如何将一个数组列表(size=1000)拆分为多个相同大小(=10)的数组列表?
ArrayList<Integer> results;
如何将一个数组列表(size=1000)拆分为多个相同大小(=10)的数组列表?
ArrayList<Integer> results;
当前回答
您可以将Guava库添加到项目中并使用列表。划分方法,例如:
List<Integer> bigList = ...
List<List<Integer>> smallerLists = Lists.partition(bigList, 10);
其他回答
使用StreamEx库,您可以使用StreamEx。ofSubLists(List<T> source, int length)方法:
返回一个新的StreamEx,它由给定源列表的不重叠子列表组成,具有指定的长度(最后一个子列表可能更短)。
// Assuming you don't actually care that the lists are of type ArrayList
List<List<Integer>> sublists = StreamEx.ofSubLists(result, 10).toList();
// If you actually want them to be of type ArrayList, per your question
List<List<Integer>> sublists = StreamEx.ofSubLists(result, 10).toCollection(ArrayList::new);
我猜你遇到的问题是命名100个数组列表并填充它们。您可以创建一个数组列表数组,并使用循环填充每个数组列表。
最简单(也是最愚蠢的)的方法是这样的:
ArrayList results = new ArrayList(1000);
// populate results here
for (int i = 0; i < 1000; i++) {
results.add(i);
}
ArrayList[] resultGroups = new ArrayList[100];
// initialize all your small ArrayList groups
for (int i = 0; i < 100; i++) {
resultGroups[i] = new ArrayList();
}
// put your results into those arrays
for (int i = 0; i < 1000; i++) {
resultGroups[i/10].add(results.get(i));
}
**Divide a list to lists of n size**
import java.util.AbstractList;
import java.util.ArrayList;
import java.util.List;
public final class PartitionUtil<T> extends AbstractList<List<T>> {
private final List<T> list;
private final int chunkSize;
private PartitionUtil(List<T> list, int chunkSize) {
this.list = new ArrayList<>(list);
this.chunkSize = chunkSize;
}
public static <T> PartitionUtil<T> ofSize(List<T> list, int chunkSize) {
return new PartitionUtil<>(list, chunkSize);
}
@Override
public List<T> get(int index) {
int start = index * chunkSize;
int end = Math.min(start + chunkSize, list.size());
if (start > end) {
throw new IndexOutOfBoundsException("Index " + index + " is out of the list range <0," + (size() - 1) + ">");
}
return new ArrayList<>(list.subList(start, end));
}
@Override
public int size() {
return (int) Math.ceil((double) list.size() / (double) chunkSize);
}
}
Function call :
List<List<String>> containerNumChunks = PartitionUtil.ofSize(list, 999)
详情:https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/
polygenelubricants提供的答案将基于给定数组的大小。我正在寻找将数组分割成给定数量的部分的代码。以下是我对代码所做的修改:
public static <T>List<List<T>> chopIntoParts( final List<T> ls, final int iParts )
{
final List<List<T>> lsParts = new ArrayList<List<T>>();
final int iChunkSize = ls.size() / iParts;
int iLeftOver = ls.size() % iParts;
int iTake = iChunkSize;
for( int i = 0, iT = ls.size(); i < iT; i += iTake )
{
if( iLeftOver > 0 )
{
iLeftOver--;
iTake = iChunkSize + 1;
}
else
{
iTake = iChunkSize;
}
lsParts.add( new ArrayList<T>( ls.subList( i, Math.min( iT, i + iTake ) ) ) );
}
return lsParts;
}
希望它能帮助到别人。
Java8流,一个表达式,没有其他库:
List<String> input = ...
int partitionSize = ...
Collection<List<String>> partitionedList = IntStream.range(0, input.size())
.boxed()
.collect(Collectors.groupingBy(partition -> (partition / partitionSize), Collectors.mapping(elementIndex -> input.get(elementIndex), Collectors.toList())))
.values();
测试:
List<String> input = Arrays.asList("a", "b", "c", "d", "e", "f", "g", "h" ,"i");
partitionSize = 1 - > [[a], [b], [c], [d], [e], [f], [g], [h],[我]] partitionSize = 2 - > [[a, b], c, d, e, f, g, h,[我]] partitionSize = 3 - > [a, b, c, d, e, f, g, h,我]] partitionSize = 7 - > [[a, b, c, d, e, f, g], [h,我]] partitionSize = 100 -> [[a, b, c, d, e, f, g, h, i]]