intersection;
}

JDK8+ (Probably Best Performance)

public static <T> Set<T> intersectCollections(Collection<T> col1, Collection<T> col2) {
    boolean isCol1Larger = col1.size() > col2.size();
    Set<T> largerSet;
    Collection<T> smallerCol;

    if (isCol1Larger) {
        if (col1 instanceof Set) {
            largerSet = (Set<T>) col1;
        } else {
            largerSet = new HashSet<>(col1);
        }
        smallerCol = col2;
    } else {
        if (col2 instanceof Set) {
            largerSet = (Set<T>) col2;
        } else {
            largerSet = new HashSet<>(col2);
        }
        smallerCol = col1;
    }

    return smallerCol.stream()
            .filter(largerSet::contains)
            .collect(Collectors.toSet());
}

If you don't care about performance and prefer smaller code just use:

col1.stream().filter(col2::contains).collect(Collectors.toList());

Answer 18

First, I am copying all values of arrays into a single array then I am removing duplicates values into the array. Line 12, explaining if same number occur more than time then put some extra garbage value into "j" position. At the end, traverse from start-end and check if same garbage value occur then discard.

public class Union {
public static void main(String[] args){

    int arr1[]={1,3,3,2,4,2,3,3,5,2,1,99};
    int arr2[]={1,3,2,1,3,2,4,6,3,4};
    int arr3[]=new int[arr1.length+arr2.length];

    for(int i=0;i<arr1.length;i++)
        arr3[i]=arr1[i];

    for(int i=0;i<arr2.length;i++)
        arr3[arr1.length+i]=arr2[i];
    System.out.println(Arrays.toString(arr3));

    for(int i=0;i<arr3.length;i++)
    {
        for(int j=i+1;j<arr3.length;j++)
        {
            if(arr3[i]==arr3[j])
                arr3[j]=99999999;          //line  12
        }
    }
    for(int i=0;i<arr3.length;i++)
    {
        if(arr3[i]!=99999999)
            System.out.print(arr3[i]+" ");
    }
}   
}

Answer 19

After testing, here is my best intersection approach.

Faster speed compared to pure HashSet Approach. HashSet and HashMap below has similar performance for arrays with more than 1 million records.

As for Java 8 Stream approach, speed is quite slow for array size larger then 10k.

Hope this can help.

public static List<String> hashMapIntersection(List<String> target, List<String> support) {
    List<String> r = new ArrayList<String>();
    Map<String, Integer> map = new HashMap<String, Integer>();
    for (String s : support) {
        map.put(s, 0);
    }
    for (String s : target) {
        if (map.containsKey(s)) {
            r.add(s);
        }
    }
    return r;
}
public static List<String> hashSetIntersection(List<String> a, List<String> b) {
    Long start = System.currentTimeMillis();

    List<String> r = new ArrayList<String>();
    Set<String> set = new HashSet<String>(b);

    for (String s : a) {
        if (set.contains(s)) {
            r.add(s);
        }
    }
    print("intersection:" + r.size() + "-" + String.valueOf(System.currentTimeMillis() - start));
    return r;
}

public static void union(List<String> a, List<String> b) {
    Long start = System.currentTimeMillis();
    Set<String> r= new HashSet<String>(a);
    r.addAll(b);
    print("union:" + r.size() + "-" + String.valueOf(System.currentTimeMillis() - start));
}

Answer 20

retainAll() method use for finding common element..i.e;intersection list1.retainAll(list2)

Answer 21

You can use the methods:

CollectionUtils.containsAny and CollectionUtils.containsAll

from Apache Commons.

Answer 22

Final solution:

//all sorted items from both
public <T> List<T> getListReunion(List<T> list1, List<T> list2) {
    Set<T> set = new HashSet<T>();
    set.addAll(list1);
    set.addAll(list2);
    return new ArrayList<T>(set);
}

//common items from both
public <T> List<T> getListIntersection(List<T> list1, List<T> list2) {
    list1.retainAll(list2);
    return list1;
}

//common items from list1 not present in list2
public <T> List<T> getListDifference(List<T> list1, List<T> list2) {
    list1.removeAll(list2);
    return list1;
}

Answer 23

If the number matches than I am checking it's occur first time or not with help of "indexOf()" if the number matches first time then print and save into in a string so, that when the next time same number matches then it's won't print because due to "indexOf()" condition will be false.

class Intersection
{
public static void main(String[] args)
 {
  String s="";
    int[] array1 = {1, 2, 5, 5, 8, 9, 7,2,3512451,4,4,5 ,10};
    int[] array2 = {1, 0, 6, 15, 6, 5,4, 1,7, 0,5,4,5,2,3,8,5,3512451};


       for (int i = 0; i < array1.length; i++)
       {
           for (int j = 0; j < array2.length; j++)
           {
               char c=(char)(array1[i]);
               if(array1[i] == (array2[j])&&s.indexOf(c)==-1)
               {    
                System.out.println("Common element is : "+(array1[i]));
                s+=c;
                }
           }
       }    
}

}

Answer 24

If you had your data in Sets you could use Guava's Sets class.