Java 8, Streams to find the duplicate elements

问题:

I am trying to list out duplicate elements in the integer list say for eg,

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

using Streams of jdk 8. Has anybody tried out. To remove the duplicates we can use the distinct() api. But what about finding the duplicated elements ? Anybody can help me out ?

回答1:

You can use Collections.frequency:

numbers.stream().filter(i -> Collections.frequency(numbers, i) >1)
                .collect(Collectors.toSet()).forEach(System.out::println);

回答2:

You need a set (allItems below) to hold the entire array contents, but this is O(n):

Integer[] numbers = new Integer[] { 1, 2, 1, 3, 4, 4 };
Set<Integer> allItems = new HashSet<>();
Set<Integer> duplicates = Arrays.stream(numbers)
        .filter(n -> !allItems.add(n)) //Set.add() returns false if the item was already in the set.
        .collect(Collectors.toSet());
System.out.println(duplicates); // [1, 4]

回答3:

Basic example. First half builds frequency-map, second half reduces it to a filtered list. Propably not as efficient as Dave's answer, but more versatile (like if you want to detect exactly two etc.)

    List<Integer> duplicates = IntStream.of( 1, 2, 3, 2, 1, 2, 3, 4, 2, 2, 2 )
                                            .boxed()
                                            .collect( Collectors.groupingBy( c -> c, Collectors.counting() ) )
                                            .entrySet()
                                            .stream()
                                            .filter( p -> p.getValue() > 1 )
                                            .map( e -> e.getKey() )
                                            .collect( Collectors.toList() );

回答4:

My StreamEx library which enhances the Java 8 streams provides a special operation distinct(atLeast) which can retain only elements appearing at least the specified number of times. So your problem can be solved like this:

List<Integer> repeatingNumbers = StreamEx.of(numbers).distinct(2).toList();

Internally it's similar to @Dave solution, it counts objects, to support other wanted quantities and it's parallel-friendly (it uses ConcurrentHashMap for parallelized stream, but HashMap for sequential). For big amounts of data you can get a speed-up using .parallel().distinct(2).

回答5:

An O(n) way would be as below:

List<Integer> numbers = Arrays.asList(1, 2, 1, 3, 4, 4);
Set<Integer> duplicatedNumbersRemovedSet = new HashSet<>();
Set<Integer> duplicatedNumbersSet = numbers.stream().filter(n -> !duplicatedNumbersRemovedSet.add(n)).collect(Collectors.toSet());

The space complexity would go double in this approach, but that space is not a waste; in-fact, we now have the duplicated alone only as a Set as well as another Set with all the duplicates removed too.

回答6:

You can get the duplicated like this :

List<Integer> numbers = Arrays.asList(1, 2, 1, 3, 4, 4);
Set<Integer> duplicated = numbers.stream().filter(n -> numbers.stream().filter(x -> x == n).count() > 1).collect(Collectors.toSet());

回答7:

I think basic solutions to the question should be as below:

Supplier supplier=HashSet::new; 
HashSet has=ls.stream().collect(Collectors.toCollection(supplier));

List lst = (List) ls.stream().filter(e->Collections.frequency(ls,e)>1).distinct().collect(Collectors.toList());

well, it is not recommended to perform a filter operation, but for better understanding, i have used it, moreover, there should be some custom filtration in future versions.

回答8:

A multiset is a structure maintaining the number of occurrences for each element. Using Guava implementation:

Set<Integer> duplicated =
        ImmutableMultiset.copyOf(numbers).entrySet().stream()
                .filter(entry -> entry.getCount() > 1)
                .map(Multiset.Entry::getElement)
                .collect(Collectors.toSet());

回答9:

I think I have good solution how to fix problem like this - List => List with grouping by Something.a & Something.b. There is extended definition:

public class Test {

    public static void test() {

        class A {
            private int a;
            private int b;
            private float c;
            private float d;

            public A(int a, int b, float c, float d) {
                this.a = a;
                this.b = b;
                this.c = c;
                this.d = d;
            }
        }


        List<A> list1 = new ArrayList<A>();

        list1.addAll(Arrays.asList(new A(1, 2, 3, 4),
                new A(2, 3, 4, 5),
                new A(1, 2, 3, 4),
                new A(2, 3, 4, 5),
                new A(1, 2, 3, 4)));

        Map<Integer, A> map = list1.stream()
                .collect(HashMap::new, (m, v) -> m.put(
                        Objects.hash(v.a, v.b, v.c, v.d), v),
                        HashMap::putAll);

        list1.clear();
        list1.addAll(map.values());

        System.out.println(list1);
    }

}

class A, list1 it's just incoming data - magic is in the Objects.hash(...) :)

回答10:

Do you have to use the java 8 idioms (steams)? Perphaps a simple solution would be to move the complexity to a map alike data structure that holds numbers as key (without repeating) and the times it ocurrs as a value. You could them iterate that map an only do something with those numbers that are ocurrs > 1.

import java.lang.Math;
import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.HashMap;
import java.util.Iterator;

public class RemoveDuplicates
{
  public static void main(String[] args)
  {
   List<Integer> numbers = Arrays.asList(new Integer[]{1,2,1,3,4,4});
   Map<Integer,Integer> countByNumber = new HashMap<Integer,Integer>();
   for(Integer n:numbers)
   {
     Integer count = countByNumber.get(n);
     if (count != null) {
       countByNumber.put(n,count + 1);
     } else {
       countByNumber.put(n,1);
     }
   }
   System.out.println(countByNumber);
   Iterator it = countByNumber.entrySet().iterator();
    while (it.hasNext()) {
        Map.Entry pair = (Map.Entry)it.next();
        System.out.println(pair.getKey() + " = " + pair.getValue());
    }
  }
}

回答11:

Try this solution:

public class Anagramm {

public static boolean isAnagramLetters(String word, String anagramm) {
    if (anagramm.isEmpty()) {
        return false;
    }

    Map<Character, Integer> mapExistString = CharCountMap(word);
    Map<Character, Integer> mapCheckString = CharCountMap(anagramm);
    return enoughLetters(mapExistString, mapCheckString);
}

private static Map<Character, Integer> CharCountMap(String chars) {
    HashMap<Character, Integer> charCountMap = new HashMap<Character, Integer>();
    for (char c : chars.toCharArray()) {
        if (charCountMap.containsKey(c)) {
            charCountMap.put(c, charCountMap.get(c) + 1);
        } else {
            charCountMap.put(c, 1);
        }
    }
    return charCountMap;
}

static boolean enoughLetters(Map<Character, Integer> mapExistString, Map<Character,Integer> mapCheckString) {
    for( Entry<Character, Integer> e : mapCheckString.entrySet() ) {
        Character letter = e.getKey();
        Integer available = mapExistString.get(letter);
        if (available == null || e.getValue() > available) return false;
    }
    return true;
}

}

回答12:

What about checking of indexes?

        numbers.stream()
            .filter(integer -> numbers.indexOf(integer) != numbers.lastIndexOf(integer))
            .collect(Collectors.toSet())
            .forEach(System.out::println);