Java 8 lambda filtering based on condition as well as order











up vote
16
down vote

favorite
2












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



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

public Student(int age, String className, String name) {
Age = age;
this.className = className;
Name = name;
}

public int getAge() {
return Age;
}

public void setAge(int age) {
Age = age;
}

public String getClassName() {
return className;
}

public void setClassName(String className) {
this.className = className;
}

public String getName() {
return Name;
}

public void setName(String name) {
Name = name;
}
}


Now if I have a list of that, say



List<Student> students = new ArrayList<>();
students.add(new Student(24, "A", "Smith"));
students.add(new Student(24, "A", "John"));
students.add(new Student(30, "A", "John"));
students.add(new Student(20, "B", "John"));
students.add(new Student(24, "B", "Prince"));


How would I be able to get a list of the oldest students with a distinct name?
In C# this would be quite simple by using System.Linq GroupBy then comparing and then flattening with select, I'm not too sure how I could achieve the same in Java.










share|improve this question


















  • 7




    students.stream().collect(groupingBy(...)).
    – Andy Turner
    Dec 5 at 14:56










  • Do you only want a list of the oldest and only the oldest ones?
    – michaeak
    Dec 5 at 15:07










  • Please tell what is the desired result. I.e. How the map/list/whatever looks like.
    – michaeak
    Dec 5 at 15:38

















up vote
16
down vote

favorite
2












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



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

public Student(int age, String className, String name) {
Age = age;
this.className = className;
Name = name;
}

public int getAge() {
return Age;
}

public void setAge(int age) {
Age = age;
}

public String getClassName() {
return className;
}

public void setClassName(String className) {
this.className = className;
}

public String getName() {
return Name;
}

public void setName(String name) {
Name = name;
}
}


Now if I have a list of that, say



List<Student> students = new ArrayList<>();
students.add(new Student(24, "A", "Smith"));
students.add(new Student(24, "A", "John"));
students.add(new Student(30, "A", "John"));
students.add(new Student(20, "B", "John"));
students.add(new Student(24, "B", "Prince"));


How would I be able to get a list of the oldest students with a distinct name?
In C# this would be quite simple by using System.Linq GroupBy then comparing and then flattening with select, I'm not too sure how I could achieve the same in Java.










share|improve this question


















  • 7




    students.stream().collect(groupingBy(...)).
    – Andy Turner
    Dec 5 at 14:56










  • Do you only want a list of the oldest and only the oldest ones?
    – michaeak
    Dec 5 at 15:07










  • Please tell what is the desired result. I.e. How the map/list/whatever looks like.
    – michaeak
    Dec 5 at 15:38















up vote
16
down vote

favorite
2









up vote
16
down vote

favorite
2






2





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



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

public Student(int age, String className, String name) {
Age = age;
this.className = className;
Name = name;
}

public int getAge() {
return Age;
}

public void setAge(int age) {
Age = age;
}

public String getClassName() {
return className;
}

public void setClassName(String className) {
this.className = className;
}

public String getName() {
return Name;
}

public void setName(String name) {
Name = name;
}
}


Now if I have a list of that, say



List<Student> students = new ArrayList<>();
students.add(new Student(24, "A", "Smith"));
students.add(new Student(24, "A", "John"));
students.add(new Student(30, "A", "John"));
students.add(new Student(20, "B", "John"));
students.add(new Student(24, "B", "Prince"));


How would I be able to get a list of the oldest students with a distinct name?
In C# this would be quite simple by using System.Linq GroupBy then comparing and then flattening with select, I'm not too sure how I could achieve the same in Java.










share|improve this question













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



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

public Student(int age, String className, String name) {
Age = age;
this.className = className;
Name = name;
}

public int getAge() {
return Age;
}

public void setAge(int age) {
Age = age;
}

public String getClassName() {
return className;
}

public void setClassName(String className) {
this.className = className;
}

public String getName() {
return Name;
}

public void setName(String name) {
Name = name;
}
}


Now if I have a list of that, say



List<Student> students = new ArrayList<>();
students.add(new Student(24, "A", "Smith"));
students.add(new Student(24, "A", "John"));
students.add(new Student(30, "A", "John"));
students.add(new Student(20, "B", "John"));
students.add(new Student(24, "B", "Prince"));


How would I be able to get a list of the oldest students with a distinct name?
In C# this would be quite simple by using System.Linq GroupBy then comparing and then flattening with select, I'm not too sure how I could achieve the same in Java.







java lambda java-8 java-stream






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share|improve this question










asked Dec 5 at 14:55









The 0bserver

17710




17710








  • 7




    students.stream().collect(groupingBy(...)).
    – Andy Turner
    Dec 5 at 14:56










  • Do you only want a list of the oldest and only the oldest ones?
    – michaeak
    Dec 5 at 15:07










  • Please tell what is the desired result. I.e. How the map/list/whatever looks like.
    – michaeak
    Dec 5 at 15:38
















  • 7




    students.stream().collect(groupingBy(...)).
    – Andy Turner
    Dec 5 at 14:56










  • Do you only want a list of the oldest and only the oldest ones?
    – michaeak
    Dec 5 at 15:07










  • Please tell what is the desired result. I.e. How the map/list/whatever looks like.
    – michaeak
    Dec 5 at 15:38










