How to return N consecutive elements from a Collection?

Question

I am passed a collection of objects (some Contact class in my case) and need to return a page from that collection. My code feels much longer than it needs to be. Am I missing some libraries that could perform that more elegantly than iterating over each element one at a time like I do below?

protected Collection<Contact> getPageOfContacts(
  Collection<Contact> contacts, int pageIndex, int pageSize) {
  if (pageIndex < 0 || pageSize <= 0
    || pageSize > contacts.size()) {
    return contacts;
  }
  int firstElement = pageIndex * pageSize;
  int lastElement = (pageIndex + 1) * pageSize - 1;
  Collection<Contact> pagedContacts = new ArrayList<Contact>();
  int index = -1;
  for (Contact contact : contacts) {
    index++;
    if (index < firstElement) {
      continue;
    }
    if (index > lastElement) {
      break;
    }
    pagedContacts.add(contact);
  }
  return pagedContacts;
}

Answer 1

You could use Guava Iterables.partition:

protected <T> Collection<T> getPageOfContacts(
        Collection<T> contacts, int pageIndex, int pageSize) {
    return Lists.newArrayList(
        Iterables.partition(contacts, pageSize)).get(pageIndex);
}

A more complex version does not create all pages to pick the right one, but stops when the right page is found.

protected <T> Collection<T> getPageOfContacts(
        Collection<T> contacts, int pageIndex, int pageSize) {
    Iterator<List<T>> partitions = Iterators.partition(contacts.iterator(), pageSize);

    for(int page = 0; page<pageSize && partitions.hasNext(); page++){
        List<T> partition = partitions.next();
        if(page == pageIndex) return partition;
    }
    return Collections. <T> emptyList(); //or fail
}

Update:

Thanks to ColinD to point out that:

Iterables.get(Iterables.partition(contacts, pageSize), pageIndex)

is a simpler implementation.

Answer 2

If you can require the data to be paged to be a List, you can get a sublist view of a single page easily using Guava:

public <T> List<T> getPage(List<T> list, int pageIndex, int pageSize) {
  return Lists.partition(list, pageSize).get(pageIndex);
}

This involves no copying or iteration (it uses sublist views of the original list) and handles a final page that has fewer than pageSize elements transparently.

For an arbitrary Iterable or Collection, I'd do this:

public <T> List<T> getPage(Iterable<T> iterable, int pageIndex, int pageSize) {
  return Iterables.get(Iterables.partition(iterable, pageSize), pageIndex);
}

By providing both these methods, you'd be able to handle objects that are known to be lists at compile-time efficiently and any other type of Iterable as efficiently as you can.

Answer 3

If you want a defined order to your elements, you should be using a List, not a collection. The basic difference between List and Collection is that List has a fixed order to the elements. It also defines the very convenient method subList(int start, int end) which creates a sub-list which is an alias of the original list containing just the elements you want without the overhead of copying them to a new list.

Answer 4

The List interface provides a subList method, that takes a start index and an end index. See http://download.oracle.com/javase/6/docs/api/java/util/List.html#subList(int,%20int). The returned sublist is backed by the original list, so you probably want to do something like

protected Collection<Contact> getPageOfContacts(...) {
    return new ArrayList<Contact>(original.subList(start,end));
}

Answer 5

Iterables.partition(contacts, pageSize).forEachRemaining(paginatedContacts->{/*Operation here*/});

Answer 6

return new ArrayList<Contact>(new ArrayList<Contact>(contacts).subList(firstElement, lastElement));

Note: this will return the sublist exclusive lastElement

Note 2: The result are copied to another list for the reasons mentioned by Kevin.