Java 8 lambda filtering based on condition as well as order

Question

I was trying to filter a list based on multiple conditions, sorting.

class Student{
        private int Age;
        private String className;
        privat         


        

Answer 1

If you need a grouping only sorted, it is quite simple:

Map<String, List<Student>> collect = students.stream() // stream capabilities
        .sorted(Comparator.comparingInt(Student::getAge).reversed()) // sort by age, descending
        .collect(Collectors.groupingBy(Student::getName)); // group by name.

Output in collect:

• Prince=[Student [Age=24, className=B, Name=Prince]],
• Smith=[Student [Age=24, className=A, Name=Smith]],
• John=[Student [Age=30, className=A, Name=John], Student [Age=24, className=A, Name=John], Student [Age=20, className=B, Name=John]]

Answer 2

Use the toMap collector:

Collection<Student> values = students.stream()
                .collect(toMap(Student::getName,
                        Function.identity(),
                        BinaryOperator.maxBy(Comparator.comparingInt(Student::getAge))))
                .values();

Explanation

We're using this overload of toMap:

toMap​(Function<? super T,? extends K> keyMapper,
      Function<? super T,? extends U> valueMapper,
      BinaryOperator<U> mergeFunction)

• Student::getName above is the keyMapper function used to extract the values for the map keys.
• Function.identity() above is the valueMapper function used to extract the values for the map values where Function.identity() simply returns the elements in the source them selves i.e. the Student objects.
• BinaryOperator.maxBy(Comparator.comparingInt(Student::getAge)) above is the merge function used to "decide which Student object to return in the case of a key collission i.e. when two given students have the same name" in this case taking the oldest Student .
• Finally, invoking values() returns us a collection of students.

The equivalent C# code being:

var values = students.GroupBy(s => s.Name, v => v,
                          (a, b) => b.OrderByDescending(e => e.Age).Take(1))
                      .SelectMany(x => x);

Explanation (for those unfamiliar with .NET)

We're using this extension method of GroupBy:

System.Collections.Generic.IEnumerable<TResult> GroupBy<TSource,TKey,TElement,TResult> 
       (this System.Collections.Generic.IEnumerable<TSource> source, 
         Func<TSource,TKey> keySelector, 
         Func<TSource,TElement> elementSelector, 
     Func<TKey,System.Collections.Generic.IEnumerable<TElement>,TResult> resultSelector);

• s => s.Name above is the keySelector function used to extract the value to group by.
• v => v above is the elementSelector function used to extract the values i.e. the Student objects them selves.
• b.OrderByDescending(e => e.Age).Take(1) above is the resultSelector which given an IEnumerable<Student> represented as b takes the oldest student.
• Finally, we apply .SelectMany(x => x); to collapse the resulting IEnumerable<IEnumerable<Student>> into a IEnumerable<Student>.

Answer 3

Or without streams:

Map<String, Student> map = new HashMap<>();
students.forEach(x -> map.merge(x.getName(), x, (oldV, newV) -> oldV.getAge() > newV.getAge() ? oldV : newV));
Collection<Student> max = map.values();

Answer 4

Just to mix and merge the other solutions, you could alternatively do :

Map<String, Student> nameToStudentMap = new HashMap<>();
Set<Student> finalListOfStudents = students.stream()
        .map(x -> nameToStudentMap.merge(x.getName(), x, (a, b) -> a.getAge() > b.getAge() ? a : b))
        .collect(Collectors.toSet());