Find first element by predicate

JavaJava 8Java Stream

Java Problem Overview


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?

Java Solutions


Solution 1 - Java

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.

Solution 2 - Java

> 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.

Solution 3 - Java

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.

Solution 4 - Java

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.

Solution 5 - Java

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)

Solution 6 - Java


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);
        }
    }
}

Solution 7 - Java

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();

Solution 8 - Java

A generic utility function with looping seems a lot cleaner to me:

static public <T> T find(List<T> elements, Predicate<T> p) {
    for (T item : elements) if (p.test(item)) return item;
    return null;
}

static public <T> T find(T[] elements, Predicate<T> p) {
    for (T item : elements) if (p.test(item)) return item;
    return null;
}

In use:

List<Integer> intList = Arrays.asList(1, 2, 3, 4, 5);
Integer[] intArr = new Integer[]{1, 2, 3, 4, 5};

System.out.println(find(intList, i -> i % 2 == 0)); // 2
System.out.println(find(intArr, i -> i % 2 != 0)); // 1
System.out.println(find(intList, i -> i > 5)); // null

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionsikiView Question on Stackoverflow
Solution 1 - JavaAlexis C.View Answer on Stackoverflow
Solution 2 - Javawha'eve'View Answer on Stackoverflow
Solution 3 - JavaCodeShadowView Answer on Stackoverflow
Solution 4 - JavaIfesinachi BryanView Answer on Stackoverflow
Solution 5 - JavaGrigory KislinView Answer on Stackoverflow
Solution 6 - JavaaillusionsView Answer on Stackoverflow
Solution 7 - Javashreedhar bhatView Answer on Stackoverflow
Solution 8 - Javalawrence-wittView Answer on Stackoverflow