Is there a common Java utility to break a list into batches?

Question

I wrote myself a utility to break a list into batches of given size. I just wanted to know if there is already any apache commons util for this.

public static <T> List<List<T>> getBatches(List<T> collection,int batchSize){
    int i = 0;
    List<List<T>> batches = new ArrayList<List<T>>();
    while(i<collection.size()){
        int nextInc = Math.min(collection.size()-i,batchSize);
        List<T> batch = collection.subList(i,i+nextInc);
        batches.add(batch);
        i = i + nextInc;
    }

    return batches;
}

Please let me know if there any existing utility already for the same.

Answer 1

Check out Lists.partition(java.util.List, int) from Google Guava:

Returns consecutive sublists of a list, each of the same size (the final list may be smaller). For example, partitioning a list containing [a, b, c, d, e] with a partition size of 3 yields [[a, b, c], [d, e]] -- an outer list containing two inner lists of three and two elements, all in the original order.

Answer 2

In case you want to produce a Java-8 stream of batches, you can try the following code:

public static <T> Stream<List<T>> batches(List<T> source, int length) {
    if (length <= 0)
        throw new IllegalArgumentException("length = " + length);
    int size = source.size();
    if (size <= 0)
        return Stream.empty();
    int fullChunks = (size - 1) / length;
    return IntStream.range(0, fullChunks + 1).mapToObj(
        n -> source.subList(n * length, n == fullChunks ? size : (n + 1) * length));
}

public static void main(String[] args) {
    List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14);

    System.out.println("By 3:");
    batches(list, 3).forEach(System.out::println);
    
    System.out.println("By 4:");
    batches(list, 4).forEach(System.out::println);
}

Output:

By 3:
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
[10, 11, 12]
[13, 14]
By 4:
[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10, 11, 12]
[13, 14]

Answer 3

Another approach is to use Collectors.groupingBy of indices and then map the grouped indices to the actual elements:

    final List<Integer> numbers = range(1, 12)
            .boxed()
            .collect(toList());
    System.out.println(numbers);

    final List<List<Integer>> groups = range(0, numbers.size())
            .boxed()
            .collect(groupingBy(index -> index / 4))
            .values()
            .stream()
            .map(indices -> indices
                    .stream()
                    .map(numbers::get)
                    .collect(toList()))
            .collect(toList());
    System.out.println(groups);

Output:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]

Answer 4

With Java 9 you can use IntStream.iterate() with hasNext condition. So you can simplify the code of your method to this:

public static <T> List<List<T>> getBatches(List<T> collection, int batchSize) {
    return IntStream.iterate(0, i -> i < collection.size(), i -> i + batchSize)
            .mapToObj(i -> collection.subList(i, Math.min(i + batchSize, collection.size())))
            .collect(Collectors.toList());
}

Using {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, the result of getBatches(numbers, 4) will be:

[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9]]

Answer 5

Here is a simple solution for Java 8+:

public static <T> Collection<List<T>> prepareChunks(List<T> inputList, int chunkSize) {
    AtomicInteger counter = new AtomicInteger();
    return inputList.stream().collect(Collectors.groupingBy(it -> counter.getAndIncrement() / chunkSize)).values();
}

Answer 6

Use Apache Commons ListUtils.partition.

org.apache.commons.collections4.ListUtils.partition(final List<T> list, final int size)

Answer 7

I came up with this one:

private static <T> List<List<T>> partition(Collection<T> members, int maxSize)
{
    List<List<T>> res = new ArrayList<>();

    List<T> internal = new ArrayList<>();

    for (T member : members)
    {
        internal.add(member);

        if (internal.size() == maxSize)
        {
            res.add(internal);
            internal = new ArrayList<>();
        }
    }
    if (internal.isEmpty() == false)
    {
        res.add(internal);
    }
    return res;
}

Answer 8

The following example demonstrates chunking of a List:

package de.thomasdarimont.labs;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class SplitIntoChunks {

    public static void main(String[] args) {

        List<Integer> ints = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11);

        List<List<Integer>> chunks = chunk(ints, 4);

        System.out.printf("Ints:   %s%n", ints);
        System.out.printf("Chunks: %s%n", chunks);
    }

    public static <T> List<List<T>> chunk(List<T> input, int chunkSize) {

        int inputSize = input.size();
        int chunkCount = (int) Math.ceil(inputSize / (double) chunkSize);

        Map<Integer, List<T>> map = new HashMap<>(chunkCount);
        List<List<T>> chunks = new ArrayList<>(chunkCount);

        for (int i = 0; i < inputSize; i++) {

            map.computeIfAbsent(i / chunkSize, (ignore) -> {

                List<T> chunk = new ArrayList<>();
                chunks.add(chunk);
                return chunk;

            }).add(input.get(i));
        }

        return chunks;
    }
}

Output:

Ints:   [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
Chunks: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]]

Answer 9

Here an example:

final AtomicInteger counter = new AtomicInteger();
final int partitionSize=3;
final List<Object> list=new ArrayList<>();
            list.add("A");
            list.add("B");
            list.add("C");
            list.add("D");
            list.add("E");
       
        
final Collection<List<Object>> subLists=list.stream().collect(Collectors.groupingBy
                (it->counter.getAndIncrement() / partitionSize))
                .values();
        System.out.println(subLists);

Input: [A, B, C, D, E]

Output: [[A, B, C], [D, E]]

You can find examples here: https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/

Answer 10

There was another question that was closed as being a duplicate of this one, but if you read it closely, it's subtly different. So in case someone (like me) actually wants to split a list into a given number of almost equally sized sublists, then read on.

