Java 8流减少了保留最新条目的删除重复项

Question

I have a Java bean, like

class EmployeeContract {
    Long id;
    Date date;
    getter/setter
}

If a have a long list of these, in which we have duplicates by id but with different date, such as:

1, 2015/07/07
1, 2018/07/08
2, 2015/07/08
2, 2018/07/09

How can I reduce such a list keeping only the entries with the most recent date, such as:

1, 2018/07/08
2, 2018/07/09

? Preferably using Java 8...

I've started with something like:

contract.stream()
         .collect(Collectors.groupingBy(EmployeeContract::getId, Collectors.mapping(EmployeeContract::getId, Collectors.toList())))
                    .entrySet().stream().findFirst();

That gives me the mapping within individual groups, but I'm stuck as to how to collect that into a result list - my streams are not too strong I'm afraid...

Answer 1

Well, I am just going to put my comment here in the shape of an answer:

 yourList.stream()
         .collect(Collectors.toMap(
                  EmployeeContract::getId,
                  Function.identity(),
                  BinaryOperator.maxBy(Comparator.comparing(EmployeeContract::getDate)))
            )
         .values();

This will give you a Collection instead of a List , if you really care about this.

Answer 2

You can do it in two steps as follows :

List<EmployeeContract> finalContract = contract.stream() // Stream<EmployeeContract>
        .collect(Collectors.toMap(EmployeeContract::getId, 
                EmployeeContract::getDate, (a, b) -> a.after(b) ? a : b)) // Map<Long, Date> (Step 1)
        .entrySet().stream() // Stream<Entry<Long, Date>>
        .map(a -> new EmployeeContract(a.getKey(), a.getValue())) // Stream<EmployeeContract>
        .collect(Collectors.toList()); // Step 2

First step : ensures the comparison of date s with the most recent one mapped to an id .

Second step : maps these key, value pairs to a final List<EmployeeContract> as a result.

Answer 3

With vavr.io you can do it like this:

var finalContract = Stream.ofAll(contract) //create io.vavr.collection.Stream
            .groupBy(EmployeeContract::getId)
            .map(tuple -> tuple._2.maxBy(EmployeeContract::getDate))
            .collect(Collectors.toList()); //result is list from java.util package

Answer 4

Just to complement the existing answers, as you're asking:

how to collect that into a result list

Here are some options:

• Wrap the values() into an ArrayList :

   List<EmployeeContract> list1 = new ArrayList<>(list.stream() .collect(toMap(EmployeeContract::getId, identity(), maxBy(comparing(EmployeeContract::getDate)))) .values()); 

• Wrap the toMap collector into collectingAndThen :

   List<EmployeeContract> list2 = list.stream() .collect(collectingAndThen(toMap(EmployeeContract::getId, identity(), maxBy(comparing(EmployeeContract::getDate))), c -> new ArrayList<>(c.values()))); 

• Collect the values to a new List using another stream:

   List<EmployeeContract> list3 = list.stream() .collect(toMap(EmployeeContract::getId, identity(), maxBy(comparing(EmployeeContract::getDate)))) .values() .stream() .collect(toList());