Java Streams - Get a “symmetric difference list” from two other lists

Question

Im trying to use Java 8 streams to combine lists. How can I get a \"symmetric difference list\" (all object that only exist in one list) from two existing lists. I know how

Answer 1

Based on your own code, there is a straight-forward solution:

List<Car> disjoint = Stream.concat(
    bigCarList.stream().filter(c->!smallCarList.contains(c)),
    smallCarList.stream().filter(c->!bigCarList.contains(c))
).collect(Collectors.toList());

Just filter one list for all items not contained in the other and vice versa and concatenate both results. That works fairly well for small lists and before consider optimized solutions like hashing or making the result distinct() you should ask yourself why you are using lists if you don’t want neither, duplicates nor a specific order.

It seems like you actually want Sets, not Lists. If you use Sets, Tagir Valeev’s solution is appropriate. But it is not working with the actual semantics of Lists, i.e. doesn’t work if the source lists contain duplicates.

---

But if you are using Sets, the code can be even simpler:

Set<Car> disjoint = Stream.concat(bigCarSet.stream(), smallCarSet.stream())
  .collect(Collectors.toMap(Function.identity(), t->true, (a,b)->null))
  .keySet();

This uses the toMap collector which creates a Map (the value is irrelevant, we simply map to true here) and uses a merge function to handle duplicates. Since for two sets, duplicates can only occur when an item is contained in both sets, these are the items we want remove.

The documentation of Collectors.toMap says that the merge function is treated “as supplied to Map.merge(Object, Object, BiFunction)” and we can learn from there, that simply mapping the duplicate pair to null will remove the entry.

So afterwards, the keySet() of the map contains the disjoint set.

Answer 2

Something like this may work:

Stream.concat(bigCarList.stream(), smallCarList.stream())
      .collect(groupingBy(Function.identity(), counting()))
      .entrySet().stream()
      .filter(e -> e.getValue().equals(1L))
      .map(Map.Entry::getKey)
      .collect(toList());

Here we first collect all the cars to the Map<Car, Long> where value is the number of such cars encountered. After that, we filter this Map leaving only cars that are encountered exactly once, drop the counts and collect to the final List.

Answer 3

A little bit math

disjoint = A and B are disjoint if their intersect is empty.

A disjoint is not a set, it is an indicator showing if two sets are disjoint or not. From your description I think you where searching the symmetric difference.

Symmetric Difference

But anyhow, if you only want to collect to new Lists then all you need is a collector.

I made a method that creates an Collector. This Collector only "collects" values, where the predicate is evaluated to true. So if you are searching for the symmetric difference, than you only need a predicate.

  public void testDisjointLists() {
    List<Car> bigCarList = get5DefaultCars();
    List<Car> smallCarList = get3DefaultCars();

    Collector<Car, ArrayList<Car>, ArrayList<Car>> inter
        = produceCollector(car -> {
          return bigCarList.contains(car) && smallCarList.contains(car);
        });

    Collector<Car, ArrayList<Car>, ArrayList<Car>> symDiff
        = produceCollector(car -> {
          return bigCarList.contains(car) ^ smallCarList.contains(car);
        });

    //Get all cars in both list as one list
    List<Car> union
        = Stream.concat(bigCarList.stream(), smallCarList.stream()).distinct().collect(Collectors.toList());

    List<Car> intersect = union.stream().collect(inter);

    //Get all cars that only exist not exists in both Lists
    List<Car> symmetricDifference = union.stream().collect(symDiff);

    System.out.println("Union Cars:");
    union.stream().forEach(car -> System.out.println("Car: " + car));
    System.out.println("");

    System.out.println("Intersect Cars: ");
    intersect.stream().forEach(car -> System.out.println("Car: " + car));
    System.out.println("");

