Java 8 Streams - Filter More Than One Condition

Question

Based on some sports results data, I have a Fixture object which has getHome() and getAway() method. I'd like to shorten this method which I've written to only use a single lambda function (instead of creating a new list and two lambdas), is this possible?

    private Collection<FixtureResult> finalResults(Team team) {

    List<FixtureResult>finalResults = new ArrayList<>();

    List<FixtureResult> homeResults = resultList.stream().filter(fixture ->
            fixture.getHome().equals(team))
            .collect(toList());

    List<FixtureResult> awayResults = resultList.stream().filter(fixture ->
            fixture.getAway().equals(team))
            .collect(toList());

    finalResults.addAll(homeResults);
    finalResults.addAll(awayResults);

    return finalResults;
}

Thiyagu · Accepted Answer · 2018-03-19 17:04:00Z

22

Simple enough

resultList.stream()
        .filter(fixture -> fixture.getHome().equals(team) || fixture.getAway().equals(team)))
        .collect(toList());

EDIT: This is on the assumption that order does not matter to you. If your final list needs to have home result and then away, have a look at Elliott Frisch's answer.

edited Mar 19, 2018 at 17:04

answered Mar 19, 2018 at 16:52

Thiyagu

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Comments

Chirlo · Accepted Answer · 2022-01-17 10:00:32Z

16

If you wan to get fancy with lambdas:

Predicate<FixtureResult> isHome = fr -> fr.getHome().equals(team)
Predicate<FixtureResult> isAway = fr -> fr.getAway().equals(team)

resultList.stream()
  .filter(isHome.or(isAway))
  .collect(toList()));

You could even extract the compose predicate to test it in isolation, with no streams involved, which is good for more complex predicates:

Predicate<FixtureResult> isHomeOrAway = isHome.or(isAway)

assertTrue(isHomeOrAway(homeFixture)); 
...

edited Jan 17, 2022 at 10:00

answered Mar 19, 2018 at 17:11

Chirlo

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Comments

Elliott Frisch · Accepted Answer · 2018-03-19 16:57:20Z

Assuming the order doesn't matter, you can do it on one line. Like,

private Collection<FixtureResult> finalResults(Team team) {
    return resultList.stream()
            .filter(fixture -> fixture.getHome().equals(team) 
                    || fixture.getAway().equals(team))
            .collect(toList());
}

If the order matters (home results and then away), you can do it with a single List like

private Collection<FixtureResult> finalResults(Team team) {
    List<FixtureResult> al = new ArrayList<>(resultList.stream()
            .filter(fixture -> fixture.getHome().equals(team)).collect(toList()));
    al.addAll(resultList.stream()
            .filter(fixture -> fixture.getAway().equals(team)).collect(toList()));
    return al;
}

Marian Klühspies · Accepted Answer · 2021-08-31 07:49:29Z

4

You can do the following

someStream.filter(((Predicate<SomeClass>) someObject-> someCondition).or(someObject-> someOtherCondition))

Or you can define your own "or" function that won't cause such a deep hierarchy

@SuppressWarnings("unchecked")
<R> Predicate<R> or(Predicate<R> ...predicates) {
   return r -> Arrays.stream(predicates).anyMatch(p -> p.test(r));
}

That gives you a cleaner interface without casting and the nesting

.filter(or(
      yourObject -> {
          return false;
      },
      yourObject -> {
          return false;
      },
      yourObject -> {
          return false;
      },
      yourObject -> {
          return false;
      }
 ))

answered Aug 31, 2021 at 7:49

Marian Klühspies

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1 Comment

ATutorMe Over a year ago

This is a gem hidden at bottom of the page! Hopefully more developers realise the elegance of these types of solutions.

David Geirola · Accepted Answer · 2018-03-19 17:11:56Z

2

You can simply create a conditions concatenations or can concatenate multiple filter call

Conditions concatenations

myList.stream()
      .filter(element -> (condition1 && condition2 && condition3))

Multiple filter call

myList.stream()
      .filter(element -> condition1)
      .filter(element -> condition2)
      .filter(element -> condition3)

edited Mar 19, 2018 at 17:11

answered Mar 19, 2018 at 16:58

David Geirola

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1 Comment

liberborn Over a year ago

Each next filter call applies to filtered results. The author most probably wanted OR filter conditions.

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