我为自己编写了一个实用程序，将列表分解为给定大小的批次。我只是想知道是否已经有任何apache commons util用于此。
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;
}
请让我知道是否有任何现有的公用事业已经相同。
当前回答
如果有人正在寻找Kotlin版本，这里是
list.chunked(size)
or
list.windowed(size)
曾经有一个面试问题，我在下面写了一个=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
}
其他回答
另一种方法是使用收集器。索引的groupingBy，然后将分组索引映射到实际元素:
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);
输出:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11]
解决这个问题的另一个方法是:
public class CollectionUtils {
/**
* Splits the collection into lists with given batch size
* @param collection to split in to batches
* @param batchsize size of the batch
* @param <T> it maintains the input type to output type
* @return nested list
*/
public static <T> List<List<T>> makeBatch(Collection<T> collection, int batchsize) {
List<List<T>> totalArrayList = new ArrayList<>();
List<T> tempItems = new ArrayList<>();
Iterator<T> iterator = collection.iterator();
for (int i = 0; i < collection.size(); i++) {
tempItems.add(iterator.next());
if ((i+1) % batchsize == 0) {
totalArrayList.add(tempItems);
tempItems = new ArrayList<>();
}
}
if (tempItems.size() > 0) {
totalArrayList.add(tempItems);
}
return totalArrayList;
}
}
这里有一个例子:
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);
输入: [a, b, c, d, e]
输出: [a, b, c]， [d, e]]
你可以在这里找到例子: https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/
Java 8中的一行代码是:
import static java.util.function.Function.identity;
import static java.util.stream.Collectors.*;
private static <T> Collection<List<T>> partition(List<T> xs, int size) {
return IntStream.range(0, xs.size())
.boxed()
.collect(collectingAndThen(toMap(identity(), xs::get), Map::entrySet))
.stream()
.collect(groupingBy(x -> x.getKey() / size, mapping(Map.Entry::getValue, toList())))
.values();
}
还有一个问题和这个问题完全一样，但如果你仔细阅读，你会发现它有微妙的不同。因此，如果有人(比如我)真的想将一个列表分割成给定数量的几乎相同大小的子列表，那么请继续阅读。
我只是简单地将这里描述的算法移植到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);
}
