How to check if one ArrayList contains objects stores in another ArrayList

[Raymond Gillespie]

I have two ArrayLists and I want to check if the items stored in one list is also in the other and if it is, store it in a third ArrayList. My problem is that each object from list two is getting stored in list three.

The lists contain:

List one: 50 items
List two: 15 items

Of the 15 items in list two, 10 of them are in List one and these are the ten that I want to store in List three.

Here is the code I have came up with but something is not right. I didn't see it necessary to show the rest of the code since I know storing the objects in list one and two is working properly.

for (Items item : listTwo){
     listThree.add(listOne.contains(item) ? 1 : 0, item);
		    
}

[Piet Souris]

hi Raymond,

your method should work, if you can get your indices right.

But there is an easier way. For your intended third List, have a
look at the methods 'addAll' and 'retainAll', in the Collection API.

Greetz,
Piet

[Jeanne Boyarsky]

 listThree.add(listOne.contains(item) ? 1 : 0, item);  

This code adds the item to listThree unconditionally. What changes is that it inserts it at index 0 or index 1 depending on whether it is in listOne. That's not what you want. Try creating an if statement around the add call.

[Jeanne Boyarsky]

Piet,
addAll() would allow duplicates in the list. If this were a set, I see how that would work. I'm puzzled for a list.

[Raymond Gillespie]

Jeanne Boyarsky wrote:Piet,
addAll() would allow duplicates in the list. If this were a set, I see how that would work. I'm puzzled for a list.

How does a set work?

[Paweł Baczyński]

Set is a collection that holds elements and it does not allow duplicates.
However it does not (generally) promise you that it will keep them in any particular order.
Some implementations of Set interface do promise you that like LinkedHashSet.

See this.

[Raymond Gillespie]

Jeanne Boyarsky wrote:

 listThree.add(listOne.contains(item) ? 1 : 0, item);  

This code adds the item to listThree unconditionally. What changes is that it inserts it at index 0 or index 1 depending on whether it is in listOne. That's not what you want. Try creating an if statement around the add call.

I tried this and got the same outcome. It still adds every object from list two into list three

    for (Items item : listTwo){  
           if(listOne.contains(item)){
               listThree.add(item);  
            }      
    }  

[Jeanne Boyarsky]

What type of objects are in the list? Do they implement equals()?

[Raymond Gillespie]

They are String objects.

[Henry Wong]

Raymond Gillespie wrote:They are String objects.

Your code shows them as Item objects -- see your previous post...

Raymond Gillespie wrote:
I tried this and got the same outcome. It still adds every object from list two into list three

    for (Items item : listTwo){  
           if(listOne.contains(item)){
               listThree.add(item);  
            }      
    }  

Henry

[Jeanne Boyarsky]

That's not what I meant. They appear to be the Items class from your for loop. Does the Items class have an equals() method? (If not, it needs one for this to work). Can you show the implementation of that method?

[R. Jain]

Raymond Gillespie wrote:They are String objects.

But your for loops says, it's Items.

[Raymond Gillespie]

I did not have the equals() method in my Items class and that was the issue. It works fine now. Thanks!!

[Henry Wong]

Raymond Gillespie wrote:It is an items object. What I mean is that Strings are in the objects. I do have an equals() method in the items class so perhaps that is the issue.

Whether it is an issue or not would depend... well, perhaps you can show us the implementation of the equals() method from the Items class.

Henry

[Paweł Baczyński]

And hashCode(), of course.

[Henry Wong]

Paweł Baczyński wrote:And hashCode(), of course.

Well, IMO, it *should* have a corresponding hashCode() method, but it is not needed here. The code is only using array lists, and not a hashing collection.

Henry

[Raymond Gillespie]

I also added the hasCode() method as well.

[Raymond Gillespie]

Henry Wong wrote:

Raymond Gillespie wrote:It is an items object. What I mean is that Strings are in the objects. I do have an equals() method in the items class so perhaps that is the issue.

Whether it is an issue or not would depend... well, perhaps you can show us the implementation of the equals() method from the Items class.

Henry

This is the equals() method. I didn't write it. I just used the generate feature in Eclipse.

@Override
	public boolean equals(Object obj) {
		if (this == obj)
			return true;
		if (obj == null)
			return false;
		if (getClass() != obj.getClass())
			return false;
		Items other = (Items) obj;
		if (item== null) {
			if (other.item != null)
				return false;
		} else if (!item.equals(other.item))
			return false;
		return true;
	}

[Piet Souris]

Jeanne Boyarsky wrote:Piet,
addAll() would allow duplicates in the list. If this were a set, I see how that would work. I'm puzzled for a list.

Sorry, I missed this reply.

The opening post did not mention anything about duplicates or not. So what I suggested
simply mirrors what was in the arrays already.

But anyway, problem is solved now, I didn't know that today's IDE's are able
to generate an 'equals' method (I knew they can generate a hashcode).

Greetz,
Piet

[Winston Gutkowski]

Raymond Gillespie wrote:This is the equals() method. I didn't write it. I just used the generate feature in Eclipse.

Well, it's produced a lot of redundant code because, assuming your 'item' field's equals() method is written the same way (or uses Object's), the second null check is not needed, viz:

@Override
public boolean equals(Object obj) {
	if (this == obj)
		return true;
	if (obj == null)
		return false;
	if (getClass() != obj.getClass())
		return false;
	// This may require an @SuppressWarnings annotation
	// to remove warnings, but so would your original.
	return item.equals(((Items) obj).item);
}

I also dislike "class-based" equals() methods intensely, because they violate LSP by definition.
The following:

public boolean equals(Object obj) {
	if (this == obj)
		return true;
	return obj instanceof Items && item.equals(((Items) obj).item);
}

will work just as well, and doesn't require annotation.

HIH

Winston