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
**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/
其他回答
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);
你也可以使用FunctionalJava库- List有分区方法。这个库有自己的集合类型,你可以将它们来回转换为java集合。
import fj.data.List;
java.util.List<String> javaList = Arrays.asList("a", "b", "c", "d" );
List<String> fList = Java.<String>Collection_List().f(javaList);
List<List<String> partitions = fList.partition(2);
你可以使用Eclipse Collections中的chunk方法:
ArrayList<Integer> list = new ArrayList<>(Interval.oneTo(1000));
RichIterable<RichIterable<Integer>> chunks = Iterate.chunk(list, 10);
Verify.assertSize(100, chunks);
这篇DZone文章中还包含了一些块方法的示例。
注意:我是Eclipse Collections的提交者。
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]]
您需要知道您划分列表的块大小。假设您有一个包含108个条目的列表,您需要25个块大小。因此,你最终会得到5个列表:
4项各有25项; 有8个元素的。
代码:
public static void main(String[] args) {
List<Integer> list = new ArrayList<Integer>();
for (int i=0; i<108; i++){
list.add(i);
}
int size= list.size();
int j=0;
List< List<Integer> > splittedList = new ArrayList<List<Integer>>() ;
List<Integer> tempList = new ArrayList<Integer>();
for(j=0;j<size;j++){
tempList.add(list.get(j));
if((j+1)%25==0){
// chunk of 25 created and clearing tempList
splittedList.add(tempList);
tempList = null;
//intializing it again for new chunk
tempList = new ArrayList<Integer>();
}
}
if(size%25!=0){
//adding the remaining enteries
splittedList.add(tempList);
}
for (int k=0;k<splittedList.size(); k++){
//(k+1) because we started from k=0
System.out.println("Chunk number: "+(k+1)+" has elements = "+splittedList.get(k).size());
}
}