Find first element by predicate

574
votes

I've just started playing with Java 8 lambdas and I'm trying to implement some of the things that I'm used to in functional languages.

For example, most functional languages have some kind of find function that operates on sequences, or lists that returns the first element, for which the predicate is true. The only way I can see to achieve this in Java 8 is:

lst.stream()
    .filter(x -> x > 5)
    .findFirst()

However this seems inefficient to me, as the filter will scan the whole list, at least to my understanding (which could be wrong). Is there a better way?

7
javajava-8java-stream
It's not inefficient, Java 8 Stream implementation is lazy evaluated, so filter is applied only to terminal operation. Same question here: stackoverflow.com/questions/21219667/stream-and-lazy-evaluation – Marek Gregor
Cool. That's what I hoped it'd do. It would've been a major design flop otherwise. – siki
If your intention is really to check whether the list contains such an element at all (not single out the first of possibly several), .findAny() can theoretically be more efficient in a parallell setting, and of course communicates that intent more clearly. – Joachim Lous
Compared to a simple forEach cycle, this would create lots of objects on the heap and dozens of dynamic method calls. While this might not always affect the bottom line in your performance tests, in the hot spots it makes a difference to abstain from the trivial use of Stream and similar heavyweight constructs. – Agoston Horvath

7 Answers

802
votes

No, filter does not scan the whole stream. It's an intermediate operation, which returns a lazy stream (actually all intermediate operations return a lazy stream). To convince you, you can simply do the following test:

List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);
int a = list.stream()
            .peek(num -> System.out.println("will filter " + num))
            .filter(x -> x > 5)
            .findFirst()
            .get();
System.out.println(a);

Which outputs:

will filter 1
will filter 10
10

You see that only the two first elements of the stream are actually processed.

So you can go with your approach which is perfectly fine.

113
votes

However this seems inefficient to me, as the filter will scan the whole list

No it won't - it will "break" as soon as the first element satisfying the predicate is found. You can read more about laziness in the stream package javadoc, in particular (emphasis mine):

Many stream operations, such as filtering, mapping, or duplicate removal, can be implemented lazily, exposing opportunities for optimization. For example, "find the first String with three consecutive vowels" need not examine all the input strings. Stream operations are divided into intermediate (Stream-producing) operations and terminal (value- or side-effect-producing) operations. Intermediate operations are always lazy.

43
votes
return dataSource.getParkingLots()
                 .stream()
                 .filter(parkingLot -> Objects.equals(parkingLot.getId(), id))
                 .findFirst()
                 .orElse(null);

I had to filter out only one object from a list of objects. So i used this, hope it helps.

18
votes

In addition to Alexis C's answer, If you are working with an array list, in which you are not sure whether the element you are searching for exists, use this.

Integer a = list.stream()
                .peek(num -> System.out.println("will filter " + num))
                .filter(x -> x > 5)
                .findFirst()
                .orElse(null);

Then you could simply check whether a is null.

4
votes

Already answered by @AjaxLeung, but in comments and hard to find.
For check only

lst.stream()
    .filter(x -> x > 5)
    .findFirst()
    .isPresent()

is simplified to

lst.stream()
    .anyMatch(x -> x > 5)
0
votes

import org.junit.Test;

import java.util.Arrays;
import java.util.List;
import java.util.Optional;

// Stream is ~30 times slower for same operation...
public class StreamPerfTest {

    int iterations = 100;
    List<Integer> list = Arrays.asList(1, 10, 3, 7, 5);


    // 55 ms
    @Test
    public void stream() {

        for (int i = 0; i < iterations; i++) {
            Optional<Integer> result = list.stream()
                    .filter(x -> x > 5)
                    .findFirst();

            System.out.println(result.orElse(null));
        }
    }

    // 2 ms
    @Test
    public void loop() {

        for (int i = 0; i < iterations; i++) {
            Integer result = null;
            for (Integer walk : list) {
                if (walk > 5) {
                    result = walk;
                    break;
                }
            }
            System.out.println(result);
        }
    }
}
0
votes

Improved One-Liner answer: If you are looking for a boolean return value, we can do it better by adding isPresent: 

return dataSource.getParkingLots().stream().filter(parkingLot -> Objects.equals(parkingLot.getId(), id)).findFirst().isPresent();