Java 8 Lambda filter by Lists

Question

I have two list and i want filter thoose elements which are both list contains. And i want to do this with lambda expression.

Users getName and Clients getUserName both are return with String.

Here is my sample code:

List<Client> clients = new ArrayList<>();
List<User> users = new ArrayList<>();
List<Client> results = new ArrayList<>();

for (Client user : users) {
    for(Client client: clients){
        if(user.getName().equals(client.getUserName())){
            result.add(client);
        }
    }
}

Your code won't compile; a User is not a Client, unless User inherits Client. — fge
– fge, Commented May 3, 2015 at 9:45

JB Nizet · Accepted Answer · 2015-05-03 10:04:21Z

103

Predicate<Client> hasSameNameAsOneUser = 
    c -> users.stream().anyMatch(u -> u.getName().equals(c.getName()));

return clients.stream()
              .filter(hasSameNameAsOneUser)
              .collect(Collectors.toList());

But this is quite inefficient, because it's O(m * n). You'd better create a Set of acceptable names:

Set<String> acceptableNames = 
    users.stream()
         .map(User::getName)
         .collect(Collectors.toSet());

return clients.stream()
              .filter(c -> acceptableNames.contains(c.getName()))
              .collect(Collectors.toList());

Also note that it's not strictly equivalent to the code you have (if it compiled), which adds the same client twice to the list if several users have the same name as the client.

edited May 3, 2015 at 10:04

answered May 3, 2015 at 9:58

JB Nizet

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3 Comments

Jeremy G Over a year ago

I'm curious how the second implementation is more efficient? It looks likei it is still O(m * n) to me. When we do acceptableNames.contains(...) we are traversing that set. Is the idea that the Set will weed out duplicates so it is a slightly better O(m * n)?

JB Nizet Over a year ago

No. Calling contains() on a List does traverse the list. But calling contains on a HashSet just computes the hashCode and traverses the bucket corresponding to that hashCode, which typically contains 0 or 1 element. A HashSet contains lookup is O(1).

Jeremy G Over a year ago

Ah, I understand. Calling .contains() on a list would be O(n), but O(1) on HashSet. Thanks for the clarification!

LeoMandrak · Accepted Answer · 2015-11-23 15:08:41Z

8

Look this:

List<Client> result = clients
    .stream()
    .filter(c -> 
        (users.stream().map(User::getName).collect(Collectors.toList())).contains(c.getName()))
        .collect(Collectors.toList());

answered Nov 23, 2015 at 15:08

LeoMandrak

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1 Comment

Lam Le Over a year ago

users.stream().map(User::getName).collect(Collectors.toList()) It is more efficient if you use Collectors.toSet() for users to distinct user.name

Eswaran Venkatesan · Accepted Answer · 2017-11-11 14:53:04Z

I would like share an example to understand the usage of stream().filter

Code Snippet: Sample program to identify even number.

import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;

public void fetchEvenNumber(){
        List<Integer> numberList = new ArrayList<>();
        numberList.add(10);
        numberList.add(11);
        numberList.add(12);
        numberList.add(13);
        numberList.add(14);
        numberList.add(15);

        List<Integer> evenNumberListObj = numberList.stream().filter(i -> i%2 == 0).collect(Collectors.toList());
        System.out.println(evenNumberListObj);
}

Output will be : [10, 12, 14]

List evenNumberListObj = numberList.stream().filter(i -> i%2 == 0).collect(Collectors.toList());

numberList: it is an ArrayList object contains list of numbers.

java.util.Collection.stream() : stream() will get the stream of collection, which will return the Stream of Integer.

filter: Returns a stream that match the given predicate. i.e based on given condition (i -> i%2 != 0) returns the matching stream.

collect: whatever the stream of Integer filter based in the filter condition, those integer will be put in a list.

daphshez · Accepted Answer · 2015-05-03 10:00:50Z

2

Something like:

clients.stream.filter(c->{
   users.stream.filter(u->u.getName().equals(c.getName()).count()>0
}).collect(Collectors.toList());

This is however not an awfully efficient way to do it. Unless the collections are very small, you will be better of building a set of user names and using that in the condition.

edited May 3, 2015 at 10:00

answered May 3, 2015 at 9:52

daphshez

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2 Comments

Pshemo Over a year ago

We are not comparing objects by their equals method here. Please reread the question.

Pshemo Over a year ago

I see you corrected approach in your answer but there are still minor problems with it. Please don't post answer if you didn't test it first. There is no shame in deleting it to avoid farther downvoting and confusion, correcting it in the shadows and undeleting it when it is correct.

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