compare java element from a findAll()

Question

I have to remove duplicates from an object list retrieved from a CrudRepository.

I done that :

if (!prospectionRepository.findAll().isEmpty()) {
    List<Prospection> all = prospectionRepository.findAll();
    for (int i = 0; i < all.size()-1; i++) {
        for (int k = i+1;k < all.size(); k++) {
            if (all.get(i).getProspectNumber() == all.get(k).getProspectNumber()) {
                all.remove(all.get(i));
            }
        }
    }
    prospectionRepository.save(all);
}

However, duplicates are not removed from the list and they are then persisted when I do not want to.

What Java type is returned by getProspectNumber()? If it's not a primative your "==" comparison is not going to work. If it returns a Java Object, use .equals(...) instead. — Gerry Mantha
– Gerry Mantha, Commented Oct 12, 2017 at 17:31

Al-un · Accepted Answer · 2017-10-12 21:48:05Z

2

Problem edit

After conversation in chat, additional parameters have to be taken into account:

multiples Prospect may have the same prospect number but they do have an unique primary key in the database. Consequently, the duplicate filter cannot rely on the Prospect equality

a Prospect has a visited status which defined whether the Prospect has been contacted by the company or not. The two main status are NEW and MET. Only one Prospect can be MET. Other duplicates (with the same prospect number) can only be NEW

Algorithm

The problem needs to an additional step to be solved:

The prospects need to be grouped by prospect number. At this stage, we will a <ProspectNumber, List<Prospect>> mapping. However, the List<Prospect> must end up with a single element according to the rules defined earlier
Within a list, then if the prospect has not be met AND another prospect is found with a met status, then the first prospect is to be discarded

Consequently, the list will be generated with the following rules:

If a prospect has no duplicate in terms of prospect numbers, it is kept regardless its status
If a prospect has duplicate in terms of prospect number, only the met one is kept
If multiple prospects have the same prospect number but no one is met, then an arbitrary one is met: Stream does not guarantee to loop in the list order.

Code

The trick is to go through a Map as the key will hold the unicity. If your propect number is a specific type, this will assume that equals() and hashCode() are properly defined.

disclaimer: code is untested

List<Prospection> all = prospectionRepository.findAll().stream()
        // we instantiate here a Map<ProspectNumber, Prospect>
        // There is no need to have a Map<ProspectNumber, List<Propect>> 
        // as the merge function will do the sorting for us
        .collect(Collectors.toMap(
                // Key: use the prospect number
                prospect -> prospect.getProspectNumber(),
                // Value: use the propect object itself
                prospect -> prospect,
                // Merge function: two prospects with the same prospect number
                // are found: keep the one with the MET status or the first one
                (oldProspect, newProspect) -> {
                    if(oldProspect.getStatus() == MET){
                        return oldProspect;
                    } else if (newProspect.getStatus() == MET){
                        return newProspect;
                    } else{
                        // return the first one, arbitrary decision
                        return oldProspect;
                    }
                }
        ))
        // get map values only
        .values()
        // stream it in order to collect its as a List
        .stream()
        .collect(Collectors.toList());
prospectionRepository.save(all);

Map.values() actually return a Collection. So if your prospectionRepository.save(...) can accept a Collection (not only List), you can go faster. I also use the following synonym:

static method reference: Prospect::getProspectNumber is the Function equivalent to prospect -> prospect.getProspectNumber()
Function.identity(): is equivalent to prospect -> prospect
Ternary operator: it returns the same thing but written differently

Collection<Prospection> all = prospectionRepository.findAll().stream()
        .collect(Collectors.toMap(
                Prospect::getProspectNumber,
                Function.identity(),
                (oldProspect, newProspect) -> newProspect.getStatus() == MET ? newProspect : oldProspect
        )).values();
prospectionRepository.save(all);

_{For your information, if two Prospection having the same ProspectNumber are equals, then a simple distinct() would have been enough:}

List<Prospection> all = prospectionRepository.findAll()
        .stream()
        .distinct()
        .collect(Collectors.toList());
prospectionRepository.save(all);

edited Oct 12, 2017 at 21:48

answered Oct 12, 2017 at 17:40

Al-un

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9 Comments

Fosfor Over a year ago

The problem is that will note delete duplicates because whatever happens, the Id's will be unique and different. I have to compare field of my object

Al-un Over a year ago

I understand, I'll edit my answer but first I have a question: if you find two Prospection with the same ProspectNumber, how do you know which one you'll remove?

Fosfor Over a year ago

A status (Enumeration) that notifies whether the prospect has been modified : if I have prospection1 and prospection2 with same prospectNumber but prospection2 has a status "MEET" involve that I will delete prospection1

Al-un Over a year ago

And what is two prospection have been modified? How can you make the difference? E.g: Prospect1 (number 1), Prospect2 (number 2), Prospect3 (number 1). How do you know if you discard Prospect1 or Prospect3? If you have a status, what if they have the same status?

Fosfor Over a year ago

It is impossible because after insertion, the duplicates have by default a status "NEW"

|

Aleh Maksimovich · Accepted Answer · 2017-10-12 17:47:40Z

1

You can use something like

List<Prospection> all = prospectionRepository.findAll();
Set<Object> prospectNumbers = new HashSet<Object>();
Iterator<Prospection> it = all.iterator();
while (it.hasNext()) {
    Prospection item = iterator.next();
    Object itemNumer = item.getProspectNumber();
    if (prospectNumbers.contains(itemNumber)) {
        it.remove();
    } else {
        prospectNumbers.add(itemNumber);
    }
}

edited Oct 12, 2017 at 17:47

answered Oct 12, 2017 at 17:31

Aleh Maksimovich

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1 Comment

dspano Over a year ago

Hrm... prospectionRepository.remove(item); could potentially delete data when the user just wants to remove the duplicates from the resultSet.

Dina Bogdan · Accepted Answer · 2017-10-12 17:31:50Z

0

Why don't you use a HashSet with the prospectNumber as value. The HashSet cannot contains duplicate keys. After that you can iterate over your List and removing all your duplicate elements. Hope this will help you!

edited Oct 12, 2017 at 17:31

answered Oct 12, 2017 at 17:27

Dina Bogdan

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3 Comments

Pshemo Over a year ago

HashSet doesn't have keys. Did you mean HashMap?

Dina Bogdan Over a year ago

Wow, I'm sorry! I've made a confusion. Yea, HashSet doesn't have keys, but he can use a HashSet for getting all the no-duplicate keys and after that, use that HashSet for iterating over the List and remove the duplicates. Thx!

Al-un Over a year ago

@DinaBogdan actually, if the only constraint is I need a list for my save() method, he can just create a list and use list.addAll(theHashSet);. In this way, there is no need to do any loop

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