如何检查一个列表的 object 是否包含另一个列表的任何 object

Question

We have two lists one is of type Question and the other of type Tag.

The Question class has those attributes

private String id;
private String header;
private String content;
private List<Tag> tags;
private Long timeStamp;

The Question list has all questions in it and the Tag list has all tags in it. We wanna check if One question contains any tag of the tag list. I want to do this for all questions.

With question.getTags, I get the list of tags.

The Tag class has an attribute called counter.

I will give some pseudocode to actually show you what I wanna do

List<Tag> allTags = ...
List<Question> allQuestions = ...
Map<Tag,Integer> map = new Hashmap<>();

if(one question contains any tag of allTags) {
     tag.setCounter(counter+1);
     map.put(tag,tag.getCounter);
}

In the end I wanna have a map where the key is the tag and the value the counter of that tag.

How can I actually do this?

EDIT here is my Tag.java

@NoArgsConstructor
@Data
@Getter
@AllArgsConstructor
@Setter
@Document(collection = "tag")
public class Tag {
    private String id;
    private String name;
    private Long timeStamp;

public Tag(String name) {
        this.id = UUID.randomUUID().toString();
        this.name = name;
        this.timeStamp = Instant.now().getEpochSecond()*1000;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Tag tag = (Tag) o;
        return Objects.equals(getId(), tag.getId());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getId());
    }
}

Answer 1

I think a counter attribute in the Tag class to count how often a Tag occurs in a list of questions is a design error. You don't need such a counter there.

A possibly plausible example: Imagine you have a class Student and a class Course . To keep track of how many students are in a course, there is no need for a counter in the Student class. In the same way a class Tag should only contain the attributes of a tag. That said, you can achieve your goal without the counter with either one of the two following approachs (provided you have java 8 or higher and your Tag class overrides the equals and hashcode methods):

Approach 1 using Streams & Collectors.groupingBy

Map<Tag,Long> map = allQuestions.stream()
            .flatMap(q -> q.getTags().stream())
            .filter(allTags::contains)
            .collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));

Approach 2 using List.forEach & Map.compute

Map<Tag, Integer> map2 = new HashMap<>();
    allQuestions.forEach(q -> {
        q.getTags()
                .stream()
                .filter(allTags::contains)
                .forEach(t -> map2.compute(t, (k, v) -> v == null ? 1 : v + 1));
    });

Or using a traditional for-loop and if-else block

Map<Tag, Integer> map3 = new HashMap<>();

for (Question q : allQuestions) {
    for (Tag t : q.getTags()) {
        if (allTags.contains(t)) {
            if (map3.containsKey(t)) {
                map3.put(t, map3.get(t) + 1);
            } else {
                map3.put(t, 1);
            }
        }
    }
}

Answer 2

I think this will work perfectly

    int count=0; 
    for(Question q : allQuestions){
     for(Tag t : q.getTags()){
     if(allTags.contains(t)){ 
        count=map.containsKey(t) ? map.get(t)+1 : 0;
        map.put(t,count);
       }
     }
    }