Select some objects from a list of objects based on list of one field

Question

I have an arraylist of objects which has a field "name" and I have a list of names. I have to select all the objects which consist of any of the name from the list. How can I achieve this in java most efficiently. Any code snippet will be highly appreciated.

Example:

Let the class name be A

A consists of fields such as id, name, subject etc.

So let's assume I have a list
List<A> objectsA
whose length let's say is 10. In this list, the name field consists of various names for eg alpha, beta, gama, delta etc etc.

So suppose I have a second list

List<String> namesList = List.of("alpha","beta");

So I want to select only those objects from objectsA list whose name fields consist of the names from namesList i.e. I want to select only those objects whose name fields consists of "alpha" or "beta".

English is a terrible way to write Java code. Post some code. Show us something. Because I want to help you. But I have no idea what you are asking. — Elliott Frisch
– Elliott Frisch, Commented Sep 5, 2021 at 10:49
You're welcome. You seem to be a beginner on StackOverflow, we have to give you some time to learn. — gouessej
– gouessej, Commented Sep 5, 2021 at 11:16
By the way, List.of is available since Java 9: docs.oracle.com/en/java/javase/11/docs/api/java.base/java/util/… — gouessej
– gouessej, Commented Sep 5, 2021 at 11:20

gouessej · Accepted Answer · 2021-09-05 14:33:51Z

2

final ArrayList<MyNamedObject> namedObjectList = ...;
final List<String> nameList = ...;

Use

namedObjectList.stream()
   .filter((final MyNamedObject o) -> nameList.stream().anyMatch((final String name) -> Objects.equals(name, o.getName())))
   .collect(Collectors.toList());

If there are some duplicate values in the name list, either keep the list but use .distinct():

namedObjectList.stream()
    .filter((final MyNamedObject o) -> nameList.stream().distinct().anyMatch((final String name) -> Objects.equals(name, o.getName())))
    .collect(Collectors.toList());

or create a Set:

final Set<String> nameSet = nameList.stream()
    .distinct()
    .collect(Collectors.toSet());
namedObjectList.stream()
    .filter((final MyNamedObject o) -> nameSet.stream().anyMatch((final String name) -> Objects.equals(name, o.getName())))
    .collect(Collectors.toList());

If nameList or nameSet can contain null, use .filter(Objects::nonNull). If neither name nor o.getName() can be null, replace Objects.equals(name, o.getName()) by name.equals(o.getName()).

edited Sep 5, 2021 at 14:33

answered Sep 5, 2021 at 10:57

gouessej

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8 Comments

Ishan Arora Over a year ago

thanks. I was looking for this, by the way what is time complexity of this code?

gouessej Over a year ago

I don't know but I don't see how to suggest something more efficient in your case.

gouessej Over a year ago

If you're sure that there is no null values, you can simplify a very little bit my example.

Ishan Arora Over a year ago

okay. I think as @Alberto Sinigaglia using set can be more efficient

gouessej Over a year ago

If o.getName() can return null, it will be enough. If nameList can contain null values, just add .filter(Objects::nonNull).

|

Alberto · Accepted Answer · 2021-09-05 10:58:54Z

2

If you have a List of names, then it's probably convenient to transform it to a Set so that you can easily check if it contains a name or not:

final Set<String> names = new HashSet<>(namesList);
List<Person> people = ...;
List<Person> result = people.stream()
   .filter(p -> names.contains(p.getName()))
   .collect(Collectors.toList());

You can avoid using the Set and just use the namesList, because there is a List.contains method, however it would be fairly inefficient

edited Sep 5, 2021 at 10:58

answered Sep 5, 2021 at 10:56

Alberto

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4 Comments

gouessej Over a year ago

You expect a list, rather use Collectors.toList().

Alberto Over a year ago

@gouessej I don't understand

gouessej Over a year ago

You fixed it several minutes ago, you replaced Collectors.toString() by Collectors.toList(), it's ok now.

Alberto Over a year ago

@gouessej oh ok, yes sorry, it was a typo

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