Remove Duplicates from ArrayList<ArrayList<String>> Object

Question

I have an ArrayList that has a nested ArrayList of Strings and I want to remove duplicates from. I know if I always wanted to remove duplicates, I shouldn't use an ArrayList, but in some cases duplicates are valid. What is the best way to remove duplicates from the nested ArrayList?

For example, I would like to execute some Java that converts:

[[duplicate], [duplicate], [duplicate], [unique1], [unique2]]

to:

[[duplicate], [unique1], [unique2]]

Isnt making a new HashSet<Object>(List<Object> ... ) the answer? Then convert to List again — kamil
– kamil, Commented Mar 27, 2012 at 19:17
You say that it's a list of lists. Do you want { {1, 2}, {1, 3} } to be collapsed to { { 1, 2 }, { 3 } }? or should each inner list be treated in isolation? — aioobe
– aioobe, Commented Mar 27, 2012 at 19:20
I am a bit confused too. Are you looking for [[1,2,3], [1,2], [3], [1,2]] to become [[1,2,3], [1,2], [3]], or [1, 2, 3] or what aioobe said? — ltfishie
– ltfishie, Commented Mar 27, 2012 at 19:26
@Itfishie - [[1,2,3], [1,2], [3], [1,2]] to become become [[1,2,3], [1,2], [3]] — c12
– c12, Commented Mar 27, 2012 at 20:08
@Itfishie - sorry misunderstood i want it to become [1,2,3], your solution worked for me. — c12
– c12, Commented Mar 27, 2012 at 20:17

aioobe · Accepted Answer · 2012-03-27 20:19:21Z

11

To remove duplicates from an ArrayList you could do

yourList = new ArrayList<String>(new LinkedHashSet<String>(yourList));

Using LinkedHashSet instead of a HashSet ensures that the order of the original lists are preserved.

Regarding your comment:

Here's a solution that transforms [[1,2,3], [1,2], [3], [1,2]] to [[1,2,3], [1,2], [3]].

Set<String> seen = new HashSet<String>();
for (List<String> l : strLists) {
    for (Iterator<String> iter = l.iterator(); iter.hasNext(); )
        if (!seen.add(iter.next()))
            iter.remove();

    // If you want to remove lists that end up empty:
    if (l.isEmpty())
        strLists.remove(l);
}

edited Mar 27, 2012 at 20:19

answered Mar 27, 2012 at 19:17

aioobe

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SG 86 · Accepted Answer · 2012-03-27 19:21:57Z

3

import java.util.Arrays;

public class RunDuplicate
{
public RunDuplicate()
{
super();
}

public static void main(String[] args)
{
String[] duplicates = new String[] {"duplicate","duplicate","duplicate","unique1","unique2"};

Arrays.sort(duplicates);

int k = 0;

for (int i = 0; i < duplicates.length; i++)
{
if (i > 0 && duplicates[i].equals(duplicates[i -1]))
continue;

duplicates[k++] = duplicates[i];
}

String[] unique = new String[k];

System.arraycopy(duplicates, 0, unique, 0, k);

//test that unique contains no duplicate strings
for (int i = 0; i < unique.length; i++)
System.out.println(unique[i]);
}
}

answered Mar 27, 2012 at 19:21

SG 86

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Comments

ltfishie · Accepted Answer · 2012-03-27 19:21:11Z

2

What aioobe said, using a set, except you will put it in a loop since you have a two demisional array:

Set<String> set = new LinkedHashSet<String>();
for (ArrayList<String> list:yourList) {
    set.addAll (list);
}
ArrayList<String> uniqueList = new ArrayList<String>(set);

answered Mar 27, 2012 at 19:21

ltfishie

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Comments

dexametason · Accepted Answer · 2012-03-27 19:20:56Z

0

You can do that like this:




for (Iterator<ArrayList<String>> it = myListInList.iterator(); it.hasNext();)
{ 
   ArrayList<String> current = it.next();
   HashSet h = new HashSet(current);
   current.clear();
   current.addAll(h);
}

answered Mar 27, 2012 at 19:20

dexametason

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2 Comments

Natix Over a year ago

-1 for using iterator (instead of a for-each loop) and raw HashSet (which also doesn't preserve ordering of the original list)

dexametason Over a year ago

I think using iterator pattern is useful. Hashset will cause that the ordering will be corrupt but it was irrelevant in this case [i can't see it as prerequisite].

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