I simply ported the algorithm described here to Java.

@Test
public void shouldPartitionListIntoAlmostEquallySizedSublists() {

    List<String> list = Arrays.asList("a", "b", "c", "d", "e", "f", "g");
    int numberOfPartitions = 3;

    List<List<String>> split = IntStream.range(0, numberOfPartitions).boxed()
            .map(i -> list.subList(
                    partitionOffset(list.size(), numberOfPartitions, i),
                    partitionOffset(list.size(), numberOfPartitions, i + 1)))
            .collect(toList());

    assertThat(split, hasSize(numberOfPartitions));
    assertEquals(list.size(), split.stream().flatMap(Collection::stream).count());
    assertThat(split, hasItems(Arrays.asList("a", "b", "c"), Arrays.asList("d", "e"), Arrays.asList("f", "g")));
}

private static int partitionOffset(int length, int numberOfPartitions, int partitionIndex) {
    return partitionIndex * (length / numberOfPartitions) + Math.min(partitionIndex, length % numberOfPartitions);
}

Answer 11

Using various cheats from the web, I came to this solution:

int[] count = new int[1];
final int CHUNK_SIZE = 500;
Map<Integer, List<Long>> chunkedUsers = users.stream().collect( Collectors.groupingBy( 
    user -> {
        count[0]++;
        return Math.floorDiv( count[0], CHUNK_SIZE );
    } )
);

We use count to mimic a normal collection index.
Then, we group the collection elements in buckets, using the algebraic quotient as bucket number.
The final map contains as key the bucket number, as value the bucket itself.

You can then easily do an operation on each of the buckets with:

chunkedUsers.values().forEach( ... );

Answer 12

Similar to OP without streams and libs, but conciser:

public <T> List<List<T>> getBatches(List<T> collection, int batchSize) {
    List<List<T>> batches = new ArrayList<>();
    for (int i = 0; i < collection.size(); i += batchSize) {
        batches.add(collection.subList(i, Math.min(i + batchSize, collection.size())));
    }
    return batches;
}

Answer 13

List<T> batch = collection.subList(i,i+nextInc);
->
List<T> batch = collection.subList(i, i = i + nextInc);

Answer 14

Here's a solution using vanilla java and the super secret modulo operator :)

Given the content/order of the chunks doesn't matter, this would be the easiest approach. (When preparing stuff for multi-threading it usually doesn't matter, which elements are processed on which thread for example, just need an equal distribution).

public static <T> List<T>[] chunk(List<T> input, int chunkCount) {
    List<T>[] chunks = new List[chunkCount];

    for (int i = 0; i < chunkCount; i++) {
        chunks[i] = new LinkedList<T>();
    }

    for (int i = 0; i < input.size(); i++) {
        chunks[i % chunkCount].add(input.get(i));
    }

    return chunks;
}

Usage:

    List<String> list = Arrays.asList("a", "b", "c", "d", "e", "f", "g", "h", "i", "j");

    List<String>[] chunks = chunk(list, 4);

    for (List<String> chunk : chunks) {
        System.out.println(chunk);
    }

Output:

[a, e, i]
[b, f, j]
[c, g]
[d, h]

Answer 15

Note that List#subList() returns a view of the underlying collection, which can result in unexpected consequences when editing the smaller lists - the edits will reflect in the original collection or may throw ConcurrentModificationException.

Answer 16

Below solution using Java 8 Streams:

        //Sample Input
        List<String> input = new ArrayList<String>();
        IntStream.range(1,999).forEach((num) -> {
            input.add(""+num);
        });
        
        //Identify no. of batches
        int BATCH_SIZE = 10;
        int multiples = input.size() /  BATCH_SIZE;
        if(input.size()%BATCH_SIZE!=0) {
            multiples = multiples + 1;
        }
        
        //Process each batch
        IntStream.range(0, multiples).forEach((indx)->{
            List<String> batch = input.stream().skip(indx * BATCH_SIZE).limit(BATCH_SIZE).collect(Collectors.toList());
            System.out.println("Batch Items:"+batch);
        });

Answer 17

if someone is looking for Kotlin version, here is

list.chunked(size)

or

list.windowed(size)

once had an interview question and I wrote below one =D

fun <T> batch(list: List<T>, limit: Int): List<List<T>> {
    val result = ArrayList<List<T>>()

    var batch = ArrayList<T>()

    for (i in list) {
        batch.add(i)
        if (batch.size == limit) {
            result.add(batch)
            batch = ArrayList()
        }
    }
    if (batch.isNotEmpty()) {
        result.add(batch)
    }
    return result
}

Answer 18

You can use below code to get the batch of list.

Iterable<List<T>> batchIds = Iterables.partition(list, batchSize);

You need to import Google Guava library to use above code.