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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()));
866
In Java 8 Stream, filter with Set. Add() is the fastest algorithm to find duplicate elements, because it loops only one time. Set<T> items = new HashSet<>(); return list. stream() .
Get the stream of elements in which the duplicates are to be found. For each element in the stream, count the frequency of each element, using Collections. frequency() method. Then for each element in the collection list, if the frequency of any element is more than one, then this element is a duplicate element.
To remove the duplicates from the arraylist, we can use the java 8 stream api as well. Use steam's distinct() method which returns a stream consisting of the distinct elements comparing by object's equals() method. Collect all district elements as List using Collectors. toList() .
With Eclipse Collections (formerly GS Collections), you can make use of a data structure called Bag that can hold the number of occurrences of each element. Using IntBag , the following will work: MutableList<Person> personsEC = ListAdapter. adapt(persons); IntBag intBag = personsEC.
Most repeated char and its value is retrieved from getKey () and getValue () from Pair instance. 3. Java - Find Most Repeated Character In String Using ASCII sized Array
Java Find duplicate objects in list using Stream Group by In Java Stream perform group by operation based on that we can find duplicate object from collection or list. java.util.List<String> list = Arrays.asList("A", "B", "B", "C", "D", "D", "Z", "E", "E");
Second, there's at least one repetition of a substring. This is best illustrated with some examples by checking out a few repeated substrings: And a few non-repeated ones:
Input string : hello world l is the most repeated character for 3 times. The above program compiles and run without any errors. But this is lengthy program using HashMap. Here, we have captured the most repeated character and its count in Pair object. Most repeated char and its value is retrieved from getKey () and getValue () from Pair instance.
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());
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);
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