How to check if element exists using a lambda expression?

Question

Specifically, I have TabPane, and I would like to know if there is element with specific ID in it.

So, I would like to do this with lambda expression in Java:

boolean idExists = false;
String idToCheck = "someId";

for (Tab t : tabPane.getTabs()){
    if(t.getId().equals(idToCheck)) {
        idExists = true;
    }
}

I would be writing with simple way and clean with List.contains method. Example: return tabPane.getTabs().contains(idToCheck); — soyphea
– soyphea, Commented Jan 4, 2021 at 11:46

Mark Hurd · Accepted Answer · 2014-05-05 14:15:30Z

316

Try to use anyMatch of Lambda Expression. It is much better approach.

 boolean idExists = tabPane.getTabs().stream()
            .anyMatch(t -> t.getId().equals(idToCheck));

edited May 5, 2014 at 14:15

Mark Hurd

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answered Apr 11, 2014 at 6:24

Masudul

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2 Comments

Blacklight Over a year ago

Also worth noting: if you want to negate the check use noneMatch instead of anyMatch.

FabioStein Over a year ago

Call requires API level 24

jFrenetic · Accepted Answer · 2016-12-12 19:45:31Z

55

While the accepted answer is correct, I'll add a more elegant version (in my opinion):

boolean idExists = tabPane.getTabs().stream()
    .map(Tab::getId)
    .anyMatch(idToCheck::equals);

Don't neglect using Stream#map() which allows to flatten the data structure before applying the Predicate.

answered Dec 12, 2016 at 19:45

jFrenetic

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6 Comments

TecHunter Over a year ago

what is better here? I only see one more operation. Sorry I'm new to this lamba thing.

Malte Hartwig Over a year ago

@TecHunter it is more explicit. Imagine you read this code the first time, or again after a while. There are several advantages: First, we immediately show that we are not actually interested in the tab, but some mapping of it. Second, by using method references (which is only possible because we break the initial lambda into two steps) we show that there are no surprises hidden in the code. Third, by using method references, we do not create a new Predicate, but really just re-use equals. Though, granted, the example here is very simple, but I hope you get what I mean.

TecHunter Over a year ago

@MalteHartwig thanks! yeah I get your 3 points but I was asking about the flattening with map, it does another processing step no? I will try to compare the 2 methods :)

TecHunter Over a year ago

@MalteHartwig tested in a 10kk ArrayList with a simple object trying to find the last element. gives a 2ms difference 131ms against 133ms for your. on a 1kk array list yours if faster by 2ms (55ms to 53ms). So we can say that yours is better :)

jFrenetic Over a year ago

@TecHunter getters are super-cheap. Always prefer code clarity over saving extra 2 milliseconds (even though, I doubt the results are accurate, it may fluctuate on every run). Besides, remember that intermediate operations on streams (such as map) are lazy by nature. That means that getId method is not applied to each element of the collection. It's evaluated lazily until anyMatch returns true.

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kevinarpe · Accepted Answer · 2018-02-23 09:58:42Z

3

The above answers require you to malloc a new stream object.

public <T>
boolean containsByLambda(Collection<? extends T> c, Predicate<? super T> p) {

    for (final T z : c) {
        if (p.test(z)) {
            return true;
        }
    }
    return false;
}

public boolean containsTabById(TabPane tabPane, String id) {
    return containsByLambda(tabPane.getTabs(), z -> z.getId().equals(id));
}
...
if (containsTabById(tabPane, idToCheck))) {
   ...
}

answered Feb 23, 2018 at 9:58

kevinarpe

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