根据自定义对象属性从arraylist中删除元素

Question

I have an ArrayList with custom objects. What i want is to remove the duplicates from the array depending on the name property of the custom object. I have tried to accomplish this with Set person = new TreeSet(); but it's not working. I guess because the set is comparing addresses or something else than the name property. So i'm now trying to use an iterator which is not removing the duplicates either. This is what i got;

ArrayList<Person> people = new ArrayList<Person>();
Iterator<Person> iterator = people.iterator();
while (iterator.hasNext()) {
   Person person = iterator.next();

   if (person.getName().equals(iterator.next().getName())) {
       iterator.remove();
   }
}
for (Person person : people) {
    System.out.println(person.getName());
}

The ArrayList is not being modified though i see duplicates in it. I need some help. Thanks!

Answer 1

I had the same situation, and I came up with this solution to use SortedSet . In this case, those objects which cause set's comparator to return 0, will only be inserted once in the Set.

Here is an example:

SortedSet<Person> persons = new TreeSet<Person>(new Comparator<Person>() {
    @Override
    public int compare(Person arg0, Person arg1) {
        return arg0.getName().compareTo(arg1.getName());
    }
});

And now if you insert a Person into your persons , the duplicates (based on their name property) will not be inserted.

So you can iterator over your list<Person> and insert every item of it into your persons set, and be sure that you will not have any duplicates. So the rest would be like:

Iterator<Person> iterator = people.iterator();
while(iterator.hasNext()) {
    persons.add(iterator.next());
}
people.clear();
people.addAll(persons); //Now, your people does not contain duplicate names

Answer 2

Your code is currently broken because you only compare objects with the next object in the list. To correct your current approach, you would need to have another sub-loop which compares each object with all others in the list. That might get messy with nested iterators.

One alternative is to define a new list and populate it with items once you've confirmed they are not duplicates. This avoids nesting iterators.

Finally, another option would be to define an equals method that compares based on this property and throw the objects in a Set . Don't forget hashCode too.

Answer 3

Its not removing because you are only comparing each element with the next element. You can store the names in a HashSet which can hold only 1 of each String then remove the item if its name is already in the set.

HashSet<String> seen = new HashSet<String>();
while (iterator.hasNext()) {
     Person p = iterator.next();
     if (seen.contains(p.getName())) {
           iterator.remove();
     } else { 
           seen.add(p.getName());
     }
}

Answer 4

Everytime you write .next(), the iteration is taken 1 step forward. So, say you have 10 person in your list. You pick the 1st person, and check the next person using iterator.next(). Though you fetch the 2nd person, the iterator is now at person 2. So in the next run, the 3rd person is fetched and compared with the 4th person.

What you ought to do is, pick 1 person and compare his name with the names of all the 10 person in the list and then remove all the instances of duplicate objects from the list.