How to loop through multiple Lists and create another list with merging of unique values from both the lists using Java 8 lambda expressions

Question

I am writing a code to loop through multiple Lists and create another list with merging of unique values from both the lists using Java 8 lambda expressions.

Model class:

class ServiceMap{
    Integer serviceMapId;
    Integer seviceId;
}

Code logic:

List<ServiceMap> listA = getServiceMaps();//Will get from database
List<Integer> listB = Arrays.asList(1, 10, 9);//Will get from client
List<ServiceMap> listC = new ArrayList<>();//Building it merging of both lists above

listA.stream().forEach(e -> {
    if (listB.parallelStream().noneMatch(x -> x == e.getServiceId())) {
        listC.add(new ServiceMap(e.getServiceId()));
        return;
    }

    listB.stream().forEach(x -> {
        if (listC.stream().anyMatch(e2->e2.getServiceId() == x)) {
            return;
        }
        if (x == e.getServiceId()) {
            listC.add(new ServiceMap(e.getServiceId()));
        } else {
            listC.add(new ServiceMap(x));
        }
    });

});
listC.stream().forEach(x -> System.out.println(x));

Is it efficient way writing code using java lambda expressions?

Stream.of(listA, listB).map(Collection::stream).reduce(Stream.empty(), Stream::concat).distinct().collect(Collectors.toList()). Your solution is certainly not correct. — Boris the Spider
– Boris the Spider, Commented Sep 25, 2018 at 19:41
isn't this better ? Set<Integer> set = new HashSet<>(listA); listB.stream().filter(i -> !set.contains(i)).forEach(set::add); — Hadi
– Hadi, Commented Sep 25, 2018 at 19:49
@BoristheSpider I actually misread your code, my bad. A minor improvement for you version could be Stream.of(listA, listB) .map(List::stream) .reduce(Stream::concat) .orElse(Stream.empty()) .distinct() .collect(Collectors.toList()); but, still, why not Stream.concat(listA, listB) directly... — Eugene
– Eugene, Commented Sep 25, 2018 at 20:08
@Krish you posted code, received comments and answers that we took time over - reading and understanding your code and posting. Changing your code after the fact is not acceptable. Please roll back your edit. — Boris the Spider
– Boris the Spider, Commented Sep 25, 2018 at 20:08

Mureinik · Accepted Answer · 2018-09-25 19:43:56Z

3

You could stream each list, apply distinct to them, and collect:

List<Integer> result = 
    Stream.concat(listA.stream(), listB.stream())
          .distinct()
          .collect(Collectors.toList());

answered Sep 25, 2018 at 19:43

Mureinik

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Comments

drowny · Accepted Answer · 2018-09-25 20:31:10Z

3

You should use also like this ;

Stream<Integer> streamOfServiceMapIds = listA.stream().map(ServiceMap::getSeviceId);
List<ServiceMap> collectedList = Stream.concat(streamOfServiceMapIds, listB.stream())
        .distinct()
        .map(ServiceMap::new)
        .collect(Collectors.toList());

edited Sep 25, 2018 at 20:31

answered Sep 25, 2018 at 20:13

drowny

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