    System.out.println("Symmetric Difference: ");
    symmetricDifference.stream().forEach(car -> System.out.println("Car: " + car));
    System.out.println("");
  }

  public Collector<Car, ArrayList<Car>, ArrayList<Car>> produceCollector(Predicate<Car> predicate) {
    Collector<Car, ArrayList<Car>, ArrayList<Car>> collector = Collector.of(
        ArrayList::new,
        (al, car) -> {
          if (predicate.test(car)) {
            al.add(car);
          }
        },
        (al1, al2) -> {
          al1.addAll(al2);
          return al1;
        }
    );
    return collector;
  }

For performance freaks

After doing some research, it seems that the collector is about 14 times faster than a first filter solution.

long before2 = System.nanoTime();
List<Car> intersect2 = union.stream().filter(car -> {
  return bigCarList.contains(car) && smallCarList.contains(car);
}).collect(Collectors.toList());
long after2 = System.nanoTime();
System.out.println("Time for first filter solution: " + (after2 - before2));


long before = System.nanoTime();
List<Car> intersect = union.stream().collect(inter);
long after = System.nanoTime();
System.out.println("Time for collector solution: " + (after - before));

Time for first filter solution: 540906

Time for collector solution: 37543

Answer 4

the lambda solution with groupingBy:
the map values with the true-key are in both lists
the map values with the false-key are disjoint

Map<Boolean,List<Car>> map = Stream.concat(bigCarList.stream(),
    smallCarList.stream()).collect(
        groupingBy( b -> bigCarList.stream().anyMatch( s -> b.equals( s ) )
            && smallCarList.stream().anyMatch( s -> b.equals( s ) ) ) );
List<Car> disjoint = map.get( false );  // [Toyota Corolla, Ford Focus]

same principle but shorter w/o inline streams:

Map<Boolean,List<Car>> map = Stream.concat(bigCarList.stream(),
    smallCarList.stream()).collect(
        groupingBy( b -> bigCarList.contains( b )
            && smallCarList.contains( b ) ) );
List<Car> disjoint = map.get( false );  // [Toyota Corolla, Ford Focus]

both are working with duplicates as well
means: duplicates in one list that are not contained in the other list
If the amount of data is not so huge that you are running into disk space issues, a simple groupingBy ‑ without filtering or additional queries to reduce the result set ‑ should be the clearest and fastest solution.

Answer 5

OP is asking for the symmetric difference. And the symmetric difference can be expressed as:

1. Either the difference between the union and the intersection:

   A △ B = (A ∪ B) - (B ∩ A)

2. Or the union of the differences:

   A △ B = (A – B) ∪ (B – A)

The first part of this answer achieves it by #2, while the second part achieves it by #1. Here I'll show a variation of approach #1:

List<Car> result = new ArrayList<>(bigCarList);
result.addAll(smallCarList); // (A ∪ B)

result.removeIf(c -> bigCarList.contains(c) && smallCarList.contains(c)); // (B ∩ A)

This can be optimized if lists are converted to sets, so that using contains is O(1):

List<Car> bigCarList = get5DefaultCars();
List<Car> smallCarList = get3DefaultCars();

Set<Car> bigCarSet = new HashSet<>(bigCarList);
Set<Car> smallCarSet = new HashSet<>(smallCarList);

Set<Car> result = new LinkedHashSet<>(bigCarList);
result.addAll(smallCarList); // (A ∪ B)

result.removeIf(c -> bigCarSet.contains(c) && smallCarSet.contains(c)); // (B ∩ A)

Answer 6

An alternative approach, albeit not as elegant as one line streams:

    HashMap<Integer, Boolean> y = new HashMap<>();
    bigCarSet ().forEach(i -> y.put(i, !y.containsKey(i)));
    bigCarList().forEach(i -> y.put(i, !y.containsKey(i)));
    y.entrySet().stream().filter(Map.Entry::getValue).map(Map.Entry::getKey)
     .collect(Collectors.toList());

which can be simplified to at least:

    HashMap<Integer, Boolean> y = new HashMap<>();
    Stream.concat(list1.stream(), list2.stream()).forEach(i -> y.put(i, !y.containsKey(i)));
    y.entrySet().stream().filter(Map.Entry::getValue)
                 .map(Map.Entry::getKey).collect(Collectors.toList());