Group by field name in Java

Question

I'm trying to group Java objects by their field, i.e Person.java

public class Person {
    String name;
    String surname;
    ....
}

So if I have n Person objects, what would be the easiest way to get all people name "David" into a map like Map<String, List<Person>> map;?

I found this on Google (but it doesn't compile), it seems to be the thing I'm looking for: http://www.anzaan.com/2010/06/grouping-objects-using-objects-property/

what does the compile error say? – MBU Mar 21 '11 at 23:50 — MBU, Mar 21 '11 at 23:50

score 58 · Accepted Answer · answered Mar 21 '11 at 23:54

58

There's probably a library that can do this more simply, but it's not too hard to do it manually:

List<Person> allPeople; // your list of all people
Map<String, List<Person>> map = new HashMap<String, List<Person>>();
for (Person person : allPeople) {
   String key = person.getName();
   if (map.get(key) == null) {
      map.put(key, new ArrayList<Person>());
   }
   map.get(key).add(person);
}

List<Person> davids = map.get("David");

answered Mar 21 '11 at 23:54

Kaleb Brasee

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But the question was about a general solution, where you just specify the name of a class, so you would pass a persons:=List of Person, and "name" as the name of the field, and the function would do the rest. In the next case you pass a list of cars, and a fieldname "manufactor". In scala, you would just call persons.filter (p => p.name == "David") or group on the list. – user unknown Mar 22 '11 at 01:27
5

poor Java people... they will never know the joys of LINQ :) – Ilia G Jul 25 '12 at 16:58
2

Unless they use Groovy (or Scala) and do the following with closures, which I assume is as simple as LINQ. `def davids = allPeople.findAll { p -> p.name == 'David' }` – Kaleb Brasee Jul 25 '12 at 19:24
6

@lliaG, with Java 8 now they [can](http://zeroturnaround.com/rebellabs/java-8-explained-applying-lambdas-to-java-collections/) :) – Jason Sperske May 26 '14 at 22:17

Alexis C. · Answer 2 · 2016-07-19T17:47:11.670

28

Using java-8 with the Collectors class and streams, you can do this:

Map<String, List<Person>> mapByName = 
    allPeople.stream().collect(Collectors.groupingBy(Person::getName));

List<Person> allDavids = mapByName.getOrDefault("David", Collections.emptyList());

Here I used getOrDefault so that you get an empty immutable list instead of a null reference if there is no "David" in the original list, but you can use get if you prefer to have a null value.

Hope it helps! :)

edited Jul 19 '16 at 17:47

answered Jan 04 '14 at 21:35

Alexis C.

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Is it possible not to use a getter, but directly use the field name? – Meng Zhang Nov 12 '15 at 12:13
1

@ZhangMeng Yes, in this case you would use the lambda `p -> p.name`, assuming you can access the `name` field. – Alexis C. Nov 12 '15 at 13:14

score 11 · Answer 3 · answered Apr 19 '13 at 14:21

11

Google Guava's Multimap does exactly what you need and much more, avoiding much boilerplate code:

ListMultimap<String, Person> peopleByFirstName = ArrayListMultimap.create();
for (Person person : getAllPeople()) {
    peopleByFirstName.put(person.getName(), person);
}

Source: http://code.google.com/p/guava-libraries/wiki/NewCollectionTypesExplained#Multimap

answered Apr 19 '13 at 14:21

GFonte

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10

1

A really helpful fact would be to tell people to call `asMap()` on the result. `keys()`, `entries()` and `values()` does not seem to group as you'd expect it to. – Daniel Kaplan Jul 29 '13 at 23:31

Nazarii Bardiuk · Answer 4 · 2011-03-22T00:07:29.193

4

something like that (i didn't compile)

void addPerson(Person p, Map<String, List<Person>> map){
  ArrayList<Person> lst = map.get(p.name);
  if(lst == null){
     lst = new ArrayList<Person>();
  }
  lst.add(p);
  map.put(p.name, lst);
}
...
for(Person p:personsCollection>){
   addPerson(p, map);
}

edited Mar 22 '11 at 00:07

answered Mar 21 '11 at 23:55

Nazarii Bardiuk

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score 1 · Answer 5 · answered Mar 25 '16 at 21:59

You can use Multimap and groupBy() from Eclipse Collections

MutableList<Person> allPeople = Lists.mutable.empty();
MutableListMultimap<String, Person> multimapByName = allPeople.groupBy(Person::getName);

If you can't change the type of allPeople from List

List<Person> allPeople = new ArrayList<>();
MutableListMultimap<String, Person> multimapByName = 
     ListAdapter.adapt(allPeople).groupBy(Person::getName);

Note: I am a contributor to Eclipse Collections.

score 1 · Answer 6 · answered Mar 22 '11 at 01:40

In Scala, this is a feature of the class List already:

class Person (val name: String, val surname: String ="Smith") 
val li = List (new Person ("David"), new Person ("Joe"), new Person ("Sue"), new Person ("David", "Miller")) 
li.groupBy (_.name)

res87: scala.collection.immutable.Map[String,List[Person]] = Map((David,List(Person@1c3f810, Person@139ba37)), (Sue,List(Person@11471c6)), (Joe,List(Person@d320e4)))

Since Scala is bytecode compatible to Java, you should be able to call that method from Java, if you include the scala-jars.

score -1 · Answer 7 · edited Jun 20 '20 at 09:12

-1

First of all you should add objects of Person.java in the arraylist later we can get back those details in below manner

public void addPerson(){

ArrayList presonList = new ArrayList();

for( int i=0;i<=presonList.size() ; i++){

Person obj = presonList.get(i);

//now do the stuff of getting individual variable here

}

}

edited Jun 20 '20 at 09:12

Community

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answered Mar 22 '11 at 00:19

developer

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Group by field name in Java

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