Filtering ArrayList of Strings

Question

I have an array list of strings and I would like to remove strings which are below a certain length. I was thinking about this method:

        for (int i = 0; i < result.size(); i++) {
            if (result.get(i).split("\\s").length != maxLength) {
                System.out.println(result.get(i));
                result.remove(i);
}
        }

But it is only removing few entries because when it removes one then it shifts the next one in the place of removed one. What is the other way to do that

What are you trying to do? You are testing the length of the array that is returned by a string that is split by the literal "\s". I haven't seen your list of strings, but I expect that the test statement will always be evaluating 1 != maxLength. — Zach
– Zach, Commented Jul 18, 2011 at 17:30

Zach · Accepted Answer · 2011-07-18 17:30:39Z

3

Never remove from an ArrayList in that way. Use an Iterator instead:

Iterator<String> stringIterator = result.iterator();
while (stringIterator.hasNext()) {
    String string = stringIterator.next();
    if (string.split("\\s").length != maxLength) {
        System.out.println(string);
        stringIterator.remove();
    }
}

edited Jul 18, 2011 at 17:30

Zach

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answered Jul 18, 2011 at 16:28

Ryan Stewart

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Comments

Jeff Foster · Accepted Answer · 2011-07-18 16:28:34Z

3

Use an iterator, for example:

final Iterator<String> x = list.iterator();
while (iterator.hasNext()) {
    if (someCondition) {
        iterator.remove();
    }
}

answered Jul 18, 2011 at 16:28

Jeff Foster

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Comments

pingw33n · Accepted Answer · 2011-07-18 16:32:12Z

3

If you happen to use Commons Collections:

CollectionUtils.filter(result, new Predicate()
{
    @Override
    public boolean evaluate(Object object)
    {
        return ((String) object).split("\\s").length >= maxLength;
    }
}));

answered Jul 18, 2011 at 16:32

pingw33n

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Comments

MRAB · Accepted Answer · 2011-07-18 16:26:11Z

2

Try iterating from the last to the first:

    for (int i = result.size() - 1; i >= 0; i--) {
        if (result.get(i).split("\\s").length != maxLength) {
            System.out.println(result.get(i));
            result.remove(i);
        }
    }

answered Jul 18, 2011 at 16:26

MRAB

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Comments

antlersoft · Accepted Answer · 2011-07-18 16:26:19Z

2

When you remove one you should decrement the index to take into account that the ones following have shifted, as in result.remove(i--);

answered Jul 18, 2011 at 16:26

antlersoft

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1 Comment

Fattie Over a year ago

it's traditional, in all programming, to just go backwards from the end.

Kainsin · Accepted Answer · 2011-07-18 16:28:59Z

1

You need to start from the end and work your way to the front.

for (int i = result.size() - 1; i >= 0; i--) {
    if (result.get(i).length < maxLength) {
        System.out.println(result.get(i));
        result.remove(i);
    }
}

answered Jul 18, 2011 at 16:28

Kainsin

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Comments

Kal · Accepted Answer · 2011-07-18 16:29:00Z

1

To remove elements from a collection, you should use an iterator.

for (Iterator<String> iter = list.iterator(); iter.hasNext();) {
      String s = iter.next();
      if (s.split("\\s").length != maxLength) {
        iter.remove();
      }
      else {
        System.out.println(s);
      }
    }

answered Jul 18, 2011 at 16:29

Kal

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Comments

Charles Goodwin · Accepted Answer · 2011-07-18 16:47:03Z

You need to modify i again to make sure it covers the full sequence of the Array List or go last->first as shown in other answers.

for (int i = 0; i < result.size(); i++) {
    if (result.get(i).split("\\s").length != maxLength) {
        result.remove(i);
        i--; // everything has been moved up in the arraylist
    }
}

Also, ArrayList takes linear time to remove a single element, so repeated removal is a bad idea.

Either use a LinkedList, which can remove in constant time during iteration, or first collect the elements in a HashSet and then remove them at the end using Collection.removeAll(Collection).

For ArrayList, the removeAll method takes time proportional to the size of the list times the lookup time for the argument collection. Using a HashSet as argument should minimize the time it takes.

If you only remove a few values, any collection will probably suffice.

LinkedList tmpLinkedList = new LinkedList(result);
for (Iterator iter = tmpLinkedList.iterator(); iter.hasNext() {
    if (iter.next().split("\\s").length != maxLength))
        iter.remove();
}
result.clear();
result.addAll(tmpLinkedList);

Or:

HashSet toRemove = new HashSet();
//for (Iterator iter = myarraylist.iterator(); iter.hasNext() {
for (String s : result) {
    if (s.split("\\s").length != maxLength)
        toRemove.add(elem);
}
result.removeAll(toRemove);

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