7




7




students.stream().collect(groupingBy(...)).
– Andy Turner
Dec 5 at 14:56




students.stream().collect(groupingBy(...)).
– Andy Turner
Dec 5 at 14:56












Do you only want a list of the oldest and only the oldest ones?
– michaeak
Dec 5 at 15:07




Do you only want a list of the oldest and only the oldest ones?
– michaeak
Dec 5 at 15:07












Please tell what is the desired result. I.e. How the map/list/whatever looks like.
– michaeak
Dec 5 at 15:38






Please tell what is the desired result. I.e. How the map/list/whatever looks like.
– michaeak
Dec 5 at 15:38














4 Answers
4






active

oldest

votes

















up vote
15
down vote













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>.






share|improve this answer























  • Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
    – Luis G.
    Dec 5 at 17:08


















up vote
5
down vote













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();





share|improve this answer























  • Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
    – nullpointer
    Dec 5 at 15:27






  • 3




    @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
    – Federico Peralta Schaffner
    Dec 5 at 15:45




















up vote
1
down vote













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]]






share|improve this answer






























    up vote
    1
    down vote













    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());





    share|improve this answer



















    • 3




      This violates the requirement of statelessness on the parameter of Stream.map.
      – Andy Turner
      Dec 5 at 22:00













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    15
    down vote













    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>.






    share|improve this answer























    • Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
      – Luis G.
      Dec 5 at 17:08















    up vote
    15
    down vote













    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>.






    share|improve this answer























    • Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
      – Luis G.
      Dec 5 at 17:08













    up vote
    15
    down vote










    up vote
    15
    down vote









    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>.






    share|improve this answer














    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>.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 5 at 15:37

























    answered Dec 5 at 14:56









    Aomine

    35.8k62960




    35.8k62960












    • Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
      – Luis G.
      Dec 5 at 17:08


















    • Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
      – Luis G.
      Dec 5 at 17:08
















    Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
    – Luis G.
    Dec 5 at 17:08




    Nice answer. Just wanted to say that using v -> v instead of Function.identity() would make the code more similar to the C# equivalent and so easier to compare.
    – Luis G.
    Dec 5 at 17:08












    up vote
    5
    down vote













    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();





    share|improve this answer























    • Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
      – nullpointer
      Dec 5 at 15:27






    • 3




      @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
      – Federico Peralta Schaffner
      Dec 5 at 15:45

















    up vote
    5
    down vote













    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();





    share|improve this answer























    • Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
      – nullpointer
      Dec 5 at 15:27






    • 3




      @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
      – Federico Peralta Schaffner
      Dec 5 at 15:45















    up vote
    5
    down vote










    up vote
    5
    down vote









    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();





    share|improve this answer














    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();






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 5 at 15:18

























    answered Dec 5 at 15:09









    Eugene

    67.6k997160




    67.6k997160












    • Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
      – nullpointer
      Dec 5 at 15:27






    • 3




      @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
      – Federico Peralta Schaffner
      Dec 5 at 15:45




















    • Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
      – nullpointer
      Dec 5 at 15:27






    • 3




      @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
      – Federico Peralta Schaffner
      Dec 5 at 15:45


















    Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
    – nullpointer
    Dec 5 at 15:27




    Thought of this while starting of from your initial suggestion, but then an additional Map here, but does map.merge save us compared to toMap?
    – nullpointer
    Dec 5 at 15:27




    3




    3




    @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
    – Federico Peralta Schaffner
    Dec 5 at 15:45






    @nullpointer This should have been my answer (I'm supposedly the without streams guy here), so I feel I can answer your question... Actually, Collectors.toMap uses Map.merge under the hood. It even says so in the docs: mergeFunction - a merge function, used to resolve collisions between values associated with the same key, as supplied to Map.merge(Object, Object, BiFunction).
    – Federico Peralta Schaffner
    Dec 5 at 15:45












    up vote
    1
    down vote













    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]]






    share|improve this answer



























      up vote
      1
      down vote













      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]]






      share|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        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]]






        share|improve this answer














        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]]







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Dec 5 at 15:45

























        answered Dec 5 at 14:57









        michaeak

        739315




        739315






















            up vote
            1
            down vote













            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());





            share|improve this answer



















            • 3




              This violates the requirement of statelessness on the parameter of Stream.map.
              – Andy Turner
              Dec 5 at 22:00

















            up vote
            1
            down vote













            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());





            share|improve this answer



















            • 3




              This violates the requirement of statelessness on the parameter of Stream.map.
              – Andy Turner
              Dec 5 at 22:00















            up vote
            1
            down vote










            up vote
            1
            down vote









            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());





            share|improve this answer














            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());






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 5 at 16:17

























            answered Dec 5 at 16:08









            nullpointer

            38.2k1073146




            38.2k1073146








            • 3




              This violates the requirement of statelessness on the parameter of Stream.map.
              – Andy Turner
              Dec 5 at 22:00
















            • 3




              This violates the requirement of statelessness on the parameter of Stream.map.
              – Andy Turner
              Dec 5 at 22:00










            3




            3




            This violates the requirement of statelessness on the parameter of Stream.map.
            – Andy Turner
            Dec 5 at 22:00






            This violates the requirement of statelessness on the parameter of Stream.map.
            – Andy Turner
            Dec 5 at 22:00




















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