Fastest way to get the first n elements of a List into an Array

Question

What is the fastest way to get the first n elements of a list stored in an array?

Considering this as the scenario:

int n = 10;
ArrayList<String> in = new ArrayList<>();
for(int i = 0; i < (n+10); i++)
  in.add("foobar");

Option 1:

String[] out = new String[n];
for(int i = 0; i< n; i++)
    out[i]=in.get(i);

Option 2:

String[] out = (String[]) (in.subList(0, n)).toArray();

Option 3: Is there a faster way? Maybe with Java8-streams?

I expect it varies by platform and JVM, you should try benchmarking it yourself. — Elliott Frisch, Jul 10 '15 at 00:01
Why should it depent on the jvm? How could I calculate the Complexity? — Joel, Jul 10 '15 at 00:02
The complexity is easy. It's `O(n)`. The actual performance varies. — Elliott Frisch, Jul 10 '15 at 00:03
both are basically doing the same thing, if you look at the source for `subList` and `toArray` — njzk2, Jul 10 '15 at 00:17

src3369 · Answer 1 · 2018-01-19T05:40:47.387

142

Assumption:

list - List<String>

Using Java 8 Streams,

to get first N elements from a list into a list,

List<String> firstNElementsList = list.stream().limit(n).collect(Collectors.toList());
to get first N elements from a list into an Array,

String[] firstNElementsArray = list.stream().limit(n).collect(Collectors.toList()).toArray(new String[n]);

edited Jan 19 '18 at 05:40

answered Apr 11 '17 at 03:58

src3369

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Why should this be faster than #1? – Joel Apr 11 '17 at 06:02
If I only have 1 element, and I use the above streaming way, I get 1 element and null – jagamot Oct 12 '18 at 16:00
1

@jagamot If N is less than the array size use `toArray(String[]::new)` instead. – Björn Lindqvist Feb 12 '19 at 11:20
2

`.limit(n).toList()` works just fine. – BadPiggie Nov 30 '21 at 05:10

Elliott Frisch · Accepted Answer · 2015-07-10T00:18:44.837

12

Option 1 Faster Than Option 2

Because Option 2 creates a new List reference, and then creates an n element array from the List (option 1 perfectly sizes the output array). However, first you need to fix the off by one bug. Use < (not <=). Like,

String[] out = new String[n];
for(int i = 0; i < n; i++) {
    out[i] = in.get(i);
}

edited Jul 10 '15 at 00:18

answered Jul 10 '15 at 00:10

Elliott Frisch

198,278
20
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3

Surprisingly enough, `subList` does not copy. It is a view on the existing list. – njzk2 Jul 10 '15 at 00:15
@njzk2 Unless that view is a no-op, I'd expect option 1 to be faster anyway. – Elliott Frisch Jul 10 '15 at 00:17
creating the view is basically creating an object with a few variables. No loop, no copy, not much, really. – njzk2 Jul 10 '15 at 00:19
The `toArray` is a bit more involved. – Elliott Frisch Jul 10 '15 at 00:19
1

and also safer, with actual bound checks (like if `n >= in.size()`, or if someone changes the List in the mean time). – njzk2 Jul 10 '15 at 00:22

score 3 · Answer 3 · answered Jul 10 '15 at 00:59

3

It mostly depends on how big n is.

If n==0, nothing beats option#1 :)

If n is very large, toArray(new String[n]) is faster.

answered Jul 10 '15 at 00:59

ZhongYu

19,446
5
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the range n is about 1-5, where as the size of the `in`-List is about double the size – Joel Jul 10 '15 at 01:01
1

very sure option#1 then. – ZhongYu Jul 10 '15 at 01:03

score 1 · Answer 4 · answered Aug 04 '20 at 07:01

Option 3

Iterators are faster than using the get operation, since the get operation has to start from the beginning if it has to do some traversal. It probably wouldn't make a difference in an ArrayList, but other data structures could see a noticeable speed difference. This is also compatible with things that aren't lists, like sets.

String[] out = new String[n];
Iterator<String> iterator = in.iterator();
for (int i = 0; i < n && iterator.hasNext(); i++)
    out[i] = iterator.next();

score 0 · Answer 5 · answered Jul 29 '22 at 09:33

0

Use .take(n) operator on your list

answered Jul 29 '22 at 09:33

FarVoyager

49
4

Can you please explain in detail which method you mean? In the [List interface](https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/List.html) there is no method called like this. – vanje Jul 29 '22 at 16:23
1

This can be a comment – Jimale Abdi Jul 31 '22 at 07:16
@vanje looks like this extension is only available on kotlin, sorry for confusing. However I would like this answer to be here because I being a kotlin user ended up here in search of this function. So other kotlin users can come to this topic and get the answer – FarVoyager Aug 02 '22 at 13:11

score 0 · Answer 6 · edited Aug 20 '23 at 09:56

0

arrayList.stream().limit(n).toArray();

n = maxSize in length

This will help you to get the max size of the Array that you require.

edited Aug 20 '23 at 09:56

toyota Supra

3,181
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answered Aug 20 '23 at 07:52

Sai Nikhil A

1
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1

Why is this faster than the other options? – Joel Aug 21 '23 at 00:09

score -6 · Answer 7 · answered Mar 28 '18 at 20:54

-6

Use: Arrays.copyOf(yourArray,n);

answered Mar 28 '18 at 20:54

user2280949

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Fastest way to get the first n elements of a List into an Array

7 Answers7

Option 1 Faster Than Option 2

Linked

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