724

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?

4
  • 81
    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 Commented May 16, 2014 at 13:35
  • 2
    Cool. That's what I hoped it'd do. It would've been a major design flop otherwise. Commented May 16, 2014 at 13:52
  • 2
    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. Commented Apr 20, 2016 at 9:01
  • 2
    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. Commented Sep 22, 2016 at 14:07

9 Answers 9

951

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.

Sign up to request clarification or add additional context in comments.

7 Comments

As a note, I used get(); here because I know which values I feed to the stream pipeline and hence that there will be a result. In practice, you should not use get();, but orElse() / orElseGet() / orElseThrow() (for a more meaningful error instead of a NSEE) as you might not know if the operations applied to the stream pipeline will result in an element.
.findFirst().orElse(null); for example
Don't use orElse null. That should be an anti-pattern. It is all included in the Optional so why should you risk a NPE? I think dealing with Optional is the better way. Just test the Optional with isPresent() before you use it.
@BeJay i don't understand. what should I use instead of orElse ?
@JohnHenckel I think what BeJay means is that you should leave it as an Optional type, which is what .findFirst returns. One of the uses of Optional is to help developers avoid having to deal with nulls. e.g. instead of checking myObject != null, you can check myOptional.isPresent(), or use other parts of the Optional interface. Did that make it clearer?
|
132

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.

1 Comment

This answer was more informative to me, and explains the why, not only how. I never new intermediate operations are always lazy; Java streams continue to surprise me.
67
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.

5 Comments

BETTER: since we looking for a boolean return value, we can do it better by adding null check: return dataSource.getParkingLots().stream().filter(parkingLot -> Objects.equals(parkingLot.getId(), id)).findFirst().orElse(null) != null;
@shreedharbhat You should not need to do .orElse(null) != null. Instead, make use of the Optional API's .isPresent i.e. .findFirst().isPresent().
@shreedharbhat first of all OP was not looking for a boolean return value. Second of all if they were, it would be been cleaner to write .stream().map(ParkingLot::getId).anyMatch(Predicate.isEqual(id))
This one is good and works well. You can get a whole object within a list however, I would like to make a corrections. You wrote equals but it is equal so as follows: .filter(parkingLot -> Objects.equal(parkingLot.getId(), id))
flawless answer!
23

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.

3 Comments

You should fix your example. You can not assign null to a plain int. stackoverflow.com/questions/2254435/can-an-int-be-null-in-java
I've edited your post. 0 (zero) may be a valid result when you are searching in a list of integers. Replaced variable type by Integer and default value by null.
flawless answer!
18

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) 

Comments

5
 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); } } } 

1 Comment

That's the reason I refrain from using streams for simple tasks. It is usually wayyy slower than using simple iteration. (It is even worse if you use array operations. But who does that, anyways... )
1

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 

Comments

0

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(); 

1 Comment

If you want a boolean return value you should use anyMatch
0

@BeJay is right, it's a bad practice to use null values in that context. NPE should only appear in the context of a code error like any forgotten or failed instantiation, null cannot be use to hold a meaningful value ( data not found ). It's a bit like the benefits of banning the use of magic numbers.

public static void main(String[] args) { List<Integer> List1 = List.of( 3, 14, 15, 9, 26); List<Integer> List2 = List.of( 5, 16, 3, 2, 1); // intend to be use as a stack of "to scan" Lists ArrayList<List<Integer>> parentList = new ArrayList<>( List.of(List1,List2)); Optional<Integer> multipleOfFour = Optional.empty(); //looping while parentList still contains sub-list(s) and no multiple of 4 was found while ( ! parentList.isEmpty() && multipleOfFour.isEmpty()) { //popping out first element of parentList List<Integer> currentList = parentList.remove(0); multipleOfFour = currentList.stream().filter(num->num%4 == 0).findFirst(); } if ( ! multipleOfFour.isPresent()) { //cheating in case no multiple of 4 was found in any of the 2 lists multipleOfFour = Optional.of(Integer.valueOf(4)); } // never use get() unless absolutely sure isPresent() // would return true or an exception could be thrown System.out.println("first multiple of 4 found: " + multipleOfFour.get()); } 

Now if I didn't instantiate multipleOfFour with Optional.empty() when declared, a NPE will occur when attempting to call isEmpty() in the while loop condition, and I'll know for sure the error comes from the code itself.

On the other hand if multipleOfFour.isPresent() doesn't return true when I expect, I'll know for sure the error comes either from the data provided or from the stream lambda that doesn't operate properly.

Easier to debug when you know where to look...

That Opened my eyes, I rewrote the method of recursive search I was working on, using Optional.empty() instead of null. Thank you @BeJay

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.