lambda -- Filter Java Stream to 1 and only 1 element

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I am trying to use Java 8 Streams to find elements in a LinkedList. I want to guarantee, however, that there is 1 and only 1 match to the filter criteria.

Take this code:

public static void main(String[] args) {    LinkedList<User> users = new LinkedList<>();
    users.add(new User(1, "User1"));
    users.add(new User(2, "User2"));
    users.add(new User(3, "User3"));    User match = users.stream().filter((user) -> user.getId() == 1).findAny().get();
    System.out.println(match.toString());
}static class User {    @Override
    public String toString() {
        return id + " - " + username;
    }    int id;
    String username;    public User() {
    }    public User(int id, String username) {
        this.id = id;
        this.username = username;
    }    public void setUsername(String username) {
        this.username = username;
    }    public void setId(int id) {
        this.id = id;
    }    public String getUsername() {
        return username;
    }    public int getId() {
        return id;
    }
}

This code finds a User based on their ID. But there are no guarantees how many Users matched the filter.

Changing the filter line to:

User match = users.stream().filter((user) -> user.getId() < 0).findAny().get();

Will throw a NoSuchElementException (good!)

I would like it to throw an error if there are multiple matches, though. Is there a way to do this?

java lambda stream java-8

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edited Mar 27 at 17:40

asked Mar 27 at 17:25

lambda — Filter Java Stream to 1 and only 1 element

ryvantage

2,58921335

count() is a terminal operation so you can’t do that. The stream can’t be used after. – ZouZou Mar 27 at 17:42

Ok, thanks @ZouZou. I wasn’t entirely certain what that method did. Why is there noStream::size ? – ryvantage Mar 27 at 17:44

@ryvantage Because a stream can only be used once: calculating its size means "iterating" over it and after that you can’t use the stream any longer. – assylias Mar 27 at 17:45

Wow. That one comment helped me understand Streams so much more than I did before… – ryvantage Mar 27 at 17:50

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edited Mar 27 at 20:20

answered Mar 27 at 17:33

skiwi

9,37332767

@ryvantage I updated my answer with how I would do it with writing the own custom Collector, which I believe is the correct method. – skiwi Mar 27 at 18:03

what do you think of my reduction? – assylias Mar 27 at 18:32

@assylias I commented on your answer, I think this one i smore concise and more straight forward though and less error prone. – skiwi Mar 27 at 18:38

The difference main between the two collectors is that the first will stop as soon as it finds a second matching item whereas the second will iterate through the whole list. – assylias Mar 27 at 22:57

Louis Wasserman’s, +1), but if you want brevity, I’d suggest the following:

List<User> result = users.stream()
    .filter(user -> user.getId() == 1)
    .limit(2)
    .collect(Collectors.toList());

Then verify the size of the result list.

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answered Mar 28 at 15:39

Stuart Marks

12.5k11850

What’s the point of limit(2) in this solution? What difference would it make whether the resulting list was 2 or 100? If it’s greater than 1. – ryvantage Mar 28 at 18:31

It stops immediately if it finds a second match. This is what all the fancy collectors do, just using more code. :-) – Stuart Marks Mar 29 at 3:24

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answered Mar 27 at 17:51

Louis Wasserman

71k894155

@skiwi’s singletonCollector was smaller and easier to follow than this, that’s why I gave him the check. But good to see consensus in the answer: a custom Collector was the way to go. – ryvantage Mar 27 at 20:37

Fair enough. I was primarily aiming for speed, not conciseness. – Louis Wasserman Mar 27 at 20:40

Yeah? Why is yours faster? – ryvantage Mar 27 at 20:45

Mostly because allocating an all-up List is more expensive than a single mutable reference. – Louis Wasserman Mar 27 at 20:52

I was unable to get yours to compile – ryvantage Mar 28 at 18:28

share|improve this answer edited Oct 3 at 17:32

David Conrad

3,8651220

answered May 24 at 19:26

Brian Goetz

8,86341733

share|improve this answer edited Mar 27 at 18:31

answered Mar 27 at 17:37

assylias

110k10153289

I would add an identity element (null) to prevent using get(). Sadly your reduce is not working as you think it does, consider a Stream that has null elements in it, maybe you think that you covered it, but I can be [User#1, null, User#2, null, User#3], now it will not throw an exception I think, unless I’m mistaken here. – skiwi Mar 27 at 18:36

@Skiwi if there are null elements the filter will throw a NPE first. – assylias Mar 27 at 18:37

share|improve this answer answered Mar 28 at 7:03