Bdtyktl

I know below is the code to find out the occurrence of each String attributes in list , how can I filter this list with only duplicates item i.e having more than 1 occurrence. Sorry I am new to java 8 .

Map<String, Long> result = list.stream()

            .collect(Collectors.groupingBy(Function.identity(),

                                              Collectors.counting()));

create a stream from the entrySet and filter:

List<Map.Entry<String, Long>> result =  list.stream()

                   .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))

                   .entrySet()

                   .stream()

                   .filter(s -> s.getValue() >= 2)

                   .collect(Collectors.toList());

or if you want to maintain a map then:

Map<String, Long> result = stringList().stream()

                .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))

                .entrySet()

                .stream()

                .filter(s -> s.getValue() >= 2)

                .collect(toMap(Map.Entry::getKey, Map.Entry::getValue));

on another note, if you just want the individual numbers that have more than or equal to 2 occurrences then you can do:

List<String> result = list.stream()

                .collect(Collectors.groupingBy(Function.identity(),

                        Collectors.counting()))

                .entrySet()

                .stream()

                .filter(x -> x.getValue() >= 2)

                .map(Map.Entry::getKey)

                .collect(toList());

another option being:

List<String> result = 

         list.stream()

             .filter(x -> list.stream().filter(x::equals).limit(2).count() == 2)

             .distinct()

             .collect(toList());

If your List is mutable, you can directly remove all elements except their second occurrence:

// example list

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

Collections.addAll(example, "foo", "bar", "baz", "bar", "bar", "baz");



// actual operation

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

example.removeIf(s -> temp.merge(s, 1, Integer::sum)!=2);



// example output

example.forEach(System.out::println);// prints bar baz

The solution above keeps only one copy for each string having multiple occurrences while removing all strings having no duplicates. If you want to keep all duplicates and just remove those string not having duplicates, there is no way around determining the duplicate status first.

// same example input as above



// actual operation

Map<String,Boolean> temp = new HashMap<>();

example.forEach(s -> temp.merge(s, true, (a,b) -> false));

example.removeIf(temp::get);



// example output

example.forEach(System.out::println);// prints bar baz bar bar baz

Here, the temporary map can be created with a Stream operation with the same logic:

Map<String,Boolean> temp = example.stream()

    .collect(Collectors.toMap(Function.identity(), s -> true, (a,b) -> false));

example.removeIf(temp::get);

The other way would be like this. after groupBy then remove entry with value=1;

result = list.stream()

            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

result.values().removeIf(v->v.intValue() == 1);

A simpler way to find that out could be

List<String> recurringItems = list.stream()

        .filter(item -> list.lastIndexOf(item) != list.indexOf(item))

        .collect(Collectors.toList());

Since for items occurring more than once, lastIndex wouldn't be equal to the first index.

Alternatively, you can use Collectors.toSet() to ensure the items are listed only once in case you are not interested in their order of recurrence.

Set<String> recurringItemsOnce = list.stream()

        .filter(item -> list.lastIndexOf(item) != list.indexOf(item))

        .collect(Collectors.toSet());

Or using Collections.frequency as:

Set<String> recurringItems = list.stream()

        .filter(item -> Collections.frequency(list, item) >= 2)

        .collect(Collectors.toSet());