Java split Arraylist into smaller ArrayLists

Question

I have an ArrayList with a size of 258

Now i wish to split this into three different ArrayLists for this i have created the following code:

    Integer start = (int) Math.floor((sitesToSearch.size() / 3));
    Integer middle = (int) Math.floor((sitesToSearch.size() / 2));
    Integer end = (int) Math.floor((sitesToSearch.size() / 1));

    ArrayList<String> crawl_list1 = (ArrayList<String>)tmp.subList(0, start);
    ArrayList<String> crawl_list2 = (ArrayList<String>)tmp.subList(start+1, middle);
    ArrayList<String> crawl_list3 = (ArrayList<String>)tmp.subList(middle+1, end);

Sadly this throws the following error:

Exception in thread "main" java.lang.ClassCastException: java.util.ArrayList$SubList cannot be cast to java.util.ArrayList

So how can i devide it into three smaller ArrayList

tmp declaration:

public ArrayList<String> getExternalLinks(ArrayList<String> rootDomains){
ArrayList<String> result = new ArrayList<String>();

Document doc = null;
/*
 * Check if root is valid
 * Find search the site for internal links
 */
for (String root : rootDomains) {
    if (!(root == null) || !root.isEmpty() ||!root.contains("#")) {
        try {
            doc = Jsoup.connect(root)
                    .userAgent("Mozilla")
                    .get();
            result.addAll(findExternalLinks(findInternalLinks(doc,root),root)); 
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
}
System.out.println(result.size());
return result;

}

possible duplicate of casting Arrays.asList causing exception: java.util.Arrays$ArrayList cannot be cast to java.util.ArrayList — Suresh Atta
– Suresh Atta, Commented Nov 27, 2013 at 14:00

Community · Accepted Answer · 2017-05-23 12:00:10Z

7

Simply use List<String> as the type of crawl_list1.

That's just an application of the general rule of "code against the interface".

There's really no good reason to require the return value of subList to be an ArrayList (and it doesn't make sense as subList returns a view onto the original ArrayList).

If you absolutely need ArrayList objects, then you need to copy the content into new ArrayList objects:

ArrayList<String> crawl_list1 = new ArrayList<String>(tmp.subList(0, start));
ArrayList<String> crawl_list2 = new ArrayList<String>(tmp.subList(start+1, middle));
ArrayList<String> crawl_list3 = new ArrayList<String>(tmp.subList(middle+1, end));

edited May 23, 2017 at 12:00

CommunityBot

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answered Nov 27, 2013 at 13:59

Joachim Sauer

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8 Comments

Marc Rasmussen Over a year ago

i need to split it into three Arraylists because i have to iterate over each of the three lists

Joachim Sauer Over a year ago

@MarcRasmussen: so? you can iterate over a List just as easily as over an ArrayList.

Marc Rasmussen Over a year ago

i can't do foreach can i?

jalynn2 Over a year ago

Yes, you should code to interfaces as much as possible: Use List everywhere that a method returns the List. Only use ArrayList when you need to instantiate a new one yourself.

Joachim Sauer Over a year ago

@MarcRasmussen: Haven't I just said that you can do exactly that? Have you tried it? You can use foreach with every Iterable (which includes Collection which includes List).

|

JB Nizet · Accepted Answer · 2013-11-27 14:00:50Z

2

Look at the javadoc for ArrayList.subList(). It doesn't return an ArrayList. It returns a List. There's no reason to cast this list to an ArrayList. It's a list, and that's all you need to know:

List<String> crawl_list1 = tmp.subList(0, start);
List<String> crawl_list2 = tmp.subList(start+1, middle);
List<String> crawl_list3 = tmp.subList(middle+1, end);

Also, you should check your indices, because the end index passed to subList is exclusive.

answered Nov 27, 2013 at 14:00

JB Nizet

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Comments

Mason T. · Accepted Answer · 2013-11-27 14:02:34Z

1

.subList(fromIndex, toIndex) is inclusive for fromIndex and exclusive for toIndex. With the current code middle and start will never populate an ArrayList.

The below two are assumptions of the logic:

.subList(start, middle)

.subList(middle, end+1)

answered Nov 27, 2013 at 14:02

Mason T.

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1 Comment

Andrew Grothe Over a year ago

"exclusive for toIndex". that is really good to know. Probably should have read the documentation on that one :p

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